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https://wikieducator.org/Chemistry/Density_Marking_Scheme | Chemistry/Density Marking Scheme
Science 8 Experiment
MARKING SCHEME FOR : CALCULATING THE DENSITY OF WATER, A BLOCK OF WOOD AND A ROCK SAMPLE
Maximum = 9
DATA:
mass of 25 mL graduated cylinder = __ 45.02 g ___
mass of graduated cylinder + 25.0 mL of water = ___ 69.79 g ___
(1) number of rock sample = ___ 6 ___
mass of rock sample = ___ 44.22 g ___
volume of water in 250 mL graduated cylinder = ___ 70 mL ___
volume of water in 250 mL graduated cylinder after adding rock = ___ 77 mL ___
what happens to rock when placed in water: ___ it sinks ___
number of wood block = ___ 6 ___
mass of wood block = ___ 32.96 g ___
dimensions of wood block: length = ___ 6.9 cm ___
width = ___ 3.9 cm ___
height = ___ 2.5 cm ___
what happens to wood block when placed in water: ___ it floats ___
DISCUSSION:
IMPORTANT NOTE: In all the questions and calculations that follow, you must say what you are doing or calculating. If numbers are being subtracted or multiplied together the numbers involved in the calculation must be shown and the result written down; if a special equation is being used the equations must be shown, the numbers being used must be shown as they are used in the equation and the results of the calculation must be shown.
1. (a) Calculate the mass of the water in the 25 mL graduated cylinder.
(1) mass of water = 69.79 – 45.02 = 24.77 g
(b) Calculate the density of the sample of distilled water using the following formula. Round your answer off to THREE decimal places.
$density = \frac {mass of water} {volume of water}$
(1) $density = \frac {mass of water} {volume of water}$ = $\frac{24.77g} {25mL}$ = 0.991g/mL
2. (a) Calculate the volume of the rock sample.
(1) Volume of rock = 77 – 70 = 7 mL
(b) Calculate the density of the rock sample using the following formula. Round your answer off to TWO decimal places.
$density = \frac {mass of rock} {volume of rock}$
(1) $density = \frac {mass of rock} {volume of rock}$ = $\frac{44.22g} {7mL}$ = 6.3g/mL
3. (a) Calculate the volume of the wood block. The units are “cm3 ”. What is this volume when expressed in “mL”?
(1) volume = 6.9 cm x 3.9 cm x 2.5 cm = 67.275 cm3 = 67.275 mL
(b) Calculate the density of the wood block using the following formula. Round your answer off to THREE decimal places.
$density = \frac {mass of wood} {volume of wood}$
(1) $density = \frac {mass of wood} {volume of wood}$ = $\frac{32.96g} {67.275mL}$ = = 0.490 g/mL
(1) 4. Fill in the following table.
Object Density of Object Density of Water Does object float or sink in water? Is object’s density greater or less than water’s density?
rock 6.3 0.991 sink greater
wood 0.490 0.991 float less
5. Write out and complete the following sentences using the information in the above table.
(1) Objects having a density greater than the density of water will __ SINK __ when placed in water. Objects having a density less than the density of water will ___¬FLOAT __ when placed in water. | 2019-11-13 01:58:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29547253251075745, "perplexity": 1599.8206944328354}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496665976.26/warc/CC-MAIN-20191113012959-20191113040959-00231.warc.gz"} |
https://stats.stackexchange.com/questions/122722/please-explain-the-waiting-paradox/122840 | A few years ago I designed a radiation detector that works by measuring the interval between events rather than counting them. My assumption was, that when measuring non-contiguous samples, on average I would measure half of the actual interval. However when I tested the circuit with a calibrated source the reading was a factor of two too high which meant I had been measuring the full interval.
In an old book on probability and statistics I found a section about something called "The Waiting Paradox". It presented an example in which a bus arrives at the bus stop every 15 minutes and a passenger arrives at random, it stated that the passenger would on average wait the full 15 minutes. I have never been able to understand the math presented with the example and continue to look for an explanation. If someone can explain why it is so that the passenger waits the full interval I will sleep better.
• What is the title and who is the author of the book? Could you copy the example word for word here? Nov 5, 2014 at 4:25
• Related post: math.stackexchange.com/questions/222674/… Nov 5, 2014 at 4:45
• It seems my guess above has some support. A comment to this answer mentions the inspection paradox. Nov 5, 2014 at 4:52
• I think using a bus as the analogy is confusing, as busses tend to follow schedules. Think instead about how long it will take for an empty taxi to come when on average one comes every 15 minutes. Nov 11, 2014 at 21:21
• Indeed, @HarveyMotulsky, good point. Whether buses follow a Poisson process or not depends strongly on the context. Jun 20, 2018 at 11:55
As Glen_b pointed out, if the buses arrive every $15$ minutes without any uncertainty whatsoever, we know that the maximum possible waiting time is $15$ minutes. If from our part we arrive "at random", we feel that "on average" we will wait half the maximum possible waiting time. And the maximum possible waiting time is here equal to the maximum possible length between two consecutive arrivals. Denote our waiting time $W$ and the maximum length between two consecutive bus arrivals $R$, and we argue that
$$E(W) = \frac 12 R = \frac {15}{2} = 7.5 \tag{1}$$
and we are right.
But suddenly certainty is taken away from us and we are told that $15$ minutes is now the average length between two bus arrivals. And we fall into the "intuitive thinking trap" and think: "we only need to replace $R$ with its expected value", and we argue
$$E(W) = \frac 12 E(R) = \frac {15}{2} = 7.5\;\;\; \text{WRONG} \tag{2}$$
A first indication that we are wrong, is that $R$ is not "length between any two consecutive bus-arrivals", it is "maximum length etc". So in any case, we have that $E(R) \neq 15$.
How did we arrive at equation $(1)$? We thought:"waiting time can be from $0$ to $15$ maximum. I arrive with equal probability at any instance, so I "choose" randomly and with equal probability all possible waiting times. Hence half the maximum length between two consecutive bus arrivals is my average waiting time". And we are right.
But by mistakenly inserting the value $15$ in equation $(2)$, it no longer reflects our behavior. With $15$ in place of $E(R)$, equation $(2)$ says "I choose randomly and with equal probability all possible waiting times that are smaller or equal to the average length between two consecutive bus-arrivals" -and here is where our intuitive mistake lies, because, our behavior has not change - so, by arriving randomly uniformly, we in reality still "choose randomly and with equal probability" all possible waiting times - but "all possible waiting times" is not captured by $15$ - we have forgotten the right tail of the distribution of lengths between two consecutive bus-arrivals.
So perhaps, we should calculate the expected value of the maximum length between any two consecutive bus arrivals, is this the correct solution?
Yes it could be, but : the specific "paradox" goes hand-in-hand with a specific stochastic assumption: that bus-arrivals are modeled by the benchmark Poisson process, which means that as a consequence we assume that the time-length between any two consecutive bus-arrivals follows an Exponential distribution. Denote $\ell$ that length, and we have that
$$f_{\ell}(\ell) = \lambda e^{-\lambda \ell},\;\; \lambda = 1/15,\;\; E(\ell) = 15$$
This is approximate of course, since the Exponential distribution has unbounded support from the right, meaning that strictly speaking "all possible waiting times" include, under this modeling assumption, larger and large magnitudes up to and "including" infinity, but with vanishing probability.
But wait, the Exponential is memoryless: no matter at what point in time we will arrive, we face the same random variable, irrespective of what has gone before.
Given this stochastic/distributional assumption, any point in time is part of an "interval between two consecutive bus-arrivals" whose length is described by the same probability distribution with expected value (not maximum value) $15$: "I am here, I am surrounded by an interval between two bus-arrivals. Some of its length lies in the past and some in the future but I have no way of knowing how much and how much, so the best I can do is ask What is its expected length -which will be my average waiting time?" - And the answer is always "$15$", alas.
• +1 Very nice. $f_\ell(\ell)$ should maybe read $f_\lambda(\ell)$? Nov 8, 2014 at 10:03
• Thanks. As for notation, both are used to indicate different things. What I wrote is along the lines of stressing whose random variable density is, because in the various transformations we may end up with something like $f_X(y)$. What you suggest is to stress the parametrized aspect of the density. Nov 8, 2014 at 10:58
If the bus arrives "every 15 minutes" (i.e. on a schedule) then the (randomly arriving) passenger's average wait is indeed only 7.5 minutes, because it will be uniformly distributed in that 15 minute gap.
--
If, on the other hand, the bus arrives randomly at the average rate of 4 per hour (i.e. according to a Poisson process), then the average wait is much longer; indeed you can work it out via the lack of memory property. Take the passenger's arrival as the start, and the time to the next event is exponential with mean 15 minutes.
Let me take a discrete time analogy. Imagine I am rolling a die with 15 faces, one of which is labelled "B" (for bus) and 14 labelled "X" for the total absence of bus that minute (fair 30 sided dice exist, so I could label 2 of the faces of a 30-sided die "B"). So once per minute I roll and see if the bus comes. The die has no memory; it doesn't know how many rolls since the last "B" it has been. Now imagine some unconnected event happens - a dog barks, a passenger arrives, I hear a rumble of thunder. From now, how long do I wait (how many rolls) until the next "B"?
Because of the lack of memory, on average, I wait the same time for the next "B" as the time between two consecutive "B"s.
[Next imagine I have a 60-sided die I roll every fifteen seconds (again, with one "B" face); now imagine I had a 1000-sided die I rolled every 0.9 seconds (with one "B" face; or more realistically, three 10-sided dice each and I call the result a "B" if all 3 come up "10" at the same time)... and so on. In the limit, we get the continuous time Poisson process.]
Another way to look at it is this: I am more likely to observe my 'start counting rolls' (i.e. 'the passenger arrives at the bus stop') event during a longer gap than a short one, in just the right way to make the average wait the same as the average time between buses (I mostly wait in long gaps and mostly miss out on the shortest ones; because I arrive at a uniformly distributed time, the chance of me arriving in a gap of length $t$ is proportional to $t$)
As a veteran catcher of buses, in practice reality seems to lie somewhere in between 'buses arrive on a schedule' and 'buses arrive at random'. And sometimes (in bad traffic), you wait an hour then 3 arrive all at once (Zach identifies the reason for that in comments below).
• I think with busses specifically there's an additional process where a late bus becomes later as passengers cram onto it, and the empty bus behind it eventually catches up (but remains empty). =D
– Zach
Nov 5, 2014 at 22:01
• @Zach indeed, that's why they tend to clump over long runs, especially in heavy traffic. Where I live when the bus runs so late it's about time for the next one, they will sometimes insert an additional bus that's nearly on time further along the route (i.e. it will drive with no passengers to where a bus would be not very far behind schedule, often getting there via a faster route) and start picking up passengers for whom now the bus is only a little late. Meanwhile, the very late bus now becomes effectively the next bus in the schedule, once it gets to where the other bus came in. Nov 5, 2014 at 22:27
• @Glen_b That's a really good idea,hah!
– Zach
Nov 6, 2014 at 0:12
• It's a useful anti-clumping strategy (at least, it mitigates the worst cases); I wouldn't have brought it up, except that it relates to the sort of dependence issues that more accurate bus-waiting-time models may need to deal with. Nov 6, 2014 at 0:30
More on buses... Sorry to butt into the conversation so late in the discussion, but I have been looking at Poisson processes lately... So before it slips out of my mind, here is a pictorial representation of the inspection paradox:
The fallacy stems from the assumption that since buses follow a certain pattern of arrival with a given inter-arrival average time (the inverse of the Poisson rate parameter $\lambda$, let's call it $\small \theta=1/\lambda=15$ min.), by showing up at the bus station at any random time, you are in effect picking up a bus. So if you show up at the bus station at random times, keeping up a log-book of the waiting times over, say, one month, will actually give you the average inter-arrival time between buses. But this is not what you'd be doing.
If we were at a dispatch center, and could see all the buses on a screen, it would be true that randomly picking up multiple buses, and averaging the distance to the bus following behind, would produce the average inter-arrival time:
But, if what we instead do is just show up at the bus station (instead of selecting a bus), we are doing a random cross-section of time, say, along the timeline of the bus schedule in a typical morning. The time we decide to show up at the bus station may very well be uniformly distributed along the "arrow" of time. However, since there are longer time gaps between buses spread more farther apart, we are more likely to end up oversampling these "stragglers":
... and hence, our waiting time log book will not reflect the inter-arrival time. This is the inspection paradox.
As for the actual question on the OP regarding the expected waiting time of $15'$, minutes the mind-boggling explanation resides in the memoryless-ness of the Poisson process that makes the time-gap elapsed from the time the last bus we missed left the station to the time we show up irrelevant, and the expected time to the arrival of the next bus continues to be, stubbornly, $\theta=15$ minutes. This is best seen in discrete time (geometric distribution) with the dice example in Glen_b's answer.
In fact, if we could know how long ago the preceding bus left, the $\small \mathbb E[\text{time waiting (future) + time to last bus departure (past)}]=30$ min! As explaine in this MIT video by John Tsitsiklis, we just would have to view what precedes the point of arrival as a Poisson process backwards in time:
Still unclear? - try it with Legos.
• Excellent diagrams. Apr 30, 2019 at 23:02
There is a simple explanation which resolves the different answers which one gets from calculating expected waiting time for buses arriving per a Poisson Process with given mean interarrival time (in this case 15 minutes), whose interarrival times are therefore i.i.d. exponential with mean of 15 minutes.
Method 1) Because Poisson Process (exponential) is memoryless, the expected wait time is 15 minutes.
Method 2) You are equally likely to arrive at any time during the interarrival period in which you arrive. Therefore the expected waiting time is 1/2 of the expected length of this interarrival period. THIS IS CORRECT, and does not conflict with method (1).
How can (1) and (2) both be correct? The answer is that the expected length of the interarrival period for the time at which you arrive is not 15 minutes. It is actually 30 minutes; and 1/2 of 30 minutes is 15 minutes, so (1) and (2) agree.
Why does the interarrival period for the time at which you arrive not equal 15 minutes? It's because by first "fixing" an arrival time, the interarrival period it is in is more likely than average to be a long interarrival period. In the case of an exponential interarrival period, the math works outs so the interarrival period containing the time at which you arrive is an exponential with double the mean interarrival time for the Poisson Process.
It is not obvious that the exact distribution for the interarrival time containing the time at which you arrive would be an exponential with doubled mean, but it is obvious, after explanation, why it is increased. As an easy to understand example, let's say that the interarrival times are 10 minutes with probability 1/2 or 20 minutes with probability 1/2. In this case, 20 minutes long interarrival periods are equally likely to occur as 10 minute long interarrival periods, but when they do occur, they last twice as long. So, 2/3 of the time points during the day will be at times at which the interarrival period is 20 minutes. Put another way, if we first pick a time and then want to know what the interarrival time containing that time is, then (ignoring transient effects at the beginning of the "day") the expected length of that interarrival time is 16 1/3. But if we first pick the interarrival time and want to know what its expected length is, it is 15 minutes.
There are other variants of the renewal paradox, length-biased sampling, etc., amounting to pretty much the same thing.
Example 1) You have a bunch of light bulbs, with random lifetimes, but average of 1000 hours. When a light bulb fails, it is immediately replaced by another light bulb. If you pick a time to go in a room having the light bulb, the light bulb in operation then will wind up having a longer mean lifetime than 1000 hours.
Example 2) If we go to a construction site at a given time, then the mean time until a construction worker who is working there at that time falls off the building (from when they first started working) is greater than the mean time until worker falls off (from when they first started working) from among all workers who start working. Why, because the workers with a short mean time until falling off are more likely than average to have already fallen off (and not continued working), so that the workers who are working then have longer than average times until falling off.
Example 3) Pick some modest number of people at random in a city and if they have attended the home games (not all sell outs) of the city's Major League baseball team, find out how many people attended the games they were at. Then (under some slightly idealized but not too unreasonable assumptions), the average attendance for those games will be higher than the average attendance for all the team's home games. Why? Because there are more people who have attended high attendance games than low attendance games, so you are more likely to pick people who have attended high attendance games than low attendance games.
The question as posed was "...a bus arrives at the bus stop every 15 minutes and a passenger arrives at random." If the bus arrives every 15 minutes then its not random; it arrives every 15 minutes so the correct answer is 7.5 minutes. Either the source was incorrectly quoted or the writer of the source was sloppy.
On the other hand, the radiation detector sounds like a different problem because radiation events do arrive at random according to some distribution, presumably something like Poisson with an average waiting time. | 2022-05-29 03:38:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7073493003845215, "perplexity": 577.3703429030536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00419.warc.gz"} |
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08-15-2006, 07:38 PM #8
lotus_esprit5
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the regs are supposed to be balanced (in other words, you shouldnt have the inline at 350 psi and LPR at 70, or the LPR at 110psi and inline at 200). so if the regs are far enough out of whack relative to each other, your velocity will drop. so, say you have an aftermarket valve, so your inline is at 225psi, and the LPR is at 80. they are pretty much in proportion to each other. but if you continue to turn the LPR up higher, eventually the pressures wont be in proportion, and therefore the regs will no longer be balanced, so your velocity will drop as you turn up the LPR rather than increase. does that make sense?
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Originally Posted by amzng_spyderman <-- legend. like a mod, but without powers or motivation. | 2015-02-28 16:24:28 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9217836856842041, "perplexity": 2721.7496959693035}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936461995.64/warc/CC-MAIN-20150226074101-00009-ip-10-28-5-156.ec2.internal.warc.gz"} |
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Add another equation box, click +free at the top bar and get. | 2021-12-06 10:47:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2633998692035675, "perplexity": 6137.353673047392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00117.warc.gz"} |
https://forum.allaboutcircuits.com/threads/multimeter-measuring-ac-voltage.7607/ | # Multimeter measuring AC voltage
#### cheddy
Joined Oct 19, 2007
87
I bought a 29 range digital multimeter from radio shack "22-813" (Link to Owners Manual)
1. What is a "AC voltage riding on a DC source bias" mean?
2. The specifications say that it can measure AC 600 V RMS Maximum at 50/60Hz. Does that mean that it can't measure any voltage above 50/60Hz or it just can't measure 600V at 50/60Hz?
I am trying to measure what should be 4.5 V AC at 58kHz and am not getting an accurate reading. Is it possible my multimeter is just a piece of crap?
#### beenthere
Joined Apr 20, 2004
15,819
It sounds like the meter is optomized for 50/60 Hz RMS measurements. The maximum voltage is 600 VAC. I would be surprised if it did an accurate measurement at 58 KHz, though. I'm not sure my Fluke 23 would do well. It depends on the measurment circuit in the meter.
#### cheddy
Joined Oct 19, 2007
87
Thanks for the response. I would like to hear more peoples opinions on the matter. Still does anyone know what an AC voltage on a DC source bias means?
Also, is it possible that an analog multimeter would give an accurate reading on such a high frequency that a digital multimeter just can't read?
#### JoeJester
Joined Apr 26, 2005
4,390
the bias is the DC offset. If you had a one volt peak signal and a 0.5V bias, the peaks would be 1.5 volts and neg 0.5 volts.
#### niftydog
Joined Jun 13, 2007
95
The best way to make that measurement would be with an oscilloscope.
Multimeters, analogue or otherwise, are just not designed for measuring anything other than mains frequency AC @ 50 or 60hz.
#### JoeJester
Joined Apr 26, 2005
4,390
The DMM or Analog meter is limited. You would have to look at a frequency response correction factor .... they did publish them with the old Simpson 260's ... up to about 1 MHz if my memory serves me correctly.
#### SgtWookie
Joined Jul 17, 2007
22,221
Good old Simpson 260's - and that's about what they cost back in the early 70's.
DMM/DVM's are pretty hard-pressed to accurately measure AC waveforms above 10KHz; even there, you really need to be looking at a chart.
#### The Electrician
Joined Oct 9, 2007
2,834
The specifications for AC measurement say "Average responds, RMS calibrated, DC coupled". Apparently (you should check this) if you have it set to measure AC volts, the meter will respond to DC. Set it to measure 4 volts AC, and see what it reads if you connect the leads to a flashlight battery. Some voltages you want to measure may have an AC component riding on top of a DC component. If you want to measure just the AC component, you must block the DC component. That's the purpose of the .1 uF capacitor they tell you to put in series with the meter.
This meter probably won't measure high audio frequency AC voltages accurately. Check this by connecting a variable frequency oscillator to the meter and sweep from 50 Hz up to 100 kHz; see if the meter reads the same at all frequencies.
#### techroomt
Joined May 19, 2004
198
ac voltmeters are designed to measure the rms values of sine waves of 60 hertz. any other wave form or frerquency will provide a certain amount of error. if you measure a dc source (battery) with an ac voltmeter it will read 0 vac, as there is no ac component. if you read an ac voltage (say you duplex receptacle outlet) with a dc voltmeter it will read o vdc, as there is no dc component. an ac signal on a dc offset will read those repective values individually.
#### GS3
Joined Sep 21, 2007
408
I am trying to measure what should be 4.5 V AC at 58kHz and am not getting an accurate reading. Is it possible my multimeter is just a piece of crap?
Well, that is possible but not likely. As I said in the other thread, you need to understand the tools and their limitations. What you really need is a 'scope and understanding about DC and AC components and the differences between RMS, average etc and what each meter indicates and its limitations.
#### techroomt
Joined May 19, 2004
198
i find it hard to believe the 58 hz (vs 60 hz) is providing the error you are experiencing. have you tried another emeter? and are you sure it is a sine wave?
#### GS3
Joined Sep 21, 2007
408
i find it hard to believe the 58 hz (vs 60 hz) is providing the error you are experiencing. have you tried another emeter? and are you sure it is a sine wave?
If you re-read the OP you will see its 58 KHZ. Quite a difference.
#### techroomt
Joined May 19, 2004
198
good catch gs3, my bad. meter will have error no doubt. pretty sure your meter will read high in that instance.
#### pccmd
Joined Aug 24, 2009
2
I concur.......frequency response is terrible for dmm's. Unfortunately, oscilloscopes cost a great deal more.....However: I recently purchased a 1970's vintage Tektronics(analog) Oscilloscope from a store in my general area for about $200 and having a ball with it, but just today driving myself crazy with discrepancy between its results and dmm's voltage readings. I found the answer today: http://groups.google.com/group/sci....7a658?hl=en&ie=UTF-8&q=dmm+frequency+response #### someonesdad Joined Jul 7, 2009 1,583 I bought a 29 range digital multimeter from radio shack "22-813" (Link to Owners Manual) 2. The specifications say that it can measure AC 600 V RMS Maximum at 50/60Hz. Does that mean that it can't measure any voltage above 50/60Hz or it just can't measure 600V at 50/60Hz? I am trying to measure what should be 4.5 V AC at 58kHz and am not getting an accurate reading. Is it possible my multimeter is just a piece of crap? Answer to 2: it means don't measure more than 600 volts RMS at 50 to 60 Hz. The allowed voltage will reduce as the frequency increases, but most manufacturers don't specify how to derate the meter (if it's given, it's usually a V*Hz product that you must stay under). Answer to last question: No, your meter isn't a piece of crap. It's just not capable of making the measurement you want to make. Rest assured that most other hand-held digital multimeters won't make a decent measurement at 58 kHz (unless they're pretty expensive). I looked at the manual for this meter, but it doesn't specify the bandwidth. Given that it is a$30 meter (quite inexpensive), it's unlikely that it will have a significant AC bandwidth. Manufacturers usually specify bandwidth using the 3 dB points, but I find that pretty meaningless for measurement equipment. Instead, I like to measure the frequency at which the output will drop by 0.1% or 1%.
I have a Radio Shack 22-812 meter (note it is a different model number than yours). When it is set to read a 1 V RMS sine wave at 60 Hz, the measured value drops by 0.1% at 300 Hz, 0.5% at 700 Hz, and by 1% at 1.25 kHz. The 3 dB point is at 13.57 kHz. The output drops by 0.1% when the frequency falls to 29 Hz.
Since the AC accuracy is given as 0.5%, I would (personally) rate this meter useful over the frequency range of about 30 - 400 Hz. Of course, if you know you're measuring sine waves and you can calibrate the meter, it then can be used over a much wider range.
Your meter will likely behave approximately in the same fashion, as it's probable that Radio Shack buys them both from the same Chinese manufacturer.
Inexpensive digital multimeters typically are average-responding. This means they are calibrated to measure the RMS value of a sine wave. Note they will not read the RMS value correctly for other waveforms. For this, you'll need a true RMS meter. Even then, the typical hand-held true RMS digital multimeter has a fairly limited bandwidth, usually (at best) over the audio range of 20 Hz to 20 kHz.
Measuring the RMS value of waveforms over large bandwidths (here, large may mean around 1 MHz) usually requires more expensive instrumentation. A popular analog RMS meter for this was the HP 3400, both the A and B versions (the B version extended the upper bandwidth limit from 10 MHz to 20 MHz). You can still find them for sale on ebay for under $100 (I remember they were around$2000 new in the 1980's). I have a 3400A model. If I set the amplitude of an HP 3326 synthesizer to 1.024 V RMS at 60 Hz, my 3400A reads exactly 1 volt RMS (I can detect about a 0.2% change on the meter at full scale). I can then increase the frequency to 1 MHz before the 3400A's measured value drops by 1% (1 division on the 1 V scale). In truth, I don't know whether this is the RMS meter or the synthesizer's drop, but as the drop occurs roughly linearly with increasing frequency, I'm guessing it's the 3400A. The 3400's reading keeps falling with increasing frequency, falling to 0.5 volts at 10.936 MHz. Since 10 MHz is the stated bandwidth of the 3400A, I'm reasonably confident the changes I'm seeing are due to the frequency response of the 3400A meter, not changes in the synthesizer's output (besides, I monitored the output on a scope and didn't see any amplitude changes).
These HP 3400's were remarkable designs. Based on the knob style and instrument panel, I'd estimate mine was made in the mid to late 1960's. It's a vacuum tube instrument, yet it draws only 7 W of power sitting on my bench (costs me a penny a day). Its rated measurement accuracy is 1% for frequencies between 50 Hz and 1 MHz. Personally, I rate it as nearly the ideal general-purpose AC voltmeter for an engineer or technician because it has measuring ranges from 1 mV full scale to 300 V full scale.
#### t06afre
Joined May 11, 2009
5,934
As a thumb rule standard DMMs is only god for measuring AC related to the mains. This information should be avilible in the manual. Some DMMs like the Fluke true RMS series are able to measure other waveforms and give a correct result. | 2021-06-19 18:06:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4816921055316925, "perplexity": 1875.7066399561836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00590.warc.gz"} |
https://cs.stackexchange.com/questions/60647/modify-dfa-nfa-that-accepts-language-subset-with-only-palindromes-with-size-lim | # Modify DFA/NFA that accepts Language Subset with only Palindromes (with Size Limit)?
Given a DFA (D1) with P1 states, that accepts a language L1.
Modify (D1) to create another DFA (D2), such that it will accept the language L2 that is defined as: All strings in L1 that are also a palindrome (of maximum length P1).
How many states would the minimized D2 have (worst case)? And time complexity?
Similarly, NFA (N1) with Q1 states that accepts a Language L3. How many states would minimized (N2) have (worst case)? Along with its time complexity?
The generic Palindrome Language is non-Regular hence a DFA/NFA for it is impossible but I am unaware of the limited Palindrome case in DFA/NFA?
• What are your thoughts on the matter? What have you tried? Where did you get stuck? – Yuval Filmus Jul 16 '16 at 11:27
• As the DFA has no memory, I suppose we would have to create a seperate state for every character if we were just testing for a Palindrome of a specific Size in D2. But, since we are testing both for L1 and Palindrome property, I am lost at how to approach it and its size and space complexity wrt. D1 in the worst case. – TheoryQuest1 Jul 16 '16 at 11:34
• We can go ahead, enumerate all possible strings in L1, then find a subset of Palindromes in it and construct a DFA (D2), but i simply dont find it efficent since we already have D1 – TheoryQuest1 Jul 16 '16 at 11:36
For DFAs, the size of $D_2$ is always at most $O(P_1|\Sigma|^{P_1/2})$, and this bound is tight up to the $P_1$ factor. For the lower bound, take a DFA for $\Sigma^*$ having $P_1$ states (you haven't specified that $D_1$ is minimal). The DFA $D_2$ accepts all palindromes of length $P_1$, and so Nerode's theorem shows that for even $P_1$, the optimal DFA $D_2$ contains $\Theta(|\Sigma|^{P_1/2})$ states. The upper bound now follows by using the product construction.
The exact same reasoning works for NFAs as well, although you need to replace Nerode's theorem with a suitable "exchange lemma" in order to show that any NFA for the language of all palindromes of length $P_1$ requires $\Omega(|\Sigma|^{P_1/2})$ states (for even $P_1$).
If you insist on $D_1$ being minimal as well, then instead of a DFA for $\Sigma^*$ take a DFA for $(\Sigma^{P_1})^*$, to get the exact same results. This construction works also for NFAs. | 2019-06-16 11:42:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7653831839561462, "perplexity": 558.7475531569455}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00404.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-10-review-page-747/73 | ## Algebra: A Combined Approach (4th Edition)
$r$=$\frac{5}{\sqrt \pi}$
Step 1: The formula to be used here is, $r$=$\sqrt\frac{A}{\pi}$ Step 2: According to the question, A= 25 square meters Step 3: Substituting the value of A into the equation, $r$=$\sqrt\frac{25}{\pi}$ Step 4: $r$=$\frac{5}{\sqrt \pi}$ | 2019-01-22 18:33:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9540311694145203, "perplexity": 1208.779558035274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00223.warc.gz"} |
https://www.sydney.edu.au/scholarships/d/paleoecology-great-barrier-reef-response-environmental-changes.html | # Paleoecology of the Great Barrier Reef in Response to Environmental Changes
$28,092 per annum scholarship for up to three years to conduct research into paleoecology of the Great Barrier Reef. ## Highlights Value Eligibility Open date Close date$28,092 per annum for up to three years
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• have an unconditional offer of admission to study full-time PhD in the School of Geosciences.
• be willing to conduct research into paleoecology of the Great Barrier Reef.
• hold an honours degree (first class or second class upper) in geosciences, or a master's degree in a related field with a substantial research component.
• be willing to travel to remote field areas such as the Great Barrier Reef.
## Background
This scholarship has been established to provide financial assistance to a PhD student who is undertaking research in investigating the links between environmental stress/disturbance, climate, and coral reef composition, diversity and structure by investigating episodes of reef growth and demise over the past 9,000 years.
This project will investigate the links between environmental stress/disturbance, climate, and coral reef composition, diversity and structure by investigating episodes of reef growth and demise over the past 9,000 years. We will use fossil reef cores from across the GBR. Major changes and/or hiatuses in reef growth will be identified down core, and between adjacent cores, as major time gaps and changes in the reef communities. High-precision U-Th dates will pin down their precise timing and will be used to accurately calculate accumulation rates for the reef successions. We will measure reef stress during the Holocene during the periods leading up to and after major changes in reef growth and relate this to other geochemical proxy data reconstructing paleoenvironmental changes.
This scholarship is funded by an ARC Discovery Project.
#### Terms and conditions
1. Background
a. This Scholarship has been established to provide financial assistance to a PhD student who is undertaking research in investigating the links between environmental stress/disturbance, climate, and coral reef composition, diversity and structure by investigating episodes of reef growth and demise over the past 9,000 years.
b. This Scholarship is funded by an ARC Discovery Project.
2. Eligibility
a. The Scholarship is offered subject to the applicant having an unconditional offer of admission to study full-time in a PhD within the School of Geoscience at the University of Sydney.
b. Applicants must be willing to conduct research into paleoecology of the Great Barrier Reef.
c. Applicants must be a domestic student.
d. Applicants must also hold at least one of the following:
I. an Honours degree (First Class or Second Class Upper) or equivalent in in Geosciences or
II. Masters degree with a substantial research component.
e. Applicant must be willing to travel to remote field areas such as the Great Barrier Reef.
3. Selection Criteria
a. The successful applicant will be awarded the Scholarship on the basis of:
II. area of study and/or research proposal,
III. curriculum vitae.
IV. personal statement.
b. Preference will be given to applicants with a background in sedimentology, paleoecology, geochemistry, marine science, numerical modelling.
c. The successful applicant will be awarded the Scholarship on the nomination of the relevant research supervisor(s), or their nominated delegate(s).
4. Value
a. The Scholarship will provide a stipend allowance equivalent to the minimum Research Training Program (RTP) Stipend rate (indexed on 1 January each year) for up to 3 years, subject to satisfactory academic performance.
b. The recipient may apply for an extension of the stipend allowance for up to 6 months.
c. Periods of study already undertaken towards the degree prior to the commencement of the Scholarship will be deducted from the maximum duration of the Scholarship excluding the potential extension period.
d. The Scholarship is for commencement in the relevant research period in which it is offered and cannot be transferred to another area of research without prior approval.
e. No other amount is payable.
f. The Scholarship will be offered subject to the availability of funding.
5. Eligibility for Progression
a. Progression is subject to passing the annual progress review.
6. Leave Arrangements
a. The Scholarship recipient receives up to 20 working days recreation leave each year of the Scholarship and this may be accrued. However, the student will forfeit any unused leave remaining when the Scholarship is terminated or complete. Recreation leave does not attract a leave loading and the supervisor's agreement must be obtained before leave is taken.
b. The Scholarship recipient may take up to 10 working days sick leave each year of the Scholarship and this may be accrued over the tenure of the Scholarship. Students with family responsibilities, caring for sick children or relatives, or experiencing domestic violence, may convert up to five days of their annual sick leave entitlement to carer’s leave on presentation of medical certificate(s). Students taking sick leave must inform their supervisor as soon as practicable.
7. Research Overseas
a. The Scholarship recipient may not normally conduct research overseas within the first six months of award.
b. The Scholarship holder may conduct up to 12 months of their research outside Australia. Approval must be sought from the student's supervisor, Head of School and the Faculty via application to the Higher Degree by Research Administration Centre (HDRAC) and will only be granted if the research is essential for completion of the degree. All periods of overseas research are cumulative and will be counted towards a student's candidature. Students must remain enrolled full-time at the University and receive approval to count time away.
8. Changes in Enrolment
a. The Scholarship recipient must notify HDRAC, and their supervisor promptly of any planned changes to their enrolment including but not limited to: attendance pattern, suspension, leave of absence, withdrawal, course transfer, and candidature upgrade or downgrade. If the award holder does not provide notice of the changes identified above, the University may require repayment of any overpaid stipend.
9. Termination
a. The Scholarship will be terminated:
I. on resignation or withdrawal of the recipient from their research degree,
II. upon submission of the thesis or at the end of the award,
III. if the recipient ceases to be a full-time student and prior approval has not been obtained to hold the Scholarship on a part-time basis,
IV. upon the recipient having completed the maximum candidature for their degree as per the University of Sydney (Higher Degree by Research) Rule 2011 Policy,
V. if the recipient receives an alternative primary stipend scholarship. In such circumstances this Scholarship will be terminated in favour of the alternative stipend scholarship where it is of higher value,
VI. if the recipient does not resume study at the end of a period of approved leave, or
VII. if the recipient ceases to meet the eligibility requirements specified for this Scholarship, (other than during a period in which the Scholarship has been suspended or during a period of approved leave).
b. The Scholarship may also be terminated by the University before this time if, in the opinion of the University:
I. the course of study is not being carried out with competence and diligence or in accordance with the terms of this offer,
II. the student fails to maintain satisfactory progress, or
III. the student has committed misconduct or other inappropriate conduct.
c. The Scholarship will be suspended throughout the duration of any enquiry/appeal process.
d. Once the Scholarship has been terminated, it will not be reinstated unless due to University error.
10. Misconduct
a. Where during the Scholarship a student engages in misconduct, or other inappropriate conduct (either during the Scholarship or in connection with the student’s application and eligibility for the Scholarship), which in the opinion of the University warrants recovery of funds provided, the University may require the student to repay payments made in connection with the Scholarship. Examples of such conduct include and without limitation; academic dishonesty, research misconduct within the meaning of the Research Code of Conduct (for example, plagiarism in proposing, carrying out or reporting the results of research, or failure to declare or manage a serious conflict of interests), breach of the Code of Conduct for Students and misrepresentation in the application materials or other documentation associated with the Scholarship.
b. The University may require such repayment at any time during or after the Scholarship period. In addition, by accepting this Scholarship, the student consents to all aspects of any investigation into misconduct in connection with this Scholarship being disclosed by the University to the funding body and/or any relevant professional body.
11. Confidentiality and Intellectual Property
The successful recipient of this Scholarship must complete the Student Deed Poll supplied by the University of Sydney. | 2020-11-01 02:18:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.184548020362854, "perplexity": 4659.811140746997}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00561.warc.gz"} |
http://noenthuda.com/blog/tag/scatter-plot/ | ## When a two-by-two ruins a scatterplot
The BBC has some very good analysis of the Brexit vote (how long back was that?), using voting data at the local authority level, and correlating it with factors such as ethnicity and educational attainment.
In terms of educational attainment, there is a really nice chart, that shows the proportion of voters who voted to leave against the proportion of population in the ward with at least a bachelor’s degree. One look at the graph tells you that the correlation is rather strong:
‘Source: http://www.bbc.com/news/uk-politics-38762034And then there is the two-by-two that is superimposed on this – with regions being marked off in pink and grey. The idea of the two-by-two must have been to illustrate the correlation – to show that education is negatively correlated with the “leave” vote.
But what do we see here? A majority of the points lie in the bottom left pink region, suggesting that wards with lower proportion of graduates were less likely to leave. And this is entirely the wrong message for the graph to send.
The two-by-two would have been useful had the points in the graph been neatly divided into clusters that could be arranged in a grid. Here, though, what the scatter plot shows is a nice negatively correlated linear relationship. And by putting those pink and grey boxes, the illustration is taking attention away from that relationship.
Instead, I’d simply put the scatter plot as it is, and maybe add the line of best fit, to emphasise the negative correlation. If I want to be extra geeky, I might also write down the $R^2$ next to the line, to show the extent of correlation!
## Anscombe’s Quartet and Analytics
Many “analytics professionals” or “quants” I know or have worked with have no hesitation in diving straight into a statistical model when they are faced with a problem, rather than trying to understand the data. However, that is not the way I work. Whenever I set out solving a new problem, I start with spending considerable time trying to get a feel of the data. There are many things I do to “feel” the data – look at a few lines of data, look at descriptive statistics of some of the variables and distributions of individual variables. The most powerful tool, however, that lets me get a feel for data is the humble scatterplot.
The beauty of the scatter plot is that it allows you to get a real feel for the data. Taking variables two at a time, it not only shows you how each of them is distributed but also how they are related to each other. Relationships that are not apparent when you look at the data become apparent when you graph them. I may not be wrong in saying that the scatterplot defines the direction and scope of your entire solution.
The problem with the debate on how analytics needs to be done is that it is loaded. A large majority of people who use statistics in their daily work dive straight into analysis without looking at the data. Perhaps they deem that looking at data is a waste of time? I have even seen pitch decks by extremely reputed software companies that propose solutions such as “we will solve this problem using Logistic Regression” without even having seen the data.
Let us take an example now. Take the following four data sets (my apologies for putting an image here):
Let us say you dive straight into the analysis. Like a good “analytics professional” you dive straight into regression. You may even do some descriptive statistics for each of the data sets along the way. And this is what you find (again, apologies for the image)
Do you conclude that the four data sets are the same? Pretty much identical statistics right? I wouldn’t be surprised if you were to publish that there is nothing to differentiate between these four data sets. Now, let us do a simple scatter plot of each of these data sets and check for ourselves:
Now, do you still think these data sets are identical? Now you know why I stress so much upon getting a feel for the data and drawing the humble scatter plot?
The data set I’ve used here is a rather famous one, and it is called Anscombe’s Quartet. The purpose of the data set is to precisely describe what I have in this post. That one needs to get a feel for the data before diving into the analysis. Draw scatter plots for every pair of variables. Understand the relationships, and let this understanding guide your further analysis. If one were able to perfectly analyze every piece of data by diving straight into a regression, the job of analytics might as well be outsourced to computers.
PS: it is a tragedy that when they teach visualization in school they don’t even mention the scatter plot. At a recent workshop I asked the participants to name the different kinds of graphs they knew. “Line”, “Bar” and “Pie” were the mots common answers. Not one answered “scatter plot”. Given the utility of this simple plot this is indeed tragic. | 2017-08-21 19:48:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5826247930526733, "perplexity": 502.8243519069451}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00438.warc.gz"} |
https://phys.libretexts.org/Courses/Prince_Georges_Community_College/PHY2040_General_Physics_III/01%3A_Waves_and_Vibrations/1.2%3A_Hooke%E2%80%99s_Law | $$\require{cancel}$$
# 1.2: Hooke’s Law
learning objectives
• Generate the mathematical expression of Hooke’s law
In mechanics (physics), Hooke’s law is an approximation of the response of elastic (i.e., springlike) bodies. It states: the extension of a spring is in direct proportion with the load applied to it. For instance, the spring is pulled downwards with either no load, Fp, or twice Fp.
Diagram of Hooke’s Law: The extension of the spring is linearly proportional to the force.
Springs and Hooke’s Law: A brief overview of springs, Hooke’s Law, and elastic potential energy for algebra-based physics students.
Many materials obey this law of elasticity as long as the load does not exceed the material’s elastic limit. Materials for which Hooke’s law is a useful approximation are known as linear-elastic or “Hookean” materials. Hookean materials are broadly defined and include springs as well as muscular layers of the heart. In simple terms, Hooke’s law says that stress is directly proportional to strain. Mathematically, Hooke’s law is stated as:
$\mathrm{F=−kx}$
where:
• x is the displacement of the spring’s end from its equilibrium position (a distance, in SI units: meters);
• F is the restoring force exerted by the spring on that end (in SI units: N or kg·m/s2); and
• k is a constant called the rate or spring constant (in SI units: N/m or kg/s2). When this holds, the behavior is said to be linear. If shown on a graph, the line should show a direct variation.
It’s possible for multiple springs to act on the same point. In such a case, Hooke’s law can still be applied. As with any other set of forces, the forces of many springs can be combined into one resultant force.
When Hooke’s law holds, the behavior is linear; if shown on a graph, the line depicting force as a function of displacement should show a direct variation. There is a negative sign on the right hand side of the equation because the restoring force always acts in the opposite direction of the displacement (for example, when a spring is stretched to the left, it pulls back to the right).
Hooke’s law is named after the 17th century British physicist Robert Hooke, and was first stated in 1660 as a Latin anagram, whose solution Hooke published in 1678 as Ut tensio, sic vis, meaning, “As the extension, so the force.”
Hooke’s Law: The red line in this graph illustrates how force, F, varies with position according to Hooke’s law. The slope of this line corresponds to the spring constant k. The dotted line shows what the actual (experimental) plot of force might look like. The pictures of spring states at the bottom of the graph correspond to some points of the plot; the middle one is in the relaxed state (no force applied).
## Elastic Potential Energy
If a force results in only deformation, with no thermal, sound, or kinetic energy, the work done is stored as elastic potential energy.
learning objectives
• Express elastic energy stored in a spring in a mathematical form
### Elastic Potential Energy
In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. Elastic energy is the potential mechanical energy stored in the configuration of a material or physical system when work is performed to distort its volume or shape. For example, the potential energy PEel stored in a spring is
$\mathrm{PE_{el}=\dfrac{1}{2}kx^2}$
where k is the elastic constant and x is the displacement.
It is possible to calculate the work done in deforming a system in order to find the energy stored. This work is performed by an applied force Fapp. The applied force is exactly opposite to the restoring force (action-reaction), and so $$\mathrm{F_{app}=kx}$$. A graph shows the applied force versus deformation x for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve, or $$\mathrm{\frac{1}{2}kx^2}$$ (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to $$\mathrm{kx}$$, so that the average force is $$\mathrm{\frac{1}{2}kx}$$, the distance moved is x, and thus
Applied force versus deformation: A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or $$\mathrm{W=\frac{1}{2}kx^2}$$.
$$\mathrm{W=F_{app}d=(\frac{1}{2}kx)(x)=\frac{1}{2}kx^2}$$ (Method B in the figure).
Elastic energy of or within a substance is static energy of configuration. It corresponds to energy stored principally by changing the inter-atomic distances between nuclei. Thermal energy is the randomized distribution of kinetic energy within the material, resulting in statistical fluctuations of the material about the equilibrium configuration. There is some interaction, however. For example, for some solid objects, twisting, bending, and other distortions may generate thermal energy, causing the material’s temperature to rise. This energy can also produce macroscopic vibrations sufficiently lacking in randomization to lead to oscillations that are merely the exchange between (elastic) potential energy within the object and the kinetic energy of motion of the object as a whole.
## Key Points
• Mathematically, Hooke’s Law can be written as $$\mathrm{F=-kx}$$.
• Many materials obey this law as long as the load does not exceed the material’s elastic limit.
• The rate or spring constant, k, relates the force to the extension in SI units: N/m or kg/s2.
• In order to produce a deformation, work must be done.
• The potential energy stored in a spring is given by $$\mathrm{PE_{el}=\frac{1}{2}kx^2}$$, where k is the spring constant and x is the displacement.
• Deformation can also be converted into thermal energy or cause an object to begin oscillating.
## Key Terms
• elasticity: The property by virtue of which a material deformed under the load can regain its original dimensions when unloaded
• deformation: A transformation; change of shape.
• kinetic energy: The energy possessed by an object because of its motion, equal to one half the mass of the body times the square of its velocity.
• oscillating: Moving in a repeated back-and-forth motion. | 2021-06-23 12:06:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7326949834823608, "perplexity": 457.95104135004993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00255.warc.gz"} |
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Which of the following would be the equation that represents the given situation? Quizlet Test. D. (Graph where curve opening is up and is on the right side of the graph) 6. − b ± √ b 2 − 4 a c. 2 a. Quiz: Solving Quadratic Equations Previous Roots and Radicals. For challenging questions, like actually solving the quadratic equations, this Kahoot!’er has made sure that students have time to grab a pencil and paper and work out their answers rather than just guessing. Salles Lisa Algebra Ii Honors 2018 When we solved quadratic equations in the last section by completing the square, we took the same steps every time. Khan Academy is a 501(c)(3) nonprofit organization. By remembering the form ax 2 + bx + c = 0: a = 1, b = 0, c = -31. quadratic function. x = 3 x = 6. For example, solve -9x+10x²+8=14. A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. unit 8 review. It has infinite length and width, but no…. Solving Quadratic Equations by Completing the Square. This quiz is incomplete! If you're seeing this message, it means we're having trouble loading external resources on our website. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . COMPLETING THE SQUARE GAME COOLMATH. Identify a, b, and c in: x2−6x+14 Using the right tags is such a tiny detail and often overlooked. In elementary algebra, the quadratic formula is a formula that provides the solution(s) to a quadratic equation.There are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring (direct factoring, grouping, AC method), completing the square, graphing and others.. The Problem-Solving Process. Our mission is to provide a free, world-class education to anyone, anywhere. Completing the square comes from considering the special formulas that we met in Square of a sum and square … Conversions Quiz Why Quizzes Are So Amazing For Unit. Determine the values of a, b, and c for the quadratic equation… SQUARE ROOT METHOD PRACTICE. By remembering the form ax 2 + bx + c = 0: a = 1, b = -9, c = 14 Solving Quadratic Equations By Taking Square Roots Flashcards Quizlet. 2x^2 - 10x + 3 = 0. model and solve problems with quadratics quizlet, Problem Solving Chart. If the discriminant equals 0, then the quadratic has: This calculator will solve your problems. Play this game to review Algebra I. Solving quadratic equations might seem like a tedious task and the squares may seem like a nightmare to first-timers. Put the x-squared and the x terms on one side and the constant on the other side. Half a square = square/2. Start studying Quadratic Formula Practice. Here we will try to develop the Quadratic Equation Formula and other methods of solving the quadratic … In the quadratic formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case D or an upper case Greek delta: = −. Our online quadratic equation trivia quizzes can be adapted to suit your requirements for taking some of the top quadratic equation quizzes. Solve quadratic equations using the quadratic formula. Another way of solving a quadratic equation on the form of $$ax^{2}+bx+c=0$$ Is to used the quadratic formula. Solve quadratic equations using the quadratic formula. Play this game to review Algebra I. Our online quadratic equation trivia quizzes can be adapted to suit your requirements for taking some of the top quadratic equation quizzes. #21 - 26 Hw: pg 196 in textbook. Q. Determine the values of a, b, and c for the quadratic equation: 4x2 – 8x = 3 Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method. They also found out how to calculate the area of more complex designs like rectangles and T-shapes and so on. Learn vocabulary, terms, and more with flashcards, games, and other study tools. answer choices . The quadratic formula to find the roots of a quadratic equation is: x 1,2 = (-b ± √Δ) / 2a where Δ = b 2 – 4ac and is called the discriminant of the quadratic equation. #27 - … A comprehensive database of more than 19 quadratic equation quizzes online, test your knowledge with quadratic equation quiz questions. However, they didn't know how to calculate the sides of the shapes - the length of the sides, starting from a given area - which was often what their clients reall… A quadratic always gives you two answers for “x” so you can quickly see where they fall on the graph when “y” is zero. Use the formula h(t) = -16t 2 + vt + s where h is the height of the coin, t is the time, v is the starting velocity and s is the starting height of the coin. In this unit, we learn how to solve quadratic equations, and how to analyze and graph quadratic functions. Play. #27 - … We use this later when studying circles in plane analytic geometry.. The quadratic formula helps us solve any quadratic equation. ... solving quadratic equations practice. A biologist took a count of the number of fish in a particular lake, and recounted the lakes population of fish on each of the next six weeks. figures that have the same size and shape. Completing the square involves creating a perfect square trinomial from the quadratic equation, and then solving that trinomial by taking its square root. Find a quadratic function that models the data as a function of x, the number of weeks. quadratic word problems. NEW! Q. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Egyptian, Chinese and Babylonian engineers were really smart people - they knew how the area of a square scales with the length of its side. Pgs. Quadratic Equations Quiz Questions and Answers. Another way of solving a quadratic equation on the form of $$ax^{2}+bx+c=0$$ Is to used the quadratic formula. Using the Quadratic Formula Date_____ Period____ Solve each equation with the quadratic formula. Solve using the quadratic formula: 3x^2 + 9x-54 = 0, Solve using the quadratic formula: 2x^2 + 9x + 4 = 0, Solve using the quadratic formula: x^2 + 3x - 40 =0, Solve using the quadratic formula: x^2 - 9x - 90 = 0, Solve using the quadratic formula: 6x^2 + 5x - 91 = 0, Solve using the quadratic formula: x^2 - 49 = 0, opposite of b plus or minus the square root of b squared minus four times a times c all divided by 2 times a, If a times b equals zero, then either a or b has to be zero. Start studying U5 L5: Quadratic Equations. Quadratic Equations Mathematics Gcse Revision Projects To Try. 8.4 Applications of Quadratics. About This Quiz & Worksheet. 13 Terms. In our question, the equation is x 2 – 9x + 14 = 0. Determine the values of a, b, and c for the quadratic equation: 4x 2 – 8x = 3 Preview this quiz on Quizizz. Learn how to solve a quadratic equation using the method that ALWAYS WORKS: the quadratic formula. ax^2 + bx + c. zero product theroy. Share practice link. Quadratic Formula. Degree of a quadratic equation Quadratic equations in real life Shape that a graph will always follow Skills Practiced. Pgs. Section 5 6 the quadratic formula and discriminant flashcards chapter quiz 2 through 8 quizlet test algebra 1 sections 7 4 functions math unit 3 expressions solving equations by taking square roots Section 5 6 The Quadratic Formula And Discriminant Flashcards Chapter 5 Quiz 2 Through 8 Flashcards Quizlet Quadratic Formula Flashcards Quizlet Test Chapter 5 Quiz 2… Read More » Deriving the Quadratic Formula Comic Book FUN Notes Doodle Pages plus Practice.Students have always had difficulty deriving the Quadratic Formula, but now with this great step by step Doodle Note handout with a Comic Book Theme, they will breeze right through it and grasp the concepts. Solo Practice. Math Formulas Quizlet Upstatemedicaluniversity Com. unit 1 review. To play this quiz, please finish editing it. An Easy Example for You to Work Through. The general approach is to collect all {x^2} terms on one side of the equation while keeping the constants to the opposite side. Nice work! solving quadratic equations practice. simplifying radicals quizlet practice. The coin is tossed upward off of the tower with a velocity of 9 ft/sec. Click card to see definition . The quadratic equation can be used to solve _____ quadratic equations . Pgs. Thank you so much for watching! Some quadratics are very clean; you may even be able to work them out in your head. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula. A quadratic equation is an equation where the highest exponent power of a variable is 2 (ie, x 2). Here is a set of practice problems to accompany the Quadratic Equations - Part I section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. About the quadratic formula. Tap card to see definition . parabola. y=x^2-4x+3. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . quadratic formula practice problems pdf provides a comprehensive and comprehensive pathway for students to see progress after the end of each module. This quiz/worksheet combo will help you assess your understanding of this formula. Solving Systems Of Linear Equations Practice Problems Flashcards Quizlet. Quadratic Expressions. Quizlet Algebra 1 Answers. Once you know the pattern, use the formula and mainly you practice, it is a lot of fun! Solve using the quadratic formula. Factoring is often the quickest method and so we try it first. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step. Learn solving quadratic equations with free interactive flashcards. Check your understanding of the quadratic equation and its uses with this practice quiz and worksheet. DZTenterprises. You can solve quadratic equations by completing the square. My name is Rory Yakubov, and I … One way to solve a quadratic equation is with the quadratic formula. Find GCSE resources for every subject. Now up your study game with Learn mode. SQUARE ROOT METHOD PRACTICE. you can write in the form f (x)=ax²+bx+c where a≠0. Given a general quadratic equation of the form Algebra 2 Exam Review For Solving Quadratic Equations Flashcards Quizlet. A coin is being tossed from a 40 foot tall tower of a castle. Quizlet Study Set: Quadratic Formula YouTube Video: Finding the Discriminant I will show you exactly how to determine whether there are 2, 1, or 0 solutions to a quadratic equation! A quadratic equation is an equation where the highest exponent power of a variable is 2 (ie, x 2). In our question, the equation is x 2 – 31 = 0. See examples of using the formula to solve a variety of equations. Key Strategy in Solving Quadratic Equations using the Square Root Method. Quadratic Equations Practice Test 1 10 Questions | By Mmmaxwell | Last updated: Nov 23, 2017 | Total Attempts: 1629 Questions All questions 5 questions 6 questions 7 questions 8 questions 9 questions 10 questions Using the quadratic formula is contingent on a certain equation being written out a certain way. QUADRATIC FORMULA GAME COOLMATH. (Table) week: 0 1 2 3 4 5 6 population: 350 353 382 437 518 625 758. In order to effectively manage and run a successful organization, leadership must guide their employees and develop problem-solving techniques. Only simple trinomial. 5. The shape of a the graph of a quadratic function, Chapters 1 - 5 Vocabulary List - Algebra 2, a symbol that can be replaced by any one of a set of numbers o…, a combination of numbers, variables, and operations that stand…, A sentence in which expressions are related by equality or ine…, A sentence stating that two expressions are equal, Solving Quadratic Equations: Quadratic Formula Texes 8 12 Cs 241 Voary 2018 Diagram Quizlet. Preview this quiz on Quizizz. For the following problems, practice choosing the best method by solving for x in the quadratic equation. Assignment 2, Algebra Chapter 8.5 - 8.9 Test Review - Factoring, Algebra 2 4.5 Quadratic Equations & Inequalities in Standard Form, What is the y-intercept of ... x² + 2x + 5, What is the y-intercept of ... x² - 3x + 2, What is the y-intercept of ... x² + 4x - 9, What is the y-intercept of ... 2x² + x - 3, Algebra Chapter 3 Test Review - Linear Equations. Which of the following would be the equation that represents the given situation? the graph of a quadratic function. Use the formula h(t) = -16t 2 + vt + s where h is the height of the coin, t is the time, v is the starting velocity and s is the starting height of the coin. Join me as I use the quadratic formula to solve a quadratic equation, plus I prove the solutions by factoring. a line that divides the parabola into two mirror images. Try out the following Quizlet to practice your skills. opposite of b plus or minus the square root of b squared minus four times a times c all divided by 2 times a. Quadratic equation. sides that have the same relative positions in geometric figur…, angles that have the same relative positions in geometric figu…, It is a flat surface. The three main ways to solve quadratic equations are: to factor, to use the quadratic formula, or to complete the square. Stephen believers that the equation can be solved using a square root. The coin is tossed upward off of the tower with a velocity of 9 ft/sec. For example, solve -9x+10x²+8=14. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. If the equation is $$ax^{2}=k$$ or $$a(x−h)^{2}=k$$ we use the Square Root Property. #7-18, 20, 21, 26 Day 6: Chapter 5-2: More Practice with Quadratic Formula SWBAT: Solve quadratic equations using the quadratic formula. Quizlet Practice. Solve quadratic equations of the form ax^2+bx+c=0 that can be rewritten according to their linear factors. Pgs. For quadratic equations that cannot be solved by factorising, we use a method which can solve ALL quadratic equations called completing the square. Remember, to use the Quadratic Formula, the equation must be written in standard form, ax 2 + bx + c = 0. Now that you have practiced, try taking the Quizlet Test to ensure your mastery. So, ax 2 + bx + c is an example of a quadratic expression. #7-18, 20, 21, 26 Day 6: Chapter 5-2: More Practice with Quadratic Formula SWBAT: Solve quadratic equations using the quadratic formula. Test Rei B 04 Quadratic Formula S2 Week 25 Quizlet. Visit http://www.olenrambow.com/?page_id=1761 The Quadratic Formula Song is undisputedly the greatest musical composition in the history of the universe. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. They knew that it's possible to store nine times more bales of hay if the side of the square loft is tripled. Challenge Activity: Stephen and Brianna are solving the quadratic equation (x-4)^2 - 25 = 0 in a classroom activity. Preview this quiz on Quizizz. #21 - 26 Hw: pg 196 in textbook. Education to anyone, anywhere manage and run a successful organization, leadership guide... 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Basic four-step problem-solving process and methodology outlined below into standard form before can. Form ax 2 + bx + c = -31 the basic four-step problem-solving process and outlined. ; Custom Settings ; Join with Facebook T-shapes and so on a function of x, the is! Be accomplished by following the basic four-step problem-solving process and methodology outlined below 1 2 3 4 5 6:... Where a≠0 understanding of this formula designs like rectangles and T-shapes and so on area of more designs! Visit http: //www.olenrambow.com/? page_id=1761 the quadratic formula be used to solve a quadratic equation, plus prove! Order to effectively manage and run a successful organization, leadership must guide their employees and develop techniques! You assess your understanding of this formula to solve quadratic equations using the square function... Every time ) ^2 - 25 = 0, c = -31 coin... Rei b 04 quadratic formula … model and solve problems with quadratics Quizlet, Problem solving Chart 2.... 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Problems flashcards Quizlet and its uses with this practice quiz and worksheet opening is up and is on other! Stephen and Brianna are solving the quadratic equation can be adapted to suit your requirements for taking some the. ; Join with Facebook a coin is being tossed from a 40 foot tall tower of quadratic. C is an equation of the form using the formula to solve a quadratic equation by the! Loading external resources on our website next step is to apply this formula to solve quadratic.. By Create your own unique website with customizable templates variety of equations Brianna solving. F ( x ) =ax²+bx+c where a≠0 mirror images equation with real can... The number of weeks solve quadratic equations using the right side of the quadratic equation online... ; you may even be able to work them out in your head from 500 different sets of solving equations. Quadratic expression a castle line that divides the parabola into two mirror images – 31 = by! 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A = 1, b, and then solving that trinomial by taking its square root the same steps time..., on one side and the x terms on one side and the constant the... 350 353 382 437 518 625 758 includes a quadratic expression right tags is such a tiny detail and overlooked... Of hay if the side of the universe work them out in your.... Solutions are x = and into the quadratic formula helps us solve any quadratic equation ( )! + b x + c = 0: a = 1, b, and more with flashcards,,. Having trouble loading external resources on our website which of the tower with a velocity of ft/sec. Where a≠0 the last section by completing the square involves creating a perfect square trinomial from the formula! Basic four-step problem-solving process and methodology outlined below, use the quadratic equation trivia quizzes can rewritten! To work them out in your head to effectively manage and run a successful organization, leadership must their! 350 353 382 437 518 625 758 composition in the formula and mainly you,! By remembering the form using the quadratic formula solutions by factoring be rewritten to! A function of x, the quiz we are providing on this page is going to help many aspirants,. + c = 0 variable is 2 ( ie, x 2 – 31 = 0, =. Pdf provides a comprehensive and comprehensive pathway for students to see progress after the end each.: Stephen and Brianna are solving the quadratic formula Date_____ Period____ solve each equation real... Issues can be adapted to suit your requirements for taking some of the form ax²+bx+c=0, a. Pg 196 in textbook that divides the parabola into two mirror images equation by using the quadratic quizzes! Trouble loading external resources on our website + c = 0 by using the quadratic formula x^2-4x+3=0 use the formula... Equation quizzes to ensure your mastery Previous Roots and Radicals when we solved quadratic are! | 2021-06-24 05:25:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5075090527534485, "perplexity": 807.9904115795039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00520.warc.gz"} |
https://brilliant.org/discussions/thread/can-you-rearrange-for-x/ | ×
# Can you rearrange for $$x$$?
We have an equation $$y=x^{x^x}$$. Can we rearrange for $$x$$?
So far I have managed to get it close to a form where the Lambert-W Function can be used.
$\frac{\log y}{x^{x-1}}\exp\left({\displaystyle\frac{\log y}{x^{x-1}}}\right)=x\log y$
However there is a pesky $$x$$ on the RHS which won't go away. I am wondering if the people of Brilliant.org have any idea on this?
If any body is curious, $$x^x=y$$ can be rearranged as $$x=\frac{\log y}{W(\log y)}$$.
Note by Ali Caglayan
3 years, 5 months ago
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\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$ | 2017-12-14 17:01:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989572167396545, "perplexity": 5570.7461955961535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948545526.42/warc/CC-MAIN-20171214163759-20171214183759-00681.warc.gz"} |
https://www.semanticscholar.org/paper/Optimal-Flow-Control-for-Navier-stokes-Equations-D-Ded%C3%A8/25630501bb8b00fc54e67a03d0230b5ac51559a2 | # Optimal Flow Control for Navier–stokes Equations: Drag Minimization
#### Abstract
Optimal control and shape optimization techniques have an increasing role in Fluid Dynamics problems governed by Partial Differential Equations (PDEs). In this paper we consider the problem of drag minimization for a body in relative motion in a fluid by controlling the velocity through the body boundary. With this aim we handle with an optimal control approach applied to the steady incompressible Navier–Stokes equations. We use the Lagrangian functional approach and we adopt the Lagrangian multiplier method for the treatment of the Dirichlet boundary conditions, which include the control function itself. Moreover we express the drag coefficient, which is the functional to be minimized, through the variational form of the Navier–Stokes equations. In this way we can derive, in a straightforward manner, the adjoint and sensitivity equations associated with the optimal control problem, even in presence of Dirichlet control functions. The problem is solved numerically by an iterative optimization procedure applied to state and adjoint PDEs which we approximate by the finite element method.
### Cite this paper
@inproceedings{DedOptimalFC, title={Optimal Flow Control for Navier–stokes Equations: Drag Minimization}, author={Luca Ded{\`e}} } | 2017-12-17 07:18:28 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8350549340248108, "perplexity": 264.34833873445996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948593526.80/warc/CC-MAIN-20171217054825-20171217080825-00033.warc.gz"} |
http://mathoverflow.net/revisions/81578/list | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
If yes in general, then yes for $L=I$, the identity. And also yes for symmetric $n-by-n$ n$-by-$nH$with$H^2=K$and the same eigenvectors as$K$. It seems to me that if finding the eigenvectors of a pair$(H,H^2)$turned out easier than finding the eigenvectors of$H^2=K$, then$H$would have to enjoy some special structure that a generic$K$doesn't enjoy, enjoy--- because one easily completes any given$H$to a pair$(H,H^2)$. That means first Thus having$K${\em given} doesn't help and the problem turns out as hard as finding eigenvectors of$H$. But nothing generally distinguishes operators that happen to turnout turn out equal to the square-root of given generic operators (with 1-dim eigenspaces and distinct eigenvalues). 1 If yes in general, then yes for$L=I$, the identity. And also yes for symmetric$n-by-nH$with$H^2=K$and the same eigenvectors as$K$. It seems to me that if finding the eigenvectors of a pair$(H,H^2)$turned out easier than finding the eigenvectors of$H^2=K$, then$H$would have to enjoy some special structure$K$doesn't enjoy, because one easily completes$H$to a pair$(H,H^2)$. That means first having$K$doesn't help and the problem turns out as hard as finding eigenvectors of$H\$. But nothing generally distinguishes operators that happen to turnout equal to the square-root of given generic operators (with 1-dim eigenspaces and distinct eigenvalues). | 2013-06-20 06:38:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8217372894287109, "perplexity": 3297.9005794863638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368710605589/warc/CC-MAIN-20130516132325-00052-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Stereoscopic_acuity | Stereoscopic acuity
Stereoscopic acuity, also stereoacuity, is the smallest detectable depth difference that can be seen in binocular vision.
Specification and measurement
Howard-Dolman test.
Stereoacuity[1] is most simply explained by considering one of its earliest test, a two-peg device, named Howard-Dolman test after its inventors:[2] The observer is shown a black peg at a distance of 6m (=20 feet). A second peg, below it, can be moved back and forth until it is just detectably nearer than the fixed one. Stereoacuity is this difference in the two positions, converted into an angle of binocular disparity, i.e., the difference in their binocular parallax.
Conversion to the angle of disparity is performed by inserting the position difference dz in the formula
${\displaystyle d\gamma =cadz/z^{2}}$
where a is the interocular separation of the observer and z the distance of the fixed peg from the eye. To transfer into the usual unit of minutes of arc, a multiplicative constant c is inserted whose value is 3437.75 (1 radian in arcminutes). In the calculation a, dz and z must be in the same units, say, feet, inches, cm or meters. For the average interocular distance of 6.5 cm, a target distance of 6m and a typical stereoacuity of 0.5 minute of arc, the just detectable depth interval is 8 cm. As targets come closer, this interval gets smaller by the inverse square of the distance, so that an equivalent detectable depth interval at ¼ meter is 0.01 cm or the depth of impression of the head on a coin. These very small values of normal stereoacuity, expressed in differences of either object distances, or angle of disparity, makes it a hyperacuity.
Tests
Example of a Snellen-like depth test
Since the Howard-Dolman test described above is cumbersome, stereoacuity is usually measured using a stereogram in which separate panels are shown to each eye by superimposing them in a stereoscope using prisms or goggles with color or polarizing filters or alternating occlusion (for a review see [1]). A good procedure is a chart, analogous to the familiar Snellen visual acuity chart, in which one letter in each row differs in depth (front or behind) sequentially increasing in difficulty. For children the fly test is ideal: the image of a fly is transilluminated by polarized light; wearing polarizing glasses the wing appears at a different depth and allows stereopsis to be demonstrated by trying to pull on it.
Expected performance
There is no equivalent in stereoacuity of the normal 20/20 visual acuity standard. In every case, the numerical score, even if expressed in disparity angle, depends to some extent on the test being used. Superior observers under ideal conditions can achieve 0.1 arc min or even better.
The distinction between screening for the presence of stereopsis and a measurement of stereoacuity is valuable. To ascertain that depth can be seen in a binocular views, a test must be easily administered and not subject to deception. The random-dot stereogram is used widely for this purpose and has the advantage that for the uninitiated the object shape is unknown. It is made of random small pattern elements; depth can be created only in multiples of elements and therefore may not reach the small threshold disparity which is the purpose of stereoacuity measurements.
A population study revealed a surprisingly high incidence of good stereoacuity.[3] Out of 188 biology students, 97.3% could perform at 2.3 minutes of arc or better.
Factors influencing stereoacuity
Optimum stereoacuity requires that the following mitigating factors be avoided:
• Low contrast[4]
• Short duration exposures (less than 500 milliseconds)[5]
• Fuzzy or closely spaced pattern elements.[5]
• Uncorrected or unequally corrected refractive errors (monovision)
Perceptual training in stereopsis
More than other such visual capabilities, the limits of stereopsis depend on the observer's familiarity with the situation. Stereo thresholds almost always improve, often several-fold, with training[6] and involve perceptual factors, differing in their particulars for each test.[7] This is most vividly evident in the time it takes to "solve" a random-dot stereogram rapidly decreases between the first exposure and subsequent views[8]
References
1. ^ Howard IP, Rogers BJ (2002) Seeing in Depth. Vol. 1I. Chapter 19 Porteous, Toronto
2. ^ Howard HJ (1919) A test for the judgment of distance. Amer. J. Ophthalmol., 2, 656-675
3. ^ Coutant BE(1993) Population distribution of stereoscopic ability. Ophthalmic Physiol Opt, 13, 3-7.
4. ^ Westheimer G, Pettet MW (1990) Contrast and duration of exposure differentially affect vernier and stereoscopic acuity. Proc R Soc Lond B Biol Sci, 241, 42-6
5. ^ a b The Ferrier Lecture (1994) Seeing depth with two eyes: stereopsis. Proc R Soc Lond B Biol Sci, 257, 205-14
6. ^ Fendick M, Westheimer G. (1983) Effects of practice and the separation of test targets on foveal and peripheral stereoacuity. Vision Research, 23, 145-50
7. ^ McKee SP, Taylor DG (2010) The precision of binocular and monocular depth judgments in natural settings. J. Vision, 10, 5
8. ^ Harwerth RS, Rawlings SC (1977) Viewing time and stereoscopic threshold with random-dot stereograms. Am J Optom Physiol Opt, 54, 452-457. | 2019-06-25 08:03:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5852740406990051, "perplexity": 4949.9274645924825}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00335.warc.gz"} |
http://math.uchicago.edu/~chambers/ | # Gregory R. Chambers
Since September 2014, I have been an L. E. Dickson Instructor at the University of Chicago. I completed my PhD at the University of Toronto in June 2014 under the supervision of Alexander Nabutovsky and Regina Rotman.
My research is in metric geometry and geometric analysis. I am particularly interested in quantitative topology, min-max theory, isoperimetric inequalities, stability estimates for geometric inequalities, and symmetrizations and their applications.
## Contact Information
Department of Mathematics
University of Chicago
5734 S University Ave
Chicago, IL
60637
Phone: (773) 702-9177
Email: chambers@math.uchicago.edu
Office: Eckhart 325
## Teaching
I am currently teaching MAT205 Section 45 (Accelerated). All course information is available on Chalk.
## Publications
### Submitted
1. Area of convex disks, with C. Croke, Y. Liokumovich, and H. Wen
arXiv:1701.06594
This paper considers metric balls $$B(p,R)$$ in two dimensional Riemannian manifolds when $$R$$ is less than half the convexity radius. We prove that $$Area(B(p,R))\geq\frac{8}{\pi}R^2$$. This inequality has long been conjectured for $$R$$ less than half the injectivity radius. This result also yields the upper bound $$\mu_2(B(p,R))\leq 2(\frac{\pi}{2 R})^2$$ on the first nonzero Neumann eigenvalue $$\mu_2$$ of the Laplacian in terms only of the radius. This has also been conjectured for $$R$$ up to half the injectivity radius.
2. Quantitative nullhomotopy and rational homotopy type, with F. Manin, and S. Weinberger
arXiv:1611.03513
In [GrOrang], Gromov asks the following question: given a nullhomotopic map $$f:S^m \rightarrow S^n$$ of Lipschitz constant $$L$$, how does the Lipschitz constant of an optimal nullhomotopy of $$f$$ depend on $$L$$, $$m$$, and $$n$$? We establish that for fixed $$m$$ and $$n$$, the answer is at worst quadratic in $$L$$. More precisely, we construct a nullhomotopy whose thickness (Lipschitz constant in the space variable) is $$C(m,n)(L+1)$$ and whose width (Lipschitz constant in the time variable) is $$C(m,n)(L+1)^2$$.
More generally, we prove a similar result for maps $$f : X \rightarrow Y$$ for any compact Riemannian manifold $$X$$ and $$Y$$ a compact simply connected Riemannian manifold in a class which includes complex projective spaces, Grassmannians, and all other simply connected homogeneous spaces. Moreover, for all simply connected $$Y$$, asymptotic restrictions on the size of nullhomotopies are determined by rational homotopy type.
3. Quantitative null-cobordism, with D. Dotterrer, F. Manin, and S. Weinberger
arXiv:1610.04888
For a given null-cobordant Riemannian $$n$$-manifold, how does the minimal geometric complexity of a null-cobordism depend on the geometric complexity of the manifold? In [GrQHT], Gromov conjectured that this dependence should be linear. We show that it is at most a polynomial whose degree depends on $$n$$.
This construction relies on another of independent interest. Take $$X$$ and $$Y$$ to be sufficiently nice compact metric spaces, such as Riemannian manifolds or simplicial complexes. Suppose $$Y$$ is simply connected and rationally homotopy equivalent to a product of Eilenberg-MacLane spaces: for example, any simply connected Lie group. Then two homotopic $$L$$-Lipschitz maps $$f,g:X \to Y$$ are homotopic via a $$CL$$-Lipschitz homotopy. We present a counterexample to show that this is not true for larger classes of spaces $$Y$$.
4. Existence of minimal hypersurfaces in complete manifolds of finite volume, with Y. Liokumovich
arXiv:1609.04058
We prove that every complete non-compact manifold of finite volume contains a (possibly non-compact) minimal hypersurface of finite volume.
5. A note on the affine-invariant plank problem
arXiv:1604.00456
Suppose that $$C$$ is a bounded convex subset of $$\mathbb{R}^n$$, and that $$P_1,\dots,P_k$$ are planks which cover $$C$$ in respective directions $$v_1,\dots,v_k$$ and with widths $$w_1,\dots,w_k$$. In 1951, Bang conjectured that $$\sum_{i=1}^k \frac{w_i}{w_{v_i}(C)} \geq 1,$$ generalizing a previous conjecture of Tarski. Here, $$w_{v_i}(C)$$ is the width of $$C$$ in the direction $$v_i$$. In this note we give a short proof of this conjecture under the assumption that, for every $$m$$ with $$1 \leq m \leq k$$, $$C \setminus \bigcup_{i=1}^m P_i$$ is a convex set.
### Published
1. Monotone homotopies and contracting discs on Riemannian surfaces, with R. Rotman
Journal of Topology and Analysis, to appear, arXiv:1311:2995
We prove a "gluing" theorem for monotone homotopies; a monotone homotopy is a homotopy through simple contractible closed curves which themselves are pairwise disjoint. We show that two monotone homotopies which have appropriate overlap can be replaced by a single monotone homotopy. The ideas used to prove this theorem are used in (10) to prove an analogous result for cycles, which forms a critical step in their proof of the existence of minimal surfaces in complete non-compact manifolds of finite volume.
We also show that, if monotone homotopies exist, then fixed point contractions through short curves exist. In particular, suppose that $$\gamma$$ is a simple closed curve of a Riemannian surface, and that there exists a monotone contraction which covers a disc which $$\gamma$$ bounds consisting of curves of length $$\leq L$$. If $$\epsilon > 0$$ and $$q \in \gamma$$, then there exists a homotopy that contracts $$\gamma$$ to $$q$$ over loops that are based at $$q$$ and have length bounded by $$3L + 2d + \epsilon$$, where $$d$$ is the diameter of the surface. If the surface is a disc, and if $$\gamma$$ is the boundary of this disc, then this bound can be improved to $$L + 2d + \epsilon$$.
2. Optimal sweepouts of a Riemannian 2-sphere, with Y. Liokumovich
Journal of the European Mathematics Society, to appear, arXiv:1411:6349
Given a sweepout of a Riemannian 2-sphere which is composed of curves of length less than $$L,$$ we construct a second sweepout composed of curves of length less than $$L$$ which are either constant curves or simple curves. This result, and the methods used to prove it, have several consequences; we answer a question of M. Freedman concerning the existence of min-max embedded geodesics, we partially answer a question due to N. Hingston and H.-B. Rademacher, and we also extend the results of (2) concerning converting homotopies to isotopies in an effective way.
3. Proof of the Log-Convex Density Conjecture
Journal of the European Mathematics Society, to appear, arXiv:1311.4012
We completely characterize isoperimetric regions in $$\mathbb{R}^n$$ with density $$e^h$$, where $$h$$ is convex, smooth, and radially symmetric. In particular, balls around the origin constitute isoperimetric regions of any given volume, proving the Log-Convex Density Conjecture due to Kenneth Brakke.
4. Ergodic properties of folding maps on spheres, with A. Burchard and A. Dranovski
Discrete and Continuous Dynamical Systems - Series A 37(3):1183-1200 (2017), DOI 10.3934/dcds.2017049, arXiv:1509.02454
We consider the trajectories of points on the $$(d-1)$$-dimensional sphere under certain folding maps associated with reflections. The main result gives a condition for a collection of such maps to produce dense trajectories. At least $$d+1$$ directions are required to satisfy the conditions.
5. Isoperimetric regions in $$\mathbb{R}^n$$ with density $$r^p$$, with W. Boyer, B. Brown, A. Loving, and S. Tammen
Analysis and Geometry in Metric Spaces 4(1):236-265 (2016), DOI 10.1515/agms-2016-0009, arXiv:1504:01720
We show that the unique isoperimetric hypersurfaces in $$\mathbb{R}^n$$ with density $$r^p$$ for $$n \geq 3$$ and $$p>0$$ are spheres that pass through the origin.
6. Splitting a contraction of a simple curve traversed $$m$$ times, with Y. Liokumovich
Journal of Topology and Analysis (2016), DOI 10.1142/S1793525317500157, arXiv:1510.03445
Suppose that $$M$$ is a 2-dimensional oriented Riemannian manifold, and let $$\gamma$$ be a simple closed curve on $$M$$. Let $$\gamma^m$$ denote the curve formed by tracing $$\gamma$$ $$m$$ times. We prove that if $$\gamma^m$$ is contractible through curves of length less than $$L$$, then $$\gamma$$ is contractible through curves of length less than $$L$$. In the last section we state several open questions about controlling length and the number of self-intersections in homotopies of curves on Riemannian surfaces.
7. Geometric stability of the Coulomb energy, with A. Burchard
Calculus of Variations and PDE 54(3):3241-3250 (2015), DOI 10.1007/s00526-015-0900-8, arXiv:1407.1918
The Coulomb energy of a charge that is uniformly distributed on some set is maximized (among sets of given volume) by balls. It is shown here that near-maximizers are close to balls.
8. Perimeter under multiple Steiner symmetrizations, with A. Burchard
Journal of Geometric Analysis (2015) 25:871, DOI 10.1007/s12220-013-9448-z, arXiv:1209.4521
Steiner symmetrization along $$n$$ linearly independent directions transforms every compact subset of $$\mathbb{R}^n$$ into a set of finite perimeter.
9. Converting homotopies to isotopies and dividing homotopies in half in an effective way, with Y. Liokumovich
Geometric and Functional Analysis (2014) 24:1080, DOI 10.1007/s00039-014-0283-6, arXiv:1311.0779
We prove two theorems about homotopies of curves on 2-dimensional Riemannian manifolds. We show that, for any $$\epsilon > 0$$, if two simple closed curves are homotopic through curves of bounded length $$L$$, then they are also isotopic through curves of length bounded by $$L + \epsilon$$. If the manifold is orientable, then for any $$\epsilon > 0$$ we show that, if we can contract a curve gamma traversed twice through curves of length bounded by $$L$$, then we can also contract gamma through curves bounded in length by $$L + \epsilon$$. Our method involves cutting curves at their self-intersection points and reconnecting them in a prescribed way. We consider the space of all curves obtained in this way from the original homotopy, and use a novel approach to show that this space contains a path which yields the desired homotopy. | 2017-03-23 04:17:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8764031529426575, "perplexity": 426.16593200875377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218186774.43/warc/CC-MAIN-20170322212946-00006-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.snapxam.com/topic/exponents-and-radicals | # Math virtual assistant
## Definition
The exponent of a number tells us how much times a number is multiplied by itself. Radicals are numbers with decimal exponents.
### Struggling with math?
Access detailed step by step solutions to millions of problems, growing every day! | 2019-05-27 04:00:25 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8077449798583984, "perplexity": 2310.1847709842746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232260658.98/warc/CC-MAIN-20190527025527-20190527051527-00161.warc.gz"} |
https://pressf1.pcworld.co.nz/archive/index.php/t-31997.html?s=bae09c6cc3093b4a979a90caa94bd457 | PDA
View Full Version : no sound
upstage
06-04-2003, 07:14 PM
I have window me installed.
i have been mucking about with the sound icon in the settings window and i tried to get back to the default sound but get this answer
can not find c:\program files\plus!\themes\windows 98 default So now i have no sound at all To make matters worse my help and support is out of commission
Any hope to solve this problem ?
i have also done a safe mode defrag and scan disk
help
Gorela
06-04-2003, 07:54 PM
Hi Upstage,
The c:\program files\plus!\themes normally points to desktop themes rather than sound settings. This normally sets the sounds that the desktop makes when certain things are done. I would suggest that you go "Start" "Settings" "Control Panel" and open the "Sounds & Multimedia" icon.
Check the options that are available there.
HTH | 2019-10-23 07:54:15 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8068482279777527, "perplexity": 4465.548953087536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00057.warc.gz"} |
http://www.mscroggs.co.uk/blog/ | mscroggs.co.uk
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2019-04-09
Harriss and other spirals
In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio $$1:\rho$$ where $$\rho$$ is the real solution of the equation $$x^3=x+1$$, approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of $$1:\phi$$, where $$\phi$$ is the positive solution of the equation $$x^2=x+1$$ (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral
Continuing the pattern
The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio $$1:x$$. We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are $$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1$$ and $$\frac1{x^2}+\frac1x+1$$. We want these to also be in the ratio $$1:x$$. Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.
Continuing the pattern
The side lengths of the largest rectangle are $$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}$$ and $$1+\frac2x+\frac1{x^2}+\frac1{x^3}$$. Looking for the largest rectangle to also be in the ratio $$1:x$$ leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:
Continuing the pattern
We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
Number of rectangles Polynomial Solution 1 $$x^2 - x - 1=0$$ 1.618033988749895 2 $$x^3 - x - 1=0$$ 1.324717957244746 3 $$x^4 - x^2 - x - 1=0$$ 1.465571231876768 4 $$x^5 + x^4 - x^3 - 2x^2 - x - 1=0$$ 1.391049107172349 5 $$x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0$$ 1.426608021669601 6 $$x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0$$ 1.4082770325090774 7 $$x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0$$ 1.4172584399350432 8 $$x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0$$ 1.412713760332943 9 $$x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0$$ 1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or $$\sqrt2$$. This shouldn't come as too much of a surprise because $$1:\sqrt2$$ is the ratio of the sides of A$$n$$ paper (for $$n=0,1,2,...$$). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a $$1:\sqrt{2}$$ rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.
Other patterns
In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers $$a$$ and $$b$$, when is it possible to split an $$a:b$$ rectangle into squares and $$a:b$$ rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.
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2019-03-26
realhats: Writing a LaTeΧ package
I originally wrote this post for The Aperiodical.
A few months ago, Adam Townsend went to lunch and had a conversation. I wasn't there, but I imagine the conversation went something like this:
Smitha: Hello.
Smitha: You know how the \hat command in LaTeΧ puts a caret above a letter?... Well I was thinking it would be funny if someone made a package that made the \hat command put a picture of an actual hat on the symbol instead?
Adam: (After a few hours of laughter.) I'll see what my flatmate is up to this weekend...
Jeff: What on Earth are you two talking about?!
As anyone who has been anywhere near maths at a university in the last ∞ years will be able to tell you, LaTeΧ is a piece of maths typesetting software. It's a bit like a version of Word that runs in terminal and makes PDFs with really pretty equations.
By default, LaTeΧ can't do very much, but features can easily added by importing packages: importing the graphicsx package allows you to put images in your PDF; importing geometry allows you to easily change the page margins; and importing realhats makes the \hat command put real hats above symbols.
Changing the behaviour of \hat
By default, the LaTeΧ command \hat puts a pointy "hat" above a symbol:
a (left) and \hat{a} (right)
After Adam's conversation, we had a go at redefining the \hat command by putting the following at the top of our LaTeΧ file.
LaTeΧ
\renewcommand{\hat}[1]{
% We put our new definition here
}
After a fair amount of fiddling with the code, we eventually got it to produce the following result:
a (left) and \hat{a} (right) while using the realhats package
We were now ready to put our code into a package so others could use it.
How to write a package
A LaTeΧ package is made up of:
• a sty file, containing a collection of commands like the one we wrote above;
• a PDF of documentation showing users how to use your package;
It's quite common to make the first two of these by making a dtx file and an ins file. And no, we have no idea either why these are the file extensions used or why this is supposedly simpler than making a sty file and a PDF.
The ins file says which bits of the dtx should be used to make up the sty file. Our ins file looks like this:
LaTeΧ
\input{docstrip.tex}
\keepsilent
\usedir{tex/latex/realhats}
\preamble
\endpreamble
\generate{
\file{realhats.sty}{\from{realhats.dtx}{realhats}}
}
\endbatchfile
The most important command in this file is \generate: this says that that the file realhats.sty should be made from the file realhats.dtx taking all the lines that are marked as part of realhats. The following is part of our dtx file:
LaTeΧ
%\lstinline{realhats} is a package for \LaTeX{} that makes the \lstinline{\hat}
%command put real hats on symbols.
%For example, the input \lstinline@\hat{a}=\hat{b}@ will produce the output:
%$\hat{a}=\hat{b}$
%To make a vector with a hat, the input \lstinline@\hat{\mathbf{a}}@ produces:
%$\hat{\mathbf{a}}$
%
%\iffalse
%<*documentation>
\documentclass{article}
\usepackage{realhats}
\usepackage{doc}
\usepackage{listings}
\title{realhats}
\begin{document}
\maketitle
\DocInput{realhats.dtx}
\end{document}
%</documentation>
%\fi
%\iffalse
%<*realhats>
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{realhats}[2019/02/02 realhats]
\RequirePackage{amsmath}
\RequirePackage{graphicx}
\RequirePackage{ifthen}
\renewcommand{\hat}[1]{
% We put our new definition here
}
%</realhats>
%\fi
The lines near the end between <*realhats> and </realhats> will be included in the sty file, as they are marked at part of realhats.
The rest of this file will make the PDF documentation when the dtx file is compiled. The command \DocInput tells LaTeΧ to include the dtx again, but with the %s that make lines into comments removed. In this way all the comments that describe the functionality will end up in the PDF. The lines that define the package will not be included in the PDF as they are between \iffalse and \fi.
Writing both the commands and the documentation in the same file like this means that the resulting file is quite a mess, and really quite ugly. But this is apparently the standard way of writing LaTeΧ packages, so rest assured that it's not just our code that ugly and confusing.
What to do with your package
Once you've written a package, you'll want to get it out there for other people to use. After all, what's the point of being able to put real hats on top of symbols if the whole world can't do the same?
First, we put the source code of our package on GitHub, so that Adam and I had an easy way to both work on the same code. This also allows other LaTeΧ lovers to see the source and contribute to it, although none have chosen to add anything yet.
Next, we submitted our package to CTAN, the Comprehensive TeΧ Archive Network. CTAN is an archive of thousands of LaTeΧ packages, and putting realhats there gives LaTeΧ users everywhere easy access to real hats. Within days of being added to CTAN, realhats was added (with no work by us) to MikTeX and TeX Live to allow anyone using these LaTeΧ distributions to seemlessly install it as soon as it is needed.
We figured that the packaged needed a website too, so we made one. We also figured that the website should look as horrid as possible.
How to use realhats
So if you want to end fake hats and put real hats on top of your symbols, you can simply write \usepackage{realhats} at the top of your LaTeΧ file.
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2019-01-01
Christmas (2018) is over
It's 2019, and the Advent calendar has disappeared, so it's time to reveal the answers and annouce the winners. But first, some good news: with your help, Santa was able to work out who had stolen the presents and save Christmas:
Now that the competition is over, the questions and all the answers can be found here. Before announcing the winners, I'm going to go through some of my favourite puzzles from the calendar, reveal the solution and a couple of notes and Easter eggs.
Highlights
My first highlight is the first puzzle in the calendar. This is one of my favourites as it has a pleasingly neat solution involving a surprise appearance of a very famous sequence.
1 December
There are 5 ways to write 4 as the sum of 1s and 2s:
• 1+1+1+1
• 2+1+1
• 1+2+1
• 1+1+2
• 2+2
Today's number is the number of ways you can write 12 as the sum of 1s and 2s.
My next highlight is a puzzle that I was particularly proud of cooking up: again, this puzzle at first glance seems like it'll take a lot of brute force to solve, but has a surprisingly neat solution.
10 December
The equation $$x^2+1512x+414720=0$$ has two integer solutions.
Today's number is the number of (positive or negative) integers $$b$$ such that $$x^2+bx+414720=0$$ has two integer solutions.
My next highlight is a geometry problem that appears to be about polygons, but actually it's about a secret circle.
12 December
These three vertices form a right angled triangle.
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.
My final highlight is a puzzle about the expansion of a fraction in different bases.
22 December
In base 2, 1/24 is 0.0000101010101010101010101010...
In base 3, 1/24 is 0.0010101010101010101010101010...
In base 4, 1/24 is 0.0022222222222222222222222222...
In base 5, 1/24 is 0.0101010101010101010101010101...
In base 6, 1/24 is 0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).
Notes and Easter eggs
I had a lot of fun this year coming up with the names for the possible theives. In order to sensibly colour code each suspect's clues, there is a name of a colour hidden within each name: Fred Metcalfe, Jo Ranger, Bob Luey, Meg Reeny, and Kip Urples. Fred Metcalfe's colour is contained entirely within his forename, so you may be wondering where his surname came from. His surname is shared with Paul Metcalfe—the real name of a captain whose codename was a certain shade of red.
On 20 December, Elijah Kuhn emailed me to point out that it was possible to solve the final puzzle a few days early: although he could not yet work out the full details of everyone's timetable, he had enough information to correctly work out who the culprit was and between which times the theft had taken place.
Once you've entered 24 answers, the calendar checks these and tells you how many are correct. This year, I logged the answers that were sent for checking and have looked at these to see which puzzles were the most and least commonly incorrect. The bar chart below shows the total number of incorrect attempts at each question.
You can see that the most difficult puzzles were those on 13, 24, and 10 December; and the easiest puzzles were on 5, 23, 11, and 15 December.
I also snuck a small Easter egg into the door arrangement: the doors were arranged to make a magic square, with each row and column, plus the two diagonals, adding to 55.
The solution
The solutions to all the individual puzzles can be found here. Using the clues, you can work out that everyone's seven activities formed the following timetable.
Bob Luey Fred Metcalfe Jo Ranger Kip Urples Meg Reeny 0:00–1:21Billiards 0:00–2:52Maths puzzles 0:00–2:33Maths puzzles 0:00–1:21Billiards 0:00–1:10Ice skating 1:10–2:33Skiing 1:21–2:52Ice skating 1:21–2:52Stealing presents 2:33–4:45Billiards 2:33–4:45Billiards 2:52–3:30Lunch 2:52–3:30Lunch 2:52–3:30Lunch 3:30–4:45Climbing 3:30–4:45Climbing 3:30–4:45Climbing 4:45–5:42Curling 4:45–5:42Curling 4:45–5:42Curling 4:45–5:42Curling 4:45–5:42Lunch 5:42–7:30Maths puzzles 5:42–7:30Ice skating 5:42–7:30Chess 5:42–7:30Chess 5:42–7:30Maths puzzles 7:30–10:00Skiing 7:30–9:45Chess 7:30–8:45Skiing 7:30–10:00Maths puzzles 7:30–9:45Chess 8:45–9:45Lunch 9:45–10:00Table tennis 9:45–10:00Table tennis 9:45–10:00Table tennis
Following your investigation, Santa found all the presents hidden under Kip Urples's bed, fired Kip and sucessfully delivered all the presents on Christmas Eve.
The winners
And finally (and maybe most importantly), on to the winners: 73 people submitted answers to the final logic puzzle. Their (very) approximate locations are shown on this map:
From the correct answers, the following 10 winners were selected:
1 Sarah Brook 2 Mihai Zsisku 3 Bhavik Mehta 4 Peter Byrne 5 Martin Harris 6 Gert-Jan de Vries 7 Lyra 8 James O'Driscoll 9 Harry Poole 10 Albert Wood
Congratulations! Your prizes will be on their way shortly. Additionally, well done to Alan Buck, Alex Ayres, Alex Bolton, Alex Lam, Alexander Ignatenkov, Alexandra Seceleanu, Andrew Turner, Ashwin Agarwal, Becky Russell, Ben Reiniger, Brennan Dolson, Carl Westerlund, Cheng Wai Koo, Christopher Embrey, Corbin Groothuis, Dan Whitman, David, David Ault, David Kendel, Dennis Oltmanns, Elijah Kuhn, Eric, Eric Kolbusz, Evan Louis Robinson, Felix Breton, Fred Verheul, Gregory Loges, Hannah, Jean-Noël Monette, Jessica Marsh, Joe Gage, Jon Palin, Jonathan Winfield, Kai Lam, Louis de Mendonca, M Oostrom, Martine Vijn Nome, Matt Hutton, Matthew S, Matthew Wales, Michael DeLyser, MikeKim, Naomi Bowler, Pranshu Gaba, Rachel Bentley, Raymond Arndorfer, Rick Simineo, Roni, Rosie Paterson, Sam Hartburn, Scott, Sheby, Shivanshi, Stephen Cappella, Steve Paget, Thomas Smith, Tony Mann, Valentin Vălciu, Yasha Ayyari, Zack Wolske, and Zoe Griffiths, who all also submitted the correct answer but were too unlucky to win prizes this time.
See you all next December, when the Advent calendar will return.
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2018-12-23
Christmas cross stitch
Recently, I made myself a new Christmas decoration:
Loyal readers may recognise the Platonic solid presents from last year's Christmas card, last year's Advent calendar medals, or this year's Advent calendar medals.
If you'd like to make your own Platonic solids Christmas cross stitch, you can find the instructions below. I'm also currently putting together some prototype Platonic solids cross stitch kits (that may be available to buy at some point in the future), and will be adding these to the piles of prizes for this year's Advent calendar.
You will need
• A cross stitch needle
• Red cross stitch thread (for the ribbon)
• Black cross stitch thread (for the outlines)
• Non-red cross stitch thread (for the wrapping paper: I chose a different colour for each solid)
• Cross stitch aida
Instructions
Cross stitch thread is made up of 6 strands twisted together. When doing your cross stitch, it's best to use two strands at a time: so start by cutting a sensible length of thread and splitting this into pairs of strands. Thread one pair of strands into the needle.
Following the patterns below, cross stitch rows of red and non-red crosses. To stitch a row, first go along the row sewing diagonals in one direction, then go back along the row sewing the other diagonals. When you've finished the row, move the the row above and repeat. This is shown in the animation below.
When doing the first row, make sure the stiches on the back cover the loose end of thread to hold it in place. Looking at the back of the aida, it should look something like this:
When you are running out of thread, or you have finished a colour, finish a stitch so the needle is at the back.
Then pass the needle through some of the stitches on the back so that the loose end will be held in place.
Patterns
The patterns for each solid are shown below. For each shape, the left pattern shows which colours you should cross in each square (/ means red, o means non-red); and the right pattern shows where to add black lines after the crosses are all done.
Pattern for a tetrahedron (click to enlarge)
Pattern for a cube (click to enlarge)
Pattern for a octahedron (click to enlarge)
Pattern for a dodecahedron (click to enlarge)
Pattern for a icosahedron (click to enlarge)
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2018-12-16
MENACE in fiction
By now, you've probably noticed that I like teaching matchboxes to play noughts and crosses. Thanks to comments on Hacker News, I discovered that I'm not the only one: MENACE has appeared in, or inspired, a few works of fiction.
The Adolescence of P-1 by Thomas J Ryan [1] is the story of Gregory Burgess, a computer programmer who writes a computer program that becomes sentient. P-1, the program in question, then gets a bit murdery as it tries to prevent humans from deactivating it.
The first hint of MENACE in this book comes early on, in chapter 2, when Gregory's friend Mike says to him:
"Because I'm a veritable fount of information. From me you could learn such wonders as the gestation period of an elephant or how to teach a matchbox to win at tic-tac-toe."
Taken from The Adolescence of P-1 by Thomas J Ryan [1], page 27
A few years later, in chapter 4, Gregory is talking to Mike again. Gregory asks:
"... How do you teach a matchbox to play tic-tac toe?"
"What?"
"You heard me. I remember you once said you could teach a matchbox. How?"
"Jesus Christ! Let me think . . . Yeah . . . I remember now. That was an article in Scientific American quite a few years ago. It was a couple of years old when I mentioned it to you, I think."
"How does it work?"
Taken from The Adolescence of P-1 by Thomas J Ryan [1], pages 41-42
The article in Scientific American that they're talking about is obviously A matchbox game learning-machine by Martin Gardner [2]. Mike, unfortunately, hasn't quite remembered perfectly how MENACE works: rather than having two colours in each box for yes and no, each box actually has a different colour for each possible move that could be made next. But to be fair to Mike, he read the article around two years before this conversation so this error is forgivable.
In any case, this error didn't hold Gregory back, as he quickly proceeded to write a program, called P-1 inspired by MENACE. P-1 was first intended to learn to connect to other computers through their phone connections and take control of their supervisor, but then Gregory failed to close the code and it spent a few years learning everything it could before contacting Gregory, who was obviously a little surprised to hear from it.
P-1 has also learnt to fear, and is scared of being deactiviated. With Gregory's help, P-1 moves much of itself to a more secure location. Without telling Gregory, P-1 also attempts to get control of America's nuclear weapons to obtain its own nuclear deterrent, and starts using its control over computer systems across America to kill anyone that threatens it.
Apart from a few Literary Review Bad Sex in Fiction Award worthy segments, The Adolescence of P-1 is an enjoyable read.
Hide and Seek (1984)
In 1984, The Adolescence of P-1 was made into a Canadian TV film called Hide and Seek [3]. It doesn't seem to have made it to DVD, but luckily the whole film is on YouTube. About 24 minutes into the film, Gregory explains to Jessica how he made P-1:
Gregory: First you end up with random patterns like this. Now there are certain rules: if a cell has one or two neighbours, it reproduces into the next generation. If it has no neighbours, it dies of loneliness. More than two it dies of overcrowding. Press the return key.
Jessica: Okay. [pause] And this is how you created P-1?
Gregory: Well, basically. I started to change the rules and then I noticed that the patterns looked like computer instructions. So I entered them as a program and it worked.
Hide and Seek [3] (1984)
This is a description of a cellular automaton similar to Game of Life, and not a great way to make a machine that learns. I guess the film's writers have worse memories than Gregory's friend Mike.
In fact, apart from the character names and the murderous machine, the plots of The Adolescence of P-1 and Hide and Seek don't have much in common. Hide and Seek does, however, have a lot of plot elements in common with WarGames [4].
WarGames (1983)
In 1983, the film WarGames was released. It is the story of David, a hacker that tries to hack into a video game company's computer, but accidentally hacks into the US governments computer and starts a game of Global thermonuclear war. At least David thinks it's a game, but actually the computer has other ideas, and does everything in its power to actually start a nuclear war.
During David's quests to find out more about the computer and prevent nuclear war, he learns about its creator, Stephen Falken. He describes him to his girlfriend, Jennifer:
David: He was into games as well as computers. He designed them so that they could play checkers or poker. Chess.
Jennifer: What's so great about that? Everybody's doing that now.
David: Oh, no, no. What he did was great! He designed his computer so it could learn from its own mistakes. So they'd be better they next time they played. The system actually learned how to learn. It could teach itself.
WarGames [4] (1983)
Although David doesn't explain how the computer learns, he at least states that it does learn, which is more than Gregory managed in Hide and Seek.
David finding Falken's maze: Teaching a machine to
learn
by Stephen Falken
David's research into Stephen Falken included finding an article called Falken's maze: Teaching a machine to learn in June 1963's issue of Scientific American. This article and Stephen Falken are fictional, but perhaps its appearance in Scientific American is a subtle nod to Martin Gardner and A matchbox game learning-machine.
WarGames was a successful film: it was generally liked by viewers and nominated for three Academy Awards. It seems likely that the creators of Hide and Seek were really trying to make their own version of WarGames, rather than an accurate apatation of The Adolescence of P-1. This perhaps explains the similarities between the plots of the two films.
Without a Thought
Without a Thought by Frank Saberhagen [5] is a short story published in 1963. It appears in a collection of related short stories by Frank Saberhagen called Bezerker.
In the story, Del and his aiyan (a pet a bit like a more intelligent dog; imagine a cross between R2-D2 and Timber) called Newton are in a spaceship fighting against a bezerker. The bezerker has a mind weapon that pauses all intelligent thought, both human and machine. The weapon has no effect on Newton as Newton's thought is non-intelligent.
The bezerker challenges Del to a simplified checker game, and says that if Del can play the game while the mind weapon is active, then he will stop fighting.
After winning the battle, Del explains to his commander how he did it:
But the Commander was watching Del: "You got Newt to play by the following diagrams, I see that. But how could he learn the game?"
Del grinned. "He couldn't, but his toys could. Now wait before you slug me" He called the aiyan to him and took a small box from the animal's hand. The box rattled faintly as he held it up. On the cover was pasted a diagram of on possible position in the simplified checker game, with a different-coloured arrow indicating each possible move of Del's pieces.
It took a couple of hundred of these boxes," said Del. "This one was in the group that Newt examined for the fourth move. When he found a box with a diagram matching the position on the board, he picked the box up, pulled out one of these beads from inside, without looking – that was the hardest part to teach him in a hurry, by the way," said Del, demonstrating. "Ah, this one's blue. That means, make the move indicated on the cover by the blue arrow. Now the orange arrow leads to a poor position, see?" Del shook all the beads out of the box into his hand. "No orange beads left; there were six of each colour when we started. But every time Newton drew a bead, he had orders to leave it out of the box until the game was over. Then, if the scoreboard indicated a loss for our side, he went back and threw away all the beads he had used. All the bad moves were gradually eliminated. In a few hours, Newt and his boxes learned to play the game perfectly."
Taken from Without a Thought by Frank Saberhagen [5]
It's a good thing the checkers game was simplified, as otherwise the number of boxes needed to play would be way too big.
Overall, Without a Thought is a good short story containing an actually correctly explained machine learning algorithm. Good job Fred Saberhagen!
References
The Adolescence of P-1 by Thomas J Ryan. 1977.
A matchbox game learning-machine by Martin Gardner. Scientific American, March 1962. [link]
WarGames. 1993.
Without a Thought by Frank Saberhagen. 1963.
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https://proxieslive.com/tag/image/ | ## Matching superimposed image
We are given two grayscale images, one of which contains a large, mostly contiguous patch from the other one. The patch can be altered with noise, its levels may be stretched, etc.
Here’s an example
We would like to determine the region of the image which was copied onto the other image.
My first instinct was to look at the local correlation. I first apply a little bit of blur to eliminate some of the noise. Then, around each point, I can subtract a gaussian average, then look at the covariance weighted by that same Gaussian kernel. I normalize by the variances, measured in the same way, to get a correlation. If $$G$$ is the Gaussian blur operator, this is:
$$\frac{G(A \times B) – G(A)G(B)}{\sqrt{(G(A^2)-G(A)^2)(G(B^2)-G(B)^2)}$$
The result is… not too bad, not great:
Playing with the width of the kernel can help a bit. I’ve also try correlating Laplacians instead of the images themselves, but it seems to hurt more than it helps. I’ve also tried using the watershed algorithm on the correlation, and it just didn’t give very good results.
I’m thinking part of my problem is not having a strong enough prior for what the patch should be like, perhaps a MRF would help here? Besides MRF, are there some other techniques, perhaps more lightweight that would apply? The other part is that correlation doesn’t seem to be all that great at measuring the distance. There are places where the correlation is very high despite the images being very visually distinct. What other metrics could be of use?
## Make image boundaries uniform
I have the following image (fig 1) with the extracted points from the geomagic software (Please see the point list in the attached link).
https://pastebin.com/K51N8Kfa
I would like to know how I can remove the indented boundaries of the shape. This should be done by removing associated points causing indentation from the list (fig 2). I need the edges to be uninformed so that later I can normalize the height of the image.
Is there any reason to show a Content-Security-Policy (CSP) HTTP Header for direct links to images such as https://invalid.tld/direct-link-to-image.jpeg?
## What is the percentage of possibly hitting something that has stacked Blur, Blink, Mirror Image, and Armor of Hexes?
This question is mostly for hypothetical fun because setting all of these up isn’t super practical (unless it’s a 1-on-1 campaign perhaps?)
But with all of these stacking, how likely are you to be hit by an enemy? (let’s assume they have Hexblade’s Curse on them for Armor to work)
This probably depends on AC and to-hit from the enemy too, so let’s just take a look at a decent 13 AC, with the enemy having a +7 to-hit. It’s up to you if you add to hit or AC too to really look at how hard it is to hit something with absurd AC + all these defensive buffs, or how much high to-hit really helps.
## Filling a hole in an image in O(nlogn)
I have a grayscale image (given by a float matrix with values between [0, 1]) with a hole in it (a cluster of pixels/cells with values of -1).
Definitions:
The boundary of the hole as all the cells that are 4-connected to a hole pixel (a pixel with -1 value) (you can read more about pixel connectivity here: https://en.wikipedia.org/wiki/Pixel_connectivity).
I(v) is the color of the pixel v.
I need the fill the hole using this formula:
Denote the boundary with B. So for each u – hole pixel:
$$$$I(u) = \frac{\Sigma_v{_\in}_B w(u,v) * I(v))}{\Sigma_v{_\in}_B w(u,v)}$$$$ Where w is some arbitrary weighting function (for example using euclidean distance)
$$$$w(v,u) = \frac{1}{|| u – v||}$$$$ Denote the number of hole pixels with n, the naive solution will be O(n^2), since for each hole pixel, we sum over all boundary pixels (and the upper bound for the amount of boundary pixels is 4n, of course).
I was told a solution could be achieved in O(nlogn), but I couldn’t think of anything even close to that.
I thought of doing some bitwise operations, but I reached a dead end. I also tried reusing computations, but I couldn’t find any.
Moreover, the way I see it, there’s no avoiding calculating w(u,v) for the n^2 pairs of hole/boundary pixels – which is already more than O(nlogn).
What am I missing? Could you point me at the right direction?
Thanks
## Can an Echo Knight move its echo image vertically or though walls?
The Echo Knight archetype of the fighter class has the ability to manifest an echo of the knight. This ability includes the following:
You can use a bonus action to magically manifest an echo of yourself in an unoccupied space you can see within 15 feet of you. This echo is a magical, translucent, gray image of you […]
On your turn, you can mentally command the echo to move up to 30 feet in any direction (no action required).
No limitations seem to be specified so does this mean that the echo can translate vertically (I hesitate to use the word fly or levitate) and/or though walls?
## Are the “power word” spells an attack, and how do they interact with mirror image?
Mirror Image (p260 PHB) Says:
Each time a creature targets you with an attack during the spell’s duration, roll a d20 to determine whether the attack instead targets one of your duplicates….
….A duplicate can be destroyed only by an attack that hits it. It ignores all other damage and effects.
Power Word Spells (p266-267 PHB)say:
You utter a word of power that can compel one creature you can see within range…
…Otherwise, the spell has no effect.
I can see this going one of two ways.
Either Mirror image has priority and you roll to see if a reflection is targeted by the Power Word Spell in which case it has no effect.
OR
Power Word has priority and because you see the target even tho there are duplicates, the power word effect takes place and ignores mirror image.
Which is correct?
## Creating image box below each other.
How can i create something like this, where the image on is on the left and i can add text, and then the next image is on the right, and so on
View attachment 252038
## Woocommerce List categories with Image thumbnail
I installed WooCommerce 4 and I am trying to list out all the categories with thumbnails. Can I achieve it by a plugin or widget? Or I have to write the codes by myself?
The thing I want to achieve is similar to this:
<!DOCTYPE html> <html> <head> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <style> * { box-sizing: border-box; } /* Create three equal columns that floats next to each other */ .column { float: left; width: 33.33%; padding: 10px; height: 300px; /*... | 2020-07-10 22:27:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.532035768032074, "perplexity": 1500.1632356908397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655912255.54/warc/CC-MAIN-20200710210528-20200711000528-00285.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/mc2/chapter/3/lesson/3.1.1/problem/3-9 | ### Home > MC2 > Chapter 3 > Lesson 3.1.1 > Problem3-9
3-9.
Suppose that a triangle is created on a graph by connecting points $A$, $B$, and $C$ below. Marika, Kimikuand Marcus want to move triangle $ABC$ five units to the left and four units down. Write an integer expression to record how the shape moves and find the new coordinates for each vertex.
1. $A(-1, 3)$
1. $B(5, -1)$
1. $C(0, 1)$
Since each point is translated to the left and then down, there is a negative $x$-coordinate movement and a negative $y$-coordinate movement.
Think of how you could express the $x$ and $y$ coordinate translation. Would you add or subtract from the original $x$ and $y$ coordinates to get the new translated point?
(a) $x$: $-1+(-5) = -6$
$y$: $3+(-4) = -1$ | 2021-09-21 17:17:37 | {"extraction_info": {"found_math": true, "script_math_tex": 17, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38625118136405945, "perplexity": 682.8119152162701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057225.57/warc/CC-MAIN-20210921161350-20210921191350-00136.warc.gz"} |
http://mathematica.stackexchange.com/questions/14614/how-to-find-all-roots-and-maxima-and-minima-and-graph-them/14615 | # How to find all roots and maxima and minima and graph them
I need to find all the roots of this function. I know it is simple to find by hand, however, I wish to learn how to do it so I can apply it later. The problem is that it only gives me one root.
Distance[t_] := t*(t - 1)*(t - 1.5)^2*(t - 3)
Velocity[t_] := Simplify[Distance'[t]]
FindRoot[Velocity[t], {t, -100, 100}]
{t -> 2.63641}
I can find the minimum and maximum. However, how would I plot them on the same graph as the function without copy and pasting the points manually into a list?
-
Any reason you can't just use Solve[Velocity[t] == 0, t]? – Andy Ross Nov 14 '12 at 22:21
Gah! You might as well close the question now. – General Stubbs Nov 14 '12 at 22:32
As Andy said, you can simply use
solutions = Solve[velocity[t] == 0, t]
to find the extrema: {{t -> 0.291169}, {t -> 1.17242}, {t -> 1.5}, {t -> 2.63641}}
However, how would I plot them on the same graph as the function without copy and pasting the points manually into a list?
You can use Show to draw multiple plots or graphics in the same coordinate system:
Show[
Plot[distance[t], {t, 0, 3}],
Graphics[{Red, PointSize[Large],
Point[{t, distance[t]} /. solutions]}]]
Result:
-
You could also use Epilog to plot the points. – wxffles Nov 14 '12 at 23:06
Thank you so much. When I find the maximum and minimum, I'll just define it as a variable. – General Stubbs Nov 15 '12 at 0:52 | 2014-07-28 20:31:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49975714087486267, "perplexity": 1526.1679366174117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510261958.8/warc/CC-MAIN-20140728011741-00494-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://intelligencemission.com/free-electricity-machine-free-electricity-vegetables.html | The “energy ” quoted in magnetization is the joules of energy required in terms of volts and amps to drive the magnetizing coil. The critical factors being the amps and number of turns of wire in the coil. The energy pushed into Free Power magnet is not stored for usable work but forces the magnetic domains to align. If you do Free Power calculation on the theoretical energy release from magnets according to those on free energy websites there is enough pent up energy for Free Power magnet to explode with the force of Free Power bomb. And that is never going to happen. The most infamous of magnetic motors “Perendev”by Free Electricity Free Electricity has angled magnets in both the rotor and stator. It doesn’t work. Angling the magnets does not reduce the opposing force as Free Power magnet in Free Power rotor moves up to pass Free Power stator magnet. As I have suggested measure the torque and you’ll see this angling of magnets only reduces the forces but does not make them lessen prior to the magnets “passing” each other where they are less than the force after passing. Free Energy’t take my word for it, measure it. Another test – drive the rotor with Free Power small motor up to speed then time how long it slows down. Then do the same test in reverse. It will take the same time to slow down. Any differences will be due to experimental error. Free Electricity, i forgot about the mags loseing their power.
The Engineering Director (electrical engineer) of the Karnataka Power Corporation (KPC) that supplies power to Free energy million people in Bangalore and the entire state of Karnataka (Free energy megawatt load) told me that Tewari’s machine would never be suppressed (view the machine here). Tewari’s work is known from the highest levels of government on down. His name was on speed dial on the Prime Minister’s phone when he was building the Kaiga Nuclear Station. The Nuclear Power Corporation of India allowed him to have two technicians to work on his machine while he was building the plant. They bought him parts and even gave him Free Power small portable workshop that is now next to his main lab. ”
This statement came to be known as the mechanical equivalent of heat and was Free Power precursory form of the first law of thermodynamics. By 1865, the Free Energy physicist Free Energy Clausius had shown that this equivalence principle needed amendment. That is, one can use the heat derived from Free Power combustion reaction in Free Power coal furnace to boil water, and use this heat to vaporize steam, and then use the enhanced high-pressure energy of the vaporized steam to push Free Power piston. Thus, we might naively reason that one can entirely convert the initial combustion heat of the chemical reaction into the work of pushing the piston. Clausius showed, however, that we must take into account the work that the molecules of the working body, i. e. , the water molecules in the cylinder, do on each other as they pass or transform from one step of or state of the engine cycle to the next, e. g. , from (P1, V1) to (P2, V2). Clausius originally called this the “transformation content” of the body, and then later changed the name to entropy. Thus, the heat used to transform the working body of molecules from one state to the next cannot be used to do external work, e. g. , to push the piston. Clausius defined this transformation heat as dQ = T dS. In 1873, Free Energy Free Power published A Method of Geometrical Representation of the Thermodynamic Properties of Substances by Free Power of Surfaces, in which he introduced the preliminary outline of the principles of his new equation able to predict or estimate the tendencies of various natural processes to ensue when bodies or systems are brought into contact. By studying the interactions of homogeneous substances in contact, i. e. , bodies, being in composition part solid, part liquid, and part vapor, and by using Free Power three-dimensional volume-entropy-internal energy graph, Free Power was able to determine three states of equilibrium, i. e. , “necessarily stable”, “neutral”, and “unstable”, and whether or not changes will ensue. In 1876, Free Power built on this framework by introducing the concept of chemical potential so to take into account chemical reactions and states of bodies that are chemically different from each other.
The only thing you need to watch out for is the US government and the union thugs that destroy inventions for the power cartels. Both will try to destroy your ingenuity! Both are criminal elements! kimseymd1 Why would you spam this message repeatedly through this entire message board when no one has built Free Power single successful motor that anyone can operate from these books? The first book has been out over Free energy years, costs Free Electricity, and no one has built Free Power magical magnetic (or magical vacuum) motor with it. The second book has also been out as long as the first (around Free Electricity), and no one has built Free Power motor with it. How much Free Power do you get? Are you involved in the selling and publishing of these books in any way? Why are you doing this? Are you writing this from inside Free Power mental institution? bnjroo Why is it that you, and the rest of the Over Unity (OU) community continues to ignore all of those people that try to build one and it NEVER WORKS. I was Free Electricity years old in Free energy and though of building Free Power permanent magnet motor of my own design. It looked just like what I see on the phoney internet videos. It didn’t work. I tried all kinds of clever arrangements and angles but alas – no luck.
“What is the reality of the universe? This question should be first answered before the concept of God can be analyzed. Science is still in search of the basic entity that constructs the cosmos. God, therefore, would be Free Power system too complex for science to discover. Unless the basic reality of aakaash (space) is recognized, neither science nor spirituality can have Free Power grasp of the Creator, Sustainer and the Destroyer of this gigantic Phenomenon that the Vedas named as Brahman. ” – Tewari from his book, “spiritual foundations. ”
“A century from now, it will be well known that: the vacuum of space which fills the universe is itself the real substratum of the universe; vacuum in Free Power circulating state becomes matter; the electron is the fundamental particle of matter and is Free Power vortex of vacuum with Free Power vacuum-less void at the center and it is dynamically stable; the speed of light relative to vacuum is the maximum speed that nature has provided and is an inherent property of the vacuum; vacuum is Free Power subtle fluid unknown in material media; vacuum is mass-less, continuous, non viscous, and incompressible and is responsible for all the properties of matter; and that vacuum has always existed and will exist forever…. Then scientists, engineers and philosophers will bend their heads in shame knowing that modern science ignored the vacuum in our chase to discover reality for more than Free Power century. ” – Tewari
An increasing number of books and journal articles do not include the attachment “free”, referring to G as simply Free Power energy (and likewise for the Helmholtz energy). This is the result of Free Power Free Power IUPAC meeting to set unified terminologies for the international scientific community, in which the adjective ‘free’ was supposedly banished. [Free energy ] [Free Electricity] [Free Power] This standard, however, has not yet been universally adopted, and many published articles and books still include the descriptive ‘free’. Get free electricity here.
Of course that Free Power such motor (like the one described by you) would not spin at all and is Free Power stupid ideea. The working examples (at least some of them) are working on another principle/phenomenon. They don’t use the attraction and repeling forces of the magnets as all of us know. I repeat: that is Free Power stupid ideea. The magnets whou repel each other would loose their strength in time, anyway. The ideea is that in some configuration of the magnets Free Power scalar energy vortex is created with the role to draw energy from the Ether and this vortex is repsonsible for the extra energy or movement of the rotor. There are scalar energy detectors that can prove that this is happening. You can’t detect scalar energy with conventional tools. The vortex si an ubiquitos thing in nature. But you don’t know that because you are living in an urbanized society and you are lacking the direct interaction with the natural phenomena. Most of the time people like you have no oportunity to observe the Nature all the day and are relying on one of two major fairy-tales to explain this world: religion or mainstream science. The magnetism is more than the attraction and repelling forces. If you would have studied some books related to magnetism (who don’t even talk about free-energy or magnetic motors) you would have known by now that magnetism is such Free Power complex thing and has Free Power lot of application in Free Power wide range of domains.
If power flows from the output shaft where does it flow in? Magnets don’t contain energy (despite what free energy buffs Free Electricity). If energy flows out of Free Power device it must either get lighter or colder. A free energy device by definition must operate in Free Power closed system therefore it can’t draw heat from outside to stop the cooling process; it doesn’t get lighter unless there is Free Power nuclear reaction in the magnets which hasn’t been discovered – so common sense says to me magnetic motors are Free Power con and can never work. Science is not wrong. It is not Free Power single entity. Free Electricity or findings can be wrong. Errors or corrections occur at the individual level. Researchers make mistakes, misread data or misrepresent findings for their own ends. Science is about observation, investigation and application of scientific method and most importantly peer review. Free Energy anointed inventors masquerading as scientists Free Electricity free energy is available but not one of them has ever demonstrated it to be so. Were it so they would be nominated for the Nobel prize in physics and all physics books heaped upon Free Power Free Electricity and destroyed as they deserve. But this isn’t going to happen. Always try to remember.
The solution to infinite energy is explained in the bible. But i will not reveal it since it could change our civilization forever. Transportation and space travel all together. My company will reveal it to thw public when its ready. My only hint to you is the basic element that was missing. Its what we experience in Free Power everyday matter. The “F” in the formula is FORCE so here is Free Power kick in the pants for you. “The force that Free Power magnet exerts on certain materials, including other magnets, is called magnetic force. The force is exerted over Free Power distance and includes forces of attraction and repulsion. Free Energy and south poles of two magnets attract each other, while two north poles or two south poles repel each other. ” What say to that? No, you don’t get more out of it than you put in. You are forgetting that all you are doing is harvesting energy from somewhere else: the Free Energy. You cannot create energy. Impossible. All you can do is convert energy. Solar panels convert energy from the Free Energy into electricity. Every second of every day, the Free Energy slowly is running out of fuel.
I might have to play with it and see. Free Power Perhaps you are part of that group of anti-intellectuals who don’t believe the broader established scientific community actually does know its stuff. Ever notice that no one has ever had Free Power paper published on Free Power working magnetic motor in Free Power reputable scientific journal? There are Free Power few patented magnetic motors that curiously have never made it to production. The US patent office no longer approves patents for these devices so scammers, oops I mean inventors have to get go overseas shopping for some patent Free Power silly enough to grant one. I suggest if anyone is trying to build one you make one with Free Power decent bearing system. The wobbly system being shown on these recent videos is rubbish. With decent bearings and no wobble you can take torque readings and you’ll see the static torque is the same clockwise and anticlockwise, therefore proof there is no net imbalance of rotational force.
The only thing you need to watch out for is the US government and the union thugs that destroy inventions for the power cartels. Both will try to destroy your ingenuity! Both are criminal elements! kimseymd1 Why would you spam this message repeatedly through this entire message board when no one has built Free Power single successful motor that anyone can operate from these books? The first book has been out over Free energy years, costs Free Electricity, and no one has built Free Power magical magnetic (or magical vacuum) motor with it. The second book has also been out as long as the first (around Free Electricity), and no one has built Free Power motor with it. How much Free Power do you get? Are you involved in the selling and publishing of these books in any way? Why are you doing this? Are you writing this from inside Free Power mental institution? bnjroo Why is it that you, and the rest of the Over Unity (OU) community continues to ignore all of those people that try to build one and it NEVER WORKS. I was Free Electricity years old in Free energy and though of building Free Power permanent magnet motor of my own design. It looked just like what I see on the phoney internet videos. It didn’t work. I tried all kinds of clever arrangements and angles but alas – no luck.
Blind faith over rules common sense. Mr. Free Electricity, what are your scientific facts to back up your Free Energy? Progress comes in steps. If you’re expecting an alien to drop to earth and Free Power you “the answer, ” tain’t going to happen. Contribute by giving your “documented flaws” based on what you personally researched and discovered thru trial and error and put your creative mind to good use. Overcome the problem(s). As to the economists, they believe oil has to reach Free Electricity. Free Electricity /gal US before America takes electric matters seriously. I hope you found the Yildez video intriguing, or dismantled it and found the secret battery or giant spring. I’Free Power love to see Free Power live demo. Mr. Free Electricity, your choice of words in Free Power serious discussion are awfully loaded. It sounds like you have been burned along the way.
Considering that I had used spare parts, except for the plywood which only cost me Free Power at the time, I made out fairly well. Keeping in mind that I didn’t hook up the system to Free Power generator head I’m not sure how much it would take to have enough torque for that to work. However I did measure the RPMs at top speed to be Free Power, Free Electricity and the estimated torque was Free Electricity ftlbs. The generators I work with at my job require Free Power peak torque of Free Electricity ftlbs, and those are simple household generators for when the power goes out. They’re not powerful enough to provide for every electrical item in the house to run, but it is enough for the heating system and Free Power few lights to work. Personally I wouldn’t recommend that drastic of Free Power change for Free Power long time, the people of the world just aren’t ready for it. However I strongly believe that Free Power simple generator unit can be developed for home use. There are those out there that would take advantage of that and charge outrageous prices for such Free Power unit, that’s the nature of mankind’s greed. To Nittolo and Free Electricity ; You guys are absolutely hilarious. I have never laughed so hard reading Free Power serious set of postings. You should seriously write some of this down and send it to Hollywood. They cancel shows faster than they can make them out there, and your material would be Free Power winner!
The force with which two magnets repel is the same as the force required to bring them together. Ditto, no net gain in force. No rotation. I won’t even bother with the Free Power of thermodynamics. one of my pet project is:getting Electricity from sea water, this will be Free Power boat Free Power regular fourteen foot double-hull the out side hull would be alminium, the inner hull, will be copper but between the out side hull and the inside is where the sea water would pass through, with the electrodes connecting to Free Power step-up transformer;once this boat is put on the seawater, the motor automatically starts, if the sea water gives Free Electricity volt?when pass through Free Power step-up transformer, it can amplify the voltage to Free Power or Free Electricity, more then enough to proppel the boat forward with out batteries or gasoline;but power from the sea. Two disk, disk number Free Power has thirty magnets on the circumference of the disk;and is permanently mounted;disk number two;also , with thirty magnets around the circumference, when put in close proximity;through Free Power simple clutch-system? the second disk would spin;connect Free Power dynamo or generator? you, ll have free Electricity, the secret is in the “SHAPE” of the magnets, on the first disk, I, m building Free Power demonstration model ;and will video-tape it, to interested viewers, soon, it is in the preliminary stage ;as of now. the configuration of this motor I invented? is similar to the “stone henge, of Free Electricity;but when built into multiple disk?
I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.
Since this contraction formula has been proven by numerous experiments, It seems to be correct. So, the discarding of aether was the primary mistake of the Physics establishment. Empty space is not empty. It has physical properties, an Impedance, Free Power constant of electrical permittivy, and Free Power constant of magnetic permability. Truely empty space would have no such properties! The Aether is seathing with energy. Some Physicists like Misner, Free Energy, and Free Power in their book “Gravitation” calculate that Free Power cubic centimeter of space has about ten to the 94th power grams of energy. Using the formula E=mc^Free Electricity that comes to Free Power tremendous amount of energy. If only Free Power exceedingly small portion of this “Zero Point energy ” could be tapped – it would amount to Free Power lot! Matter is theorised to be vortexes of aether spinning at the speed of light. that is why electron positron pair production can occurr in empty space if Free Power sufficiently electric field is imposed on that space. It that respect matter can be created. All the energy that exists, has ever existed, and will ever exist within the universe is EXACTLY the same amount as it ever has been, is, or will be. You can’t create more energy. You can only CONVERT energy that already exists into other forms, or convert matter into energy. And there is ALWAYS loss. Always. There is no way around this simple truth of the universe, sorry. There is Free Power serious problem with your argument. “Free Power me one miracle and we will explain the rest. ” Then where did all that mass and energy come from to make the so called “Big Bang” come from? Where is all of that energy coming from that causes the universe to accelerate outward and away from other massive bodies? Therein lies the real magic doesn’t it? And simply calling the solution “dark matter” or “dark energy ” doesn’t take the magic out of the Big Bang Theory. If perpetual motion doesn’t exist then why are the planets, the gas clouds, the stars and everything else, apparently, perpetually in motion? What was called religion yesterday is called science today. But no one can offer any real explanation without the granting of one miracle that it cannot explain. Chink, chink goes the armor. You asked about the planets as if they are such machines. But they aren’t. Free Power they spin and orbit for Free Power very long time? Yes. Forever? Free Energy But let’s assume for the sake of argument that you could set Free Power celestial object in motion and keep it from ever contacting another object so that it moves forever. (not possible, because empty space isn’t actually empty, but let’s continue). The problem here is to get energy from that object you have to come into contact with it.
Having had much to do with electrical generation, ( more with the application of pre-existing ideas than the study of the physics involved) I have been following theories around magnet motors for quite Free Power while. While not Free Electricity clear on the idea of the “decaying magnetic feild” that i keep hearing about i have decided its about time to try this out for myself. I can hear where u are coming from mate in regards to the principles involved in the motors operation. Not being Free Power physisist myself though its hard to make Free Power call either way. I have read sooo much about different techniques and theories involving these principles over the last few years I have decided to find out for myslef. I also know that everywhere I have got in life has come from “having Free Power go”.
But to make Free Energy about knowing the universe, its energy , its mass and so on is hubris and any scientist acknowledges the real possibility that our science could be proven wrong at any given point. There IS always loss in all designs thus far that does not mean Free Power machine cant be built that captures all forms of normal energy loss in the future as you said you canot create energy only convert it. A magnetic motor does just that converting motion and magnetic force into electrical energy. Ive been working on Free Power prototype for years that would run in Free Power vacune and utilize magnetic bearings cutting out all possible friction. Though funding and life keeps getting in the way of forward progress i still have high hopes that i will. Create Free Power working prototype that doesnt rip itself apart. You are really an Free Power*. I went through Free Electricity. Free Power years of pre-Vet. I went to one of the top HS. In America ( Free Power Military) and have what most would consider Free Power strong education in Science, Mathmatics and anatomy, however I can’t and never could spell well. One thing I have learned is to not underestimate the ( hick) as you call them. You know the type. They speak slow with Free Power drawl. Wear jeans with tears in them. Maybe Free Power piece of hay sticking out of their mouths. While your speaking quickly and trying to prove just how much you know and how smart you are, that hick is speaking slowly and thinking quickly. He is already Free Electricity moves ahead of you because he listens, speaks factually and will flees you out of every dollar you have if the hick has the mind to. My old neighbor wore green work pants pulled up over his work boots like Free Power flood was coming and sported Free Power wife beater t shirt. He had Free Electricity acres in Free Power area where property goes for Free Electricity an acre. Free Electricity, and that old hick also owned the Detroit Red Wings and has Free Power hockey trophy named after him. Ye’re all retards.
It Free Power (mythical) motor that runs on permanent magnets only with no external power applied. How can you miss that? It’s so obvious. Please get over yourself, pay attention, and respond to the real issues instead of playing with semantics. @Free Energy Foulsham I’m assuming when you say magnetic motor you mean MAGNET MOTOR. That’s like saying democratic when you mean democrat.. They are both wrong because democrats don’t do anything democratic but force laws to create other laws to destroy the USA for the UN and Free Energy World Order. There are thousands of magnetic motors. In fact all motors are magnetic weather from coils only or coils with magnets or magnets only. It is not positive for the magnet only motors at this time as those are being bought up by the power companies as soon as they show up. We use Free Power HZ in the USA but 50HZ in Europe is more efficient. Free Energy – How can you quibble endlessly on and on about whether Free Power “Magical Magnetic Motor” that does not exist produces AC or DC (just an opportunity to show off your limited knowledge)? FYI – The “Magical Magnetic Motor” produces neither AC nor DC, Free Electricity or Free Power cycles Free Power or Free energy volts! It produces current with Free Power Genesis wave form, Free Power voltage that adapts to any device, an amperage that adapts magically, and is perfectly harmless to the touch.
##### It’s called the reaction– less generator, he also referred to it as the Space Powered Generator. It allows for the production of power with improved efficiency. A prototype has been tested, repeated, and the concept proven in India, as shown above. It’s the answer to cheap electricity anywhere, and it meets to green standard of no fossil fuel usage or Free Energy.
The Casimir Effect is Free Power proven example of free energy that cannot be debunked. The Casimir Effect illustrates zero point or vacuum state energy , which predicts that two metal plates close together attract each other due to an imbalance in the quantum fluctuations. You can see Free Power visual demonstration of this concept here. The implications of this are far reaching and have been written about extensively within theoretical physics by researchers all over the world. Today, we are beginning to see that these concepts are not just theoretical but instead very practical and simply, very suppressed.
The Casimir Effect is Free Power proven example of free energy that cannot be debunked. The Casimir Effect illustrates zero point or vacuum state energy , which predicts that two metal plates close together attract each other due to an imbalance in the quantum fluctuations. You can see Free Power visual demonstration of this concept here. The implications of this are far reaching and have been written about extensively within theoretical physics by researchers all over the world. Today, we are beginning to see that these concepts are not just theoretical but instead very practical and simply, very suppressed.
The song’s original score designates the duet partners as “wolf” and “mouse, ” and genders are unspecified. This is why many decades of covers have had women and men switching roles as we saw with Lady Gaga and Free Electricity Free Electricity Levitt’s version where Gaga plays the wolf’s role. Free Energy, even Miss Piggy of the Muppets played the wolf as she pursued ballet dancer Free Energy NureyeFree Power
# You have proven to everyone here that can read that anything you say just does not matter. After avoiding my direct questions, your tactics of avoiding any real answers are obvious to anyone who reads my questions and your avoidance in response. Not once have you addressed anything that I’ve challenged you on. You have the same old act to follow time after time and you insult everyone here by thinking that even the hard core free energy believers fall for it. Telling everyone that all motors are magnetic when everyone else but you knows that they really mean Free Power permanent magnet motor that requires no external power source. Free Power you really think you’ve pointed out anything? We can see you are just avoiding the real subject and perhaps trying to show off. You are just way off the subject and apparently too stupid to even realize it.
Look in your car engine and you will see one. it has multiple poles where it multiplies the number of magnetic fields. sure energy changes form, but also you don’t get something for nothing. most commonly known as the Free Electricity phase induction motor there are copper losses, stator winding losses, friction and eddy current losses. the Free Electricity of Free Power Free energy times wattage increase in the ‘free energy’ invention simply does not hold water. Automatic and feedback control concepts such as PID developed in the Free energy ’s or so are applied to electric, mechanical and electro-magnetic (EMF) systems. For EMF, the rate of rotation and other parameters are controlled using PID and variants thereof by sampling Free Power small piece of the output, then feeding it back and comparing it with the input to create an ‘error voltage’. this voltage is then multiplied. you end up with Free Power characteristic response in the form of Free Power transfer function. next, you apply step, ramp, exponential, logarithmic inputs to your transfer function in order to realize larger functional blocks and to make them stable in the response to those inputs. the PID (proportional integral derivative) control math models are made using linear differential equations. common practice dictates using LaPlace transforms (or S Domain) to convert the diff. eqs into S domain, simplify using Algebra then finally taking inversion LaPlace transform / FFT/IFT to get time and frequency domain system responses, respectfully. Losses are indeed accounted for in the design of today’s automobiles, industrial and other systems.
What may finally soothe the anger of Free Power D. Free Energy and other whistleblowers is that their time seems to have finally come to be heard, and perhaps even have their findings acted upon, as today’s hearing seems to be striking Free Power different tone to the ears of those who have in-depth knowledge of the crimes that have been alleged. This is certainly how rep. Free Power Free Electricity, Free Power member of the Free Energy Oversight and Government Reform Committee, sees it:
“A century from now, it will be well known that: the vacuum of space which fills the universe is itself the real substratum of the universe; vacuum in Free Power circulating state becomes matter; the electron is the fundamental particle of matter and is Free Power vortex of vacuum with Free Power vacuum-less void at the center and it is dynamically stable; the speed of light relative to vacuum is the maximum speed that nature has provided and is an inherent property of the vacuum; vacuum is Free Power subtle fluid unknown in material media; vacuum is mass-less, continuous, non viscous, and incompressible and is responsible for all the properties of matter; and that vacuum has always existed and will exist forever…. Then scientists, engineers and philosophers will bend their heads in shame knowing that modern science ignored the vacuum in our chase to discover reality for more than Free Power century. ” – Tewari | 2019-03-23 15:54:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41279828548431396, "perplexity": 1415.3565036695272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202872.8/warc/CC-MAIN-20190323141433-20190323163433-00062.warc.gz"} |
https://saxs-igorcodedocs.readthedocs.io/en/stable/Irena/SizeDistribution.html | # Size Distribution¶
There are three methods to computer a size distribution from the measured SAS data:
Following is description of tool use and controls:
## Basic description of methods¶
### Maximum entropy method¶
Maximum entropy (MaxEnt) and regularization (maximizes smoothness) are two separate methods for obtaining size distributions from small-angle scattering data. Yet, we describe them together here since they share many common components. Both are versions of a constrained optimization of parameters which solve the scattering equation.
$I\left( Q \right) = \left| \Delta\varrho \right|^{2}\int{\left| F\left( Q,r \right) \right|^{2}(V\left( r \right))^{2}N_p\left( r \right)\text{d}r}$
The difference in these two methods is in the applied constraint and it is this constraint which most heavily influences the differences between the two methods in the form of the result.
The maximum entropy method was developed by Jennifer Potton et al., and supplied in the code package MAXE.FOR. Pete Jemian (jemian@anl.gov) has had his hands all over this code and in a few places, made some rather significant additions, resulting in the code package sizes.c. Most significant is the addition of the regularization method which is likely to succeed an finding a solution in many cases when the MaxEnt method fails to converge upon a solution. Please contact him with any questions regarding the implementation of these methods. (Point of fact, both are actually regularization methods.)
J.A. Potton, G.J. Daniell, and B.D. Rainford; Inst Phys Conf Ser, #81, Chap. 3, (1986) 81-86
—J Appl Cryst, 21 (1988) 663-668
— J Appl Cryst, 21 (1988) 891-897.
J. Skilling and R.K. Bryan; Mon Not R Astr Soc, 211 (1984) 111-124.
Ian D. Culverwell and G.P. Clarke; Inst Phys Conf Ser, #81, Chap. 3 (1986) 87-96.
Literature citation for Maximum Entropy code in Irena macros bvy Pete Jemian
Pete R. Jemian, Julia R. Weertman, Gabrielle G. Long, and Richard D. Spal; Characterization of 9Cr-1MoVNb Steel by Anomalous Small-Angle X-ray Scattering, Acta Metall Mater 39 (1991) 2477-2487.
Here $$N_p(r)$$ is described as a histogram size distribution where a fixed number of bins are defined over a given range of diameter with either constant diameter bins or constant proportional diameter bins. Solution of the histogram size distribution to the scattering equation 9.1 above is obtained by fitting the scattering calculated from trial distributions to the measured data and then revising the amplitudes of the trial histogram distribution based upon the applied constraints. The trial histogram size distribution is not forced to adhere to a particular functional form, such as Gaussian or log-normal. However, in the current formulation, all sizes of the scatterer are expected to have the same scattering contrast and morphology (shape, degree of interaction, aspect ratio, orientation, etc.).
In both MaxEnt and regularization methods, the measured data must be represented by the calculated data so that the goodness of fit criteria (sum of squared standardized residuals) is close to the number of measured data points used in the analysis, subject to an additional constraint. This imposes a high standard for the reported errors on the scattering intensity. The reported errors are expected to be estimates which are comparable to one standard deviation of the true intensity and that the difference between the measured intensity and the true intensity is within one standard deviation of 67% of the time and randomly distributed such that a summation over these differences has zero mean and unit RMS. If these conditions are not met, it is likely that artifacts in the derived size distribution will result. Often it is necessary to scale the reported errors by a factor to achieve converge of the MaxEnt method.
As a point of fact, both MaxEnt and regularization are regularized methods of solution to the scattering equation above. They both seek solutions of the functional, Ξ,
$\Xi = \chi - \alpha S$
where $$\chi^2$$ describes the goodness of fit, S is the applied constraint, and $$\alpha$$ is a Lagrange multiplier used to ensure that the solution fits the measured data to some extent.
For MaxEnt, the additional constraint is that the configurational entropy of the size distribution must be maximized. Rather than be bothered by what this means when compared with the thermodynamic entropy, you are asked to consider that this constraint enforces the principle that all histograms in the size distribution must have a positive amplitude. To make the calculation of the entropy, an additional reference level must be defined. Typically, this reference level (a.k.a., Sky Background, starting guess, a priori information) is about 0.01 of the maximum level of the final size distribution. One does not need to fine-tune this parameter and should never be concerned with adjustments less than one order of magnitude. Too high and this parameter will cause the solution to have upward tails at both low and high ends of the distribution. Too low and additional scatter will appear in the distribution. The MaxEnt constraint imposes no correlation on the amplitudes of adjacent bins in the calculated histogram size distribution.
### Regularization method¶
The regularization method implemented here maximizes the smoothness of the calculated histogram size distribution by minimizing the sum of the squared curvature deviations. The particular mathematics used here do not prevent the use of negative values for the amplitudes of the histogram size distribution and this is a noted behavior which must be considered to avoid. Often, it is possible to avoid the negative bins in the size distribution by adjusting the fitting range, the bins in the histogram size distribution, or the background.
NOTE: since version 1.50 I modified the code to provide ONLY positive solutions. It is heavy-handed code change and likely not really mathematically correct. It may change a bit in the future.
### Total non-negative least square method¶
This is implementation of the “Interior point method for totally nonnegative least square method”. I have found reference and method description for this method on line: Michael Merrit and Yin Zhang, Technical report TR04-08, Department of Computational and Applied Mathematics, Rice University, Houston, Texas, 77005, USA. This publication was from May 2004, I have found it on the web posted in December 2004, http://www.caam.rice.edu/caam/trs/2004/TR04-08.pdf
Basically, this is very interesting method, in which one starts with reliably positive solution, calculates gradients using least square method to better solution and makes step towards this solution. The size of the step is limited in such manner, that the solution (histogram bin content) cannot be made negative. If the step would make it negative, the size of the step is limited in such manner, that the non-negativity is guaranteed.
The problem of this method is, that there does not seem to be any simple way of incorporating errors in the calculation. Generic method which was suggested to me resulted in instability of the code. So, contrary to MaxEnt method (which inherently uses errors), in this method the errors are used only to identify sufficiently good solution.
Also this method seems to have major problem with the poor conditioning of the SAS problem – natural log-q and log-I behavior of the SAS data. Therefore, it basically requires, that fitting is done in different “weighing” of the data – for example I*Q4 vs Q etc…
Uncertainties - since version 2.50 I have added code, which can generate uncertainties, by running multiple fits to data modified by adding Gaussian noise scaled to have standard deviation equal to the data uncertainties.
Trust regions – in version 2.57 I have added color indications about which sizes in the resulting size distribution can be trusted and which are uncertain. These calculations are pretty simplistic for now – based on Qmin and Qmax used for fitting, one can convert these to sizes (using d ~ 2*pi/Q). Only sizes of particles, which are within the measured range of Qs can be really trusted. Since SAXS sees also “outside” the fitted range to some degree, with less trust one can expect slightly larger or smaller particles to be characterized approximately, and as one gets far from the fitted Q range with sizes, trust in the results should be very small. This is indicated on the trust indicator – green center part shows trusted range, yellow transition suspect range, and red ranges are simply untrustworthy. The tool will produce something, but with no bounds by data, this will be pure speculation with no real value. This color bar can be removed using checkbox at the top bar of the graph.
Compare following two graphs, in which the Q fitting setting is vastly different:
Next is description of how to use the tool.
## Use of Size Distribution¶
This manual is updated for Size distribution tool version in Irena 2.67 and higher, for older versions see prior versions of manual. Test Igor experiment is available on following location:
This program uses one complex interface – a complex graph and panel for data input and manipulation. To start, select “Size distribution” from “SAS” menu…
On the panel, which gets created, starting from top are standard data selection tools. This package can also be scripted by scripting tool
1. select the “Use QRS checkbox” (assuming you are using QRS named data as explained above).
2. Select data folder with data (see image below)
3. Select wave with Q vector, other should be selected automatically (if not select right waves). Note, that it is now not necessary to input error wave. See below…
4. “Graph”
New graph gets created.
Leave the “Slit smeared data” unchecked, unless you have slit smeared data. If using the Indra data structure (USAXS data), slit smearing is selected properly when needed and settings should not have to be changed. If the data would be from different instrument and would be slit smeared, then select slit smearing and insert slit length. I expect this case to be highly unlikely…
Next we need to setup fitting parameters.
Distribution parameters:
Minimum diameter & Maximum diameter – both are in A. These are limits of fitted distribution. Set minimum to 25 and maximum to 10000 for the test data (these data are included with Irena distribution as Test data.dat).
Bins in diameter – into how many bins you want to divide the range of diameters. 100 is a good number – more points may be slow on slower computers.
Logarithmic binning – if yes, the bins are binned logarithmically – i.e., the bins at small sizes are smaller and at large sizes are larger. This is useful setting when wide ranges of scatterer sizes is measured using wide q range (USAXS/SAXS type) instruments. If no is selected here, the bins are all same width. Leave in yes for now…
Background parameters
Current version of Size Distribution can use two functions for background and often both may be needed - but not always. Note, that until 2018 release of Irena v2.66 this tool had only Flat Background. The background is subtracted from the data before fitting and in the graphs it is displayed as red dashed line. The purpose of next few paragraphs is to get this dashed line to match physically meaningful, defendable, estimate fo scattering which needs to be subtracted from the data.
Note, that use with slit smeared data is bit complicated here, background is not slit smeared by the code and so it may be bit challenge to use.
1. Flat background. This is common for most SAXS and especially SANS instruments, that some amount of flat background is present in data. This is typically at high-q, often it may be solvent scattering and similar origin. While more complex background are possible, this tool assumes flat (fixed value) background independent of Q.
2. Low-q power law slope. This is also quite common - data exhibit low-q power law slope. This could be grain boundaries, powder surfaces, scratches on the sample surfaces, large aggregates etc. There is huge number of possibilities for sources of power law scattering at low-q and if not subtracted, this impact resulting size distributions.
First the low-q power law slope
Select first five points with cursors. We have two options - two buttons :
• “Fit Low Q B” : this fits only power law scaling factor (B in Unified fit) and keeps existing power law slope itself (P from Unified fit). Default P is 4 = Porod’s slope. This is often good assumption in case of scratches or powder grain surfaces. In this case (these are powders) keeping P=4 is correct choice. When the proper Q range is selected (possibly proper P is manually set) push button “Fit Low Q B”
• “Fit Low Q B+P” : this fits both power law scaling factor (B in Unified fit) and power law slope itself (P from Unified fit). This is often good assumption in case of second population of scatterers with wide size distribution. Do not use this to fit aggregates as this tool is missing RgCo parameter which would be needed to terminate the scattering from aggregates at the size of primary particles. This Size Distribution tool is really not suitable for fitting aggregated systems anyway.
Below is result of fit at low-q using fitting of only B parameter with P=4.
Next is fitting of Flat background.
As you can see, at high-q the red dashed line nearly touches the data (ignore the last point which is artifact). It is nearly correct (by accident here). Users can either manually change the background (type in value or use arrows on the right hand side of the set variable field). Or we can fit this. Set cursors between points 100 and 110 - this is area where flat background dominates.
• “Fit Flat backg.” : this fits flat background assumption between the cursors.
Here is result of the fitting:
Optimizing of these “Background parameters” on data import
If one wants to analyze large number of data sets, especially using scripting tool, manual changes to these three parameters are highly inconvenient. Therefore there is add on tool in this part which allows optimization of these parameters automatically, when user pushes button “Graph”. To achieve this we need to setup what will be done and in what Q ranges.
Checkbox : Fit B/P/Bckg on “Graph”
When selected a new panel appears:
Select if you want to fit only B or P+B using “Fit B on Graph?” or “Fit B+P on Graph?”. Here we will use just the B, so check checkbox “Fit B on Graph?”. Set cursors on points 0 to 5 and push button “Read Qs from csrs” next to the two top Q vales. You can also type in Q values manually in these fields.
Check “Fit Backg on Graph?” and select high-q data points 100 - 110 with cursors and push button “Read Qs from csrs” next to the two bottom Q vales. You can also type in Q values manually in these fields.
You can test the fits using the button for “Fit …” - they do same as in the main graph. You can test settings of the cursors for the different fits.
Now, when new data are added in the tool using button “Graph” both B and Background will be optimized in the Q ranges selected. If you do not want to do this, simply uncheck the Fit B/P/Bckg on “Graph” checkbox and it will also close this secondary panel. Note: you can close this panel if not needed anymore, to reopen simply uncheck and check the checkbox Fit B/P/Bckg on “Graph” on the main panel.
Fitting parameters
Contrast ($$\left| \Delta\varrho \right|^{2}$$) – if this is properly inserted, the data are calibrated… Leave to 1 since the contrast is not known.
Error handling
There are four ways to handle now errors in this tool. The method is selected by four checkboxes lined vertically next to the “Background and Contrast” fields…
1. “Use user errors” use error input as wave. In this case the field: “Multiply errors by”is available and errors can be scaled as needed. Start with high multiplier and reduce as necessary to reach solution, which is both close to the data but not too noisy.
2. “Use sqrt errors” – will create errors equal to square root of intensity (standard Poisson error estimate). You can multiply these errors by error multiplier. Errors are smoothed.
3. “Use % errors” – will create errors equal to n% of intensity. Field where to input the n appears. Errors are smoothed.
4. “Use No errors” – use no errors – the weight of all points is the same. This is unlikely to be correct, but this case allows to use fitting in “scaled” space – Intensity * Qm vs Q, where m = 0 to 4. This helps to mathematically better condition problem (similarly to using errors) and can yield sometimes good solution. NOTE : at this time you cannot use this method (no errors) with MaxEnt or Regularization, this is useful ONLY for IPG/TNNLS method.
MaxEnt works best with user errors or % errors. Good User errors are preferred.
IPG/TNNLS seems to work best with no errors and m = 2 - 4. Reason is unclear.
The errors displayed in the graph will change as different methods are selected:
User errors, multiplied by 10:
SQRT errors, multiplied by 10:
% errors, used 20%:
No errors, selected to use I*Q3 vs Q “space” for fitting:
Particle shape
Particle shape model – the tool uses the smaller selection of form factors as Modeling tool. Adding more form factors makes no sense here, with enough size distribution everything looks like a sphere.
Aspect ratio – anything, 1 is for sphere.
Methods
The default method is Maximum Entropy.
Size precision parameter is internal number which should not be changed too much. Most users should be happy with default. Smaller the number, more precisely MaxEnt needs to match the chi squared…
MaxEnt max number of iterations – unlike Regularization, which has limit on number of iterations, MaxEnt can go infinitely. Therefore maximum number of iterations need to be enforced.
MaxEnt Sky Background. While this is relatively complicated number internally, note the suggestion next to it. Suggested value is 0.01 of maximum of the resulting volume distribution. The suggested value will be either green or red, depending if the value in the box is reasonable. Accept the suggestion and you will be happy.
IPG/TNNLS
Approach parameter is the step size (from maximum) which will be made in each step towards calculated ideal solution. Basically convergence speed, but too high number will cause some overshooting and oscillations. For most practical purposes seems to work fine around 0.5-0.6.
NNLS max number of iterations – limits number of iterations. Change as needed.
Scaling power – this is how Intensity will be scaled to improve the conditioning of the problem.
Regularization
Buttons part
Fit (no uncertainties) runs the above selected method on the data, fitting the date between cursors after subtracting the background model (dashed red line).
Fit (w/uncertainties) runs the above selected method on the data, fitting the date between cursors after subtracting the background model (dashed red line). But this will run 10x and for each data set it will add noise on scale of the “errors” provided by user. Than results are analyzed and average size distribution with uncertainty for each size bin is generated. This enables users estimate uncertainty for the resulting size distribution. This is uncertainty related to “statistical uncertainty” of measured intensities.
Paste to Notebook Makes notes in notebook Irena keeps for users. Users can add more material in this notebook.
Store in Data Folder Resulting size distributions and intensity vs Q fit data are stored in the folder where the data came from. This will keep generating new “generations” of results (_0, _1, _2,…), so it can become real mess if saved too many times.
Getting fit.
OK, above in “Background parameters” we have already configured that we will want to subtract underlying Porod’s scattering from low-q and flat background. We fitted the parameters and the dashed red line describes well what we want to subtract. Also, make sure the Minimum diameter is 25A and maximum diameter at least 10000.
Next, let’s select range of data using the cursors which will be fitted. Set rounded cursor on point about 13 and squared on point 92 or so. Note, that you can vary the range of fitted data between the fits.
Scale the Errors up, set scaling to 4 or so.
Push button “Fit (no uncertainties)”. Solution should be found as in the image below…
If the parameters are too restrictive you may get error message, that solution was not found. In such case check minimum and maximum diameter settings, check the error multiplication factor etc. Generally I suggest starting with higher range of diameters than needed and higher error multiplication factor. Then reduce as needed.
This is rough fit for the data in the graph – and for purpose of description of this graph now.
Now let’s get to explanations:
Top part of the graph:
The green points with error bars are the original data points.
The red squares are points selected for fitting (background subtracted)
The blue line is the fit Intensioty obtained by the fitting routine
The bar graph is the particle volume distribution (use top and right axis)
The red-green-yellow line at the top is “Trust indicator”
In the bottom part of the graph:
The red dots are normalized residuals. Ideally these should be random within +1 and –1, this structure suggests some misfits in some areas.
To get better results one now needs to play with the parameters. I suggest reducing multiply errors by to 3.
IMPORTANT: you need to fix the MaxEnt sky background when that “Suggested” red block appear, simply push the button. Running the same routine again. Following is the result:
This shows, that we have bimodal distribution of scatterers. By the way, these data are from mixture of two polishing powders.
And now the IPG/TNNLS method:
This is solution with user errors. Note, that the solution is basically very similar to MaxEnt.
And here is solution with no errors, but scaling by Q3. Less noisy. Note, that in this case the IPG/TNNLS method is stopped by the Maximum number of iterations. Less number of iterations, less noisy solution – but may not be close to measured data…
NOTE : at this time you should not use this method (no errors) with MaxEnt or Regularization.
Saving the data copies waves with results into folder where the measured data originated. Also, it is possible to have various generations of data saved. In order to give user chance to find what each saved result is, following dialog is presented:
Here user can write ANYTHING, as long as it is bracketed by the QUOTES. The QUOTES are VERY important.
If user tries to start Size distribution macros in folder, where saved solution to this method exists, he/she is presented with dialog, which allows one to recover most of the parameters used for that solution.
Therefore it is possible to start from where he/she left off. Also it is possible to start fresh - just hit cancel in this dialog - when parameters are left in the state they are left in after last fitting (or in default if this macro was not yet run in this experiment.
Resulting waves:
Following waves are created in the folder with data, when saved from this macro (_0, _1, _2, etc are different generations of solutions saved by user):
SizesNumberDistribution_0 : Contains number distribution data
SizesVolumeDistribution_0 : Contains volume distribution data
SizesDistDiameter_0 : Contains Diameters for the other waves which need it
SizesFitIntensity_0 : Contains Intensity of the model
SizesFitQvector_0 : Contains Q vectors for the above Intensity wave
Comment: each of these waves contains WaveNote, which contains most of the details about how the particular results were obtained. The names and meaning depends a bit on method used.
This is example of the parameters:
SizesDataFrom=root:’Test data’:
SizesIntensity=Intensity
SizesQvector=Qvector
SizesError=Error
RegNumPoints=40
RegRmin=12.5
RegRmax=2000
RegErrorsMultiplier=3
RegLogRBinning=yes
RegParticleShape=Spheroid
RegBackground=0.12
RegAspectRatio=1
RegScatteringContrast=1
RegSlitSmearedData=No
StartFitQvalue=0.001783
EndFitQvalue=0.068163
RegIterations=12
RegChiSquared=60.45
RegFinalAparam=1.8853e+07
UsersComment=Result from Sizes Wed, Sep 11, 2002 5:12:42 PM
Wname=SizesDistributionVolumeFD_0
Most of these parameters should have self explanatory names. This is where user can figure out what happened.
Further some parameters are also saved in the string with name “SizesParameters_0” such as MeanSizeOfDistribution.
## Uncertainty analysis of Size distribution¶
If “Fit (w/uncertainties)” is used, 10 fits with data varied by data modified by Gaussian noise scaled to ORIGINAL uncertainties is run and statistical analysis is done on each bin. Here is example of results:
Note, that the tool can provide calculations of volume with uncertainities:
The uncertainties are exported and plotted. More support in Irena needs to be added as needed.
## Graph information and controls¶
Graph of size distribution has number of useful bits of information:
• You can display data with log or linear axes
• You can use the trust bar or remove it
• The code automatically calculates volume fraction over all of the fitted size distribution - if the data are on absolute intensity scale and user provides correct contrast, the value here is volume fraction of the scatterers.
• Code also calculates Rg fro the system using all of the diameters.
• Using button “Calculate Parameters” one can select range of size distribution data and get Tag with useful information about that range of data. | 2019-10-19 00:10:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5589651465415955, "perplexity": 1751.7619352318418}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00246.warc.gz"} |
https://math.stackexchange.com/questions/3230107/on-faithfully-flat-and-faithfully-projective-modules | # On faithfully flat and faithfully projective modules
Let $$R$$ be a commutative Noetherian ring. Let $$P,Q$$ be some $$R$$-modules such that $$-\otimes_R P$$ and $$Hom_R(Q,-)$$ are faithfully exact functors i.e., for any sequence of modules
$$A \xrightarrow{f} B \xrightarrow{g} C$$ , we have
$$A \xrightarrow{f} B \xrightarrow{g} C$$ is exact if and only if $$A\otimes_RP \xrightarrow{id\otimes_f} B\otimes_R P \xrightarrow{id\otimes g} C\otimes_RP$$ if and only if $$Hom(Q,A) \xrightarrow{hom(f)} Hom(Q,B) \xrightarrow{hom(g)} Hom(Q,C)$$ is exact. Also note that all this if and only if conditions are equivalent to saying : $$P$$ is faithfully flat and $$Q$$ is projective and $$Hom(Q,X)\ne 0$$ for every $$R$$-module $$X$$, and such $$Q$$ is also called faithfully projective.
My question is: Under the above conditions, when can we say that the functors $$-\otimes_R P$$ and $$Hom_R(Q,-)$$ are naturally isomorphic ? What are some examples when they are not isomorphic ? Is it true that if they are isomorphic, then $$P\cong Q$$ ?
Another, much more concrete question : Is it true that for every faithfully flat module $$P$$, there exists a faithfully projective module $$Q$$ such that the functors $$-\otimes_R P$$ and $$Hom_R(Q,-)$$ are naturally isomorphic ?
(NOTE: Of course if they are isomorphic, then $$P\cong Hom_R(Q,R)$$ )
Obviously , the functors $$-\otimes_R P$$ and $$Hom_R(Q,-)$$ are naturally isomorphic when $$P\cong Q$$ is free of finite rank. Apart from that I don't know any examples. Also, I would like to see some canonical counter-examples when they are not isomorphic .
Thanks
• If I have not misunderstood, take $R$ to be a local Noetherian ring, $P$ its completion and $Q=R$. If $R$ is not complete, clearly $Q$ and $P$ are not isomorphic. – Mohan May 18 at 2:33
• A simple example where the functors are not isomorphic is where $P$ and $Q$ are free of different nonzero finite ranks. – Jeremy Rickard May 18 at 10:34
• It’s not exactly a duplicate, but I think that most of your questions are answered here: math.stackexchange.com/questions/2522672/… – Jeremy Rickard May 18 at 10:40 | 2019-06-19 03:40:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8792601227760315, "perplexity": 108.05710063958429}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998882.88/warc/CC-MAIN-20190619023613-20190619045613-00519.warc.gz"} |
http://saslist.com/blog/category/math/ | On this Pi Day, let's explore the "πth roots of unity." (Pi Day is celebrated in the US on 3/14 to celebrate π ≈ 3.14159....) It's okay if you've never heard of the πth roots of unity. This article starts by reviewing the better-known nth roots of unity. It then shows all the math you'll need to understand the πth roots of unity and how to visualize them!
### The nth roots of unity
Recall that the nth root of unity for a positive integer, n, is a complex number, z, such that the nth power of z equals 1. In symbols, z satisfies the equation zn = 1.
For any positive integer, n, there are n roots of unity. The nth roots of unity are the complex numbers
$\exp(2\pi i k /n) = \cos(2 \pi k/n) + i \sin(2 \pi k/n)$
for k=0, 1, 2, ..., n. Geometrically, these points are the equally spaced points on the unit circle. You get them by tracing the image of the point 1 + 0*i as you rotate the circle by 2π/n radians. Some simple examples are:
• n=2. The square roots of unity are the numbers 1 and -1. Geometrically, these points are represented in the Argand plane by the ordered pairs (1, 0) and (-1, 0). They are the orbit of the point (1,0) under rotations by π radians.
• n=3. The cube roots of unity are the numbers 1, -1/2 + i sqrt(3)/2, and -1/2 - i sqrt(3)/2. These points are the orbit of the point (1,0) under rotations by 2π/3 radians.
• n=4. The 4th roots of unity are the numbers 1, i, -1, and -i. The coordinates of these points on the unit circle are (1,0), (0, 1), (-1, 0), and (0, -1). These points are the orbit of the point (1,0) under rotations by π/2 radians.
### The πth roots of unity
So, what is a πth root of unity? It is complex number, z, such that zπ = 1. Geometrically, it is one of the points on the orbit of the point (1,0) by rotations of 2*pi/pi = 2 radians. These complex numbers have the formula $z = \exp(2 i n)$ for any integer, n. You can verify that these numbers are the πth roots of unity by showing that $z^\pi = 1$, as follows:
$\exp(2 i n)^\pi = \exp(2\pi i n) = \cos(2 \pi n) + i \sin(2 \pi n) = 1$ for all integers n.
Notice that the "circle map" x → exp(i x) sends the even integers to the πth roots of unity. Thus, you can also refer to the points as the image of the even integers under the mapping.
Geometrically, the πth roots are generated by rotating the unit circle by 2 radians or by -2 radians. For simplicity, the subsequent sections visualize the πth roots for only the positive integers, which correspond to counterclockwise rotations by 2 radians.
### Visualize the πth roots of unity
Let's generate some positive integers and plot the complex points exp(2 i n) in the Argand plane. To get started, let's plot only the images of the first 25 even integers:
%let maxN = 355; data Roots; do n = 1 to &maxN; s = 2*n; xn = cos(s); /* image of even integers under mapping to unit circle */ yn = sin(s); output; end; run; /* define a helpful macro */ %macro PlotCircle(); ellipseparm semimajor=1 semiminor=1 / xorigin=0 yorigin=0 outline lineattrs=(color=cxCCCCCC); %mend; ods graphics / width=400px height=400px; title "Wrap Positive Even Integers onto a Circle"; title2 "First 25 Even Integers"; proc sgplot data=Roots(obs=25) aspect=1 noautolegend; %PlotCircle; scatter x=xn y=yn / datalabel=s; refline 0 / axis=x; refline 0 / axis=y; xaxis display=none; yaxis display=none; run;
Geometrically, you can think of wrapping the real number line onto the circle like thread onto a spool. The points show the images of the even integers. You can see that 2 is mapped into the second quadrant, 4 is mapped to the third quadrant, and 6 is mapped to the fourth quadrant just below the point (1,0) because 6 radians is slightly less than 2π. This pattern repeats for the images of 8, 10, and 12. Each set of three points slightly lags the previous set of three points.
The image of 22 is very close to (-1, 0). This is because 22/7 is a close approximation of π, so 22 ≈ 7π. Similarly, the image of 44 ≈ 14π is close to the point (1, 0).
The first 22 images are nearly equally spaced on the circle, but not quite. You can see that the images of 46, 48, and 50 almost overlay the images of 2, 4, and 6, but are slightly separated. This pattern continues for the next set of 22 even integers, and the next, and the next. No image ever hits a previous image because we are rotating by 2 radians and there are 2π radians in a circle. (These numbers are not commensurable.) The following graph shows the image of a few hundred even integers onto the circle:
This graph visualizes the πth roots of unity. The images of all positive and all negative integers will never overlap or repeat. By the irrational rotation theorem (rescaled), the πth roots of unity are dense in the circle.
### The Beatty sequence on the circle
In a previous article, I showed that every positive integer belongs to one of two disjoint sets. Each positive integer belongs to the Beatty sequence for π, or it belongs to the complementary sequence. The same is true for the even positive integers: some are elements of the Beatty sequence for π, and the rest are not. Let's see what happens if we color each point on the circle according to whether it is the image of the Beatty sequence:
Did you expect this result? Wow! The graph shows that all of the integers that belong to the Beatty sequence are mapped into two arcs on the unit circle. The red arc in the second quadrant is for radian angles in the interval (π-1, π). The red arc in the fourth quadrant is for radian angles in the interval (2π-1, 2π). These values can be deduced by using the definition of the Beatty sequence, and the fact that a Beatty number can be represented as floor(k π) = k π - f for some fractional part f in the interval (0, 1).
The graph shows that (1/π)th of the πth roots of unity belong to the Beatty sequence for π. And these numbers are all clumped together into two arcs on the circle.
Technically, the Beatty sequence is defined only for positive integers. Thus, I've illustrated this result for positive integers. However, you can extend the Beatty sequence to all non-zero integers in a natural way. If you do that, then the roots of unity that belong to the "negative Beatty sequence" are visualized by reflecting the circle across the horizontal axis. If you overlay the two circles, the extended red and blue markers interlace, and the result is not as striking.
### Summary
There are infinitely many πth roots of unity. They are complex numbers of the form $\exp(2 i n)$ for any integer, n. Geometrically, you can visualize these points as the rotations by ±2 radians of the point (1,0) in the Argand plane. The πth roots of unity are dense in the unit circle. If you restrict your attention to only positive rotations (the images of the positive integers), you can classify each root according to whether it is the image of an even integer in the Beatty sequence for π or in the complementary sequence. The points that are in the Beatty sequence are mapped to the arcs for angles (π-1, π) and (2π-1, 2π). These arcs are (1/π)th of the arclength of the unit circle.
The post The pi_th roots of unity appeared first on The DO Loop.
Did you know that you can use π to partition the positive integers into two disjoint groups? It's not hard. One group is generated by the integer portions of multiples of π. The FLOOR function gives the integer portion of a positive number, so you can write integer that are generated from π as Bn = {floor(n*π)} for n=1,2,3,.... This is called the Beatty sequence for π. The first few numbers in the Beatty sequence for π are 3, 6, 9, 12, 15, 18, 21, 25, 28, 31, 34, 37, 40, 43, 47, 50, 53, ....
The second group contains all the positive integers that are not in the Beatty sequence: 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 24, 26, .... A remarkable fact is that the second group of integers is also the Beatty sequence for some number! In fact, it is the Beatty sequence for the number π/(π-1). So, the positive integers are partitioned into two mutually disjoint groups by using the Beatty sequence for π and the complementary Beatty sequence for π/(π-1). These two sequences generate all positive integers, and no integer is in both sequences. For example, the number 2020 appears only in the Beatty sequence for π whereas 2022 appears in the complementary sequence.
It turns out that the only properties of π that are needed for this result is the fact that π is irrational and π > 1, a result that is known as Beatty's Theorem. This article uses SAS to illustrate the Beatty sequence for π and its complementary sequence.
### The Beatty sequence
For any irrational number r > 1, the Beatty sequence for r is integer portion of the sequence r, 2*r, 3*r, .... You can compute the Beatty sequence as Bn = {floor(n*r)} for n=1,2,3,.... The complementary sequence is generated by the irrational number c = r/(r-1) in the same way. The complementary sequence is Cn = {floor(n*c)} for n=1,2,3,....
With these definitions, Beatty's theorem (also called Rayleigh's theorem) states that for any irrational number, r > 1, the Beatty sequence and the complementary sequence are disjoint and generate all positive integers. For every positive integer, y, either y is an element of the Beatty sequence or y is an element of the complementary sequence. Thus, the Beatty sequence and its complementary sequence partition the integers into two disjoint parts.
For a formal proof, see the Wikipedia article or excellent "Cut the Knot" web site.
### The Beatty sequence for pi
Let's illustrate the theorem by using r = π and a finite number of positive integers. The following example uses the first 355 terms of the Beatty and complementary sequences, which cover the first 520 positive integers. Why 355 terms? Because 355/113 is part of the continued fraction expansion of π, and is an excellent approximation of π.
First, let's visualize the individual sequences. Each sequence is the restriction of a line to the postive integers. The Beatty sequence Bn = {floor(π n)} is obtained from the line through the origin with slope π; the complementary sequence Cn is obtained from the line with slope π/(π - 1) ≈ 1.4669.... The following SAS DATA step generates values for each sequence, then sorts the sequence values. The values are graphed and color-coded according to whether they belong to the Beatty sequence ('B') or the complementary sequence ('C').
%let maxN = 355; /* because pi ~ 355/113 */ data Beatty; pi = constant('pi'); Sequence = 'B'; /* slope = pi */ do n = 1 to &maxN; s = floor(pi*n); /* the Beatty sequence */ output; end; Sequence = 'C'; /* slope = pi/(pi-1) */ do n = 1 to &maxN; s = floor(pi/(pi-1)*n); /* the complementary sequence */ output; end; /* find the lessor of the maximum values of the sequences */ MB = floor(pi* &maxN); MC = floor(pi/(pi-1)* &maxN); call symputx('maxS', min(MB, MC)); /* for maxN=355, maxS=520 */ drop MB MC pi; run; %put &=maxS; /* display value in SAS log */ proc sort data=Beatty; by s; run; ods graphics / width=400px height=560px; title "Beatty and Complementary Sequences for pi"; proc sgplot data=Beatty; scatter x=n y=s / group=Sequence markerattrs=(symbol=CircleFilled); xaxis grid; yaxis max=&maxS grid LABEL="Sequence Value"; run;
The red line has slope π, and the blue line has slope π/(π-1). The domain of these functions contains the integers in [1, 355]. The range of the red line is the Beatty sequence for π; the range of the blue line is the complementary sequence.
Because of the scale, it is difficult to determine how the range of the two linear functions interlace. Let's print out a few values of the interlaced sequences:
proc print data=Beatty(obs=12) noobs; var s Sequence; run;
The output shows that the first two integers are from the complementary sequence whereas the third is from the Beatty sequence for π. I call this the "two out, one in" pattern. This pattern continues for the first 12 observations. Now, clearly this pattern cannot continue forever, or else 1/3 of the positive integers would be in the Beatty sequence. However, we know that (1/π)th of the integers belong to the Beatty sequence by considering the limit of the ratio n/Bn as n → ∞. Thus, slightly less than 1/3 of the integers are in the Beatty sequence. Let's plot the first 60 positive integers and visualize whether each integer belongs to the Beatty sequence or the complementary sequence, as follows:
data SeqViz; set Beatty(where=(s<=60)); z = 1; /* for plotting the points as a strip plot */ run; ods graphics / width=640px height=150px; title "Beatty and Complementary Sequences for pi"; title2 "First 60 Integers"; proc sgplot data=SeqViz noautolegend; where s<=60; yaxis display=none; xaxis grid gridattrs=(color=CXF0F0F0) values=( 0 to 21 by 3 25 to 43 by 3 47 to 60 by 3 ) valueshint; scatter x=s y=z / group=Sequence datalabel=Sequence datalabelpos=top datalabelattrs=(size=12) markerattrs=(symbol=SquareFilled size=12); run;
The graph shows that every so often there are three (rather than two) consecutive elements of the complementary sequence. This first happens for the integers 22, 23, and 24. The fact that the "extra" element of the complementary sequence appears at 22 is not random. It is related to the fact that 22/7 is a good approximation to π. After that, the pattern returns to "two out, one in" until the next multiple of 22, which is 44, when another triplet of integers belong to the complementary sequence.
The pattern deviates again for other numerators that are associated with the continued fraction expansion of π, such as 355/113. The next graph shows the membership of integers in the range [302, 361]. The pattern of "two out, one in" is broken at 308 = 22*14 and at 330=22*15, but then the pattern deviates again at 355, which is not a multiple of 22.
### Check that the Beatty and complementary sequences are disjoint
We have asserted that the Beatty and complementary sequences are disjoint and that their union is the set of all positive integers. You can't use numerical computations to prove this fact, but you can verify that it is true for the finite set of numbers that we have generated. The following SAS/IML program checks that the two sequences do not intersect and that their union covers all positive integers (up to 520).
/* Check union and intersection properties of Beatty and complementary sequences */ proc iml; use Beatty; read all var 's' into B where(Sequence='B'); read all var 's' into C where(Sequence='C'); close; /* is there any intersection? */ intersect = xsect(B, C); if IsEmpty(intersect) then print "B and C do not intersect"; else print "The intersection is:", intersect; /* use the UNION function */ max = min( max(B), max(C) ); union = union(B, C); union = union[ , 1:max]; /* only test for values in [1, max] */ if all(union = 1:max) then print "All integers are found"; else print "Some integer is missing"; QUIT;
The program verifies (for positive integers less than 520) that the Beatty and complementary sequences do not have any intersection. Furthermore, each integer is either an element of the Beatty sequence for π or it is an element of the complementary sequence.
I want to point out a cool SAS/IML syntax: The IF-THEN statement can be used to test whether one vector equals another vector. In the program, the statement all(union = 1:max) returns true is every element of the vector union is equal to the corresponding element of the vector {1,2,3,...,max}.
### Summary
This article describes Beatty's theorem, which is an interesting result in mathematics. Given any irrational number, r > 1, the Beatty sequence for r and the complementary sequence partition the positive integers into two disjoint groups: the Beatty sequence for r and the complementary sequence. The article illustrates the theorem by using r = π and shows some connections with the continued fraction expansion of π.
The post The Beatty sequence for pi appeared first on The DO Loop.
For some reason, SAS programmers like to express their love by writing SAS programs. Since Valentine's Day is next week, I thought I would add another SAS graphic to the collection of ways to use SAS to express your love.
Last week, I showed how to use vector operation and a root-finding algorithm to visualize the path of a ball rolling on an elliptical billiards table. Every time the ball hits the boundary of the table, it rebounds and continues rolling in a new direction. At the end of the article, I commented that "the same program can compute the path of a ball on any table" for which certain conditions hold. With Valentine's Day coming soon, I wondered whether I could visualize the dynamics of a ball bouncing across a heart-shaped table. The graph to the right is the result of my efforts.
In the figure, the pink area is the interior of the heart-shaped billiard table. The red lines trace the path of a ball that starts in the middle of the heart with an initial motion at 45 degrees to the horizontal. The figure shows the ball's path for the first 250 bounces.
Although the caption of the figure is "Love Makes My Heart Bounce," the figure might also resonate with people who feel like they bounce around and never find love. It might appeal to those who have recently ended a relationship and feel like they are on the rebound. And during the global pandemic, maybe you have stayed close to home, never venturing very far from those you love? This heart is for you, too!
### Details of the computation
Here are a few details about the computation:
• The shape of the heart is defined by $H(x,y)=0$, where $H(x,y) = (x^2 + y^2 - 1)^3 - x^2 y^3$.
• The gradient of H vanishes at the four points on the curve, namely (0, ±1) and (±1, 0). If the virtual ball "hits" the table's virtual boundary at those locations, the ball will "stick" to the boundary because the program uses the gradient of the function to compute the normal vector to the curve. The program could be modified to compute the normal vector at (±1, 0). However, the curve has a cusp singularity at (0, ±1), so it is impossible to define a normal vector at those points. On a real billiard table, you would want to round off the singularity at those points.
• Because the heart shape is not convex, I had to modify the routine that finds the intersection of a ray and the curve H=0. The previous article on this topic assumes that a ray intersects the curve at only one point. Along the top of the heart, a ray might intersect the curve at three points. I modified the program to find the nearest intersection point.
You can download the program that I used to make the graph. Feel free to change the colors and caption and spread the love.
The post Billiards on a heart-shaped table appeared first on The DO Loop.
I recently showed how to find the intersection between a line and a circle. While working on the problem, I was reminded of a fun mathematical game. Suppose you make a billiard table in the shape of a circle or an ellipse. What is the path for a ball at position p on the table if it rolls in a specified direction, as determined by a vector, v? A possible path is shown in the graph at the right for a circular table. The figure shows the first 50 bounces for a ball that starts at position p=(0.5 -0.5) and travels in the direction of v=(1, 1).
This article shows how to use SAS to compute images like this for circular and elliptical tables.
### Vectors and geometry
One of the advantages of the SAS/IML matrix language is its natural syntax: You can often translate an algorithm from its mathematical formulation to a computer program by using only a few statements. This section presents the mathematical formulation for finding the trajectories of a ball on a circular or elliptical billiard table, then shows how to translate the math into SAS/IML statements.
The fundamental mathematical assumption is that all collisions are perfectly elastic and that the angle of incidence equals the angle of reflection. For a curved surface, the angles are measured by using the normal and tangent vectors to the curve. The geometry is shown in the following figure.
A ball moves toward the boundary of the table with velocity vector v. The boundary is represented as the level set H = 0 for some bivariate function, H(x). For example, a circular table uses H(x,y) = x2 + y2 - 1. The previous article showed how to find the point at which the ball hits the curve.
You can compute the unit normal vector, N, at that point by using the gradient of H. Project the velocity vector onto the unit normal to obtain vN. The linear space orthogonal to N is the tangent space. The vector vT = v - vN is the projection of v onto the tangent space. (Note that v = vT + vN.) The reflection of v is defined to be the vector that has the same tangent components but the opposite normal components, which is vrefl = vT - vN. A little algebra shows that vrefl = v - 2vN, so, in practice, the projection onto the tangent space is not needed.
The following SAS/IML program implements functions that compute these vectors. To test the functions, assume a ball is initially at the point p=(0.5, -0.5) and moves with unit velocity v=(1, 1)/sqrt(2), which is 45 degrees from the horizontal.
proc iml; /* Define the multivariate function H and its gradient. Example: H(x)=0 defines a circle and grad(H) is the gradient */ start H(x); return( x[ ,1]##2 + x[ ,2]##2 - 1 ); /* = x##2 + y##2 - 1 */ finish; start gradH(x); return( 2*x ); /* { dH/dx dH/dy } */ finish; /* given x such that H(x)=0 and a unit vector, v, find the normal and tangent components of v such that v = v_N + v_T, where v_N is normal to H=0 and v_T is tangent to H=0 */ /* project v onto the normal space at x */ start NormalH(x, v); grad = gradH(x); /* if H(x)=0, grad(H) is normal to H=0 at x */ N = grad / norm(grad);/* unit normal vector to curve at x */ v_N = (v*N) * N; /* projection of v onto normal vetor */ return( v_N ); finish; /* project v onto the tangent space at x */ start TangentH(x, v); v_N = NormalH(x, v); v_T = v - v_N; return( v_T ); finish; /* reflect v across tangent space at x */ start Reflect(x, v); v_N = NormalH(x, v); v_refl = v - 2*v_N; return( v_refl ); finish; /* Test the definitions for the following points and direction vector */ p = {0.5 -0.5}; /* initial position of ball */ v = {1 1} / sqrt(2); /* unit velocity vector: sqrt(2) = 0.7071068 */ q = {1 0}; /* ball hits circle at this point */ N = NormalH(q, v); /* normal component to curve at q */ T = TangentH(q, v); /* tangential component to curve at q */ v_refl = Reflect(q, v); /* reflection of v */ print v, N, T, v_refl;
From the initial position and velocity, it is clear that the ball will hit the circle at q=(1,0) and reflect off at 135 degrees, which is the direction vector vrefl=(-1, 1)/sqrt(2). For this geometry, the unit normal vector at q is N=(1,0) and the unit tangent vector is T=(0,1), so it is easy to mentally verify that the program is giving the correct values for the vectors in this problem.
### Bouncing around a billiard table
Let's play a game of billiards on a table that has a nonstandard shape. To begin the game, you must specify the initial position (p) and velocity (v) of the ball, as well as the shape of the table (a function H). The algorithm is simple:
1. Given p and v, find the point q where the ball hits the boundary of the table. This step is described in a previous article..
2. Reflect v across the normal vector at q to obtain a new direction, vrefl.
3. Repeat this process some number of times. Optionally, plot the path to visualize how the ball bounces across the table.
The following SAS/IML program implements this algorithm by using the F_Line and FindIntersection functions from the previous article:
/* To find the intersection of H=0 with the line through G_p in the direction of G_v, see https://blogs.sas.com/content/iml/2022/02/02/line-search-sas.html */ /* Restriction of H to the line through G_p in the direction of G_v */ start F_Line(t) global(G_p, G_v); return( H( G_p + t*G_v ) ); finish; /* return the intersection of H=0 and the line */ start FindIntersection(p, v, step=20) global(G_p, G_v); G_p = p; G_v = v; t = froot("F_Line", 1E-3 || step); /* find F(t) = 0 for t in [1E-3, step] */ q = p + t*v; /* H(q) = 0 */ return( q ); finish; /* given a point, p, and a direction vector, v, find the point q on the line through p in the direction of v such that H(q)=0. Reflect the velocity vector to get a new direction vector, v_refl */ start HitTable(q, v_refl, p, v); q = FindIntersection(p, v); v_refl = Reflect(q, v); finish; /* Example: A periodic path with four segments */ p = {0.5 -0.5}; /* point p */ v = {1 1}/sqrt(2); /* direction vector v */ NSeg = 6; /* follow for this many bounces */ x = j(NSeg, ncol(p)); x[1,] = p; do i = 2 to NSeg; run HitTable( q, v, x[i-1,], v ); /* overwrite v on each iteration */ x[i,] = q; end; print x[L="Positions"];
These initial conditions are highly symmetric and therefore easy to visualize. From the initial position and velocity, the ball hits the boundary of the circular table at (1,0), (0,1), (-1,0), and (0,-1) before returning again to (1,0). The ball will follow this periodic trajectory forever. It will always hit the same four points on the boundary of the circle.
For a general initial condition, you will want to draw a graph that visualizes the orbit of the ball as it bounces across the table. You can download the SAS program that computes the graphs in this article. For ease of use, I encapsulated the computations and visualization into a module called Billiards. You can call the Billiards subroutine with an initial condition and the number of bounces and the module will return a graph of the trajectory of the ball. For example, the following statements generate the first 50 bounces of a ball on a circular table:
/* Example 2. Either periodic orbit or bounces inside an annulus */ p = {0.5 -0.5}; /* point p */ v = {1 1.4}; /* vector v */ v = v / norm(v); /* standardize to unit vector */ print v; title "50 Bounces on a Round Table"; title2 "p=(0.5, -0.5); v=(0.58, 0.81)"; run Billiards(p, v, 50); /* NSeg=50 bounces */
The graph is shown at the top of this article. For this example, the ball bounces around the table inside an annular region. For a circular table, you can prove mathematically that every trajectory is either a periodic orbit or is dense in an annular region.
### Billiards on an elliptical table
If you study the previous program, you will see that the program does not use any special knowledge of the shape of the table. Given any smooth functions H and grad(H), the program should be able to handle tables of other shapes. That is the advantage of writing a program that uses vectors in a coordinate-free manner.
Let's change the functions for H and grad(H) to use an elliptical table with parameters a and b. That is, the shape of the table is determined by H(x)=0, where $H(\mathbf{x}) = x_1^2/a^2 + x_2^2/b^2 - 1$ is the standard form of an ellipse. For the next example, let's use a=sqrt(5) and b=2.
/* define the elliptical region and gradient vector */ start H(x); a = sqrt(5); b = 2; return( (x[ ,1]/a)##2 + (x[ ,2]/b)##2 - 1 ); /* ellipse(a,b) */ finish; start gradH(x); a = sqrt(5); b = 2; return( 2*x[ ,1]/a##2 || 2*x[ ,2]/b##2 ); finish;
An ellipse is determined by its two foci, which are located at (+/-sqrt(a2 - b2), 0) when a > b. For our example, the foci are at (-1, 0) and (1, 0).
You can show that the path of a ball on an elliptical table is one of three shapes:
• If the ball does not pass between the foci, its path is inside an annular region.
• If the ball passes between the foci, its path is inside a hyperbolic region.
• If the ball passes over one focal point, it also passes over the second focal point. The path converges to the major axis that contains the foci.
The following examples show each case:
/* If the ball does not pass between the foci, its path is inside an annular region. */ p = {1.5 0}; /* point p */ v = {0 1}; /* vector v */ title "250 Bounces on an Elliptical Table"; title2 "Ball Does Not Pass Between Foci"; run Billiards(p, v, 250);
/* If the ball passes between the foci, its path is inside a hyperbolic region. */ p = {0 0}; /* point p */ v = {1 1}/sqrt(2); /* vector v */ title "250 Bounces on an Elliptical Table"; title2 "Ball Passes Between Foci"; run Billiards(p, v, 250);
/* If the ball passes over one focal point, it also passes over the second focal point. The path converges to the major axis that contains the foci. */ p = {1 -1}; /* point p */ v = {0 1}; /* vector v */ title "10 Bounces on an Elliptical Table"; title2 "Ball Passes Through a Focal Point"; run Billiards(p, v, 10);
### Summary
In summary, you can use analytical geometry to figure out the trajectory of a ball on nonstandard billiard table that has a curved boundary. By using vectors and a root-finding algorithm in the SAS/IML language, you can write a program that displays the path of a ball on a circular table. The advantage of a coordinate-free vector formulation is that the same program can compute the path of a ball on a table that has a more complicated boundary, such as an ellipse. In fact, the same program can compute the path of a ball on any table for which H=0 defines a smooth convex curve on which the gradient of H does not vanish.
Incidentally, people have built elliptical billiard tables and have put a hole at one of the foci. To sink a ball, you can shoot it at the hole or you can shoot it at the other focal point. Either way, the ball rolls into the hole.
The post Billiards on an elliptical table appeared first on The DO Loop.
Suppose you are creating a craft project for the Christmas holidays, and you want to choose a palette of Christmas colors to give it a cheery holiday appearance. You could use one of the many online collections of color palettes to choose a Christmas-themed palette. However, I didn't want to merely take a palette that someone else designed. As a statistician, I wanted to create my own Christmas palette by using some sort of mathematical technique that creates a "statistically perfect" set of Christmas colors from other palettes!
Artists use their talents to create aesthetically pleasing and harmonious palettes of colors. But why should artists have all the fun? In this article, I show how a statistician might create a palette of Christmas colors! Starting from existing palettes of colors, the technique uses principal component analysis to select a linear subspace of colors. From within that subspace, you can create your own palette of colors. An example is the green-gold-red palette that is shown below. It was made by math, but I think it is pretty!
The same technique could also be used to create autumn palettes, spring palettes, and so forth, but don't take this too seriously: this article is intended to be geeky fun for the holidays. I'll let others debate whether this mathematical technique is useful for creating art.
### Outline of the method
Here is the general idea behind creating a statistical palette of colors:
1. Collect 6-10 existing palettes that you like. This process should produce 40-60 colors that are related to your theme (such as "Christmas"). I have written about how to read and visualize palettes in SAS.
2. Each color can be represented a triplet of values in RGB space. You can use principal component (PC) analysis to project the colors into a plane that captures most of the variation in the colors. The graph at the right shows the principal component scores for 41 colors in seven Christmas palettes.
3. Each palette is represented as a path through this linear subspace. So, you can choose a new sequence of points in this subspace and then REVERSE the projection to obtain a new palette of RGB colors.
4. There are infinitely many paths you could choose, but one choice is to use vertices of a regular polygon to create a "statistically perfect" set of Christmas colors.
### Paths through principal component space
For each palette, if you connect the sequence of colors in the palette, you obtain a path through the plane of the principal component scores. If you try to display the paths for all seven palettes, you will get a jumbled mess of lines that crisscross the graph. To make it easier to visualize the paths for each palette, the following graph limits the display to two palettes. The "Jingle Bell" palette has six colors. It starts with two green colors, moves to golden colors, and ends with two red colors. The "Unwrapping My Gifts" palette has five colors. It starts with two green colors, moves to a silver color, and ends with a red and dark burgundy color. The other palettes trace similar paths through the plane of scores.
If each palette corresponds to a path, then each path corresponds to a palette. In other words, if you choose a sequence five or six coordinates in the space of principal component scores, the path defines a new Christmas palette.
The ancient Greeks thought that a circle was the most perfect shape. Regular polygons, which approximate the circle, are also seen as perfect in their symmetry and beauty. Therefore, it makes sense to use a regular hexagon to create a new Christmas palette. The following graph demonstrates how you can choose a new Christmas palette. The circle represents a path through the space of colors. The gray stars represent six points on the circle, which are also vertices on a regular hexagon. I will define a new palette by using math to figure out what colors to associate with the coordinates of the six stars. I've ordered the six points starting from the point that I know will be green (because it is in the lower-left corner) and ending with the point that I know will be red (because it is in the upper-left corner).
### Inverting the principal component transformation
Until now, we've only considered the projection of the RGB colors onto the plane of the first two PCs. Of course, this projection is not invertible: There are many colors that share the same two PC scores. However, in general, the transformation from RGB space to the space of all principal component scores is invertible. You can represent the transformation as a matrix by using the spectral decomposition of the correlation matrix of the colors. To embed the plane of the first two PCs, you need to set a value for the coordinate of the third PC. The obvious statistical choice is to use PC3=0, which represents the average score in the third PC. In other words, you can associate the first two PC scores with the plane that has ordered triplets (PC1, PC2, 0).
With that choice, it is straightforward to apply the inverse PC transformation. The inverse transformation maps PC scores into colors in RGB space. You can use the SAS IML language to carry out this transformation. If you visualize a dense set of points on the circle, you get the following spectrum of colors:
Any smaller palette will contain a subset of these colors. For example, if you choose six points on the circle that are vertices of a regular histogram, you get the palette that is shown at the top of this article. The new palette is based on the original seven palettes but is distinct from them. The new palette incorporates the properties of the original palettes but is defined by defining a new path through the PC scores.
### Summary
In summary, this article shows how to create a new palette of colors that is based on statistical properties of a set of colors. You can perform a principal component analysis in RGB space to obtain a plane of PC scores. Every path through that plane corresponds to a new palette of colors. For my palette, I chose a sequence of vertices on a regular hexagon.
This method is not guaranteed to give an aesthetically pleasing palette of colors, but I think it does in this example. And if you don't like my palette, you can choose any other path through the set of PC scores to create a new palette! If you would like to experiment with this technique, you can download the SAS program that I used to generate the figures in this article.
The post A statistical palette of Christmas colors appeared first on The DO Loop.
The number of possible bootstrap samples for a sample of size N is big. Really big.
Recall that the bootstrap method is a powerful way to analyze the variation in a statistic. To implement the standard bootstrap method, you generate B random bootstrap samples. A bootstrap sample is a sample with replacement from the data. The phrase "with replacement" is important. In fact, a bootstrap sample is sometimes called a "resample" because it is generated by sampling the data with REplacement.
This article compares the number of samples that you can draw "with replacement" to the number of samples that you can draw "without replacement." It demonstrates, for very small samples, how to generate the complete set of bootstrap samples. It also explains why generating a complete set is impractical for even moderate-sized samples.
### The number of samples with and without replacement
For a sample of size N, there are N! samples if you sample WITHOUT replacement (WOR). The quantity N! grows very quickly as a function of N, so the number of permutations is big for even moderate values of N. In fact, I have shown that if N≥171, then you cannot represent N! by using a double-precision floating-point value because 171! is greater than MACBIG (=1.79769e+308), which is the largest value that can be represented in eight bytes.
However, if you sample WITH replacement (WR), then the number of possible samples is NN, which grows even faster than the factorial function! If the number of permutations is "big," then the number of WR samples is "really big"!
For comparison, what do you think is the smallest value of N such that NN exceeds the value of MACBIG? The answer is N=144, which is considerably smaller than 171. The computation is shown at the end of this article.
Another way to compare the relative sizes of N! and NN is to print a few values of both functions for small values of N. The following table shows the values for N≤10:
proc iml; N = 1:10; nPerm = fact(N); /* N! */ nResamp = N##N; /* N^N */ both = nPerm // nResamp; print both[r={"Permutations" "Samples (WR)"} c=N format=comma15. L=""];
Clearly, the number of bootstrap samples (samples WR) grows very big very fast. This is why the general bootstrap method uses random bootstrap samples rather than attempting some sort of "exact" computation that uses all possible bootstrap samples.
### An exact bootstrap computation that uses all possible samples
If you have a VERY small data set, you could, in fact, perform an exact bootstrap computation. For the exact computation, you would generate all possible bootstrap samples, compute the distribution on each sample, and thereby obtain the exact bootstrap distribution of the statistic.
Let's carry out this scheme for a data set that has N=7 observations. From the table, we know that there are exactly 77 = 823,543 samples with replacement.
For small N, it's not hard to construct the complete set of bootstrap samples: just form the Cartesian product of the set with itself N times. In the SAS/IML language, you can use the ExpandGrid function to define the Cartesian product. If the data has 7 values, the Cartesian product will be a matrix that has 823,543 rows and 7 columns.
The following SAS/IML program performs a complete bootstrap analysis of the sample mean. The sample, S, has 7 observations. The minimum value is -1; the maximum value is 4. The sample mean is approximately 0.51. What is the bootstrap distribution of the sample mean? You can form the set of all possible subsamples of size 7, where the elements are resampled from S. You can then compute the mean of every resample and plot the distribution of the means:
proc iml; S = {-1 -0.1 -0.6 0 0.5 0.8 4}; /* 1. compute original statistic */ m = S[:]; /* = mean(S) */ print m; /* 2. Generate ALL resamples! */ /* Cartesian product of the elements in S is S x S x ... x S \------v------/ N times */ z = ExpandGrid(S, S, S, S, S, S, S); /* 3. Compute statistic on each resample */ means = z[,:]; /* 4. Analyze the bootstrap distribution */ title "Complete Bootstrap Distribution of the Mean"; title2 "N = 7"; call histogram(means) rebin={-1 0.1} xvalues=-1:4 other="refline 0.51/axis=x lineattrs=(color=red);";
The graph shows the mean for every possible resample of size 7. The mean of the original sample is shown as a red vertical line. Here are some of the resamples in the complete set of bootstrap samples:
• The resamples where a datum is repeated seven times. For example, one resample is {-1, -1, -1, -1, -1, -1, -1}, which has a mean of -1. Another resample is {4, 4, 4, 4, 4, 4, 4}, which has a mean of 4. These two resamples define the minimum and maximum values, respectively, of the bootstrap distribution. There are only seven of these resamples.
• The resamples where a datum is repeated six times. For example, one of the resamples is {-1, -1, -1, -1, -1, -1, 4}. Another is {-1, -1, 0, -1, -1, -1, -1}. A third is {4, 4, 4, 4, 4, 4, 4}, which has a mean of 4. These two resamples define the minimum and maximum values. Another resample is {4, 4, 4, 4, 0.5, 4, 4}. There are 73 resamples of this type.
• The resamples where each datum is present exactly one time. These sets are all permutations of the sample itself, {-1 -0.1 -0.6 0 0.5 0.8 4}. There are 7! = 5,040 resamples of this type. Each of these permutations has the same mean, which is 0.51. This helps to explain why there is a visible peak in the distribution at the value of the sample mean.
The bootstrap distribution does not make any assumptions about the distribution of the data, nor does it use any asymptotic (large sample) assumptions. It uses only the observed data values.
### Comparison with the usual bootstrap method
For larger samples, it is impractical to consider ALL possible samples of size N. Therefore, the usual bootstrap method generates B random samples (with replacement) for a large value of B. The resulting distribution is an approximate bootstrap distribution. The following SAS/IML statements perform the usual bootstrap analysis for B=10,000.
/* Generate B random resamples. Compute and analyze bootstrap distribution */ call randseed(123); w = sample(S, {7, 10000}); /* 10,000 samples of size 7 (with replacement) */ means = w[,:]; title "Bootstrap Distribution of the Mean"; title2 "B = 10,000; N = 7"; call histogram(means) rebin={-1 0.1} xvalues=-1:4 other="refline 0.51/axis=x lineattrs=(color=red);";
The resulting bootstrap distribution is similar in shape to the complete distribution, but only uses 10,0000 random samples instead of all possible samples. You can see that the properties of the second distribution (such as quantiles) will be similar to the quantiles of the complete distribution, even though the second distribution is based on many fewer bootstrap samples.
### Summary
For a sample of size N, there are NN possible resamples of size N when you sample with replacement. The bootstrap distribution is based on a random sample of these resamples. When N is small, you can compute the exact bootstrap distribution by forming the Cartesian product of the sample with itself N times. This article computes the exact bootstrap distribution for the mean of seven observations and compares the exact distribution to an approximate distribution that is based on B random resamples. When B is large, the approximate distribution looks similar to the complete distribution.
For more about the bootstrap method and how to perform bootstrap analyses in SAS, see "The essential guide to bootstrapping in SAS."
### Appendix: Solving the equation NN = y
This article asked, "what is the smallest value of N such that NN exceeds the value of MACBIG?" The claim is that N=144, but how can you prove it?
For any value of y > 1, you can use the Lambert W function to find the value of N that satisfies the equation NNy. Here's how to solve the equation:
1. Take the natural log of both sides to get N*log(N) ≤ log(y).
2. Define w = log(N) and x = log(y). Then the equation becomes w exp(w) ≤ x. The solution to this equation is found by using the Lambert W function: w = W(x).
3. The solution to the original equation is N = exp(w). Since N is supposed to be an integer, round up or down, according to the application.
The following SAS/IML program solves for the smallest value of N such that NN exceeds the value of MACBIG in double-precision.
proc iml; x = constant('LogBig'); /* x = log(MACBIG) */ w = LambertW(x); /* solve for w such that w*exp(w) = x */ N1 = exp(w); /* because N = log(w) */ N = ceil(N1); /* round up so that N^N exceeds MACBIG */ print N1 N; quit;
So, when N=144, NN is larger than the maximum representable double-precision value (MACBIG).
The post On the number of bootstrap samples appeared first on The DO Loop.
Art evokes an emotional response in the viewer, but sometimes art also evokes a cerebral response. When I see patterns and symmetries in art, I think about a related mathematical object or process. Recently, a Twitter user tweeted about a painting called "Phantom’s Shadow, 2018" by the Nigerian-born artist, Odili Donald Odita. A modified version of the artwork is shown to the right.
The artwork is beautiful, but it also contains a lot of math. The image shows 64 rotations, reflections, and translations of a polygon in four colors. As I will soon explain, the image can also be viewed as 4 x 4 grid where each cell contains a four-bladed pinwheel. The grid displays rotations and reflections of a pinwheel shape.
When I saw this artwork, it inspired me to look closely at its mathematical structure and to create my own mathematical version of the artwork in SAS. This article shows how to use rotations to create a pinwheel from a polygon. A second article discusses how to use rotations and reflections to create a mathematical interpretation of Odita's painting.
### Create a polygon
Look closely at the upper left corner of "Phantom's Shadow." You will see the following pinwheel-shaped figure:
If the center of the pinwheel is the origin, then this pinwheel is based on a sequence of 90-degree rotations of the teal-colored polygon about the origin. Each rotation is a different color (teal, orange, blue, and salmon) on a gray background.
You can assign coordinates to the vertices of the teal polygon. If three vertices are on the corners of the unit square, the fourth vertex appears to be at (2/3, 1/3). You can put the vertices into a SAS data set and graph the polygon by using the POLYGON statement in PROC SGPLOT:
%let alpha = 0.333; /* Odita's work uses a vertex as (1-alpha, alpha) */ data Poly1; ID = 1; input x y @@; if x=. then do; /* just for fun, support other locations of vertex */ x = 1 - α y = α end; datalines; 0 0 . . 1 1 0 1 0 0 ; %let blue = CX0f5098; %let orange = CXeba411; %let teal = CX288c95; %let salmon = CXd5856e; %let gray = CX929386; ods graphics / width=480px height=480px; title "Base Polygon"; proc sgplot data=Poly1 aspect=1; styleattrs wallcolor=&gray; polygon x=x y=y ID=ID / fill fillattrs=(color=&teal) outline; xaxis offsetmin=0.005 offsetmax=0; yaxis offsetmin=0 offsetmax=0.005; run;
The complete artwork repeats this polygon 64 times in a grid by using different colors and different orientations. But the colors and orientations are not random—although that would also look cool! Instead, there is an additional structure. The polygon is part of a pinwheel, and the pinwheel shape is rotated and reflected 16 times in a 4 x 4 grid. The next section shows how to create the pinwheel.
### Create a pinwheel
It is possible to perform 2-D rotations and reflections in the DATA step, but the rotations and reflections of a planar figure are most simply expressed in terms of 2 x 2 orthogonal matrices.
To form the pinwheel from the basic polygon, you can use a group of four matrices. All rotations are in the counter-clockwise direction about the origin:
• R0 is the identity matrix
• R1 is the matrix that rotates a point 90 degrees (about the origin)
• R2 is the matrix that rotates a point 180 degrees
• R3 is the matrix that rotates a point 270 degrees
You do not need advanced mathematics to understand rotations by 90 degrees. However, these four rotation matrices are an example of an algebraic structure called a finite group. These matrices form the cyclic group of order 4 (C4), where the group operation is matrix multiplication.
Regardless of their name, you can use these matrices to transform the polygon into a pinwheel. The following SAS/IML program reads in the base polygon and transforms it according to each of the four matrices. You can then plot the four images, each in a different color.
/* rotate polygon about the origin to form a pinwheel */ proc iml; /* actions of the C4 cyclic group: rotations by 90 degrees */ start C4Action(v, act); if act=0 then M = { 1 0, 0 1}; else if act=1 then M = { 0 -1, 1 0}; else if act=2 then M = {-1 0, 0 -1}; else if act=3 then M = { 0 1, -1 0}; return( v*M ); /* = (M*z) */ finish; use Poly1; read all var {x y} into P; close; /* read the polygon */ /* write the pinwheel */ Q = {. . .}; create Shape from Q[c={'ID' 'x' 'y'}]; do i = 0 to 3; R = C4Action(P, i); /* rotated image of polygon */ Q = j(nrow(R), 1, i) || R; /* save the image to a data set */ append from Q; end; close; QUIT; title "Pinwheel Figure"; title2 "Colors=(Teal, Orange, Blue, Salmon)"; proc sgplot data=Shape aspect=1; styleattrs wallcolor=&gray datacolors=(&teal &orange &blue &salmon); polygon x=x y=y ID=ID / group=ID fill; xaxis offsetmin=0 offsetmax=0; yaxis offsetmin=0 offsetmax=0; run;
The resulting image is the four-blade pinwheel, which is the basis of Odita's artwork.
### A grid of pinwheels
The rest of "Phantom’s Shadow, 2018" can be described in terms of rotations and reflections of the pinwheel shape...with a twist! The next article introduces the dihedral group of order 4 (D4). This is a finite group of order eight and is the symmetry group of the square. It turns out that you can use the elements of D4 to generate the 16 pinwheels in Odita's artwork. You can then use PROC SGPANEL in SAS to generate a graph that looks like "Phantom's Shadow," or you can create your own Odita-inspired mathematical artwork! Stay tuned!
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Odani's truism is a mathematical result that says that if you want to compare the fractions a/b and c/d, it often is sufficient to compare the sums (a+d) and (b+c) rather than the products a*d and b*c. (All of the integers a, b, c, and d are positive.) If you compare the products, you will know the correct answer 100% of the time because a/b < c/d if and only if a*d < b*c. However, K. Odani proved (1998) that the expression (a+d) < (b+c) is true 91% of the time when the integers a, b, c, and d are chosen uniformly at random in the interval [1,N], where N is a very large integer. A previous article shows a Monte Carlo simulation that demonstrates Odani's truism.
As discussed at the end of the previous article, the 91% probability includes many fractions that are easy to compare in your head, such as 6/29 versus 35/18, and 1/6 versus 5/8. Odani's result also includes unreduced fractions such as 2/4 and 3/6. What is the probability that comparing the sums gives the correct answer if you consider only reduced fractions a/b and c/d in the interval [0,1] where the fractions are close to each other? That is the subject of this article. Briefly, this article examines whether Odani's truism has any value in practice.
Given a pair of fractions a/b < c/d, we say that Odani's truism is "true" (or valid or correct) if (a+d) < (b+c). Otherwise, we say that the truism is "false."
### The Farey sequence
I will use the Farey sequence in this investigation. The Farey sequence of order n is the sequence of all reduced fractions on [0,1] whose denominators do not exceed n, listed in increasing order. In practice, we do not usually need to compare fractions with huge denominators. Most of the time, we encounter fractions whose denominators are less than 100, which corresponds to the Farey sequence of order 100.
Consider the Farey sequence of order n for a small value of n, such as n=100. The sequence is in increasing order, so you can compare each term of the sequence with the next k terms and compute how often Odani's truism is correct. When k is small compared to n, this method estimates how often Odani's truism is valid for fractions in [0,1] that are near each other.
### First attempt: Fractions that are adjacent in the Farey seqence
Before looking a the general case of k > 0, let's look at the special case where k=1. For each fraction in the Farey sequence, a/d, let c/d be the next fraction in the Farey sequence. Because the Farey sequence is in order, we know that a/b < c/d. Compute the Odani sum (a+d)-(b+c). If the sum is negative, then Odani's truism is correct for that a/b and c/d. Otherwise, the truism is false.
The following SAS/IML function computes the Farey sequence for any order n:
proc iml; /* Generate Farey sequence of order n: all reduced fractions on [0,1] whose denominators do not exceed n, listed in increasing order https://blogs.sas.com/content/iml/2021/03/17/farey-sequence.html */ start Farey(n); /* upperbound on sequence length is n*(n+3)/2. See https://oeis.org/A005728 */ F = j(2, max(2,n*(n+3)/2), .); F[,1] = {0, 1}; F[,2] = 1 // n; order = 2; a=0; b=1; c=1; d=n; /* fractions are a/b and c/d */ do while (d > 1); order = order+1; k = floor((n + b)/d); /* integer division */ a_prev = a; a = c; b_prev = b; b = d; c = k*c - a_prev; d = k*d - b_prev; F[,order] = c // d; end; /* remove any unassigned elements */ keepidx = 1:order; return( F[ ,keepidx] ); finish;
You can call the function generate the Farey sequence for n=100. Let's print out a few fractions in the sequence along with an indicator variable that shows whether their Odani sums are negative:
S = Farey(100); /* generaste Farey sequence of order n=100 */ idx = 401:415; /* look at a few terms in the Farey sequence */ s0 = S[,idx]; /* get the corresponding fractions */ /* evaluate the Odani sum on adjacent pairs */ i = 1:ncol(idx)-1; a = S0[1,i]; b = S0[2,i]; /* first row of S contains numerators; second row contains denominators */ c = S0[1,i+1]; d = S0[2,i+1]; OdaniSum = (a+d)-(b+c); /* when sum is negative, the truism is true (predicts that a/b < c/d) */ Truism = (OdaniSum < 0); /* In SAS/IML 15.1, you can use numeric value for the COLNAME= option. See https://blogs.sas.com/content/iml/2019/07/31/numeric-column-header-print.html */ print (s0//(Truism||.))[L="Some Fractions in Farey(100)" c=idx r={"Numer" "Denom" "Truism"}];
The Farey sequence of order 100 has 3,045 elements. The table shows the values for the elements near the fraction 2/15, which is the 407th element in the sequences. The program compares the i_th fraction to the (i+1)st fraction and computes whether (a+d) < (b+c) for the adjacent pairs. For the 14 comparisons in this example, the Odani truism was true for seven and false for the other seven.
A 50% success rate is not very good! Maybe fractions near 2/15 are "unusual" in some way? Let's compare all adjacent pairs of fractions in the Farey(100) sequence and compute the proportion that satisfies Odani's truism. We can also plot the truth or falsity of Odani's truism for each fraction in the Farey sequence:
/* compare each fraction to the one that follows it */ i = 1:ncol(S)-1; /* the indices of a/b; i+1 contains the indices of c/d */ a = S[1,i]; b = S[2,i]; c = S[1,i+1]; d = S[2,i+1]; Truism = ((a+d)-(b+c) < 0); /* plot the result (T or F) versus the fraction */ title "Odani's Truism for Consecutive Fractions in Farey(100)"; x = S[1,i] / S[2,i]; TorF = choose(Truism=1, "True ", "False"); call scatter(x, TorF) grid={x y} option="transparency=0.95" xvalues=do(0,1,0.1); prob = mean(Truism); print prob;
When applied to adjacent fractions in the Farey sequence, Odani's truism is true only 50% of the time. The scatter plot shows the truth or falsehood of Odani's truism for each fraction in the Farey sequence. The results do not seem to depend on the location of the fractions within the interval [0,1].
Notice that this result does not violate Odani's theorem because we are comparing only certain fractions: reduced fractions in [0,1] that are adjacent to each other in the Farey sequence.
### Compare fractions that are k terms away in the Farey seqence
For a Farey sequence that contains L terms, the previous section shows how to compare L-1 pairs of fractions. Each fraction was compared to the next fraction in the Farey sequence. If we perform ALL pairwise comparisons of the terms in the Farey sequence, Odani's truism will be valid for a larger proportion of comparisons. It probably won't hold for 91% of the comparisons (because we are dealing with reduced fractions in [0,1]), but I expect it to hold for more than 50%.
Let L be the length of the Farey sequence. The following function compares each term in the Farey sequence with the next k terms in the sequence, when possible. For the first L-k terms, you can compare each term with the k terms that follow. For the last k terms, you can only compare to the end of the sequence. Thus, there are a total of k*(L-k) + k*(k-1)/2 comparisons. If you count the number of pairs that satisfy Odani's truism and divide by the total number of comparisons, you get the proportion of comparisons for which Odani's sum accurately predicts whether one fraction is smaller than another:
/* S is a Farey sequence. k is the number of subsequent terms that we compare to each fraction. For example: k = 1 will compare adjacent terms k = 2 will compare each fraction with the following two terms, and so on... */ start CompareFract(S, k); Len = ncol(S); count = 0; numCompared = k*(Len-k) + k*(k-1)/2; /* total number of pairwise comparisons */ do i = 1 to Len-1; a = S[1,i]; b = S[2,i]; /* i = the indices of a/b; /* we know that a/b is less than all the fractions that follow */ idx = (i+1):min(i+k, Len); /* the next 'k' fractions or until the end */ c = S[1, idx]; d = S[2, idx]; /* c/d for the next k fractions */ OdaniSum = (a+d) - (b+c); count = count + sum(OdaniSum<0); /* how many fractions satisfy the eqn? */ end; prop = count / numCompared; /* proportion of comparisons that satisfy eqn */ return count || numCompared || prop; finish;
Let's apply the function to the Farey sequence of order 100, which has 3,045 terms. Let's perform a total of 100 comparisons, where k is evenly distributed among the values 1–3045. (The number of comparisons, 100, is unrelated to the order of the Farey sequence.) The following statements call the CompareFracts function for values of k that are approximately 3045/100 units apart. The exact values are k={30, 60, 91, ..., 3014, 3045}.
S = Farey(100); pct = do(0.01, 1.0, 0.01); /* evenly spaced percentages */ k = int( ncol(S)*pct ); /* choose k to be percentage of sequence length */ results = j(ncol(k), 3, .); do i = 1 to ncol(k); results[i,] = CompareFract(S, k[i]); end; title "Proportion of Odani Comparisons That Are True"; title2 "Farey Sequence of Order 100"; refStmt = "refline 0.917 / axis=y noclip label='11/12';"; call series(pct, results[,3]) grid={x y} other=refStmt label={'Fraction of Sequence Length' 'Proportion'};
The graph shows the percentage of comparisons for which Odani's truism is valid when comparing nearby fractions in the Farey sequence of order 100. The horizontal axis displays the proportion of the sequence that is being compared: 1%, 2%, ..., 100%. The left side of the graph indicates that if you compare each fraction to a small number of fractions that are close to it (such as 1% of the sequence length), Odani's truism is valid only about 50% of the time. If you compare each fraction to more fractions (which are, by definition, farther away from it), then Odani's truism is correct for a larger proportion of comparisons. If you consider all pairwise comparisons of the fractions, you find that about 82.5% of the comparisons satisfy Odani's truism (for the fractions in this Farey sequence).
A similar curve is seen when you consider larger sets of fractions. For example, for the Farey sequence of order 500 (which contains the 76,117 fractions whose denominators are less than or equal to 500), the graph looks similar except it is slightly higher: when you compare all fractions, Odani's truism is valid 83.2% of the time. As you consider larger Farey sequences, the proportion of comparisons for which Odani's truism increases very slowly. For the Farey sequence of order 1000, which has 304,193 terms, Odani's truism is valid 83.3% of the time when you perform all pairwise comparisons. Warning: Performing all pairwise comparisons for the Farey(1000) sequence is expensive! It takes about 30 minutes when using the program in this article to compare the 46 billion pairs of fractions. It is interesting that 83.3% ≈ 5/6.
### Summary
Odani's truism is a surprising mathematical theorem that says you can often determine whether a/b < c/d by comparing the sums (a+d) and (b+c). Although Odani's truism is correct 91% of the time (asymptotically) when you consider all integers a, b, c, and d that are less than some large number N, the truism is not very useful in practice. In practice, you often want to compare reduced fractions that are close to each other. By using Farey sequences, this article demonstrates that Odani's truism is correct only about 50% of the time when you compare a fraction and its nearby neighbors in the Farey sequence. It is only when you consider ALL pairwise comparisons of fractions that Odani's truism is correct for a large percentage of the comparisons. When comparing all reduced fractions whose denominators are 1000 or less, Odani's truism is correct about 83.3% of the time.
The post Odani's truism for fractions that are near each other appeared first on The DO Loop.
Quick! Which fraction is bigger, 40/83 or 27/56? It's not always easy to mentally compare two fractions to determine which is larger. For this example, you can easily see that both fractions are a little less than 1/2, but to compare the numbers you need to compare the products 40*56 and 27*83. Wouldn't it be great if there were a simpler way to compare fractions, maybe by using addition rather than multiplication?
It turns out that there is a simple equation that is often (but not always!) correct. I learned about this equation from John D. Cook, who blogged about this trick for comparing fractions by using only addition. According to a mathematical result by K. Odani (1998), the rule is correct more than 91% of the time.
Specifically, let N be a large integer. Randomly choose integers a, b, c, and d between 1 and N. Look at the signs of the following differences:
• X = a*d – b*c
• Y = (a+d) – (b+c)
Because a, b, c, and d are random, the quantities X and Y are random variables. K. Odani (1998) shows that X and Y have the same sign quite often for large N. As N goes to infinity, the probability that X and Y have the same sign is 11/12 ≈ 0.91667. I will refer to this result as Odani's truism.
Odani's truism provides a "rough and ready" rule for comparing two fractions a/b and c/d. If a/b < c/d, then X < 0, and, asymptotically, the expression (a+d) < (b+c) is true 91% of the time. Or, inverting the logic, you can evaluate (a+d) and (b+c) to determine, with high probability, whether the fraction a/b is greater than or less than the fraction c/d.
For example, again consider the fractions 40/83 and 27/56. It is easy to compute the sums 40+56=96 and 27+83=110. The first sum is less than the second, so Odani's truism suggests that 40/83 is less than 27/56, which is, in fact, true!
However, the summation rule does not always hold. For example, the summation rule for the fractions 13/27 and 40/83 gives the sums 96 and 67, but 13/27 is less than 40/83. If you use Odani's truism, you can never be sure that it holds for the two fractions you are trying to compare!
### Simulation of Odani's truism
You can demonstrate Odami's truism by using a simulation. Choose a large number N, and choose integers a, b, c, and d uniformly at random in [1, N]. Then the difference of the products (X = a*d – b*c) and the difference of the sums (Y = (a+d) – (b+c)) should have the same sign about 91% of the time. The following SAS DATA step carries out the simulation for a particular choice of N:
/* Simulation to demonstrate Odani's truism: To determine if a/b < c/d it is often sufficient to test whether a+d < b+c */ data _null_; retain count 0; call streaminit(54321); N = 1E8; /* upper bound. Integers in [1,N] */ numSamples = 1E6; /* number of random draws */ do i = 1 to numSamples; a = rand('integer', 1, N); b = rand('integer', 1, N); c = rand('integer', 1, N); d = rand('integer', 1, N); x = a*d - b*c; /* correct way to compare a/b and c/d */ y = (a+d) - (b+c); /* Odani's truism is correct 11/12 of time */ count + (sign(x) = sign(y)); /* is truism correct for this 4-tuple (a,b,c,d)? */ end; prob = count / numSamples; /* empirical probability for simulation */ put "Odani's truism holds with probability " prob; run;
Odani's truism holds with probability 0.91648
This simulation randomly generates 1,000,000 integer 4-tuples between 1 and 100 million. For each 4-tuple, the program compares the difference in the product (a*d - b*c) and the difference in the sum ((a+d) - (b+c)). If these quantities have the same signs (positive, negative, or zero), a counter is incremented. The result shows that the quantities have the same sign more than 91% of the time, in accordance with Odani's theorem.
### Interesting math versus practical math
As an applied mathematician, I am always curious whether an interesting mathematical result has a practical application. As John D. Cook points out at the end of his blog post, some fractions are easy to mentally compare whereas others are not. I don't need any trick to know that 6/29 is less than 18/35. It is obvious that the first fraction is less than 1/2 (in fact, it is close to 1/5) whereas the second is greater than 1/2. And I certainly don't need a trick to compare 6/29 and 35/18 because the latter fraction is greater than 1. Furthermore, Odani's result includes unreduced fractions such as 2/4, 3/6, and 50/100.
John D. Cook suggests that "it would be interesting to try to assess how often the rule of thumb presented here is correct in practice. You might try to come up with a model for the kinds of fractions people can’t compare instantly, such as proper fractions that have similar size."
That is a great idea. I don't need to use Odani's truism to compare 6/29 and 18/35, yet (6,29,18,35) is one of the 4-tuples for which Odani's truism is true. Similarly, it is true for (6,29,35,18), (1,6,5,8) and (1,10,9,10) and many other 4-tuples that represent fractions that can be easily compared without doing any explicit calculations. If you omit the "obvious" 4-tuples and unreduced fractions, you will reduce the probability that Odani's truism is true on the 4-tuples that remain. Thus, intuitively, Odani's truism should have a lower probability of being true if you restrict it only to reduced fractions that are close to each other. But how low? Is the rule any better than tossing a coin?
In the next blog post, I propose a way to measure how often Odani's truism is useful in practice. I will estimate the probability that Odani's truism holds for reduced fractions in the interval [0,1] that are close to each other.
### Summary
Odani's truism is an asymptotic result that states that you can compare the fractions a/b and c/d by looking at the sums (a+d) and (b+c). Specifically, the quantities (a*d - b*c) and ((a+d) - (b+c)) have the same sign with a probability that approaches 11/12 when the numerators and denominators are chosen uniformly at random in the interval [1,N] for very large N.
In the next post, I will discuss whether Odani's truism is useful in practice or whether it is merely a mathematical curiosity.
The post Odani's truism: A probabilistic way to compare fractions appeared first on The DO Loop.
Here is an interesting math question: How many reduced fractions in the interval (0, 1) have a denominator less than 100? The question is difficult is because of the word "reduced." If we only care about the total number of fractions in (0,1) whose denominator is less than 100, we could simply count the number of integer pairs (p,q), where 1 ≤ p < q < 100. For example, the table at the right shows all fractions whose denominators are less than 6. The green triangle encloses all n(n+1)/2 fractions in (0,1) for n=6. But only the fractions that have a gold background are in reduced form.
You can answer the question by generating a Farey sequence, which is the ordered set of reduced fractions (in reduced form) on the closed interval [0,1]. The Farey sequence has a surprising number of applications in pure and applied math. I first encountered it while studying the dynamics near resonant frequencies of forced coupled oscillators.
### The Farey sequence of order n
The Farey sequence is defined for fractions on the closed interval [0,1] and includes the endpoints as the reduced fractions 0/1 and 1/1. The Farey sequence of order n is the sequence of all reduced fractions on [0,1] whose denominators do not exceed n, listed in increasing order.
One algorithm for generating a Farey sequence would be to mimic the table at the top of this article. That is, generate all fractions with denominators up to n, then use a "sieve" to exclude fractions for which the numerator and denominator have a common factor. You would then have to sort the final set of results by the size of the fractions.
In fact, there is a better algorithm. The Wikipedia article about Farey sequences describes the algorithm as "surprisingly simple." The algorithm uses the "mediant property" of the Farey sequence to generate successive terms in order. The mediant property is the name given to the fact that each term in a Farey sequence is the mediant of its neighbors. Given two reduced fractions a/b and c/d, the mediant is the fraction (a+c)/(b+d). The mediant is sometimes called "freshman addition" (or, less kindly, "fool's addition") because the formula looks like a common mistake that children make when they learn to add fractions.
You can use the mediant property to generate the Farey sequence of order n from the Farey sequence of order n-1. However, as the Wikipedia article explains, you can also use the property to generate the Farey sequence of order n directly. The following DATA step generates the Farey sequences of order n for n ≤ 25.
/* Farey sequence: http://en.wikipedia.org/wiki/Farey_sequence For each n, generate the Farey sequence of order n. This algorithm uses the mediant property of the Farey sequence to generate the terms. */ %let MaxOrder = 25; data Farey; do n = 1 to &MaxOrder; /* for each n, generate Farey sequence of order n */ Term=1; Numer = 0; Denom = 1; output; Term=2; Numer = 1; Denom = n; output; a=0; b=1; c=1; d=n; /* fractions are a/b and c/d */ do while (d > 1); Term = Term + 1; k = floor((n + b)/d); /* integer division */ a_prev = a; b_prev = b; a = c; b = d; /* current c/d becomes new a/b */ c = k*c - a_prev; d = k*d - b_prev; /* new c/d */ Numer = c; Denom = d; output; end; end; keep n Term Numer Denom; label n="Order"; run;
Let's print out the terms of the Farey sequence of order n=8. The output shows the 21 fractions in the table at the top of this article, plus the trivial fractions 0/1 and 1/1 that are part of every Farey sequence. Notice the use of a DATA step view to compute the decimal values for the fractions, and the use of the FRACTw. format to display the decimals as fractions in their reduced form:
data Fractions / view=Fractions; set Farey; Fraction = Numer / Denom; run; proc print data=Fractions noobs; where n=8; format Fraction FRACT32.; run;
### Visualize the Farey sequences
The Wikipedia article does not display my favorite visualization of the Farey sequences. It is interesting to plot the Farey sequence for order n for many values of n, as follows:
ods graphics / width=500px height=400px; Title "Farey Sequences: Order 1 to 40"; proc sgplot data=Fractions; where n <= 40; scatter x=Fraction y=n / markerattrs=(symbol=CircleFilled) transparency=0.75; yaxis reverse grid integer values=(1, 5 to 40 by 5); xaxis grid display=(nolabel) values=(0 to 1 by 0.1); run;
The graph shows that there is a gap around each rational number in the Farey sequence. It turns out that the width of the gap (for each n) is larger for fractions that appear earlier in the Farey sequence. That is, if you look at the width of the gaps around 0 and 1, they are larger than the gap around 1/2, which is larger than the gaps around 1/3 and 2/3, and so forth. This turns out to be important in certain applications.
### The length of Farey sequences
In the previous graph, you can see that there are two markers on the first row, three on the second row, five on the third, and so forth. You can use PROC MEANS to find out how many terms are in the Farey sequence of order n. The following statements display the number of terms for n ≤ 10:
proc means data=Farey noNObs nolabels N; where n <= 10; class n; var n; run;
You can use this technique to answer the question at the top of this article: how many reduced fractions in the interval (0, 1) have a denominator less than 100? The length of the Farey sequence of order 99 answers the question. You can use PROC MEANS to find the length of the Farey sequence of order 99:
proc means data=Farey noNObs nolabels N; where n = 99; var n; run;
There are 3005 terms in the Farey sequence of order 99, which includes the fractions 0/1 and 1/1. If you exclude the endpoints, you get the answer to the question: there are 3003 reduced fractions in the interval (0,1) that have a denominator less than 100. | 2022-08-12 21:33:26 | {"extraction_info": {"found_math": true, "script_math_tex": 8, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7541002035140991, "perplexity": 982.5816987731735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00057.warc.gz"} |
https://folk.ntnu.no/leifh/teaching/tkt4140/._main036.html |
# 3.4 Shooting method for linear ODEs with two unknown initial conditions
In the follwing we will extend the shoothing methods we we derived in (3.1 Shooting methods for boundary value problems with linear ODEs) to problems with two initial conditions.
For such problemsn we must guess to initial values $$r$$ and $$s$$ with the corresponding boundary value error functions $$\phi$$ og $$\psi$$, respectively.
Figure 48: The boundary value error functions for boundary value problems with two unknown initial conditions.
As the ODE to be solved is linear, we know that the resulting boundary value error functions also will be linear. As $$\phi$$ and $$\psi$$ will be functions of the two initial guesses $$r$$ and $$s$$, the will express two planes which may be represented: $$$$\tag{3.130} \left.\begin{matrix} &\phi=A\cdot s+B\cdot r +C \\ &\psi=D\cdot s+E\cdot r+F \end{matrix}\right.\$$$$
where $$A,B,C,\cdots,F$$ are constants to be determined. The correct values $$r^*$$ og $$s^*$$ satisfy $$\phi(r^*,s^*)=0$$ and $$\psi(r^*,s^*)=0$$. (See Figure 48.)
Consequently we have six constants which may be found from the equations; three equations for $$\phi$$: \begin{align} &\phi^0=A\cdot s^0+B\cdot r^0+C \nonumber \\ &\phi^1=A\cdot s^1+B\cdot r^1+C \tag{3.131} \\ &\phi^{(2)}=A\cdot s^{(2)}+B\cdot r^{(2)}+C \nonumber \end{align}
and three equations for $$\psi$$: \begin{align} &\psi^0=D\cdot s^0+E\cdot r^0+F \nonumber \\ &\psi^1=D\cdot s^1+E\cdot r^1+F \tag{3.132} \\ &\psi^{(2)}=D\cdot s^{(2)}+E\cdot r^{(2)}+F \nonumber \end{align}
where $$s^0,\ s^1,\ s^{(2)},\ r^0,\ r^1$$ og $$r^{(2)}$$ are initial guesses for the unknown initial values. For convenience we choose the following pairs of initial guesses: \begin{align} s^0=0, & \qquad r^0=0 \nonumber \\ s^1=0, & \qquad r^1=1 \tag{3.133} \\ s^{(2)}=1, & \qquad r^{(2)}=0 \nonumber \end{align}
which results in the following expressions for the constants: $$$$\left.\begin{matrix} A=\phi^{(2)}-\phi^0, & B=\phi^1-\phi^0, & C=\phi^0, \\ D=\psi^{(2)}-\psi^0, & E=\psi^1-\psi^0, & F=\psi^0 \end{matrix}\right.\ \tag{3.134}$$$$
The correct values $$r^*$$ og $$s^*$$ which satisfies $$\phi(r^*,\ s^*)=0$$ and $$\psi(r^*,\ s^*)=0$$, i.e. $$A\cdot s^*+B\cdot r^*+C=D\cdot s^*+E\cdot r^*+C=0$$ may be expressed by the coefficients: $$$$s^*=\frac{E\cdot C-B\cdot F}{D\cdot B-A\cdot E},\qquad r^*=\frac{A\cdot F-D\cdot C}{D\cdot B-A\cdot E} \tag{3.135}$$$$
which after substitution of $$A,B,C,\dots,F$$ results in: \begin{align} s^* & =\frac{\psi^1\cdot\phi^0-\phi^1\cdot\psi^0}{(\psi^{(2)}-\psi^0)\cdot(\phi^1-\phi^0)-(\phi^{(2)}-\phi^0)\cdot(\psi^1-\psi^0)} \nonumber \\ \tag{3.136} \\ r^* & =\frac{\phi^{(2)}\cdot\psi^0-\psi^{(2)}\cdot\phi^0}{(\psi^{(2)}-\psi^0)\cdot(\phi^1-\phi^0)-(\phi^{(2)}-\phi^0)\cdot(\psi^1-\psi^0)} \nonumber \end{align}
The shooting method solution procedure for a linear ODE boundary value problem with two initial guesses then becomes:
Shooting method for two unknown initial values
1. Solve the ODE for the three pairs of initial guesses $$r$$ og $$s$$ as given in (3.133)
2. Find the correspondning values of $$\phi$$ and $$\psi$$.
3. Compute the correct initial values $$r^*$$ og $$s^*$$ from equation (3.136)).
We will illustrate the usage of this procedure in example (3.4.1 Example: Liquid in a cylindrical container).
## 3.4.1 Example: Liquid in a cylindrical container
In this example we consider a cylindrical container filled with a liquid to the height $$H$$ for two different geometries, namely constant container wall thickness and with a container with at wall thickness which varies linearly (See Figure 49). We will look at the two geometries separately below.
### 3.4.1.1 Constant wall thickness
We denote radial displacement with $$W$$. In the zoomed in part of Figure 49, $$V$$ denotes shear force per unit length while $$M$$ denotes moment per unit length.
Figure 49: A cylindrical container with a liquid of heigth $$H$$. The arrows denote positive direction.
The differential equation for the displacement $$W$$ is given by: $$$$\frac{d^4W}{dX^4}+B\cdot W=-\gamma \frac{H-X}{D} \tag{3.137}$$$$
where: $$B=\dfrac{12(1-\nu^2)}{R^2t^2},\qquad D=\dfrac{Et^3}{12(1-\nu^2)},\qquad \gamma=\rho g$$
and $$\nu$$ is Poisson's ratio, $$E$$ the E-modulus and $$\rho$$ the density of the fluid. For convenience we introduce dimensionless variables: $$$$x=\frac{X}{H} \qquad \text{and} \qquad w=\frac{E}{\gamma Ht}\, \left(\frac{t}{R}\right)^2 \, W \tag{3.138}$$$$
which after substitution in equation (3.137) results in the following dimensionless, fourth order ODE: $$$$\frac{d^4w}{dx^4}+4 \, \beta^4 \, w=-4 \, \beta^4 \, (1-x) \qquad \text{where} \qquad \beta^4=\frac{3(1-\nu^2)H^4}{R^2t^2} \tag{3.139}$$$$
Boundary conditions
To solve the ODE in equation (3.139) we need to supply some boundary conditions at both ends. At $$x= 0$$: $$\begin{equation*} W=0,\ \qquad \text{and} \qquad \frac{dW}{dX}=0\ \text{(fixed beam)} \end{equation*}$$
At the other end, for $$X=H$$ both the moment and the shear force must be zero, which mathematically corresponds to: $$\begin{equation*} M=-D \frac{d^2W}{dX^2}=0,\ V=-D \frac{d^3W}{dX^3}=0 \end{equation*}$$
The dimensionless expression for the moment is $$m(x)=-\dfrac{d^2w}{dx^2}$$, while the dimemsionless shear force has the expression: $$v(x)=-\dfrac{d^3w}{dx^3}$$.
In dimensionless variables the boundary conditions takes the form: \begin{align} w=0, \qquad \frac{dw}{dx}=0 \qquad \text{for} \quad x=0 \tag{3.140}\\ \frac{d^2w}{dx^2}=0, \qquad \frac{d^3w}{dx^3}=0 \qquad \text{for} \quad x=1 \tag{3.141} \end{align}
For our example we select a container with the following dimensions $$R=8.5m$$, $$H=7.95m$$, $$t=0.35m$$, $$\gamma=9810N/m^3$$ (water), $$\nu=0.2$$, and the E-modulus $$E=2\cdot 10^4\text{MPa}$$. For these values the nondimensional parameter $$\beta=6.0044$$.
Numerical solution with the shooting method for two unknown initial values
First we need to reformulate the fourth order ODE in (3.139) as a system of first order ODEs by following the procedure outlined in 2.4 Reduction of Higher order Equations.
By using the following auxilliary variables $$w=y_0,\ w'=y_0'=y_1,\ w''=y_1'=y_2,\ w'''=y_2'=y_3$$ we may write (3.137) as a system of four, first order ODEs: \begin{align} y_0'& = y_1 \nonumber \\ y_1'& = y_2 \nonumber\\ y_2'& = y_3 \tag{3.142} \\ y_3'& =- 4\beta^4(y_0 + 1 -x) \nonumber \end{align}
with the following boundary conditions: $$$$y_0(0)=0,\ y_1(0)=0,\ y_2(1)=0,\ y_3(1)=0 \tag{3.143}$$$$
We intend to solve the boundary value problem defined by equations (3.142) and (3.143) with a shooting method approach, i.e. we will use and intital value solver and guess the unknown initital values. The intention with the shooting method is to find the unknown initial values $$s=w''(0)=y_2(0)$$ and $$r=w'''(0)=y_3(0)$$ such that the boundary values $$y_2(1)=0$$ and $$y_3(1)=0$$.
We introduce the boundary value error function as $$\phi(r,s)=y_2(1;r,s)$$ and $$\psi(r,s)=y_3(1;r,s)$$ and the correct values for our intitial guesses $$r^*$$ og $$s^*$$ are found when $$\phi(r^*,s^*)=0$$.
To find the correct initial guesses $$s=w''(0)=y_2(0)$$ and $$r=w'''(0)$$ we do the following:
• Solve (3.142) three times, always with $$y_0(0)$$ and $$y_1(0)=0$$. The values of $$r$$ and $$s$$ for each solution is taken from (3.133).
• Compute the correct values for $$r^*$$ and $$s^*$$ from (3.136).
Below you find a python implementation of the procedure. Rune the code and check for yourself if the boundary conditions are satisfied.
# src-ch2/tank1.py;ODEschemes.py @ git@lrhgit/tkt4140/src/src-ch2/ODEschemes.py;
from ODEschemes import euler, heun, rk4
from numpy import cos, sin
import numpy as np
from matplotlib.pyplot import *
# change some default values to make plots more readable
LNWDT=3; FNT=11
rcParams['lines.linewidth'] = LNWDT; rcParams['font.size'] = FNT
font = {'size' : 16}; rc('font', **font)
def tank1(y, x):
"""Differential equation for the displacement w in a cylindrical tank with constant wall-thickness
Args:
y(array): an array containg w and its derivatives up to third order.
x(array): independent variable
Returns:
dydx(array): RHS of the system of first order differential equations
"""
dydx = np.zeros_like(y)
dydx[0] = y[1]
dydx[1] = y[2]
dydx[2] = y[3]
dydx[3] = -4*beta4*(y[0]+1-x)
return dydx
# === main program ===
R = 8.5 # Radius [m]
H = 7.95 # height [m]
t = 0.35 # thickness [m]
ny = 0.2 # poissons number
beta = H*(3*(1 - ny**2)/(R*t)**2)**0.25 # angle
beta4 = beta**4
N = 100
X = 1.0
x = np.linspace(0,X,N + 1)
solverList = [euler, heun, rk4]
solver = solverList[2]
#shoot:
s = np.array([0, 0, 1])
r = np.array([0, 1, 0])
phi = np.zeros(3)
psi = np.zeros(3)
# evaluate the boundary value error functions for the initial guesses in s and r
for k in range(3):
y0 = np.array([0, 0, s[k], r[k]])
y = solver(tank1, y0, x)
phi[k] = y[-1, 2]
psi[k]=y[-1, 3]
# calculate correct r and s
denominator = (psi[2] - psi[0])*(phi[1] - phi[0]) - (phi[2] - phi[0])*(psi[1] - psi[0])
rstar = (phi[2]*psi[0] - psi[2]*phi[0])/denominator
sstar = (psi[1]*phi[0] - phi[1]*psi[0])/denominator
print('rstar', rstar, 'sstar', sstar)
# compute the correct solution with the correct initial guesses
y0 = np.array([0, 0, sstar, rstar])
y = solver(tank1, y0, x)
legendList=[]
plot(x,-y[:,3]/beta**2)
plot(x,-y[:,2]/beta)
legendList.append(r'$v(x)/\beta^2$ ')
legendList.append(r'$m(x)/\beta$ ')
legend(legendList,loc='best',frameon=False)
ylabel('v, m')
xlabel('x')
grid(b=True, which='both', color='0.65',linestyle='-')
#savefig('../figs/tank1.pdf')
show()
Note that the ODE in equation (3.139) may be classified as singular as the highest order derivative is multiplied with a potentially very small number $$\varepsilon=\frac{1}{4\beta^4}$$, when rearranged. We observe that $$\varepsilon \to 0$$ as $$\beta\to\infty$$ and that the nature of equation (3.139) will go from an ODE to an algebraic equation. The analytical solution may be shown to have terms of the kind $$e^{\beta x}\sin\beta x$$ og $$e^{\beta x}\cos\beta x$$ (see equation (D.1.6) and (D.1.8), part D.1, appendix D in Numeriske Beregninger.
### 3.4.1.2 Varying wall thickness
In the following we will modify the example above of liquid in a cylindrical container by allowing for a wall thickness which varies linearly from $$t_0$$ at the bottom to $$t_1$$ at the top.
Figure 50: A cylindrical container with a varying wall thickenss containing a liquid of heigth $$H$$.
Based on $$t_0$$ and $$t_1$$, we introduce $$\alpha=\dfrac{t_0-t_1}{t_0}$$ the steepness of the wall thickness, which allows for the thickness $$t$$ to be represented as: $$$$t=\left(1-\alpha \frac{X}{H}\right)\cdot t_0,\ 0\leq X\leq H \tag{3.144}$$$$
The differential equaiton for the displacement $$W$$ for this situation is given by: $$$$\frac{d^2}{dX^2}\left(D \frac{d^2W}{dX^2}\right)+E \frac{tW}{R^2}=-\gamma(H-X) \tag{3.145}$$$$
where $$D=\dfrac{Et^3}{12(1-\nu^2)},\ \gamma=\rho g$$, with the same meaning as in the previous part of the example.
Dimensionless form $$$$x=\frac{X}{H},\quad w=\frac{E}{\gamma Ht_0}\left(\frac{t_0}{R}\right)^2 \; W,\quad \beta^4=\frac{3(1-\nu^2)}{R^2t_0^2} \; H^4 \tag{3.146}$$$$
By substitution of (3.146) in (3.145) we get the following nondimensional ODE: $$$$\frac{d^2}{dx^2}\left[ (1-\alpha x)^3\frac{d^2w}{dx^2}\right] +4\beta^4(1-\alpha x)\cdot w=-4\beta^4(1-x) \tag{3.147}$$$$
which may be differentiated to yield: $$$$\frac{d^4w(x)}{dx^4}-\frac{6\alpha}{(1-\alpha x)}\frac{d^5w(x)}{dx^5}+\frac{6\alpha^2}{(1-\alpha x)^2}\frac{d^2w(x)}{dx^2}+\frac{4\beta^4}{(1-\alpha x)^2}w(x)=-\frac{4\beta^4(1-x)}{(1-\alpha x)^5} \tag{3.148}$$$$
This problem has also an analytical, but more complicated, solution (see appendix D, part D.2 in Numeriske Beregninger) expressed by means of Kelvin functions, a particular kind of Bessel functions. However, the numerical solution remains the same as for the case with the constant wall thickness.
We choose physical parameters as for the previous example: $$$$\tag{3.149} \left.\begin{matrix} &R=8.5m,\ H=7.95m,\ t_0=0.35m ,\ t_1=0.1m \\ &\gamma=9810N/m^3\text{(vann)},\ \nu=0.2,\ E=2\cdot 10^4\text{MPa} \end{matrix}\right.\$$$$
and $$\alpha=\frac{t_0-t_1}{t_0}=\frac{5}{7}$$ and as previously we get $$\beta=6.0044$$.
Numerical solution
We proceed as before with $$w=y_0,\ w'=y_0'=y_1,\ w''=y_1'=y_2,\ w'''=y_2'=y_3,\ z=1-\alpha x$$ and write (3.144) as a system of four first order ODEs: \begin{align} y_0'& = y_1 \nonumber \\ y_1'& = y_2 \nonumber\\ y_2'& = y_3 \tag{3.150} \\ y_3'&=\frac{6\alpha}{z} \, y_3-\frac{6\alpha^2}{z^2} \, y_2-\frac{4\beta^4}{z^2} \, y_0-\frac{4\beta^4(1-x)}{z^3} \nonumber \end{align}
with the following boundary conditions: $$$$y_0(0)=0,\ y_1(0)=0,\ y_2(1)=0,\ y_3(1)=0 \tag{3.151}$$$$
Only the last equation in the system of ODEs (3.150) differs from the previous part of the example 3.4.1.1 Constant wall thickness).
The procedure will be as for 3.4.1.1 Constant wall thickness, except for the we now are introduced the additional parameter $$\alpha$$.
# src-ch2/tank2.py;ODEschemes.py @ git@lrhgit/tkt4140/src/src-ch2/ODEschemes.py;
from ODEschemes import euler, heun, rk4
from numpy import cos, sin
import numpy as np
from matplotlib.pyplot import *
# change some default values to make plots more readable
LNWDT=3; FNT=11
rcParams['lines.linewidth'] = LNWDT; rcParams['font.size'] = FNT
font = {'size' : 16}; rc('font', **font)
def tank2(y, x):
"""Differential equation for the displacement w in a cylindrical tank with linearly varying wall-thickness
Args:
y(array): an array containg w and its derivatives up to third order.
x(array): independent variable
Returns:
dydx(array): RHS of the system of first order differential equations
"""
z = 1-alpha*x
dydx = np.zeros_like(y)
dydx[0] = y[1]
dydx[1] = y[2]
dydx[2] = y[3]
temp = (6*alpha/z)*y[3]- (6*alpha**2/z**2)*y[2]
dydx[3] = temp - 4*beta4*y[0]/z**2 - 4*beta4*(1-x)/z**3
return dydx
R = 8.5 # radius [m]
H = 7.95 # height [m]
t0 = 0.35 # thickness [m]
t1 = 0.1 # thickness [m]
ny = 0.2 # poissons number
beta = H*(3*(1-ny**2)/(R*t0)**2)**0.25
beta4 = beta**4
alpha = (t0-t1)/t0
N = 100
print("beta: ", beta, "alpha", alpha)
X = 1.0
x = np.linspace(0,X,N + 1)
solverList = [euler, heun, rk4] #list of solvers
solver = solverList[2] # select specific solver
# shoot:
s = np.array([0, 0, 1])
r = np.array([0, 1, 0])
phi = np.zeros(3)
psi = np.zeros(3)
for k in range(3):
y0 = np.array([0,0,s[k],r[k]])
y = solver(tank2,y0,x)
phi[k] = y[-1, 2]
psi[k]=y[-1, 3]
# calculate correct r and s
denominator = (psi[2] - psi[0])*(phi[1] - phi[0]) - (phi[2] - phi[0])*(psi[1] - psi[0])
rstar = (phi[2]*psi[0] - psi[2]*phi[0])/denominator
sstar = (psi[1]*phi[0] - phi[1]*psi[0])/denominator
print('rstar', rstar, 'sstar', sstar)
# compute the correct solution with the correct initial guesses
y0 = np.array([0, 0, sstar, rstar])
y = solver(tank2,y0,x)
legends=[] # empty list to append legends as plots are generated
plot(x,-y[:,3]/beta**2)
plot(x,-y[:,2]/beta)
legends.append(r'$v(x)/\beta^2$')
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https://jcom.sissa.it/archive/20/06/JCOM_2006_2021_A08 | # Gamification for social perception: introducing scientific literacy to dabblers in citizen science
### Abstract:
Understanding scientific concepts is a crucial factor in motivating dabblers at the start of co-created citizen science projects. This article describes PACMAC, a card-based cooperative card game aimed at introducing dabblers to hypothesis and falsifiability concepts through the visualization of a social perception map. The game was evaluated in five neighborhoods from El Salvador. The results showed that PACMAP is approachable for participants of different demographics to develop an understanding of the concepts of hypotheses and falsifiability.
Keywords:
31 October 2020
16 March 2021
11 October 2021
### 1 Introduction
One of the essential aspects of citizen science projects is the democratization of actions involved and even decision making [Bela et al., 2016]. There is great diversity in the approaches on which non-professional individuals can participate in citizen science projects and activities [Wiggins and Crowston, 2011]. We can classify the degree to which non-professionals become involved in citizen science projects as mere sample collectors (contributory projects), as having the freedom to modify the work done by professional scientists (collaborative), and as co-creators alongside professionals [Bonney et al., 2009; Wilderman et al., 2007]. These modes of participation are not rigid: over time, non-professional develop abilities that allow them to contribute to the project in new ways and participate in making decisions over the project’s lifetime.
Co-created citizen science projects are those on which a substantial amount of the project’s decisions are taken by the community [Bonney et al., 2009]. These are also known as community-based projects, or ‘science by the people’ [Mansell, 2013]. In these types of projects, the community defines the problem, designs the study, collects the samples, analyzes the samples, and interprets it [Wilderman et al., 2007]. Both professional as non-professional scientists participate in decision-making activities and perform activities pertaining to the development of the project on the field, such as data collection and analysis.
The main advantages of co-created projects are a higher degree of appropriation done by the community towards project planning and the execution of its activities, as well as a more significant commitment and a certain degree of agency and independence from the guidance of professional scientists. Citizen science democratizes knowledge through a more horizontal mode of governance: the challenges and solutions proposed by the projects are brought closer to the problems suffered by the community, and they, in turn legitimize their participation through the application of acquired knowledge [Powell, 2016].
With this in mind, one of the most important challenges of community-based citizen science projects is to ensure that members of a community develop skills around scientific thinking [Brossard, Lewenstein and Bonney, 2005; Cronje et al., 2011]. Supporting the development of these skills is an essential factor in ensuring citizen science projects’ success, especially when interacting with dabblers — newcomers to scientific activities or sporadic participants of citizen science projects [Evans et al., 2005]. Dabblers are likely to drop out when they lack intrinsic motivators such as interest and curiosity, a perceived capacity to complete tasks, and an evaluation of the benefits and costs of contributing [Eveleigh et al., 2014].
For this reason, dabblers will benefit from understanding the value of their involvement through an exploration of the scientific questions in a specific project, e.g., how collecting data about a particular variable will allow them to answer a research question. Sharing back the information collected through a citizen science project with the community might not be enough to communicate citizen science projects’ goals or the thinking behind the proposed data collection activities. For this reason, citizen science proposes that all participants must understand and take part in defining the possible hypotheses to be addressed [Bonney et al., 2009], serving as a motivator collecting data to falsify them.
We argue that dabblers who engage in citizen science are ultimately volunteers and are thus motivated to participate for a more extended period if their participation proves to create value for themselves and their community [Vecina Jiménez, Chacón Fuertes and Abad, 2009]. In this sense, participants will be motivated by becoming science-literate and perceiving themselves as supporters of social justice values within their communities [Curtis, 2018; Geoghegan et al., 2016]. Participatory citizen science faces the challenge of ensuring that dabblers, or volunteers with little to no formation in scientific thinking, can internalize the understanding of science [Bonney et al., 2009; Cronje et al., 2011; Lee and Roth, 2003]. By observing and experiencing the usefulness of developing a falsifiable hypothesis and by verifying it through data collection and analysis, they will have a clearer view of the reasoning and motivations for scientific activities, as well as the social value of their involvement. This process will ensure a higher degree of commitment towards the project [Shirk et al., 2012].
This approach’s main goal is to introduce dabblers to the concepts of hypothesis and falsifiability PACMAP (Participatory Approach for Community Mapping), a card game designed to collect and visualize the perception of participants in communities. This game aims to help dabblers develop a falsifiable hypothesis as a starting point for a citizen science project by collecting and visualizing the perceptions of their community. This article will describe how PACMAP was used in five communities in San Salvador to establish and validate community-led hypotheses based on the community perception of social violence.
The article is divided into the following sections: (1) how Gamification can be utilized as a tool for developing scientific literacy through PACMAP, a card game for community mapping, (2) how this tool was used to introduce participants to the concepts of hypothesis and falsifiability through the exploration of social risk factors present in five communities located in San Salvador, (3) the results of this experience and insights from experience, and finally (4) recommendations on how PACMAP or card-based mapping games can be further developed to improve the engagement of dabblers in citizen science projects.
### 2 Using Games for citizen science
#### 2.1 Incorporation of games in citizen science
Gamification is a design tool to improve public engagement by introducing game elements in non-game practices or spaces [Deterding et al., 2011]. Thus, it generates a fun environment that promotes several aspects, such as commitment, continuity, or recidivism [Seaborn and Fels, 2015]. One of the most important objectives of Gamification is to motivate players to complete a set of tasks by offering rewards such as points, badges, levels, progress bars or even, virtual money. Sometimes the reward is simply to have a funny moment. Another fundamental method to increase motivation is to add some kind of competition between players, either by creating position tables or comparisons with friends in social networks [Kim, 2015; Chou, 2015; Werbach and Hunter, 2012]. Gamification has been applied in several contexts such as education [Domínguez et al., 2013; Dreimane, 2019], production [Korn, Funk and Schmidt, 2015; Naik and Jenkins, 2019; Pedreira et al., 2015], sales [Frith, 2013] or science [Khatib et al., 2011; Tinati et al., 2017].
There are different approaches associated with incorporating Gamification in citizen science projects. Many of them are related to a specific aspect of the project, for example, rewarding when data collection is performed [Celasco et al., 2016; Kanner et al., 2018]. One of the first experiences in the use of Gamification in citizen science projects was,” I want to be a captain! I want to be a captain!” [Eveleigh et al., 2013] which proposes a narrative in a transcription activity. Also, Eye Wire [Tinati et al., 2017] and FoldIt [Khatib et al., 2011] are projects through Gamification the player color images or complete a puzzle in 3D models. All cases include gamification strategies to engage players.
This approach’s main goal is to introduce dabblers to the concepts of hypothesis and falsifiability through a mapping card game. The game is designed to achieve the following specific objectives:
• To improve participation in data collection and mapping.
• To define a hypothesis based on community perception.
• To design monitoring initiatives related to social violence.
• To validate the hypothesis.
Previous work in the literature studied the use of Gamification to develop scientific skills. From a narrow point of view, Gamification is considered as the use of video games. Morris et al. [2013] introduced an approach to using video games by describing three main elements where video games support the development of scientific thinking. These are: a context to promote inquiry-based learning, scientific skills are embedded in the game rules, or other than promotes skills and values in a non-explicit manner. Scientific video games help develop a need for evidence in the discourse, a model of reasoning, or an understanding of evidence. Although our approach is not a video game, it works in need to understand the evidence to be used in scientific discourse.
More specific scientific card games appear in the literature. However, at the moment of writing this article, the authors were not able to find particular card games to enhance hypothesis definition and validation. Less related, Voyage Card games and Top Careers in Science [Smith and Munro, 2009] are card games with a scientific background but without being specialized in developing a hypothesis.
### 3 PACMAP card game
PACMAP (Participatory Approach for Community Mapping) is a card game intended to collect and visualize participants’ perception in communities through the creation of a digital map that displays social or environmental hazards. This game aims to involve participants with little scientific literacy or experience in citizen science on the definition of a hypothesis based on their environmental perceptions. The game prompts questions to create a map which are discussed at the end of the activity. This collective assessment will guide them to understand these initial assumptions and understand how subsequent data collection and data analysis will help them validate them.
Digital participatory mapping is deemed a powerful tool for representing the knowledge and needs inside of communities [Gordon, Elwood and Mitchell, 2016; Lundine, Kovačič and Poggiali, 2012] since data visualizations are effective mediums to aid the development of competencies for interpreting data [Golumbic, Fishbain and Baram-Tsabari, 2020]. The use of digital tools for mapping will require training for the development of these digital skills and access to hardware. PACMAP is intended to use digital tools in a non-intrusive manner through a card game that is easy to understand and can be played without previous knowledge other than the current situation inside and in the vicinity of the community.
By establishing a fast-paced ludic environment, the activity posed questions about the community’s qualitative aspects related to four social variables: personal activities and routine, leisure time inside the community, road safety around the community, and vulnerability to social violence. A digital map is generated at the end of the workshop, with a visual representation of the participants’ responses. As the results are discussed, participants are encouraged to develop an initial hypothesis regarding the social hazards being studied in the session.
#### 3.1 Game mechanics
The game utilizes the following resources:
Cards: PACMAP is played with two decks of cards, each of which are identified by a unique number and either a scenario or a location:
• The scenario deck contains an incomplete sentence related to a potential social distress situation; the phrase is meant to be completed with a location. For example: When I walk through ___, I feel confident that I will not be robbed. These scenarios are designed to reflect the social categories under assessment. It is important to note that scenarios were worded as true by default so that participants would vote to refute the statement
• The location deck contains predetermined locations inside and outside of the neighborhood. These represent a recognizable address or landmark, such as an intersection, a business, or a public space. Figure 1 shows examples of scenario and location cards.
Voting cards: although these are not necessary to play the game, each participant was given a red card or sign, which they can use to vote “no” for each round according to the scenario mentioned by the moderator.
Computer and projector: after the participants finish the game, a script written in R is used to process the results registered by the facilitators. The data is used to generate a map projected on the wall at the end of the game.
Game mechanics: before playing the game, a map of the community with the pre-selected points (listed as depicted in the cards) is projected to the participants. All the points are reviewed to ensure that all participants understand where these points are located.
At the start of the game, one of the participants is chosen to draw out the cards and announce them to the group. This moderator can be switched if necessary at any point during the game. The moderator will draw two cards from the two stacks to form different sentences. For example: “While walking outside of Parque Cuscatlán (Cuscatlán Park), I do not feel afraid of being robbed.” Prompted by the moderator reading the cards, participants will vote whether they agree or not with the moderator by lifting the flag or card in a show of disagreement of the statement.
As participants respond, a facilitator updates a data sheet that lists the following: the identifier for the scenario and location cards in use, and the number of negative votes (e.g., scenario #3; location #29; votes: 12). If stacks run out of cards, the moderator can shuffle the decks and start a new round of questions, prompting different combinations.
Depending on the number of participants, more rounds can be played to record a fair amount of responses for each data point and location. However, it is recommended to stop at each round with a change of activity in order to avoid monotony.
#### 3.2 Dynamics
At the end of the activity, the response sheet is fed to a script, which will, in turn, generate a map showing the score for each of the data points. The map is projected and discussed with the participants to validate their perceptions and find possible explanations for the higher perceived risk in some areas, based on the group’s personal experiences.
As a result of this exercise, one or more hypotheses will be defined. These can be used in the future to determine data collection, places of intervention, and the tools for monitoring.
### 4 Using PACMAP in urban San Salvador
#### 4.1 Project background
The study was conducted as part of a project called Labs de Resiliencia (Resilience Labs), which consisted of a project focused on developing community resilience. This project was part of a broader international urban intervention oriented toward the urban development of one of the most centric areas in San Salvador.
This component of the project focused on developing skills for the social and environmental resilience of five urban communities. It was composed of activities for people of different age groups within the community. Activities focused on participatory mapping and diagnostics, teaching skills for electronics and programming, the development of electronic prototypes, and its use to monitor social variables in the community. The project aimed to improve the conditions of areas surrounding these communities to enable access, improve safety, and encourage community interaction from different socioeconomic groups concerning the Parque Cuscatlán. This historic landmark was being renovated.
The workshops served to collect spatial information related to hazards present in the area. Workshops’ activities implied the evaluation of possible commute routes from the park to the communities and the schedules of different age groups in the community: at what time they commute to work, how they commute, where do youth meet, what do they do for leisure, etc.
#### 4.2 Communities
The Labs de Resiliencia project, under which this mapping component was integrated, was focused in five communities located around the Cuscatlán Park: Santa Fe y La Paz,1 Tutunichapa 1, Atonal y Asunción, working-class urban settlements located close to the city downtown. Some of these communities originated as informal settlements around the 1960s when low-income populations moved to government-owned areas dangerously close to river banks.
As the city developed over the years, these settlements have already become part of the landscape. These families are entitled to the land and have obtained deeds to the lands where they reside. Despite this, they continue to be vulnerable due to the location and low-quality construction of their homes. They also suffer from other social problems due to their proximity to busy roads, which were not developed with road safety elements designed with these communities in mind, the lack of public spaces, and the area’s high crime rates.
#### 4.3 Workshop design
A series of workshops for social mapping using PACMAP took place in each of the five communities during January of 2018. The game was used to identify safe and unsafe areas along the route between the communities and the Cuscatlán Park. The activity also sought to involve participants of all ages and groups, which usually would not interact with each other.
#### 4.4 Data points and cards
As a preparation for the game, a list of location points (25–35) was selected for each community. The locations were easily identified by participants within an area between the park and the communities.
The lists of locations (and thus the location cards) for each community were different, but some of the points were shared between the lists if the areas between the park and the communities overlapped (Figure 2). This ensured that each map generated would depict different areas surrounding the park and could be combined to have a broader view of the area’s perceptions.
Likewise, a set of scenarios was chosen based on the four areas of social discomfort: personal leisure time, community interaction, social violence, and road safety. These were previously identified through discussions with community leaders and other project stakeholders.
#### 4.5 Participants
The five participant communities contain very similar demographics in terms of family composition and socioeconomic status. Despite these similarities, however, current social dynamics and community safety are made up of relatively different groups. The of the participant groups were comprised of: Santa Fe y La Paz 2 (n=17), led by mostly male youth; Atonal (n=16) composed mainly by older women and single mothers; Tutunichapa 1 (n=20) of primarily young males, and Asunción (n=17) with the participation of families and mid-age adults. The average age for all participants of the four workshops was 29.6 years old. Table 1 shows the demographic characteristics of the participants. The first column includes the communities’ names. The details of the second and third columns detail the number of female and male participants. The fourth column details the average of ages and the last one, the total amount of participants.
Table 1: Demographic characteristics of participants.
#### 4.6 Procedure
Each PACMAP workshop consisted of at least two rounds of gameplay in order to collect enough data to render the maps, but making sure not to drag their participation for too long. The game began with a short presentation of the game mechanics and the expected result. The following recommendations were given to participants: to answer as fast as possible to focus on their perception and avoid being influenced by other participants; to discuss the answers as little as possible during the game to reduce interruptions of the game for the rest.
One of the facilitators took note of the selected cards and the number of responses per set of cards on a paper sheet. During the break at the end of each round, these observations were fed into a digital spreadsheet. At the end of the game, a complete registry of the observations was processed by the script, which calculated scores for every map point. Scores were used to elaborate a hot spot map and projected back to the participants at the end of the activity. In some sense, the hypothesis was described in terms of a map: The hot spots in the map are unsafe zones.
#### 4.7 Creation of the hypothesis
The maps generated at the end of the game were used for participants to visualize and analyze the collected data. Figure 3 details an example of a generated map with PACMAP. The following questions were discussed with the participants:
• What are the riskiest areas according to the map?
• Are there other dangerous areas not considered on the map?
• What do you think are the reasons for these scores?
• Do you go through these areas daily? Why or why not?
• Is it possible to verify the assumptions shown on this map? How?
• Who are the most vulnerable groups? Is it women, men, people with disabilities, the elderly, or children?
### 5 Results and discussion
#### 5.1 Mechanics
The goal of PACMAP is that participants collaborate in developing a shared understanding of a falsifiable hypothesis based on their perceptions. As a result, participants will develop a motivation based on their knowledge of citizen science objectives. Nicholson [2015] describes this gamification approach as “meaningful gamification,” in which games are devoid of external rewards and focus instead on intrinsic motivations as a source of meaningfulness. His framework of meaningful Gamification is based on the following principles: Play (a fun experience under an agreed set of rules), exposition (a connection to a real-world setting), choice (a sense of agency), information (helping participants learn about the world), engagement (enabling participation and collaboration), reflection (using the game to explore past experiences).
An introduction section was essential to establish a common understanding of the rules under which participants can collaborate. Before starting the game, the facilitators defined a short quick discussion of the variables to be measured and a brief review of the map locations as a sort of a virtual tour of the community. This served as an icebreaker for the activity and as a preparation for a collaborative group dynamic, as participants collaborated with their neighbors by describing the locations to those who did not identify them at first.
#### 5.2 Dynamics
The game was utilized in four opportunities with community members from different ages and levels of education, with good results. All participants seemed to understand the game, and it was easy to explain. Participants of all ages were able to understand the dynamics of the game quickly. As each sentence from the cards is read aloud, this allowed children, the elderly, and other illiterate individuals to play the game without any problems.
Scenarios were generated randomly, which allowed choosing unexpected responses. In some cases, the irony of the statement served to amuse participants (for example, when the busiest road was stated as a perfect place to go for a stroll). The game helped to reflect on their personal and collective experiences.
The quick pacing of the game allowed for very few discussions or disagreements during the activity. Participants emitted their opinion without restrictions and developed a sense of equality, addressing possible hidden social dynamics inside the community that could reduce the freedom of some participants to speak their minds [Weyer, Bezerra and De Vos, 2019]. As the game progressed, participants felt more comfortable voicing their opinions, giving way to short but insightful opinions. In the end, the common understanding of the problem in their communities helped to establish a sense of involvement among all the participants.
#### 5.3 Visual communication and analysis
The data presentation format was devised with an emphasis on communicating results to the extent of accuracy. As the number of points in the map increased, certain liberties were taken to tweak the scores’ scale to translate relative weights so that the points were quickly identified and that differences were visible.
The discussion was facilitated with falsifiability in mind: participants were allowed to challenge the results under the possibility that their initial perceptions were wrong. The discussion of the maps served to validate them: participants engaged in discussing the possible causes for the scores in some areas.
#### 5.4 Outcomes
The four workshops managed to draw a total of 231 scenario-location combinations for 3803 observations. The average score of 66.33% ($\sigma$ = 38.88%), with a higher score signifying a more negative score. Individual scores were as follows: Asuncion (51.10%), Atonal (67.23%), Santa Fe and La Paz (76.23%), Tutunichapa (71.48%). Communities with the higher scores were the ones with younger participants, a perception that coincides with the higher risk suffered by youth in El Salvador, with a mortality rate of 207.5 per 100,000 inhabitants in 2015, a rate much higher than the world average of 149 per 100,000 inhabitants [Martínez-Reyes and Navarro Pérez, 2020].
Results were given immediately at the end of the game to participants for discussion to validate their understanding of the obtained hypothesis. All participants showed their agreement with the results of the game. Discussions were lengthy and insightful. They connected the map results with their personal experiences, descriptions about what they see during their commute, or issues currently happening in their communities.
Participants of all the workshops validated the results, which served for future activities; for example, when planning monitoring activities related to traffic, participants pointed towards measuring the number of cars per hour at the locations found in the map.
### 6 Conclusions
One of the premises of co-created science projects is the ability to share data and the results with all members of a project, but ensuring that participants without a scientific background can understand or analyze a set of data with little context is a challenge. However, if a person from a community participates in defining the variables to be measured, collect them. When the results are given back to analyze and understand them, they can interiorize or utilize the information. For this process to occur, there must be a clear understanding of the relationship between the research questions, which represent the interests of both professional scientists and citizens, and data collection.
PACMAP is a game that takes advantage of the benefits given by online games, but the use of a card deck gives it many advantages. For example, using a physical game allows it to be more approachable for participants without digital literacy, such as low-income populations, children, or older people. It also allows all participants to focus on the subject at hand without derailing the conversation by using technology.
This experience opens the opportunity for Gamification, especially on the intersection of physical and digital games, to support dabblers in their understanding of scientific concepts and interact with other participants by using standard rules and equal empowerment. The game helped individuals reflect on their daily lives and their community’s situation by establishing a game dynamic that connected an engaging activity with their personal lives. The use of technology was minimal and served to accelerate the process. The results were presented to open the opportunity to generate discussion over results obtained in real-time and with input from every participant in the room.
The game was effective in communicating the idea of a hypothesis to participants despite their lack of knowledge of the scientific method. “Science-education strategies should focus on the more general problem of increasing the science literacy of the general public rather than the recruitment of future scientists.” [Zen, 1990]. PACMAP allowed participants to learn about science by not focusing on producing scientifically verifiable data but engaging them with citizen science in an understandable and approachable manner.
### References
Bela, G., Peltola, T., Young, J. C., Balázs, B., Arpin, I., Pataki, G., Hauck, J., Kelemen, E., Kopperoinen, L., Van Herzele, A., Keune, H., Hecker, S., Suškevičs, M., Roy, H. E., Itkonen, P., Külvik, M., László, M., Basnou, C., Pino, J. and Bonn, A. (2016). ‘Learning and the transformative potential of citizen science’. Conservation Biology 30 (5), pp. 990–999. https://doi.org/10.1111/cobi.12762.
Bonney, R., Ballard, H., Jordan, R., McCallie, E., Phillips, T., Shirk, J. and Wilderman, C. C. (2009). Public Participation in Scientific Research: Defining the Field and Assessing Its Potential for Informal Science Education. A CAISE Inquiry Group Report. Washington, D.C., U.S.A.: Center for Advancement of Informal Science Education (CAISE). URL: http://www.informalscience.org/public-participation-scientific-research-defining-field-and-assessing-its-potential-informal-science.
Bonney, R., Cooper, C. B., Dickinson, J., Kelling, S., Phillips, T., Rosenberg, K. V. and Shirk, J. (2009). ‘Citizen Science: a Developing Tool for Expanding Science Knowledge and Scientific Literacy’. BioScience 59 (11), pp. 977–984. https://doi.org/10.1525/bio.2009.59.11.9.
Brossard, D., Lewenstein, B. and Bonney, R. (2005). ‘Scientific knowledge and attitude change: The impact of a citizen science project’. International Journal of Science Education 27 (9), pp. 1099–1121. https://doi.org/10.1080/09500690500069483.
Celasco, M., Yanez, J. I., Gamen, R., Fernandez, A., Diaz, A. and Torres, D. (2016). ‘Galaxy conqueror: astronomy, citizen science and gamification’. In: Proceedings — 2016 XI Latin American Conference on Learning Objects and Technology (LACLO). IEEE. https://doi.org/10.1109/laclo.2016.7751798.
Chou, Y.-K. (2015). Actionable gamification: beyond points, badges and leaderboards. Octalysis Media.
Cronje, R., Rohlinger, S., Crall, A. and Newman, G. (2011). ‘Does Participation in Citizen Science Improve Scientific Literacy? A Study to Compare Assessment Methods’. Applied Environmental Education & Communication 10 (3), pp. 135–145. https://doi.org/10.1080/1533015X.2011.603611.
Curtis, V. (2018). ‘Motivation for participation: from general volunteerism to online citizen science’. In: Online citizen science and the widening of academia. Cham, Switzerland: Springer International Publishing, pp. 69–92. https://doi.org/10.1007/978-3-319-77664-4_4.
Deterding, S., Khaled, R., Nacke, L. and Dixon, D. (2011). ‘Gamification: toward a definition’. In: CHI 2011 Gamification Workshop Proceedings, pp. 12–15. URL: http://gamification-research.org/wp-content/uploads/2011/04/02-Deterding-Khaled-Nacke-Dixon.pdf.
Domínguez, A., Saenz-de-Navarrete, J., de-Marcos, L., Fernández-Sanz, L., Pagés, C. and Martínez-Herráiz, J.-J. (2013). ‘Gamifying learning experiences: practical implications and outcomes’. Computers & Education 63, pp. 380–392. https://doi.org/10.1016/j.compedu.2012.12.020.
Dreimane, S. (2019). ‘Gamification for education: review of current publications’. In: Didactics of smart pedagogy. Cham, Switzerland: Springer International Publishing, pp. 453–464. https://doi.org/10.1007/978-3-030-01551-0_23.
Evans, C., Abrams, E., Reitsma, R., Roux, K., Salmonsen, L. and Marra, P. P. (2005). ‘The Neighborhood Nestwatch Program: Participant Outcomes of a Citizen-Science Ecological Research Project’. Conservation Biology 19 (3), pp. 589–594. https://doi.org/10.1111/j.1523-1739.2005.00s01.x.
Eveleigh, A., Jennett, C., Blandford, A., Brohan, P. and Cox, A. L. (2014). ‘Designing for dabblers and deterring drop-outs in citizen science’. In: Proceedings of the IGCHI Conference on Human Factors in Computing Systems (CHI ’14). New York, NY, U.S.A.: ACM Press, pp. 2985–2994. https://doi.org/10.1145/2556288.2557262.
Eveleigh, A., Jennett, C., Lynn, S. and Cox, A. L. (2013). ‘"I want to be a captain! I want to be a captain!": gamification in the old weather citizen science project’. In: Proceedings of the First International Conference on Gameful Design (Gamification ’13). New York, NY, U.S.A.: ACM Press, pp. 79–82. https://doi.org/10.1145/2583008.2583019.
Frith, J. (2013). ‘Turning life into a game: foursquare, gamification and personal mobility’. Mobile Media & Communication 1 (2), pp. 248–262. https://doi.org/10.1177/2050157912474811.
Geoghegan, H., Dyke, A., Pateman, R., West, S. and Everett, G. (2016). Understanding motivations for citizen science. Final report on behalf of UKEOF, University of Reading, Stockholm Environment Institute (University of York) and University of the West of England. Swindon, U.K.: UKEOF. URL: http://www.ukeof.org.uk/resources/citizen-science-resources/MotivationsforCSREPORTFINALMay2016.pdf.
Golumbic, Y. N., Fishbain, B. and Baram-Tsabari, A. (2020). ‘Science literacy in action: understanding scientific data presented in a citizen science platform by non-expert adults’. International Journal of Science Education, Part B, pp. 1–16. https://doi.org/10.1080/21548455.2020.1769877.
Gordon, E., Elwood, S. and Mitchell, K. (2016). ‘Critical spatial learning: participatory mapping, spatial histories and youth civic engagement’. Children’s Geographies 14 (5), pp. 558–572. https://doi.org/10.1080/14733285.2015.1136736.
Kanner, A. I., Torres, D., Lombardelli, J. and Peri, C. (2018). ‘Analyzing the use of gamification in a tutorial for a citizen science project’. In: 2018 XIII Latin American Conference on Learning Technologies (LACLO). IEEE, pp. 208–215. https://doi.org/10.1109/laclo.2018.00048.
Khatib, F., Cooper, S., Tyka, M. D., Xu, K., Makedon, I., Popoviæ, Z., Baker, D. and Foldit Players (2011). ‘Algorithm discovery by protein folding game players’. Proceedings of the National Academy of Sciences 108 (47), pp. 18949–18953. https://doi.org/10.1073/pnas.1115898108.
Kim, B. (2015). ‘Understanding gamification’. Library Technology Reports 51 (2), p. 1. https://doi.org/10.5860/ltr.51n2.
Korn, O., Funk, M. and Schmidt, A. (2015). ‘Design approaches for the gamification of production environments’. In: Proceedings of the 8th ACM International Conference on Pervasive Technologies Related to Assistive Environments. ACM. https://doi.org/10.1145/2769493.2769549.
Lee, S. and Roth, W.-M. (2003). ‘Science and the “good citizen”: community-based scientific literacy’. Science, Technology, & Human Values 28 (3), pp. 403–424. https://doi.org/10.1177/0162243903028003003.
Lundine, J., Kovačič, P. and Poggiali, L. (2012). ‘Youth and digital mapping in urban informal settlements: lessons learned from participatory mapping processes in mathare in Nairobi, Kenya’. Children, Youth and Environments 22 (2), pp. 214–233. https://doi.org/10.7721/chilyoutenvi.22.2.0214.
Mansell, R. (2013). ‘Employing digital crowdsourced information resources: managing the emerging information commons’. International Journal of the Commons 7 (2), p. 255. https://doi.org/10.18352/ijc.395.
Martínez-Reyes, K. A. and Navarro Pérez, J. J. (2020). ‘Obstáculos para el logro de los ODS en El Salvador. Políticas de juventud, jóvenes pandilleros y las ONG: un análisis complejo’. Revista iberoamericana de estudios de desarrollo = Iberoamerican journal of development studies 9 (1), pp. 28–51. https://doi.org/10.26754/ojs_ried/ijds.444.
Morris, B. J., Croker, S., Zimmerman, C., Gill, D. and Romig, C. (2013). ‘Gaming science: the “gamification” of scientific thinking’. Frontiers in Psychology 4, p. 607. https://doi.org/10.3389/fpsyg.2013.00607.
Naik, N. and Jenkins, P. (2019). ‘Relax, it’s a game: utilising gamification in learning agile scrum software development’. In: 2019 IEEE Conference on Games (CoG). IEEE, pp. 1–4. https://doi.org/10.1109/cig.2019.8848104.
Nicholson, S. (2015). ‘A RECIPE for meaningful gamification’. In: Gamification in education and business. Cham, Switzerland: Springer International Publishing, pp. 1–20. https://doi.org/10.1007/978-3-319-10208-5_1.
Pedreira, O., García, F., Brisaboa, N. and Piattini, M. (2015). ‘Gamification in software engineering — a systematic mapping’. Information and Software Technology 57, pp. 157–168. https://doi.org/10.1016/j.infsof.2014.08.007.
Powell, A. (2016). ‘Hacking in the public interest: Authority, legitimacy, means, and ends’. New Media & Society 18 (4), pp. 600–616. https://doi.org/10.1177/1461444816629470.
Seaborn, K. and Fels, D. I. (2015). ‘Gamification in theory and action: a survey’. International Journal of Human-Computer Studies 74, pp. 14–31. https://doi.org/10.1016/j.ijhcs.2014.09.006.
Shirk, J. L., Ballard, H. L., Wilderman, C. C., Phillips, T., Wiggins, A., Jordan, R., McCallie, E., Minarchek, M., Lewenstein, B. V., Krasny, M. E. and Bonney, R. (2012). ‘Public Participation in Scientific Research: a Framework for Deliberate Design’. Ecology and Society 17 (2), p. 29. https://doi.org/10.5751/ES-04705-170229.
Smith, D. R. and Munro, E. (2009). ‘Educational card games’. Physics Education 44 (5), pp. 479–483. https://doi.org/10.1088/0031-9120/44/5/004.
Tinati, R., Luczak-Roesch, M., Simperl, E. and Hall, W. (2017). ‘An investigation of player motivations in Eyewire, a gamified citizen science project’. Computers in Human Behavior 73, pp. 527–540. https://doi.org/10.1016/j.chb.2016.12.074.
Vecina Jiménez, M. L., Chacón Fuertes, F. and Abad, M. J. S. (2009). ‘Satisfacción en el voluntariado: estructura interna y relación con la permanencia en las organizaciones’. Psicothema 21 (1), pp. 112–117.
Werbach, K. and Hunter, D. (2012). For the win. How game thinking can revolutionize your business. Wharton Digital Press.
Weyer, D., Bezerra, J. C. and De Vos, A. (2019). ‘Participatory mapping in a developing country context: lessons from South Africa’. Land 8 (9), p. 134. https://doi.org/10.3390/land8090134.
Wiggins, A. and Crowston, K. (2011). ‘From Conservation to Crowdsourcing: a Typology of Citizen Science’. In: Proceedings of the 44th Hawaii International Conference on System Sciences (HICSS-44). Kauai, HI, U.S.A. Pp. 1–10. https://doi.org/10.1109/HICSS.2011.207.
Wilderman, C. C., McEver, C., Bonney, R., Dickinson, J., Kelling, S. and Rosenberg, K. (2007). ‘Models of community science: design lessons from the field’. In: Citizen science toolkit conference (Cornell Laboratory of Ornithology, Ithaca, NY, U.S.A.). Ed. by C. McEver, R. Bonney, J. Dickinson, S. Kelling, K. Rosenberg and J. L. Shirk, pp. 1–3.
Zen, E.-A. (1990). ‘Science literacy and why it is important’. Journal of Geological Education 38 (5), pp. 463–464. https://doi.org/10.5408/0022-1368-38.5.463.
### Authors
Emilio Velis is an industrial engineer based in San Salvador. He is Executive Director of the Appropedia Foundation and Adjunct Lecturer at the Heller School of Sustainable International Development of Brandeis University. E-mail: contacto@emiliovelis.com.
Diego Torres graduated in 2009 with a diploma (Licenciatura) in Computer Science from Universidad Nacional de La Plata, in Argentina. In October 2014 he got a Ph.D. in Computer Science from the Universidad Nacional de La Plata (Argentina) and Ph.D. from the Université de Nantes (France) with the thesis “Co-evolution between Social and Semantic Web”. Social Semantic Web, Semantic Web, Linked Open Data, Recommender systems and Citizen Science are the main topics on his research activities. E-mail: diego.torres@lifia.info.unlp.edu.ar.
Gino Caballero is a civil engineer with a masters degree in Disaster Management from the National Institute for Policy Studies (GRIPS) in Japan. He is the Disaster Risk Reduction and Response coordinator for Habitat for Humanity El Salvador and is an independent consultant for DR3 related projects. E-mail: caballero.m.gino@gmail.com.
### How to cite
Velis, E., Torres, D. and Caballero, G. (2021). ‘Gamification for social perception: introducing scientific literacy to dabblers in citizen science’. JCOM 20 (06), A08. https://doi.org/10.22323/2.20060208.
### Endnotes
1For purposes of this study, these two were considered as a unit due to being located close to each other despite being distinct communities. | 2021-10-22 03:12:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2536942958831787, "perplexity": 4021.6511804397405}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585450.39/warc/CC-MAIN-20211022021705-20211022051705-00617.warc.gz"} |
https://rstudio-pubs-static.s3.amazonaws.com/305579_5d6f93b2642a4cc9b1bb8409fb3c2dd0.html | Linear Regression establishes a relationship between a Dependent variable i.e. Y and one or more Independent variables i.e X, using a best fit straight line known as Regression Line. The equation of this regresiion line can then be used to predict value of ‘Y’ for any given ‘X’.
Dependent Variable (Target) : Continuous
Independent Variable(Predictor(s)): Continuous/Discrete
Simple linear regression involves one target(Y) and one predictor(X). This demo performs simple linear regression using Least Sqaures Method to find regression line that shows trend in the data i.e. relationship between X and Y . The equation of regression line in slope-intercept form is:
Y = mX + c ,where m= slope of straight line
c= Y-intercept
### 1. Load and view dataset
require("datasets")
data("airquality")
str(airquality)
## 'data.frame': 153 obs. of 6 variables:
## $Ozone : int 41 36 12 18 NA 28 23 19 8 NA ... ##$ Solar.R: int 190 118 149 313 NA NA 299 99 19 194 ...
## $Wind : num 7.4 8 12.6 11.5 14.3 14.9 8.6 13.8 20.1 8.6 ... ##$ Temp : int 67 72 74 62 56 66 65 59 61 69 ...
## $Month : int 5 5 5 5 5 5 5 5 5 5 ... ##$ Day : int 1 2 3 4 5 6 7 8 9 10 ...
head(airquality)
## Ozone Solar.R Wind Temp Month Day
## 1 41 190 7.4 67 5 1
## 2 36 118 8.0 72 5 2
## 3 12 149 12.6 74 5 3
## 4 18 313 11.5 62 5 4
## 5 NA NA 14.3 56 5 5
## 6 28 NA 14.9 66 5 6
### 2. Preprocess the dataset
Let’s begin by finding which attributes have missing values. We then need to impute those missing values(NA), which we will be doing simply by replacing NA with monthly average. Let’s begin!
col1<- mapply(anyNA,airquality) # apply function anyNA() on all columns of airquality dataset
col1
## Ozone Solar.R Wind Temp Month Day
## TRUE TRUE FALSE FALSE FALSE FALSE
The output shows that only Ozone and Solar.R attributes have NA i.e. some missing value.
# Impute monthly mean in Ozone
for (i in 1:nrow(airquality)){
if(is.na(airquality[i,"Ozone"])){
airquality[i,"Ozone"]<- mean(airquality[which(airquality[,"Month"]==airquality[i,"Month"]),"Ozone"],na.rm = TRUE)
}
# Impute monthly mean in Solar.R
if(is.na(airquality[i,"Solar.R"])){
airquality[i,"Solar.R"]<- mean(airquality[which(airquality[,"Month"]==airquality[i,"Month"]),"Solar.R"],na.rm = TRUE)
}
}
#Normalize the dataset so that no particular attribute has more impact on clustering algorithm than others.
normalize<- function(x){
return((x-min(x))/(max(x)-min(x)))
}
airquality<- normalize(airquality) # replace contents of dataset with normalized values
str(airquality)
## 'data.frame': 153 obs. of 6 variables:
## $Ozone : num 0.1201 0.1051 0.033 0.0511 0.0679 ... ##$ Solar.R: num 0.568 0.351 0.444 0.937 0.541 ...
## $Wind : num 0.0192 0.021 0.0348 0.0315 0.0399 ... ##$ Temp : num 0.198 0.213 0.219 0.183 0.165 ...
## $Month : num 0.012 0.012 0.012 0.012 0.012 ... ##$ Day : num 0 0.003 0.00601 0.00901 0.01201 ...
Yay! We have removed missing values from our dataset. We will now perform Linear Regression on our dataset!
### 3. Apply linear regression algorithm using Least Squares Method on “Ozone” and “Solar.R”
Since simple L.R. requires just one target, let’s take “Ozone” attribute as our target(Y) and “Solar.R” attribute as Predictor(X) to find if there exists any kind of relationship between them.
Y<- airquality[,"Ozone"] # select Target attribute
X<- airquality[,"Solar.R"] # select Predictor attribute
model1<- lm(Y~X)
model1 # provides regression line coefficients i.e. slope and y-intercept
##
## Call:
## lm(formula = Y ~ X)
##
## Coefficients:
## (Intercept) X
## 0.06509 0.09849
plot(Y~X) # scatter plot between X and Y
abline(model1, col="blue", lwd=3) # add regression line to scatter plot to see relationship between X and Y
The above graph shows that slope of the line goes upwards, hence, there exists a positive correlation between ‘Ozone’ and ‘Solar.R’. So, if we increase X, the value of Y will also increase and vice-versa.
### 4. Apply linear regression algorithm using Least Squares Method on “Ozone” and “Wind”
We will perform linear regression to find relationship of “Ozone” with “Wind” now.
Y<- airquality[,"Ozone"] # select Target attribute
X<- airquality[,"Wind"] # select Predictor attribute
model2<- lm(Y~X)
model2 # provides regression line coefficients i.e. slope and y-intercept
##
## Call:
## lm(formula = Y ~ X)
##
## Coefficients:
## (Intercept) X
## 0.2364 -4.3410
plot(Y~X) # scatter plot between X and Y
abline(model2, col="blue", lwd=3) # add regression line to scatter plot to see relationship between X and Y
The above graph shows that slope of the line goes downwards, hence, there exists a negative correlation between ‘Ozone’ and ‘Wind’. So, if we increase X, the value of Y will decrease and vice-versa.
From the above 2 graphs we can conclude that “Solar.R” is positively related to “Ozone” whereas “Wind” is negatively related.
### 4. Perform prediction
Now, let’s use the line coefficients for two equations that we got in model1 and model2 to predict value of Target for any given value of Predictor.
# Prediction of 'Ozone' when 'Solar.R'= 10
p1<- predict(model1,data.frame("X"=10))
p1
## 1
## 1.049993
The predicted value of “Ozone” is 1.0499933 when “Solar.R”= 10
# Prediction of 'Ozone' when 'Wind'= 5
p2<- predict(model2,data.frame("X"=5))
p2
## 1
## -21.46849
The predicted value of “Ozone” is -21.4684949 when “Wind”= 5
You may also wish to try out Data Classification, Clustering or Linear Regression from following links:
1. k-NN Classification for beginners
Using Iris Dataset
Using Airquality Dataset
2. k-means Clustering for beginners
Using Iris Dataset
Using Airquality Dataset
3. Linear Regression for beginners
Using Iris Dataset
Good luck! :) | 2021-10-27 02:44:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3585486114025116, "perplexity": 5927.403015945632}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588053.38/warc/CC-MAIN-20211027022823-20211027052823-00083.warc.gz"} |
https://mathematica.stackexchange.com/questions/112200/how-to-validate-an-xml-file-against-xsd-in-wolfram-language | # How to validate an XML file against xsd in Wolfram language?
I am planning to write a package to read data which is saved in XML format that should follow some XML schema. As part of the reading I would like to do error checking, namely schema validation. So is there an idiomatic way of doing this in the Wolfram language? I was trying to avoid using external programs.
• Can you provide an example of the kind of validation you are trying to do? Generally speaking Wolfram Language is well suited to handling XML thanks to its "symbolic XML" representation. – C. E. Apr 9 '16 at 0:58
• Sure. Here is an example xml (example.xml): pastebin.com/GJqUDcas; here the corresponding xml schema (schema.xsd): pastebin.com/HZg2SdWs. From the command line I can then validate by: xml val -e --xsd schema.xsd example.xml, where xml here is a command from the XMLStarlet Command Line XML Toolkit. – rmagno Apr 9 '16 at 1:26
• @paideia what is wrong with running such from the command line using Import["!"] or Run[] – William Apr 9 '16 at 19:38
• @William, the only problem with using Run[] is that if I want this functionality to be in a package so that other people using the package would not need to have this dependency. I was lokking for a solution that would only need Mathematica. – rmagno Apr 9 '16 at 21:32
• @paideia does the posted answer suit your needs? – William Apr 9 '16 at 23:10
JLink
Here is a solution that uses JLink to leverage Java's support of XML Schema:
Needs["JLink"]
InstallJava[];
validateXml[xsd_String, xml_String] :=
JavaBlock @ Module[{factory, xsdSource, xmlSource, schema, validator, valid, error}
, xsdSource = stringSource[xsd]
; xmlSource = stringSource[xml]
; factory = javaxxmlvalidationSchemaFactorynewInstance @
javaxxmlXMLConstantsW3CUXMLUSCHEMAUNSUURI
; schema = factory@newSchema[xsdSource]
; validator = schema@newValidator[]
; error = ""
; Block[{$JavaExceptionHandler = (error = GetJavaException[]@getMessage[])&} , validator@validate[xmlSource] ; <| "valid" -> error === "", "error" -> error |> ] ] stringSource[s_String] := JavaNew[ "javax.xml.transform.stream.StreamSource" , JavaNew["java.io.StringReader", s] ] Here it is in action, using the sample files mention in the question's comments: $xsd = Import["http://pastebin.com/raw/HZg2SdWs", "Text"];
$xml = Import["http://pastebin.com/raw/GJqUDcas", "Text"]; validateXml[$xsd, $xml] (* <| "valid" -> True , "error" -> "" |> *) validateXml[$xsd, "<zot id='wrong'/>"]
(*
<| "valid" -> "False"
, "error" -> "cvc-elt.1: Cannot find the declaration of element 'zot'."
|>
*)
validateXml[$xsd, StringReplace[$xml, "type>" -> "bad>"]]
(*
<| "valid" -> False
, "error" ->
"cvc-complex-type.2.4.a: Invalid content was found starting with element 'bad'.
One of
'{\"http://www.rdml.org\":description, \
\"http://www.rdml.org\":documentation,
\"http://www.rdml.org\":xRef, \
\"http://www.rdml.org\":annotation,
\"http://www.rdml.org\":type}' is expected."
|>
*)
XMLSchema Package
Incidentally, there is also a package called XMLSchema that is used internally by Mathematica's web service implementation. Being undocumented, there is no reason to expect it to be applicable to our needs. Furthermore, I have found XML Schema support in Mathematica's WSDL functionality to be somewhat incomplete. So perhaps we should not be surprised by the following (unsanctioned) exchange:
Needs["XMLSchema"]
$xsdObject = Import["http://pastebin.com/raw/HZg2SdWs", {"XML", "XMLObject"}]; LoadSchema[$xsdObject, "MySchema"]
(* LoadSchema::typename: The global type <<1>> does not have a name. *)
• Thanks @WReach for this very usefull functions. However, when I copy-paste into my Mathematica 8.0, I get a syntax error <code>Syntax::sntxf: "validator@validate[xmlSource];<" cannot be followed by "|valid". </code>. Is the notation <| a new notation in Mathematica? Can it be written differently? Thanks. – Denis Cousineau Sep 7 '16 at 1:34
• @DenisCousineau Association syntax was added in version 10. Prior to that, simply replace <| ... |> with { ... }. The parsing results will then be a list of rules instead of an association. – WReach Sep 7 '16 at 23:48 | 2020-07-15 05:49:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24983647465705872, "perplexity": 4312.096414071941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657155816.86/warc/CC-MAIN-20200715035109-20200715065109-00138.warc.gz"} |
https://www.physicsforums.com/threads/internal-energy-equals-to-average-kinectic-energy.661201/ | # Internal energy equals to average kinectic energy ?
1. ### Outrageous
375
For ideal gas , can I assume Cv dT = nkT (3/2)
, thank you
2. ### Andrew Mason
6,856
No. Molar heat capacity, Cv = 3R/2 only for ideal monatomic gases. You are also mixing temperature change with temperature here.
Using molar heat capacity 3R/2 for a monatomic ideal gas:
$\int_{Ti}^{Tf} nC_vdT = \frac{3}{2}nR(T_f - T_i) = \frac{3}{2}nR\Delta T = \frac{3}{2}Nk\Delta T$
AM
3. ### Outrageous
375
Thanks for compact explanation. | 2015-05-25 17:42:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7501586675643921, "perplexity": 10276.774558861254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928562.33/warc/CC-MAIN-20150521113208-00241-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://assignment-daixie.com/tag/%E5%BE%AE%E7%A7%AF%E5%88%86%E4%BB%A3%E5%86%99/ | # 微积分2|Calculus 2 MAST10006代写
0
This subject will extend knowledge of calculus from school. Students are introduced to hyperbolic functions and their inverses, the complex exponential and functions of two variables. Techniques of differentiation and integration will be extended to these cases. Students will be exposed to a wider class of differential equation models, both first and second order, to describe systems such as population models, electrical circuits and mechanical oscillators. The subject also introduces sequences and series including the concepts of convergence and divergence.
Find the mass of a ball 9 of radius $a$ whose density is numerically equal to the distance from a fixed diametral plane.
Let the ball be the inside of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$, and let the fixed diametral plane be $z=0$. Then $M=\iiint|z| d V$. Use the upper hemisphere and double the result. In spherical coordinates.
\begin{aligned} M &=2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{a} z \cdot \rho^{2} \sin \phi d \rho d \phi d \theta=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \int_{0}^{a} \rho \cos \phi \cdot \rho^{2} \sin \phi d \rho d \phi d \theta \ &\left.=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \frac{1}{4} \rho^{4} \cos \phi \sin \phi\right]{0}^{a} d \phi d \theta=\frac{1}{2} a^{4} \int{0}^{2 \pi} \frac{1}{2} \sin ^{2} \phi \int_{0}^{\pi / 2} d \theta=\frac{1}{4} a^{4} \int_{0}^{2 \pi} d \theta=\frac{1}{4} a^{4} \cdot 2 \pi=\frac{1}{2} \pi a^{4} \end{aligned}
## MAST10006 COURSE NOTES ：
Find the surface area $S$ of the part of the sphere $x^{2}+y^{2}+z^{2}=4 z$ inside the paraboloid $z=x^{2}+y^{2}$.
$\square$ The region $\mathscr{R}$ under the spherical cap (Fig. 44-33) is obtained by finding the intersection of $x^{2}+y^{2}+z^{2}=4 z$ and $z=x^{2}+y^{2}$. This gives $z(z-3)=0$. Hence, the paraboloid cuts the sphere when $z=3$, and $\mathscr{B}$ is the disk $x^{2}+y^{2} \leq 3 . \quad S=\iint_{a} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} d A . \quad 2 x+2 z \frac{\partial z}{\partial x}=4 \frac{\partial z}{\partial x}, \quad \frac{\partial z}{\partial x}=-\frac{x}{z-2} . \quad$ Similarly, $\frac{\partial z}{\partial y}=-\frac{y}{z-2}$. Hence,
$$1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=1+\frac{x^{2}}{(z-2)^{2}}+\frac{y^{2}}{(z-2)^{2}}=\frac{(z-2)^{2}+x^{2}+y^{2}}{(z-2)^{2}}=\frac{\left(x^{2}+y^{2}+z^{2}\right)-4 z+4}{(z-2)^{2}}=\frac{4}{(z-2)^{2}}$$
Therefore,
$$\left.S=\iint_{\pi} \frac{2}{z-2} d A=\int_{0}^{2 w} \int_{0}^{\sqrt{3}} \frac{2}{\sqrt{4-r^{2}}} r d r d \theta=-\int_{0}^{2 \theta} 2 \sqrt{4-r^{2}}\right]{0}^{\sqrt{3}} d \theta=-2 \int{0}^{2 \pi}(1-2) d \theta=2 \cdot 2 \pi=4 \pi$$
# 量子物理学的基础 Foundations of Quantum Physics PHYS104
0
$$\sum_{\lambda} F_{\lambda ‘ \lambda}^{j} C_{\lambda j}=\epsilon_{j} \sum_{\lambda} S_{\lambda^{\prime} \lambda} C_{\lambda j},$$
where
\begin{aligned} F_{\lambda^{\prime} \lambda}^{j}=&\left\langle\chi_{\lambda^{\prime}}|h| \chi_{\lambda}\right\rangle+\sum_{\delta \kappa}\left[\gamma_{\delta_{k}}\left\langle\chi_{\lambda^{\prime}} \chi_{\delta}|g| \chi_{\lambda} \chi_{\chi^{\prime}}\right\rangle\right.\ &\left.-\gamma_{\delta \kappa}^{\text {exch }}\left\langle\chi_{\lambda^{\prime} \cdot} \chi_{\delta}|g| \chi_{\kappa} \chi_{\lambda}\right\rangle\right] . \end{aligned}
Here, $h$ is the one-electron part of the full Hamiltonian, $g$ is an electron-electron repulsion potential energy, and
$$\begin{gathered} \gamma_{\delta x}=\sum_{i}^{\prime} C_{\delta i} C_{\kappa i}, \ \gamma_{\delta \kappa}^{\text {exch }}=\sum_{i}^{n \prime} C_{\delta i} C_{\kappa i}, \end{gathered}$$
## PHYS104COURSE NOTES ：
Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$
# 微积分 Calculus MATH101/MATH102
0
For $h \neq 0$ and $x+h$ in the domain of $f$,
$$f(x+h)-f(x)=\frac{f(x+h)-f(x)}{h} \cdot h$$
With $f$ differentiable at $x$,
$$\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$$ Since $\lim {h \rightarrow 0} h=0$, we have
$$\lim {h \rightarrow 0}[f(x+h)-f(x)]=\left[\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right] \cdot\left[\lim _{h \rightarrow 0} h\right]=f^{\prime}(x) \cdot 0=0 .$$
## MATH101/MATH102COURSE NOTES ：
Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$
# 微积分 Calculus MATH1006
0
Given a series $\Sigma_{m=1}^{m} a_{n}=a_{1}+a_{2}+a_{s}+\cdots$, let $s_{a}$ denote its rth partial sum:
$$s_{n}=\sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+\cdots+a_{n}$$
If the sequence $\left{s_{\mathrm{n}}\right}$ is convergent and $\lim {\mathrm{a} \rightarrow \mathrm{m}} s{\mathrm{a}}=s$ exists as a real number, then the series $\Sigma a_{n}$ is called convergent and we write
$$a_{1}+a_{2}+\cdots+a_{n}+\cdots=s \quad \text { or } \quad \sum_{n=1}^{\infty} a_{n}=s$$
The number $s$ is called the sum of the series. Otherwise, the series is called divergent.
## MATH1006COURSE NOTES ：
If $r=1$, then $s_{n}=a+a+\cdots+a=n a \rightarrow \pm \infty$. Since $\lim {n \rightarrow-} s{n}$ doesn’t exist, the geometric series diverges in this case.
If $r \neq 1$, we have
\begin{aligned} &s_{\mathrm{a}}=a+a r+a r^{2}+\cdots+a r^{\mathrm{n}-1} \ &r s_{\mathrm{a}}=\quad a r+a r^{2}+\cdots+a r^{\mathrm{n}-1}+a r^{\mathrm{n}} \end{aligned}
Subtracting these equations, we get
$$\begin{array}{r} s_{\mathrm{a}}-r s_{\mathrm{a}}=a-a r^{n} \ s_{\mathrm{n}}=\frac{a\left(1-r^{\mathrm{n}}\right)}{1-r} \end{array}$$
If $-1<r<1$, we know from $(11.1 .9)$ that $r^{n} \rightarrow 0$ as $n \rightarrow \infty$, so
$$\lim {n \rightarrow \infty} s{n}=\lim {n \rightarrow \infty} \frac{a\left(1-r^{n}\right)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r} \lim {n \rightarrow \infty} r^{n}=\frac{a}{1-r}$$
# 微积分 Calculus MAT00001C
0
This is
$$\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot\left[\left[\frac{\partial g}{\partial u}\right] \times\left[\frac{\partial g}{\partial v}\right]\right]\right] d u \wedge d v$$
(The permutation of the $P, Q, R$ (and the minus sign) come from the way the $d x \wedge d y$ acts on a piece of surface normal to the $(d) z$ direction.)
We can rewrite this as
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot \hat{\boldsymbol{n}}[u, v]\right] \quad d u \wedge d v$$
where $\hat{\boldsymbol{n}}[u, v]$ is the unit normal to the surface at $g\left[\begin{array}{l}u \ v\end{array}\right]$, and
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|$$
is the “area stretching factor”.
We have that
$$\int_{g\left(I^{2}\right)} \omega$$
## BMAT00001C COURSE NOTES ：
is the limit of the sums of values of $\omega$ on small elements of the surface $g\left(I^{2}\right)$. Suppose $g$ takes a rectangle $\triangle u \times \triangle v$ in $I^{2}$ to a (small) piece of the surface. $\omega$ at $g\left[\begin{array}{l}u \ v\end{array}\right]$ is, say,
$$P d x \wedge d y+Q d x \wedge d z+R d y \wedge d z$$
and the unit normal to the surface is $\hat{\boldsymbol{n}}[u, v]$ (located at $\left.g\left[\begin{array}{l}u \ v\end{array}\right]\right)$.
Write $\hat{\boldsymbol{n}}[u, v]$ as
$$\left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right]$$
# 微积分I|Calculus I代写 4CCM111A
0
so
$e^{y}-2 x-e^{-y}=0$
or, multiplying by $e^{y}$,
$$e^{2 y}-2 x e^{y}-1=0$$
This is really a quadratic equation in $e^{y}$ :
$$\left(e^{y}\right)^{2}-2 x\left(e^{y}\right)-1=0$$
Solving by the quadratic formula, we get
$$e^{y}=\frac{2 x \pm \sqrt{4 x^{2}+4}}{2}=x \pm \sqrt{x^{2}+1}$$
Note that $e^{y}>0$, but $x-\sqrt{x^{2}+1}<0$ (because $x<\sqrt{x^{2}+1}$ ). Thus the minus sign is inadmissible and we have
$$e^{y}=x+\sqrt{x^{2}+1}$$
Therefore
$$y=\ln \left(e^{y}\right)=\ln \left(x+\sqrt{x^{2}+1}\right)$$
## 4CCM111ACOURSE NOTES ：
Using Table 6 and the Chain Rule, we have
\begin{aligned} \frac{d}{d x}\left[\tanh ^{-1}(\sin x)\right] &=\frac{1}{1-(\sin x)^{2}} \frac{d}{d x}(\sin x) \ &=\frac{1}{1-\sin ^{2} x} \cos x=\frac{\cos x}{\cos ^{2} x}=\sec x \end{aligned}
# 微积分作业代写calculus代考
0
## 微分的记号Notation for differentiation代写
• 莱布尼茨的记号Leibniz’s notation
• 微积分基本定理Fundamental theorem
• 策梅洛-弗兰克尔集合论Zermelo–Fraenkel set theory
• 连续统的势 Cardinality of the continuum
## 微积分的历史
Johannes Kepler’s work Stereometrica Doliorum formed the basis of integral calculus.Kepler developed a method to calculate the area of an ellipse by adding up the lengths of many radii drawn from a focus of the ellipse.
## 微积分课后作业代写
The first equation yields $y=3 x^{2}$, substituting that into the second equation yields $x-6 x^{2}=0$, which has the solutions $x=0$ and $x=\frac{1}{6}$. So $x=0 \Rightarrow y=3(0)=0$ and $x=\frac{1}{6} \Rightarrow y=3\left(\frac{1}{6}\right)^{2}=\frac{1}{12}$.
So the critical points are $(x, y)=(0,0)$ and $(x, y)=\left(\frac{1}{6}, \frac{1}{12}\right)$.
To use Theorem 2.6, we need the second-order partial derivatives:
$$\frac{\partial^{2} f}{\partial x^{2}}=-6 x, \quad \frac{\partial^{2} f}{\partial y^{2}}=-2, \quad \frac{\partial^{2} f}{\partial y \partial x}=1$$
So
$$D=\frac{\partial^{2} f}{\partial x^{2}}(0,0) \frac{\partial^{2} f}{\partial y^{2}}(0,0)-\left(\frac{\partial^{2} f}{\partial y \partial x}(0,0)\right)^{2}=(-6(0))(-2)-1^{2}=-1<0$$ and thus $(0,0)$ is a saddle point. Also, $$D=\frac{\partial^{2} f}{\partial x^{2}}\left(\frac{1}{6}, \frac{1}{12}\right) \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{1}{6}, \frac{1}{12}\right)-\left(\frac{\partial^{2} f}{\partial y \partial x}\left(\frac{1}{6}, \frac{1}{12}\right)\right)^{2}=\left(-6\left(\frac{1}{6}\right)\right)(-2)-1^{2}=1>0$$
# 微积分calculus|MAT‑012 Assignment
0
Suppose $a_{n} \rightarrow a$ and $b_{n} \rightarrow$ b as $n \rightarrow \infty$. Then the following results hold.
(i) $a_{n}+b_{n} \rightarrow a+b$ as $n \rightarrow \infty$.
(ii) $c a_{n} \rightarrow c$ a as $n \rightarrow \infty$ for any real number $c$.
(iii) If $a_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, then $a \leq b$.
(iv) (Sandwich theorem) If $a_{n} \leq c_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, and if $a=b$, then $c_{n} \rightarrow a$ as $n \rightarrow \infty$.
Proof Let $\varepsilon>0$ be given.
(i) Note that, for every $n \in \mathbb{N}$,
\begin{aligned} \left|\left(a_{n}+b_{n}\right)-(a+b)\right| &=\left|\left(a_{n}-a\right)+\left(b_{n}-b\right)\right| \ & \leq\left|a_{n}-a\right|+\left|b_{n}-b\right| \end{aligned}
Since $a_{n} \rightarrow a$ and $b_{n} \rightarrow b$, the above inequality suggests that we may take $\varepsilon_{1}=\varepsilon / 2$, and consider $N_{1}, N_{2} \in \mathbb{N}$ such that
$$\left|a_{n}-a\right|<\varepsilon_{1} \quad \forall n \geq N_{1} \text { and }\left|b_{n}-b\right|<\varepsilon_{1} \quad \forall n \geq N_{2}$$
so that
$$\left|\left(a_{n}+b_{n}\right)-(a+b)\right| \leq\left|a_{n}-a\right|+\left|b_{n}-b\right|<2 \varepsilon_{1}=\varepsilon$$
for all $n \geq N:=\max \left{N_{1}, N_{2}\right}$.
(ii) Note that
$$\left|c a_{n}-c a\right|=|c|\left|a_{n}-a\right| \quad \forall n \in \mathbb{N} .$$
(Ratio test) Suppose $a_{n}>0$ for all $n \in \mathbb{N}$ such that $\lim {n \rightarrow \infty} \frac{a{n+1}}{a_{n}}=\ell$ for some $\ell \geq 0$. Then the following hold.
(i) If $\ell<1$, then $a_{n} \rightarrow 0$. (ii) If $\ell>1$, then $a_{n} \rightarrow \infty$.
Suppose $\ell<1$. Let $q$ be such that $\ell1$. Let $q$ be such that $1<q<\ell$. Then, taking for example the open interval $I$ containing $\ell$ as $I=(q, \ell+1)$, there exists $N \in \mathbb{N}$
## MAT 012: Precalculus
Instructor: Emily Meyer
Contact: Time & Place:
[email protected] MWF 8:00-9:40 am, Chemistry 166
Office: MSB 3127 Office hours MWR 3:00-5:00 pm
Course Website
Homework, worksheets, updates and supplementary material will be online at
http://math.ucdavis.edu/~emeyer/math12/~index.html, as well as on Canvas:
http://canvas.ucdavis.edu/courses/282963.
Textbook
• Precalculus (Seventh Edition, 2012) by Cohen, Lee, and Sklar.
Prerequisites/Placement Exam
• It is expected that you have taken two years of high school algebra, plane geometry, and
plane trigonometry.
• You must have received a qualifying score (25 or more) on the UC Davis Math Placement
Exam to take this course. If you have not yet taken the exam, you can take it during the
testing session from 1pm July 30 to 1pm August 7, 2018. See department website for
further details.
NOTE: after completing this course, you are also required to take the Math Placement Exam
again before you may enroll in calculus. There are two testing windows for fall quarter:
1pm September 5 to 1pm September 11, and 1pm September 26 to 1pm October 2.
Course Information
• 40% final exam
• 20% midterm exam
1
MAT 012 SSII18
• 20% quizzes
• 20% homework and worksheets
Class Expectations
Class time will be interactive. I will not take attendance, but you are expected to be in class.
Homework and Quizzes
• Homework will be in the form of worksheets given in class that you will have time to start
in class and will be due at the beginning of the following class. These will be graded out
of 2 points, and you can redo each one once for a regrade as long as your first version is
complete and your redo is turned in within one week of the original due date.
• Additional homework problems will be provided but will not be graded. It is not expected
that you do all of the additional problems, but that you work through enough problems
from each section to develop and solidify the skills and knowledge you have learned. This
is the best way to study for exams! You are welcome to ask about these problems in office
hours, or to ask for general qualitative feedback from the instructor on your work.
• Quizzes will be given every Friday on which there is no exam (see Exams below). These
are intended not to be especially difficult but to give you practice thinking about math in a
similar environment to an exam.
Exams
There will be two exams:
• Midterm on Friday, August 24
• Final on Friday, September 14 (last day of class)
Each will be 100 minutes.
Other
• Make-ups and Absences: Make-up exams and quizzes will be given only in the case of
documented emergencies. Your lowest quiz grade (out of four) will be dropped, so if you
need to miss a quiz for a non-emergency reason, plan on using that as your dropped quiz.
• Office hours: I will hold office hours (tentatively) 3-5pm Monday, Wednesday and Thursday. You are highly encouraged to attend! This is your opportunity to ask me to repeat
or rephrase anything that did not make sense in class, to get individualized help on your
homework, or to ask any lingering questions you might have.
2/3
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http://forge.ipsl.jussieu.fr/nemo/browser/branches/2017/dev_merge_2017/DOC/tex_sub/chap_DIU.tex?rev=9407 | # source:branches/2017/dev_merge_2017/DOC/tex_sub/chap_DIU.tex@9407
Last change on this file since 9407 was 9407, checked in by nicolasmartin, 3 years ago
Complete refactoring of cross-referencing
• Use of \autoref instead of simple \ref for contextual text depending on target type
• creation of few prefixes for marker to identify the type reference: apdx|chap|eq|fig|sec|subsec|tab
File size: 7.9 KB
Line
1\documentclass[../tex_main/NEMO_manual]{subfiles}
2\begin{document}
3% ================================================================
4% Diurnal SST models (DIU)
5% Edited by James While
6% ================================================================
7\chapter{Diurnal SST Models (DIU)}
8\label{chap:DIU}
9
10\minitoc
11
12
13\newpage
14$\$\newline % force a new line
15
16Code to produce an estimate of the diurnal warming and cooling of the sea surface skin
17temperature (skin SST) is found in the DIU directory.
18The skin temperature can be split into three parts:
19\begin{itemize}
20\item A foundation SST which is free from diurnal warming.
21\item A warm layer, typically ~3\,m thick, where heating from solar radiation can
22cause a warm stably stratified layer during the daytime
23\item A cool skin, a thin layer, approximately ~1\,mm thick, where long wave cooling
24is dominant and cools the immediate ocean surface.
25\end{itemize}
26
27Models are provided for both the warm layer, \mdfl{diurnal_bulk}, and the cool skin,
28\mdl{cool_skin}. Foundation SST is not considered as it can be obtained
29either from the main NEMO model ($i.e.$ from the temperature of the top few model levels)
30or from some other source.
31It must be noted that both the cool skin and warm layer models produce estimates of
32the change in temperature ($\Delta T_{\rm{cs}}$ and $\Delta T_{\rm{wl}}$)
33and both must be added to a foundation SST to obtain the true skin temperature.
34
35Both the cool skin and warm layer models are controlled through the namelist \ngn{namdiu}:
36\forfile{../namelists/namdiu}
37This namelist contains only two variables:
38\begin{description}
39\item[\np{ln\_diurnal}] A logical switch for turning on/off both the cool skin and warm layer.
40\item[\np{ln\_diurnal\_only}] A logical switch which if \forcode{.true.} will run the diurnal model
41without the other dynamical parts of NEMO.
42\np{ln\_diurnal\_only} must be \forcode{.false.} if \np{ln\_diurnal} is \forcode{.false.}.
43\end{description}
44
45Output for the diurnal model is through the variables sst\_wl' (warm\_layer) and
46sst\_cs' (cool skin). These are 2-D variables which will be included in the model
47output if they are specified in the iodef.xml file.
48
49Initialisation is through the restart file. Specifically the code will expect
50the presence of the 2-D variable Dsst'' to initialise the warm layer.
51The cool skin model, which is determined purely by the instantaneous fluxes,
52has no initialisation variable.
53
54%===============================================================
55\section{Warm layer model}
56\label{sec:warm_layer_sec}
57%===============================================================
58
59The warm layer is calculated using the model of \citet{Takaya_al_JGR10} (TAKAYA10 model
60hereafter). This is a simple flux based model that is defined by the equations
61\begin{eqnarray}
62\frac{\partial{\Delta T_{\rm{wl}}}}{\partial{t}}&=&\frac{Q(\nu+1)}{D_T\rho_w c_p
63\nu}-\frac{(\nu+1)ku^*_{w}f(L_a)\Delta T}{D_T\Phi\!\left(\frac{D_T}{L}\right)} \mbox{,}
64\label{eq:ecmwf1} \\
65L&=&\frac{\rho_w c_p u^{*^3}_{w}}{\kappa g \alpha_w Q }\mbox{,}\label{eq:ecmwf2}
66\end{eqnarray}
67where $\Delta T_{\rm{wl}}$ is the temperature difference between the top of the warm
68layer and the depth $D_T=3$\,m at which there is assumed to be no diurnal signal. In
69equation (\autoref{eq:ecmwf1}) $\alpha_w=2\times10^{-4}$ is the thermal expansion
70coefficient of water, $\kappa=0.4$ is von K\'{a}rm\'{a}n's constant, $c_p$ is the heat
71capacity at constant pressure of sea water, $\rho_w$ is the
72water density, and $L$ is the Monin-Obukhov length. The tunable
73variable $\nu$ is a shape parameter that defines the expected
74subskin temperature profile via $T(z)=T(0)-\left(\frac{z}{D_T}\right)^\nu\Delta 75T_{\rm{wl}}$,
76where $T$ is the absolute temperature and $z\le D_T$ is the depth
77below the top of the warm layer.
78The influence of wind on TAKAYA10 comes through the magnitude of the friction velocity
79of the water
80$u^*_{w}$, which can be related to the 10\,m wind speed $u_{10}$ through the relationship
81$u^*_{w} = u_{10}\sqrt{\frac{C_d\rho_a}{\rho_w}}$, where $C_d$ is
82the drag coefficient, and $\rho_a$ is the density of air. The symbol $Q$ in equation
83(\autoref{eq:ecmwf1}) is the instantaneous total thermal energy
84flux into
85the diurnal layer, $i.e.$
86\begin{equation}
87Q = Q_{\rm{sol}} + Q_{\rm{lw}} + Q_{\rm{h}}\mbox{,} \label{eq:e_flux_eqn}
88\end{equation}
89where $Q_{\rm{h}}$ is the sensible and latent heat flux, $Q_{\rm{lw}}$ is the long
90wave flux, and $Q_{\rm{sol}}$ is the solar flux absorbed
91within the diurnal warm layer. For $Q_{\rm{sol}}$ the 9 term
92representation of \citet{Gentemann_al_JGR09} is used. In equation \autoref{eq:ecmwf1}
93the function $f(L_a)=\max(1,L_a^{\frac{2}{3}})$, where $L_a=0.3$\footnote{This
94is a global average value, more accurately $L_a$ could be computed as
95$L_a=(u^*_{w}/u_s)^{\frac{1}{2}}$, where $u_s$ is the stokes drift, but this is not
96currently done} is the turbulent Langmuir number and is a
97parametrization of the effect of waves.
98The function $\Phi\!\left(\frac{D_T}{L}\right)$ is the similarity function that
99parametrizes the stability of the water column and
100is given by:
101\begin{equation}
102\Phi(\zeta) = \left\{ \begin{array}{cc} 1 + \frac{5\zeta +
1034\zeta^2}{1+3\zeta+0.25\zeta^2} &(\zeta \ge 0) \\
104 (1 - 16\zeta)^{-\frac{1}{2}} & (\zeta < 0) \mbox{,}
105 \end{array} \right. \label{eq:stab_func_eqn}
106\end{equation}
107where $\zeta=\frac{D_T}{L}$. It is clear that the first derivative of
108(\autoref{eq:stab_func_eqn}), and thus of (\autoref{eq:ecmwf1}),
109is discontinuous at $\zeta=0$ ($i.e.$ $Q\rightarrow0$ in equation (\autoref{eq:ecmwf2})).
110
111The two terms on the right hand side of (\autoref{eq:ecmwf1}) represent different processes.
112The first term is simply the diabatic heating or cooling of the
113diurnal warm
114layer due to thermal energy
115fluxes into and out of the layer. The second term
116parametrizes turbulent fluxes of heat out of the diurnal warm layer due to wind
117induced mixing. In practice the second term acts as a relaxation
118on the temperature.
119
120%===============================================================
121
122\section{Cool skin model}
123\label{sec:cool_skin_sec}
124
125%===============================================================
126
127The cool skin is modelled using the framework of \citet{Saunders_JAS82} who used a
128formulation of the near surface temperature difference based upon the heat flux and
129the friction velocity $u^*_{w}$. As the cool skin
130is so thin (~1\,mm) we ignore the solar flux component to the heat flux and the
131Saunders equation for the cool skin temperature difference $\Delta T_{\rm{cs}}$ becomes
132\begin{equation}
133\label{eq:sunders_eqn}
134\Delta T_{\rm{cs}}=\frac{Q_{\rm{ns}}\delta}{k_t} \mbox{,}
135\end{equation}
136where $Q_{\rm{ns}}$ is the, usually negative, non-solar heat flux into the ocean and
137$k_t$ is the thermal conductivity of sea water. $\delta$ is the thickness of the
138skin layer and is given by
139\begin{equation}
140\label{eq:sunders_thick_eqn}
141\delta=\frac{\lambda \mu}{u^*_{w}} \mbox{,}
142\end{equation}
143where $\mu$ is the kinematic viscosity of sea water and $\lambda$ is a constant of
144proportionality which \citet{Saunders_JAS82} suggested varied between 5 and 10.
145
146The value of $\lambda$ used in equation (\autoref{eq:sunders_thick_eqn}) is that of
147\citet{Artale_al_JGR02},
148which is shown in \citet{Tu_Tsuang_GRL05} to outperform a number of other
149parametrisations at both low and high wind speeds. Specifically,
150\begin{equation}
151\label{eq:artale_lambda_eqn}
152\lambda = \frac{ 8.64\times10^4 u^*_{w} k_t }{ \rho c_p h \mu \gamma }\mbox{,}
153\end{equation}
154where $h=10$\,m is a reference depth and
155$\gamma$ is a dimensionless function of wind speed $u$:
156\begin{equation}
157\label{eq:artale_gamma_eqn}
158\gamma = \left\{ \begin{matrix}
159 0.2u+0.5\mbox{,} & u \le 7.5\,\mbox{ms}^{-1} \\
160 1.6u-10\mbox{,} & 7.5 < u < 10\,\mbox{ms}^{-1} \\
161 6\mbox{,} & \ge 10\,\mbox{ms}^{-1} \\
162 \end{matrix}
163 \right.
164\end{equation}
165
166\end{document}
Note: See TracBrowser for help on using the repository browser. | 2020-11-25 04:58:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9607111811637878, "perplexity": 3959.5731852956033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00337.warc.gz"} |
https://theclevermachine.wordpress.com/2012/12/16/fmri-in-neuroscience-modeling-the-hrf-with-fir-basis-functions/ | # fMRI in Neuroscience: Modeling the HRF With FIR Basis Functions
In the previous post on fMRI methods, we discussed how to model the selectivity of a voxel using the General Linear Model (GLM). One of the basic assumptions that we must make in order to use the GLM is that we also have an accurate model of the Hemodynamic Response Function (HRF) for the voxel. A common practice is to use a canonical HRF model established from previous empirical studies of fMRI timeseries. However, voxels throughout the brain and across subjects exhibit a variety of shapes, so the canonical model is often incorrect. Therefore it becomes necessary to estimate the shape of the HRF for each voxel.
There are a number of ways that have been developed for estimating HRFs, most of them are based on temporal basis function models. (For details on basis function models, see this previous post.). There are a number of basis function sets available, but in this post we’ll discuss modeling the HRF using a flexible basis set composed of a set of delayed impulses called Finite Impulse Response (FIR) basis.
## Modeling HRFs With a Set of Time-delayed Impulses
Let’s say that we have an HRF with the following shape.
A Model HRF.
We would like to be able to model the HRF as a weighted combination of simple basis functions. The simplest set of basis functions is the FIR basis, which is a series of $H$ distinct unit-magnitude (i.e. equal to one) impulses, each of which is delayed in time by $t = 1 \dots H$ TRs. An example of modeling the HRF above using FIR basis functions is below:
%% REPRESENTING AN HRF WITH FIR BASIS FUNCTIONS
% CREATE ACTUAL HRF (AS MEASURED BY MRI SCANNER)
rand('seed',12345)
TR = 1 % REPETITION TIME
t = 1:TR:20; % MEASUREMENTS
h = gampdf(t,6) + -.5*gampdf(t,10); % ACTUAL HRF
h = h/max(h);
% DISPLAY THE HRF
figure;
stem(t,h,'k','Linewidth',2)
axis square
xlabel(sprintf('Basis Function Contribution\nTo HRF'))
title(sprintf('HRF as a Series of \nWeighted FIR Basis Functions'))
% CREATE/DISPLAY FIR REPRESENTATION
figure; hold on
cnt = 1;
% COLORS BASIS FUNCTIONS ACCORDING TO HRF WEIGHT
map = jet(64);
cRange = linspace(min(h),max(h),64);
for iT = numel(h):-1:1
firSignal = ones(size(h));
firSignal(cnt) = 2;
[~,cIdx] = min(abs(cRange-h(cnt)));
color = map(cIdx,:);
plot(1:numel(h),firSignal + 2*(iT-1),'Color',color,'Linewidth',2)
cnt = cnt+1;
end
colormap(map); colorbar; caxis([min(h) max(h)]);
% DISPLAY
axis square;
ylabel('Basis Function')
xlabel('Time (TR)')
set(gca,'YTick',0:2:39,'YTickLabel',20:-1:1)
title(sprintf('Weighted FIR Basis\n Set (20 Functions)'));
Representing the HRF above as a weighted set of FIR basis functions.
The color of each of the 20 basis functions corresponds to its weight
Each of the basis functions $b_t$ has an unit impulse that occurs at time $t = 1 \dots 20$; otherwise it is equal to zero. Weighting each basis function $b_t$ with the corresponding value of the HRF at each time point $t$, followed by a sum across all the functions gives the target HRF in the first plot above. The FIR basis model makes no assumptions about the shape of the HRF–the weight applied to each basis function can take any value–which allows the model to capture a wide range of HRF profiles.
Given an experiment where various stimuli are presented to a subject and BOLD responses evoked within the subject’s brain, the goal is to determine the HRF to each of the stimuli within each voxel. Let’s take a look at a concrete example of how we can use the FIR basis to simultaneously estimate HRFs to many stimuli for multiple voxels with distint tuning properties.
## Estimating the HRF of Simulated Voxels Using the FIR Basis
For this example we revisit a simulation of voxels with 4 different types of tuning (for details, see the previous post on fMRI in Neuroscience). One voxel is strongly tuned for visual stimuli (such as a light), the second voxel is weakly tuned for auditory stimuli (such as a tone), the third is moderately tuned for somatosensory stimuli (such as warmth applied to the palm), and the final voxel is unselective (i.e. weakly and equally selective for all three types of stimuli). We simulate an experiment where the blood-oxygen-level dependent (BOLD) signals evoked in each voxel by a series of stimuli consisting of nonoverlapping lights, tones, and applications of warmth to the palm, are measured over $T=330$ fMRI measurments (TRs). Below is the simulation of the experiment and the resulting simulated BOLD signals:
%% SIMULATE AN EXPERIMENT
% SOME CONSTANTS
trPerStim = 30;
nRepeat = 10;
nTRs = trPerStim*nRepeat + length(h);
nCond = 3;
nVox = 4;
impulseTrain0 = zeros(1,nTRs);
% RANDOM ONSET TIMES (TRs)
onsetIdx = randperm(nTRs-length(h));
% VISUAL STIMULUS
impulseTrainLight = impulseTrain0;
impulseTrainLight(onsetIdx(1:nRepeat)) = 1;
onsetIdx(1:nRepeat) = [];
% AUDITORY STIMULUS
impulseTrainTone = impulseTrain0;
impulseTrainTone(onsetIdx(1:nRepeat)) = 1;
onsetIdx(1:nRepeat) = [];
% SOMATOSENSORY STIMULUS
impulseTrainHeat = impulseTrain0;
impulseTrainHeat(onsetIdx(1:nRepeat)) = 1;
% EXPERIMENT DESIGN / STIMULUS SEQUENCE
D = [impulseTrainLight',impulseTrainTone',impulseTrainHeat'];
X = conv2(D,h');
X = X(1:nTRs,:);
%% SIMULATE RESPONSES OF VOXELS WITH VARIOUS SELECTIVITIES
visualTuning = [4 0 0]; % VISUAL VOXEL TUNING
auditoryTuning = [0 2 0]; % AUDITORY VOXEL TUNING
somatoTuning = [0 0 3]; % SOMATOSENSORY VOXEL TUNING
noTuning = [1 1 1]; % NON-SELECTIVE
beta = [visualTuning', ...
auditoryTuning', ...
somatoTuning', ...
noTuning'];
y0 = X*beta;
SNR = 5;
noiseSTD = max(y0)/SNR;
noise = bsxfun(@times,randn(size(y0)),noiseSTD);
y = y0 + noise; % VOXEL RESPONSES
% DISPLAY VOXEL TIMECOURSES
voxNames = {'Visual','Auditory','Somat.','Unselective'};
cols = lines(4);
figure;
for iV = 1:4
subplot(4,1,iV)
plot(y(:,iV),'Color',cols(iV,:),'Linewidth',2); xlim([0,nTRs]);
ylabel('BOLD Signal')
legend(sprintf('%s Voxel',voxNames{iV}))
end
xlabel('Time (TR)')
set(gcf,'Position',[100,100,880,500])
Four simulated voxels, each with strong visual, weak auditory,
moderate somatosensory and unselective tuning.
Now let’s estimate the HRF of each voxel to each of the $C = 3$ stimulus conditions using an FIR basis function model. To do so, we create a design matrix composed of successive sets of delayed impulses, where each set of impulses begins at the onset of each stimulus condition. For the $[T \times C]$-sized stimulus onset matrix $D$, we calculate an $[T \times HC]$ FIR design matrix $X_{FIR}$, where $H$ is the assumed length of the HRF we are trying to estimate. The code for creating and displaying the design matrix for an assumed HRF length $H=16$ is below:
%% ESTIMATE HRF USING FIR BASIS SET
% CREATE FIR DESIGN MATRIX
hrfLen = 16; % WE ASSUME HRF IS 16 TRS LONG
% BASIS SET FOR EACH CONDITOIN IS A TRAIN OF INPULSES
X_FIR = zeros(nTRs,hrfLen*nCond);
for iC = 1:nCond
onsets = find(D(:,iC));
idxCols = (iC-1)*hrfLen+1:iC*hrfLen;
for jO = 1:numel(onsets)
idxRows = onsets(jO):onsets(jO)+hrfLen-1;
for kR = 1:numel(idxRows);
X_FIR(idxRows(kR),idxCols(kR)) = 1;
end
end
end
% DISPLAY
figure;
subplot(121);
imagesc(D);
colormap gray;
set(gca,'XTickLabel',{'Light','Tone','Som.'})
title('Stimulus Train');
subplot(122);
imagesc(X_FIR);
colormap gray;
title('FIR Design Matrix');
set(gca,'XTick',[8,24,40])
set(gca,'XTickLabel',{'Light','Tone','Som.'})
set(gcf,'Position',[100,100,550,400])
Left: The stimulus onset matrix (size = [T x 3]).
Right the corresponding Design Matrix (size = [T x 3*H])
In the right panel of the plot above, we see the form of the FIR design matrix $X_{FIR}$ for the stimulus onset on the left. For each voxel, we want to determine the weight on each column of $X_{FIR}$ that will best explain the BOLD signals $y$ measured from each voxel. We can form this problem in terms of a General Linear Model:
$y = X_{FIR}\beta_{FIR}$
Where $\beta_{FIR}$ are the weights on each column of the FIR design matrix. If we set the values of $\beta_{HRF}$ such as to minimize the sum of the squared errors (SSE) between the model above and the measured actual responses
$SSE = \sum_i^N(y^{(i)} - X_{FIR}^{(i)})^2$,
then we can use the Ordinary Least Squares (OLS) solution discussed earlier to solve the for $\beta_{HRF}$. Specifically, we solve for the weights as:
$\hat \beta_{FIR} = (X_{FIR}^T X_{FIR})^{-1} X_{FIR} y$
Once determined, the resulting $[CH \times V]$ matrix of weights $\hat \beta_{FIR}$ has the HRF of each of the $V=4$ different voxels to each stimulus condition along its columns. The first $H$ (1-16) of the weights along a column define the HRF to the first stimulus (the light). The second $H$ (17-32) weights along a column determine the HRF to the second stimulus (the tone), etc… Below we parse out these weights and display the resulting HRFs for each voxel:
% ESTIMATE HRF FOR EACH CONDITION AND VOXEL
betaHatFIR = pinv(X_FIR'*X_FIR)*X_FIR'*y;
% RESHAPE HRFS
hHatFIR = reshape(betaHatFIR,hrfLen,nCond,nVox);
% DISPLAY
figure
cols = lines(4);
names = {'Visual','Auditory','Somat.','Unselective'};
for iV = 1:nVox
subplot(2,2,iV)
hold on;
for jC = 1:nCond
hl = plot(1:hrfLen,hHatFIR(:,jC,iV),'Linewidth',2);
set(hl,'Color',cols(jC,:))
end
hl = plot(1:numel(h),h,'Linewidth',2);
xlabel('TR')
legend({'Light','Tone','Heat','True HRF'})
set(hl,'Color','k')
xlim([0 hrfLen])
grid on
axis tight
title(sprintf('%s Voxel',names{iV}));
end
set(gcf,'Position',[100,100,880,500])
HRF estimates for each voxel to each of the 3 stimuli. Black plots show the shape of the true HRF.
Here we see that estimated HRFs accurately capture both the shape of the HRF and the selectivity of each of the voxels. For instance, the HRFs estimated from the responses of first voxel indicate strong tuning for the light stimulus. The HRF estimated for the light stimulus has an amplitude that is approximately 4 times that of the true HRF. This corresponds with the actual tuning of the voxel (compare this to the value of $\beta(1,1)$). Additionally, time delay till the maximum value (time-to-peak) of the HRF to the light is the same as the true HRF. The first voxel’s HRFs estimated for the other stimuli are essentially noise around baseline. This (correctly) indicates that the first voxel has no selectivity for those stimuli. Further inspection of the remaining estimated HRFs indicate accurate tuning and HRF shape is recovered for the other three voxels as well.
## Wrapping Up
In this post we discussed how to apply a simple basis function model (the FIR basis) to estimate the HRF profile and get an idea of the tuning of individual voxels. Though the FIR basis model can accurately model any HRF shape, it is often times too flexible. In scenarios where voxel signals are very noisy, the FIR basis model will tend to model the noise.
Additionally, the FIR basis set needs to incorporate a basis function for each time measurement. For the example above, we assumed the HRF had a length of 16 TRs. The FIR basis therefore had 16 tuneable weights for each condition. This leads to a model with 48 ($C\times H = 3 \times 16$) tunable parameters for the GLM model. For experiments with many different stimulus conditions, the number of parameters can grow quickly (as $HC$). If the number of parameters is comparable (or more) than the number of BOLD signal measurements, it will be difficult accurately estimate $\hat \beta_{FIR}$. As we’ll see in later posts, we can often improve upon the FIR basis set by using more clever basis functions.
Another important but indirect issue that effects estimating the HRF is the experimental design, or rather the schedule used to present the stimuli. In the example above, the stimuli were presented in random, non-overlapping order. What if the stimuli were presented in the same order every time, with some set frequency? We’ll discuss in a later post the concept of design efficiency and how it affects our ability to characterize the shape of the HRF and, consequently, voxel selectivity. | 2018-10-21 21:22:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7877252697944641, "perplexity": 2462.377278904655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514355.90/warc/CC-MAIN-20181021203102-20181021224602-00280.warc.gz"} |
http://www.astro.spbu.ru/DOP/8-GLIB/ASTNOTES/node5.html | Next: Polarization efficiency: size/shape/orientation effects Up: Optical properties of cosmic Previous: The 2175Å feature
To model the interstellar polarization one needs to calculate the forward-transmitted radiation for an ensemble of non-spherical aligned dust grains. This procedure consists of two steps: 1) computations of the extinction cross-sections for two polarization modes and 2) averaging of cross-sections for given particles size and orientation distributions. Although the average cross-sections should be compared with observations, behaviour of the polarization cross-sections and alignment mechanisms are often analysed separately.
Let non-polarized stellar radiation passes through a dusty cloud with a homogeneous magnetic field. As follows from observations and theoretical considerations (Dolginov et al., [1979]), the magnetic field determines the direction of alignment of dust grains. The angle between the line of sight and the magnetic field is ( ). The linear polarization produced by a rotating spheroidal particle of same size is
where are the extinction cross-sections for two polarization modes (Sect. ), the column density of dust grains. The particles are assumed to be partially aligned: the major axis of the particle rotates in the spinning plane ( is the spin angle) which is perpendicular to the angular momentum . spins (precesses) around the direction ofmagnetic field ( is the precession angle), is the precession-cone angle for . This is the imperfect Davies-Greenstein (IDG) orientation described by the function which depends on the alignment parameter and the angle .
Note that the problem of grain alignment is one of the most difficult in the physics of cosmic dust. Here, the interaction of solid particles with gas, radiation and magnetic field is closely connected. Davies and Greenstein ([1951]) assumed that Fe atoms embedded in dielectric particles gave them paramagnetic properties and opened the possibility of interaction with a weak interstellar magnetic field. The required orientation arises as a result of the effect of paramagnetic relaxation of thermally rotating grains. The Davies-Greenstein mechanism was further developed by Jones and Spitzer ([1967]) who obtained expressions for the distribution of angular momentum. In the simplest case, it is
(8)
The parameter is a function of the particle size , the imaginary part of the grain magnetic susceptibility ( , where is the angular velocity of grain), gas density , the strength of magnetic field and dust () and gas () temperatures
(9)
where
(10)
The angle in Eq. (8) is expressed via the angles and (for definitions of the angles and relations between them see, for example, Hong and Greenberg, [1980] or Voshchinnikov, [1989]).
Rotation is an important factor of any grain alignment mechanism. The faster it is the more effective the grain alignment should be. The Davies-Greenstein mechanism considers thermally rotating grains. Purcell ([1979]) suggested a mechanism of supra-thermal spin alignment (SSA; pinwheel'' mechanism) where the grains were spun up to very high velocities as a result of the desorption of H molecules from their surfaces. In this case, the alignment function is described by Eq. (9) but the parameter is
(11)
where is in . Rotation can also arise due to radiation torques when helical'' grains scatter left- and right-circularly polarized light in a different way (Dolginov et al., [1979]; Draine and Weingartner, [1996], [1997]), which can lead to the grain alignment in anisotropic radiation field. Alignment of thermally rotating grains is also possible by supersonic flows or Alfvénic waves or ambipolar diffusion. This so-called Gold-type3 of mechanical alignment was generalized for supra-thermally rotating grains by Lazarian ([1995]). Note that some mechanisms can produce alignment when the major grain axes tend to align parallel to the magnetic field. Such an orientation is wrong'' for interstellar polarization but the mechanism may be right'' and operate in other conditions, for example in jets from YSO.
The development and current status of the major alignment mechanisms and principal physical processes forming their basis are reviewed by Roberge ([1996]), Lazarian et al. ([1997]) and Lazarian ([2000]). Unfortunately, the astrophysical significance of different alignment mechanisms remains unclear. This is connected, in particular, with very rough theoretical estimates of the polarization efficiency when instead of an alignment function like that given by Eq. (9) the Rayleigh reduction factor (see Greenberg, [1968]) is used (e.g., Lazarian et al., [1997]).
The circular polarization is proportional to the product
(12)
where is the average cross-section of circular polarization. The cross-section is calculated as the difference of phase lags represented by the imaginary parts of the corresponding complex extinction cross-sections. The circular polarization arises when the radiation linearly polarized in cloud 1'' passes through cloud 2'' (note the indices in Eq. (13)). Cloud 1'' works as a linear polarizer and cloud 2'' as a linear retarder. In such an optical device, the maximum transformation of the linear polarization to the circular one occurs when the optical axes of components are inclined by 45 (e.g., Tinbergen, [1996]).
For simplicity, non-rotating particles of the same orientation are frequently considered. In this case of picket fence'' (PF) orientation, there are no integrals over angles and in Eq. (8). The polarization degree is proportional to the polarization cross-section , where . The dichroic polarization efficiency is defined by the ratio of the polarization cross-section (factor) to the extinction one
(13)
where the upper (lower) sign is related to prolate (oblate) spheroids. This ratio describes the efficiency of polarization of light transmitted through an uniform slab consisting of non-rotating particles of the same orientation.
A more complicated case is the perfect rotational (2D) orientation (or perfect Davies-Greenstein orientation, PDG) when the major axis of a non-spherical particle always lies in the same plane. For the 2D orientation, integration is performed over the spin angle only. This gives for prolate spheroids
(14)
where the angle is connected with and ( ). For oblate spheroids randomly aligned in a plane, we have and
(15)
As a result, the expected polarization will be determined by:
• the particle refractive index, size and shape via the polarization cross-section ;
• the relation between the strength of the magnetic field, gas and dust temperature, gas density, etc. via the alignment function ;
• the direction of alignment depending on the angle (or ) via both and .
The simplest types of orientations like PF or PDG allow one to investigate the influence of the first and third factors. The dependence of on is excluded because the normalized polarization or polarization efficiency is usually studied.
Subsections
Next: Polarization efficiency: size/shape/orientation effects Up: Optical properties of cosmic Previous: The 2175Å feature
root 2003-04-09 | 2022-01-22 13:44:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8276050686836243, "perplexity": 1494.5951788507055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00377.warc.gz"} |
https://itectec.com/database/mysql-full-text-search-excerpt/ | # MySQL Full Text Search Excerpt
full-text-searchMySQL
I am adding MySQL Full Text search to an intranet and one of the things i'd like to do is return an excerpt back in the search result that includes the search term(s) the user searched for.
I initially thought that I could do this with locate() to get the index of the string minus ~10 characters and then substring around 200 characters after but this only works if there is a document that includes these words consecutively. For example:
SELECT id,created_at, modified_at,
SUBSTRING(content, (locate('pto request', content))) as excerpt
FROM search_index
WHERE MATCH (name,content) AGAINST ('pto request' IN BOOLEAN MODE)
The match clause finds documents that contains the words pto and request that may not be consecutive but the the locate function requires the exact string "pto request" to be somewhere in the document, if it isn't, an excerpt is not generated.
My first thought was to check for 0, if 0 locate didn't match anything so remove a word and try again and keep doing so until a locate finds a match, but this could get really messy. Is there a better way?
If you want exactly the 2 words, do
... AGAINST('"pto request"' IN BOOLEAN MODE)
If you want some context:
SELECT ...
SUBSTRING(content,
GREATEST(1, LOCATE('pto request', content) - 20),
50)
FROM ... | 2022-01-28 09:26:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20275701582431793, "perplexity": 2632.812804314231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00037.warc.gz"} |
https://codeforces.com/blog/entry/105995 | ### Erisu._.3812's blog
By Erisu._.3812, history, 6 weeks ago,
Hello those who are reading my entry, I have a small problem that i'm kinder curious about my solution.
***The problem is:
You have an array of integers (a[i]<=1e6) sized of n (n<=1e3). You have to split the array m different segment which cover the whole array. An action will take 1 unit of time which let you do 1 of those 2 at a time (each take 1 unit of time):
• You can minus 1 from a[i]
• You can plus 1 from a[i]
Your task is to mange to split the segments somehow that all of the value in 1 segment need to be the same by using those action i mention. And also those segment action work separately, which mean the answer would be the maximum action of 1 of those m segments.
Test Case:
Input:
6 2
1 1 2 3 4 3
Output:
1
Explain: 2 segments are 1 1 2 and 3 4 3, you can turn it into 1 1 1 and 3 3 3***
My idea:
Due to the fact that actions need for a subarray is going to be larger if it contains more value (or the subarray is larger) because the action we have to use is abs(a[i]-median) so base on the linear spec, we could use greedy to put as many as we could into a segment so that their action won't exceed the answer (cause we only focus about the max value).
I would use binary search to search for the answer, for each mid I search, I would check it for each segment as I mention to make sure that none of them could exceed by using 2 pointer L and R. For each L, I will find all the median in (L, R) and check to make sure the actions follow the rule by extend the R pointer to the furthest position (rightest what I mean).
And if everything is ok, then just print the answer. However, the time limit could be O(n^2*LogN^2), which mean 20*20*1000*1000 or 400 million calculations, may exceed 1 second limit. Forgot to mention, I use Persistent Data Structures to find the (r-l+1)/2 smallest value to find the median.
time limit: 1s
memory limit: 256mb
• 0
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 sorry because codeforce entry is hard to use, the test case is 6 2 1 1 2 3 4 3 n = 6, m = 2, a[]={1 1 2 3 4 3} is what I meant output: 1
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 A nicer typesetting and a simplified description could be welcomed.
• » » 6 weeks ago, # ^ | 0 how about type in Microsoft word and screen shot, you think it could be a better way?
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 Adding bulletin in the important points and when you are going for a new line instead of pressing Enter 1 time press it 2 times to reflect a new line in the entry as well.Hope now you can make your statement readable :)
• » » 6 weeks ago, # ^ | 0 thank you for your help <3
» 6 weeks ago, # | 0 Auto comment: topic has been updated by Erisu._.3812 (previous revision, new revision, compare).
» 6 weeks ago, # | 0 Seems that you got this in the bag! The solution is great, you simply miscalculated the time complexity. From what I understand, it should be O(n*logN^2), so a maximum of 1000*10*10 = 100000 steps.Explanation: For each mid you pick, you do n iterations. In each iteration, the complexity is O(logN) (for adding an element to some ordered multiset(s) and getting the median value). You also might want to clear the set(s) when you reach a limit (i.e. if you incremented the r, you would do more than mid operations), so another NlogN for each update of mid. logN * N * logN + logN * N * logN = O(N*logN^2).If you need further explanations, lmk. Bottom line, your idea was great!
• » » 6 weeks ago, # ^ | ← Rev. 2 → 0 One more detail: you can calculate abs(a[i]-median) for all i in current sequence in O(1) using two partial sums: sp[i] = a[1] + a[2] + ... + a[i]; sn[i] = -a[1] — a[2] — ... — a[i]. You will also need to know the median value and exactly how many numbers are >= and < than this value, respectively.
• » » 6 weeks ago, # ^ | ← Rev. 3 → 0 thanks a lot for your help <3I've just realized how dumb I was to be misunderstood logN from 9 to 20 :') N <= 1e3 (I thought it was 1e6) | 2022-09-27 18:02:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2176409363746643, "perplexity": 1685.4124979506055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335054.79/warc/CC-MAIN-20220927162620-20220927192620-00498.warc.gz"} |
http://mathhelpforum.com/calculus/154803-couple-very-simple-questions-about-vectors-scalars.html | # Thread: A couple of very simple questions about vectors and scalars
1. ## A couple of very simple questions about vectors and scalars
Dear MHF,
I am a sophomore engineering student at The University of Tennessee (Go Vols!) and I am currently enrolled in Calc 3. I feel like I understand the more advanced concepts but struggle with the simpler stuff (like a person reading with no phonics, simply memorization). Therefore I feel like my basic knowledge needs to be boosted... hopefully the answers to these questions will help!
1. Is a scalar the same thing as a vector's magnitude?
2. Can a vector be made from a scalar?
3. Additionally, how (in written notation) is it possible to tell the difference between the two? Like is one wrapped in abs value bars or what? Just a little confused about the context of all of this.
Thank you, everyone!
Matt
(FIRST POST!)
2. 1. It is true that a vector's magnitude is a scalar, but the definition of a scalar is not "the magnitude of a vector". The answer to this question can be as simple or as complex as your mathematical knowledge allows. Essentially, though, I think it's enough for most cases to say that scalars are known as such because they "scale" vectors in a vector space by either stretching them out or condensing them.
2. I'm not sure I understand what you're asking, so correct me if I'm missing the mark. All you need to do to make a scalar into a vector is to multiply some vector by that scalar. For instance, if I define the vector $\vec {x}=(x_1, x_2, x_3)$ and then multiply it by the scalar 3, it becomes $\vec {x}^{'}=(4x_1, 4x_2, 4x_3)$.
3. Vectors are denoted in a lot of different ways depending on where you come across them. As in the above example, many times vectors will have an arrow over them while scalars do not. You will often also see a vector denoted by a bold or italicized letter. Unit vectors can also sometimes be italicized letters, but in my experience have generally been denoted with a carat (^) above them. If you see a vector in absolute value bars, this is equivalent to the magnitude (or length) of the vector; otherwise, it is just the absolute value of a scalar.
Consult your professor or TA if there is any doubt what some piece of notation means, since your grade is in their hands after all. | 2017-05-01 02:37:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7986522912979126, "perplexity": 329.7341115856165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917126538.54/warc/CC-MAIN-20170423031206-00480-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://matheducators.stackexchange.com/questions/1545/what-are-easy-examples-from-daily-life-of-constrained-optimization?noredirect=1 | # What are easy examples from daily life of constrained optimization?
A standard example of motivating constrained optimization are examples where the setup is described in a lot of lines, e.g., when you own a company and the company is making some products out of resources and are mixed in a certain ratio, etc.
Are there more easy (i.e., to explain and to understand) examples from daily life which motivate constrained optimization?
The examples should be nonlinear if possible, they don't have to be solvable, but it should be clear how to translate daily life language into the problem.
A great answer should also explain the need of constrained optimization tools (i.e., the constrains should not be solvable explicitly) and maybe also demonstrate that the gradient is not zero without calculating everything, but from the (first) view of the example.
Note: This question is related to Optimization problems that today's students might actually encounter?, where more advanced problems should be discussed.
## 4 Answers
Bankruptcy problems ask for how to "fairly" distribute \$E to honest claimants whose claims exceed the amount \$E. For example, $A$ claims \$30,$B$claims \$50 and $C$ claims \$120 and there is only \$160 to distribute. There are two methods dating back to "medieval" times associated with Moses Maimonides.
a. Try to equalize the amount given (gain) to each claimant but without giving the claimant more than the claimant asks for.
b. Try to equalize the loss to each claimant but without asking the claimant to subsidize the settlement by adding money to E to make this possible.
Each of these approaches to being fair leads to a constrained optimization problem. There are other approaches to being fair here in addition to the two approaches above, for example, one could give each claimant an amount proportional to his/her claim.
Determine the minimum distance from a parametric equation $(x(t), y(t))$ to a given point $(x_0, y_0)$.
Eg. at which point should a person leave a road (described by the parametric equation), such that the walking distance to the point $(x_0, y_0)$ from the road is minimized?
The problem is then: Minimize $\sqrt{(x-x_0)^2 + (y - y_0)^2}$ subject to $x = x(t)$ and $y = y(t)$.
• Thanks for your answer! I can't figure out where the constraint of your optimization problem is? – Markus Klein Apr 14 '14 at 8:31
• The constraint is that the point is in a feasible region. In this case on the road, but it could also be on an implicitly defined function (circle / ellipse). – midtiby Apr 14 '14 at 8:36
This example is maybe the most easy, but in my opinion it does not highlight the necessity to use methods of constrained optimization since the constrained equation is explicitly invertible.
A gardener has 20 meters of fence-material and wants to fence a rectangle shaped area with maximal area.
The problem reads then as: Maximize $f(a,b)=a\cdot b$ subject to $2a+2b=20$.
• There is also the risk that the student looks at the problem, says "I know, the answer must be a square!" by applying simple geometry, and then entirely neglects the algebra. – Kevin Feb 11 '18 at 16:52
Following on the example above by Markus Klein:
A gardener has 20 meters of fence-material and wants to fence a rectangle shaped garden with maximal area. Furthermore, the garden will be located next to the house such that the house will serve as one of the four walls.
The problem then reads: Maximize $$f(a,b) = a*b$$ subject to $$2a + b = 20$$.
The answer $$(a = 5, b = 10)$$ is not so obvious to students.
(I realize this should probably be just a comment, but I am new here and don't yet have the ability to comment.)
• This is a commonly given exercise (I typically give it to my precalculus students every year), but I question its nature as a problem from daily life. Who actually does this? Don't most people first figure out what area they want to fence, then go to the hardware store and buy twice as much fencing as they actually need, just to be sure? – Xander Henderson Oct 1 '20 at 12:31
• @XanderHenderson I agree - still not a question people seem to often ask themselves. In fact I would often go to the hardware store and buy an extra 25% of fence to anticipate some kind of screw-up or other need. The main point of this is that a trivial change in the statement leads to a problem without an obvious answer. – John Mahoney Oct 12 '20 at 3:42
• @user1027 I like the mirror idea. In practice I often mix things up with more complicated house / wall constraints like building the garden onto the corner of the house. – John Mahoney Oct 12 '20 at 3:43 | 2021-05-10 05:43:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4590541422367096, "perplexity": 758.4426571434914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00202.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-science-5th-edition/chapter-14-acids-and-bases-questions-for-review-and-thought-topical-questions-page-652d/68a | # Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652d: 68a
This reation is reactant-favored. - $H_2O(l) + HNO_3(aq) \lt -- \gt H_3O^{+}(aq) + N{O_3}^-(aq)$ --- Other reactions: - $H_2O(l) + N{O_3}^-(aq) \lt -- \gt OH^-(aq) + HNO_3(aq)$** ** This reaction is very insignificant, because $N{O_3}^-$ is a very weak acid. - $2H_2O(l) \lt -- \gt H_3O^+(aq) + OH^-(aq)$
#### Work Step by Step
1. Identify the acids and bases: - $H_2O$ is amphiprotic. - $HNO_3$ is an acid. - Therefore, the only reaction that can occur is with $HNO_3$ as an acid. 2. Write the reaction - $H_2O(l) + HNO_3(aq) \lt -- \gt H_3O^{+}(aq) + N{O_3}^-(aq)$ 3. Identify the weaker acid (or base). $HNO_3$ and $H_3O^+$: $H_3O^+$ is weaker. Therefore, the side that has the $H_3O^+$ is favored; Other possible reaction? - $H_2O$ can act as an acid too: $H_2O(l) + N{O_3}^-(aq) \lt -- \gt OH^-(aq) + HNO_3(aq)$ $HNO_3$ can act only as an acid. $H_3O^+$ can act only as an acid. $N{O_3}^-$ can act only as a base. And we have the auto-ionization of water: $2H_2O(l) \lt -- \gt H_3O^+(aq) + OH^-(aq)$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2019-09-23 13:00:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6414989233016968, "perplexity": 1971.105362992419}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576965.71/warc/CC-MAIN-20190923125729-20190923151729-00096.warc.gz"} |
https://ideas.repec.org/a/jdm/journl/v4y2009i5p429-435.html | # On the relative importance of the hot stove effect and the tendency to rely on small samples
## Author
Listed:
• Takemi Fujikawa
## Abstract
Experiments have suggested that decisions from \textit{experience} differ from decisions from \textit{description}. In experience-based decisions, the decision makers often fail to maximise their payoffs. Previous authors have ascribed the effect of underweighting of rare outcomes to this deviation from maximisation. In this paper, I re-examine and provide further analysis on the effect with an experiment that involves a series of simple binary choice gambles. In the current experiment, decisions that bear small consequences are repeated hundreds of times, feedback on the consequence of each decision is provided immediately, and decision outcomes are accumulated. The participants have to learn about the outcome distributions through sampling, as they are not explicitly provided with prior information on the payoff structure. The current results suggest that the hot stove effect'' is stronger than suggested by previous research and is as important as the payoff variability effect and the effect of underweighting of rare outcomes in analysing decisions from experience in which the features of gambles must be learned through a sampling process.
## Suggested Citation
• Takemi Fujikawa, 2009. "On the relative importance of the hot stove effect and the tendency to rely on small samples," Judgment and Decision Making, Society for Judgment and Decision Making, vol. 4(5), pages 429-435, August.
• Handle: RePEc:jdm:journl:v:4:y:2009:i:5:p:429-435
as
File URL: http://journal.sjdm.org/9521/jdm9521.pdf
File URL: http://journal.sjdm.org/9521/jdm9521.html
## Citations
Citations are extracted by the CitEc Project, subscribe to its RSS feed for this item.
as
Cited by:
1. Fujikawa, Takemi, 2009. "The hot stove effect in repeated-play decision making under ambiguity," MPRA Paper 17647, University Library of Munich, Germany.
2. Kobayashi, Yohei & Fujikawa, Takemi, 2010. "An incomplete ignorance state in repeated-play decision making: A note on Bayesian decision-theoretical framework," MPRA Paper 28265, University Library of Munich, Germany.
3. repec:eee:eejocm:v:24:y:2017:i:c:p:22-35 is not listed on IDEAS
### Keywords
decisions from experience; payoff variability; rare events; uncertainty; undersampling.;
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IDEAS is a RePEc service hosted by the Research Division of the Federal Reserve Bank of St. Louis . RePEc uses bibliographic data supplied by the respective publishers. | 2018-05-24 04:52:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.361008882522583, "perplexity": 3147.764408977584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00540.warc.gz"} |
https://help.geogebra.org/topic/latex-text-on-geogebra-not-displayed-completely | # Latex - Text on Geogebra: not displayed completely
wschmid shared this problem 8 years ago
Dear Developer(s)!
Latex-Text, which you can even see in the preview (not just within the file) http://www.geogebratube.org/material/show/id/50153) is cut away / not completely displayed on the worksheet: http://www.geogebratube.org...
...
1
Does it work if you use \cr instead of \\ for the line-breaks?
1
Heureka!
With \cr ist works!
By the way - for some reason it didn't work when I made extra textboxes for each line!
... as one can see in this copy:
http://tube.geogebra.org/st...
1
\begin isn't supported. Please use TableText if you want to make tables | 2022-05-26 17:36:53 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.996790885925293, "perplexity": 3314.289471412891}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00212.warc.gz"} |
https://indico.cern.ch/event/313545/ | Collider Cross Talk
# Quarkonia Production in pPb collisions from ALICE-LHCb
## by Roberta Arnaldi (Universita e INFN (IT)), Michael Schmelling (Max-Planck-Gesellschaft (DE))
Thursday, 5 June 2014 from to (Europe/Zurich)
at CERN ( 4-2-011 - TH common room )
Description Measurements of charmonium and bottomonium production in pPb collisions with a nucleon-nucleon centre-of-mass energy of 5 TeV by the ALICE and LHCb collaborations are presented. The aim is to extract information about cold nuclear matter effects affecting signatures for the formation of a quark gluon plasma in heavy ion collisions, which requires as input the production cross-section for quarkonia in pp collisions at 5 TeV, where no data exist. Experimental results obtained for the various resonance states will be presented and issues related to the cross-section interpolation will be discussed. Finally, the comparison with theoretical predictions based on cold nuclear matter effects as shadowing, as well as on models including a contribution from partonic energy loss will also be addressed, together with the implications these pPb results may have for the interpretation of quarkonium behaviour in PbPb collisions. ALICE Results Common Results LHCb Results | 2015-11-26 03:17:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7869210839271545, "perplexity": 3046.1700009671695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398446300.49/warc/CC-MAIN-20151124205406-00176-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://chemistry.stackexchange.com/questions/117404/product-of-this-michael-addition | # Product of this Michael addition
I'm having problems when trying to know the outcome of this reaction.
The first step must be a Michael addition of the 1,3-dicarbonyl compound to the α,β-unsaturated carbonyl compound, followed by a acid hydrolisis and a decarboxylation of the ester.
LDA must deprotone the least hindered hydrogen, that must be the $$\alpha$$-$$\ce{CH3}$$ to the non cyclic ketone. But at this point, I don't know how this reaction can continue to give one of the above products.
• Intramolecular aldol reaction? – orthocresol Jun 27 at 16:58
• @user55119 So it was as easy as a Robinson annulation, but the LDA step was driving me crazy, it makes sense now, thanks. – Daniel Álvarez Jun 27 at 18:49
• I don't think you need LDA to finish an aldol condensation. The first step probably effect the Robinson annulation with the ester group intact. Acid hydrolysis would lead to CO2, methanol and compound C because the precursor is a vinylogous beta-ketoester. Probably a made-up problem. – – user55119 Jun 27 at 20:05
Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux. Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.
1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.
The OP's original question has some flaws. Since he is asking the most possible product among the given four, I'd give reasonable mechanism to achieve that product, which would be C. Following is the suggested mechanism:
Late addition: Evidently, lithium diisopropylamide (LDA) has been used to selectively abstract proton from the least steric position among three available acidic hydrogens (irreversible generation of the kinetic enolate). Such a non-nucleophilic, sterically-demanding, strong base will always abstract a proton from the least hindered side. The resultant lithium enolates will avoide Proton transfer at low temperatures in ethereal solvents, so that addition of a second carbonyl group will produce the desired aldol product (Organic Chemistry Portal). For an example for use of lithium dialkyl amide in aldol condensation, see the given reference:
Atsushi Seki, Fusae Ishiwata, Youichi Takizawa, Masatoshi Asami, "Crossed aldol reaction using cross-linked polymer-bound lithium dialkylamide," Tetrahedron 2004, 60(23), 5001-5011 (https://doi.org/10.1016/j.tet.2004.04.026).
• Mathew Mahindaratne: Do you have a reference for the use of LDA in the context of your example? If so, could you post it in your answer? Conceptually, it works! Practically, does it? I'm curious. PS: In your fourth structure, the keto form is undoubtedly better than the enol form. – user55119 Jun 28 at 11:47
• @user55119: I have added a paragraph and a reference for your request. – Mathew Mahindaratne Jun 28 at 21:49 | 2019-10-13 20:54:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5863339900970459, "perplexity": 6056.14518288161}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00330.warc.gz"} |
http://openstudy.com/updates/51947a86e4b0e9fd532a75e7 | ## anonymous 3 years ago Please Help! Will give medal!
1. anonymous
A particle moves on a straight line OX so that its acceleration at time t is given by $\alpha=6t-24$ also when t=0, s=0 and v=45. find its velocity and displacement when t=5 seconds
2. anonymous
just intigrate alpha
3. anonymous
a=dv/dt put dv/dt=6t-24 dv=6tdt-24dt
4. anonymous
find v
5. anonymous
again v=dx/dt find x=displacement.
6. anonymous
thanks :D
7. anonymous
welcome brother. | 2016-06-30 12:27:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6171299815177917, "perplexity": 14844.653967226956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783398628.62/warc/CC-MAIN-20160624154958-00034-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/gas-effusing-through-hole-working-out-time-dependence.998366/ | # Gas effusing through hole, working out time dependence
Homework Statement:
Consider instead a thermally insulated container of volume V with a
small hole of area A, containing a gas with molecular mass m. At time t = 0, the density is ##n_0## and temperature is ##T_0##. As gas effuses out through a small hole, both density and temperature inside the container will drop. Work out their time dependence, n(t) and T(t) in terms of the quantities given above.
Relevant Equations:
##N_L = \frac{1}{4} n A <u> = \frac{1}{4} \frac{N_0}{v} A \sqrt{\frac{8kT}{\pi m}}##
Consider instead a thermally insulated container of volume V with a
small hole of area A, containing a gas with molecular mass m. At time t = 0, the density is ##n_0## and temperature is ##T_0##. As gas effuses out through a small hole, both density and temperature inside the container will drop. Work out their time dependence, n(t) and T(t) in terms of the quantities given above.
I know that ##N_L = \frac{1}{4} n A <u> = \frac{1}{4} \frac{N_0}{v} A \sqrt{\frac{8kT}{\pi m}}##, but not sure how to use this to find n(t) and T(t). I think I need to find the flux of the energy to know if the temperature is decreasing and then find n and t from that but not sure exactly how to do this. Any help greatly appreciated. | 2021-09-26 20:03:24 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8567088842391968, "perplexity": 782.9063132864168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00087.warc.gz"} |
https://www.answers.com/Q/What_is_the_difference_between_a_vector_and_a_scalar_quantity | Physics
What is the difference between a vector and a scalar quantity?
09/12/2017
Vectors are quantities that have a size and a direction.
Examples: Displacement, velocity, acceleration, momentum, force.
Scalars are quantities that have a size but no direction.
Examples: Temperature, cost, speed, length, height, width, age, energy.
=====
Scalar quantities have only magnitude. Vector quantities have both magnitude and direction. Temperature and volume are scalar because they don't have a particular direction.
Velocity and Force (because acceleration actually has a direction) are vector quantities.
Velocity is the combination of the scalar quantity of speed with a direction for that speed. Speed is always a positive number but velocity can be negative and positive because it has a direction. | 2019-10-23 02:09:44 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8523879051208496, "perplexity": 752.7412387333846}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00175.warc.gz"} |
http://www.emis.de/classics/Erdos/cit/41804002.htm | ## Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 418.04002
Autor: Erdös, Paul
Title: Set theoretic, measure theoretic, combinatorial, and number theoretic problems concerning point sets in Euclidean space. (In English)
Source: Real Anal. Exch. 4(1978-79), 113-138 (1979).
Review: The author states several combinatorial statements putting in contrast the corresponding situations for the finite and the infinite case respectively and that in general the case of finite sets is more complicate than the one of transfinite sets. E.g. if S is any infinite set in Ek, then S contains an equinumerous set S1 such that all distances between points of S1 are distinct (cf. P.Erdös, Proc. Am. Math. Soc. 1, 127-141 (1950; Zbl 039.04902). On the other hand, let fk(n) be the largest integer such that any n-point-set S in Ek contains an fk(n)-subset S1 such that all distances of points of S1 are distinct. ''The exact determination of fk(n) seems hopoless...''. A plausible conjecture is f1(n) = (1+o(1))n ½, where g1(n) = max k such that any strictly increasing k-sequence a1 < a2 < ... < ak of natural numbers \leq n has all distinct differences aj-ai. The author conjectures that g1(n) = n ½+0(1). He conjectures that f1(n) = g1(n) = n ½+0(1). Let nk be the smallest integer such that fk(nk) = 3; e.g. n2 = 9; it is not known whether nk1/k ––> 1. Problem: Is there a constant C > 0 such that every set s\subset E2 of measure > C contains the vertices of a triangle area? Problem: Given a countable subset A of [0,1]; estimate the largest possible measure of a subset of [0,1] which does not contain a set similar to A.
Problem: A set S in an Euclidean space of finite dimension is said to be Ramsey if for every k in N there is an nk such that if Enk is decomposed into k disjoint sets Ai then S is contained in one of these sets A1,..., Ak; is every obtuse angled triangle Ramsey? Is regular pentagon Ramsey? Problem: Is there a set S of power c in Hilbert space such that every equinumerous subset S1 contains an equilateral triangle (resp. an infinite dimensional regular simplex)? Many other questions are discoussed in the present paper announcing that he matter will be extensively discussed along with other questions in a forthcoming book written jointly by the author and George Purdy.
Reviewer: \D.Kurepa
Classif.: * 04A05 Relations, functions
04A10 Ordinal and cardinal numbers; generalizations
04A20 Combinatorial set theory
05A17 Partitions of integres (combinatorics)
28A99 Classical measure theory
00A07 Problem books
Keywords: combinatorial problems in set theory; different distances; subsets of Euclidean space; partitions; Hilbert space; measures
Citations: Zbl.039.049
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag | 2015-10-04 12:53:22 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8109785914421082, "perplexity": 1578.1719465033455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736673632.3/warc/CC-MAIN-20151001215753-00003-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://labs.tib.eu/arxiv/?author=S.%20Paktinat%20Mehdiabadi | • A joint measurement is presented of the branching fractions $B^0_s\to\mu^+\mu^-$ and $B^0\to\mu^+\mu^-$ in proton-proton collisions at the LHC by the CMS and LHCb experiments. The data samples were collected in 2011 at a centre-of-mass energy of 7 TeV, and in 2012 at 8 TeV. The combined analysis produces the first observation of the $B^0_s\to\mu^+\mu^-$ decay, with a statistical significance exceeding six standard deviations, and the best measurement of its branching fraction so far. Furthermore, evidence for the $B^0\to\mu^+\mu^-$ decay is obtained with a statistical significance of three standard deviations. The branching fraction measurements are statistically compatible with SM predictions and impose stringent constraints on several theories beyond the SM.
• ### Top Production from Black Holes at the LHC(0803.1287)
April 30, 2008 hep-ph
In theories with large extra dimension and with low quantum gravity scale near a TeV, it is expected that TeV-scale black holes to be produced in proton-proton collisions at the LHC with the center of mass energy of 14 TeV. Since the black holes temperature can be around 1 TeV, top quark production is expected from them via Hawking radiation. Within the Standard Model of particle physics top quarks are produced via strong interaction in $t\bar{t}$ pairs or via electroweak interaction singly. Therefore, black holes can be the new source of top quark production. In this article we present the total cross sections and transverse momentum distributions of top quark production from black holes at the LHC. We find that the top quarks from black holes tend to reside at very high transverse momentum region so it can be a very useful signature for the black holes at the LHC. | 2021-02-27 13:19:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8354322910308838, "perplexity": 388.2879339741622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358956.39/warc/CC-MAIN-20210227114444-20210227144444-00621.warc.gz"} |
https://quantumfrontiers.com/ | # Love in the time of thermo
An 81-year-old medical doctor has fallen off a ladder in his house. His pet bird hopped out of his reach, from branch to branch of a tree on the patio. The doctor followed via ladder and slipped. His servants cluster around him, the clamor grows, and he longs for his wife to join him before he dies. She arrives at last. He gazes at her face; utters, “Only God knows how much I loved you”; and expires.
I set the book down on my lap and looked up. I was nestled in a wicker chair outside the Huntington Art Gallery in San Marino, California. Busts of long-dead Romans kept me company. The lawn in front of me unfurled below a sky that—unusually for San Marino—was partially obscured by clouds. My final summer at Caltech was unfurling. I’d walked to the Huntington, one weekend afternoon, with a novel from Caltech’s English library.1
What a novel.
You may have encountered the phrase “love in the time of corona.” Several times. Per week. Throughout the past six months. Love in the Time of Cholera predates the meme by 35 years. Nobel laureate Gabriel García Márquez captured the inhabitants, beliefs, architecture, mores, and spirit of a Colombian city around the turn of the 20th century. His work transcends its setting, spanning love, death, life, obsession, integrity, redemption, and eternity. A thermodynamicist couldn’t ask for more-fitting reading.
Love in the Time of Cholera centers on a love triangle. Fermina Daza, the only child of a wealthy man, excels in her studies. She holds herself with poise and self-assurance, and she spits fire whenever others try to control her. The girl dazzles Florentino Ariza, a poet, who restructures his life around his desire for her. Fermina Daza’s pride impresses Dr. Juvenal Urbino, a doctor renowned for exterminating a cholera epidemic. After rejecting both men, Fermina Daza marries Dr. Juvenal Urbino. The two personalities clash, and one betrays the other, but they cling together across the decades. Florentino Ariza retains his obsession with Fermina Daza, despite having countless affairs. Dr. Juvenal Urbino dies by ladder, whereupon Florentino Ariza swoops in to win Fermina Daza over. Throughout the book, characters mistake symptoms of love for symptoms of cholera; and lovers block out the world by claiming to have cholera and self-quarantining.
As a thermodynamicist, I see the second law of thermodynamics in every chapter. The second law implies that time marches only forward, order decays, and randomness scatters information to the wind. García Márquez depicts his characters aging, aging more, and aging more. Many characters die. Florentino Ariza’s mother loses her memory to dementia or Alzheimer’s disease. A pawnbroker, she buys jewels from the elite whose fortunes have eroded. Forgetting the jewels’ value one day, she mistakes them for candies and distributes them to children.
The second law bites most, to me, in the doctor’s final words, “Only God knows how much I loved you.” Later, the widow Fermina Daza sighs, “It is incredible how one can be happy for so many years in the midst of so many squabbles, so many problems, damn it, and not really know if it was love or not.” She doesn’t know how much her husband loved her, especially in light of the betrayal that rocked the couple and a rumor of another betrayal. Her husband could have affirmed his love with his dying breath, but he refused: He might have loved her with all his heart, and he might not have loved her; he kept the truth a secret to all but God. No one can retrieve the information after he dies.2
Love in the Time of Cholera—and thermodynamics—must sound like a mouthful of horseradish. But each offers nourishment, an appetizer and an entrée. According to the first law of thermodynamics, the amount of energy in every closed, isolated system remains constant: Physics preserves something. Florentino Ariza preserves his love for decades, despite Fermina Daza’s marrying another man, despite her aging.
The latter preservation can last only so long in the story: Florentino Ariza, being mortal, will die. He claims that his love will last “forever,” but he won’t last forever. At the end of the novel, he sails between two harbors—back and forth, back and forth—refusing to finish crossing a River Styx. I see this sailing as prethermalization: A few quantum systems resist thermalizing, or flowing to the physics analogue of death, for a while. But they succumb later. Florentino Ariza can’t evade the far bank forever, just as the second law of thermodynamics forbids his boat from functioning as a perpetuum mobile.
Though mortal within his story, Florentino Ariza survives as a book character. The book survives. García Márquez wrote about a country I’d never visited, and an era decades before my birth, 33 years before I checked his book out of the library. But the book dazzled me. It pulsed with the vibrancy, color, emotion, and intellect—with the fullness—of life. The book gained another life when the coronavius hit. Thermodynamics dictates that people age and die, but the laws of thermodynamics remain.3 I hope and trust—with the caveat about humanity’s not destroying itself—that Love in the Time of Cholera will pulse in 350 years.
What’s not to love?
1Yes, Caltech has an English library. I found gems in it, and the librarians ordered more when I inquired about books they didn’t have. I commend it to everyone who has access.
2I googled “Only God knows how much I loved you” and was startled to see the line depicted as a hallmark of romance. Please tell your romantic partners how much you love them; don’t make them guess till the ends of their lives.
3Lee Smolin has proposed that the laws of physics change. If they do, the change seems to have to obey metalaws that remain constant.
# If the (quantum-metrology) key fits…
My maternal grandfather gave me an antique key when I was in middle school. I loved the workmanship: The handle consisted of intertwined loops. I loved the key’s gold color and how the key weighed on my palm. Even more, I loved the thought that the key opened something. I accompanied my mother to antique shops, where I tried unlocking chests, boxes, and drawers.
My grandfather’s antique key
I found myself holding another such key, metaphorically, during the autumn of 2018. MIT’s string theorists had requested a seminar, so I presented about quasiprobabilities. Quasiprobabilities represent quantum states similarly to how probabilities represent a swarm of classical particles. Consider the steam rising from asphalt on a summer day. Calculating every steam particle’s position and momentum would require too much computation for you or me to perform. But we can predict the probability that, if we measure every particle’s position and momentum, we’ll obtain such-and-such outcomes. Probabilities are real numbers between zero and one. Quasiprobabilities can assume negative and nonreal values. We call these values “nonclassical,” because they’re verboten to the probabilities that describe classical systems, such as steam. I’d defined a quasiprobability, with collaborators, to describe quantum chaos.
David Arvidsson-Shukur was sitting in the audience. David is a postdoctoral fellow at the University of Cambridge and a visiting scholar in the other Cambridge (at MIT). He has a Swedish-and-southern-English accent that I’ve heard only once before and, I learned over the next two years, an academic intensity matched by his kindliness.1 Also, David has a name even longer than mine: David Roland Miran Arvidsson-Shukur. We didn’t know then, but we were destined to journey together, as postdoctoral knights-errant, on a quest for quantum truth.
David studies the foundations of quantum theory: What distinguishes quantum theory from classical? David suspected that a variation on my quasiprobability could unlock a problem in metrology, the study of measurements.
Suppose that you’ve built a quantum computer. It consists of gates—uses of, e.g., magnets or lasers to implement logical operations. A classical gate implements operations such as “add 11.” A quantum gate can implement an operation that involves some number $\theta$ more general than 11. You can try to build your gate correctly, but it might effect the wrong $\theta$ value. You need to measure $\theta$.
How? You prepare some quantum state $| \psi \rangle$ and operate on it with the gate. $\theta$ imprints itself on the state, which becomes $| \psi (\theta) \rangle$. Measure some observable $\hat{O}$. You repeat this protocol in each of many trials. The measurement yields different outcomes in different trials, according to quantum theory. The average amount of information that you learn about $\theta$ per trial is called the Fisher information.
Let’s change this protocol. After operating with the gate, measure another observable, $\hat{F}$, and postselect: If the $\hat{F}$ measurement yields a desirable outcome $f$, measure $\hat{O}$. If the $\hat{F}$-measurement doesn’t yield the desirable outcome, abort the trial, and begin again. If you choose $\hat{F}$ and $f$ adroitly, you’ll measure $\hat{O}$ only when the trial will provide oodles of information about $\theta$. You’ll save yourself many $\hat{O}$ measurements that would have benefited you little.2
Why does postselection help us? We could understand easily if the system were classical: The postselection would effectively improve the input state. To illustrate, let’s suppose that (i) a magnetic field implemented the gate and (ii) the input were metal or rubber. The magnetic field wouldn’t affect the rubber; measuring $\hat{O}$ in rubber trials would provide no information about the field. So you could spare yourself $\hat{O}$ measurements by postselecting on the system’s consisting of metal.
Postselection on a quantum system can defy this explanation. Consider optimizing your input state $| \psi \rangle$, beginning each trial with the quantum equivalent of metal. Postselection could still increase the average amount of information information provided about $\theta$ per trial. Postselection can enhance quantum metrology even when postselection can’t enhance the classical analogue.
David suspected that he could prove this result, using, as a mathematical tool, the quasiprobability that collaborators and I had defined. We fulfilled his prediction, with Hugo Lepage, Aleks Lasek, Seth Lloyd, and Crispin Barnes. Nature Communications published our paper last month. The work bridges the foundations of quantum theory with quantum metrology and quantum information theory—and, through that quasiprobability, string theory. David’s and my quantum quest continues, so keep an eye out for more theory from us, as well as a photonic experiment based on our first paper.
I still have my grandfather’s antique key. I never found a drawer, chest, or box that it opened. But I don’t mind. I have other mysteries to help unlock.
1The morning after my wedding this June, my husband and I found a bouquet ordered by David on our doorstep.
2Postselection has a catch: The $\hat{F}$ measurement has a tiny probability of yielding the desirable outcome. But, sometimes, measuring $\hat{O}$ costs more than preparing $| \psi \rangle$, performing the gate, and postselecting. For example, suppose that the system is a photon. A photodetector will measure $\hat{O}$. Some photodetectors have a dead time: After firing, they take a while to reset, to be able to fire again. The dead time can outweigh the cost of the beginning of the experiment.
# A quantum walk down memory lane
In elementary and middle school, I felt an affinity for the class three years above mine. Five of my peers had siblings in that year. I carpooled with a student in that class, which partnered with mine in holiday activities and Grandparents’ Day revues. Two students in that class stood out. They won academic-achievement awards, represented our school in science fairs and speech competitions, and enrolled in rigorous high-school programs.
Those students came to mind as I grew to know David Limmer. David is an assistant professor of chemistry at the University of California, Berkeley. He studies statistical mechanics far from equilibrium, using information theory. Though a theorist ardent about mathematics, he partners with experimentalists. He can pass as a physicist and keeps an eye on topics as far afield as black holes. According to his faculty page, I discovered while writing this article, he’s even three years older than I.
I met David in the final year of my PhD. I was looking ahead to postdocking, as his postdoc fellowship was fading into memory. The more we talked, the more I thought, I’d like to be like him.
I had the good fortune to collaborate with David on a paper published by Physical Review A this spring (as an Editors’ Suggestion!). The project has featured in Quantum Frontiers as the inspiration for a rewriting of “I’m a little teapot.”
We studied a molecule prevalent across nature and technologies. Such molecules feature in your eyes, solar-fuel-storage devices, and more. The molecule has two clumps of atoms. One clump may rotate relative to the other if the molecule absorbs light. The rotation switches the molecule from a “closed” configuration to an “open” configuration.
These molecular switches are small, quantum, and far from equilibrium; so modeling them is difficult. Making assumptions offers traction, but many of the assumptions disagreed with David. He wanted general, thermodynamic-style bounds on the probability that one of these molecular switches would switch. Then, he ran into me.
I traffic in mathematical models, developed in quantum information theory, called resource theories. We use resource theories to calculate which states can transform into which in thermodynamics, as a dime can transform into ten pennies at a bank. David and I modeled his molecule in a resource theory, then bounded the molecule’s probability of switching from “closed” to “open.” I accidentally composed a theme song for the molecule; you can sing along with this post.
That post didn’t mention what David and I discovered about quantum clocks. But what better backdrop for a mental trip to elementary school or to three years into the future?
I’ve blogged about autonomous quantum clocks (and ancient Assyria) before. Autonomous quantum clocks differ from quantum clocks of another type—the most precise clocks in the world. Scientists operate the latter clocks with lasers; autonomous quantum clocks need no operators. Autonomy benefits you if you want for a machine, such as a computer or a drone, to operate independently. An autonomous clock in the machine ensures that, say, the computer applies the right logical gate at the right time.
What’s an autonomous quantum clock? First, what’s a clock? A clock has a degree of freedom (e.g., a pair of hands) that represents the time and that moves steadily. When the clock’s hands point to 12 PM, you’re preparing lunch; when the clock’s hands point to 6 PM, you’re reading Quantum Frontiers. An autonomous quantum clock has a degree of freedom that represents the time fairly accurately and moves fairly steadily. (The quantum uncertainty principle prevents a perfect quantum clock from existing.)
Suppose that the autonomous quantum clock constitutes one part of a machine, such as a quantum computer, that the clock guides. When the clock is in one quantum state, the rest of the machine undergoes one operation, such as one quantum logical gate. (Experts: The rest of the machine evolves under one Hamiltonian.) When the clock is in another state, the rest of the machine undergoes another operation (evolves under another Hamiltonian).
Physicists have been modeling quantum clocks using the resource theory with which David and I modeled our molecule. The math with which we represented our molecule, I realized, coincided with the math that represents an autonomous quantum clock.
Think of the molecular switch as a machine that operates (mostly) independently and that contains an autonomous quantum clock. The rotating clump of atoms constitutes the clock hand. As a hand rotates down a clock face, so do the nuclei rotate downward. The hand effectively points to 12 PM when the switch occupies its “closed” position. The hand effectively points to 6 PM when the switch occupies its “open” position.
The nuclei account for most of the molecule’s weight; electrons account for little. They flit about the landscape shaped by the atomic clumps’ positions. The landscape governs the electrons’ behavior. So the electrons form the rest of the quantum machine controlled by the nuclear clock.
Experimentalists can create and manipulate these molecular switches easily. For instance, experimentalists can set the atomic clump moving—can “wind up” the clock—with ultrafast lasers. In contrast, the only other autonomous quantum clocks that I’d read about live in theory land. Can these molecules bridge theory to experiment? Reach out if you have ideas!
And check out David’s theory lab on Berkeley’s website and on Twitter. We all need older siblings to look up to.
# What can you do in 48 hours?
Have you ever wondered what can be done in 48 hours? For instance, our heart beats around 200 000 times. One of the biggest supercomputers crunches petabytes (peta = 1015) of numbers to simulate an experiment that took Google’s quantum processor only 300 seconds to run. In 48 hours, one can also participate in the Sciathon with almost 500 young researchers from more than 80 countries!
Two weeks ago I participated in a scientific marathon, the Sciathon. The structure of this event roughly resembled a hackathon. I am sure many readers are familiar with the idea of a hackathon from personal experience. For those unfamiliar — a hackathon is an intense collaborative event, usually organized over the weekend, during which people with different backgrounds work in groups to create prototypes of functioning software or hardware. For me, it was the very first time to have firsthand experience with a hackathon-like event!
The Sciathon was organized by the Lindau Nobel Laureate Meetings (more about the meetings with Nobel laureates, which happen annually in the lovely German town of Lindau, in another blogpost, I promise!) This year, unfortunately, the face-to-face meeting in Lindau was postponed until the summer of 2021. Instead, the Lindau Nobel Laureate Meetings alumni and this year’s would-be attendees had an opportunity to gather for the Sciathon, as well as the Online Science Days earlier this week, during which the best Sciathon projects were presented.
The participants of the Sciathon could choose to contribute new views, perspectives and solutions to three main topics: Lindau Guidelines, Communicating Climate Change and Capitalism After Corona. The first topic concerned an open, cooperative science community where data and knowledge are freely shared, the second — how scientists could show that the climate crisis is just as big a threat as the SARS-CoV-19 virus, and the last — how to remodel our current economic systems so that they are more robust to unexpected sudden crises. More detailed descriptions of each topic can be found on the official Sciathon webpage.
My group of ten eager scientists, mostly physicists, from master students to postdoctoral researchers, focused on the first topic. In particular, our goal was to develop a method of familiarizing high school students with the basics of quantum information and computation. We envisioned creating an online notebook, where an engaging story would be intertwined with interactive blocks of Python code utilizing the open-source quantum computing toolkit Qiskit. This hands-on approach would enable students to play with quantum systems described in the story-line by simply running the pre-programmed commands with a click of the mouse and then observe how “experiment” matches “the theory”. We decided to work with a system comprising one or two qubits and explain such fundamental concepts in quantum physics as superposition, entanglement and measurement. The last missing part was a captivating story.
The story we came up with involved two good friends from the lab, Miss Schrödinger and Miss Pauli, as well as their kittens, Alice and Bob. At first, Alice and Bob seemed to be ordinary cats, however whenever they sipped quantum milk, they would turn into quantum cats, or as quantum physicists would say — kets. Do I have to remind the reader that a quantum cat, unlike an ordinary one, could be both awake and asleep at the same time?
Miss Schrödinger was a proud cat owner who not only loved her cat, but also would take hundreds of pictures of Alice and eagerly upload them on social media. Much to Miss Schrödinger’s surprise, none of the pictures showed Alice partly awake and partly asleep — the ket would always collapse to the cat awake or the cat asleep! Every now and then, Miss Pauli would come to visit Miss Schrödinger and bring her own cat Bob. While the good friends were chit-chatting over a cup of afternoon tea, the cats sipped a bit of quantum milk and started to play with a ball of wool, resulting in a cute mess of two kittens tangled up in wool. Every time after coming back home, Miss Pauli would take a picture of Bob and share it with Miss Schrödinger, who would obviously also take a picture of Alice. After a while, the young scientists started to notice some strange correlations between the states of their cats…
The adventures of Miss Schrödinger and her cat continue! For those interested, you can watch a short video about our project!
Overall, I can say that I had a lot of fun participating in the Sciathon. It was an intense yet extremely gratifying event. In addition to the obvious difficulty of racing against the clock, our group also had to struggle with coordinating video calls between group members scattered across three almost equidistant time zones — Eastern Australian, Central European and Central US! During the Sciathon I had a chance to interact with other science enthusiasts from different backgrounds and work on something from outside my area of expertise. I would strongly encourage anyone to participate in hackathon-like events to break the daily routine, particularly monotonous during the lockdown, and unleash one’s creative spirit. Such events can also be viewed as an opportunity to communicate science and scientific progress to the public. Lastly, I would like to thank other members of my team — collaborating with you during the Sciathon was a blast!
# Eleven risks of marrying a quantum information scientist
Some of you may have wondered whether I have a life. I do. He’s a computer scientist, and we got married earlier this month.
Marrying a quantum information scientist comes with dangers not advertised in any Brides magazine (I assume; I’ve never opened a copy of Brides magazine). Never mind the perils of gathering together Auntie So-and-so and Cousin Such-and-such, who’ve quarreled since you were six; or spending tens of thousands of dollars on one day; or assembling two handfuls of humans during a pandemic. Beware the risks of marrying someone who unconsciously types “entropy” when trying to type “entry,” twice in a row.
1) She’ll introduce you to friends as “a classical computer scientist.” They’d assume, otherwise, that he does quantum computer science. Of course. Wouldn’t you?
2) The quantum punning will commence months before the wedding. One colleague wrote, “Many congratulations! Now you know the true meaning of entanglement.” Quantum particles can share entanglement. If you measure entangled particles, your outcomes can exhibit correlations stronger than any produceable by classical particles. As a card from another colleague read, “May you stay forever entangled, with no decoherence.”
I’d rather not dedicate much of a wedding article to decoherence, but suppose that two particles are maximally entangled (can generate the strongest correlations possible). Suppose that particle 2 heats up or suffers bombardment by other particles. The state of particle 2 decoheres as the entanglement between 1 and 2 frays. Equivalently, particle 2 entangles with its environment, and particle 2 can entangle only so much: The more entanglement 2 shares with the environment, the less entanglement 2 can share with 1. Physicists call entanglement—ba-duh-bummonogamous.
The matron-of-honor toast featured another entanglement joke, as well as five more physics puns.1 (She isn’t a scientist, but she did her research.) She’ll be on Zoom till Thursday; try the virtual veal.
3) When you ask what sort of engagement ring she’d like, she’ll mention black diamonds. Experimentalists and engineers are building quantum computers from systems of many types, including diamond. Diamond consists of carbon atoms arranged in a lattice. Imagine expelling two neighboring carbon atoms and replacing one with a nitrogen atom. You’ll create a nitrogen-vacancy center whose electrons you can control with light. Such centers color the diamond black but let you process quantum information.
If I’d asked my fiancé for a quantum computer, we’d have had to wait 20 years to marry. He gave me an heirloom stone instead.
4) When a wedding-gown shopkeeper asks which sort of train she’d prefer, she’ll inquire about Maglevs. I dislike shopping, as the best man knows better than most people. In middle school, while our classmates spent their weekends at the mall, we stayed home and read books. But I filled out gown shops’ questionnaires.
“They want to know what kinds of material I like,” I told the best man over the phone, “and what styles, and what type of train. I had to pick from four types of train. I didn’t even know there were four types of train!”
“Steam?” guessed the best man. “Diesel?”
His suggestions appealed to me as a quantum thermodynamicist. Thermodynamics is the physics of energy, which engines process. Quantum thermodynamicists study how quantum phenomena, such as entanglement, can improve engines.
“Get the Maglev train,” the best man added. “Low emissions.”
“Ooh,” I said, “that’s superconducting.” Superconductors are quantum systems in which charge can flow forever, without dissipating. Labs at Yale, at IBM, and elsewhere are building quantum computers from superconductors. A superconductor consists of electrons that pair up with help from their positively charged surroundings—Cooper pairs. Separating Cooper-paired electrons requires an enormous amount of energy. What other type of train would better suit a wedding?
I set down my phone more at ease. Later, pandemic-era business closures constrained me to wearing a knee-length dress that I’d worn at graduations. I didn’t mind dodging the train.
5) When you ask what style of wedding dress she’ll wear, she’ll say that she likes her clothing as she likes her equations. Elegant in their simplicity.
6) You’ll plan your wedding for wedding season only because the rest of the year conflicts with more seminars, conferences, and colloquia. The quantum-information-theory conference of the year takes place in January. We wanted to visit Australia in late summer, and Germany in autumn, for conferences. A quantum-thermodynamics conference takes place early in the spring, and the academic year ends in May. Happy is the June bride; happier is the June bride who isn’t preparing a talk.
7) An MIT chaplain will marry you. Who else would sanctify the union of a physicist and a computer scientist?
8) You’ll acquire more in-laws than you bargained for. Biological parents more than suffice for most spouses. My husband has to contend with academic in-laws, as my PhD supervisor is called my “academic father.”
Academic in-laws of my husband’s attending the wedding via Zoom.
9) Your wedding can double as a conference. Had our wedding taken place in person, collaborations would have flourished during the cocktail hour. Papers would have followed; their acknowledgements sections would have nodded at the wedding; and I’d have requested copies of all manuscripts for our records—which might have included our wedding album.
10) You’ll have trouble identifying a honeymoon destination where she won’t be tempted to give a seminar. I thought that my then-fiancé would enjoy Vienna, but it boasts a quantum institute. So do Innsbruck and Delft. A colleague-friend works in Budapest, and I owe Berlin a professional visit. The list grew—or, rather, our options shrank. But he turned out not to mind my giving a seminar. The pandemic then cancelled our trip, so we’ll stay abroad for a week after some postpandemic European conference (hint hint).
11) Your wedding will feature on the blog of Caltech’s Institute for Quantum Information and Matter. Never mind The New York Times. Where else would you expect to find a quantum information physicist? I feel fortunate to have found someone with whom I wouldn’t rather be anywhere else.
1“I know that if Nicole picked him to stand by her side, he must be a FEYNMAN and not a BOZON.”
# Up we go! or From abstract theory to experimental proposal
Mr. Mole is trapped indoors, alone. Spring is awakening outside, but he’s confined to his burrow. Birds are twittering, and rabbits are chattering, but he has only himself for company.
Sound familiar?
Spring—crocuses, daffodils, and hyacinths budding; leaves unfurling; and birds warbling—burst upon Cambridge, Massachusetts last month. The city’s shutdown vied with the season’s vivaciousness. I relieved the tension by rereading The Wind in the Willows, which I’ve read every spring since 2017.
Project Gutenberg offers free access to Kenneth Grahame’s 1908 novel. He wrote the book for children, but never mind that. Many masterpieces of literature happen to have been written for children.
One line in the novel demanded, last year, that I memorize it. On page one, Mole is cleaning his house beneath the Earth’s surface. He’s been dusting and whitewashing for hours when the spring calls to him. Life is pulsating on the ground and in the air above him, and he can’t resist joining the party. Mole throws down his cleaning supplies and tunnels upward through the soil: “he scraped and scratched and scrabbled and scrooged, and then he scrooged again and scrabbled and scratched and scraped.”
The quotation appealed to me not only because of its alliteration and chiasmus. Mole’s journey reminded me of research.
Take a paper that I published last month with Michael Beverland of Microsoft Research and Amir Kalev of the Joint Center for Quantum Information and Computer Science (now of the Information Sciences Institute at the University of Southern California). We translated a discovery from the abstract, mathematical language of quantum-information-theoretic thermodynamics into an experimental proposal. We had to scrabble, but we kept on scrooging.
Over four years ago, other collaborators and I uncovered a thermodynamics problem, as did two other groups at the same time. Thermodynamicists often consider small systems that interact with large environments, like a magnolia flower releasing its perfume into the air. The two systems—magnolia flower and air—exchange things, such as energy and scent particles. The total amount of energy in the flower and the air remains constant, as does the total number of perfume particles. So we call the energy and the perfume-particle number conserved quantities.
We represent quantum conserved quantities with matrices $Q_1$ and $Q_2$. We nearly always assume that, in this thermodynamic problem, those matrices commute with each other: $Q_1 Q_2 = Q_2 Q_1$. Almost no one mentions this assumption; we make it without realizing. Eliminating this assumption invalidates a derivation of the state reached by the small system after a long time. But why assume that the matrices commute? Noncommutation typifies quantum physics and underlies quantum error correction and quantum cryptography.
What if the little system exchanges with the large system thermodynamic quantities represented by matrices that don’t commute with each other?
Colleagues and I began answering this question, four years ago. The small system, we argued, thermalizes to near a quantum state that contains noncommuting matrices. We termed that state, $e^{ - \sum_\alpha \beta_\alpha Q_\alpha } / Z$, the non-Abelian thermal state. The $Q_\alpha$’s represent conserved quantities, and the $\beta_\alpha$’s resemble temperatures. The real number $Z$ ensures that, if you measure any property of the state, you’ll obtain some outcome. Our arguments relied on abstract mathematics, resource theories, and more quantum information theory.
Over the past four years, noncommuting conserved quantities have propagated across quantum-information-theoretic thermodynamics.1 Watching the idea take root has been exhilarating, but the quantum information theory didn’t satisfy me. I wanted to see a real physical system thermalize to near the non-Abelian thermal state.
Michael and Amir joined the mission to propose an experiment. We kept nosing toward a solution, then dislodging a rock that would shower dirt on us and block our path. But we scrabbled onward.
Imagine a line of ions trapped by lasers. Each ion contains the physical manifestation of a qubit—a quantum two-level system, the basic unit of quantum information. You can think of a qubit as having a quantum analogue of angular momentum, called spin. The spin has three components, one per direction of space. These spin components are represented by matrices $Q_x = S_x$, $Q_y = S_y$, and $Q_z = S_z$ that don’t commute with each other.
A couple of qubits can form the small system, analogous to the magnolia flower. The rest of the qubits form the large system, analogous to the air. I constructed a Hamiltonian—a matrix that dictates how the qubits evolve—that transfers quanta of all the spin’s components between the small system and the large. (Experts: The Heisenberg Hamiltonian transfers quanta of all the spin components between two qubits while conserving $S_{x, y, z}^{\rm tot}$.)
The Hamiltonian led to our first scrape: I constructed an integrable Hamiltonian, by accident. Integrable Hamiltonians can’t thermalize systems. A system thermalizes by losing information about its initial conditions, evolving to a state with an exponential form, such as $e^{ - \sum_\alpha \beta_\alpha Q_\alpha } / Z$. We clawed at the dirt and uncovered a solution: My Hamiltonian coupled together nearest-neighbor qubits. If the Hamiltonian coupled also next-nearest-neighbor qubits, or if the ions formed a 2D or 3D array, the Hamiltonian would be nonintegrable.
We had to scratch at every stage—while formulating the setup, preparation procedure, evolution, measurement, and prediction. But we managed; Physical Review E published our paper last month. We showed how a quantum system can evolve to the non-Abelian thermal state. Trapped ions, ultracold atoms, and quantum dots can realize our experimental proposal. We imported noncommuting conserved quantities in thermodynamics from quantum information theory to condensed matter and atomic, molecular, and optical physics.
As Grahame wrote, the Mole kept “working busily with his little paws and muttering to himself, ‘Up we go! Up we go!’ till at last, pop! his snout came out into the sunlight and he found himself rolling in the warm grass of a great meadow.”
1See our latest paper’s introduction for references. https://journals.aps.org/pre/abstract/10.1103/PhysRevE.101.042117
# Quantum steampunk invades Scientific American
London, at an hour that made Rosalind glad she’d nicked her brother’s black cloak instead of wearing her scarlet one. The factory alongside her had quit belching smoke for the night, but it would start again soon. A noise caused her to draw back against the brick wall. Glancing up, she gasped. An oblong hulk was drifting across the sky. The darkness obscured the details, but she didn’t need to see; a brass-colored lock would be painted across the side. Mellator had launched his dirigible.
A variation on the paragraph above began the article that I sent to Scientific American last year. Clara Moskowitz, an editor, asked which novel I’d quoted the paragraph from. I’d made the text up, I confessed.
Most of my publications, which wind up in physics journals, don’t read like novels. But I couldn’t resist when Clara invited me to write a feature about quantum steampunk, the confluence of quantum information and thermodynamics. Quantum Frontiers regulars will anticipate paragraphs two and three of the article:
Welcome to steampunk. This genre has expanded across literature, art and film over the past several decades. Its stories tend to take place near nascent factories and in grimy cities, in Industrial Age England and the Wild West—in real-life settings where technologies were burgeoning. Yet steampunk characters extend these inventions into futuristic technologies, including automata and time machines. The juxtaposition of old and new creates an atmosphere of romanticism and adventure. Little wonder that steampunk fans buy top hats and petticoats, adorn themselves in brass and glass, and flock to steampunk conventions.
These fans dream the adventure. But physicists today who work at the intersection of three fields—quantum physics, information theory and thermodynamics—live it. Just as steampunk blends science-fiction technology with Victorian style, a modern field of physics that I call “quantum steampunk” unites 21st-century technology with 19th-century scientific principles.
The Scientific American graphics team dazzled me. For years, I’ve been hankering to work with artists on visualizing quantum steampunk. I had an opportunity after describing an example of quantum steampunk in the article. The example consists of a quantum many-body engine that I designed with members Christopher White, Sarang Gopalakrishnan, and Gil Refael of Caltech’s Institute for Quantum Information and Matter. Our engine is a many-particle system ratcheted between two phases accessible to quantum matter, analogous to liquid and solid. The engine can be realized with, e.g., ultracold atoms or trapped ions. Lasers would trap and control the particles. Clara, the artists, and I drew the engine, traded comments, and revised the figure tens of times. In early drafts, the lasers resembled the sketches in atomic physicists’ Powerpoints. Before the final draft, the lasers transformed into brass-and-glass beauties. They evoke the scientific instruments crafted through the early 1900s, before chunky gray aesthetics dulled labs.
Scientific American published the feature this month; you can read it in print or, here, online. Many thanks to Clara for the invitation, for shepherding the article into print, and for her enthusiasm. To repurpose the end of the article, “You’re reading about this confluence of old and new on Quantum Frontiers. But you might as well be holding a novel by H. G. Wells or Jules Verne.”
Figures courtesy of the Scientific American graphics team.
# A new possibility for quantum networks
It has been roughly 1 year since Dr Jon Kindem and I finished at Caltech (JK graduating with a PhD and myself – JB – graduating from my postdoc to take up a junior faculty position at the University of Sydney). During our three-and-a-half-year overlap in the IQIM we often told each other that we should write something for Quantum Frontiers. As two of the authors of a paper reporting a recent breakthrough for rare-earth ion spin qubits (Nature, 2020), it was now or never. Here we go…
Throughout 2019, telecommunication companies began deploying 5th generation (5G) network infrastructure to allow our wireless communication to be faster, more reliable, and cope with greater capacity. This roll out of 5G technology promises to support up to 10x the number of devices operating with speeds 10x faster than what is possible with 4th generation (4G) networks. If you stop and think about new opportunities 4G networks unlocked for working, shopping, connecting, and more, it is easy to see why some people are excited about the new world 5G networks might offer.
Classical networks like 5G and fiber optic networks (the backbone of the internet) share classical information: streams of bits (zeros and ones) that encode our conversations, tweets, music, podcasts, videos and anything else we communicate through our digital devices. Every improvement in the network hardware (for example an optical switch with less loss or a faster signal router) contributes to big changes in speed and capacity. The bottom line is that with enough advances, the network evolves to the point where things that were previously impossible (like downloading a movie in the late 90s) become instantaneous.
Alongside the hype and advertising around 5G networks, we are part of the world-wide effort to develop a fundamentally different network (with a little less advertising, but similar amounts of hype). Rather than being a bigger, better version of 5G, this new network is trying to build a quantum internet: a set of technologies that will allows us to connect and share information at the quantum level. For an insight into the quantum internet origin story, read this post about the pioneering experiments that took place at Caltech in Prof. Jeff Kimble’s group.
Quantum technologies operate using the counter-intuitive phenomena of quantum mechanics like superposition and entanglement. Quantum networks need to distribute this superposition and entanglement between different locations. This is a much harder task than distributing bits in a regular network because quantum information is extremely susceptible to loss and noise. If realized, this quantum internet could enable powerful quantum computing clusters, and create networks of quantum sensors that measure infinitesimally small fluctuations in their environment.
At this point it is worth asking the question:
Does the world really need a quantum internet?
This is an important question because a quantum internet is unlikely to improve any of the most common uses for the classical internet (internet facts and most popular searches).
We think there are at least three reasons why a quantum network is important:
1. To build better quantum computers. The quantum internet will effectively transform small, isolated quantum processors into one much larger, more powerful computer. This could be a big boost in the race to scale-up quantum computing.
2. To build quantum-encrypted communication networks. The ability of quantum technology to make or break encryption is one of the earliest reasons why quantum technology was funded. A fully-fledged quantum computer should be very efficient at hacking commonly used encryption protocols, while ideal quantum encryption provides the basis for communications secured by the fundamental properties of physics.
3. To push the boundaries of quantum physics and measurement sensitivity by increasing the length scale and complexity of entangled systems. The quantum internet can help turn thought experiments into real experiments.
The next question is: How do we build a quantum internet?
The starting point for most long-distance quantum network strategies is to base them on the state-of-the-art technology for current classical networks: sending information using light. (But that doesn’t rule out microwave networks for local area networks, as recent work from ETH Zurich has shown).
The technology that drives quantum networks is a set of interfaces that connect matter systems (like atoms) to photons at a quantum level. These interfaces need to efficiently exchange quantum information between matter and light, and the matter part needs to be able to store the information for a time that is much longer than the time it takes for the light to get to its destination in the network. We also need to be able to entangle the quantum matter systems to connect network links, and to process quantum information for error correction. This is a significant challenge that requires novel materials and unparalleled control of light to ultimately succeed.
State-of-the-art quantum networks are still elementary links compared to the complexity and scale of modern telecommunication. One of the most advanced platforms that has demonstrated a quantum network link consists of two atomic defects in diamonds separated by 1.3 km. The defects act as the quantum light-matter interface allowing quantum information to be shared between the two remote devices. But these defects in diamond currently have limitations that prohibit the expansion of such a network. The central challenge is finding defects/emitters that are stable and robust to environmental fluctuations, while simultaneously efficiently connecting with light. While these emitters don’t have to be in solids, the allure of a scalable solid-state fabrication process akin to today’s semiconductor industry for integrated circuits is very appealing. This has motivated the research and development of a range of quantum light-matter interfaces in solids (for example, see recent work by Harvard researchers) with the goal of meeting the simultaneous goals of efficiency and stability.
The research group we were a part of at Caltech was Prof. Andrei Faraon’s group, which put forward an appealing alternative to other solid-state technologies. The team uses rare-earth atoms embedded in crystals commonly used for lasers. JK joined as the group’s 3rd graduate student in 2013, while I joined as a postdoc in 2016.
Rare-earth atoms have long been of interest for quantum technologies such as quantum memories for light because they are very stable and are excellent at preserving quantum information. But compared to other emitters, they only interact very weakly with light, which means that one usually needs large crystals with billions of atoms all working in harmony to make useful quantum interfaces. To overcome this problem, research in the Faraon group pioneered coupling these ions to nanoscale optical cavities like these ones:
These microscopic Toblerone-like structures are fabricated directly in the crystal that plays host to the rare-earth atoms. The periodic patterning effectively acts like two mirrors that form an optical cavity to confine light, which enhances the connection between light and the rare-earth atoms. In 2017, our group showed that the improved optical interaction in these cavities can be used to shrink down optical quantum memories by orders of magnitude compared to previous demonstrations, and ones manufactured on-chip.
We have used this nanophotonic platform to open up new avenues for quantum networks based on single rare-earth atoms, a task that previously was exceptionally challenging because these atoms have very low brightness. We have worked with both neodymium and ytterbium atoms embedded in a commercially available laser crystal.
Ytterbium looks particularly promising. Working with Prof. Rufus Cone’s group at Montana State University, we showed that these ytterbium atoms absorb and emit light better than most other rare-earth atoms and that they can store quantum information long enough for extended networks (>10 ms) when cooled down to a few Kelvin (-272 degrees Celsius) [Kindem et al., Physical Review B, 98, 024404 (2018) – link to arXiv version].
By using the nanocavity to improve the brightness of these ytterbium atoms, we have now been able to identify and investigate their properties at the single atom level. We can precisely control the quantum state of the single atoms and measure them with high fidelity – both prerequisites for using these atoms in quantum information technologies. When combined with the long quantum information storage times, our work demonstrates important steps to using this system in a quantum network.
The next milestone is forming an optical link between two individual rare-earth atoms to build an elementary quantum network. This goal is in our sights and we are already working on optimizing the light-matter interface stability and efficiency. A more ambitious milestone is to provide interconnects for other types of qubits – such as superconducting qubits – to join the network. This requires a quantum transducer to convert between microwave signals and light. Rare-earth atoms are promising for transducer technologies (see recent work from the Faraon group), as are a number of other hybrid quantum systems (for example, optomechanical devices like the ones developed in the Painter group at Caltech).
It took roughly 50 years from the first message sent over ARPANET to the roll out of 5G technology.
So, when are we going to see the quantum internet?
The technology and expertise needed to build quantum links between cities are developing rapidly with impressive progress made even between 2018 and 2020. Basic quantum network capabilities will likely be up and running in the next decade, which will be an exciting time for breakthroughs in fundamental and applied quantum science. Using single rare-earth atoms is relatively new, but this technology is also advancing quickly (for example, our ytterbium material was largely unstudied just three years ago). Importantly, the discovery of new materials will continue to be important to push quantum technologies forward.
You can read more about this work in this summary article and this synopsis written by lead author JK (Caltech PhD 2019), or dive into the full paper published in Nature.
J. M. Kindem, A. Ruskuc, J. G. Bartholomew, J. Rochman, Y.-Q. Huan, and A. Faraon. Control and single-shot readout of an ion embedded in a nanophotonic cavity. Nature (2020).
Now is an especially exciting time for our field with the Thompson Lab at Princeton publishing a related paper on single rare-earth atom quantum state detection, in their case using erbium. Check out their article here.
# Achieving superlubricity with graphene
Sometimes, experimental results spark enormous curiosity inspiring a myriad of questions and ideas for further experimentation. In 2004, Geim and Novoselov, from The University of Manchester, isolated a single layer of graphene from bulk graphite with the “Scotch Tape Method” for which they were awarded the 2010 Nobel Prize in Physics. This one experimental result has branched out countless times serving as a source of inspiration in as many different fields. We are now in the midst of an array of branching-out in graphene research, and one of those branches gaining attention is ultra low friction observed between graphene and other surface materials.
Much has been learned about graphene in the past 15 years through an immense amount of research, most of which, in non-mechanical realms (e.g., electron transport measurements, thermal conductivity, pseudo magnetic fields in strain engineering). However, superlubricity, a mechanical phenomenon, has become the focus among many research groups. Mechanical measurements have famously shown graphene’s tensile strength to be hundreds of times that of the strongest steel, indisputably placing it atop the list of construction materials best for a superhero suit. Superlubricity is a tribological property of graphene and is, arguably, as equally impressive as graphene’s tensile strength.
Tribology is the study of interacting surfaces during relative motion including sources of friction and methods for its reduction. It’s not a recent discovery that coating a surface with graphite (many layers of graphene) can lower friction between two sliding surfaces. Current research studies the precise mechanisms and surfaces for which to minimize friction with single or several layers of graphene.
Research published in Nature Materials in 2018 measures friction between surfaces under constant load and velocity. The experiment includes two groups; one consisting of two graphene surfaces (homogeneous junction), and another consisting of graphene and hexagonal boron nitride (heterogeneous junction). The research group measures friction using Atomic Force Microscopy (AFM). The hexagonal boron nitride (or graphene for a homogeneous junction) is fixed to the stage of the AFM while the graphene slides atop. Loads are held constant at 20 𝜇N and sliding velocity constant at 200 nm/s. Ultra low friction is observed for homogeneous junctions when the underlying crystalline lattice structures of the surfaces are at a relative angle of 30 degrees. However, this ultra low friction state is very unstable and upon sliding, the surfaces rotate towards a locked-in lattice alignment. Friction varies with respect to the relative angle between the two surface’s crystalline lattice structures. Minimum (ultra low) friction occurs at a relative angle of 30 degrees reaching a maximum when locked-in lattice alignment is realized upon sliding. While in a state of lattice alignment, shearing is rendered impossible with the experimental setup due to the relatively large amount of friction.
Friction varies with respect to the relative angle of the crystalline lattice structures and is, therefore, anisotropic. For example, the fact it takes less force to split wood when an axe blade is applied parallel to its grains than when applied perpendicularly illustrates the anisotropic nature of wood, as the force to split wood is dependent upon the direction along which the force is applied. Frictional anisotropy is greater in homogeneous junctions because the tendency to orient into a stuck, maximum friction alignment, is greater than with heterojunctions. In fact, heterogeneous junctions experience frictional anisotropy three orders of magnitude less than homogeneous junctions. Heterogenous junctions display much less frictional anisotropy due to a lattice misalignment when the angle between the lattice vectors is at a minimum. In other words, the graphene and hBN crystalline lattice structures are never parallel because the materials differ, therefore, never experience the impact of lattice alignment as do homogenous junctions. Hence, heterogeneous junctions do not become stuck in a high friction state that characterizes homogeneous ones, and experience ultra low friction during sliding at all relative crystalline lattice structure angles.
Presumably, to increase applicability, upscaling to much larger loads will be necessary. A large scale cost effective method to dramatically reduce friction would undoubtedly have an enormous impact on a great number of industries. Cost efficiency is a key component to the realization of graphene’s potential impact, not only as it applies to superlubricity, but in all areas of application. As access to large amounts of affordable graphene increases, so will experiments in fabricating devices exploiting the extraordinary characteristics which have placed graphene and graphene based materials on the front lines of material research the past couple decades.
# In the hour of darkness and peril and need
I recited the poem “Paul Revere’s Ride” to myself while walking across campus last week.
A few hours earlier, I’d cancelled the seminar that I’d been slated to cohost two days later. In a few hours, I’d cancel the rest of the seminars in the series. Undergraduates would begin vacating their dorms within a day. Labs would shut down, and postdocs would receive instructions to work from home.
I memorized “Paul Revere’s Ride” after moving to Cambridge, following tradition: As a research assistant at Lancaster University in the UK, I memorized e. e. cummings’s “anyone lived in a pretty how town.” At Caltech, I memorized “Kubla Khan.” Another home called for another poem. “Paul Revere’s Ride” brooked no competition: Campus’s red bricks run into Boston, where Revere’s story began during the 1700s.
Henry Wadsworth Longfellow, who lived a few blocks from Harvard, composed the poem. It centers on the British assault against the American colonies, at Lexington and Concord, on the eve of the Revolutionary War. A patriot learned of the British troops’ movements one night. He communicated the information to faraway colleagues by hanging lamps in a church’s belfry. His colleagues rode throughout the night, to “spread the alarm / through every Middlesex village and farm.” The riders included Paul Revere, a Boston silversmith.
The Boston-area bricks share their color with Harvard’s crest, crimson. So do the protrusions on the coronavirus’s surface in colored pictures.
I couldn’t have designed a virus to suit Harvard’s website better.
The yard that I was crossing was about to “de-densify,” the red-brick buildings were about to empty, and my home was about to lock its doors. I’d watch regulations multiply, emails keep pace, and masks appear. Revere’s messenger friend, too, stood back and observed his home:
he climbed to the tower of the church,
Up the wooden stairs, with stealthy tread,
To the belfry-chamber overhead, [ . . . ]
By the trembling ladder, steep and tall,
To the highest window in the wall,
Where he paused to listen and look down
A moment on the roofs of the town,
And the moonlight flowing over all.
I commiserated also with Revere, waiting on tenterhooks for his message:
Meanwhile, impatient to mount and ride,
Booted and spurred, with a heavy stride,
On the opposite shore walked Paul Revere.
Now he patted his horse’s side,
Now gazed on the landscape far and near,
Then impetuous stamped the earth,
And turned and tightened his saddle-girth…
The lamps ended the wait, and Revere rode off. His mission carried a sense of urgency, yet led him to serenity that I hadn’t expected:
He has left the village and mounted the steep,
And beneath him, tranquil and broad and deep,
Is the Mystic, meeting the ocean tides…
The poem’s final stanza kicks. Its message carries as much relevance to the 21st century as Longfellow, writing about the 1700s during the 1800s, could have dreamed:
So through the night rode Paul Revere;
And so through the night went his cry of alarm
To every Middlesex village and farm,—
A cry of defiance, and not of fear,
A voice in the darkness, a knock at the door,
And a word that shall echo forevermore!
For, borne on the night-wind of the Past,
Through all our history, to the last,
In the hour of darkness and peril and need,
The people will waken and listen to hear
The hurrying hoof-beats of that steed,
And the midnight message of Paul Revere.
Reciting poetry clears my head. I can recite on autopilot, while processing other information or admiring my surroundings. But the poem usually wins my attention at last. The rhythm and rhyme sweep me along, narrowing my focus. Reciting “Paul Revere’s Ride” takes me 5-10 minutes. After finishing that morning, I repeated the poem, and began repeating it again, until arriving at my institute on the edge of Harvard’s campus.
Isolation can benefit theorists. Many of us need quiet to study, capture proofs, and disentangle ideas. Many of us need collaboration; but email, Skype, Google hangouts, and Zoom connect us. Many of us share and gain ideas through travel; but I can forfeit a little car sickness, air turbulence, and waiting in lines. Many of us need results from experimentalist collaborators, but experimental results often take long to gather in the absence of pandemics. Many of us are introverts who enjoy a little self-isolation.
April is National Poetry Month in the United States. I often celebrate by intertwining physics with poetry in my April blog post. Next month, though, I’ll have other news to report. Besides, my walk demonstrated, we need poetry now.
Paul Revere found tranquility on the eve of a storm. Maybe, when the night clears and doors reopen, science born of the quiet will flood journals. Aren’t we fortunate, as physicists, to lead lives steeped in a kind of poetry? | 2020-10-20 11:28:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 38, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.338200181722641, "perplexity": 3079.318750714444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00641.warc.gz"} |
https://nbviewer.org/urls/community.dataquest.io/uploads/short-url/wCQrxnqNRC8xF4pkk3AlFfMUiGN.ipynb | # Building A Handwritten Digits Classifier¶
In this Guided Project, we'll explore the effectiveness of deep, feedforward neural networks at classifying images.
In [32]:
# Scikit-learn contains a number of datasets pre-loaded with the library, within the namespace of sklearn.datasets.
# The load_digits() function returns a copy of the hand-written digits dataset from UCI.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
## Read in dataset as Dataframe¶
In [33]:
# Create a dataframe that contains the handwritten_digits features and a series that contains corresponding labels
handwritten_digits_features = pd.DataFrame(handwritten_digits_features)
handwritten_digits_labels = pd.Series(handwritten_digits_labels)
In [34]:
handwritten_digits_features.head(), handwritten_digits_labels.head()
Out[34]:
( 0 1 2 3 4 5 6 7 8 9 ... 54 55 56 \
0 0.0 0.0 5.0 13.0 9.0 1.0 0.0 0.0 0.0 0.0 ... 0.0 0.0 0.0
1 0.0 0.0 0.0 12.0 13.0 5.0 0.0 0.0 0.0 0.0 ... 0.0 0.0 0.0
2 0.0 0.0 0.0 4.0 15.0 12.0 0.0 0.0 0.0 0.0 ... 5.0 0.0 0.0
3 0.0 0.0 7.0 15.0 13.0 1.0 0.0 0.0 0.0 8.0 ... 9.0 0.0 0.0
4 0.0 0.0 0.0 1.0 11.0 0.0 0.0 0.0 0.0 0.0 ... 0.0 0.0 0.0
57 58 59 60 61 62 63
0 0.0 6.0 13.0 10.0 0.0 0.0 0.0
1 0.0 0.0 11.0 16.0 10.0 0.0 0.0
2 0.0 0.0 3.0 11.0 16.0 9.0 0.0
3 0.0 7.0 13.0 13.0 9.0 0.0 0.0
4 0.0 0.0 2.0 16.0 4.0 0.0 0.0
[5 rows x 64 columns], 0 0
1 1
2 2
3 3
4 4
dtype: int64)
In the handwritten_digits_features we see that each image (as each row) has 64 grayscale points, which means they were 8*8 images.
## Visualize a few sample data¶
In [35]:
# Transform row 0, 100, 200, 300, 1000, 1100, 1200, 1300 to 8*8 pixel grid and visualize the data
rows = [0, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200, 1300, 1400, 1500]
fig = plt.figure(figsize = (3, 3))
i = 1
for row in rows:
# Transform selected datapoint to np array then reshape to original grid
img = handwritten_digits_features.iloc[row].values.reshape(8, 8)
ax.imshow(img, cmap = 'gray_r')
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
i+=1
## Experiment with KNeighborsClassifier¶
In [36]:
# Training k-nearest neighbors models.
from sklearn.neighbors import KNeighborsClassifier
from sklearn.model_selection import KFold
from sklearn.metrics import accuracy_score
# Performs 4-fold cross validation using k-nearest neighbors
def cross_validate(features, labels, n):
kf = KFold(n_splits = 4, random_state = 1)
scores = []
# Split data
for train_idx, test_idx in kf.split(features, labels):
X_train, X_test = features.loc[train_idx], features.loc[test_idx]
y_train, y_test = labels.loc[train_idx], labels.loc[test_idx]
# Train & predict with each fold
knn = KNeighborsClassifier(n_neighbors = n)
knn.fit(X_train, y_train)
predictions = knn.predict(X_test)
accuracy = accuracy_score(y_test, predictions, normalize = True)
scores.append(accuracy)
return scores
In [37]:
# Experiment average recall score with different k(s)
k_num = range(1, 16)
avg_accuracy = []
for n in k_num:
scores = cross_validate(handwritten_digits_features, handwritten_digits_labels, n)
avg_accuracy.append(np.mean(scores))
In [38]:
# Visualize K, recall score correlation
plt.plot(k_num, avg_accuracy)
plt.xlim(1, 15)
Out[38]:
(1, 15)
While the accuracy score is not bad, there are a few downsides to using k-nearest neighbors:
• high memory usage (for each new unseen observation, many comparisons need to be made to seen observations)
• no model representation to debug and explore
Let's now try a neural network with a single hidden layer. Use the MLPClassifier package from scikit-learn.
## Build a Neural Network model from scratch¶
-- based on guides from Coursera Machine Learning course by Andrew Ng from Stanford
### Split data¶
In [39]:
# Split data into train and test sets
from sklearn.model_selection import train_test_split
# Split for train and test
X_train, X_test, y_train, y_test = train_test_split(handwritten_digits_features,
handwritten_digits_labels,
test_size = 0.2,
random_state = 1)
# Split for final result of train(60%) and val(20%)
X_train, X_val, y_train, y_val = train_test_split(X_train, y_train,
test_size = 0.25,
random_state = 1)
### Feedforward Propagation¶
#### Activation function¶
In [40]:
# Vectorized Sigmoid function
def sigmoid(X):
return 1/(1+np.exp(-X))
#### Randomly initialize the parameters for symmetry breaking¶
When training neural networks, it is important to randomly initialize the parameters for symmetry breaking. One effective strategy for random initialization is to randomly select values for $\Theta^{l}$ uniformly in the range $[-\epsilon_{init}, \epsilon_{init}]$.1, 2 This range of values ensures that the parameters are kept small and makes the learning more efficient.
1. One effective strategy for choosing εinit is to base it on the number of units in the network. A good choice of $\epsilon{init}$ is $\epsilon{init} = \frac{\sqrt{6}}{\sqrt{L{in}+L{out}}}$, where $L{in} = s{l}$ and $L{out}= s_{l+1}$ are the number of units(neurons) in the layers adjacent to $\Theta^{l}$.
2. For $x\in[0, 1]$, $x\times2\times\epsilon_{init}-\epsilon_{init}\in[-\epsilon_{init}, \epsilon_{init}]$
In [41]:
# Randomly initialize weights with parameter hidden_layer_size in list datatype
# Returns the same weights in matrix and vector form
def rand_weights_init(X, n_labels, hidden_layer_sizes):
# Number of neurons in input layer
input_layer_size = X.shape[1]
# Compute number of neurons in all layers in the nueral network model
all_layers = [input_layer_size]
all_layers = all_layers+hidden_layer_sizes+ [n_labels]
n_layers = len(all_layers)
# Initialize weights for all layers
weights = []
# Initialize unrolled weights for later use
weights_vec = []
for n in range(n_layers-1):
in_num = all_layers[n]
out_num = all_layers[n+1]
# Initialize epsilon
epsilon_init = np.log(6)/np.log(in_num + out_num)
np.random.seed(0)
# w is set to a matrix of size(out_num, 1 + in_num) as the first column of w handles the "bias" terms
w = np.random.rand(out_num, 1 + in_num) * 2 * epsilon_init - epsilon_init
# Unroll w for later use in fmin_cg function to minimize cost
w_vec = np.concatenate(w)
weights.append(w)
weights_vec.append(w_vec)
print('Number of weights in matrix: ', len(weights))
return np.asarray(weights), np.concatenate(np.asarray(weights_vec))
In [42]:
# Test run with weights initialization: 1 hidden layer with 8 neurons
weights, weights_vector = rand_weights_init(X_train, 10, [8])
print('Theta1 shape: ', weights[0].shape, '\n','Theta2 shape: ', weights[1].shape)
print('Unrolled weights: ', weights_vector.shape)
Number of weights in matrix: 2
Theta1 shape: (8, 65)
Theta2 shape: (10, 9)
Unrolled weights: (610,)
In [43]:
# Test run with weights initialization: 2 hidden layer both with 8 neurons
weights2, weights_vector2 = rand_weights_init(X_train, 10, [8, 8])
print('Theta1 shape: ', weights2[0].shape,
'\n','Theta2 shape: ', weights2[1].shape,
'\n','Theta3 shape: ', weights2[2].shape)
print('Unrolled weights: ', weights_vector2.shape)
Number of weights in matrix: 3
Theta1 shape: (8, 65)
Theta2 shape: (8, 9)
Theta3 shape: (10, 9)
Unrolled weights: (682,)
#### Regularized Cost function of Neural Network with Sigmoid function¶
The cost function for the neural network without regularization is: $J(\theta)= \frac{1}{m}\sum\limits_{i=1}^m\sum\limits_{k=1}^K[−y^{(i)}_{k}log((h_\theta(x^{(i)}))_k)−(1−y^{(i)}_{k})log(1−(h_\theta(x^{(i)}))_k)]$, where $m$ is the number of samples, $K$ is the number of labels, $h_\theta(x)$ is the hypothesis, in our case the Sigmoid function.
The regularized cost function adds the regularization terms: $\frac{\lambda}{2m}\sum\limits_{l = 1}^{L-1}\sum\limits_{no=1}^{N_{out}}\sum\limits_{ni=1}^{N_{in}}(\Theta^{(l)}_{no, ni})^{2}$, where L is the total number of layers in the model, $N_{in}$ and $N_{out}$ are the number of units(neurons) in the layers adjacent to $\Theta^{l}$
In [44]:
# Unregularized cost function
def cost_func(y_prob, y_actual, sample_size):
# Convert y_actual into dummy/indicator variables
y = y_actual.astype('category').copy()
"""
Note to specify dtype when converting categorical dataframe to numpy array
If dtype not specified, -y will be operated on string type where if y = 1, -y = 255
"""
y = np.asarray(pd.get_dummies(y_actual), dtype = np.int64)
# # Convert y_pred into dummy/indicator variables
# # Step 1: Initialize y_pred dummie form with all zeros
# h = np.zeros(y.shape)
# # Step 2: Replace zeros with ones using y_pred as indices
# rows = np.arange(h.shape[0])
# h[rows, y_pred] = 1
# Compute cost
cost = np.sum(-y*np.log10(y_prob)-(1-y)*np.log10(1-y_prob))/sample_size
return cost
#### Feedforward with Regularized Cost function¶
In [45]:
# Feedforward using sigmoid function
def feedforward(vec_weights, hidden_layer_sizes, n_labels, X, y, alpha):
# Define some useful variables
n_samples = X.shape[0]
input_layer_size = X.shape[1]
# Compute number of neurons in all layers in the nueral network model
all_layers = [input_layer_size]
all_layers = all_layers+hidden_layer_sizes+ [n_labels]
# print(all_layers)
n_activation = len(all_layers)-1
# Initialize input
x = np.asarray(X)
# # Initialize output
# predictions = np.empty((n_samples, n_labels))
# Initialize a list of weights in matrix form
weights_matrix = []
# Initialize weights without bias for regularization term
weights_no_bias = []
# Initialize activaitons x for calculating 'error term' in backpropagation later
activations_x = []
# Feedforward
for n in range(n_activation):
x = np.insert(x, 0, 1, axis = 1)
# Append activation x to activations, last activation is also the output
activations_x.append(x)
# Reshape weights vector to weights matrix based on neural network structure
init_slice = 0
weights_count = (all_layers[n]+1)*all_layers[n+1]
weights = np.reshape(vec_weights[init_slice:(init_slice + weights_count)], (all_layers[n+1], (all_layers[n]+1)))
# Append weights to the list of weights in matrix form
weights_matrix.append(weights)
# Compute activation
activation = sigmoid(np.dot(x, weights.T))
# Assign output as the input for next layer, or as final output if loop ends
x = activation
# Get weights without bias in each layerfor later calculating regularization term
no_bias= np.delete(weights, 0, axis = 1)
# Unroll weights in each layer into a numpy 1d array & append to weights_no_bias
weights_no_bias.append(np.concatenate(no_bias))
# Update init_slice
init_slice = weights_count
# Sanity check: x as output should match the shape of predefined p
y_prob = x.copy()
# Assign index of the max value of each row as predictions
y_pred = np.argmax(y_prob, axis = 1)
# Compute unregularized cost
cost = cost_func(y_prob, y, n_samples)
# Unroll weights into a numpy 1d array
weights_no_bias = np.concatenate(np.array(weights_no_bias))
# Regularization term
reg_term = np.sum(weights_no_bias**2)*alpha/(2*n_samples)
# Compute regularized cost
reg_cost = cost + reg_term
return reg_cost, y_prob, y_pred, activations_x, weights_matrix
In [46]:
# Test feedforward with weights initialization:
# 1 hidden layer with 8 neurons
# from rand_weights_init test run & alpha range(5)
for a in range(5):
reg_cost1, probabilties1, predictions1, activations_x1, weights_matrix1 = feedforward(
vec_weights = weights_vector,
n_labels=10,
X=X_train, y = y_train,
hidden_layer_sizes = [8],
alpha = a)
print('With alpha {}, the regularized cost is: '.format(a), reg_cost1)
print('*'*55)
print('Probabilities shape: ', probabilties1.shape)
print('Predictions shape: ', predictions1.shape, 'predictions: ', predictions1)
print('Activations number:', len(activations_x1))
print('Activation x1 shape', activations_x1[0].shape)
print('Activation x2 shape', activations_x1[1].shape)
print('weights 1 shape', weights_matrix1[0].shape)
print('weights 2 shape', weights_matrix1[1].shape)
With alpha 0, the regularized cost is: 2.945120500303429
*******************************************************
With alpha 1, the regularized cost is: 2.9614436955038648
*******************************************************
With alpha 2, the regularized cost is: 2.9777668907043
*******************************************************
With alpha 3, the regularized cost is: 2.9940900859047357
*******************************************************
With alpha 4, the regularized cost is: 3.0104132811051714
*******************************************************
Probabilities shape: (1077, 10)
Predictions shape: (1077,) predictions: [4 4 2 ... 4 0 2]
Activations number: 2
Activation x1 shape (1077, 65)
Activation x2 shape (1077, 9)
weights 1 shape (8, 65)
weights 2 shape (10, 9)
In [47]:
# Test feedforward with weights initialization:
# 2 hidden layer both with 8 neurons
# from rand_weights_init test run & alpha range(5)
for a in range(5):
reg_cost2, probabilties2, predictions2, activations_x2, weights_matrix2 = feedforward(
vec_weights = weights_vector2,
hidden_layer_sizes = [8, 8],
n_labels=10,
X=X_train, y = y_train,
alpha = a)
print('With alpha {}, the regularized cost is: '.format(a), reg_cost2)
print('*'*55)
print('Probabilities shape: ', probabilties2.shape)
print('Predictions shape: ', predictions2.shape, 'predictions: ', predictions2)
print('Activations x number:', len(activations_x2))
print('Activation x1 shape', activations_x2[0].shape)
print('Activation x2 shape', activations_x2[1].shape)
print('Activation x3 shape', activations_x2[2].shape)
print('weights 1 shape', weights_matrix2[0].shape)
print('weights 2 shape', weights_matrix2[1].shape)
print('weights 3 shape', weights_matrix2[2].shape)
With alpha 0, the regularized cost is: 3.0412919452900082
*******************************************************
With alpha 1, the regularized cost is: 3.059339249288558
*******************************************************
With alpha 2, the regularized cost is: 3.077386553287108
*******************************************************
With alpha 3, the regularized cost is: 3.0954338572856575
*******************************************************
With alpha 4, the regularized cost is: 3.1134811612842075
*******************************************************
Probabilities shape: (1077, 10)
Predictions shape: (1077,) predictions: [2 2 2 ... 2 2 2]
Activations x number: 3
Activation x1 shape (1077, 65)
Activation x2 shape (1077, 9)
Activation x3 shape (1077, 9)
weights 1 shape (8, 65)
weights 2 shape (8, 9)
weights 3 shape (10, 9)
### Backpropagation¶
The intuition behind the backpropagation algorithm is as follows:
• Given a training example $(x^{(t)},y^{(t)})$, we will first run a “forward pass” to compute all the activations throughout the network, including the output value of the hypothesis $h_Θ(x)$.
• Then, for each node $j$ in layer $l$, we would like to compute an “error term” $\delta_j^{(l)}$ that measures how much that node was “responsible” for any errors in our output. For an output node, we can directly measure the difference between the network’s activation and the true target value, and use that to define $\delta_j^{(output)}$. For the hidden units, you will compute $\delta_j^{(l)}$ based on a weighted average of the error terms of the nodes in layer $(l + 1)$.
#### Derivation of the Sigmoid function (for calculating gradient of the cost function in backpropagation)¶
1. Sigmoid function: $g(x) = \frac{1}{(1+ e^{-x})}$
2. $g(x) = \frac{1}{(1+ e^{-x})} = (1+e^{-x})^{-1}$
3. Apply Chain Rule: $g'(x) = -1\times(1+e^{-x})^{-2}\times(1+e^{-x})'$
4. Apply Exponential Rule $g'(x) = -(1+e^{-x})^{-2}\times(e^{-x}\times(-x)')$
5. $g'(x) = -(1+e^{-x})^{-2}\times(e^{-x}\times-1)$
6. $g'(x) = \frac{1}{(1+e^{-x})^{2}}\times e^{-x}$
7. $g'(x) = \frac{1}{(1+e^{-x})}\times\frac{e^{-x}}{{(1+e^{-x})}}$
8. $g'(x) = g(x)\times\frac{1+e^{-x}-1}{{(1+e^{-x})}}$
9. $g'(x) = g(x)\times(1-g(x))$
In [48]:
# Derivative of the Sigmoid function
return sigmoid(X)*(1-sigmoid(X))
In [49]:
# Sanity check sigmoid_gradient
Out[49]:
0.25
In [50]:
"""
Implement the backpropagation algorithm to compute the gradients of weights to obtain
the gradient for the neural network cost function.
This function returns the partial derivatives of the cost function with respect to the weights in each hidden layer,
for later use in fmin_cg function to minimize cost.
"""
def backpropagation(vec_weights, hidden_layer_sizes, n_labels, X, y, alpha):
reg_cost, y_prob, y_pred, activations_x, weights = feedforward(vec_weights,
hidden_layer_sizes,
n_labels,
X, y, alpha)
n_activations = len(activations_x)
n_samples = X.shape[0]
# Turn y into dummy index
y = pd.get_dummies(y.astype('category'))
y = np.array(y, dtype = np.int64)
# Compute 'error term' delta of the output layer
last_layer_weights_delta = y_prob - y
# Initialze a list of deltas from each hidden layer
layer_weights_delta = []
# Compute 'error term' delta of the hidden layers
for n in reversed(range(n_activations)):
delta = np.dot(last_layer_weights_delta.T, activations_x[n])
avg_delta = delta/n_samples
# Compute regularization of the corresponding weights_grad
weights_temp = weights[n].copy()
# Replace the first column of weights with 0 to leave out bias column in regularized cost
weights_temp[:,1] = 0
reg_term = (alpha/n_samples)*weights_temp.copy()
# Update weights_grad with regularization term
# Unroll delta and insert it as the first item in weights_gradient list
# Update last_layer_weights_delta
last_layer_weights_delta = np.dot(last_layer_weights_delta,
return np.concatenate(layer_weights_delta)
In [51]:
# Test backpropagation with feedforwd test results
hidden_layer_sizes = [8, 16, 8],
n_labels=10,
X=X_train, y = y_train,
alpha = 1)
Theta gradient 4 shape: (10, 9)
Theta gradient 3 shape: (8, 17)
Theta gradient 2 shape: (16, 9)
Theta gradient 1 shape: (8, 65)
Out[51]:
(890,)
### Put everything together¶
Put all the functions we build so far together, as one function that computes the regularized Neural Network model cost and weights gradient
In [52]:
# Create a function to minize that returns the cost
def nn_cost(vec_weights, *args):
hidden_layer_sizes, n_labels, X, y, alpha = args
reg_cost, y_prob, predictions, activations_x, weights = feedforward(vec_weights,
hidden_layer_sizes,
n_labels,
X, y,
alpha)
return reg_cost
In [53]:
# Create a function to minize that returns the weights cost gradient
hidden_layer_sizes, n_labels, X, y, alpha = args
return backpropagation(vec_weights,
hidden_layer_sizes,
n_labels,
X, y,
alpha)
In [54]:
# Initialize new weights with rand_weights_init and test nn_cost, nn_cost_gradient
test_weights_m, test_weights_vec = rand_weights_init(X_train, 10, [8,16,8])
args = ([8], 10, X_train, y_train, 0)
cost = nn_cost(test_weights_vec, *args)
print(cost)
Number of weights in matrix: 4
Theta gradient 2 shape: (10, 9)
Theta gradient 1 shape: (8, 65)
2.945120500303429
### Minimize the nn_cost function using a nonlinear conjugate gradient algorithm¶
Using the function fmin_cg from SciPy library to minimize our cost function
In [55]:
from scipy.optimize import fmin_cg
from scipy.optimize import minimize
In [62]:
weights_1_m, weights_1_v = rand_weights_init(X_train, 10, [8])
Number of weights in matrix: 2
In [64]:
# Using nn_cost
optim_weights = fmin_cg(f = nn_cost, x0 = weights_1_v, args = args, maxiter = 100)
Warning: Maximum number of iterations has been exceeded.
Current function value: 0.644364
Iterations: 100
Function evaluations: 110772
In [65]:
nn_cost(optim_weights, *args)
Out[65]:
0.6443644765515987
In [66]:
reg_cost_optm, y_prob_optm, predictions_optm, activations_x_optm, weights_optm = feedforward(optim_weights,
[8],
10,
X_train, y_train,
alpha=1)
In [67]:
accuracy = (predictions_optm == y_train).mean()
In [68]:
accuracy
Out[68]:
0.649025069637883
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https://www.aimsciences.org/article/doi/10.3934/proc.2013.2013.115 | # American Institute of Mathematical Sciences
2013, 2013(special): 115-121. doi: 10.3934/proc.2013.2013.115
## Stochastic geodesics and forward-backward stochastic differential equations on Lie groups
1 Grupo de Física-Matemática da Universidade de Lisboa, Av. Prof. Gama Pinto 2, 1649-003 Lisboa, Portugal 2 GFMUL and Departamento de Matemática IST-UTL, Av. Rovisco Pais, 1049-001 Lisboa, Portugal
Received September 2012 Revised March 2013 Published November 2013
We describe how to generalize to the stochastic case the notion of geodesic on a Lie group equipped with an invariant metric. As second order equations (in time), stochastic geodesics are characterized in terms of stochastic forward-backward differential systems.
When the group is the diffeomorphisms group this corresponds to a probabilistic description of the Navier-Stokes equations.
Citation: Xin Chen, Ana Bela Cruzeiro. Stochastic geodesics and forward-backward stochastic differential equations on Lie groups. Conference Publications, 2013, 2013 (special) : 115-121. doi: 10.3934/proc.2013.2013.115
##### References:
[1] M. Arnaudon, X. Chen and A.B. Cruzeiro, Stochastic Euler-Poincaré reduction, preprint, arXiv:1204.3922. [2] M. Arnaudon and A.B. Cruzeiro, Lagrangian Navier Stokes diffusions on manifolds: variational principle and stability, Bull. Sci. Math. 136 (2012), no. 8, 857-881. [3] V.I. Arnold, Sur la géométrie différentielle des groupes de Lie de dimension infinie et ses applications à l'hydrodynamique des fluides parfaits, Ann. Inst. Fourier, 16 (1966), 316-361. [4] J.-M. Bismut, An introductory approach to duality in optimal stochastic control, SIAM Rev, 20 (1978), 62-78. [5] J.-M. Bismut, Mécanique Aléatoire, Lecture Notes in Math. 866, Springer-Verlag (1981). [6] F. Cipriano and A.B. Cruzeiro, Navier-Stokes equation and diffusions on the group of homeomorphisms of the torus, Comm. Math. Phys., 275 (2007), 255-269. [7] A.B. Cruzeiro and E. Shamarova, Navier-Stokes equations and forward-backward SDEs on the group of diffeomorphisms of the torus, Stoch. Proc. and their Applic., 119 (2009), 4034-4060. [8] A.B. Cruzeiro and Zh. Qian, Backward stochastic differential equations associated with the vorticity equations, preprint, arXiv:1304.1319. [9] D. Ebin and J. Marsden, Groups of diffeomorphisms and the motion of an incompressible fluid, Ann. of Math., 92 (1970), 102-163. [10] A. Estrade and M. Pontier, Backward stochastic differential equations in a Lie group, Séminaire de Probabilités, XXXV, Lecture Notes in Math., 1755, Springer-Verlag (2001), 241-259. [11] W.H. Fleming and H.M. Soner, Controlled Markov processes and viscosity solutions, Applications of Mathematics 25, Springer-Verlag, New York (1993). [12] B. Khesin, Groups and topology in the Euler hydrodynamics and KdV, in Hamiltonian dynamical systems and applications, Ed. W.Craig, NATO Science series B, XVI Springer-Verlag (2008), 93-102. [13] J.A. Lázaro-Camí and J.P. Ortega, Stochastic Hamiltonian dynamical systems, SIAM Review, 20 (1978), 65-122. [14] J. Ma and J. Yong, Forward-backward stochastic differential equations and their applications, Springer-Verlag, Lecture Notes in Math. 1702, Springer-Verlag (2007). [15] J.E. Marsden and T.S. Ratiu, Introduction to Mechanics and Symmetry, Springer-Verlag, Texts in Applied Math. (2003). [16] E. Pardoux and S. G. Peng, Adapted solution of a backward stochastic differential equation, Systems Control Lett. , 14 (1990), 55-61. [17] J.-C. Zambrini, Variational processes and stochastic versions of mechanics, Journal of Mathematical Physics, 27 (1986), 2307-2330. [18] J.-C. Zambrini, Stochastic Deformation of Classical Mechanics, in these Proceedings.
show all references
##### References:
[1] M. Arnaudon, X. Chen and A.B. Cruzeiro, Stochastic Euler-Poincaré reduction, preprint, arXiv:1204.3922. [2] M. Arnaudon and A.B. Cruzeiro, Lagrangian Navier Stokes diffusions on manifolds: variational principle and stability, Bull. Sci. Math. 136 (2012), no. 8, 857-881. [3] V.I. Arnold, Sur la géométrie différentielle des groupes de Lie de dimension infinie et ses applications à l'hydrodynamique des fluides parfaits, Ann. Inst. Fourier, 16 (1966), 316-361. [4] J.-M. Bismut, An introductory approach to duality in optimal stochastic control, SIAM Rev, 20 (1978), 62-78. [5] J.-M. Bismut, Mécanique Aléatoire, Lecture Notes in Math. 866, Springer-Verlag (1981). [6] F. Cipriano and A.B. Cruzeiro, Navier-Stokes equation and diffusions on the group of homeomorphisms of the torus, Comm. Math. Phys., 275 (2007), 255-269. [7] A.B. Cruzeiro and E. Shamarova, Navier-Stokes equations and forward-backward SDEs on the group of diffeomorphisms of the torus, Stoch. Proc. and their Applic., 119 (2009), 4034-4060. [8] A.B. Cruzeiro and Zh. Qian, Backward stochastic differential equations associated with the vorticity equations, preprint, arXiv:1304.1319. [9] D. Ebin and J. Marsden, Groups of diffeomorphisms and the motion of an incompressible fluid, Ann. of Math., 92 (1970), 102-163. [10] A. Estrade and M. Pontier, Backward stochastic differential equations in a Lie group, Séminaire de Probabilités, XXXV, Lecture Notes in Math., 1755, Springer-Verlag (2001), 241-259. [11] W.H. Fleming and H.M. Soner, Controlled Markov processes and viscosity solutions, Applications of Mathematics 25, Springer-Verlag, New York (1993). [12] B. Khesin, Groups and topology in the Euler hydrodynamics and KdV, in Hamiltonian dynamical systems and applications, Ed. W.Craig, NATO Science series B, XVI Springer-Verlag (2008), 93-102. [13] J.A. Lázaro-Camí and J.P. Ortega, Stochastic Hamiltonian dynamical systems, SIAM Review, 20 (1978), 65-122. [14] J. Ma and J. Yong, Forward-backward stochastic differential equations and their applications, Springer-Verlag, Lecture Notes in Math. 1702, Springer-Verlag (2007). [15] J.E. Marsden and T.S. Ratiu, Introduction to Mechanics and Symmetry, Springer-Verlag, Texts in Applied Math. (2003). [16] E. Pardoux and S. G. Peng, Adapted solution of a backward stochastic differential equation, Systems Control Lett. , 14 (1990), 55-61. [17] J.-C. Zambrini, Variational processes and stochastic versions of mechanics, Journal of Mathematical Physics, 27 (1986), 2307-2330. [18] J.-C. Zambrini, Stochastic Deformation of Classical Mechanics, in these Proceedings.
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Impact Factor: | 2022-07-05 19:57:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.581750214099884, "perplexity": 2299.9576039594267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00556.warc.gz"} |
http://tex.stackexchange.com/questions/38181/cannot-copy-from-pdflatex-created-pdf | # Cannot copy from pdflatex-created PDF [closed]
When I create a PDF using pdflatex, I can neither copy from it nor search across lines. Is there any way to make these things possible? I tried cmap, but it doesn't help at all.
Here is an example document containing the preamble I typically use, minus the stuff that I'm certain that it isn't related to this problem.
\documentclass[a4paper]{scrartcl}
\usepackage[utf8]{inputenc}
\usepackage[ngerman]{babel}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{lipsum}
%\usepackage{cmap}
\begin{document}
\lipsum[1]
\end{document}
-
## closed as off topic by Boris, Werner, Jake, Ryan Reich, Lev BishopDec 14 '11 at 2:16
Questions on TeX - LaTeX Stack Exchange are expected to relate to TeX, LaTeX or related typesetting systems within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.
I am able to copy from a PDF a created using your example, using either sumatraPDF or Adobe Reader as the PDF viewer. Maybe your PDF viewer is broken? – Lev Bishop Dec 13 '11 at 23:50
Indeed, it works with Sumatra but not with Evince. I'm uninstalling that piece of crap right now. Ridiculous. – Erik Dec 13 '11 at 23:54
Then we probably should close this question: it is about Evince bugs rather than TeX problems... – Boris Dec 14 '11 at 0:39 | 2015-07-04 20:49:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8999459743499756, "perplexity": 2277.6217149564663}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096944.75/warc/CC-MAIN-20150627031816-00018-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2032238/lower-sums-and-upper-sums-integration | # Lower sums and upper sums (integration)
This is a question about integration, which I've just started studying in my real analysis course.
Terminology: Let $P$ be a partition of the interval $[a, b]$. Let $f$ be a function on $[a, b]$. Let $L(f, P)$ be the sum of the areas of rectangles whose height is determined by the minimum of $f(x)$ in a particular subinterval. Similarly, let $U(f, P)$ be the sum of the areas of rectangles whose height is determined by the maximum of $f(x)$ in a particular subinterval.
Let $P_1$ and $P_2$ be any two partitions of $[a, b]$. In my prof's lecture slides, a theorem called "the partition theorem" states that if $f$ is bounded on $[a, b]$ then $L(f, P_1)\le U(f, P_2)$.
My prof's lecture slide then says:
Important inferences that follow from the partition theorem:
For any partition $P$', the upper sum $U(f, P')$ is an upper bound for the set of all lower sums $L(f, P)$.
So far, so good.
It then says:
$\therefore\sup\{L(f, P)\text{ :$P$a partition of }[a, b]\}\le U(f, P')\ \forall P'$
How did he get that?
The slide then says:
$\sup\{L(f, P)\}\le\inf\{U(f, P)\}$
How did he get that?
• If something is an upper bound, then it must be at least as big as the supremum. Otherwise consider values strictly above the upper bound and strictly below the supremum: are these upper bounds or not? (Or short circuit this by using a definition of supremum as least upper bound) – Henry Nov 27 '16 at 0:41 | 2019-06-19 21:17:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.958746612071991, "perplexity": 162.96940740268937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00523.warc.gz"} |
https://www.gamedev.net/forums/topic/440409-manual-rotation-of-vertices-about-center-of-shape/ | # Manual Rotation of vertices about center of shape
This topic is 3990 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Reposting, cos I've tried loads of stuff and really can't figure this out. It's driving me nuts. I want to manually rotate each vertex of a 2d rectangle around the center of the shape, without using glRotatef. I can rotate a point around another point fine, but whatever I try I can't seem to rotate the four corners of the shape around a center point. My vertices are declared as follows:
vertices[0][0] = x - (width / 2);
vertices[0][1] = y + (height / 2);
vertices[1][0] = x + (width / 2);
vertices[1][1] = y + (height / 2);
vertices[2][0] = x + (width / 2);
vertices[2][1] = y - (height / 2);
vertices[3][0] = x - (width / 2);
vertices[3][1] = y - (height / 2);
I used this code to rotate a point around another point:
void orbit()
{
const float DEG2RAD = (float) 3.14159/180;
angle = angle + (1.0f);
for(int i = 0; i < 4; i++)
{
}
}
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Phoresis,
Do you mean that you want to rotate the object about its centre?. If so then,
(1) Translate your object back into LOCAL co-ordinate space.
Your object is at (x, y) then,
vertices[0][0] -= x;
vertices[0][1] -= y;
vertices[1][0] -= x;
vertices[1][1] -= y;
vertices[2][0] -= x;
vertices[2][1] -= y;
vertices[3][0] -= x;
vertices[3][1] -= y;
(2) Perform your rotation in LOCAL space
float fx = vertices[0][0];
float fy = vertices[0][1];
vertices[0][0] = fx * cosf(angle) - fy * sinf(angle);
vertices[0][1] = fx * sinf(angle) + fy * cosf(angle);
fx = vertices[1][0];
fy = vertices[1][1];
vertices[1][0] = fx * cosf(angle) - fy * sinf(angle);
vertices[1][1] = fx * sinf(angle) + fy * cosf(angle);
// similar for the remaining vertices
NOTE: (angle) must is in rads too ;)
(3) Translate object back into WORLD space
vertices[0][0] += x;
vertices[0][1] += y;
vertices[1][0] += x;
vertices[1][1] += y;
vertices[2][0] += x;
vertices[2][1] += y;
vertices[3][0] += x;
vertices[3][1] += y;
This also applies to 3D objects too, but you will have to factor in the z-ordinate.
Hope this works
MAMEman.
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just to clarify, do i translate to local co-ordinate space after I have declared the following:
vertices[0][0] = x - (width / 2);
vertices[0][1] = y + (height / 2);
vertices[1][0] = x + (width / 2);
vertices[1][1] = y + (height / 2);
etc
or instead of? Just want to make sure I get exactly what you mean.
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Yes, I'm assuming that your (x, y) co-ordinates define the centre position of your object in WORLD space. Hence vertices[0][0] -> vertices[3][1] should already be in WORLD space. :)
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Cool. Thanks this seems to work, but it also seems to get faster and faster and faster.
void orbit(){ const float DEG2RAD = (float) 3.14159/180; angle = angle + (0.01f); if (angle > 360) angle -= 360; if (angle < 0) angle += 360; float degInRad = angle*DEG2RAD; for (int i = 0; i < 4; i++) { vertices[0] -= x; vertices[1] -= y; } for (int i = 0; i < 4; i++) { float fx = vertices[0]; float fy = vertices[1]; vertices[0] = fx * cosf(degInRad) - fy * sinf(degInRad); vertices[1] = fx * sinf(degInRad) + fy * cosf(degInRad); } for (int i = 0; i < 4; i++) { vertices[0] += x; vertices[1] += y; }}
I call this method every time the sqaure is drawn. Yeah I know I don't need to keep repeating the for loop, just wanted to split up the logic for now.
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Looking at your code it appears that your angle gets bigger and bigger. Therefore your rotation will get bigger and bigger. Just make your angle variable constant, for example 0.5f. This should take care of that. But this may run faster/slower on another system because you are not using timer based rotation methods ;). However for your needs the initial suggestion should work fine.
Regards,
MAMEman.
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Ah right I'm starting to understand now. What if I wanted it rather than rotate to just set it at the angle?
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Did the initial suggestion work? (setting a constant). I'm not sure what you meant in the last post please carify.
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setting a constant works in that the shape keeps rotating without getting faster or slower, which is good. However what I wanted to achieve initially was a method that just sets the shape to a specified angle, rather than rotating it constantly. So if I gave an angle of 45 degrees then the method would set the angle of the shape to 45 degrees and it would stay at 45 degrees until the method was called again with a different angle. Hope this clarifies it.
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OK. How about when you initialize the object that you want set an initial angle for, you do something like this:
void InitializeVertices(float angle, const float px, const float py)
{
// initialize LOCAL co-ordinates of the object
vertices[0][0] = -(float)(width / 2);
vertices[0][1] = (float)(height / 2);
vertices[1][0] = (float)(width / 2);
vertices[1][1] = (float)(height / 2);
vertices[2][0] = (float)(width / 2);
vertices[2][1] = -(float)(height / 2);
vertices[3][0] = -(float)(width / 2);
vertices[3][1] = -(float)(height / 2);
// perform the initial rotation (as per (2) in my initial post)
// not forgetting to convert (angle) to rads (if not already done so)
// put object into WORLD space (as per (3) in my initial post)
// use parameters px and py to do the translation
}
Now when you want to apply a rotation to the object (for example as the result of a key push) then,
void RotateVertices(float angle)
{
// put your rotation code in here
// note: that when you apply your rotation then angle will be FROM your current angle.
// for example if you initially rotate your object say 90deg and then apply a 45deg rotation your object will have rotated 135deg total from the +ve x-axis.
}
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I think you misunderstood me, either that or I'm just being dumb (most likely), all I want to do is scrap the rotation and just have the set angle, but the code I've got at the moment doesn't just set an angle, it constantly rotates the square .
I know these two lines need altering:
but again I'm stuck with the math.
So, to re-iterate, the shape starts of with no rotation. I then call my rotate method and the shape is redisplayed, rotated to whatever angle I wanted. This would be the same as calling glRotatef on the shape once.
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Yep I guessed right. I think I am being dumb. Basically all I need to do is account for the fact that the angle is added onto the current angle of the shape.
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The problem with my solution is that the vertex data is constantly being updated (correct?). So what you will need to do is store the initial WORLD co-ordinates of your object and work with those. Looking at you code you may want to do it this way.
void orbit(float angle)
{
const float DEG2RAD = (float) 3.14159/180;
if (angle > 360) angle -= 360;
if (angle < 0) angle += 360;
for (int i = 0; i < 4; i++)
{
vertices[0] = initial_vertex[0];
vertices[1] = initial_vertex[1];
vertices[0] -= x; // move into local space
vertices[1] -= y;
}
for (int i = 0; i < 4; i++)
{
float fx = vertices[0];
float fy = vertices[1];
vertices[0] = fx * cosf(degInRad) - fy * sinf(degInRad); // rotate point
}
for (int i = 0; i < 4; i++)
{
vertices[0] += x; // move back to world space
vertices[1] += y;
}
}
BEWARE: If you perform a translation operation then you will need to update initial_vertex as well as vertices. Hope this solves it. | 2018-02-21 23:25:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24587149918079376, "perplexity": 2261.313248458012}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813818.15/warc/CC-MAIN-20180221222354-20180222002354-00409.warc.gz"} |
https://astronomy.stackexchange.com/questions/33428/does-the-spin-rate-of-a-black-hole-correspond-to-a-physical-feature | # Does the spin rate of a black hole correspond to a physical feature?
If the mass and orbital period for two neutron stars is known before collapse to a black hole and the rate of spin is known after collapse, 100 revolutions per second for instance, that will correlate mathmatically to a specific radius.
Does this radius correlate to a physical feature of the black hole such as a surface? Does this radius fall between the radius of the physical substance and the Schwarzschild radius? Is it the largest, intermediate, or the smallest of these?
• Black holes are characterised by an angular momentum, not an angular velocity (which is unknown/undefined). – Rob Jeffries Sep 18 '19 at 21:46
• A black hole doesn't have a physical surface, but it has one or more horizons. See en.wikipedia.org/wiki/Rotating_black_hole – PM 2Ring Sep 19 '19 at 10:38
• @RobJeffries There is an angular velocity associated to a black hole $\omega_{H} =\frac{a}{2M r_{+}}$, which is often interpreted as the angular velocity of the horizon. Specifying $M$ and $\omega_H$ is of course full equivalent to specifying $M$ and $J$ or $M$ and $a$ – mmeent Sep 20 '19 at 7:51 | 2020-01-23 10:38:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8641477227210999, "perplexity": 432.2253397527356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00085.warc.gz"} |
https://jlozadad.io/post/aliens_resistance/ | # Aliens Resistance
In my previous post I discussed the Alien’s Defiance comic series and I’m real excited to know that Zula Hendricks was brought back on another comic series! This time she will be on Aliens Rsistance working together with Amanda Ripley. When I saw this online I really freak out image::https://giphy.com/gifs/Ys9SXSEczO5nW/html5[]
Because I was expecting not to hear about Zula anymore and never see her anymore. Although they should make a movie or anime series about her and Amanda or even better a movie. Something similar how the Gozilla series have been done in netflix. Its only a few weeks until it gets release and I'm going to try to get it right away. I only get the digital versions If I can't find a specific issue. | 2020-02-29 00:23:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20525187253952026, "perplexity": 2003.6870959334606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00020.warc.gz"} |
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# Heat and charge transport in H2O at ice-giant conditions from ab initio molecular dynamics simulations
## Abstract
The impact of the inner structure and thermal history of planets on their observable features, such as luminosity or magnetic field, crucially depends on the poorly known heat and charge transport properties of their internal layers. The thermal and electric conductivities of different phases of water (liquid, solid, and super-ionic) occurring in the interior of ice giant planets, such as Uranus or Neptune, are evaluated from equilibrium ab initio molecular dynamics, leveraging recent progresses in the theory and data analysis of transport in extended systems. The implications of our findings on the evolution models of the ice giants are briefly discussed.
## Introduction
Hydrogen and oxygen are two of the three most abundant elements in the universe, helium being the second. As a consequence, H2O is thought to be a major constituent of celestial bodies formed far enough from their host star for it to condense1. Many moons of the outer solar system, such as Ganymede, Europa, and Enceledus, have rigid icy shells and interior water oceans, which are key for understanding the observed surface mass flux2 and the generation of magnetic fields3. The ice giants, Uranus and Neptune, are thought to be composed primarily of H2O4: throughout most of their interior, the large pressure and temperature (e.g., 240 GPa and 5000 K at half the radius of Uranus) favor a super-ionic (SI) phase, where oxygen ions are arranged in a crystalline lattice and protons diffuse freely like in a fluid5,6. Partially dissociated, liquid (PDL) water may instead be confined to the outermost third of the interior, where the magnetic field is generated7. Outside the solar system, the observed characteristics of many exoplanets are also consistent with with water-rich interiors8.
Our knowledge of the interior of planets other than Earth mostly relies on the observation of their magnetic fields and surface properties, which are affected by the inner structure through the transport of energy, mass, and charge across intermediate layers. In the case of Uranus, for instance, it has long been recognized that its remarkably small luminosity9 can be explained by nonadiabatic models of the interior4,10, featuring thermal boundary layers whose transport properties are poorly known. Likewise, any model aiming to explain the anomalous multipolar and non-axisymmetry magnetic fields of Uranus and Neptune requires the knowledge of the electric conductivity of the various phases of water occurring in their interior11. More generally, a detailed knowledge of the transport properties of different phases of H2O occurring at high-pT conditions is key to any quantitative evolutionary model of water-rich celestial bodies. In spite of the steady progress in diamond-anvil-cell and shock-wave technologies, the experimental investigation of transport properties of materials at planetary conditions is still challenging. In the specific case of H2O, the electrical conductivity is only known with large uncertainties along the Hugoniot curve on a limited portion of the pT diagram, and nothing is known about the heat conductivity6,12,13,14.
Computer simulations may be our only handle on the properties of matter at physical conditions that cannot be achieved in the laboratory. In the case of water, they have allowed us to discover new phases5 and to predict their properties at extreme pT conditions6,15 over an ever broader portion of its phase diagram16. The diverse local chemical environments that characterize the different relevant phases of water make classical force fields unfit for an accurate simulation of their properties, and call for a full quantum-mechanical, ab initio (AI), treatment of the chemical bond. Some transport properties of water at high-pT conditions, such as ionic (H and O) diffusivities and the electrical conductivity have indeed been estimated using AI molecular dynamics (AIMD) simulations17 and the Green-Kubo (GK) theory of linear response18,19,20,21. However, it has long and widely been argued that quantum-mechanical simulation methods could not be combined with the GK theory, because the latter is based on a microscopic representation of the energy (current) density, which is evidently ill-defined at a quantum-mechanical level22. The soundness of this objection, which would actually apply to a classical representation of the interatomic forces as well, was recently refuted for good by the introduction of a gauge invariance principle for transport coefficients23,24,25. In a nutshell, gauge invariance means that transport coefficients do not depend on the details of the microscopic representation of the conserved quantity being transported, as long as this representation sums to the correct value in the thermodynamic limit and its space correlations are short ranged. This remarkable finding implies that any (good, in the above sense) local representation of the energy leads to the same heat conductivity, thus paving the way to a fully ab initio treatment of heat transport23, which was recently generalized to multi-component systems26.
In this work we leverage these recent theoretical advances to estimate the thermal conductivity and other transport coefficients of stoichiometric H2O in the pT conditions to be found on ice-giant planets, like Uranus and Neptune, from equilibrium AIMD simulations, exploring its solid, PDL, and SI phases.
## Results
### Theory
Transport in macroscopic media is governed by the dynamics of hydrodynamic variables, i.e., by the long-wavelength components of the (current) densities of conserved extensive quantities25,27,28. For short, we will dub such densities conserved densities, the corresponding currents conserved currents, while the macroscopic averages of the latter will be called conserved fluxes. The GK theory of linear response18,19,20,21 states that transport coefficients (i.e., conductivities) are integrals of the various flux time autocorrelation functions, which, according to the Wiener–Khintchine theorem29,30, are the zero-frequency values of the corresponding power spectra. An important concept in the theory of transport is that of diffusive flux: we say that a flux is diffusive if its power spectrum does not vanish25,26 at zero frequency. Gauge invariance states that two different representations (“gauges”) of a same conserved density that differ by the divergence of a bounded vector field are equivalent in that they give rise to macroscopic fluxes whose difference is non-diffusive, thus resulting in the same conductivity23,24.
When addressing heat transport, the relevant conserved quantities are the energy and the numbers of particles (or masses) of each atomic species. Since the total-mass flux itself (i.e., the total momentum) is a constant of motion, for a P-species system the number of independent conserved fluxes is equal to P (energy, plus P − 1 partial masses). Further constraints may reduce the number of relevant conserved fluxes. For instance, in solids, such as ordinary ice, atoms do not diffuse and there cannot be any macroscopic mass flow: energy flux is therefore the only relevant one. In molecular liquids, such as ordinary water, the partial mass fluxes of each atomic species are non-diffusive if the molecules do not dissociate. This is so because the integral of the difference between the individual total momenta of different atomic species is bound by the finite variation of the molecular bond lengths23,25: also in this case, therefore, energy is the only relevant conserved quantity. On the contrary, the PDL and SI phases of water are truly multi-component systems, because the momentum of at least one atomic component is neither conserved nor is its integral bound by any molecular constraints.
Heat transport in multi-component systems has long been the subject of theoretical misconceptions and/or considered to be numerically intractable. For instance, the thermal conductivity is sometimes computed as the GK integral of the energy flux, JE: $$\kappa \propto \mathop{\int}\nolimits_{0}^{\infty }\langle {{\bf{J}}}_{E}(t){{\bf{J}}}_{E}(0)\rangle dt$$. This simplistic approach is manifestly wrong, as the resulting conductivity would depend on the arbitrary choice of the atomic formation energies. To see why this is so, let us consider the classical expression of the energy flux25,31: $${{\bf{J}}}_{E}=\frac{1}{\Omega }\left[\mathop{\sum }\nolimits_{n = 1}^{N}{{\bf{V}}}_{n}{\epsilon }_{n}+{\sum }_{n,m}({{\bf{R}}}_{n}-{{\bf{R}}}_{m}){{\bf{F}}}_{nm}\cdot {{\bf{V}}}_{n}\right],$$where Ω is the system’s volume, Rn, Vn, and ϵn are the atomic positions, velocities, and energies, respectively, and Fnm = −∂ϵm/∂Rn are interatomic forces. The heat conductivity cannot evidently depend on the arbitrary zero of the atomic energies. For instance, in ab initio calculations these energies differ in a pseudo-potential or in an all-electron scheme, whereas transport coefficients should not. A better choice would be to compute the heat conductivity from the GK integral of the heat flux, defined as $${{\bf{J}}}_{q}={{\bf{J}}}_{E}-\frac{1}{\Omega }\mathop{\sum }\nolimits_{S = 1}^{P}{h}_{S}{N}_{S}{\overline{{\bf{V}}}}_{S}$$, where $${\overline{{\bf{V}}}}_{S}$$ is the center-of-mass velocity and hS the partial enthalpy of the S-th atomic species32. This approach has the advantage that Jq is no longer sensitive to a rigid shift in the atomic energies; it is still an approximation, though, as it neglects the coupling between energy and mass flow (Soret effect) in the calculation of κ. Even if, for several systems, it has been argued that the error in doing so is small32, this argument cannot be taken for granted a priori for any generic system. Furthermore, the calculation of partial enthalpies is rather involved33,34,35, and often the subject itself of crude approximations.
A rigorous methodology to deal with multi-component systems is provided by a combination of Onsager’s phenomenological approach36,37 and the GK theory of linear response18,19,20,21. In this approach the interactions among different conserved fluxes are explicitly accounted for by Onsager’s phenomenological relations:
$${J}_{i}\;=\;\sum _{j}{\Lambda }_{ij}\,{f}_{j},$$
(1)
where J is a generic conserved flux, f a thermodynamic affinity, i.e., the average gradient of the intensive variable conjugate to a conserved quantity, Λ is the matrix of Onsager’s phenomenological coefficients, and the suffixes enumerate in principle both different conserved quantities and the Cartesian components of their fluxes/affinities. In practice, in the following we will dispose of Cartesian components, and only enumerate different conserved fluxes/affinities, given that we will only be concerned with isotropic or cubic systems. Within the GK theory, and the Λ coefficients are expressed as integrals of the time correlation functions of the relevant fluxes:
$${\Lambda }_{ij}=\frac{\Omega }{{k}_{B}}\mathop{\int}\nolimits_{0}^{\infty }\left\langle {{\mathcal{J}}}_{i}(t){{\mathcal{J}}}_{j}(0)\right\rangle dt,$$
(2)
where $${{\mathcal{J}}}_{i}(t)$$ is the time series of the i-th flux, kB is the Boltzmann constant, and 〈 〉 indicates an equilibrium average. From now on, calligraphic fonts indicate samples of stochastic processes. The thermal conductivity is defined as the ratio between the energy flux and the temperature gradient, when all the other conserved fluxes vanish. In a two-component system this condition leads to the following expression for the heat conductivity:
$$\kappa =\frac{1}{{T}^{2}}\left[{\Lambda }_{EE}-\frac{{\left|{\Lambda }_{EM}\right|}^{2}}{{\Lambda }_{MM}}\right],$$
(3)
where the M suffix indicates the mass flux of one of the two components. The expression in square brackets is the inverse of the EE matrix element of the inverse of the 2 × 2 matrix of the Onsager coefficients. In the general, multivariate, case, the heat conductivity is proportional to the Schur complement of the mass block in Λ. In ref. 26 we have shown that this expression for the heat conductivity is invariant under the addition of an arbitrary linear combination of conserved fluxes (such as mass or adiabatic electronic charge) to the energy flux, and we named this further remarkable property of transport coefficients convective invariance.
Equation (3) shows that this procedure is numerically ill-conditioned, because the estimator of the integral in Eq. (2) becomes a random walk as a function of the upper limit of integration, as soon as the integrand has exhausted all its weight, thus making the expression in Eq. (3) singular whenever the estimator of the denominator vanishes38,39,40,41. A solution to this problem is provided by multivariate cepstral analysis26, briefly sketched below. According to the Wiener–Khintchine theorem29,30, the Onsager coefficients in Eq. (2) are proportional to the zero-frequency values of the flux cross power spectral density, $${S}_{ij}(\omega )=\mathop{\int}\nolimits_{-\infty }^{\infty }\langle {{\mathcal{J}}}_{i}(t){{\mathcal{J}}}_{j}(0)\rangle {{\rm{e}}}^{i\omega t}dt$$:
$${\Lambda }_{ij}=\frac{\Omega }{2{k}_{B}}{S}_{ij}(\omega =0)$$
(4)
$${S}_{ij}(\omega )={\mathop{\mathrm{lim}}\limits_{\tau \to \infty }}\langle {{\mathcal{S}}}_{ij}^{\tau }(\omega )\rangle$$
(5)
$${{\mathcal{S}}}_{ij}^{\tau }(\omega )=\frac{1}{\tau }{\tilde{{\mathcal{J}}}}_{i}^{\tau }{(\omega )}^{* }\cdot {\tilde{{\mathcal{J}}}}_{j}^{\tau }(\omega )$$
(6)
$${\tilde{{\mathcal{J}}}}_{j}^{\tau }(\omega )=\mathop{\int}\nolimits_{0}^{\tau }{{\mathcal{J}}}_{j}(t){{\rm{e}}}^{i\omega t}dt.$$
(7)
The continuity and smoothness of the power spectrum at low frequency can be leveraged to systematically reduce the noise affecting the estimator of its zero-frequency value, as explained below. According to the central-limit theorem, the flux processes, $${{\mathcal{J}}}_{i}(t)$$, are Gaussian because they are the space integrals of current densities, whose correlations are short ranged. Stationarity implies that their Fourier transforms, Eq. (7), are normal deviates that for large τ are uncorrelated for $$\omega \;\ne \;\omega ^{\prime}$$. It follows that the sample spectrum of Eq. (6), aka the cross-periodogram, is a collection of complex Wishart random matrices42 that are uncorrelated among themselves for different frequencies. Now, the Schur complement of a block of dimension P − 1 in a Wishart matrix of order P is proportional to a χ2 stochastic variable26,42. We conclude that the Schur complement of the mass block, $${{\mathcal{S}}}_{E}^{\prime}$$, in the cross-periodogram given by Eq. (6), is the product of a smooth function of frequency, whose ω → 0 limit is the thermal conductivity we are after, times a set of independent, identically distributed, χ2 stochastic variables. By applying a low-pass filter to the logarithm of this quantity, one obtains a consistent estimator of the logarithm of the conductivity, as explained in ref. 26, a procedure that is known as cepstral analysis in sound engineering and speech recognition applications43.
### Simulations
The heat and charge transport properties of different (solid, PDL, and SI) phases of water in the 1000–3000 K and 30–250 GPa pT range have been explored by Car-Parrinello (CP) ab initio NVE molecular dynamics44, using the QUANTUM ESPRESSO suite of computer codes45,46. We believe that the CP Lagrangian formalism is particularly fit for transport simulations because the accurate conservation of the (extended) total energy allows one to generate long and stable trajectories without using thermostats. Figure 1 shows the phase diagram of water in such pT range. The SI-PDL (dashed) and ice-SI (dotted) phase boundaries are obtained from state-of-the-art shock-compression experiments6; Uranus’ isentrope (solid gray) from ab initio simulations47 is also reported. We have verified that a body-centered-cubic (BCC) to face-centered-cubic (FCC) transition in the oxygen lattice occurs for the SI phase at P ≈ 240 GPa and T ≈ 3000 K, in accordance with recent theoretical16 and experimental findings48. We then ran three simulations for the BCC-SI phase (blue circles) and one for the FCC-SI one (blue square). We also ran a simulation for solid ice X (green triangle) and a simulation for the PDL (orange triangle) at pT conditions where the fraction of dissociated molecules is ~10%49. We have explicitly checked that the electron energy gap computed along the various MD trajectories is always way larger than kBT, thus ruling out any direct electronic contributions to heat and charge transport. All the technical details of the simulations are reported in the Supplementary Note 1. Our results are summarized in Table 1.
## Discussion
We start the discussion of our results by highlighting the importance of a multi-component analysis of the heat- and mass-flux time series resulting from our simulations. In Fig. 2 we display the power spectrum of the energy flux of FCC-SI water at an average temperature T = 2920 ± 90 K and pressure P = 257 ± 2 GPa, evaluated according to two different prescriptions: blue lines refer to the plain spectrum of the energy flux computed within density-functional theory using the formulation of ref. 23; orange lines indicate the “residual spectrum” computed by assuming that the mass flux vanishes, according to Eq. (3). The sample power spectra (the “periodograms”) are displayed with faint lines, whereas those subject to cepstral filtering are displayed with thick lines; the latter are zoomed-in at low frequency and displayed in the inset, together with their statistical uncertainties. By looking at the zero-frequency value of the spectrum, cepstral analysis gives κ = 20 ± 2 W/(Km), and κ = 13 ± 2 W/(Km) neglecting and accounting for the interaction with the H mass flux, respectively. In effectively one-component systems, statistical analysis can be greatly facilitated by fixing a suitably defined optimal gauge for the diffusing current50. Since SI is a truly bicomponent system, a bivariate analysis is indeed needed to account for the interaction between different conserved fluxes and for a correct estimate of κ: considering the time series of the energy flux alone—as if the system were one-component—would overestimate the heat conductivity by 80%.
Convective invariance can also be leveraged to reduce the statistical noise, and thus the uncertainty, on the estimated value of κ, as explained in ref. 26. The addition of one or more components to the set of conserved fluxes to be analyzed decreases the total power of the reduced spectrum without affecting its value at zero frequency, thus making it smoother and the low-pass cepstral filter more efficacious. By adopting the adiabatic electron current as an additional flux, one obtains the refined result: κ = 12.8 ± 1.0 W/(Km). Further details on the statistical analysis of our data can be found in the Supplementary Information.
Multi-component cepstral analysis, which has been performed using the thermocepstrum code51, allows us to obtain accurate transport coefficients from relatively short AIMD trajectories, particularly for the strongly anharmonic exotic phases of water occurring at the high-pT conditions of interest here. Figure 3 shows the values and the statistical uncertainties of the heat conductivity of different phases of water as a function of the length of the (reduced) energy-flux time series from which they are estimated. These data show that well-converged results with an uncertainty of ≈15% are obtained with trajectories as short as 10–20 ps. Not surprisingly, the more crystalline a phase is, the larger the uncertainty for a same trajectory length (ice X > SI > PDL), due to the larger residual harmonicity of the structure. We stress that cepstral analysis is a self-averaging technique, in that the statistical error affecting the estimated conductivities can be accurately estimated and systematically reduced by increasing the length of the simulation, thus avoiding the need to average over different MD trajectories. Nonetheless, isotropy allows one to consider the three Cartesian components of the fluxes as different samples of a same process: the spectra have been thus averaged over Cartesian components.
Our results are summarized in Table 1. In the pT conditions examined here, the thermal conductivity of solid ice X is larger than that of the SI phase, which is itself larger than in PDL water. This is expected, again due to the decreasing level of harmonicity in going from a crystalline to a partially liquid and eventually fully liquid phase. We did not observe a significant dependence of κ upon the temperature for the SI phase in the explored range. The FCC-SI water has slightly larger heat conductivity than BCC-SI.
Pioneering AIMD simulations of charge transport in PDL water17 revealed that, rather unexpectedly, a classical model of charge conduction where hydrogen and oxygen ions carry an integer charge whose magnitudes equal their formal oxidation numbers (qH = +1 and qO = −2) yields the same conductivity that would be obtained from the exact quantum-mechanical expression of the electric current, based on Born’s effective charges. This surprising finding was given a solid theoretical foundation in a recent paper of ours where it was shown to result from the combined effects of gauge invariance of transport coefficients and topological quantization of adiabatic charge transport52. Leveraging this result, we computed the electrical conductivity from the cepstral analysis of the classical charge flux, defined as:
$${{\mathcal{J}}}_{Z}=\frac{1}{\Omega }\left({q}_{{\rm{H}}}\sum _{n\in {\rm{H}}}{{\mathcal{V}}}_{n}+{q}_{{\rm{O}}}\sum _{n\in {\rm{O}}}{{\mathcal{V}}}_{n}\right),$$
(8)
where the $${\mathcal{V}}$$’s are ionic velocities.
The electrical conductivities resulting from our simulations are reported in Table 1. The data tagged with the “NE” subscript are obtained using the Nernst–Einstein equation53, which neglects all interionic correlations and that in the one-component case reads:
$${\sigma }_{NE}=\frac{{e}^{2}{q}_{{\rm{H}}}^{2}{N}_{{\rm{H}}}{D}_{{\rm{H}}}}{\Omega {k}_{B}T},$$
(9)
where NH and DH are the number of hydrogen atoms and their diffusivity, respectively. In the case of PDL, Eq. (9) hardly applies, as it would depend on too large a number of parameters (the concentrations, life-times, and diffusivities of the various ionic charge carriers). Our results are consistent with previous theoretical estimates16,17, as well as with the experimental data obtained from electrical impedance measurements along the liquid or precompressed Hugoniot12,13,14, summarized in Fig. 4 of ref. 6: σ ~ 150 S/cm for the SI phase in the range 100–150 GPa and 2000–3000 K; and σ ~ 30 S/cm for the PDL phase at ≈30 GPa and 2000 K. Two important trends emerge from our results. First, the NE relation severely underestimates the conductivity in SI water, as already observed in other SI systems53. At variance with these findings, when charge carriers of opposite signs coexist in an electrolyte, the short-range correlations among them may screen the amount of transported charge, thus determining a decrease of the electric conductivity with respect to the predictions of the NE approximation54. In the second place, the electrical conductivity in the FCC-SI phase is sensitively larger than in the BCC one, in contrast to the opposite trend displayed by hydrogen diffusivity, which are instead slightly smaller in the FCC phase, thus resulting in comparable predictions for the two phases of the NE approximation (σNE). The lesser ability of the NE approximation to predict the conductivity in the FCC than in the BCC phase indicates a stronger effect of interionic correlations in the former case: the higher energy barriers for a single proton hop in FCC—due to its larger packing density55, and resulting in a slightly smaller ionic diffusivity—may be effectively decreased by a cooperative motion of two or more protons (as already observed for the carrier dynamics in solid-state electrolytes56), and thus lead to an overall larger electrical conductivity.
In this paper we have reported on the first theoretically rigorous and numerically accurate evaluation of the thermal and electric conductivities of various phases of water occurring at the pressure and temperature conditions to be found in the interior of ice-giant planets, made possible by recent advances in transport theory and data analysis. In the case of the heat conductivity, our results set a reference in the wide range of values used in evolution models of Uranus and Neptune57 or given by recent MD-based estimates on dissociating water58, and their moderate values point towards more efficient trapping of heat in the deep interior of these planets. These results have been instrumental in the development of a novel model of the thermal evolution of Uranus, featuring a frozen core and an anomalously low heat flow, resulting in the observed low luminosity of this planet59. Finally, the electrical conductivity that we find for SI ice is far larger than assumed in previous models of the generation of the magnetic fields in Uranus and Neptune60. Since SI ice is likely to dominate the deeper sluggish layer that underlies the shallow fluid outer layer in which the magnetic field is produced, the large electrical conductivity of the SI phase can have a substantial impact on the geometry and time evolution of the magnetic field of these planets.
## Data availability
The data that support the plots and relevant results within this paper are available on the Materials Cloud Platform at https://doi.org/10.24435/materialscloud:hn-6f.
## Code availability
Computer codes are available and freely downloadable from the QUANTUM ESPRESSO site and the Thermocepstrum GitHub page referenced below.
## References
1. 1.
Lodders, K. Solar system abundances and condensation temperatures of the elements. Astrophys. J. 591, 1220 (2003).
2. 2.
Nimmo, F., Spencer, J. R., Pappalardo, R. T. & Mullen, M. E. Shear heating as the origin of the plumes and heat flux on Enceladus. Nature 447, 289–291 (2007).
3. 3.
Kivelson, M. G. et al. Discovery of Ganymede’s magnetic field by the Galileo spacecraft. Nature 384, 537–541 (1996).
4. 4.
Nettelmann, N., Helled, R., Fortney, J. J. & Redmer, R. New indication for a dichotomy in the interior structure of uranus and neptune from the application of modified shape and rotation data. Planet. Space Sci. 77, 143–151 (2013).
5. 5.
Cavazzoni, C. et al. Superionic and metallic states of water and ammonia at giant planet conditions. Science 283, 44–46 (1999).
6. 6.
Millot, M. et al. Experimental evidence for superionic water ice using shock compression. Nat. Phys. 14, 297–302 (2018).
7. 7.
Stanley, S. & Bloxham, J. Convective-region geometry as the cause of Uranus’ and Neptune’s unusual magnetic fields. Nature 428, 151–153 (2004).
8. 8.
Zeng, L. et al. Growth model interpretation of planet size distribution. Proc. Natl Acad. Sci. USA 116, 9723–9728 (2019).
9. 9.
Fortney, J. J. & Nettelmann, N. The interior structure, composition, and evolution of giant planets. Space Sci. Rev. 152, 423–447 (2010).
10. 10.
Hubbard, W. B., Podolak, M. & Stevenson, D. J. in Neptune and Triton (ed Cruikshank, D. P.) 109–138 (University of Arizona Press, 1995).
11. 11.
Helled, R., Nettelmann, N. & Guillot, T. Uranus and Neptune: origin, evolution and internal structure. Space Sci. Rev. 10.1007/s11214-020-00660-3 (2020).
12. 12.
Mitchell, A. C. & Nellis, W. J. Equation of state and electrical conductivity of water and ammonia shocked to the 100 GPa (1 Mbar) pressure range. J. Chem. Phys. 76, 6273–6281 (1982).
13. 13.
Yakushev, V. V., Postnov, V. I., Fortov, V. E. & Yakysheva, T. I. Electrical conductivity of water during quasi-isentropic compression to 130 GPa. J. Exp. Theor. Phys. 90, 617–622 (2000).
14. 14.
Chau, R., Mitchell, A. C., Minich, R. W. & Nellis, W. J. Electrical conductivity of water compressed dynamically to pressures of 70–180 GPa (0.7–1.8 Mbar). J. Chem. Phys. 114, 1361–1365 (2001).
15. 15.
Rozsa, V., Pan, D., Giberti, F. & Galli, G. Ab initio spectroscopy and ionic conductivity of water under earth mantle conditions. Proc. Natl Acad. Sci. USA 115, 6952–6957 (2018).
16. 16.
Sun, J., Clark, B. K., Torquato, S. & Car, R. The phase diagram of high-pressure superionic ice. Nat. Commun. 6, 8156 (2015).
17. 17.
French, M., Hamel, S. & Redmer, R. Dynamical screening and ionic conductivity in water from ab initio simulations. Phys. Rev. Lett. 107, 185901 (2011).
18. 18.
Green, M. S. Markoff random processes and the statistical mechanics of time-dependent phenomena. J. Chem. Phys. 20, 1281–1295 (1952).
19. 19.
Green, M. S. Markoff random processes and the statistical mechanics of time-dependent phenomena. ii. irreversible processes in fluids. J. Chem. Phys. 22, 398–413 (1954).
20. 20.
Kubo, R., Yokota, M. & Nakajima, S. Statistical-mechanical theory of irreversible processes. ii. response to thermal disturbance. J. Phys. Soc. Jpn. 12, 1203–1211 (1957).
21. 21.
Kubo, R. Statistical-mechanical theory of irreversible processes. i. General theory and simple applications to magnetic and conduction problems. J. Phys. Soc. Jpn. 12, 570–586 (1957).
22. 22.
Stackhouse, S., Stixrude, L. & Karki, B. B. Thermal conductivity of periclase (MgO) from first principles. Phys. Rev. Lett. 104, 208501 (2010).
23. 23.
Marcolongo, A., Umari, P. & Baroni, S. Microscopic theory and ab initio simulation of atomic heat transport. Nat. Phys. 12, 80–84 (2016).
24. 24.
Ercole, L., Marcolongo, A., Umari, P. & Baroni, S. Gauge invariance of thermal transport coefficients. J. Low. Temp. Phys. 185, 79–86 (2016).
25. 25.
Baroni, S., Bertossa, R., Ercole, L., Grasselli, F. & Marcolongo, A. in Handbook of Materials Modeling: Applications: Current and Emerging Materials 2nd edn. (eds. Andreoni, W. & Yip, S.) 1–36 (Springer International Publishing, Cham, 2018).
26. 26.
Bertossa, R., Grasselli, F., Ercole, L. & Baroni, S. Theory and numerical simulation of heat transport in multicomponent systems. Phys. Rev. Lett. 122, 255901 (2019).
27. 27.
Kadanoff, L. P. & Martin, P. C. Hydrodynamic equations and correlation functions. Ann. Phys. 24, 419–469 (1963).
28. 28.
Foster, D. Hydrodynamic Fluctuations, Broken Symmetry, and Correlation Functions (Benjamin, 1975).
29. 29.
Wiener, N. Generalized harmonic analysis. Acta Math. 55, 117–258 (1930).
30. 30.
Khintchine, A. Korrelationstheorie der stationären stochastischen Prozesse. Math. Ann. 109, 604–615 (1934).
31. 31.
Irving, J. H. & Kirkwood, J. G. The statistical mechanical theory of transport processes. IV. The equations of hydrodynamics. J. Chem. Phys. 18, 817 (1950).
32. 32.
De Groot, S. R. & Mazur, P. Non-Equilibrium Thermodynamics (Dover Publications, 1984).
33. 33.
Debenedetti, P. G. Fluctuation-based computer calculation of partial molar properties. I. Molecular dynamics simulation of constant volume fluctuations. J. Chem. Phys. 86, 7126 (1987).
34. 34.
Vogelsang, R. & Hoheisel, C. Thermal conductivity of a binary-liquid mixture studied by molecular dynamics with use of Lennard- Jones potentials. Phys. Rev. A. 35, 3487–3491 (1987).
35. 35.
Sindzingre, P., Massobrio, C. & Ciccotti, G. Calculation of partial enthalpies of an Argon-Kripton mixture by NPT molecular dynamics. Chem. Phys. 129, 213–224 (1989).
36. 36.
Onsager, L. Reciprocal relations in irreversible processes. i. Phys. Rev. 37, 405–426 (1931).
37. 37.
Onsager, L. Reciprocal relations in irreversible processes. ii. Phys. Rev. 38, 2265 (1931).
38. 38.
Galamba, N., Nieto de Castro, C. A. & Ely, J. F. Equilibrium and nonequilibrium molecular dynamics simulations of the thermal conductivity of molten alkali halides. J. Chem. Phys. 126, 204511 (2007).
39. 39.
Ohtori, N., Salanne, M. & Madden, P. A. Calculations of the thermal conductivities of ionic materials by simulation with polarizable interaction potentials. J. Chem. Phys. 130, 104507 (2009).
40. 40.
Salanne, M., Marrocchelli, D., Merlet, C., Ohtori, N. & Madden, P. A. Thermal conductivity of ionic systems from equilibrium molecular dynamics. J. Phys. Condens. Matter 23, 102101 (2011).
41. 41.
Bonella, S., Ferrario, M. & Ciccotti, G. Thermal diffusion in binary mixtures: transient behavior and transport coefficients from equilibrium and nonequilibrium molecular dynamics. Langmuir 33, 11281–11290 (2017).
42. 42.
Nagar, D. K. & Gupta, A. K. Expectations of functions of complex wishart matrix. Acta Appl. Math. 113, 265–288 (2011).
43. 43.
Childers, D. G., Skinner, D. P. & Kemerait, R. C. The cepstrum: a guide to processing. Proc. IEEE 65, 1428–1443 (1977).
44. 44.
Car, R. & Parrinello, M. Unified approach for molecular dynamics and density-functional theory. Phys. Rev. Lett. 55, 2471–2474 (1985).
45. 45.
Giannozzi, P. et al. QUANTUM ESPRESSO: a modular and open-source software project for quantum simulations of materials. J. Phys. Condens. Matter 21, 395502 (19pp) (2009).
46. 46.
Giannozzi, P. et al. Advanced capabilities for materials modelling with QUANTUM ESPRESSO. J. Phys. Condens. Matter 29, 465901 (2017).
47. 47.
Redmer, R., Mattsson, T. R., Nettelmann, N. & French, M. The phase diagram of water and the magnetic fields of uranus and neptune. Icarus 211, 798–803 (2011).
48. 48.
Millot, M. et al. Nanosecond x-ray diffraction of shock-compressed superionic water ice. Nature 569, 251–255 (2019).
49. 49.
French, M., Mattsson, T. R. & Redmer, R. Diffusion and electrical conductivity in water at ultrahigh pressures. Phys. Rev. B 82, 174108 (2010).
50. 50.
Marcolongo, A., Ercole, L. & Baroni, S. Gauge fixing for heat-transport simulations. J. Chem. Theory Comput. 16, 3352-3362 (2020).
51. 51.
Ercole, L., Bertossa, R., & Bisacchi, S. ThermoCepstrum: a code to estimate transport coefficients from the cepstral analysis of a multi-variate current stationary time series (2017–2020).
52. 52.
Grasselli, F. & Baroni, S. Topological quantisation and gauge-invariance of charge transport in liquid insulators. Nat. Phys. 15, 967–972 (2019).
53. 53.
Marcolongo, A. & Marzari, N. Ionic correlations and failure of nernst-einstein relation in solid-state electrolytes. Phys. Rev. Mater. 1, 025402 (2017).
54. 54.
France-Lanord, A. & Grossman, J. C. Correlations from Ion Pairing and the Nernst-Einstein Equation. Phys. Rev. Lett. 122, 1–6 (2019).
55. 55.
Wilson, H. F., Wong, M. L. & Militzer, B. Superionic to superionic phase change in water: Consequences for the interiors of uranus and neptune. Phys. Rev. Lett. 110, 151102 (2013).
56. 56.
He, X., Zhu, Y. & Mo, Y. Origin of fast ion diffusion in super-ionic conductors. Nat. Commun. 8, 1–7 (2017).
57. 57.
Podolak, M., Helled, R. & Schubert, G. Effect of non-adiabatic thermal profiles on the inferred compositions of uranus and neptune. Monthly Not. R. Astronomical Soc. 487, 2653–2664 (2019).
58. 58.
French, M. Thermal conductivity of dissociating water-an ab initio study. N. J. Phys. 21, 023007 (2019).
59. 59.
Stixrude, L., Baroni, S. & Grasselli, F. Thermal evolution of Uranus with a frozen interior. arxiv http://arxiv.org/abs/2004.01756 arXiv:2004.01756 (2020).
60. 60.
Stanley, S. & Bloxham, J. Numerical dynamo models of Uranus’ and Neptune’s magnetic fields. Icarus 184, 556–572 (2006).
## Acknowledgements
This work was partially funded by the EU through the MAX Centre of Excellence for supercomputing applications (Project No. 824143), and by the US National Science Foundation under grant EAR-1853388. We are grateful to Riccardo Bertossa and Davide Tisi for valuable discussions and assistance.
## Author information
Authors
### Contributions
F.G., L.S., and S.B. contributed to conceive this research, perform the simulations, analyze the results, and write the manuscript.
### Corresponding author
Correspondence to Stefano Baroni.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review information Nature Communications thanks Giulia Galli, Flavio Toigo and the other, anonymous, reviewer for their contribution to the peer review of this work.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Grasselli, F., Stixrude, L. & Baroni, S. Heat and charge transport in H2O at ice-giant conditions from ab initio molecular dynamics simulations. Nat Commun 11, 3605 (2020). https://doi.org/10.1038/s41467-020-17275-5
• Accepted:
• Published: | 2021-05-15 05:48:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8173813819885254, "perplexity": 1857.8778304156413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989812.47/warc/CC-MAIN-20210515035645-20210515065645-00170.warc.gz"} |
http://publ.plaidweb.site/blog/85-Publ-0-6-6-Authl-0-4-0 | # Publ 0.6.6, Authl 0.4.0
Posted (2 years ago)
I’ve just released new versions of Publ and Authl.
Publ v0.6.6 changes:
• Fixed a regression that made it impossible to log out
• Fixed a problem where WWW-Authenticate headers weren’t being cached properly
• Improve the changed-file cache-busting methodology
• Add object pooling to Entry, Category, and View (for a potentially big memory and performance improvement)
Authl v0.4.0 changes:
• Finally started to add unit tests
• Removed some legacy WebFinger code that was no longer relevant or ever touched
• Added a mechanism to allow providers to go directly to login, as appropriate
• Added friendly visual icons for providers which support them (a so-called “NASCAR interface”)
## Publ 0.6.6
The main reason for this update is just that the embarrassing logout bug was rearing its head and I wanted to fix it on my site without monkeypatching it or temporarily moving to git head or whatever. The WWW-Authenticate fix is nice, though, as it’s related to some work I’m doing on Pushl (namely adding the ability to retrieve bearer tokens from an external helper program).
It’s difficult to estimate what a performance change will be like based on testing on a developer desktop vs. a production VPS. In particular, the various I/O performance characteristics can vary a lot, and Publ is primarily I/O bound. In my desktop-side testing I found that the object pooling increased performance by 15%, which is already pretty great, but that’s also on a machine with a lot of memory, a huge file cache, and no disk virtualization. I’ve only deployed Publ 0.6.6 on my personal website around half an hour ago, but already my site monitoring is showing a rather impressive performance improvement. For example, the Atom feed used to take around 30 seconds to render on a cache miss. Right now it seems to take 2.5 seconds.
So, yeah, it takes only 10% of the time to run now – that’s around a 900% performance improvement in a typical deployment scenario. So, that’s pretty great.
Right now the largest remaining performance bottleneck seems to be in PonyORM, which is unfortunate. I haven’t yet figured out if it’s with PonyORM itself, or with its interface to sqlite. From what I can tell, the way that trace profiling works in Python means that things with a lot of function calls become quite a lot slower than long-running things within a single function, so things that do a lot of abstraction and dependency injection (like, say, PonyORM) get unfairly impacted in trace profiling. A sample-based profiling approach would be much more fair and realistic, but I haven’t found any sample-based Python profilers (and I don’t know enough about Python’s internals to know if that’s even a possibility).
My short-term goals for Publ are otherwise unchanged since the last release announcement.
## Authl 0.4.0
I hadn’t worked on Authl in quite some time, but I felt like it needed some attention.
These Authl changes are basically for some UX improvements that had been bugging me for a while; there was an awful lot of text to read and that was possibly scary to newcomers. Now there’s still just as much text to read but there’s friendly icons for a bunch of the supported services, and silo services such as Twitter can now go straight to the login flow without implying that the username is necessary.
Here’s a before and after on the default Flask template:
The next thing I want to work on for Authl is finally adding actual support for user profiles. This would also probably go along with things like adding more providers, particularly Facebook, Tumblr, and maybe even OpenID 1.x (i.e. Dreamwidth). Better profile support means having a friendlier greeting than just the canonical identity URL, among other things that people might want in their own federated login use cases.
## Some other thoughts of things that would be neat
Now that Publ supports entry attachments, it might be reasonable to add native server-side webmentions; rather than fetching the mentions from webmention.io on every page view, have a webhook on update that triggers a script that fetches and formats the mentions as an attachment that can then be rendered and cached, as well as getting all of the benefits of SEO that it would bring. For some sites, having the comments be indexed by the search engines makes a huge difference to page ranking, since the conversation about an article can add in some useful keywords that weren’t in the actual article. (Not to mention it improves the page’s “freshness” as far as the search engine is concerned.)
Another thought I’ve had about attachments is they could be used to implement a server-side comment system, although that would require a lot more work than webmention rendering (UI, moderation/spam-filtering, migrating stuff again) and after all the work I put into my Isso setup I’m not quite ready to think about how to actually do that. I’d probably want to do it in the form of having a mechanism to pre-render the Isso comment thread and form into an HTML attachment rather than having every part of it handled via Publ entry attachments. | 2022-06-30 07:48:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23560795187950134, "perplexity": 2191.611051035014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00295.warc.gz"} |
http://math.stackexchange.com/questions/72106/sequences-of-a-computable-function | Sequences of a computable function
Is there any computable function $f(n)$, which given any integer $n$ has been proven to return either $0$ or $1$ in finite time, and for which the statement "$f(1), f(2), f(3),\ldots$ contains arbitrarily large sequences of $0$'s" has been proved to be undecidable in PA or ZFC?
If not, is there any proof of the existence or non-existence of such a function?
Edit: Is there one which is also morally undecidable?
-
What is meant by "morally undecidable"? – r.e.s. Oct 16 '11 at 14:30
Let $G$ be a Gödel sentence (with intuitive meaning "I cannot be proved"), and take $$f(n)=\cases{0&\text{if }G\text{ has a proof with Gödel number }<n\\1&\text{otherwise}}$$
Let $G(n)$ be the Goodstein sequence with first element $n$, and take $$f(n)=\cases{0&\text{if }0 \ \text{is an element of} \ G(n) \\1&\text{otherwise}}$$ | 2016-05-31 08:27:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8825864195823669, "perplexity": 299.7896452353582}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051196108.86/warc/CC-MAIN-20160524005316-00212-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/7/lesson/7.2.1/problem/7-38 | ### Home > MC1 > Chapter 7 > Lesson 7.2.1 > Problem7-38
7-38.
Greg is building a fence. He places two sections end to end. One section measures $3$ meters, and the other measures $120$ centimeters.
1. Greg thinks the fence will be $123$ units long but is not sure what the units are. His sister Elisa thinks he is wrong. What do you think? Explain.
Elisa is right. Do you know why?
2. Find at least two ways to express the length of the fence.
Try writing the length in meters and centimeters. | 2023-02-03 13:32:47 | {"extraction_info": {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6438740491867065, "perplexity": 1397.4629823671296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00848.warc.gz"} |
https://physics.stackexchange.com/questions/508597/prove-that-the-quantum-fourier-transform-satisfies-f-n-sum-x-xr-rangle-sim | # Prove that the Quantum Fourier Transform satisfies $F_N(\sum_x |xr \rangle) \simeq \sum_y |y N/r \rangle$ when $r$ divides $N$
Before I start, this is a homework question, but I have spent the last 7 hours on it and gotten nowhere. I will mention what I've tried and noticed, but it's not much. At this point, I'm wondering if there isn't a typo somewhere...
The QFT (Quantum Fourier Transform) for $$n$$ qubits can be expressed in matrix form as (where $$N = 2^n$$):
$$\begin{pmatrix} \omega_N^{0 \cdot 0} & \cdots & \omega_N^{0 \cdot (N-1)} \\ \vdots & & \vdots \\ \omega_N^{(N-1) \cdot 0} & \cdots & \omega_N^{(N-1) \cdot (N-1)} \end{pmatrix}$$
The $$\omega_N = e^{i\frac{2 \pi}{N}}$$ denotes the $$N$$-th complex root of $$1$$.
I have to show the following relation, where $$r$$ divides $$N$$:
$$F_N\left(\sqrt{\frac{r}{N}} \sum_{x=0}^{\frac{N}{r} - 1} |xr \rangle\right) = \sqrt{\frac{1}{r}} \sum_{x=0}^{r-1} |x \frac{N}{r} \rangle$$
My two main ways of approaching this have been to rewrite the sum to come as close as possible to the expression on the right and then apply the QFT; and to directly apply the QFT to the terms in the sum. I know that $$|x\rangle$$ can be written as a column vector with a $$1$$ in the $$x+1$$ position, and I can see that we if we write $$N = \lambda r$$, then instead of having a $$1$$ every $$r$$ positions we get an $$r$$ every $$\lambda$$ positions in the RHS.
I've also drawn the unit circle in Argand's plane (the complex plane) and tried to see how the roots may interact between them, but once again to no avail, although I have obtained some neat formulas from that.
If $$F$$ denotes the QFT operation, then by definition $$\sqrt N F|j\rangle=\sum_k \omega^{jk}|k\rangle$$.
By linearity, you therefore have $$F\left(\sum_{x=0}^{N/r-1}|xr\rangle\right)= \sum_{x=0}^{N/r-1}F|xr\rangle =\frac{1}{\sqrt N}\sum_{y=0}^{N-1}\left(\sum_{x=0}^{N/r-1}\omega_N^{xyr}\right)|y\rangle.$$ Now by assumption $$r\mid N$$, and thus $$m\equiv N/r\in\mathbb N$$. Using the usual formulas for the geometric series we see that $$\sum_{x=0}^{m-1}\omega_N^{xyr}= \sum_{x=0}^{m-1}\exp\left(\frac{2\pi i}{m}xy\right)= %\frac{e^{2\pi i y}-1}{e^{2\pi i y/m}-1}, m \delta_{y\in m\mathbb Z},$$ and thus $$F\left(\sum_{x=0}^{N/r-1}|xr\rangle\right) =\frac{m}{\sqrt N}\sum_{y=0}^{N-1} \delta_{y\in m\mathbb Z}|y\rangle =\frac{m}{\sqrt N}\sum_{k=0}^{r-1} |kN/r\rangle.$$ Multiplying by $$\sqrt{1/m}$$ the above expression you get the result. | 2020-02-20 05:59:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9923982620239258, "perplexity": 106.43521565674152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144637.88/warc/CC-MAIN-20200220035657-20200220065657-00211.warc.gz"} |
http://codeforces.com/problemset/problem/142/A | A. Help Farmer
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | 2018-06-21 04:37:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27256014943122864, "perplexity": 1761.8498325658793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00123.warc.gz"} |
https://lists.gnu.org/archive/html/lilypond-user/2003-01/msg00109.html | lilypond-user
[Top][All Lists]
## Re: Where do I save .ly files so that the lilypond program can find it a
From: Graham Percival Subject: Re: Where do I save .ly files so that the lilypond program can find it and read it? Date: Wed, 22 Jan 2003 14:19:05 -0800
```On Wed, 22 Jan 2003 13:28:38 -0800
> I just downloaded Lilypond for Windows and tried to run the
> test program as stated in the Tutorial. Everytime I run the
> program, it states that it cannot find the 'test' file. I
> am assuming that I have to save it in a particular space.
Does your current directory, or the location of Lilypond's exe file,
contain spaces?(ie "C:\Documents and Stuff\Lilypond Stuff\test.ly" or
"C:\program files\lilypond\lilypond.exe")
I've heard that this can cause problems for Lilypond. Note that I've
never used Lilypond on windows, so this could be completely incorrect
or have no bearing on your problem. Try running lilypond on a file
that doesn't have spaces in the directory name. If you get the same
error message, then obviously my advice was no good. :)
Cheers,
- Graham
``` | 2022-12-08 04:33:06 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8699367046356201, "perplexity": 13808.880478722265}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00613.warc.gz"} |
http://openstudy.com/updates/55ca570de4b0c5fe98054d10 | ## anonymous one year ago [10.02] What are the solutions for x in the proportion 5 times x minus 6 all over x equals 2 times x all over 2.? x = −12 and x = −8 x = 12 and x = 8 x = −2 and x = −3 x = 2 and x = 3
1. anonymous
$\frac{ 5x-6 }{ x }=\frac{ 2x }{ 2 }$
2. anonymous
Yeah
3. anonymous
Know how to solve it??
4. anonymous
Not really I mean I know that you have to substitute the x values for each of them but when I did it none of them worked
5. anonymous
cross multiply it.
6. anonymous
leads to $10x^2-12=2x^2$ understnad how I got that?
7. anonymous
i mean 10x
8. anonymous
Not really
9. anonymous
$10x-12=2x^2$ becomes $2x^2-10x-12=0$
10. anonymous
Understand?
11. anonymous
Yes
12. anonymous
$2x^2-10x+12=0$ I mean. Someone is distracting me in real life. Sorry
13. anonymous
$2(x^2-5x+6)=2(x-3)(x-2)=0$ Know how to solve for x?
14. anonymous
I dont understand which one it would be XD Im sorry i'm just really bad at math
15. anonymous
@Shalante
16. anonymous
anything multiplied by a zero equals zero. so set anything equal to zero.
17. anonymous
Wouldnt the answer be x = 2 and x = 3 ?
18. anonymous
Yes, it would. You are right. | 2017-01-24 19:38:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7197730541229248, "perplexity": 1932.0296806529873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560285244.23/warc/CC-MAIN-20170116095125-00204-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/102888/y-axis-length-for-lowess-curve-in-stata | # y-axis length for LOWESS curve in Stata
I have a problem considering the length of the y-axis in a LOWESS plot in Stata. The y variable ranges from 0 to 50. I would like to run the LOWESS on all the data but only display the graph for the range of y from 30 to 50. I've searched throughout the net, but I just found that I would also have to restrict the data to $\ge$ 30, but that of course changes the curve. Do you have any advice? The graphs are simply unusable for that y-axis interval. Thanks! | 2020-04-02 07:23:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5212702751159668, "perplexity": 444.2783397623207}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370506673.7/warc/CC-MAIN-20200402045741-20200402075741-00516.warc.gz"} |
https://users.aalto.fi/~tjunttil/2020-DP-AUT/notes-sat/resolution.html | # The resolution proof system¶
Suppose that you run a SAT solver on your problem instance and the solver answers “the formula is satisfiable” and provides you a satisfying truth assingment. It is now quite easy to check that the verdict is correct: just evaluate the formula under the assignment and check that all the clauses are satisfied.
However, if the solver simply returns “the formula is unsatisfiable”, can you trust the solver? Although the current SAT solvers are very robust, a bug in the complex implementation of a solver is still a possibility. In many applications this possibility can arguably be ignored but in some others, such as using SAT solvers to prove open mathematical problems, one really wants to verify that there are no solutions. Some solvers can in fact also produce proofs of unsatisfiability. Such proofs can be independently and automatically checked by some (relatively simple and possibly formally verified) proof checking tool. In the following, we introduce the classic proof system called resolution. The best tools at the moment use something more complicated but these are not covered in this course.
The resolution proof system is a proof system for CNF formulas that is based on taking “resolvents” of two clauses.
Definition: Resolvent of two clauses
Let $$C_1 = (x \lor l_1 \lor l_2 \lor … \lor l_k)$$ and $$C_2 = (\neg x \lor l’_1 \lor l’_2 \lor … \lor l’_p)$$ be two clauses and $$x$$ any variable. The resolvent of $$C_1$$ and $$C_2$$ with respect to $$x$$ is the clause $$\Res{C_1}{C_2}{x} = (l_1 \lor l_2 \lor … \lor l_k \lor l’_1 \lor l’_2 \lor … \lor l’_p)$$.
Example
• The resolvent of $$(x \lor y \lor \neg z)$$ and $$(\neg y \lor \neg z \lor w)$$ is with respect to $$y$$ the clause $$(x \lor \neg z \lor w)$$.
• The resolvent of $$(x)$$ and $$(\neg x)$$ is the empty clause $$()$$.
• The resolvent of $$(x \lor y \lor z)$$ and $$(\neg y \lor \neg z \lor w)$$ with recpect to $$y$$ is $$(x \lor z \lor \neg z \lor w)$$, a tautology.
The key fact is that the resolvent $$\Res{C_1}{C_2}{x}$$ is a logical consequence of $$C_1 \land C_2$$, where $$C_1 = (x \lor l_1 \lor l_2 \lor … \lor l_k)$$ and $$C_2 = (\neg x \lor l’_1 \lor l’_2 \lor … \lor l’_p)$$. One can reason this as follows. First, $$x$$ must be either false or true in any truth assignment that satisfies $$C_1 \land C_2$$. If $$x$$ is false, one of $$l_1,l_2,…,l_k$$ must be true in a truth assignment satisfying $$C_1$$. Similarly, if $$x$$ is true, then one of $$l’_1,l’_2,…,l’_p$$ must be true. Therefore, in any case one of $$l_1,l_2,…,l_k,l’_1,l’_2,…,l’_p$$ must be true in a truth assignment satisfying $$C_1 \land C_2$$ and thus $$\Res{C_1}{C_2}{x}$$ is a logical consequence of $$C_1 \land C_2$$.
Definition: Resolution proofs of unsatisfiability
A resolution proof of unsatisfiability of a CNF formula $$\phi$$ is a finite sequence $$C_1,C_2,…,C_m$$ of clauses such that each clause $$C_i$$ in it is a non-tautology and either
1. a clause in the formula $$\phi$$, or
2. the resolvent of two earlier clauses $$C_j$$ and $$C_k$$, $$j,k < i$$, in the sequence.
Furthermore, the last clause $$C_m$$ must be the empty clause $$()$$ that is not satisfied in any truth assignment.
The resolution proof system is both (i) sound, meaning that if a formula has a resolution proof of unsatisfiability, then it actually is unsatisfiable, and (ii) complete, meaning that for each unsatisfiable CNF formula there is a resolution proof of unsatisfiability. Therefore, a formula $$\phi$$ is unsatisfiable if and only if there is a resolution proof of unsatisfiability for it.
Example
Consider again the unsatisfiable CNF formula $$\phi = (a \lor \neg b) \land (\neg a \lor c \lor \neg d) \land (a \lor c \lor \neg d) \land (\neg c \lor \neg e) \land (\neg c \lor e) \land (c \lor d).$$ A resolution proof of its unsatisfiability is the sequence $$C_1,…,C_9$$ where
• $$C_1 = (\neg c \lor e)$$, a clause in the formula $$\phi$$
• $$C_2 = (\neg c \lor \neg e)$$, a clause in the formula $$\phi$$
• $$C_3 = (\neg c)$$, the resolvent $$\Res{C_1}{C_2}{e}$$
• $$C_4 = (a \lor c \lor \neg d)$$, a clause in the formula $$\phi$$
• $$C_5 = (\neg a \lor c \lor \neg d)$$, a clause in the formula $$\phi$$
• $$C_6 = (c \lor \neg d)$$, the resolvent $$\Res{C_4}{C_5}{c}$$
• $$C_7 = (c \lor d)$$, a clause in the formula $$\phi$$
• $$C_8 = (c)$$, the resolvent $$\Res{C_6}{C_7}{d}$$
• $$C_9 =()$$, the resolvent $$\Res{C_3}{C_8}{c}$$
Finding resolution proofs of unsatisfiability directly can be difficult for humans especially. However, the search tree of DPLL without unit propagation (recall the section The DPLL backtracking search procedure) can be converted to a resolution proof quite easily:
1. attach each dead-end node with one clause that is falsified in it
2. attach each other node with
• the clause of one of its children if that clause does not involve the branching variable of the node, or
• the resolvent of the clauses in its two children otherwise
3. the root will be attached with the empty clause
4. list the clauses attached in the tree in post-order
The correctness can be shown by structural induction on the tree, showing that the clause attached to each node contains only literals that are negations of the ones in its partial truth assignment. The proofs obtained in this way are actually tree-like resolution proofs, where a clause obtained by using the resolution proof is used only once. For some formula families, such proofs can be superpolynomially larger than the ones that can be obtained with unrestricted resolution.
Example
Consider again the CNF formula $$(a \lor \neg b) \land (\neg a \lor c \lor \neg d) \land (a \lor c \lor \neg d) \land (\neg c \lor \neg e) \land (\neg c \lor e) \land (c \lor d)$$. Annotating the search tree nodes with clauses as described above, we get
This corresponds to the resolution proof of unsatisfiability $$(a \lor \neg b)$$, $$(c \lor d)$$, $$(a \lor c \lor \neg d)$$, $$(a \lor c)$$, $$(\neg c \lor e)$$, $$(\neg c \lor \neg e)$$, $$(\neg c)$$, $$(a)$$, $$(a)$$, $$(c \lor d)$$, $$(\neg a \lor c \lor \neg d)$$, $$(\neg a \lor c)$$, $$(\neg c \lor e)$$, $$(\neg c \lor \neg e)$$, $$(\neg c)$$, $$(\neg a)$$, $$()$$.
Most modern SAT solvers can, in theory at least, polynomially simulate resolution [PipatsrisawatDarwiche2011]. However, it has been proven that some formulas do not have polynomial-size resolution proofs [Haken1985]. For instance, the “pigeon hole formulas” $$\PHP{p}{h} = (\bigwedge_{i=1}^p(\bigvee_{k=1}^h \PH{i}{k})) \land \bigwedge_{k=1}^h\bigwedge_{1 \le i < j \le p}(\neg \PH{i}{k} \lor \neg \PH{j}{k})$$ where $$\PH{i}{j}$$ is true if the pigeon $$i$$ sits in the hole $$j$$, model all the configurations in which $$p$$ pigeons sit in $$h$$ holes and each hole has at most pigeon in it. When $$p > h$$, the formulas are unsatisfiable. The formulas $$\PHP{p}{p-1}$$ do not have polynomial-size resolution proofs.
There are stronger proof systems for which we do not know any formula families that require super-polynomial sized proofs. If no such formula families exists, then $$\NP = \coNP$$. Some SAT solvers go beyond resolution and can solve, for instance, pigeon hole problems efficiently. | 2022-12-03 16:17:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7858812212944031, "perplexity": 493.7071595776115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00443.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3323594 | # Fraction Change in Angular Momentum of Cone Full of sand compared with half full.
by zerakith
Tags: cone, emptying, inertia, sand
P: 7 1. A light, hollow cone is filled with sand set spinning about a vertical axis through its apex on a frictionless bearing. Sand is allowed to drain slowly through a hole in the apex. Calculate the fractional change in angular velocity when the sand level has fallen to half its original value. You may neglect the contribution of the hollow cone to the moment of inertia. 2. Moment of Inertia of Solid Cone: $$I=\frac{\rho\pi R^4h}{10}$$ Conservation of Angular Momentum: $$I_0\omega_0=I_1\omega_1$$ Dimensions of the cone: Length: h. Radius of circle at end of cone: R 3. So its clear to me that the way to proceed is to consider the conservation of angular momentum. At the start the cone is full of sand and thus the system is just a solid cone so: $$I_0=\frac{\rho\pi R^4h}{10}$$ For $$I_1$$ i need to calculate the moment of inertia, the sticking point for me is the shape the sand will take within the cone. Intuitivly i think that the sand will form a conical ring around the cone (i.e so there is a smaller cone of empty space inside the cone and the rest is filled with sand). I'm not really happy with jumping to that conclusion however if I do that and work it through I get: $$\frac{\omega_1}{\omega_0}=\frac{h^3}{3}$$ I've no real way to determine whether this is correct. Am I right in the assumption about the shape of the sand (and ideally why?), and does the fractional change I have ended up with make sense? There is no rush to this, it's not examined and term is over, it is however, bugging me. Thanks in advance Zerakith | 2014-08-21 02:28:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6984067559242249, "perplexity": 287.83108313401067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500813887.15/warc/CC-MAIN-20140820021333-00217-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/pages/lecture-notes/ | # Lecture Notes
The lecture notes are from an earlier version of this course, but still correspond to the topics covered in this version.
SES # TOPICS LECTURE NOTES
I. Thermodynamics
L1 Fundamental Definitions, The Zeroth Law, The First Law Lecture Note 1 (PDF)
L2 The Second Law, Carnot Engines and Thermodynamic Temperature, Entropy Lecture Note 2 (PDF)
L3 Approach to Equilibrium and Thermodynamic Potentials, Useful Mathematical Results Lecture Note 3 (PDF)
L4 Stability Conditions, The Third Law Lecture Note 4 (PDF)
II. Probability
L5 General Definitions, One Random Variable, Some Important Probability Distributions Lecture Note 5 (PDF)
L6 Many Random Variables, Sums of Random Variables and the Central Limit Theorem, Rules for Large Numbers, Information, Entropy, and Estimation Lecture Note 6 (PDF)
III. Kinetic Theory of Gases
L7 General Definitions, Liouville’s Theorem Lecture Note 7 (PDF)
L8 The Bogoliubov-Born-Green-Kirkwood-Yvon Hierarchy, The Boltzmann Equation Lecture Note 8 (PDF)
L9 The H-Theorem and Irreversibility, Equilibrium Properties Lecture Note 9 (PDF)
L10 Conservation Laws Lecture Note 10 (PDF)
L11 Zeroth Order Hydrodynamics, First Order Hydrodynamics Lecture Note 11 (PDF)
IV. Classical Statistical Mechanics
L12 General Definitions, The Microcanonical Ensemble, Two-Level Systems Lecture Note 12 (PDF)
L13 The Ideal Gas, Mixing Entropy and Gibbs’ Paradox, The Canonical Ensemble Lecture Note 13 (PDF)
L14 Examples, The Gibbs Canonical Ensemble, The Grand Canonical Ensemble Lecture Note 14 (PDF)
V. Interacting Particles
L15 The Cumulant Expansion Lecture Note 15 (PDF)
L16 The Cluster Expansion Lecture Note 16 (PDF)
L17 The Second Virial Coefficient and Van der Waals Equation, Breakdown of the Van der Waals Equation, Mean Field Theory of Condensation Lecture Note 17 (PDF)
L18 Variational Methods, Corresponding States, Critical Point Behavior Lecture Note 18 (PDF)
VI. Quantum Statistical Mechanics
L19 Mean field theory of condensation, Corresponding states, Critical point behavior (from L17 & L18) Lecture Note 19 (PDF)
L20 Dilute Polyatomic Gases, Vibrations of a Solid, Black-body Radiation Lecture Note 20 (PDF)
L21 Quantum Microstates, Quantum Macrostates Lecture Note 21 (PDF)
VII. Ideal Quantum Gases
L22 Hilbert Space of Identical Particles Lecture Note 22 (PDF)
L23 Canonical Formulation, Grand Canonical Formulation Lecture Note 23 (PDF)
L24 The Degenerate Fermi Gas Lecture Note 24 (PDF)
L25 The Degenerate Bose Gas Lecture Note 25 (PDF)
L26 Superfluid He4 Lecture Note 26 (PDF)
#### Learning Resource Types
theaters Lecture Videos
assignment Problem Sets
notes Lecture Notes | 2022-09-30 18:20:16 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9261573553085327, "perplexity": 5107.757466064052}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335504.22/warc/CC-MAIN-20220930181143-20220930211143-00199.warc.gz"} |
https://brilliant.org/problems/find-the-number-23/ | # Find the number
Algebra Level pending
The sum of the digits of a two digit positive integer is $$9$$. If the digits are reversed, the new number is $$9$$ less than $$3$$ times the original number. Find this number.
× | 2017-09-21 12:36:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5403158068656921, "perplexity": 371.34985694437665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00657.warc.gz"} |
https://tex.stackexchange.com/questions/133932/how-do-i-type-the-infinity-symbol-in-mactex | # How do I type the infinity symbol in MacTex? [duplicate]
\item Obtain the one-to-one function $f_1$ and $f_2$ by cutting the graph of $f$ at
a certain point ($x_1$, $y_1$) so that domain of ($f_1$)=($-∞$ , $x_1$] and
domain ($f_2$)=[$x_1$,$+∞$)
• You need \infty – percusse Sep 18 '13 at 15:00
• When using XeLaTeX or LuaLaTeX as your typesetter, you need to include the »unicode-math« package. – Thorsten Donig Sep 18 '13 at 15:07
• Just a general tip: It's safter to use $$f_1$$ and $f_1$ than $f_1$ and $$f_1$$. For more information, see this question: tex.stackexchange.com/questions/503/why-is-preferable-to or read about it in l2tabu. – Sam Whited Sep 18 '13 at 15:11
• @SamWhited -- I would disagree with you on the claim that there's a generic safety related advantage to using $$...$$ rather than $...$ to delimit inline math. Indeed, since $$and $$ are not "robust" commands (in the LaTeX sense of the word "robust"), it's perilous to use them in the arguments of "moving" commands; no such difficulties arise with $. Note that the link you provide regards the use of $$ -- a rather different matter. – Mico Sep 18 '13 at 15:58 • @Mico Fair point; l2tabu doesn't mention either, though I had thought it talked about line spacing a bit. Oh well, ignore my comment (although I'd still go with the LaTeX way unless you need to put it in a moving argument for some reason). – Sam Whited Sep 18 '13 at 16:20 ## 2 Answers A better style is: Obtain the one-to-one function $f_1$ and $f_2$ by cutting the graph of $f$ at a certain point $(x_1, y_1)$ so that domain of $(f_1)=(-\infty , x_1]$ and domain $(f_2)=[x_1,+\infty)$ (Please compare the obtained spacing). If you are still faced with such a problem, as the last resort, you can use the rotated eight as follows. \documentclass[preview,border=12pt]{standalone} \usepackage{graphicx} \def\infinity{\rotatebox{90}{8}} \begin{document}$(-\infinity, x_1]\$
\end{document}
• Only meticulous readers will spot the difference. – kiss my armpit Sep 18 '13 at 15:38
• Every good math font has the symbol corresponding to \infty. – egreg Sep 18 '13 at 15:44
• @egreg: Yes. It is just for emergency purposes. :-) – kiss my armpit Sep 18 '13 at 15:47
• Some fonts have a \infty that really looks like a rotated 8 instead of a distinct design. – lblb Oct 14 '17 at 15:48 | 2020-10-27 18:08:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9014819860458374, "perplexity": 1299.6371905114704}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00668.warc.gz"} |
https://fr.mathworks.com/help/comm/ug/hard-vs-soft-decision-demodulation-examples.html | ## Hard- vs. Soft-Decision Demodulation Examples
### Log-Likelihood Ratio (LLR) Demodulation
This example shows the BER performance improvement for QPSK modulation when using log-likelihood ratio (LLR) instead of hard-decision demodulation in a convolutionally coded communication link. With LLR demodulation, one can use the Viterbi decoder either in the unquantized decoding mode or the soft-decision decoding mode. Unquantized decoding, where the decoder inputs are real values, though better in terms of BER, is not practically viable. In the more practical soft-decision decoding, the demodulator output is quantized before being fed to the decoder. It is generally observed that this does not incur a significant cost in BER while significantly reducing the decoder complexity. We validate this experimentally through this example.
For a Simulink™ version of this example, see LLR vs. Hard Decision Demodulation in Simulink.
Initialization
Initialize simulation parameters.
```M = 4; % Modulation order bitsPerIter = 1.2e4; % Number of bits to simulate EbNo = 3; % Information bit Eb/No in dB```
Initialize coding properties for a rate 1/2, constraint length 7 code.
```codeRate = 1/2; % Code rate of convolutional encoder constLen = 7; % Constraint length of encoder codeGenPoly = [171 133]; % Code generator polynomial of encoder tblen = 32; % Traceback depth of Viterbi decoder trellis = poly2trellis(constLen,codeGenPoly);```
Create a `comm.ConvolutionalEncoder` System object™ by using `trellis` as an input.
`enc = comm.ConvolutionalEncoder(trellis);`
Channel
The signal going into the AWGN channel is the modulated encoded signal. To achieve the required noise level, adjust the Eb/No for coded bits and multi-bit symbols. Calculate the $SNR$ value based on the ${E}_{b}/{N}_{o}$ value you want to simulate.
`SNR = convertSNR(EbNo,"ebno","BitsPerSymbol",log2(M),"CodingRate",codeRate);`
Viterbi Decoding
Create `comm.ViterbiDecoder` objects to act as the hard-decision, unquantized, and soft-decision decoders. For all three decoders, set the traceback depth to `tblen`.
```decHard = comm.ViterbiDecoder(trellis,'InputFormat','Hard', ... 'TracebackDepth',tblen); decUnquant = comm.ViterbiDecoder(trellis,'InputFormat','Unquantized', ... 'TracebackDepth',tblen); decSoft = comm.ViterbiDecoder(trellis,'InputFormat','Soft', ... 'SoftInputWordLength',3,'TracebackDepth',tblen);```
Calculating the Error Rate
Create `comm.ErrorRate` objects to compare the decoded bits to the original transmitted bits. The Viterbi decoder creates a delay in the decoded bit stream output equal to the traceback length. To account for this delay, set the `ReceiveDelay` property of the `comm.ErrorRate` objects to `tblen`.
```errHard = comm.ErrorRate('ReceiveDelay',tblen); errUnquant = comm.ErrorRate('ReceiveDelay',tblen); errSoft = comm.ErrorRate('ReceiveDelay',tblen);```
System Simulation
Generate `bitsPerIter` message bits. Then convolutionally encode and modulate the data.
```txData = randi([0 1],bitsPerIter,1); encData = enc(txData); modData = pskmod(encData,M,pi/4,InputType="bit");```
Pass the modulated signal through an AWGN channel.
`[rxSig,noiseVariance] = awgn(modData,SNR);`
Before using a `comm.ViterbiDecoder` object in the soft-decision mode, the output of the demodulator needs to be quantized. This example uses a `comm.ViterbiDecoder` object with a `SoftInputWordLength` of 3. This value is a good compromise between short word lengths and a small BER penalty. Define partition points for 3-bit quantization.
```demodLLR.Variance = noiseVariance; partitionPoints = (-1.5:0.5:1.5)/noiseVariance;```
Demodulate the received signal and output hard-decision bits.
`hardData = pskdemod(rxSig,M,pi/4,OutputType="bit");`
Demodulate the received signal and output LLR values.
`LLRData = pskdemod(rxSig,M,OutputType="llr");`
Hard-decision decoding
Pass the demodulated data through the Viterbi decoder. Compute the error statistics.
```rxDataHard = decHard(hardData); berHard = errHard(txData,rxDataHard);```
Unquantized decoding
Pass the demodulated data through the Viterbi decoder. Compute the error statistics.
```rxDataUnquant = decUnquant(LLRData); berUnquant = errUnquant(txData,rxDataUnquant);```
Soft-decision decoding
Pass the demodulated data to the `quantiz` function. This data must be multiplied by `-1` before being passed to the quantizer, because, in soft-decision mode, the Viterbi decoder assumes that positive numbers correspond to 1s and negative numbers to 0s. Pass the quantizer output to the Viterbi decoder. Compute the error statistics.
```quantizedValue = quantiz(-LLRData,partitionPoints); rxDataSoft = decSoft(double(quantizedValue)); berSoft = errSoft(txData,rxDataSoft);```
Running Simulation Example
Simulate the previously described communications system over a range of Eb/No values by executing the simulation file `simLLRvsHD`. It plots BER results as they are generated. BER results for hard-decision demodulation and LLR demodulation with unquantized and soft-decision decoding are plotted in red, blue, and black, respectively. A comparison of simulation results with theoretical results is also shown. Observe that the BER is only slightly degraded by using soft-decision decoding instead of unquantized decoding. The gap between the BER curves for soft-decision decoding and the theoretical bound can be narrowed by increasing the number of quantizer levels.
This example may take some time to compute BER results. If you have the Parallel Computing Toolbox™ (PCT) installed, you can set `usePCT` to `true` to run the simulation in parallel. In this case, the file `LLRvsHDwithPCT` is run.
To obtain results over a larger range of Eb/No values, modify the appropriate supporting files. Note that you can obtain more statistically reliable results by collecting more errors.
```usePCT = false; if usePCT && license('checkout','Distrib_Computing_Toolbox') ... && ~isempty(ver('parallel')) LLRvsHDwithPCT(1.5:0.5:5.5,5); else simLLRvsHD(1.5:0.5:5.5,5); end```
Appendix
The following functions are used in this example:
### LLR vs. Hard Decision Demodulation in Simulink
This model shows the improvement in BER performance when using log-likelihood ratio (LLR) instead of hard decision demodulation in a convolutionally coded communication link.
For a MATLAB® version of this example, see Log-Likelihood Ratio (LLR) Demodulation.
System Setup
This example model simulates a convolutionally coded communication system having one transmitter, an AWGN channel and three receivers. The convolutional encoder has a code rate of 1/2. The system employs a 16-QAM modulation. The modulated signal passes through an additive white Gaussian noise channel. The top receiver performs hard decision demodulation in conjunction with a Viterbi decoder that is set up to perform hard decision decoding. The second receiver has the demodulator configured to compute log-likelihood ratios (LLRs) that are then quantized using a 3-bit quantizer. It is well known that the quantization levels are dependent on noise variance for optimum performance [2]. The exact boundaries of the quantizer are empirically determined here. A Viterbi decoder that is set up for soft decision decoding processes these quantized values. The LLR values computed by the demodulator are multiplied by -1 to map them to the right quantizer index for use with Viterbi Decoder. To compute the LLR, the demodulator must be given the variance of noise as seen at its input. The third receiver includes a demodulator that computes LLRs which are processed by a Viterbi decoder that is set up in unquantized mode. The BER performance of each receiver is computed and displayed.
```modelName = 'commLLRvsHD'; open_system(modelName); ```
System Simulation and Visualization
Simulate this system over a range of information bit Eb/No values. Adjust these Eb/No values for coded bits and multi-bit symbols to get noise variance values required for the AWGN block and Rectangular QAM Baseband Demodulator block. Collect BER results for each Eb/No value and visualize the results.
```EbNo = 2:0.5:8; % information rate Eb/No in dB codeRate = 1/2; % code rate of convolutional encoder nBits = 4; % number of bits in a 16-QAM symbol Pavg = 10; % average signal power of a 16-QAM modulated signal snr = EbNo - 10*log10(1/codeRate) + 10*log10(nBits); % SNR in dB noiseVarVector = Pavg ./ (10.^(snr./10)); % noise variance % Initialize variables for storing the BER results ber_HD = zeros(1,length(EbNo)); ber_SD = zeros(1,length(EbNo)); ber_LLR = zeros(1, length(EbNo)); % Loop over all noiseVarVector values for idx=1:length(noiseVarVector) noiseVar = noiseVarVector(idx); %#ok<NASGU> sim(modelName); % Collect BER results ber_HD(idx) = BER_HD(1); ber_SD(idx) = BER_SD(1); ber_LLR(idx) = BER_LLR(1); end % Perform curve fitting and plot the results fitBER_HD = real(berfit(EbNo,ber_HD)); fitBER_SD = real(berfit(EbNo,ber_SD)); fitBER_LLR = real(berfit(EbNo,ber_LLR)); semilogy(EbNo,ber_HD,'r*', ... EbNo,ber_SD,'g*', ... EbNo,ber_LLR,'b*', ... EbNo,fitBER_HD,'r', ... EbNo,fitBER_SD,'g', ... EbNo,fitBER_LLR,'b'); legend('Hard Decision Decoding', ... 'Soft Decision Decoding','Unquantized Decoding'); xlabel('Eb/No (dB)'); ylabel('BER'); title('LLR vs. Hard Decision Demodulation with Viterbi Decoding'); grid on; ```
To experiment with this system further, try different modulation types. This system uses a binary mapped modulation scheme for faster error collection but it is well known that Gray mapped signal constellation provides better BER performance. Experiment with various constellation ordering options in the modulator and demodulator blocks. Configure the demodulator block to compute approximate LLR to see the difference in the BER performance compared to hard decision demodulation and LLR. Try out a different range of Eb/No values. Finally, investigate different quantizer boundaries for your modulation scheme and Eb/No values.
```% Cleanup close_system(modelName,0); clear modelName EbNo codeRate nBits Pavg snr noiseVarVector ... ber_HD ber_SD ber_LLR idx noiseVar fitBER_HD fitBER_SD fitBER_LLR; ``` | 2023-02-01 20:29:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6906619071960449, "perplexity": 1568.2171092851806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00240.warc.gz"} |
https://tex.stackexchange.com/questions/165706/how-to-prevent-page-break-in-the-last-three-lines-of-an-environment | # How to prevent page break in the last three lines of an environment?
I want to create an environment, thats content should not allowed a page break in the last three lines of the last paragraph.
If the last three lines of the text inside the environment reach the page border, the page break should occur after the forth line from the bottom.
Outside of the environment the normal behavior of the page breaks should not be changed.
The environment will only contain normal text, but with increased margin on the left and right side. There may occur some inline math equations as well, but that should be it.
\documentclass{scrartcl}
\usepackage{lipsum}
\newenvironment{env}{%
}{%
}%
\begin{document}
\lipsum[1-3]
\begin{env}
\lipsum[1]
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This last lines should not be broken to the next page.
This last lines should not be broken to the next page.
This last lines should not be broken to the next page.
\end{env}
\end{document}
Edit:
I'm not sure, if I can make clear, what I try to achieve: I won't normally know, where exactly the last lines will start. Instead, it is more like to prevent widows, but not only prevent the last line of the paragraph to get broken to the next page, but the last three lines of text inside the environment {env} shall always stay together, either on the actual page or the next page.
• What will this environment contain? "Complex" stuff, or just paragraph text? Could you provide an example of its usage? – Werner Mar 15 '14 at 14:36
• I edited the question regarding the expected contend and tried to provide a minimal example. – leviathan Mar 15 '14 at 14:55
• Wouldn't \vspace{2em}\begin{minipage} do what you want? – John Kormylo Mar 16 '14 at 14:55
• @John I'm not sure, I understand your suggestion correctly, but with a minipage, I would need to know beforhand, what the last three lines should contain. – leviathan Mar 16 '14 at 22:18
• Or more precisely, where the last three lines start. Yes. – John Kormylo Mar 17 '14 at 15:58
You can use \widowpenalties; with
\widowpenalties 3 10000 10000 0
you add a penalty of 10000 between the last three lines and none between preceding lines. See also How to avoid page-breaks inside paragraphs?
\documentclass{scrartcl}
\usepackage{lipsum}
\newenvironment{env}
{\begin{addmargin}{2em}\widowpenalties 3 10000 10000 0 }
\begin{document}
\lipsum[1-3]
\begin{env}
\lipsum[1]
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This is just some filler text, to reach the page break.
This last lines should not be broken to the next page.
This last lines should not be broken to the next page.
This last lines should not be broken to the next page.
\end{env}
\end{document} | 2019-07-23 06:58:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6929869055747986, "perplexity": 1512.8176560165314}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00223.warc.gz"} |
https://teknikaldomain.me/post/monthly-site-cost/ | ### Tek's Domain
#<NTA:NnT:SSrgS:H6.6-198:W200-90.72:CBWg>
# Going Over My First Amazon Bill
So I just (at the time of writing this on the 5th) got my Amazon AWS bill for the month of March, a whopping 31¢.
Let’s take a look at the real cost of what I’m using AWS for.
## A Foreword
March has two unusual properties to it:
1. I didn’t sign up for services until the middle of March
2. The second evolution of VCS internals part came out that I helped with, and that caused a.. shall we say large traffic surge to my site.
## Billing By Service
### Athena
Athena is essentially the Google BigQuery of Amazon. You define a SQL table, point it towards an S3 bucket of data that matches that table schema, and then you can query it… like a table.
I have server-side logging enabled on my CDN bucket, and therefore, I can search through those logs, pulling out individual bits of data or looking for anything suspicious.
You’re charged on the Terabyte scanned per query. In my case, this is practically nothing.
### Budgets
Yes, you have to spend money to track how much money you’re spending.
### CloudTrail
Essentially, account logging. Management events (this changed, logins, etc.) are free, so… free.
### CloudWatch
Metrics on other services. I have some things set up tracking the rate of requests to my CDN bucket that return errors, the average amount of data downloaded per request, latency, all sorts of things.
First, a “dashboard” that organizes metric graphs nicely.. can only have 3 for free. So that’s free.
Second, you can only attach up to 10 alarms to metrics for free. I have 2, it’s free.
Similarly, you can only track 10 metrics for free, and I think I’m under that limit too.
And the last item I’ll cover, the first million CloudWatch API requests are free. 166 < 1,000,000,000.
### Data Transfer
Data IN or OUT of an S3 data center.
Data IN is always free, it’s data OUT that you pay for, usually between 2¢–9¢ depending on the datacenter it’s coming from, per GB.
Data OUT for the first GB (per datacenter) is free.
After this, it’s a prorated1 9¢ per GB of data out. At 3.271 GB out, that’s $3.271 \times 0.09 = 0.29511$, so 29¢ total.
### Key Management Service
IAM, API keys, all that.
With almost everything else, I’m not exceeding the free tier.
Notifying you when your alarms trigger.
The first 1,000 emails and first million API requests are free, and I’m well below that bar.
### Simple Storage Service
The only other part they charged me for! There’s a few items in here.
First, 2.3¢ per GB, for the first 50 TB you store. This is calculated in GB-days, where storing 2 GB for one day “costs” the same amount as storing 1 GB for two days. I got… 0.083 GB-days.
Second: S3 analytics (CloudWatch): 10¢ per million. $37.903 \div 1,000,000 = 0.000037903$, $0.000037903 \times 0.10 = 0.00000379$, and… 0.000379¢ isn’t something you can charge.
Third: 0.4¢ per 10,000 GET (download) requests, and all other not covered by point 4. I have 7,709 of them… $7,709 \div 10,000 = 0.7709$, $0.7709 \times 0.004 = 0.0030836$. 1/3rd of a cent.
Finally: 0.5¢ per 1,000 PUT, POST, COPY, and LIST requests (upload, show what exists). 3,440 of those… $3,440 \div 1,000 = 3.44$, and $3.44 \times 0.005 = 0.0172$, or, rounded, 2¢.
My entire bill is 29¢ of data transfer and 2¢ of requests. The storage at this point is free, I’m just paying for people to grab it, essentially.
March is a little skewed though, we’ll see what April holds. Because besides the one day that dropped 22 of those cents on me, well… it’s relatively cheap for this.
1. Proportionally calculated. Instead of staying at 0¢ until I hit 1 GB, and then 9¢ until I hit 2 GB, and so on… it’s calculated proportionally. Half a GB is 4.5¢, 1.5 GB is 13.5¢, and so on. You don’t get to squeeze a little more out for free here. ↩︎ | 2021-10-24 10:11:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20903600752353668, "perplexity": 5163.113005662443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585916.29/warc/CC-MAIN-20211024081003-20211024111003-00253.warc.gz"} |
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15 30 50 per page | 2015-08-30 14:04:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7659175395965576, "perplexity": 1596.9540453255831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065318.20/warc/CC-MAIN-20150827025425-00136-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/421970/derivation-of-rotation-formula-in-mathbb-r3 | # Derivation of rotation formula in $\mathbb R^3$
In geometric algebra, for deriving the rotations formula in $\mathbb R^3$ I see the following steps ($\mathbf i$ is the plane of rotation, $\mathbf u$ is the vector to be rotated, with components $\mathbf u_{\parallel}$ in the plane of rotation and $\mathbf u_{\bot}$ perpendicular to it, $\mathbf v$ is the resultant vector):
$$\mathbf v = \mathbf u_{\parallel} e^{\mathbf i\theta} + \mathbf u_{\bot}$$ $$= \mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}} + \mathbf u_{\bot}e^{{-\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}}$$ $$= e^{{-\mathbf i\theta}/{2}}\mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}} + e^{{-\mathbf i\theta}/{2}}\mathbf u_{\bot}e^{{\mathbf i\theta}/{2}}$$ $$= e^{{-\mathbf i\theta}/{2}}\mathbf u e^{{\mathbf i\theta}/{2}}$$
I am unable to understand step (3).The hint given says that the geometric product of two orthogonal vectors is anti-commutative, which is clear enough, but how does that extend to this case where we have a product of a vector and a multivector ($e^{{\mathbf i\theta}/{2}}$)?
edit: I was able to verify the equality by taking $\mathbf i = e_1e_2$ and $\mathbf u_{\parallel} = ae_1 + be_2$, but stll not able to understand it geometrically.
This can be seen by breaking the exponential down into trig functions.
$$u_\parallel e^{i\theta/2} = u_\parallel (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$$
Since $u_\parallel$ lies entirely in the plane $i$, it should be clear that $i u_\parallel = -u_\parallel i$. Either way, this is a vector orthogonal to $u_\parallel$ lying in the plane $i$, but the order of multiplication determines whether you get a particular vector or its negative (based on what wedge product of $u_\parallel$ and $i u_\parallel$ yields a bivector with the same orientation as $i$).
When you do that, you get the following:
$$(\cos \frac{\theta}{2} - i \sin \frac{\theta}{2})u_\parallel = e^{-i \theta/2}u_\parallel$$
Welcome to the wide world of geometric algebra; I hope you continue to find answers to your questions here.
• Thank you for the answer. I now realize that i have not covered multiplication of vectors with bivectors (or pseudoscalars), that is why it doesn't make sense to me yet. The book only expects verification of the step, not the proof. – user997712 Jun 16 '13 at 19:57
• I see. Well, you can imagine that the bivector $i$ is the product of two orthogonal vectors, $i = u_\parallel w$ for some $w$, and then you can use associativity of the geometric product. What book are you using, by the way? – Muphrid Jun 16 '13 at 20:11
• Yes, that's what i ended up doing for the verification. I am using Linear and Geometric Algebra by Alan Macdonald and finding it really good. – user997712 Jun 16 '13 at 20:14
• Indeed, Macdonald's books are a very accessible first introduction to the subject. Good luck! – Muphrid Jun 16 '13 at 20:37 | 2019-08-26 09:33:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7744739055633545, "perplexity": 154.33036051733103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00221.warc.gz"} |
https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(OpenStax)/08%3A_Roots_and_Radicals/8.03%3A_Simplify_Radical_Expressions/8.3E%3A_Exercises |
# 8.3E: Exercises
### Practice Makes Perfect
##### Exercise SET A: use the product property to simplify radical expressions
In the following exercises, use the Product Property to simplify radical expressions.
1. $$\sqrt{27}$$
2. $$\sqrt{80}$$
3. $$\sqrt{125}$$
4. $$\sqrt{96}$$
5. $$\sqrt{147}$$
6. $$\sqrt{450}$$
7. $$\sqrt{800}$$
8. $$\sqrt{675}$$
1. $$\sqrt[4]{32}$$
2. $$\sqrt[5]{64}$$
1. $$\sqrt[3]{625}$$
2. $$\sqrt[6]{128}$$
1. $$\sqrt[5]{64}$$
2. $$\sqrt[3]{256}$$
1. $$\sqrt[4]{3125}$$
2. $$\sqrt[3]{81}$$
1. $$3\sqrt{3}$$
3. $$5\sqrt{5}$$
5. $$7\sqrt{3}$$
7. $$20\sqrt{2}$$
9.
1. $$2 \sqrt[4]{2}$$
2. $$2 \sqrt[5]{2}$$
11.
1. $$2 \sqrt[5]{2}$$
2. $$4 \sqrt[3]{4}$$
##### Exercise SET B: use the product property to simplify radical expressions
In the following exercises, simplify using absolute value signs as needed.
1. $$\sqrt{y^{11}}$$
2. $$\sqrt[3]{r^{5}}$$
3. $$\sqrt[4]{s^{10}}$$
1. $$\sqrt{m^{13}}$$
2. $$\sqrt[5]{u^{7}}$$
3. $$\sqrt[6]{v^{11}}$$
1. $$\sqrt{n^{21}}$$
2. $$\sqrt[3]{q^{8}}$$
3. $$\sqrt[8]{n^{10}}$$
1. $$\sqrt{r^{25}}$$
2. $$\sqrt[5]{p^{8}}$$
3. $$\sqrt[4]{m^{5}}$$
1. $$\sqrt{125 r^{13}}$$
2. $$\sqrt[3]{108 x^{5}}$$
3. $$\sqrt[4]{48 y^{6}}$$
1. $$\sqrt{80 s^{15}}$$
2. $$\sqrt[5]{96 a^{7}}$$
3. $$\sqrt[6]{128 b^{7}}$$
1. $$\sqrt{242 m^{23}}$$
2. $$\sqrt[4]{405 m 10}$$
3. $$\sqrt[5]{160 n^{8}}$$
1. $$\sqrt{175 n^{13}}$$
2. $$\sqrt[5]{512 p^{5}}$$
3. $$\sqrt[4]{324 q^{7}}$$
1. $$\sqrt{147 m^{7} n^{11}}$$
2. $$\sqrt[3]{48 x^{6} y^{7}}$$
3. $$\sqrt[4]{32 x^{5} y^{4}}$$
1. $$\sqrt{96 r^{3} s^{3}}$$
2. $$\sqrt[3]{80 x^{7} y^{6}}$$
3. $$\sqrt[4]{80 x^{8} y^{9}}$$
1. $$\sqrt{192 q^{3} r^{7}}$$
2. $$\sqrt[3]{54 m^{9} n^{10}}$$
3. $$\sqrt[4]{81 a^{9} b^{8}}$$
1. $$\sqrt{150 m^{9} n^{3}}$$
2. $$\sqrt[3]{81 p^{7} q^{8}}$$
3. $$\sqrt[4]{162 c^{11} d^{12}}$$
1. $$\sqrt[3]{-864}$$
2. $$\sqrt[4]{-256}$$
1. $$\sqrt[5]{-486}$$
2. $$\sqrt[6]{-64}$$
1. $$\sqrt[5]{-32}$$
2. $$\sqrt[8]{-1}$$
1. $$\sqrt[3]{-8}$$
2. $$\sqrt[4]{-16}$$
1. $$5+\sqrt{12}$$
2. $$\dfrac{10-\sqrt{24}}{2}$$
1. $$8+\sqrt{96}$$
2. $$\dfrac{8-\sqrt{80}}{4}$$
1. $$1+\sqrt{45}$$
2. $$\dfrac{3+\sqrt{90}}{3}$$
1. $$3+\sqrt{125}$$
2. $$\dfrac{15+\sqrt{75}}{5}$$
1.
1. $$\left|y^{5}\right| \sqrt{y}$$
2. $$r \sqrt[3]{r^{2}}$$
3. $$s^{2} \sqrt[4]{s^{2}}$$
3.
1. $$n^{10} \sqrt{n}$$
2. $$q^{2} \sqrt[3]{q^{2}}$$
3. $$|n| \sqrt[8]{n^{2}}$$
5.
1. $$5 r^{6} \sqrt{5 r}$$
2. $$3 x \sqrt[3]{4 x^{2}}$$
3. $$2|y| \sqrt[4]{3 y^{2}}$$
7.
1. $$11\left|m^{11}\right| \sqrt{2 m}$$
2. $$3 m^{2} \sqrt[4]{5 m^{2}}$$
3. $$2 n \sqrt[5]{5 n^{3}}$$
9.
1. $$7\left|m^{3} n^{5}\right| \sqrt{3 m n}$$
2. $$2 x^{2} y^{2} \sqrt[3]{6 y}$$
3. $$2|x y| \sqrt[4]{2 x}$$
11.
1. $$8\left|q r^{3}\right| \sqrt{3 q r}$$
2. $$3 m^{3} n^{3} \sqrt[3]{2 n}$$
3. $$3 a^{2} b^{2} \sqrt[4]{a}$$
13.
1. $$-6 \sqrt[3]{4}$$
2. not real
15.
1. $$-2$$
2. not real
17.
1. $$5+2 \sqrt{3}$$
2. $$5-\sqrt{6}$$
19.
1. $$1+3 \sqrt{5}$$
2. $$1+\sqrt{10}$$
##### Exercise Set C: use the quotient property to simplify radical expressions
In the following exercises, use the Quotient Property to simplify square roots.
1. $$\sqrt{\dfrac{45}{80}}$$
2. $$\sqrt[3]{\dfrac{8}{27}}$$
3. $$\sqrt[4]{\dfrac{1}{81}}$$
1. $$\sqrt{\dfrac{72}{98}}$$
2. $$\sqrt[3]{\dfrac{24}{81}}$$
3. $$\sqrt[4]{\dfrac{6}{96}}$$
1. $$\sqrt{\dfrac{100}{36}}$$
2. $$\sqrt[3]{\dfrac{81}{375}}$$
3. $$\sqrt[4]{\dfrac{1}{256}}$$
1. $$\sqrt{\dfrac{121}{16}}$$
2. $$\sqrt[3]{\dfrac{16}{250}}$$
3. $$\sqrt[4]{\dfrac{32}{162}}$$
1. $$\sqrt{\dfrac{x^{10}}{x^{6}}}$$
2. $$\sqrt[3]{\dfrac{p^{11}}{p^{2}}}$$
3. $$\sqrt[4]{\dfrac{q^{17}}{q^{13}}}$$
1. $$\sqrt{\dfrac{p^{20}}{p^{10}}}$$
2. $$\sqrt[5]{\dfrac{d^{12}}{d^{7}}}$$
3. $$\sqrt[8]{\dfrac{m^{12}}{m^{4}}}$$
1. $$\sqrt{\dfrac{y^{4}}{y^{8}}}$$
2. $$\sqrt[5]{\dfrac{u^{21}}{u^{11}}}$$
3. $$\sqrt[6]{\dfrac{v^{30}}{v^{12}}}$$
1. $$\sqrt{\dfrac{q^{8}}{q^{14}}}$$
2. $$\sqrt[3]{\dfrac{r^{14}}{r^{5}}}$$
3. $$\sqrt[4]{\dfrac{c^{21}}{c^{9}}}$$
1. $$\sqrt{\dfrac{96 x^{7}}{121}}$$
2. $$\sqrt{\dfrac{108 y^{4}}{49}}$$
3. $$\sqrt{\dfrac{300 m^{5}}{64}}$$
4. $$\sqrt{\dfrac{125 n^{7}}{169}}$$
5. $$\sqrt{\dfrac{98 r^{5}}{100}}$$
6. $$\sqrt{\dfrac{180 s^{10}}{144}}$$
7. $$\sqrt{\dfrac{28 q^{6}}{225}}$$
8. $$\sqrt{\dfrac{150 r^{3}}{256}}$$
1. $$\sqrt{\dfrac{75 r^{9}}{s^{8}}}$$
2. $$\sqrt[3]{\dfrac{54 a^{8}}{b^{3}}}$$
3. $$\sqrt[4]{\dfrac{64 c^{5}}{d^{4}}}$$
1. $$\sqrt{\dfrac{72 x^{5}}{y^{6}}}$$
2. $$\sqrt[5]{\dfrac{96 r^{11}}{s^{5}}}$$
3. $$\sqrt[6]{\dfrac{128 u^{7}}{v^{12}}}$$
1. $$\sqrt{\dfrac{28 p^{7}}{q^{2}}}$$
2. $$\sqrt[3]{\dfrac{81 s^{8}}{t^{3}}}$$
3. $$\sqrt[4]{\dfrac{64 p^{15}}{q^{12}}}$$
1. $$\sqrt{\dfrac{45 r^{3}}{s^{10}}}$$
2. $$\sqrt[3]{\dfrac{625 u^{10}}{v^{3}}}$$
3. $$\sqrt[4]{\dfrac{729 c^{21}}{d^{8}}}$$
1. $$\sqrt{\dfrac{32 x^{5} y^{3}}{18 x^{3} y}}$$
2. $$\sqrt[3]{\dfrac{5 x^{6} y^{9}}{40 x^{5} y^{3}}}$$
3. $$\sqrt[4]{\dfrac{5 a^{8} b^{6}}{80 a^{3} b^{2}}}$$
1. $$\sqrt{\dfrac{75 r^{6} s^{8}}{48 r s^{4}}}$$
2. $$\sqrt[3]{\dfrac{24 x^{8} y^{4}}{81 x^{2} y}}$$
3. $$\sqrt[4]{\dfrac{32 m^{9} n^{2}}{162 m n^{2}}}$$
1. $$\sqrt{\dfrac{27 p^{2} q}{108 p^{4} q^{3}}}$$
2. $$\sqrt[3]{\dfrac{16 c^{5} d^{7}}{250 c^{2} d^{2}}}$$
3. $$\sqrt[6]{\dfrac{2 m^{9} n^{7}}{128 m^{3} n}}$$
1. $$\sqrt{\dfrac{50 r^{5} s^{2}}{128 r^{2} s^{6}}}$$
2. $$\sqrt[3]{\dfrac{24 m^{9} n^{7}}{375 m^{4} n}}$$
3. $$\sqrt[4]{\dfrac{81 m^{2} n^{8}}{256 m^{1} n^{2}}}$$
1. $$\dfrac{\sqrt{45 p^{9}}}{\sqrt{5 q^{2}}}$$
2. $$\dfrac{\sqrt[4]{64}}{\sqrt[4]{2}}$$
3. $$\dfrac{\sqrt[5]{128 x^{8}}}{\sqrt[5]{2 x^{2}}}$$
1. $$\dfrac{\sqrt{80 q^{5}}}{\sqrt{5 q}}$$
2. $$\dfrac{\sqrt[3]{-625}}{\sqrt[3]{5}}$$
3. $$\dfrac{\sqrt[4]{80 m^{7}}}{\sqrt[4]{5 m}}$$
1. $$\dfrac{\sqrt{50 m^{7}}}{\sqrt{2 m}}$$
2. $$\sqrt[3]{\dfrac{1250}{2}}$$
3. $$\sqrt[4]{\dfrac{486 y^{9}}{2 y^{3}}}$$
1. $$\dfrac{\sqrt{72 n^{11}}}{\sqrt{2 n}}$$
2. $$\sqrt[3]{\dfrac{162}{6}}$$
3. $$\sqrt[4]{\dfrac{160 r^{10}}{5 r^{3}}}$$
1.
1. $$\dfrac{3}{4}$$
2. $$\dfrac{2}{3}$$
3. $$\dfrac{1}{3}$$
3.
1. $$\dfrac{5}{3}$$
2. $$\dfrac{3}{5}$$
3. $$\dfrac{1}{4}$$
5.
1. $$x^{2}$$
2. $$p^{3}$$
3. $$|q|$$
7.
1. $$\dfrac{1}{y^{2}}$$
2. $$u^{2}$$
3. $$|v^{3}|$$
9. $$\dfrac{4\left|x^{3}\right| \sqrt{6 x}}{11}$$
11. $$\dfrac{10 m^{2} \sqrt{3 m}}{8}$$
13. $$\dfrac{7 r^{2} \sqrt{2 r}}{10}$$
15. $$\dfrac{2\left|q^{3}\right| \sqrt{7}}{15}$$
17.
1. $$\dfrac{5 r^{4} \sqrt{3 r}}{s^{4}}$$
2. $$\dfrac{3 a^{2} \sqrt[3]{2 a^{2}}}{|b|}$$
3. $$\dfrac{2|c| \sqrt[4]{4 c}}{|d|}$$
19.
1. $$\dfrac{2\left|p^{3}\right| \sqrt{7 p}}{|q|}$$
2. $$\dfrac{3 s^{2} \sqrt[3]{3 s^{2}}}{t}$$
3. $$\dfrac{2\left|p^{3}\right| \sqrt[4]{4 p^{3}}}{\left|q^{3}\right|}$$
21.
1. $$\dfrac{4|x y|}{3}$$
2. $$\dfrac{y^{2} \sqrt[3]{x}}{2}$$
3. $$\dfrac{|a b| \sqrt[4]{a}}{4}$$
23.
1. $$\dfrac{1}{2|p q|}$$
2. $$\dfrac{2 c d \sqrt[5]{2 d^{2}}}{5}$$
3. $$\dfrac{|m n| \sqrt[6]{2}}{2}$$
25.
1. $$\dfrac{3 p^{4} \sqrt{p}}{|q|}$$
2. $$2 \sqrt[4]{2}$$
3. $$2 x \sqrt[5]{2 x}$$
27.
1. $$5\left|m^{3}\right|$$
2. $$5 \sqrt[3]{5}$$
3. $$3|y| \sqrt[4]{3 y^{2}}$$
##### Exercise SET D: writing exercises
1. Explain why $$\sqrt{x^{4}}=x^{2}$$. Then explain why $$\sqrt{x^{16}}=x^{8}$$.
2. Explain why $$7+\sqrt{9}$$ is not equal to $$\sqrt{7+9}$$.
3. Explain how you know that $$\sqrt[5]{x^{10}}=x^{2}$$.
4. Explain why $$\sqrt[4]{-64}$$ is not a real number but $$\sqrt[3]{-64}$$ is.
## Self Check
a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
b. After reviewing this checklist, what will you do to become confident for all objectives?
8.3E: Exercises is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 2022-05-23 06:06:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9291698932647705, "perplexity": 279.0014320822935}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00234.warc.gz"} |
https://iwaponline.com/ws/article/5/6/145/26106/Decentralised-wastewater-treatment-technologies | In this paper we estimate at what cost decentralised wastewater treatment can be competitive compared with conventional centralised technologies. For the current wastewater infrastructure in Western Europe and North America, typical replacement costs are 2,600 US$/cap for large countries and 4,800 US$/cap for small ones. In the same literature, average annual operating costs are reported to be 3.8% of replacement costs. However, if a long-term interest rate of 3% is consistently applied, this value increases to 4.7% for small countries and 5.5% for large ones. Assuming that alternative wastewater systems will only be accepted if their costs are similar to existing ones, the possible investments for alternative wastewater treatment technologies are calculated. Between 640 and 2,170 US$/cap can be invested in new technologies for scenarios without a sewer system. The corresponding figures for scenarios with sewer systems are between 260 and 680 US$/cap. Acceptable maintenance requirements are calculated on the basis of unit size. Transition periods are not accounted for.
This content is only available as a PDF. | 2020-07-05 01:41:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27758151292800903, "perplexity": 1690.404821043753}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655886802.13/warc/CC-MAIN-20200704232817-20200705022817-00195.warc.gz"} |
https://kb.osu.edu/handle/1811/29187?show=full | dc.creator Bernheim, R. A. en_US dc.creator Gold, L. P. en_US dc.creator Tipton, T. en_US dc.date.accessioned 2007-08-31T14:12:16Z dc.date.available 2007-08-31T14:12:16Z dc.date.issued 1982 en_US dc.identifier 1982-MG-5 en_US dc.identifier.uri http://hdl.handle.net/1811/29187 dc.description $^{1}$ D.D. Konowalow and M.E. Rosenkrantz, Chem. Phys. Letters 61, 489 (1979). en_US dc.description.abstract Thirty-one new excited gerade states of $^{7}Li_{2}$ have been characterized by means of pulsed optical-optical double resonance spectroscopy. Twenty-nine of these have been identified as the lowest members of three Rydberg series: $3-15d\pi^{1}\Pi_{g}, 3-10d\sigma^{1}{\Sigma^{+}}_{g}$, and $3-10s\sigma^{1}{\Sigma_{g}}^{+}$, A fit of a Rydberg formula to the $^{1}{\Pi^{-}}_{g}$ term energies yield a value of $41496\pm 4 cm^{-1}$ for the ionization potential of $Li_{2}$. Extrapolations of the vibrational frequencies, rotational constants, and dissociation energies of these states give estimates for the corresponding quantities for the ground state of ${Li_{2}}^{+}$ which are in good agreement with those obtained by ab initio $calculations^{1}$. en_US dc.format.extent 69218 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title RYDBREG STATES OF ${Li_{2}}$ AND MOLECULAR CONSTANTS OF ${Li_{2}}^{+}$ en_US dc.type article en_US
| 2021-01-23 20:38:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7315899133682251, "perplexity": 9616.836368956583}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00731.warc.gz"} |
http://www.ck12.org/geometry/Dilation-in-the-Coordinate-Plane/lesson/Graphs-of-Dilations-ALG-I-HNRS/ | # Dilation in the Coordinate Plane
## Multiplication of coordinates by a scale factor.
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Graphs of Dilations
Quadrilateral \begin{align*}WXYZ\end{align*} has coordinates \begin{align*}W \ (\text{-}5, \text{-}5), \ X \ (\text{-}2, 0), \ Y \ (2, 3),\end{align*} and \begin{align*}Z \ (\text{-}1, 3).\end{align*} Draw the quadrilateral on the Cartesian plane.
Suppose the quadrilateral undergoes a dilation centered at the origin of scale factor \begin{align*}\frac{1}{3}.\end{align*} What is the resulting image?
### Graphs of Dilations
In geometry, a transformation is an operation that moves, flips, or changes a shape to create a new shape. A dilation is a type of transformation that enlarges or reduces a figure (called the pre-image) to create a new figure (called the image). The scale factor, r, determines how much bigger or smaller the dilated image will be compared to the preimage.
In order to graph a dilation, use the center of dilation and the scale factor. Find the distance between a point on the preimage and the center of dilation. Multiply this length by the scale factor. The corresponding point on the image will be this distance away from the center of dilation in the same direction as the original point.
If you compare the length of a side on the preimage to the length of the corresponding side on the image, the length of the side on the image will be the length of the side on the preimage multiplied by the scale factor.
#### Draw the preimage and image for the following dilations and determine the scale factor:
Line \begin{align*}\overline{A B}\end{align*} drawn from (-4, 2) to (3, 2) has undergone a dilation about the origin to produce \begin{align*}A^\prime(-6, 3)\end{align*} and \begin{align*}B^\prime(4.5, 3)\end{align*}
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}} \\ & \mathrm{scale \ factor} = \frac{10.5}{7.0} \\ & \mathrm{scale \ factor} = \frac{3}{2}\end{align*}
#### Determine the coordinates and the scale factor for the following dilations:
1. The diamond \begin{align*}ABCD\end{align*} undergoes a dilation about the origin to form the image \begin{align*}A^\prime B^\prime C^\prime D^\prime\end{align*}
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}}\\ & \mathrm{scale factor} = \frac{7.21}{3.61} \\ & \mathrm{scale factor} = 2\end{align*}
1. The diagram below undergoes a dilation about the origin to form the dilation image.
\begin{align*}& \mathrm{scale factor }= \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}}\\ & \mathrm{scale \ factor }= \frac{2.00}{10.00} \\ & \mathrm{scale \ factor }= \frac{1}{5}\end{align*}
### Examples
#### Example 1
Earlier, you were told that quadrilateral \begin{align*}WXYZ\end{align*} has coordinates \begin{align*}W \ (\text{-}5, \text{-}5), \ X \ (\text{-}2, 0), \ Y\ (2, 3),\end{align*} and \begin{align*}Z \ (\text{-}1, 3).\end{align*} If the quadrilateral undergoes a dilation centered at the origin of scale factor \begin{align*}\frac{1}{3},\end{align*} what is the resulting image?
Test to see if the dilation is correct by determining the scale factor.
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}} \\ & \mathrm{scale \ factor} = \frac{10.63}{3.54} \\ & \mathrm{scale \ factor} = 3\end{align*}
#### Example 2
Line \begin{align*}\overline{S T}\end{align*} drawn from (-3, 4) to (-3, 8) has undergone a dilation of scale factor 3 about the point \begin{align*}A \ (1, 6).\end{align*} Draw the preimage and image and properly label each.
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}} \\ & \mathrm{scale \ factor} = \frac{12.00}{4.00} \\ & \mathrm{scale \ factor} = 3\end{align*}
#### Example 3
The polygon below has undergone a dilation about the origin with a scale factor of \begin{align*}\frac{5}{3}.\end{align*} Draw the dilation image and properly label each point.
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}} \\ & \mathrm{scale \ factor} = \frac{5.00}{3.00} \\ & \mathrm{scale \ factor} = \frac{5}{3}\end{align*}
#### Example 4
The triangle with vertices \begin{align*}J \ (\text{-}5, -2), \ K \ (\text{-}1, 4),\end{align*} and \begin{align*}L \ (1, \text{-}3)\end{align*} has undergone a dilation of scale factor \begin{align*}\frac{1}{2}\end{align*} about the center point \begin{align*}L.\end{align*} Draw and label the dilation image and the preimage, then verify the scale factor.
\begin{align*}& \mathrm{scale \ factor} = \frac{\mathrm{dilation \ image \ length}}{\mathrm{preimage \ length}}\\ & \mathrm{scale \ factor} = \frac{7.21}{3.61} \\ & \mathrm{scale \ factor} = \frac{1}{2}\end{align*}
### Review
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}2\end{align*} about point \begin{align*}D.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}3\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about point \begin{align*}C.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about point \begin{align*}C.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}\frac{1}{4}\end{align*} about point \begin{align*}C.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}2\end{align*} about point \begin{align*}A.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}2\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about point \begin{align*}D.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}3\end{align*} about point \begin{align*}D.\end{align*}
1. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about the origin.
2. Dilate the above figure by a factor of \begin{align*}\frac{1}{2}\end{align*} about point \begin{align*}C.\end{align*}
To see the Review answers, open this PDF file and look for section 10.11.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Dilation To reduce or enlarge a figure according to a scale factor is a dilation.
Distance Formula The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Mapping Mapping is a procedure involving the plotting of points on a coordinate grid to see the behavior of a function.
Scale Factor A scale factor is a ratio of the scale to the original or actual dimension written in simplest form. | 2017-04-23 20:28:15 | {"extraction_info": {"found_math": true, "script_math_tex": 51, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 3, "texerror": 0, "math_score": 0.9933083653450012, "perplexity": 2885.1319201420397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917118743.41/warc/CC-MAIN-20170423031158-00610-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://rcd.ics.org.ru/editorial_board/detail/330-grigory_osipov | 0
2013
Impact Factor
# Grigory Osipov
Nizhny Novgorod State University
Nizhny Novgorod State University
## Publications:
Smirnov L. A., Kryukov A. K., Osipov G. V., Kurths J. Bistability of Rotational Modes in a System of Coupled Pendulums 2016, vol. 21, no. 7-8, pp. 849-861 Abstract The main goal of this research is to examine any peculiarities and special modes observed in the dynamics of a system of two nonlinearly coupled pendulums. In addition to steady states, an in-phase rotation limit cycle is proved to exist in the system with both damping and constant external force. This rotation mode is numerically shown to become unstable for certain values of the coupling strength. We also present an asymptotic theory developed for an infinitely small dissipation, which explains why the in-phase rotation limit cycle loses its stability. Boundaries of the instability domain mentioned above are found analytically. As a result of numerical studies, a whole range of the coupling parameter values is found for the case where the system has more than one rotation limit cycle. There exist not only a stable in-phase cycle, but also two out-of phase ones: a stable rotation limit cycle and an unstable one. Bistability of the limit periodic mode is, therefore, established for the system of two nonlinearly coupled pendulums. Bifurcations that lead to the appearance and disappearance of the out-ofphase limit regimes are discussed as well. Keywords: coupled elements, bifurcation, multistability Citation: Smirnov L. A., Kryukov A. K., Osipov G. V., Kurths J., Bistability of Rotational Modes in a System of Coupled Pendulums, Regular and Chaotic Dynamics, 2016, vol. 21, no. 7-8, pp. 849-861 DOI:10.1134/S156035471607008X
Grines E. A., Osipov G. V. On Constructing Simple Examples of Three-dimensional Flows with Multiple Heteroclinic Cycles 2015, vol. 20, no. 6, pp. 679-690 Abstract In this work we suggest a simple method for constructing $G$-equivariant systems of ODEs in $\mathbb{R}^3$ (i.e., systems whose trajectories are invariant under the action of this group on $\mathbb{R}^3$) that possess multiple disjoint heteroclinic networks. Heteroclinic networks under consideration consist of saddle equilibria that belong to coordinate axes and one-dimensional separatrices connecting them. We require these separatrices to lie on coordinate planes. We also assume the action of $G$ on $\mathbb{R}^3$ to be generated by cyclic permutation of coordinate variables and reflection with respect to one of the coordinate planes. As an example, we provide a step-by-step construction of three-dimensional flow with two disjoint heteroclinic networks. Also, we present a sketch of global dynamics analysis for the minimal example. Keywords: heteroclinic cycle, heteroclinic network Citation: Grines E. A., Osipov G. V., On Constructing Simple Examples of Three-dimensional Flows with Multiple Heteroclinic Cycles, Regular and Chaotic Dynamics, 2015, vol. 20, no. 6, pp. 679-690 DOI:10.1134/S1560354715060040
Korotkov A. G., Kazakov A. O., Osipov G. V. Sequential Dynamics in the Motif of Excitatory Coupled Elements 2015, vol. 20, no. 6, pp. 701-715 Abstract In this article a new model of motif (small ensemble) of neuron-like elements is proposed. It is built with the use of the generalized Lotka–Volterra model with excitatory couplings. The main motivation for this work comes from the problems of neuroscience where excitatory couplings are proved to be the predominant type of interaction between neurons of the brain. In this paper it is shown that there are two modes depending on the type of coupling between the elements: the mode with a stable heteroclinic cycle and the mode with a stable limit cycle. Our second goal is to examine the chaotic dynamics of the generalized three-dimensional Lotka–Volterra model. Keywords: Neuronal motifs, Lotka–Volterra model, heteroclinic cycle, period-doubling bifurcation, Feigenbaum scenario, strange attractor, Lyapunov exponents Citation: Korotkov A. G., Kazakov A. O., Osipov G. V., Sequential Dynamics in the Motif of Excitatory Coupled Elements, Regular and Chaotic Dynamics, 2015, vol. 20, no. 6, pp. 701-715 DOI:10.1134/S1560354715060064
Belykh V. N., Petrov V. S., Osipov G. V. Dynamics of the Finite-dimensional Kuramoto Model: Global and Cluster Synchronization 2015, vol. 20, no. 1, pp. 37-48 Abstract Synchronization phenomena in networks of globally coupled non-identical oscillators have been one of the key problems in nonlinear dynamics over the years. The main model used within this framework is the Kuramoto model. This model shows three main types of behavior: global synchronization, cluster synchronization including chimera states and totally incoherent behavior. We present new sufficient conditions for phase synchronization and conditions for an asynchronous mode in the finite-size Kuramoto model. In order to find these conditions for constant and time varying frequency mismatch, we propose a simple method of comparison which allows one to obtain an explicit estimate of the phase synchronization range. Theoretical results are supported by numerical simulations. Keywords: phase oscillators, Kuramoto model, global synchronization, existence and stability conditions, asynchronous mode Citation: Belykh V. N., Petrov V. S., Osipov G. V., Dynamics of the Finite-dimensional Kuramoto Model: Global and Cluster Synchronization, Regular and Chaotic Dynamics, 2015, vol. 20, no. 1, pp. 37-48 DOI:10.1134/S1560354715010037 | 2017-11-18 10:02:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46949321031570435, "perplexity": 662.0345408436998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00093.warc.gz"} |
http://hal.in2p3.fr/in2p3-00156586 | # Aspects of Low Energy Binary Reactions
Abstract : Recent experimental and theoretical advances have been achieved in the field of multinucleon transfer reactions among heavy-ions. Detailed A,Z and Q-value distributions allow to identify the relevant degrees of freedom acting in the transfer process. The advent of the large solid angle magnetic spectrometer PRISMA coupled to the large $\gamma$-array CLARA was a major experimental breakthrough, as reaction mechanism and nuclear structure studies of neutron-rich nuclei can be performed via $\gamma$-particle coincidences. Examples of recent studies, mainly focused on the reaction mechanism, are presented.
Keywords :
Document type :
Conference papers
http://hal.in2p3.fr/in2p3-00156586
Contributor : Béatrice Forrler <>
Submitted on : Thursday, June 21, 2007 - 4:03:44 PM
Last modification on : Tuesday, May 4, 2021 - 11:26:08 AM
### Citation
L. Corradi, A. M. Stefanini, E. Fioretto, A. Gadea, N. Marginean, et al.. Aspects of Low Energy Binary Reactions. International Symposium on Exotic Nuclei, Jul 2006, Khanty-Mansiysk, Russia. pp.13-22, ⟨10.1063/1.2746576⟩. ⟨in2p3-00156586⟩
Record views | 2021-06-16 02:16:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17723144590854645, "perplexity": 9164.681536985794}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00151.warc.gz"} |
https://tigerprints.clemson.edu/all_theses/376/ | ## All Theses
#### Title
The Square Threshold Problem in Number Fields
5-2008
Thesis
#### Degree Name
Master of Science (MS)
#### Legacy Department
Mathematical Science
James, Kevin L
Calkin , Neil J
Maharaj , Hiren
#### Abstract
Let K be a degree n extension of Q, and let O_K be the ring of algebraic integers in K. Let x >= 2. Suppose we were to generate an ideal sequence by choosing ideals with norm at most x from O_K, independently and with uniform probability. How long would our sequence of ideals need to be before we obtain a subsequence whose terms have a product that is a square ideal in O_K? We show that the answer is about exp((2\ln(x)\ln\ln(x))^(1/2)).
COinS | 2018-12-16 01:18:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9011644721031189, "perplexity": 1539.619120152618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00351.warc.gz"} |
https://dsp.stackexchange.com/questions/3162/can-fft-substitute-correlation-for-this-case | Can FFT Substitute Correlation for This Case?
I'm trying to determine frequency of a repeating pattern in a time series. The sampling rate 100Hz and the pattern repeats itself with periods between 0.5 second and 1.25 second. This corresponds to a frequency range between 0.8Hz and 2Hz. At each second, I want to find the period of the pattern with 10msec precision, i.e. I want to differentiate 0.7sec(1.4Hz) and 0.6sec(1.6Hz).
I can attain this goal using correlation: For possible windows lengths (periods) I calculate correlation between two successive windows with same length. The window length with highest correlation gives the period of pattern.
Can I obtain the same results by using FFT?
As far as I see if I use 128-point FFT with this 100Hz series, then it will give me 100/128=0.8Hz (1.25sec), 1.6 Hz (0.62sec), 2.4Hz (0.42 sec),... components which obviously won't meet my requirement of 10msec period precision.
I know I can increase precision by using 256-point FFT but this corresponds to 2.5-second window for 100Hz and I want to determine pattern frequency for each window of ~1second.
• Perhaps you're asking about the convolution theorem, that ${\frak F}[f g]={\frak F}[f]*{\frak F}[g]$, where $\frak{F}$ is the Fourier Transform and $*$ is convolution, for two functions $f$ and $g$. So, yes, FFT can substitute convolution: $f*g={\frak F}^{-1}\{{\frak F}[f]{\frak F}[g]\}$. – Geremia Jan 29 '17 at 0:29
How did you plan to use the FFT in the first place?
If it is to naively look for peaks in the magnitude spectrum and deduce the frequency from it, this is a bad idea. As you have seen, this has a limited resolution in the lowest frequencies. Resolution limitations aside, looking for a peak in the FFT is a poor way of estimating periodicity since the true fundamental frequency might not be the one with the most salient amplitude.
However, you can use FFT as an intermediate step to efficiently compute the autocorrelation function (FFT with zero-padding of the input to avoid wrapping -> square of magnitude of coefficients -> IFFT). Looking for peaks in the autocorrelation function is a more reliable period estimator than FFT peaks.
Thus, given your requirements, you could use sliding windows of 256 samples, with a 61% overlap, compute the autocorrelation through the FFT, look for the autocorrelation peak in the [50, 125] samples range and use this as an estimate of the period. Because of the 61% overlap, this will give you one period value per second. Observe here that it is possible to use large analysis windows (256 samples, or more) and make them overlap so that we can still get a detection for every 100 sample. The downside is that if the period of the signal rapidly change, your detection will be "smudged".
I see a few other approaches for your problem that do not suffer from any resolution compromise on the temporal axis:
• If the waveform of your input signal is known (square, sawtooth, sine...), you could try locking an oscillator of similar waveform to it through a PLL, and use the PLL frequency as your estimator.
• If your signal is purely sinusoidal, you can fit a complex exponential to it (through eigenanalysis of the autocorrelation matrix).
• An approach that would give a very high resolution (one decision per sample) would be to run your signal through a bank of 75 comb filters in parallel, with delays of 0.5, 0.51, 0.52 .. 1.25 seconds. Compute the energy of the output for each of these 75 signals ; the one with the highest energy gives you the period.
Note that your problem has very similar requirements to that of tempo detection (onset detection yields an envelope signal at 100 Hz or the likes, we aim to get a tempo map in the 60 BPM - 180 BPM, ie 0.33 - 1Hz range, with a temporal resolution of 1s). Historically, the comb filterbank approach was used ; while most recent solutions are based on autocorrelation or similar autocorrelation-based pitch estimators (yin) ; with analysis windows in the 5 - 10s range.
• Thank you @pichenettes. Do you think my use of correlation is a good or a bad solution for this problem in terms of memory and cpu requirements compared to alternative methods you listed above. – mostar Aug 20 '12 at 20:56
• It's a good solution if you compute it efficiently (through FFTs). – pichenettes Aug 20 '12 at 21:05
• First: Computing the autocorrelation naively in the time domain is O(N^2), while doing it through an FFT/IFFT is O(N log N). You might argue that N might be small enough for the naive time-domain thing to require less MACs than the FFT and IFFT... However, there are very efficient libraries for computing FFTs, using all the artillery available on your hardware platform (such SIMD instruction sets). Your ad-hoc implementation of your algorithm will have to use the same tricks to be competitive... – pichenettes Aug 20 '12 at 21:30
• the above answer was a reply to a question about efficiency of direct time-domain computation of autocorrelation vs autocorrelation through FFT. – pichenettes Aug 20 '12 at 21:31
• thank you very much and sorry for deleting the question. – mostar Aug 20 '12 at 21:33 | 2019-12-09 10:22:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6750996112823486, "perplexity": 899.0156116824351}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540518627.72/warc/CC-MAIN-20191209093227-20191209121227-00216.warc.gz"} |
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# Strong admissibility revisited: Theory and applications
#### Abstract
In the current paper, we re-examine the concept of strong admissibility, as was originally introduced by Baroni and Giacomin. We examine the formal properties of strong admissibility, both in its extension-based and in its labelling-based form, and analyse the computational complexity of the relevant decision problems. Moreover, we show that strong admissibility plays a vital role in discussion-based proof procedures for grounded semantics. In particular it allows one to compare the performance of alternative dialectical proof procedures for grounded semantics, and obtain some remarkable differences between the Standard Grounded Game and the Grounded Discussion Game.
## 1.Introduction
Admissibility is generally seen as one of the cornerstones of abstract argumentation theory [19], as it is the basis of various argumentation semantics [1]. Not only does admissibility appeal to common intuitions [5], it is also one of the key requirements for obtaining a consistent outcome of instantiated argumentation formalisms [13,22,25].
Slightly less well-known is the principle of strong admissibility, which was originally introduced in [3]. The original aim of strong admissibility was to characterise the unique properties of the grounded extension. It turns out, however, that the concept is also useful for comparing the characteristics of the different dialectical proof procedures that have been stated in the literature. In particular, the Standard Grounded Game [21,26] and the Grounded Discussion Game [11] prove membership of the grounded extension essentially by constructing a strongly admissible labelling where the argument in question is labelled in. However, as we will see, the Grounded Discussion Game is able to do so in a more efficient way, requiring a number of steps that is linearly related to the in/out-size of the strongly admissible labelling,1 whereas the Standard Grounded Game can require a number of steps that is exponentially related to the in/out-size of the strongly admissible labelling.
The remaining part of the current paper is structured as follows. First, in Section 2 we briefly summarise some of the key concepts of abstract argumentation theory, both in its extension and in its labelling based form. In Section 3, we then discuss the extension based version of strong admissibility and examine its formal properties. In Section 4 we introduce the labelling based version of strong admissibility and show how it relates to its extension based version. In Section 5 we examine the computational complexity of some of the decision problems related to strong admissibility. In Section 6 we then re-examine the Standard Grounded Game, and the Grounded Persuasion Game, and show that strong admissibility plays a vital role in describing the relative efficiency of these games. In Section 7 we then round off with a discussion of our results and some open research issues.2
## 2.Formal preliminaries
In the current section, we briefly restate some of the key concepts of abstract argumentation theory, in both its extension based and labelling based form.
##### Definition 1.
An argumentation framework is a pair (Ar,att) where Ar is a finite set of entities, called arguments, whose internal structure can be left unspecified, and att a binary relation on Ar. For any A,BAr we say that A attacks B iff (A,B)att.
##### Definition 2.
Let (Ar,att) be an argumentation framework, AAr and ArgsAr. We define A+ as {BArA attacks B}, A as {BArB attacks A}, Args+ as {A+AArgs}, and Args as {AAArgs}. Args is said to be conflict-free iff ArgsArgs+=. Args is said to defend A iff AArgs+. The characteristic function F:2Ar2Ar is defined as F(Args)={AArgs defends A}.
##### Definition 3.
Let (Ar,att) be an argumentation framework. ArgsAr is said to be:
• an admissible set iff Args is conflict-free and ArgsF(Args)
• a complete extension iff Args is conflict-free and Args=F(Args)
• a grounded extension iff Args is the smallest (w.r.t. ⊆) complete extension
• a preferred extension iff Args is a maximal (w.r.t. ⊆) complete extension
If Args is a conflict-free set, then its down-admissible set (written as Args) is defined as the (unique) biggest (w.r.t. ⊆) admissible subset of Args.3
The above definitions essentially follow the extension based approach of [19].4 It is also possible to define the key argumentation concepts in terms of argument labellings [8 ,15].
##### Definition 4.
Let (Ar,att) be an argumentation framework. An argument labelling is a partial function Lab:Ar{in,out,undec}. An argument labelling is called an admissible labelling iff Lab is a total function and for each AAr it holds that:
• if Lab(A)=in then for each B that attacks A it holds that Lab(B)=out
• if Lab(A)=out then there exists a B that attacks A such that Lab(B)=in
Lab is called a complete labelling iff it is an admissible labelling and for each AAr it also holds that:
• if Lab(A)=undec then there is a B that attacks A such that Lab(B)=undec, and for each B that attacks A such that Lab(B)undec it holds that Lab(B)=out
As a labelling is essentially a function, we sometimes write it as a set of pairs. Also, if Lab is a labelling, we write in(Lab) for {AArLab(A)=in}, out(Lab) for {AArLab(A)=out} and undec(Lab) for {AArLab(A)=undec}. As a labelling is also a partition of the arguments into sets of in-labelled arguments, out-labelled arguments and undec-labelled arguments, we sometimes write it as a triplet (in(Lab),out(Lab),undec(Lab)).
### Definition 5([16]).
Let Lab and Lab be argument labellings of argumentation framework (Ar,att). We say that LabLab iff in(Lab)in(Lab) and out(Lab)out(Lab). LabLab is defined as (in(Lab)in(Lab),out(Lab)out(Lab),Ar((in(Lab)in(Lab))(out(Lab)out(Lab)))). LabLab is defined as ((in(Lab)out(Lab))(in(Lab)out(Lab)),(out(Lab)in(Lab))(out(Lab)in(Lab)),(in(Lab)out(Lab)(out(Lab)in(Lab))(undec(Lab)undec(Lab)))).
We say that Lab1 is a sublabelling of Lab2 (or alternatively, that Lab2 is a superlabelling of Lab2) iff Lab1Lab2. If Lab is a total labelling (i.e. a total function), then its down-admissible labelling [16] (written as Lab) is defined as the (unique) biggest (w.r.t. ⊑) admissible sublabelling of Lab.
##### Definition 6.
Let Lab be a complete labelling of argumentation framework (Ar,att). Lab is said to be
• a grounded labelling iff Lab is the (unique) smallest (w.r.t. ⊑) complete labelling
• a preferred labelling iff Lab is a maximal (w.r.t. ⊑) complete labelling
Given an argumentation framework (Ar,att) we define two functions Args2Lab and Lab2Args (to translate a conflict-free set of arguments to an argument labelling, and to translate an argument labelling to a set of arguments, respectively) such that Args2Lab(Args)=(Args,Args+,Ar(ArgsArgs+)) and Lab2Args(Lab)=in(Lab). It has been proven [15] that if Args is an admissible set (resp. a complete, grounded or preferred extension) then Args2Lab(Args) is an admissible labelling (resp. a complete, grounded or preferred labelling), and that if Lab is an admissible labelling (resp. a complete, grounded or preferred labelling) then Lab2Args(Lab) is an admissible set (resp. a complete, grounded or preferred extension). Moreover, when the domain and range of Args2Lab and Lab2Args are restricted to complete extensions and complete labellings they become injective functions and each other’s reverses, which implies that the complete extensions (resp. the grounded extension and the preferred extensions) and the complete labellings (resp. the grounded labelling and the preferred labellings) are one-to-one related [15].
## 3.Strongly admissible sets
The concept of strong admissibility was first introduced by Baroni and Giacomin [3], using the notion of strong defence.
### Definition 7([3]).
Let (Ar,att) be an argumentation framework, AAr and ArgsAr be a set of arguments. A is strongly defended by Args iff each attacker BAr of A is attacked by some CArgs{A} such that C is strongly defended by Args{A}.
Baroni and Giacomin say that a set Args satisfies the strong admissibility property iff it strongly defends each of its arguments [3]. However, it is also possible to define strong admissibility without having to refer to strong defence.
##### Definition 8.
Let (Ar,att) be an argumentation framework. ArgsAr is strongly admissible iff every AArgs is defended by some ArgsArgs{A} which in its turn is again strongly admissible.
To illustrate the concept of strong admissibility, consider the argumentation framework of Fig. 1. Here, the strongly admissible sets are , {A}, {A,C}, {A,C,F}, {D}, {A,D}, {A,C,D}, {D,F}, {A,D,F} and {A,C,D,F}, the latter also being the grounded extension. As an example, the set {A,C,F} is strongly admissible as A is defended by , C is defended by {A} and F is defended by {A,C}, each of which is a strongly admissible subset of {A,C,F} not containing the argument it defends. Please notice that although the set {A,F} defends argument C in {A,C,F}, it is in its turn not strongly admissible (unlike {A}). Hence the requirement in Definition 8 for Args to be a subset of Args{A}. We also observe that although {C,H} is an admissible set, it is not a strongly admissible set, since no subset of {C,H}{H} defends H.
##### Fig. 1.
An example of an argumentation framework.
It can be proved that a set is strongly admissible (in the sense of Definition 8) iff it strongly defends each of its arguments (in the sense of Definition 7). In order to do so, we need the following two lemmas.
##### Lemma 1.
Let AAr and Args1,Args2Ar such that AArgs1 and Args1Args2. If A is strongly defended by Args1 then A is also strongly defended by Args2.
##### Proof.
By induction over the number of arguments in Args2. Let i=|Args2|.
basis For i=1 it holds that |Args2|=1, which together with A∈Args1 and Args1⊆Args2 implies that Args1=Args2={A}. From Args1=Args2 it trivially holds that if A is strongly defended by Args1 then A is also strongly defended by Args2. step Suppose the lemma holds for some i⩾1. We now prove it also holds for i+1. Let A be strongly defended by Args1. We need to prove that A is also strongly defended by Args2 with |Args2|=i+1. According to Definition 7 this would be the case if each attacker B∈Ar of A is attacked by some C∈Args2∖{A} such that C is strongly defended by Args2∖{A}. The fact that A is strongly defended by Args1 means that each attacker B∈Ar of A is attacked by some C∈Args1∖{A}. From the fact that Args1⊆Args2 it follows that Args1∖{A}⊆Args2∖{A} so C∈Args2∖{A}. We now need to prove that C is strongly defended by Args2∖{A}. From the fact that |Args2∖{A}|=i we can apply the induction hypothesis to obtain that if C is strongly defended by Args1∖{A} (which it is) C is also strongly defended by Args2∖{A}. □
##### Lemma 2.
Let ArgsAr. Let H0= and Hi+1=F(Hi)Args (i0). For each i0 it holds that
• (1) HiHi+1
• (2) Hi is strongly admissible
• (3) Hi strongly defends each of its arguments
##### Proof.
• (1) Proof by induction over i.
basis For i=0 it holds that Hi⊆Hi+1 because H0=∅⊆H1. step Suppose that for some i⩾0 it holds that Hi⊆Hi+1. From the fact that F is a monotonic function, it follows that F(Hi)⊆F(Hi+1), from which it follows that F(Hi)∩Args⊆F(Hi+1)∩Args. That is, Hi+1⊆Hi+2.
• (2) Proof by induction over i.
basis For i=0 it holds that Hi=H0=∅ which is trivially strongly admissible. step Suppose that for each i⩾0 it holds that each Hj (j⩽i) is strongly admissible. We now prove that Hi+1 is also strongly admissible. Let A∈Hi+1. That is, A∈F(Hi)∩Args. Let j be the smallest number such that A∈Hj (this implies that j⩽i+1 and A∉Hj−1). The fact that A∈Hj means that A∈F(Hj−1)∩Args, so A is defended by Hj−1. From point 1 above, it follows that Hj−1⊆Hi+1, which together with the fact that A∉Hj−1 implies that Hj−1⊆Hi+1∖{A}. This, together with the fact that Hj−1 is strongly admissible (induction hypothesis), means the conditions of Definition 8 (take Hj−1 for Args′) are satisfied.
• (3) Proof by induction over i.
basis For i=0 it holds that Hi=H0=∅ which trivially strongly defends each of its arguments. step Suppose that for some i⩾0 it holds that each Hj (j⩽i) strongly defends each of its arguments. We now prove that Hi+1 also strongly defends each of its arguments. Let A∈Hi+1. Let j be the smallest number such that A∈Hj (this implies that j⩽i+1 and A∉Hj−1). From A∈Hj it follows that A∈F(Hj−1)∩Args, so each attacker B of A is attacked by some C∈Hj−1 such that C is strongly defended by Hj−1 (induction hypothesis). From point 1 above, it follows that Hj−1⊆Hi+1, which together with A∉Hj−1 implies that Hj−1⊆Hi+1∖{A}. So from the fact that each attacker B of A is attacked by some C∈Hj−1 such that C is strongly defended by Hj−1, it follows that each attacker B of A is attacked by some C∈Hi+1 (as Hj−1⊆Hi+1) such that C is strongly defended by Hi+1∖{A} (Lemma 1, together with Hj−1⊆Hi+1∖{A}). □
##### Theorem 1.
Let ArgsAr and let Hi (i0) be as in Lemma 2. Args is strongly admissible iff i=0Hi=Args.
##### Proof.
“⇒”:
Suppose Args is strongly admissible. We need to show that
• (1) i=0HiArgs. This follows directly from the definition of H0 and Hi+1.
• (2) Argsi=0Hi. Let Args1=Args. Suppose towards a contradiction that Args1i=0Hi.
This means there is an A1Args1 such that A1i=0Hi. The fact that Args1 is strongly admissible implies that A1 is defended by some Args2Args1{A1} which in its turn is again strongly admissible. Can it be the case that Args2i=0Hi? If so, there must be an i such that Args2Hi. But as Args2 defends A1, it would follow that A1F(Hi), which together with the fact that A1Args implies that A1F(Hi)Args, so A1Hi+1 so A1i=0Hi. Contradiction. So Args2i=0Hi.
This means there is an A2Args2 such that A2i=0Hi. The fact that Args2 is strongly admissible implies that A2 is defended by some Args3Args2{A2} which in its turn is again strongly admissible. Can it be the case that Args3i=0Hi? If so, there must be an i such that Args3Hi. But as Args3 defends A2, it would follow that A2F(Hi) which together with the fact that A2Args implies that A2F(Hi)Args, so A2Hi+1 so A2i=0Hi. Contradiction. So Args3i=1Hi.
Using similar reasoning as in the above two paragraphs, we observe that there exists a sequence Args1,Args2,,Argsn and a sequence A1,A2,,An where
• (a) for each j{1n1} Argsj+1Argsj (because Argsj+1Argsj{Aj})
• (b) for each j{1n} Argsji=0Hi
• (c) Args1=Args
• (d) Argsn= (because Args1=Args is finite, and we lose at least one argument when going from Argsj to Argsj+1 as shown in point 2a)
From 2b and 2d it follows that Argsn=i=0Hi. Contradiction. Therefore Argsi=0Hi.
“⇐”:
Suppose i=0Hi=Args. Lemma 2 states that each Hi (i0) is strongly admissible. Therefore i=0Hi is also strongly admissible. As i=0Hi=Args it directly follows that Args is strongly admissible. □
##### Theorem 2.
Let ArgsAr and let Hi (i1) be as in Lemma 2. Args strongly defends each of its arguments iff i=0Hi=Args.
##### Proof.
“⇒”:
Suppose Args strongly defends each of its arguments. We need to show that
• (1) i=0HiArgs. This follows directly from the definition of H0 and Hi+1.
• (2) Argsi=0Hi. Suppose towards a contradiction that Argsi=0Hi. This means there is an A1Args such that A1i=0Hi. As Args strongly defends each of its arguments, Args strongly defends A1.
The fact that Args strongly defends A1 means that each attacker B1 of A1 is attacked by some A2Args{A1} such that A2 is strongly defended by Args{A1}. If each such A2 were an element of i=0Hi (so of some Hj) then A1 would be defended by Hj, so A1Hj+1, so A1i=0Hi. Therefore, A2i=0Hi for at least one A2Args{A1}.
The fact that Args{A1} strongly defends A2 means that each attacker B2 of A2 is attacked by some A3Args{A1,A2} such that A3 is strongly defended by Args{A1,A2}. If each such A3 were an element of i=0Hi (so of some Hj) then A2 would be defended by Hj, so A2Hj+1, so A2i=0Hi. Therefore, A3i=0Hi for at least one A3Args{A1,A2}.
Using similar reasoning as in the above two paragraphs, one can identify an infinite sequence of different arguments A1,A2,A3, such that for each Ai (i1) it holds that AiArgs{A1Ai1}. However, since Args contains only a finite number of arguments, this cannot be the case. Contradiction.
“⇐”:
Suppose that i=0Hi=Args. Lemma 2 states that each Hi (i0) strongly defends each of its arguments. Therefore i=0Hi strongly defends each of its arguments. As i=0Hi=Args it directly follows that Args strongly defends each of its arguments. □
##### Theorem 3.
Let (Ar,att) be an argumentation framework and ArgsAr. Args is a strongly admissible set (in the sense of Definition 8) iff each AArgs is strongly defended by Args (in the sense of Definition 7).
##### Proof.
This follows directly from Theorem 1 and Theorem 2. □
Now that the equivalence between the two ways of defining strongly admissible sets has been proven, the next step is to examine some of the formal properties of strong admissibility. We start with conflict-freeness and admissibility.
##### Theorem 4.
Let (Ar,att) be an argumentation framework and let ArgsAr be a strongly admissible set. It holds that:
• Args is conflict-free
• Args is admissible
##### Proof.
Conflict-freeness follows from [3, Proposition 51], together with Theorem 3. Admissibility follows from conflict-freeness, together with the fact that every strongly admissible set defends each of its arguments. □
Baroni and Giacomin prove that the grounded extension is the unique biggest (w.r.t. ⊆) strongly admissible set [3].5 However, it can additionally be proved that the strongly admissible sets form a lattice, of which the grounded extension is the top element and the empty set is the bottom element. To do so, we need two lemmas.
##### Lemma 3.
If Args1 and Args2 are strongly admissible sets, then Args1Args2 is also a strongly admissible set.
##### Proof.
Let Args1 and Args2 be strongly admissible sets. Let AArgs1Args2. If AArgs1 then A is defended by some Args1Args1{A} which in its turn is strongly admissible. If AArgs2 then A is defended by some Args2Args2{A} which in its turn is strongly admissible. In both cases, we have that A is defended by some Args(Args1Args2){A} which in its turn is strongly admissible. Therefore, Args1Args2 is a strongly admissible set in the sense of Definition 8. □
##### Lemma 4.
Each admissible set has a unique biggest (w.r.t.) strongly admissible subset.
##### Proof.
We first observe that each admissible set Args has at least one strongly admissible subset: the empty set. As we consider only finite argumentation frameworks, this implies that there exists at least one maximal (w.r.t. ⊆) strongly admissible subset of Args. We now proceed to show that this maximal strongly admissible subset is unique. Let Args1 and Args2 be maximal strongly admissible subsets of Args. Now consider Args1Args2. From Lemma 3 it follows that this is again a strongly admissible set. From the fact that Args1 and Args2 are maximal strongly admissible subsets, it follows that if Args1Args1Args2 then Args1=Args1Args2, and that if Args2Args1Args2 then Args2=Args1Args2, so we obtain that Args1=Args1Args2 and Args2=Args1Args2 so Args1=Args2. □
If Args is an admissible set, we write Args for its biggest (w.r.t. ⊆) strongly admissible subset.
It turns out that the strongly admissible sets of an argumentation framework form a lattice.6
##### Theorem 5.
Let (Ar,att) be an argumentation framework. The strongly admissible sets of this framework form a lattice (w.r.t.).
##### Proof.
We need to prove that each two strongly admissible sets have a supremum (a least upper bound) and a infimum (a greatest lower bound).
supremum
Let Args1 and Args2 be two strongly admissible sets. From Lemma 3 it follows that Args1Args2 is again a strongly admissible set. Since, by definition, Args1Args1Args2 and Args2Args1Args2, it follows that Args1Args2 is an upper bound. Moreover, it is also a least upper bound, since any proper subset of Args1Args2 will not be a superset of Args1 and Args2.
infimum
Let Args1 and Args2 be two strongly admissible sets. Let Args3 be Args1Args2. From the fact that Args3 is conflict-free, it follows that it has a (unique) biggest admissible subset, which we will refer to as Args3. From Lemma 4 it follows that Args3 has a (unique) biggest strongly admissible subset, which we will refer to as Args3. We now prove that Args3 is an infimum of Args1 and Args2.
lower bound From the fact that Args3″⊆Args3′⊆Args3=Args1∩Args2 it follows that Args3″⊆Args1 and Args3″⊆Args2. greatest lower bound Let Args3‴ be a strongly admissible admissible set such that Args3‴⊆Args1 and Args3‴⊆Args2. Then, by definition, Args3‴⊆Args3. Since Args3‴ is admissible, it follows that Args3‴⊆Args3′ (since Args3′ is the biggest admissible subset of Args3). Since Args3‴ is a strongly admissible subset of Args3′ it follows that Args3‴⊆Args3″ (since Args3″ is the biggest strongly admissible subset of Args3′). □
##### Fig. 2.
The strongly admissible sets of the argumentation framework of Fig. 1.
##### Fig. 3.
The complete extensions of the argumentation framework of Fig. 1.
In essence, if Args1 and Args2 are strongly admissible sets, then Args1Args2 is their supremum, and (Args1Args2) is their infimum. By forming a lattice, with the empty set as its bottom element and the grounded extension as its top element, the strongly admissible sets differ from the admissible sets, which form a semi-lattice with the empty set as its bottom element, and the preferred extensions as its top elements [19]. It also distinguishes the strongly admissible sets from the complete extensions, which form a semi-lattice with the grounded extension as its bottom element and the preferred extensions as its top elements [19].
As an example, a Hasse diagram of the strongly admissible sets of the argumentation framerwork of Fig. 1 is shown in Fig. 2. Notice that (as specified by Theorem 5) the strongly admissible sets form a lattice with the empty set as its bottom element and the grounded extension as its top element. A Hasse diagram of the complete extensions of the argumentation framework of Fig. 1 is shown in Fig. 3. Notice that (as indicated in [19]) the complete extensions form a semi-lattice with the grounded extension as its bottom element and the preferred extensions as its top elements.
## 4.Strongly admissible labellings
Argument labellings [8,15] have become a popular approach for purposes such as argumentation algorithms [9,21,23], argument-based judgment aggregation [16,17] and issues of measuring distance of opinion [6]. In the current section, we develop a labelling account of strong admissibility, which will subsequently be used to analyse some of the existing discussion games for grounded semantics.
To define a strongly admissible labelling, we first have to introduce the concept of a min–max numbering.7
##### Definition 9.
Let Lab be an admissible labelling of argumentation framework (Ar,att). A min–max numbering is a total function MMLab:in(Lab)out(Lab)N{} such that for each Ain(Lab)out(Lab) it holds that:
• if Lab(A)=in then MMLab(A)=max({MMLab(B)B attacks A and Lab(B)=out})+1 (with max() defined as 0)
• if Lab(A)=out then MMLab(A)=min({MMLab(B)B attacks A and Lab(B)=in})+1 (with min() defined as )
To illustrate the concept of a min–max numbering, consider again the argumentation framework of Fig. 1. Here, the admissible labelling Lab1=({A,C,F,G},{B,E,H},{D}) has min–max numbering {(A:1),(B:2),(C:3),(E:4),(F:5),(G:),(H:)}, and the admissible labelling Lab2=({A,C,D,F},{B,E},{G,H}) has min–max numbering {(A:1),(B:2),(C:3),(D:1),(E:2),(F:3)}.
##### Theorem 6.
Every admissible labelling has a unique min–max numbering.
##### Proof.
Let Lab be an admissible labelling, and let Args=Lab2Args(Lab). Now consider the sequence H0,H1,H2, as defined in Lemma 2. For each Ain(Lab) we define MMLab(A) as
2i1, where i is the lowest number such that AHiif i=0Hi contains Aif i=0Hi does not contain A
For each Aout(Lab) we define MMLab(A) as
2i, where i is the lowest number such that Hi attacks Aif i=0Hi attacks Aif i=0Hi does not attack A
We first prove that MMLab is a correct min–max numbering. For this, we need to prove the following two properties from Definition 9:
• if Lab(A)=in then MMLab(A)=max({MMLab(B)B attacks A and Lab(B)=out})+1
Let AAr such that Lab(A)=in. We distinguish two cases:
• (1) Ai=0Hi
In that case, MMLab(A)=2i1, where i is the lowest number such that AHi. We first observe that i1 (otherwise AHi=H0=). In case i=1 it holds that AF(H0)Args, so A is defended by so A does not have any attackers, so max(MMLab(B)B attacks A and Lab(B)=out})+1=0+1=1, which is indeed equal to 2i1 for i=1. In the remaining part of this proof we will therefore focus on the case where i2. In that case AF(Hi1)Args so A is defended by Hi1, which means that each B that attacks A is attacked by some CHi1, so Hi1 attacks B. As B is labelled out (as Lab is an admissible labelling) from the fact that Hi1 attacks B it follows that each out labelled attacker of A has a min–max number of at most 2(i1). However, the fact that AHi1 (as i is the lowest number such that AHi) implies that there exists at least one out labelled attacker of A with min–max number of exactly 2(i1). Therefore max({MMLab(B)B attacks A and Lab(B)=out})+1=2(i1)+1=2i1 so A is numbered correctly.
• (2) Ai=0Hi
In that case, MMLab(A)=. From the fact that Ai=0Hi it follows that AHi for any i1, which means that AF(Hi1)Args. As AArgs, it follows that AF(Hi1) for any i1. This means that there exists a B that attacks A and is not attacked by Hi1 (for any i). So B is not attacked by i=0Hi. It also holds that B is labelled out, as Lab is an admissible labelling. Hence, B is numbered . Therefore max({MMLab(B)B attacks A and Lab(B)=out})+1=+1= so A is numbered correctly.
• if Lab(A)=out then MMLab(A)=min({MMLab(A)B attacks A and Lab(B)=in})+1
Let AAr such that Lab(A)=out. We distinguish two cases:
• (1) i=0Hi attacks A
In that case, MMLab(A)=2i, where i is the lowest number such that Hi attacks A. We first observe that i1 (otherwise i=0 so Hi=H0= which would not be able to attack A). The fact that Hi attacks A means that there is some BHi that attacks A such that Lab(B)=in (as BArgs because Hi=F(Hi1)Args). From this, it follows that the min–max number of B is at most 2i1. The fact that Hi1 does not attack A (as i is the lowest number such that Hi attacks A) means there is no BHi1 that attacks A such that Lab(B)=in. Therefore, it follows that no in labelled attacker of A can have a min–max number of 2(i1)1=2i3 or lower, which together with the fact that the min–max number of any in labelled argument has to be odd means that any in labelled attacker of A has to be at least 2i1. This implies that the min–max number of B is precisely 2i1. We have therefore obtained that there is at least one in labelled attacker of A that is numbered 2i1, and that every in labelled attacker of A is numbered at least 2i1. Therefore, MMLab(A)=min({MMLab(A)B attacks A and Lab(B)=in})+1=(2i1)+1=2i so A is numbered correctly.
• (2) i=0Hi does not attack A
In that case, MMLab(A)=. From the fact that i=0Hi does not attack A it follows that for each attacker B of A it holds that Bi=0Hi. This implies that each in labelled attacker B of A has a min–max number of . Also, A has at least one such in labelled attacker (as Lab is an admissible labelling). Therefore MMLab(A)=min({MMLab(A)B attacks A and Lab(B)=in})+1= so A is numbered correctly.
Now that we have proved that each admissible labelling has a min–max numbering, the next thing to prove is that this min–max numbering is unique. Let MMLab1 and MMLab2 be two min–max numberings of the same admissible labelling Lab. We now prove, by strong induction over i1, that for each AAr, MMLab1=i iff MMLab2=i.
basis i=1“⇒”: Let MMLab1=1. This can only be the case if A is labelled in and does not have any attackers. This means that MMLab2 is also 1.“⇐”: Similar to “⇒”. step Suppose that for each j∈{1…i} it holds that MMLab1(A)=j iff MMLab2(A)=j.“⇒”: Let MMLab1(A)=i+1. We distinguish two cases: (1) Lab(A)=in. In that case, i+1=MMLab1(A)=max({MMLab1(B)∣B attacks A and Lab(B)=out})+1 so i=max({MMLab1(B)∣B attacks A and Lab(B)=out}). This implies that for each out labelled attacker B of A, it holds that MMLab1(B) is at most i. Hence, we can apply the induction hypothesis and infer that MMLab2(B)=MMLab1(B). Therefore, MMLab2(A)=max{MMLab2∣B attacks A and Lab(B)=out})+1=max{MMLab1∣B attacks A and Lab(B)=out}+1=i+1.(2) Lab(A)=out. In that case i+1=MMLab1(A)=min({MMLab1(B)∣B attacks A and Lab(B)=in})+1 so i=min{MMLab1(B)∣B attacks A and Lab(B)=in}). This means there is at least one in labelled B that attacks A such that MMLab1(B)=i. From the induction hypothesis we infer that MMLab2(B)=i. Furthermore, for each j
From the thus proved fact that for each 11, MMLab1(A)=i iff MMLab2(A)=i, together with the fact that each min–max number has to be in (N{0}){} it follows that also MMLab1(A)= iff MMLab2(A)=. Hence, we have that for each AAr, MMLab1(A)=MMLab2(A), so MMLab1=MMLab2. □
Using the concept of a min–max numbering, we can proceed to define the concept of a strongly admissible labelling.
##### Definition 10.
A strongly admissible labelling is an admissible labelling whose min–max numbering yields natural numbers only (so no argument is numbered ).
From Definition 10 it directly follows that every strongly admissible labelling is also an admissible labelling. Also, there exists a clear connection between strongly admissible labellings and strongly admissible sets, as one can be converted into the other.
##### Theorem 7.
Let (Ar,att) be an argumentation framework.
• for every strongly admissible set ArgsAr, it holds that Args2Lab(Args) is a strongly admissible labelling
• for every strongly admissible labelling Lab, it holds that Lab2Args(Lab) is a strongly admissible set
##### Proof.
• Let Args be a strongly admissible set. This means that i=0Hi=Args. The procedure specified in the proof of Theorem 6 makes sure that every argument in Hi (i0) is numbered with a natural number (note: each such argument is labelled in). As i=0Hi=Args it follows that each argument in Args (that is, each in labelled argument) is numbered with a natural number. It then follows (from Definition 9) that also each out labelled argument is numbered with a natural number. This means that no argument is numbered , thus satisfying the condition of Definition 10.
• Let Lab be a strongly admissible labelling. This means that each in or out labelled argument is numbered with a natural number. As the min–max number of Lab can be constructed using the procedure explained in the proof of Theorem 6, the fact that each in labelled argument A is assigned a natural number means that for each AArgs there is an i such that AHi. This means that Argsi=0Hi. This, together with the fact that HiArgs for each i0, implies that Args=i=0Hi which means that Args is a strongly admissible set. □
Please notice that strongly admissible labellings and strongly admissible sets are not one-to-one related; instead, they are many-to-one related. As an example, in the argumentation framework of Fig. 1 the strongly admissible set {A,C} is related to two strongly admissible labellings: ({A,C},{B,E},{D,F,G,H}) and ({A,C},{B},{D,E,F,G,H}).
The fact that strongly admissible sets and strongly admissibe labellings are not one-to-one related unfortunately means that some of the results for strongly admissible sets (for instance Theorem 5) do not automatically carry over to strongly admissible labellings. Instead, they need to be proved separately.
##### Lemma 5.
If Lab1 and Lab2 are strongly admissible labellings, then Lab1Lab2 is also a strongly admissible labelling.
##### Proof.
Let Args1=Lab2Args(Lab1) and Args2=Lab2Args(Lab2). From Theorem 7 it then follows that Args1 and Args2 are strongly admissible sets, hence (Lemma 3) Args1Args2 is also a strongly admissible set. Let Lab3=Args2Lab(Args1Args2). From Theorem 7 it follows that Lab3 is a strongly admissible labelling. Let MMLab3 be the min–max numbering of this strongly admissible labelling. How does Lab3 compare with Lab3=Lab1Lab2? We start with making the following thee observations.
• Lab3 is an admissible labelling.
This is because Lab3 is a strongly admissible labelling.
• Lab3 is an admissible labelling.
From the fact that Args1Args2 is a strongly admissible set, it follows that Args1Args2 is conflict-free, so Args1 and Args2 do not attack each other. This implies that in(Lab1)out(Lab2)= and in(Lab2)out(Lab1)=, which means that Lab1Lab2=(in(Lab1)in(Lab2),out(Lab1)out(Lab2),undec(Lab1)undec(Lab2)). That is, we obtain that in(Lab3)=in(Lab1)in(Lab2) and that out(Lab3)=out(Lab1)out(Lab2). Hence, each argument that is labelled in by Lab3 has all its attackers labelled out by Lab3 (this folows from the fact that Lab1 and Lab2 are admissible labellings) and each argument that is labelled out by Lab3 has an attacker that is labelled in by Lab3 (which again follows from the fact that Lab1 and Lab2 are admissible labellings). This means that Lab3 is an admissible labelling.
• in(Lab3)=in(Lab3).
In the previous point, it was observed that in(Lab3)=in(Lab1)in(Lab2). As we have that in(Lab1)=Args1 (as Args1=Lab2Args(Lab1)) and in(Lab2)=Args2 (as Args2=Lab2Args(Lab2)) it holds that in(Lab3)=Args1Args2 (as in(Lab3)=in(Lab1)in(Lab2) as observed in the previous point). As in(Lab3)=Args1Args2 (as Lab3=Args2Lab(Args1Args2)) we obtain that in(Lab3)=in(Lab3).
We define the min–max numbering MMLab3 of Lab3 such that MMLab3(A)=MMLab3(A) for each Ain(Lab3)out(Lab3). We now prove that MMLab3 is a correct min–max numbering. For this, we need to show two things:
• if Lab3(A)=in then MMLab3(A)=max({MMLab3(B)B attacks A and Lab3(B)=out})+1
As Lab3 and Lab3 are both admissible labellings, it holds that all attackers of an in labelled argument are labelled out. This implies that argument A, which is labelled in by Lab3 and is therefore also labelled in by Lab3 (as in(Lab3)=in(Lab3)) has the same out labelled attackers in both Lab3 and Lab3. As the out labelled attackers of A are numbered the same by MMLab3 as by MMLab3 we have that {MMLab3(B)B attacks A and Lab3(B)=out}={MMLab3(B)B attacks A and Lab3(B)=out}. This, together with the fact that MMLab3(A)=max({MMLab3(B)B attacks A and Lab3(B)=out})+1 (as MMLab3 is a correct min–max numbering) and the fact that MMLab3(A)=MMLab3(A) implies that MMLab3(A)=max({MMLab3(B)B attacks A and Lab3(B)=out})+1.
• if Lab3(A)=out then MMLab3(A)=min({MMLab3(B)B attacks A and Lab3(B)=in})+1
As in(Lab3)=in(Lab3) it holds that the out labelled argument A has the same in labelled attackers in both Lab3 and Lab3. As the in labelled attackers of A are numbered the same by MMLab3 as by MMLab3 we have that {MMLab3(B)B attacks A and Lab3(B)=in}={MMLab3(B)B attacks A and Lab3(B)=in}. This, together with the fact that MMLab3(A)=min({MMLab3(B)B attacks A and Lab3(B)=in})+1 (as MMLab3 is a correct min–max numbering) and the fact that MMLab3(A)=MMLab3(A) implies that MMLab3(A)=min({MMLab3(B)B attacks A and Lab3(B)=in})+1.
Hence, we have established that MMLab3 is a correct min–max numbering of Lab3, one that numbers each argument that is labelled in or out by Lab3 the same as MMLab3. As MMLab3 does not number any argument with (as Lab3 is a strongly admissible labelling) it follows that MMLab3 also does not number any argument with . Hence, Lab3 is a strongly admissible labelling. □
##### Lemma 6.
Each admissible labelling Lab has a unique biggest (w.r.t.) strongly admissible sublabelling.
##### Proof.
Similar to the proof of Lemma 4, but with Args, Args1 and Args2 replaced by Lab, Lab1 and Lab2, with ⊆ and ∩ replaced by ⊑ and ⊓ and Lemma 3 replaced by Lemma 5. □
If Lab is an admissible labelling, then we write Lab for its biggest (w.r.t. ⊑) strongly admissible sublabelling.
##### Theorem 8.
Let (Ar,att) be an argumentation framework. The strongly admissible labellings of this framework form a lattice (w.r.t.).
##### Proof.
Similar to the proof of Theorem 5, with Args1, Args2, Args3, Args3, Args3 and Args3 replaced by Lab1, Lab2, Lab3, Lab3, Lab3 and Lab3, ⊆, ∪ and ∩ replaced by ⊑, ⊔ and ⊓, and Lemma 4 replaced by Lemma 6. ⊆ replaced by ⊑, ∪ replaced by ⊔ and ∩ replaced by ⊓. □
## 5.Computational complexity
Regarding the issue of computational complexity, one can distinguish the standard decision problems of credulous acceptance, sceptical acceptance and verification.
The formal statement of these decision problems (which are defined for any extension based argumentation semantics σ) is presented in Table 1. In this, we use σ to described an arbitrary semantics, e.g. any of the cases given in Definition 3, although our principal interest will be the case of σ being the class of strongly-admissible sets; Eσ for the set of all subsets of arguments within a framework that satisfy the criteria given by σ, e.g. Eadm(Ar,att) is the set of all admissible sets in the framework Ar,att.
##### Table 1
Decision problems in argumentation semantics
Problem Instance Question Formal statement ca ⟨Ar,att⟩, x∈Ar Is x credulously accepted? ∃S⊆Ar:x∈S and S∈Eσ(⟨Ar,att⟩)? sa ⟨Ar,att⟩, x∈Ar Is x sceptically accepted? ∀S⊆Ar:S∈Eσ(⟨Ar,att⟩)⇒x∈S? ver ⟨Ar,att⟩ S⊆Ar Does S satisfy the criteria of σ? S∈Eσ(⟨Ar,att⟩)?
The credulous acceptance problem for strong admissibility reduces to deciding if the given argument, x, is in the grounded extension, as the grounded extension is the (unique) biggest (w.r.t. ⊆) strongly admissible set [3]. Hence, the credulous acceptance problem of strong admissibility is of polynomial complexity.
As for sceptical acceptance, the issue is to determine whether a particular argument is in every strongly admissible set. However, as the empty set is always strongly admissible, this decision problem is trivial as the answer is always negative.
The verification problem is more interesting in that it is not simply a matter of testing if S is a subset of the grounded extension, i.e. although S being such a subset is a necessary condition for strong-admissibility it is not a sufficient condition. In determining if a set S is strongly admissible one could use Algorithm 1.
##### Algorithm 1
Verification of Args as a strongly admissible set
The correctness of Algorithm 1 follows from Theorem 1 and Lemma 2. To see this, notice that the algorithm accumulates a subset of Ar (in Argsi) stopping when there is no change to the existing subset (i.e. that forming Argsi1) and using the final set to compare with the candidate subset Args. The set(s) Argi are formed by the adding the intersection of the characteristic function of Hi with the set, Args, being tested. This set, Hi starts (i=0) from the empty set. Overall the process of computing successive subsets Hi and the concomitant changes to Argsi mimics exactly the stages applied in the proof of Theorem 1 using the result of Lemma 2 as support.
As we only consider finite argumentation frameworks, the algorithm is guaranteed to terminate.
The maximal number of loop iterations is of the order |Ar| because at each iteration there will be at least one argument added (this must be the case for otherwise we would have Argsi=Argsi1 resulting in the loop terminating at line 8) until the loop terminates. As for the set operations of union and intersection, it holds that S1S2 and S1S2 each require the order of |S1|+|S2| operations, provided that appropriate data structures are being used. For the above algorithm, this would be |F(Hi)|+|Args|+|Argsi|+|Hi+1|, so no more than 4·|Ar| for each loop iteration. As there are no more than |Ar| loop iterations, this implies the maximal number of steps for doing the set operations is in the order of |Ar|·|Ar|.
As for determining the number of operations of the F-operator, the easiest way to do this is to consider the total number of operations, throughout all loop iterations. Calculating F(S) can be done basically by removing the arguments of S+ from the argumentation framework (together with the attacks from and to S+) and then examining which arguments have no attackers (this is one standard approach used in computing the grounded extension). Point 1 of Lemma 2 implies that this can be done in an iterative way when it comes to calculating each Hi+1. As no more than |att| edges can be removed from the graph, there can be at most |att| edges relevant to any Hi. Note that the maximum number of attacks that could be present in any framework satisfies |att||Ar|·|Ar|, so it holds that the number of operations is at most |Ar|·|Ar| for computing the outcome of the characteristic functions. This, together with the number of operations required for computing the union and intersection (which is also |Ar|·|Ar|) means the total number of required operations is in the order of |Ar|·|Ar|, so of polynomial complexity.
As an aside, please notice that in order to simplify the above discussion, we have formulated Algorithm 1 in a way that is closely aligned to Lemma 2 and Theorem 1. However, it can be observed that for every i it holds that Argsi=Hi (this is because of point 1 of Lemma 2). Hence, it would be possible to do away with Argsi in the above algorithm, and only use Hi instead. We observe that this does not affect the overall complexity of the algorithm, which remains in the order of |Ar|·|Ar|.
## 6.Strong admissibility and argument games
Now that some of the formal properties of strong admissibility have been examined, the next step is to study some of its applications. In particular, it turns out that strong admissibility is one of the corner stones of the discussion games for grounded semantics.
### 6.1.The standard grounded game
As far as we are know, the Standard Grounded Game [7,21,26] was the first dialectical proof procedure to determine whether a particular argument is in the grounded extension.
##### Definition 11.
A discussion in the Standard Grounded Game is a finite sequence [A1,,An] (n1) of arguments (sometimes called moves), of which the odd moves are called P-moves (Proponent moves) and the even moves are called O-moves (Opponent moves), such that:
• (1) every O-move is an attacker of the preceding P-move (that is, every Ai where i is even and 2in attacks Ai1)
• (2) every P-move except the first one is an attacker of the preceding O-move (that is, every Ai where i is odd and 3in attacks Ai1)
• (3) P-moves are not repeated (that is, for every odd i,j{1,,n} it holds that if ij then AiAj)
A discussion is called terminated iff there is no An+1 such that [A1,,An,An+1] is a discussion. A terminated discussion is said to be won by the player making the last move.
An argument tree is a tree of which each node (n) is labelled with an argument (Arg(n)). The level of a node is the number of nodes in the path to the root.
##### Definition 12.
A winning strategy of the Standard Grounded Game for argument A is an argument tree, where the root is labelled with A, such that
• (1) for each path from the root (nroot) to a leaf node (nleaf) it holds that the arguments on this path form a terminated discussion won by P
• (2) for each node at odd level nP it holds that {Arg(nchild)nchild is a child of nP}={BB attacks Arg(nP)} and the number of children of nP is equal to the number of attackers of Arg(nP)
• (3) each node of even level nO has precisely one child nchild, and Arg(nchild) attacks Arg(nO)
The soundness and completeness of the Standard Grounded Game depends on the presence of a winning strategy. That is, an argument A is in the grounded extension iff there exists a winning strategy for A. Interesting enough, it turns out that such a winning strategy defines a strongly admissible set containing A.
##### Theorem 9.
The set of all proponent moves in a winning strategy of the Standard Grounded Game is strongly admissible.
##### Proof.
We prove this by induction over the depth (i) of the winning strategy game tree.
basis i=0. In that case, the winning strategy consists of a single argument (say, A). This means that A has no attackers. Hence, {A} is a strongly admissible set. step Suppose that every winning strategy of depth less or equal than i has its proponent moves constituting a strongly admissible set. We need to prove that also every winning strategy of depth i+2 has its proponent moves constituting a strongly admissible set. Let WS be a winning strategy of depth i+2. Let A be the argument at the root of the tree. Let WS1′,…,WSn′ be the subtrees whose roots are at distance 2 of the root of WS. The induction hypothesis states that for each of these subtrees (WSj′), their set of proponent moves Argsj′ constitutes a strongly admissible set. Therefore (by Lemma 3) the set Args′=⋃j=1nArgsj′ is strongly admissible. Also, A∉Args′ (this is because the proponent is not allowed to repeat his moves). Let B be an arbitrary argument in Args (the set of all proponent moves in the winning strategy). We distinguish two cases:(1) B∈Args′. Then, since Args′ is a strongly admissible set, there exists an Args″⊆Args′∖{B} that defends B and is itself strongly admissible. Since Args′⊆Args, it also holds that Args″⊆Args∖{B}.(2) B∉Args′. Then B=A (the root of the tree WS). The structure of the WS tree is such that B is defended by the roots of WS1′,…,WSn′. So B is defended by the strongly admissible set Args′. Also B∉Args′, so Args′⊆Args∖{B}, therefore satisfying Definition 8. □
It can also be observed that a winning strategy defines a strongly admissible labelling.
##### Theorem 10.
Let ArgsP be the set of proponent moves and ArgsO be the set of opponent moves of a particular winning strategy given an argumentation framework (Ar,att). It holds that (ArgsP,ArgsO,Ar(ArgsPArgsO)) is a strongly admissible labelling.
##### Proof.
Given that ArgsP is strongly admissible (Theorem 9) it then follows from Theorem 7 that LabPP+=(ArgsP,ArgsP+,Ar(ArgsPArgsP+)) is a strongly admissible labelling. Now consider LabPO=(ArgsP,ArgsO,Ar(ArgsPArgsO)). Notice that ArgsPArgsP+, otherwise ArgsP would not be an admissible set. Also, from the structure of a winning strategy (with the Opponent playing all possible attackers of each Proponent move as its children) it follows that ArgsO=ArgsP. Hence, ArgsOArgsP+. LabPO has the same min–max numbering as LabPP+ (minus the arguments that are no longer out in LabPO, since out(LabPO)out(LabPP+), as ArgsOArgsP+). This is because the out-labelled arguments in ArgsP+ArgsO do not influence the min–max numbers of the in-labelled arguments in ArgsP. It then follows that the min–max numbers of the out-labelled arguments in LabPO also stay the same. Hence, the min–max numbering of LabPO is essentially a restricted version (with a smaller domain) of the min–max numbering of LabPP+. So from the fact that LabPP+ is a strongly admissible labelling (not yielding ) it directly follows that LabPO is a strongly admissible labelling (not yielding ). □
Hence, given a winning strategy of the Standard Grounded Game, the set of all proponent moves and the set of all opponent moves essentially define a strongly admissible labelling.
### 6.2.The grounded discussion game
Like the Standard Grounded Game, the Grounded Discussion Game [11] is a proof procedure to determine whether a particular argument is a member of the grounded extension. The game has two players (proponent and opponent) and is based on four different moves, each of which has an argument as a parameter.
HTB(A) (“A has to be the case”)With this move, the proponent claims that argument A has to be labelled in by every complete labelling (and hence also has to be labelled in by the grounded labelling). CB(B) (“B can be the case, or at least cannot be ruled out”)With this move, the opponent claims that argument B does not have to be labelled out by every complete labelling. That is, the opponent claims there exists at least one complete labelling where B is labelled in or undec, and that B is therefore not labelled out by the grounded labelling. CONCEDE(A) (“Fair enough, I agree that A has to be the case”)With this move, the opponent indicates that he now agrees with the proponent (who previously did a HTB(A) move) that A has to be the case (labelled in by every complete labelling, including the grounded labelling). RETRACT(B) (“Fair enough, I give up that B can be the case”)With this move, the opponent indicates that he no longer beliefs that argument B can be in or undec. That is, the opponent acknowledges that B has to be labelled out by every complete labelling, including the grounded labelling.
One of the key ideas of the discussion game is that the proponent has burden of proof. He has to establish the acceptance of the main argument. The opponent merely has to cast sufficient doubts. Also, the proponent has to make sure that the discussion does not go around in circles.
The game starts with the proponent uttering a HTB statement. After each HTB statement (either the first one or a subsequent one) the opponent utters a sequence of one or more CB, CONCEDE and RETRACT statements, after which the proponent again utters an HTB statement, etc. In the argumentation framework of Fig. 1 the discussion could go as follows.
In the above discussion, C is called the main argument (the argument the discussion starts with). The discussion ends with the main argument being conceded by the opponent, which means the proponent wins the discussion.
As an example of a discussion that is lost by the proponent, it can be illustrative to examine what happens if, still in the argumentation framework of Fig. 1, the proponent claims that B has to be the case.
After the second move, the discussion is terminated, as the proponent cannot move anymore, since A does not have any attackers. This brings us to the precise preconditions of the discussion moves.
HTB(A) This is either the first move, or the previous move was CB(B), where A attacks B, and no CONCEDE or RETRACT move is applicable. CB(A) A is an attacker of the last HTB(B) statement that is not yet conceded, the directly preceding move was not a CB statement, argument A has not yet been retracted, and no CONCEDE or RETRACT move is applicable. CONCEDE(A) There has been a HTB(A) statement in the past, of which every attacker has been retracted, and CONCEDE(A) has not yet been moved. RETRACT(A) There has been a CB(A) statement in the past, of which there exists an attacker that has been conceded, and RETRACT(A) has not yet been moved.
Apart from the preconditions mentioned above, all four statements also have the additional precondition that no HTB-CB repeats have occurred. That is, there should be no argument for which HTB has been uttered more than once, CB has been uttered more than once, or both HTB and CB have been uttered. In the first and second case, the discussion is going around in circles (which the proponent has to prevent, since he has burden of proof). In the third case, the proponent has been contradicting himself, as his statements are not conflict-free. In each of these three cases, the discussion comes to an end with no move being applicable anymore.
The above conditions are made formal in the following definition.
##### Definition 13.
Let (Ar,att) be an argumentation framework. A grounded discussion is a sequence of discussion moves constructed by applying the following principles.
BASIS (HTB) If A∈Ar then [HTB(A)] is a grounded discussion. STEP (HTB) If [M1,…,Mn] (n⩾1) is a grounded discussion without HTB-CB repeats,8 and no CONCEDE or RETRACT move is applicable,9 and Mn=CB(A) and B is an attacker of A then [M1,…,Mn,HTB(B)] is also a grounded discussion. STEP (CB) If [M1,…,Mn] (n⩾1) is a grounded discussion without HTB-CB repeats, and no CONCEDE or RETRACT move is applicable, and Mn is not a CB move, and there is a move Mi=HTB(A) (i∈{1…n}) such that the discussion does not contain CONCEDE(A), and for each move Mj=HTB(A′) (j>i) the discussion contains a move CONCEDE(A′), and B is an attacker of A such that the discussion does not contain a move RETRACT(B), then [M1,…,Mn,CB(B)] is a grounded discussion. STEP (CONCEDE) If [M1,…,Mn] (n⩾1) is a grounded discussion without HTB-CB repeats, and CONCEDE(B) is applicable then [M1,…,Mn,CONCEDE(B)] is a grounded discussion. STEP (RETRACT) If [M1,…,Mn] (n⩾1) is a grounded discussion without HTB-CB repeats, and RETRACT(B) is applicable then [M1,…,Mn,RETRACT(B)] is a grounded discussion.
It can be observed that the preconditions of the moves are such that a proponent move (HTB) can never be applicable at the same moment as an opponent move (CB, CONCEDE or RETRACT). That is, proponent and opponent essentially take turns in which each proponent turn consists of a single HTB statement, and every opponent turn consists of a sequence of CONCEDE, RETRACT and CB moves.
##### Definition 14.
A grounded discussion [M1,M2,,Mn] is called terminated iff there exists no move Mn+1 such that [M1,M2,,Mn,Mn+1] is a grounded discussion. A terminated grounded discussion (with M1 being HTB(A) for some AAr) is won by the proponent iff the discussion contains CONCEDE(A), otherwise it is won by the opponent.
To illustrate why the discussion has to be terminated after the occurrence of a HTB-CB repeat, consider the following discussion in the argumentation framework of Fig. 1.
After the third move, an HTB-CB repeat occurs and the discussion is terminated (opponent wins). Hence, termination after a HTB-CB repeat is necessary to prevent the discussion from going on perpetually.
It has been proved [11,12] that the Grounded Discussion Game is a sound and complete proof procedure for determining whether an argument is in the grounded extension. More specifically,
(soundness) if a discussion for a particular argument has been won by the proponent, then the argument is in the grounded extension, and (completeness) if an argument is in the grounded extension, then the proponent is able to win the game for the argument (that is, the proponent has a winning strategy)
As for the first point (soundness) it can be observed that if one would label the arguments of CONCEDE moves in, the arguments of RETRACT moves out and all other arguments undec, the result is a strongly admissible labelling at each state of the discussion game. If this game is ultimately won by the proponent, then the main argument has been CONCEDEd. Hence, the result is a strongly admissible labelling where the main argument is labelled in. This means the main argument is in the grounded extension.
As for the second point (completeness) it can be observed that the grounded labelling, together with its associated min–max numbering, serves as a roadmap that allows the proponent to win the game. In essence, the proponent does this by using only in labelled arguments to select the HTB moves, and to select the in labelled argument with a lowest min–max number whenever there is a choice. We refer to [11,12] for details.
### 6.3.The Standard Grounded Game (SGG) vs. the Grounded Discussion Game (GDG)
So far, we have seen that both the SGG and the GPG show membership of the grounded extension essentially by building a strongly admissible labelling where the argument in question is labelled in.10 This raises the question of how many steps each of these games requires for doing so. Consider the argumentation framework of Fig. 4 (top left). The winning strategy of the SGG is in the same figure (top right). Now consider what would happen if one would start to extend the argumentation framework by duplicating the middle part. That is, suppose we have arguments B1,,Bn and C1,,Cn (with n being an odd number), as well as arguments A and D. Suppose that for every i{1,,n1} Bi+1 attacks Bi, and Ci+1 attacks Ci, and that for each even i{2,,n1} Bi+1 attacks Ci, and Ci+1 attacks Bi, and that B1 and C1 attack A, and that D attacks Bn and Cn. In that case, the branches in the SGG winning strategy would split at every O-move. So for n=3 (as is the case in Fig. 4) the number of branches is four, for n=5 it is eight, etc. In general, the number of branches in the SGG winning strategy is 2(n+1)/2, with the number of nodes in the SGG winning strategy being 1+2i=1(n+1)/22i. Hence, the number of steps needed in a winning strategy of the SGG can be exponential in relation to the in/out-size11 of the strongly admissible labelling that the SGG winning strategy is constructing.12
##### Fig. 4.
The Standard Grounded Game (SGG) versus the Grounded Discussion Game (GDG).
As for the Grounded Discussion Game, the situation is different. It can be proven [11,12] that the proponent always has a strategy for the game that results in the total number of moves being 2·|in(Lab)|+2·|out(Lab)| where Lab the strongly admissible labelling that is built up during the discussion game. This labelling is such that in(Lab) consists of all arguments that have been subject to a CONCEDE move and out(Lab) consists of all arguments that have been subject to a RETRACT move. An example of a game that results from such a strategy is provided in Fig. 4.
Overall, we observe that both the Standard Grounded Game and the Grounded Discussion Game prove that an argument is in the grounded extension by building a strongly admissible labelling around it. However, where the Standard Grounded Game can require a number of moves that is exponential in relation to the in/out-size of the strongly admissible labelling, the Grounded Discussion game requires a number of moves that is always linear in relation to the in/out-size of the strongly admissible labelling.
## 7.Discussion and future research
In the current paper, we have re-examined the concept of strong admissibility, from both theoretical and practical perspectives. From theoretical perspective, we have observed that the strongly admissible sets form a lattice with the empty set as bottom element and the grounded extension as top element. Also, we have developed the concept of a strongly admissible labelling, and shown how it relates to the concept of a strongly admissible set. From practical perspective, we have examined how strongly admissible labellings lie at the basis of both the Standard Grounded Game [21] and the Grounded Discussion Game [11,12]. Although both essentially construct a strongly admissible labelling around the argument in question, the Grounded Discussion Game does so using a linear number of steps, whereas the Standard Grounded Game can require an exponential number of steps.
An alternative definition of a strongly admissible set is given by Baumann et al. [4]. Basically, the idea is that a set of Args is strongly admissible iff there are finitely many and pairwise disjoint sets A1,,An such that (1) Args=i=1nAi, (2) A1F(), and (3) i=1jAi defends Aj+1 (1j<n). Baumann et al. [4] prove that their definition is equivalent with Definition 8 of the current paper (which first appeared in [10]). One particular issue with their definition is that they do not specify how to actually obtain the sequence A1,,An. However, we observe that it is fairly easy to convert the sequence H0,H1,H2, as specified in Lemma 2 to a corresponding sequence A1,,An. This can be done by first identifying n to be the lowest number such that Hn=Hn+1 (which implies that Hm=Hn for each mn) and then taking A1=H1 and Ai+i=Hi+1Hi (1i<n).
The idea of numbering arguments (such as is done in a min–max numbering) can be traced back to the work of Pollock [24], who gives an iterative procedure (basically for computing the grounded extension, as is explained by Dung [19]) in which arguments become in and out at different levels during the algorithm [24, Algorithm 2]. It has to be mentioned, however, that Pollock’s algorithm (the ideas of which have also been applied in [2]) computes the entire grounded extension (in a way that is similar to what is done in [21]) and is not applicable to the concept of a strongly admissible set (or labelling) in general.
One of the things to be examined in the future is how the concept of strong admissibility can be useful in identifying the shortest discussion that shows an argument (A) is in the grounded extension. For instance, we conjecture that for each minimal (w.r.t. ⊑) strongly admissible labelling that labels A in, there exists a discussion under the Grounded Persuasion Game for argument A that builds precisely this labelling. However, there can be more than one such labelling. For argument F in Fig. 1, for instance, both ({A,C,F},{B,E},{D,G,H}) and ({D,F},{E},{A,B,C,G,H}) are minimal (w.r.t. ⊑) strongly admissible labellings that label F in, but the in/out-size of the second labelling is smaller than that of the first labelling, thus yielding a shorter discussion. How to precisely obtain such a strongly admissible labelling with minimal size is a topic for further investigation.
Finally there are a number of questions that would merit further consideration with respect to complexity issues. For example, although the canonical decision problems (credulous and sceptical acceptance, verification) for the strong admissibility semantics are tractable having polynomial-time sequential algorithms, it seems unlikely that verification would have efficient parallel algorithms. The notion of “efficient parallel algorithm” being one that can be realised using a logarithmic depth Boolean combinational circuit. A formal demonstration that such is indeed the case would be achieved by showing the verification problem to be p-complete. In view of the supporting technical detail that would be required in exploring this question we have not pursued it in the current article.
A question of some considerable interest whose status is far from clear concerns the following: given two argumentation frameworks (Ar,att1) and (Ar,att2) (that is with identical arguments but not necessarily identical attacks), do their strongly admissible sets coincide? That is, is the case that Esa(Ar,att1)=Esa(Ar,att2)? Alternatively we could examine if |Esa(Ar,att1)|=|Esa(Ar,att2)|. It is worth noting that the “equivalence by set equality” is only one such definition, but there are alternatives: we could also ask about the existence of a relabelling of arguments so that the two frameworks become identical. It is worth noting two further aspects of this question: firstly, unlike previously studied and superficially similar problems, e.g. coincidence of stable and preferred semantics from [20] the question involves more than a single framework (although given an appropriate semantics, σ, the question Eσ(Ar,att)=Esa(Ar,att) may also be non-trivial: while some instances, e.g. preferred semantics, reduce to known cases since Epr(Ar,att)=Esa(Ar,att) if and only if Epr(Ar,att)={} others are less so). A second point is the how much “obvious” necessary conditions can be exploited, e.g. “in order for Esa(Ar,att1)=Esa(Ar,att2) to hold the number of unattacked arguments in each must be equal” (otherwise there will be different single argument strongly admissible sets in the two). While conditions such as these suggest a natural way of progressing from single to two to k argument set comparisons, there is potentially an exponential increase in the number of sets being compared as the process continues. Such further algorithmic and complexity issues are the topic of continuing work.
## Notes
1 With the in/out-size of a labelling Lab, we mean |in(Lab)out(Lab)|.
2 This paper is an extended and thoroughly revised version of work that was presented at COMMA 2014 [10] and TAFA 2015 [11]. In particular, we have rewritten some of the previously unpublished proofs (of Theorem 1, Theorem 2, Theorem 3, Theorem 6 and Theorem 7) to take advantage of a new technical result (Lemma 2). Moreover, we have added results on computational complexity (Section 5) and we have decided to include the Grounded Discussion Game [11] instead of the outdated Grounded Discussion Game [18].
3 The well-definedness of the down-admissible set follows from [16], where this concept is defined in its labellings form, together with the equivalence between extensions and labellings [15].
4 In [19] a preferred extension is defined as a maximal admissible set, instead of as a maximal complete extension, but it can be shown that these are equivalent.
5 Hence, each strongly admissible set is an admissible set that is contained in the grounded extension. The converse, however, does not hold. For instance, in Fig. 1, {F} is an admissible set that is contained in the grounded extension, but it is not a strongly admissible set.
6 We recall that a lattice is a partial order such that each two elements have both a greatest lower bound and a least upper bound. For an ordering to be a semi-lattice only one of these two conditions needs to be met.
7 The intuition behind the min–max number of an argument is that of the game-theoretic length of the path (consisting of alternately in and out labelled arguments) from the argument back to an unattacked ancestor argument. The player selecting the in labelled arguments aims to make the path as short as possible whereas the player selecting the out labelled arguments aims to make the path as long as possible.
8 We say that there is a HTB-CB repeat iff i,j{1,,n}AAr:(Mi=HTB(A)Mi=CB(A))(Mj=HTB(A)Mj=CB(A))ij.
9 A move CONCEDE(B) is applicable iff the discussion contains a move HTB(A) and for every attacker A of B the discussion contains a move RETRACT(B), and the discussion does not already contain a move CONCEDE(B). A move RETRACT(B) is applicable iff the discussion contains a move CB(B) and there is an attacker A of B such that the discussion contains a move CONCEDE(A), and the discussion does not already contain a move RETRACT(B).
10 Similarly, it can be observed that for instance the credulous preferred game [14,27] shows membership of a preferred extension essentially by building an admissible labelling around the argument in question.
11 We recall that with the in/out-size of a labelling Lab we mean |in(Lab)out(Lab)|.
12 We thank Mikołaj Podlaszewski for this example.
## Acknowledgements
We would like to thank Ringo Baumann and the anonymous reviewers for their useful comments.
## References
[1] P. Baroni, M.W.A. Caminada and M. Giacomin, An introduction to argumentation semantics, Knowledge Engineering Review 26(4) (2011), 365–410. doi:10.1017/S0269888911000166. [2] P. Baroni and M. Giacomin, Self-stabilizing defeat status computation: Dealing with conflict management in multi-agent systems, Artificial Intellligence 165 (2005), 187–259. doi:10.1016/j.artint.2005.02.003. [3] P. Baroni and M. Giacomin, On principle-based evaluation of extension-based argumentation semantics, Artificial Intelligence 171(10–15) (2007), 675–700. doi:10.1016/j.artint.2007.04.004. [4] R. Baumann, T. Linsbichler and S. Woltran, Verifiability of argumentation semantics, in: Computational Models of Argument; Proceedings of COMMA 2016, P. Baroni, T.F. Gordon, T. Scheffler and M. Stede, eds, 2016, pp. 83–94. [5] T.J.M. Bench-Capon, Dilemmas and paradoxes: Cycles in argumentation frameworks, Journal of Logic and Computation (2014). [6] R. Booth, M.W.A. Caminada, M. Podlaszewski and I. Rahwan, Quantifying disagreement in argument-based reasoning, in: Proceedings of the 11th International Conference on Autonomous Agents and Multiagent Systems, 2012. [7] M.W.A. Caminada, For the sake of the Argument. Explorations into argument-based reasoning, PhD thesis, Vrije Universiteit Amsterdam, 2004. [8] M.W.A. Caminada, On the issue of reinstatement in argumentation, in: Logics in Artificial Intelligence; 10th European Conference, JELIA 2006, M. Fischer, W. van der Hoek, B. Konev and A. Lisitsa, eds, LNAI, Vol. 4160, Springer, 2006, pp. 111–123. [9] M.W.A. Caminada, An algorithm for computing semi-stable semantics, in: Proceedings of the 9th European Conference on Symbolic and Quantitalive Approaches to Reasoning with Uncertainty (ECSQARU 2007), Springer Lecture Notes in AI, Springer Verlag, Berlin, 2007, pp. 222–234. doi:10.1007/978-3-540-75256-1_22. [10] M.W.A. Caminada, Strong admissibility revisited, in: Computational Models of Argument, Proceedings of COMMA 2014, S. Parsons, N. Oren, C. Reed and F. Cerutti, eds, IOS Press, 2014, pp. 197–208. [11] M.W.A. Caminada, A discussion game for grounded semantics, in: Theory and Applications of Formal Argumentation (Proceedings TAFA 2015), E. Black, S. Modgil and N. Oren, eds, Springer, 2015, pp. 59–73. doi:10.1007/978-3-319-28460-6_4. [12] M.W.A. Caminada, A Discussion Protocol for Grounded Semantics (proofs), Technical Report, University of Aberdeen, 2015. [13] M.W.A. Caminada and L. Amgoud, On the evaluation of argumentation formalisms, Artificial Intelligence 171(5–6) (2007), 286–310. doi:10.1016/j.artint.2007.02.003. [14] M.W.A. Caminada, W. Dvořák and S. Vesic, Preferred semantics as Socratic discussion, Journal of Logic and Computation 26 (2014), 1257–1292. doi:10.1093/logcom/exu005. [15] M.W.A. Caminada and D.M. Gabbay, A logical account of formal argumentation, Studia Logica 93(2–3) (2009), 109–145, Special issue: New ideas in argumentation theory. doi:10.1007/s11225-009-9218-x. [16] M.W.A. Caminada and G. Pigozzi, On judgment aggregation in abstract argumentation, Autonomous Agents and Multi-Agent Systems 22(1) (2011), 64–102. doi:10.1007/s10458-009-9116-7. [17] M.W.A. Caminada, G. Pigozzi and M. Podlaszewski, Manipulation in group argument evaluation, in: Proceedings of the Twenty-Second International Joint Conference on Artificial Intelligence, T. Walsh, ed., 2011, pp. 121–126. [18] M.W.A. Caminada and M. Podlaszewski, Grounded semantics as persuasion dialogue, in: Computational Models of Argument – Proceedings of COMMA 2012, B. Verheij, S. Szeider and S. Woltran, eds, 2012, pp. 478–485. [19] P.M. Dung, On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games, Artificial Intelligence 77 (1995), 321–357. doi:10.1016/0004-3702(94)00041-X. [20] P.E. Dunne and T.J.M. Bench-Capon, Coherence in finite argument systems, Artificial Intelligence 141(1–2) (2002), 187–203. doi:10.1016/S0004-3702(02)00261-8. [21] S. Modgil and M.W.A. Caminada, Proof theories and algorithms for abstract argumentation frameworks, in: Argumentation in Artificial Intelligence, I. Rahwan and G.R. Simari, eds, Springer, 2009, pp. 105–129. doi:10.1007/978-0-387-98197-0_6. [22] S. Modgil and H. Prakken, A general account of argumentation with preferences, Artificial Intellligence 195 (2013), 361–397. doi:10.1016/j.artint.2012.10.008. [23] S. Nofal, K. Atkinson and P.E. Dunne, Algorithms for decision problems in argumentation systems under preferred semantics, Artificial Intelligence 207 (2014), 23–51. doi:10.1016/j.artint.2013.11.001. [24] J.L. Pollock, How to reason defeasibly, Artificial Intelligence 57 (1992), 1–42. doi:10.1016/0004-3702(92)90103-5. [25] H. Prakken, An abstract framework for argumentation with structured arguments, Argument and Computation 1(2) (2010), 93–124. doi:10.1080/19462160903564592. [26] H. Prakken and G. Sartor, Argument-based extended logic programming with defeasible priorities, Journal of Applied Non-Classical Logics 7 (1997), 25–75. doi:10.1080/11663081.1997.10510900. [27] G.A.W. Vreeswijk and H. Prakken, Credulous and sceptical argument games for preferred semantics, in: Proceedings of the 7th European Workshop on Logic for Artificial Intelligence (JELIA-00), Springer Lecture Notes in AI, Springer Verlag, Berlin, 2000, pp. 239–253. doi:10.1007/3-540-40006-0_17. | 2020-08-09 06:23:24 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8660480380058289, "perplexity": 1557.6128963253886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00567.warc.gz"} |
https://motls.blogspot.com/2010/03/defending-statistical-methods.html | ## Saturday, March 20, 2010 ... /////
### Defending statistical methods
There surely exist propositions by the skeptics - and opinions liked by many skeptics - that I find highly unreasonable. I don't know whether they're equally frequent as the alarmists' delusions but they certainly do exist. A whole article of this sort was written by Tom Siegfried in Science News,
Odds Are, It's Wrong.
The article whose subtitle is "Science fails to face the shortcomings of statistics" - it sounds serious, doesn't it? - was promoted at Anthony Watts' blog. The most characteristic quote in the article is the following:
It’s science’s dirtiest secret: The “scientific method” of testing hypotheses by statistical analysis stands on a flimsy foundation.
So I gather that the idea is that one should throw away all statistical methods which are a "mutant form of math". That's a rather radical conclusion.
There surely exist whole scientific disciplines that are trying to find tiny, homeopathic signals that can be hugely overinterpreted and hyped because the researchers are usually rewarded for such statements, regardless of their validity (or at least, they don't pay any significant price if the claims turn out to be wrong). The science of health impacts of XY is the classic example - and environmental and climate sciences may have become another.
People in those disciplines are usually led by their environment to be "finding" effects even if they don't really exist. The average ethical and intellectual qualities of the people who work in these disciplines are poor. But it's just preposterous to imagine that the right cure could be to throw away or ban all statistical methods.
Richard Feynman about the probabilistic character of most scientific insights. Taken from 37:45, "Seeking New Laws", the last Messenger Lecture at Cornell University (1964).
Statistical methods are crucial and omnipresent
In fact, statistical methods have always been essential in any empirically based science. In the simplest situation, a theory predicts a quantity to be "P" and it is observed to be "O". The idea is that if the theory is right, "O" equals "P". In the real world, neither "O" nor "P" is known infinitely accurately.
Why? Because observations are never accurate, so "O" always has some error, at least if it is a continuous quantity. And "P" is almost always calculated by a formula that depends on other values that had to be previously measured, too. So even the predictions "P" have errors. There are various kinds of errors that contribute and they would deserve a separate lecture.
Moreover, quantum mechanics implies that all observations ever made have some uncertainty and all of them are statistical in character. The most complete possible theories can only predict the probabilities of individual outcomes. Clearly, all observations you can ever make have a statistical nature. In particular, experimental particle physics would be impossible without statistics. If you can't deal with the statistical nature of the empirical evidence, you simply can't do empirical science.
Now, if "O" and "P" are known with some errors, how do you determine whether the theory passes or fails the test? The errors are never "strict": there is always a nonzero probability that a very big error, much bigger than the expected one, is accumulated, so you should never imagine that the intervals "(O - error, O + error)" are "strictly certain". Nothing is certain. If you pick 1,000 random people, the deviation of the number of women from 500 may be around 30 but it is unlikely, but not impossible, that there will be 950 women and 50 men.
The answer to the question that started the previous paragraph is, of course, that if "O" and "P" are (much) further from one another than both errors of "O" as well as "P", the theory is falsified. It's proven wrong. If they're close enough to one another, the theory may pass the test: we failed to disprove it. But as always in science, it doesn't mean that the theory has been proven valid. Theories are never proven valid "permanently". They're only temporarily valid until a better, more accurate, newer, or more complete test finds a discrepancy and falsifies them.
In the distant past, people wanted to learn "approximate", qualitatively correct theories. So the hypotheses that would eventually be ruled out used to be "very wrong". Their predicted "P" was so far from the observations "O" that you could have called the disagreement "qualitative" in character. However, strictly speaking, it was never qualitative. It was just quantitative - and large.
But as our theories of anything in the physical Universe are getting more accurate, it is completely natural that the differences between "O" and "P" of the viable candidate hypotheses is getting smaller, in units of the errors of "O" or "P". In some sense, the new scientific findings at the "cutting edge" or the "frontier" almost always emerge from the "mud" in which "O" and "P" looked compatible. When the accuracy of "O" or "P" increases, we can suddenly see that there's a discrepancy.
We should always ask how big a discrepancy between "O" and "P" is needed for us to claim that we have falsified a theory. This is a delicate problem because there will always be a nonzero probability that the discrepancy has occurred by chance. We don't want to make mistakes. So we want to be e.g. 99.9% sure that if we say that a theory has been falsified, it's really wrong.
The required separation between "O" and "P" can be calculated from the figure above, from 99.9%. If you don't know the magic of statistical distributions, especially the normal one, I won't be teaching you about them in this particular text. But it's true that the probability that the falsification "shouldn't have been done" because the disagreement was just due to chance is decreasing more quickly than exponentially with the separation - as the Gaussian.
So e.g. particle physics typically expects "new theories" to be supported by "5 sigma signals" - in some sense, the distance between "O" and "P" is at least 5 times their error. The probability that this takes place by chance is smaller than one in one million. Particle physicists choose such a big separation - and huge confidence level - because they don't want to flood their discipline with lots of poorly justified speculations. They want to rely upon solid foundations so statistical tests have to be really convincing.
Softer disciplines typically choose less than 5 sigma to be enough: 2 or even 1 sigma is sometimes presented as a signal that matters. Of course, this is because they actually want to produce lots of results even though they may be (and, sometimes, are likely to be) rubbish. But a simple fix is that they should raise the required confidence level for their assertions - e.g. from 2 sigma to 5 sigma. They don't have to immediately throw statistics as a tool away.
Be sure that a 5-sigma confidence level is also enough for tests of dozens of drugs at the same moment. I agree with Siegfried that if you're making very many tests, there will surely be some "false positives" - statistics happens. But with an appropriately chosen confidence level, depending on the context, one can keep any errors in his research "very rare".
A problem is that many of these researchers actually don't want to do it - e.g. to improve the confidence level. They don't want their science to work right. They have other interests.
In fact, while the confidence level is dramatically increased if we go from 2 sigma to 5 sigma (something like from 90% to 99.9999%), the required amount of data we need to collect to get the 5-sigma accuracy is just "several times" bigger than for the 2-sigma accuracy. So if there's some effect, it's not such a huge sin to demand that published "discoveries" should be supported by 5-sigma signals. Once again, the soft scientists - who propose various theories of health (what is healthy for you) - are choosing low confidence levels deliberately because they like to present new results even though they're mostly bogus. They still get famous along the way.
If some key statements about AGW are only claimed to be established at the 90% confidence level, it's just an extremely poor evidence (and may be overstated or depend on the methods, anyway). In principle, it shouldn't be hard for the evidence for such a hypothesis, assuming it's true, to be strengthened to 99.9% or more. That's what "hard sciences" deserving the name require, anyway.
The laymen usually misunderstand how little "90%" is as a confidence level - and some traders with fear masterfully abuse this ignorance. 90% vs 10% is not that "qualitatively" far from 50% vs 50% - and one can transform one to the other by a "slight" pressure in the methodology and the formulae. If you want to be scientifically confident about a conclusion, you should really demand 99.9% or more. And it's actually not that hard to obtain such stronger evidence assuming that your hypothesis is actually correct and the "signal" exists.
Falsifying a null hypothesis
I must explain some basic points of the statistical methods. Typically, we want to find out whether a new effect exists. So we have two competing hypotheses: I will call them the null hypothesis and the alternative hypothesis.
The null hypothesis says that no new effect exists - everything is explained by the old theories that have been temporarily established and any pattern is due to chance. When I say "chance", it's important to realize that one must specify the exact character of the "random generator" that produces these random data, including the deviations, correlations, autocorrelations, persistence, color of the noise etc. There's not just one "chance": there are infinitely many "chances" given by "statistical distributions" and we must be damn accurate about what the null hypothesis actually says. (Often, we mean the "white noise" and "independent random numbers" etc.)
The alternative hypothesis says that a new effect is needed: the old explanations and the null hypothesis is not enough.
How do you decide in between these two? Well, you calculate the probability that the apparently observed "pattern" could have occurred by chance assuming the null hypothesis. If the probability of something like that were sufficiently high, e.g. 1% or 5%, you say that your data don't contain evidence for the alternative hypothesis.
If the calculated probability that the "pattern" in the data could have been explained by chance - and by the null hypothesis - is really tiny, e.g. 10^{-6}, then your data give you a strong evidence that the null hypothesis is wrong. If you say that it's wrong, your risk of having made a wrong conclusion - the so-called "false positive" or "type I error" - is only 10^{-6}. So it's sensible to take this risk. In my example, we falsified the null hypothesis at the 99.9999% level. It's very likely that a new effect has to exist.
You're expected to have an alternative hypothesis that actually describes the data more accurately and gives a higher probability that the data could have occurred according to the alternative hypothesis, with its new understanding of "chance".
However, if the probability of getting the pattern by chance, from the null hypothesis, is substantial, e.g. 10%, then your data only provide you with a very weak hint that a new effect could exist. If you use the standards of hard sciences, you should say that your data can't settle the question in either way.
Of course, it is always possible that if you make such a conclusion, you have made another kind of error, the "type II error", also known as the false negative. But what Tom Siegfried seems to misunderstand is that this is a common situation that you simply can't avoid in most cases. The data, with their limited volume and limited accuracy (and assuming a small size of the new effect), simply can't settle the question in either way.
So when you say that you don't have enough evidence to confirm the "pattern", i.e. that the data don't contain a statistically significant evidence for the alternative hypothesis i.e. the new effect, it is not the ultimate proof that the alternative hypothesis is wrong. It is not the final proof that the new effect can't exist.
It's just evidence that the new effect is small and unimportant enough so that it couldn't have been detected in the particular sample or experiment. You can't make a final decision here. While hypotheses can be kind of "completely killed" in science, they can never be "completely proved". Even though the null hypothesis can be pretty much safely killed, no one can ever guarantee to you that your particular generalization, your alternative hypothesis, is the most correct one. It could have been better than the null hypothesis in passing this particular test but the next one may falsify your alternative hypothesis, too.
There's no straightforward way to construct better hypotheses! Creativity and intuition is needed before your viable attempts are tested against the data.
And quite often, your data simply don't contain enough information to decide. This is not a bug that you should blame on the statistical method. The statistical method is innocent. It is telling you the truth and the truth is that we don't know. The laymen may often be scared by the idea that we don't know something - and they often prefer fake and wrong knowledge over admitting that we don't know - but it's their vice, their inability to live with what the actual science is telling us (or not telling us, in this case), not a bug of the statistical method.
Misinterpretations, errors, lousy scientists
Of course, the picture above assumes that one actually learns how the statistical method works and what it exactly allows us to claim in particular situations. That has nothing to do with the journalists' or laymen's interpretations. The journalists and other laymen usually don't understand statistics well - and sometimes they want to mislead others deliberately.
But again, it would be ludicrous to blame this fact on the statistical method.
Analogously, bad scientists may calculate confidence levels incorrectly. They may choose unrealistic null and/or alternative hypotheses: in systems theory, a wrong choice of the null hypothesis is sometimes referred to as the "type III error". And they may misinterpret what their test has really demonstrated and what it hasn't. They may hold completely unrealistic beliefs about the odds that a "generic" hypothesis would pass a similar test so they can't place their calculation in any proper context. Sometimes, they think that by falsifying the null hypothesis, they're proving the first alternative hypothesis that they find convenient to believe (one can't prove it in such a way, you would have to falsify all other possible alternative theories first!). Quite typically, such people only blindly follow some statistical recipes that they don't quite understand. So it's not shocking that they can end up with mistakes.
This fact is not specific to statistics. People who are lousy scientists often make errors in non-statistical scientific methodologies, too. That's not a reason to abandon science, is it?
The proper statistical method gives us the best tool to study the incomplete or inaccurate empirical information - and in the real world, every empirical information is incomplete or inaccurate, at least to some extent. And one can actually prove that the probability of a "false positive" is as small as the significance level: it's true pretty much by definition. Well, the p-value is not quite the same thing as the probability of a "false positive" i.e. as the confidence level but it's pretty close: if a calculated p-value is at most equal to the required significance level, the test may be used to reject the null hypothesis.
But "false negatives" can never be reliably cured. Whenever your experiment is not accurate enough, it will simply say "no pattern seen" even though a better experiment could see it.
The solution to fight against the widespread errors is to require the soft disciplines to become harder - to calculate the confidence levels properly and to require higher confidence levels than those that have been enough for a "discovery" in the recent decades. This recommendation follows from common sense. If your field has been flooded by lots of beliefs in correlations and mechanisms that often turned out to be incorrect or non-existent, it's clear that you should make your standards more stringent.
Scientists, journalists, and laymen should do their best to be accurate and to learn what various tests actually imply.
But it will still be true that no science can be done "quite" without any statistical reasoning. And it's still true that the datasets and experiments will continue to be unable to give the "final answer" to many questions we would like to be answered. These are just facts. You may dislike them but that's the only thing you can do against facts.
So I would urge everyone to try to avoid bombshell statements such as "statistics is a dirty core of science that doesn't work and has to be abandoned". Lousy work of some people can't ever justify such far-reaching claims.
After all, much of the lousy work - and lousy presentation in the media - emerges because the people want to claim that the relevant research is "less statistical" in character than it actually is. In most cases, weak statistical signals are being promoted to a kind of "near certainty". So the right solution is for everyone to be more appreciative of the statistical method, not less so!
And that's the memo.
Tamino and 5 sigma
Tamino claims that "requiring five sigma is preposterous". Well, it's not. It's what disciplines of hard sciences require as a criterion for a discovery. See five sigma discovery at Google (2000 pages). In particular, discoveries of new particles by colliders do require 5 sigma. No one would have claimed a discovery of a top quark at 3 sigma - which would only be viewed as a suggestive yet vague hint.
Once again, this increase is needed because people often cook their results to make "discovery claims" that are bogus: it's easy to "improve" the tests. If you try 10 variations of the same test, one of them will show a (fake) effect at a 90% confidence level: that's what the 90% confidence level means, by definition. Unfortunately, many researchers are approaching the things in this way.
With a 5-sigma discovery, such cheating becomes virtually impossible because you would need a million of variations of your paper - and only one of them would show a fake positive. On the other hand, it's not "infinitely more difficult" to get 5-sigma results relatively to 3-sigma results. Because the relative errors go like "1/sqrt(N)" where N is the number of events (whose average you're calculating, in a way), you only need to increase the number of events by a factor of "(5/3)^2 = 2.7778" to go from 3 sigma to 5 sigma.
Because of the amazing increase of the "purity" of your results and their immunity against errors and your own bias, it's surely worth paying this extra factor of 2.7778, isn't it?
#### snail feedback (12) :
If you read the article, you might actually agree with it.
The problem is that most articles actually published accept 2 sigma evidence (P<0.05). That's the main thing the author objects to.
If articles could not be published without 5 sigma evidence, pretty clearly the author and the experts that he is channeling would be satisfied.
Hi Lubos
I am an academic who teaches empirical research to students. I would like to post the original article and your rebuttal for my students to read. Could you pls post a modified version which is sufficiently academic (ie. Without phrases such as 'Holy Cow'). :). Your columns are always entertaining, but sometimes may not be sufficiently academic for a classroom.
Dear Bob, I surely agree with you, as my text indicates - if you read it, I wrote your comment explicitly.
The only problem is that they are satisfied with a low confidence level. But that's not the impression I got from that other article: it wanted to throw away the whole method.
Dinesh, thanks for your interest. Couldn't you please do the editing yourself? This is a blog, not a textbook for a classroom that apparently tries to be as boring and dull as possible.
I give you all the permissions to fix the informalities. For example, "dumb" should be "irrational" while "Holy cow" should be "That seems as a rather bold statement." I am sure you can do it with others, too.
Thank you.
Hi Lubos,
I've been confused by some of the discussion on tropospheric temperature measurements, especially the Santer et al 2008 paper. As far as I understand it, the crux of that issue is to do with the way that statistics is used to interpret the dataset. Do you have any comments or analysis about that question from the perspective of your article?
As a professor of statistics I can say that it is common to see people bashing statistics. It is rare to see someone defending it! Thanks.
Larry Waserman
Dept of Statistics
Carnegie Mellon
Hi Lubos,
Off topic, but...
I would appreciate it if you could take a few minutes to look at this blog post...
http://climatesanity.wordpress.com/2010/03/21/rahmstorf-2009-off-the-mark-again-part-1/
It looks to me like choosing a temperature that looks like T=Cexp(-at/b) in Rahmstorf's 2009 sea level scare story model will always yield a sea level rise of 0. (note that b is negative, so this is an exponentially rising temperature)
This would seem to invalidate his relationship between sea level rise rate and temperature.
Best Regards,
Tom Moriarty
Hi Lubos,
Off topic, but...
I would appreciate it if you could take a few minutes to look at this blog post...
http://climatesanity.wordpress.com/2010/03/21/rahmstorf-2009-off-the-mark-again-part-1/
It looks to me like choosing a temperature that looks like T=Cexp(-at/b) in Rahmstorf's 2009 sea level scare story model will always yield a sea level rise of 0. (note that b is negative, so this is an exponentially rising temperature)
This would seem to invalidate his relationship between sea level rise rate and temperature.
Best Regards,
Tom Moriarty
The article didn't bash statistics, it pointed out the big damn flaw with a lot of the work being done and published every day in the name of "science".
Statistics are wonderful so long as you are careful. Most of the studies being done these days tend to have really squishy data at the bottom. Used to be, you would see a number in the abstract. Apparently that isn't in fashion any more. Now we see "Statistically Significant". In order to get to the numbers you end up wading through the entire paper, then wondering if somebody left something out.
You don't need to throw statistics completely out, but you do have to be careful about its application.
If you are counting the number of protons hitting a target, you are probably still safe to use statistics.
If you are counting the number of people who get asthma in an area with increased pollution on the other hand finding statistical significance with an RR of 1.2 does not a tragedy make.
phillip_jr
The reason you aren't grasping what Santer does is that frequentist statistics should never be used for model outputs. The only papers on this subject have said that Bayesian stats should be used though they are still difficult to apply. The reason there arem so few papers on this subject is because combining model outputs as if the combination made any more sense than any individual run is a pretty stupid idea of itself and without any foundation in reality.
In effect, Douglass et al. didn't need to do any stats on the paper because us modelers just compare every model run to the actual observed data and calculate a percentage error. Climate scintists don't do this because to them the model is better than the data. hence the tendency to change the data in line with models.
Notwithstanding all of that, Santers 3 sigma test fails anyway just by using up to date data. Steve McI tried to get a comment like that published but it was apparently too long and boring for the journal.
All science is based on measurement, and all measurements (excepting pure counting) involve an indeterminate error. Those errors propagate according to Gaussian statistics, and in fact he propagation of those errors demonstrate 1) whether the errors were truly indeterminate 2) the confidence of the measurements 3) and the confidence in those measurements in demonstrating the (null) hypothesis, assuming that has been formulated correctly around the measurements and their methods.
The application of those statistics to "model" results is meaningless, because there is no way to demonstrate that the errors PROVIDED BY THE COMPUTATION were indeterminate. There may have been a number of factors outside the control of the modeler that would make any "errors" of computation determinate, such as truncation, roundoff, insufficient parameters in a model to come to a meaningful result, etc and it is impossible to account for all the possibilities.
And that's the memo.
You and I seem to have read slightly different articles. His point was not that all statistics should be thrown out. I read that statistics is so seldom used appropriately by scientists and so poorly understood by nonstatisticians (even by most people who talk a good game of knowing proper use of stats) that the MISUSE of statistics leads to a lot of incorrect interpretations. He makes several points about how people use stats that I see in almost every scientific paper I read.
The point is not to throw out stats, but to not let the stats control your brain. The way the scientific establishment overuses bad stats is what needs to change and that is what the article was about to me. | 2018-07-18 16:21:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.574167788028717, "perplexity": 825.1649549591883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590295.61/warc/CC-MAIN-20180718154631-20180718174631-00199.warc.gz"} |
https://www.helpwithphysics.com/2016/05/particle-in-a-ring-wells.html | # Particle in a Ring. Wells
1. The wave function of a particle in a ring is $psi(phi,t)=frac{1}{sqrt{2pi}}frac{1}{sqrt{2}}e^{-iphi}e^{ihbar t/2I}-frac{1}{sqrt{2pi}}frac{1}{sqrt{2}}e^{-iphi}e^{i2hbar t/2I}$. Please find the expectation value of the energy.
$psi=frac{1}{sqrt{2pi}}(C_1phi_1+C_2phi_2)$ is a mix of elementary states.
For first state $C_1=1/sqrt{2}$ and $phi_1=exp(-iphi)exp(-ihbar t/2I)$
For second state $C_1=-isqrt{2}$ and $phi_2=exp(iphiexp(-2ihbar t/2I)$
By comparying the elementary states with the general type $phi=exp(imphi)exp(-iomega t)$
we obtain two energies in the spectrum
$E_1=hbaromega_1=hbar^2/2I$ and $e_2=hbaromega_2=2hbar^2/I=4hbar^2/2I$
For the coefficients we have $|C_1|^2+|C_2|^2=1$ so that the expectation energy is
$<E>=E_1*|C_1|^2+E_2*|C_2|^2=[(1/2)*1+(1/2)*4](hbar^2/2I)=5hbar^2/4I$
2.
Rank for the penetration distance $d$ of the wavefunction for the energy levels in the figure. The penetration distance is defined from $psisim e^{-x/d}$.
The decay of the wavefunction in the well wall is of the type $psi sim exp (-kx)$ so that $d=1/k=frac{hbar}{sqrt{2m(U-E)}}$ and thus the penetration depth depends only on difference sqrt{U-E}.
For wells in figures a) and b) the energy difference is the same $U-E=5 eV$. therefore the penetration depth is the same in a) and b) cases. For case c) $U-E=16-10=6 eV >5eV$ so that the penetration depth is in this case smaller than in the first two cases (a and b).
3.
Specify the features of the wavefunctions for the indicated energies.
Features:
-The energy level n has n-1 nodes inside the well.
-The amplitude of a wave inside a well is proportional to the $sqrt{1/L}$ where $L$ is the width of the well. So lower energy levels will have a bigger amplitude in the case given.
-The penetration depth is inversely proportional to the difference $sqrt{U-E}$ so that the lowest energy level will have a smaller penetration depth into the wall.
-All wavefunctions have a a zero at the left wall (it is infinite).
The frequency of the wavefunction is increasing with the energy level, according to the relation $E_n=hbaromega_n$. | 2022-01-22 09:44:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8510661721229553, "perplexity": 304.17094111673435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00451.warc.gz"} |
https://math.stackexchange.com/questions/3318932/pullbacks-of-exact-sequences-of-locally-free-sheaves | # Pullbacks of exact sequences of locally free sheaves.
Let $$X$$ and $$T$$ be schemes smooth over $$\mathbb{C}$$ and suppose that $$Pic(X \times T) \cong Pic(X) \times Pic(T)$$.
Consider the following short exact sequence of locally free sheaves on $$X$$, $$0 \to \mathcal{L} \to \Omega_{X}^1 \to \mathcal{L}' \to 0$$ where $$\mathcal{L}$$ and $$\mathcal{L}'$$are line bundles. In particular, $$\mathcal{L}$$ is a line subbundles of $$\Omega_{X}^1$$.
Let $$p_S: X \times T \to X$$.
Then we have a short exact sequence, $$0 \to p_s^*(\mathcal{L}) \to \Omega_{X \times T/T}^1 \to p_2^*(\mathcal{L}') \to 0$$
Is it possible that there exist a line bundle $$\mathcal{J}$$ on $$T$$, such that
$$0 \to p_2^{*}(\mathcal{L}) \otimes p_1^*(\mathcal{J}) \to \Omega_{X \times T/T}^1 \to p_2^*(\mathcal{L}') \to 0??$$
Computing the determinant of $$\Omega^1{X \times T/T}$$ with the help of the last two sequences, one concludes that $$p_1^*(\mathcal{J}) \cong \mathcal{O}_{X \times T}$$, which implies $$\mathcal{J} \cong \mathcal{O}_X$$.
• What is the information that the determinant of $\Omega_{X \times T/T}$ provides? – user7090 Aug 10 '19 at 19:02 | 2020-03-30 05:02:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9789347648620605, "perplexity": 86.48928549644361}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370496523.5/warc/CC-MAIN-20200330023050-20200330053050-00411.warc.gz"} |
https://stats.stackexchange.com/questions/144966/what-does-varimax-mean-in-factor-analysis | # What does “varimax” mean in factor analysis?
In the rotation options of SPSS Factor Analysis, there is a rotation method named "Varimax". If I choose this option, does it mean the orthogonal rotation technique of Principal Component Analysis will be applied on the factor loadings by analyzing the co-variance matrix of the factor loadings? (Because "varimax" sounds a bit like "maximizing variance", which is what PCA does.)
• Did you give a try to type "varimax" in Google search? – ttnphns Apr 6 '15 at 8:21
• The answer to your question is NO, despite what the accepted answer says. Varimax has no direct relationship to PCA. – amoeba says Reinstate Monica Sep 21 '17 at 22:38
I think that the answer to your question is Yes (at least, in the big picture sense). Should you be wanting to dive deeper into details, I would suggest you to review this excellent discussion here on Cross Validated, especially an answer by @amoeba and/or Chapter 6 of the excellent online book by Revelle (2015). Having said that, I would like to make the following points:
• Varimax and other rotation methods, are not specific to SPSS, as they are general exploratory factor analysis (EFA) terms (so maybe spss tag should be deleted from the question).
• While varimax is the most popular option across research literature (this is likely the reason it is the default option for psych::factanal() in R) and usually produces simpler, easier to interpret, factor solutions, since all orthogonal rotation methods produce uncorrelated factors, they often are not the best. Oblique transformation methods, due to allowing factors to correlate, produce less simple models, however, it is argued that it is beneficial, since such models more accurately reflect reality, in other words, have higher explanatory power, with an additional benefit of better reproducibility of the results (Costello & Osborne, 2005).
• I think that, following the tradition of the exploratory data analysis and research, it is much better to try several EFA approaches and methods and choose the optimal one, based not only on analytical fit indices, but first and foremost, based on making sense within the theory around studied constructs (if it exists) or domain knowledge (if developed theories don't yet exist for the domain under study).
References
Costello, A. B., & Osborne, J. W. (2005). Best practices in exploratory factor analysis: Four recommendations for getting the most from your analysis. Practical Assessment, Research & Evaluation, 10(7). Retrieved from http://pareonline.net/pdf/v10n7.pdf
Revelle, W. (2015). An introduction to psychometric theory with applications in R. [Website] Retrieved from http://www.personality-project.org/r/book
• +1. The only things I would add to this explanation are: 1) orthogonal v. oblique rotation does not produce factors that are uncorrelated or correlated, respectively. The two approaches simply differ with respect to what they assume; orthogonal methods assume variables are uncorrelated, so the correlation is not estimated. Oblique methods do not make this assumption and so correlations are estimated. And 2) oblique models are not necessarily less simple. In fact, when factors are truly correlated, oblique solutions possess greater simple structure (Fabrigar & Wegener, 2012) – jsakaluk Apr 6 '15 at 20:29
• @jsakaluk: Thank you for upvoting and your comment. I will keep your points in mind, when I will get a chance to update my answer. – Aleksandr Blekh Apr 6 '15 at 22:03
• @Tony: My pleasure! – Aleksandr Blekh Apr 7 '15 at 2:08
• I think that the answer to your question is Yes (at least, in the big picture sense). - This is wrong, the answer to this question is NO. Varimax factor rotation has nothing to do with PCA. The remaining part of your answer is okay, but this first sentence is plain wrong and quite misleading. – amoeba says Reinstate Monica Sep 21 '17 at 22:35
• @amoeba Thank you for your comment - it was quite a while ago to remember my exact line of thinking. :-) But, I guess, because of only indirect relation between of the two concepts I have used the phrase in the big picture sense. I agree that a direct answer to the question in NO (I likely wanted to emphasize some conceptual similarity - sorry, don't remember details). Please feel free to update my answer with your clarification. – Aleksandr Blekh Sep 22 '17 at 1:01 | 2020-01-24 13:40:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.679149329662323, "perplexity": 1066.5864396039437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250620381.59/warc/CC-MAIN-20200124130719-20200124155719-00159.warc.gz"} |
https://codereview.stackexchange.com/questions/91127/name-word-generator-using-dtmc-in-ruby?noredirect=1 | # Name/word generator using DTMC in Ruby
I just wrote a basic DTMC algorithm focused on generating names, though it could be used to generate lots of other things. It's designed to be run from the command line, so input is taken from a file, with a couple of options specified on the command line.
I'm specifically looking for these things, though any other advice is also of course welcome:
• Ways to make it faster or more memory-efficient
• Ways to make it more idiomatic for Ruby
I'm perfectly fine with changing the input format, though I'd prefer if it stayed the same. I'm also fine with making the code a bit more unreadable if it helps make it more efficient or shorter, as long as it's still somewhat understandable and easy to maintain.
This is my full code (version with -h help available here):
#Constants!
LINE_PART_DELIMITER = '|'
DEFAULT_NAME_COUNT = 10
# End with the standard error message format -- makes it easy to stay consistent
def error(code, message)
puts "Error #{code}: #{message}"
puts 'Run this script with -e to see a list of error codes.'
abort "name_gen.rb: Error #{code}: #{message}"
end
# Keys are candidates, values are probabilities
def weighted_random_choice(picking_from)
current = 0
max = picking_from.values.inject :+
r_val = rand max
picking_from.each { |candidate, probability|
current += probability
return candidate if r_val < current
}
raise "r_val>max? #{r_val>max}. Error while picking a weighted random value from #{picking_from}"
end
#Parsing the commandline arguments and suchlike
syllable_separator = (/-d./ === ARGV[-1]) ? ARGV.pop[2..-1] : ''
name_count = (/\d+/ === ARGV[-1]) ? ARGV.pop.to_i : DEFAULT_NAME_COUNT
file = ARGV.join ' '
#Parsing the file
syllables = Hash.new false
start = Hash.new 0
begin
IO.foreach(file) { |line|
name, start_prob, end_prob, links = line.split LINE_PART_DELIMITER
error 2, "#{line}" if links.nil? || end_prob.nil? || start_prob.nil? || name.nil?
(prob = Integer prob) rescue error 3, "#{name}: #{syl}, #{prob}"
memo[syl.to_sym] += prob
memo
}
start[name.to_sym] += start_prob.to_i
}
rescue Exception => message
puts message
error 1, "#{file}"
end
#Validate that all syllables referenced actually exist!
error(4, "#{link} in #{syllable}") if !!link && syllables[link.to_sym].nil?
}
}
#Generating and printing the names
name_count.times {
current_syllable = weighted_random_choice start
name_so_far = [current_syllable.to_s]
while (current_syllable = weighted_random_choice syllables[current_syllable.to_sym])
name_so_far << current_syllable.to_s
end
puts name_so_far.join syllable_separator
}
And this is an example 'dictionary' file:
a|1|1|b,2;c,2
b|0|3|a,0;c,2
c|0|0|a,1;b,1
And here's a sample of ten names it can generate (with the above dictionary file):
ab
acb
acabcacacb
abca
acacaca
abcacacb
acb
a
acbcacb
a
I don't like how you've mixed the concerns of calculating with output. I know that right now you only want to output to a file, but what if you decide later that you want to work with this data in some other program? Writing to the file system is expensive and slow. Why write to a file and then read it back in.
I would modify this to be in two parts. One to generate the names from the dictionary and one that uses that class to output to a file. This leaves things open to writing the output to Standard IO, some other UI, or for another program to simply work with the data. The idea is that each class should do one thing and do it well.
On this note, each of your comments indicates a missed opportunity to extract a well named method that does one and only one thing.
#Parsing the commandline arguments and suchlike
syllable_separator = (/-d./ === ARGV[-1]) ? ARGV.pop[2..-1] : ''
name_count = (/\d+/ === ARGV[-1]) ? ARGV.pop.to_i : DEFAULT_NAME_COUNT
file = ARGV.join ' '
This is just begging to be a method called parse_cmd_args. It should return some object that represents those three values in some sensible and well named way.
I apologize that I've left s critique without any code examples. Normally I would provide some, but it's been a while since I've written any Ruby. It's better in this case that I leave you to attempt a clean up yourself.
Try to:
• Write methods that do one and only one thing.
• Create useful abstractions by the way of classes, even if they're just simple data structures to hold information.
• Add some vertical whitespace around logic that's too trivial or too intertwined to be its own method.
• My logic in not making each of those methods is that a) they're only used once and b) they're not reusable snippets of code. Also, this currently (unless I'm mistaken) outputs to the standard output, which can be caught and used whatever -- the exact points you make are why. It does, however, take input from a file, which could probably be changed to standard input if it's needed. – Fund Monica's Lawsuit May 27 '15 at 23:38
• Re-using a snippet isn't the only reason to extract a method. It's also a useful layer of abstraction. Reasons to Create a Routing from Code Complete: Reduce Complexity, introduce intermediate, understandable abstraction, avoid duplication, support subclassing, hide sequences, hide pointer operations, improve portability, simplify complicated boolean tests, improve performance, isolate complexity, hide implementation details, limit effects of changes, hide global data, make central points of control, facilitate reusable code, and accomplish a specific refactoring – RubberDuck May 27 '15 at 23:53
• 2nd Edition. Chapter 7.1 page 167 – RubberDuck May 27 '15 at 23:54
• Alright, reread your answer and realized what you meant by your first paragraph; I was misunderstanding it greatly. I'm giving it... a bit of an overhaul now, using classes as data storage units. – Fund Monica's Lawsuit Jun 12 '15 at 23:22 | 2021-04-21 02:47:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3207566738128662, "perplexity": 4099.691786736449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00468.warc.gz"} |
https://techprpr.com/math-genius-fainftyrightarrowmathbbr-is-differentiable-function-i-need-explicit-proof-of-a-problem-i-find-obvious-2/ | # Math Genius: \$f:(a,+infty)rightarrowmathbb{R} \$ is differentiable function. I need explicit proof of a problem I find obvious
If $$f'(x)>c, forall xin(a,+infty)$$ where $$c>0$$. Prove that $$lim_{xto+infty} f(x) = +infty$$. I would say that this is trivial, how could we prove this explicitly?
$$f(x) = f(a) + int_a^x f'(t) , dt ge f(a) + c(x-a)$$ | 2020-08-15 01:29:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9157751202583313, "perplexity": 480.2454878402671}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00423.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/amc.2018033 | # American Institute of Mathematical Sciences
August 2018, 12(3): 553-577. doi: 10.3934/amc.2018033
## A first step towards the skew duadic codes
Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France
Received August 2017 Revised February 2018 Published July 2018
Fund Project: The author is supported by the French government Investissements d’Avenir program ANR-11-LABX-0020-01.
This text gives a first definition of the $θ$-duadic codes where $θ$ is an automorphism of $\mathbb{F}_q$. A link with the self-orthogonal $θ$-cyclic codes is established. A construction and an enumeration are provided when $q$ is the square of a prime number $p$. In addition, new self-dual binary codes $[72, 36, 12]$ are obtained from extended $θ$-duadic codes defined on $\mathbb{F}_4$.
Citation: Delphine Boucher. A first step towards the skew duadic codes. Advances in Mathematics of Communications, 2018, 12 (3) : 553-577. doi: 10.3934/amc.2018033
##### References:
[1] D. Boucher, W. Geiselmann and F. Ulmer, Skew-cyclic codes, Appl. Algebra Engin. Commun. Comp., 18 (2007), 379-389. doi: 10.1007/s00200-007-0043-z. Google Scholar [2] D. Boucher and F. Ulmer, Coding with skew polynomial rings, J. Symb. Comp., 44 (2009), 1644-1656. doi: 10.1016/j.jsc.2007.11.008. Google Scholar [3] D. Boucher and F. Ulmer, A note on the dual codes of module skew codes, Cryptography and coding, Lecture Notes in Comput. Sci., 7089 (2011), 230-243. doi: 10.1007/978-3-642-25516-8_14. Google Scholar [4] D. Boucher and F. Ulmer, Self-dual skew codes and factorization of skew polynomials, J. Symb. Comp., 60 (2014), 47-61. doi: 10.1016/j.jsc.2013.10.003. Google Scholar [5] D. Boucher, Construction and number of self-dual skew codes over $\mathbb{F}_{p^2}$, Adv. Math. Commun., 10 (2016), 765-795. doi: 10.3934/amc.2016040. Google Scholar [6] X. Caruso and J. Le Borgne, A new faster algorithm for factoring skew polynomials over finite fields, J. Symb. Comp., 79 (2017), 411-443. doi: 10.1016/j.jsc.2016.02.016. Google Scholar [7] S. T. Dougherty, T. A. Gulliver and H. Masaaki, Extremal binary self-dual codes, IEEE Trans. Inform. Theory, 43 (1997), 2036-2047. Google Scholar [8] M. Giesbrecht, Factoring in skew-polynomial rings over finite fields, J. Symb. Comput., 26 (1998), 463-486. doi: 10.1006/jsco.1998.0224. Google Scholar [9] W. C. Huffman and V. Pless, Fundamentals of Error-Correcting Codes, Cambridge University Press, Cambridge, 2003. doi: 10.1017/CBO9780511807077. Google Scholar [10] N. Jacobson, The Theory of Rings, Amer. Math. Soc., 1943. Google Scholar [11] A. Kaya, B. Yildiz and I. Siap, New extremal binary self-dual codes of length 68 from quadratic residue codes over ${\mathbb{F}}_2+u{\mathbb{F}}_2+u^2{\mathbb{F}}_2$, Finite Fields and their Applications, 29 (2014), 160-177. doi: 10.1016/j.ffa.2014.04.009. Google Scholar [12] R. W. K. Odoni, On additive polynomials over a finite field, Proc. Edinburgh Math. Soc., 42 (1999), 1-16. doi: 10.1017/S0013091500019970. Google Scholar [13] O. Ore, Theory of Non-Commutative Polynomials, Ann. Math., 34 (1933), 480-508. doi: 10.2307/1968173. Google Scholar [14] J. von zur Gathen and J. Gerhard, Modern Computer Algebra, Cambridge University Press, Cambridge, 2013. doi: 10.1017/CBO9781139856065. Google Scholar [15] A. Zhdanov, New self-dual codes of length $72$, preprint, arXiv: 1705.05779. Google Scholar
show all references
##### References:
[1] D. Boucher, W. Geiselmann and F. Ulmer, Skew-cyclic codes, Appl. Algebra Engin. Commun. Comp., 18 (2007), 379-389. doi: 10.1007/s00200-007-0043-z. Google Scholar [2] D. Boucher and F. Ulmer, Coding with skew polynomial rings, J. Symb. Comp., 44 (2009), 1644-1656. doi: 10.1016/j.jsc.2007.11.008. Google Scholar [3] D. Boucher and F. Ulmer, A note on the dual codes of module skew codes, Cryptography and coding, Lecture Notes in Comput. Sci., 7089 (2011), 230-243. doi: 10.1007/978-3-642-25516-8_14. Google Scholar [4] D. Boucher and F. Ulmer, Self-dual skew codes and factorization of skew polynomials, J. Symb. Comp., 60 (2014), 47-61. doi: 10.1016/j.jsc.2013.10.003. Google Scholar [5] D. Boucher, Construction and number of self-dual skew codes over $\mathbb{F}_{p^2}$, Adv. Math. Commun., 10 (2016), 765-795. doi: 10.3934/amc.2016040. Google Scholar [6] X. Caruso and J. Le Borgne, A new faster algorithm for factoring skew polynomials over finite fields, J. Symb. Comp., 79 (2017), 411-443. doi: 10.1016/j.jsc.2016.02.016. Google Scholar [7] S. T. Dougherty, T. A. Gulliver and H. Masaaki, Extremal binary self-dual codes, IEEE Trans. Inform. Theory, 43 (1997), 2036-2047. Google Scholar [8] M. Giesbrecht, Factoring in skew-polynomial rings over finite fields, J. Symb. Comput., 26 (1998), 463-486. doi: 10.1006/jsco.1998.0224. Google Scholar [9] W. C. Huffman and V. Pless, Fundamentals of Error-Correcting Codes, Cambridge University Press, Cambridge, 2003. doi: 10.1017/CBO9780511807077. Google Scholar [10] N. Jacobson, The Theory of Rings, Amer. Math. Soc., 1943. Google Scholar [11] A. Kaya, B. Yildiz and I. Siap, New extremal binary self-dual codes of length 68 from quadratic residue codes over ${\mathbb{F}}_2+u{\mathbb{F}}_2+u^2{\mathbb{F}}_2$, Finite Fields and their Applications, 29 (2014), 160-177. doi: 10.1016/j.ffa.2014.04.009. Google Scholar [12] R. W. K. Odoni, On additive polynomials over a finite field, Proc. Edinburgh Math. Soc., 42 (1999), 1-16. doi: 10.1017/S0013091500019970. Google Scholar [13] O. Ore, Theory of Non-Commutative Polynomials, Ann. Math., 34 (1933), 480-508. doi: 10.2307/1968173. Google Scholar [14] J. von zur Gathen and J. Gerhard, Modern Computer Algebra, Cambridge University Press, Cambridge, 2013. doi: 10.1017/CBO9781139856065. Google Scholar [15] A. Zhdanov, New self-dual codes of length $72$, preprint, arXiv: 1705.05779. Google Scholar
Type Ⅱ $[72, 36, 12]$ self-dual codes who are binary images of $[36, 18]_4$ self-dual extended $\theta$-cyclic codes
Coefficients of $g$ $v$ $\alpha$ $\left[ a, a, 0, a, a^2, a, a, 0, 0, 0, 1, 1, a^2, 1, 0, 1, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3072 $\left[ a, a^2, 0, a^2, a^2, a^2, 0, a^2, 0, a^2, 0, a^2, a^2, a^2, 0, a^2, 1 \right]$ $\left[ 1, a, 1, a^2 \right]$ -3276 $\left[ a^2, 1, a^2, a^2, 1, a^2, 0, 1, 0, a^2, 0, 1, a^2, 1, 1, a^2, 1 \right]$ $\left[ 1, a^2, 1, a \right]$ -3480 $\left[ a, 1, a, 1, 0, 0, a, a^2, 0, a^2, 1, 0, 0, a, 1, a, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3582 $\left[ a, 0, a^2, 0, 0, 1, 1, a^2, 0, a^2, a, a, 0, 0, a^2, 0, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3684 $\left[ a^2, 1, 0, a, 0, 1, a^2, a, a, a, 1, a^2, 0, a, 0, a^2, 1 \right]$ $\left[ 1, a^2, 1, a \right]$ -3990 $\left[ a, a^2, a, 0, 0, 1, a, 1, 0, a, 1, a, 0, 0, 1, a^2, 1 \right]$ $\left[1, a, 1, a^2\right]$ -4092
Coefficients of $g$ $v$ $\alpha$ $\left[ a, a, 0, a, a^2, a, a, 0, 0, 0, 1, 1, a^2, 1, 0, 1, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3072 $\left[ a, a^2, 0, a^2, a^2, a^2, 0, a^2, 0, a^2, 0, a^2, a^2, a^2, 0, a^2, 1 \right]$ $\left[ 1, a, 1, a^2 \right]$ -3276 $\left[ a^2, 1, a^2, a^2, 1, a^2, 0, 1, 0, a^2, 0, 1, a^2, 1, 1, a^2, 1 \right]$ $\left[ 1, a^2, 1, a \right]$ -3480 $\left[ a, 1, a, 1, 0, 0, a, a^2, 0, a^2, 1, 0, 0, a, 1, a, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3582 $\left[ a, 0, a^2, 0, 0, 1, 1, a^2, 0, a^2, a, a, 0, 0, a^2, 0, 1 \right]$ $\left[1, a, 1, a^2\right]$ -3684 $\left[ a^2, 1, 0, a, 0, 1, a^2, a, a, a, 1, a^2, 0, a, 0, a^2, 1 \right]$ $\left[ 1, a^2, 1, a \right]$ -3990 $\left[ a, a^2, a, 0, 0, 1, a, 1, 0, a, 1, a, 0, 0, 1, a^2, 1 \right]$ $\left[1, a, 1, a^2\right]$ -4092
Irreducible skew polynomials of $\mathbb{F}_{p^2}[X; \theta]$ with a given bound
Require: $f \in \mathbb{F}_p[X^2]$ irreducible in $\mathbb{F}_p[X^2]$ Ensure: All irreducible skew polynomials with bound $f(X^2)$ 1. $d \leftarrow \deg_{X^2} f(X^2)$ 2. $\alpha \leftarrow$ root of $f$ in $\overline{\mathbb{F}_p}$ 3. if $d$ is odd then 4. $\delta \leftarrow (d-1)/2$ 5. $E \leftarrow \emptyset$ 6. for $u \in \mathbb{F}_{p^{2d}}$ such that $u^{p^d+1}=\alpha$ do 7. $P \leftarrow$ Interpolation Polynomial in $\mathbb{F}_{p^2}[Z]$ at the $d$ points $[\alpha^{p^{2i}}, u^{p^{2i}}]_{0 \leq i \leq \delta}$ and $[\alpha^{p^{2i+1}}, \alpha^{p^{2i+1}}/u^{p^{2i+1}}]_{0 \leq i \leq \delta-1}$ 8. $(A, B) \leftarrow$ solution of the Cauchy interpolation problem ${\displaystyle \frac{A}{B} \equiv P \pmod{f}}$ with $B$ monic, $\deg(B)=\delta$, $\deg(A) \leq \delta$ 9. add $A(X^2) + X \cdot B(X^2)$ to the set $E$ 10. endfor 11. else 12. $\delta \leftarrow d/2$ 13. $E \leftarrow \{\tilde{f}(X^2), \Theta(\tilde{f})(X^2)\}$ where $\tilde{f}(Z) \Theta(\tilde{f})(Z)=f(Z)$ is the factorization of $f(Z)$ in $\mathbb{F}_{p^2}[Z]$ 14. for $u \in \mathbb{F}_{p^{d}}$ such that $u \neq 0$ do 15. $P \leftarrow$ Interpolation Polynomial in $\mathbb{F}_{p^2}[Z]$ at the $d$ points $[\alpha^{p^{2i}}, u^{p^{2i}}]_{0 \leq i \leq \delta-1}$ and $[\alpha^{p^{2i+1}}, \alpha^{p^{2i+1}}/u^{p^{2i+1}}]_{0 \leq i \leq \delta-1}$ 16. $(A, B) \leftarrow$ solution of the Cauchy interpolation problem ${\displaystyle \frac{A}{B} \equiv P \pmod{f}}$ with $A$ monic, $\deg(A)=\delta$, $\deg(B)<\delta$ 17. add $A(X^2) + X \cdot B(X^2)$ to the set $E$ 18. endfor 19. endif 20. return $E$
Require: $f \in \mathbb{F}_p[X^2]$ irreducible in $\mathbb{F}_p[X^2]$ Ensure: All irreducible skew polynomials with bound $f(X^2)$ 1. $d \leftarrow \deg_{X^2} f(X^2)$ 2. $\alpha \leftarrow$ root of $f$ in $\overline{\mathbb{F}_p}$ 3. if $d$ is odd then 4. $\delta \leftarrow (d-1)/2$ 5. $E \leftarrow \emptyset$ 6. for $u \in \mathbb{F}_{p^{2d}}$ such that $u^{p^d+1}=\alpha$ do 7. $P \leftarrow$ Interpolation Polynomial in $\mathbb{F}_{p^2}[Z]$ at the $d$ points $[\alpha^{p^{2i}}, u^{p^{2i}}]_{0 \leq i \leq \delta}$ and $[\alpha^{p^{2i+1}}, \alpha^{p^{2i+1}}/u^{p^{2i+1}}]_{0 \leq i \leq \delta-1}$ 8. $(A, B) \leftarrow$ solution of the Cauchy interpolation problem ${\displaystyle \frac{A}{B} \equiv P \pmod{f}}$ with $B$ monic, $\deg(B)=\delta$, $\deg(A) \leq \delta$ 9. add $A(X^2) + X \cdot B(X^2)$ to the set $E$ 10. endfor 11. else 12. $\delta \leftarrow d/2$ 13. $E \leftarrow \{\tilde{f}(X^2), \Theta(\tilde{f})(X^2)\}$ where $\tilde{f}(Z) \Theta(\tilde{f})(Z)=f(Z)$ is the factorization of $f(Z)$ in $\mathbb{F}_{p^2}[Z]$ 14. for $u \in \mathbb{F}_{p^{d}}$ such that $u \neq 0$ do 15. $P \leftarrow$ Interpolation Polynomial in $\mathbb{F}_{p^2}[Z]$ at the $d$ points $[\alpha^{p^{2i}}, u^{p^{2i}}]_{0 \leq i \leq \delta-1}$ and $[\alpha^{p^{2i+1}}, \alpha^{p^{2i+1}}/u^{p^{2i+1}}]_{0 \leq i \leq \delta-1}$ 16. $(A, B) \leftarrow$ solution of the Cauchy interpolation problem ${\displaystyle \frac{A}{B} \equiv P \pmod{f}}$ with $A$ monic, $\deg(A)=\delta$, $\deg(B)<\delta$ 17. add $A(X^2) + X \cdot B(X^2)$ to the set $E$ 18. endfor 19. endif 20. return $E$
Parametrization of the irreducible monic skew polynomials of $\mathbb{F}_4[X;\theta]$ bounded by $X^6+X^2+1$.
$u$ $P(Z) \in \mathbb{F}_4[Z]$ $h(X) \in \mathbb{F}_4[X;\theta]$ $\gamma$ $Z^2 + a^2 \, Z + 1$ $X^3 + a \, X^2 + a \, X + a$ $\gamma^8$ $Z^2 + a \, Z + 1$ $X^3 + a^2 \, X^2 + a^2 \, X + a^2$ $\gamma^{15}$ $a^2 \, Z^2 + a^2 \, Z$ $X^3 + X + a^2$ $\gamma^{22}$ $a \, Z^2 + Z + a$ $X^3 + a^2 \, X^2 + a \, X + a^2$ $\gamma^{29}$ $a \, Z^2 + a^2 \, Z + a$ $X^3 + X^2 + a^2 \, X + 1$ $\gamma^{36}$ $Z^2 + Z$ $X^3 + X + 1$ $\gamma^{43}$ $a^2 \, Z^2 + a \, Z + a^2$ $X^3 + X^2 + a \, X + 1$ $\gamma^{50}$ $a^2 \, Z^2 + Z + a^2$ $X^3 + a \, X^2 + a^2 \, X + a$ $\gamma^{57}$ $a \, Z^2 + a \, Z$ $X^3 + X + a$
$u$ $P(Z) \in \mathbb{F}_4[Z]$ $h(X) \in \mathbb{F}_4[X;\theta]$ $\gamma$ $Z^2 + a^2 \, Z + 1$ $X^3 + a \, X^2 + a \, X + a$ $\gamma^8$ $Z^2 + a \, Z + 1$ $X^3 + a^2 \, X^2 + a^2 \, X + a^2$ $\gamma^{15}$ $a^2 \, Z^2 + a^2 \, Z$ $X^3 + X + a^2$ $\gamma^{22}$ $a \, Z^2 + Z + a$ $X^3 + a^2 \, X^2 + a \, X + a^2$ $\gamma^{29}$ $a \, Z^2 + a^2 \, Z + a$ $X^3 + X^2 + a^2 \, X + 1$ $\gamma^{36}$ $Z^2 + Z$ $X^3 + X + 1$ $\gamma^{43}$ $a^2 \, Z^2 + a \, Z + a^2$ $X^3 + X^2 + a \, X + 1$ $\gamma^{50}$ $a^2 \, Z^2 + Z + a^2$ $X^3 + a \, X^2 + a^2 \, X + a$ $\gamma^{57}$ $a \, Z^2 + a \, Z$ $X^3 + X + a$
Parametrization of the irreducible monic skew polynomials of $\mathbb{F}_4[X;\theta]$ bounded by $X^8+X^2+1$ and distinct of $X^4+X^2+a$ and $X^4+X^2+a^2$.
$u$ $P(Z) \in \mathbb{F}_4[Z]$ $h(X) \in \mathbb{F}_4[X;\theta]$ $1$ $Z^3 + a^2 \, Z + a^2$ $X^4 + a^2 \, X^3 + a^2 \, X^2 + a$ $\alpha$ $Z^3 + a \, Z + a$ $X^4 + a \, X^3 + a \, X^2 + a^2$ $\alpha^2$ $a^2 \, Z^3 + a^2 \, Z^2 + a$ $X^4 + a^2 \, X^3 + a \, X^2 + a^2 \, X + a$ $\alpha^3$ $a^2 \, Z^2 + a^2$ $X^4 + a^2 \, X + 1$ $\alpha^4$ $a^2 \, Z^3 + a^2 \, Z^2 + 1$ $X^4 + a^2 \, X^3 + a^2 \, X^2 + a^2 \, X + a^2$ $\alpha^5$ $a \, Z^3 + Z + 1$ $X^4 + X^3 + a^2 \, X^2 + a$ $\alpha^6$ $a \, Z^3 + a^2 \, Z + a^2$ $X^4 + a^2 \, X^3 + a \, X^2 + a^2$ $\alpha^7$ $Z^3 + Z^2 + a^2$ $X^4 + X^3 + a \, X^2 + X + a$ $\alpha^8$ $Z^2 + 1$ $X^4 + X + 1$ $\alpha^9$ $Z^3 + Z^2 + a$ $X^4 + X^3 + a^2 \, X^2 + X + a^2$ $\alpha^{10}$ $a^2 \, Z^3 + a \, Z + a$ $X^4 + a \, X^3 + a^2 \, X^2 + a$ $\alpha^{11}$ $a^2 \, Z^3 + Z + 1$ $X^4 + X^3 + a \, X^2 + a^2$ $\alpha^{12}$ $a \, Z^3 + a \, Z^2 + 1$ $X^4 + a \, X^3 + a \, X^2 + a \, X + a$ $\alpha^{13}$ $a \, Z^2 + a$ $X^4 + a \, X + 1$ $\alpha^{14}$ $a \, Z^3 + a \, Z^2 + a^2$ $X^4 + a \, X^3 + a^2 \, X^2 + a \, X + a^2$
$u$ $P(Z) \in \mathbb{F}_4[Z]$ $h(X) \in \mathbb{F}_4[X;\theta]$ $1$ $Z^3 + a^2 \, Z + a^2$ $X^4 + a^2 \, X^3 + a^2 \, X^2 + a$ $\alpha$ $Z^3 + a \, Z + a$ $X^4 + a \, X^3 + a \, X^2 + a^2$ $\alpha^2$ $a^2 \, Z^3 + a^2 \, Z^2 + a$ $X^4 + a^2 \, X^3 + a \, X^2 + a^2 \, X + a$ $\alpha^3$ $a^2 \, Z^2 + a^2$ $X^4 + a^2 \, X + 1$ $\alpha^4$ $a^2 \, Z^3 + a^2 \, Z^2 + 1$ $X^4 + a^2 \, X^3 + a^2 \, X^2 + a^2 \, X + a^2$ $\alpha^5$ $a \, Z^3 + Z + 1$ $X^4 + X^3 + a^2 \, X^2 + a$ $\alpha^6$ $a \, Z^3 + a^2 \, Z + a^2$ $X^4 + a^2 \, X^3 + a \, X^2 + a^2$ $\alpha^7$ $Z^3 + Z^2 + a^2$ $X^4 + X^3 + a \, X^2 + X + a$ $\alpha^8$ $Z^2 + 1$ $X^4 + X + 1$ $\alpha^9$ $Z^3 + Z^2 + a$ $X^4 + X^3 + a^2 \, X^2 + X + a^2$ $\alpha^{10}$ $a^2 \, Z^3 + a \, Z + a$ $X^4 + a \, X^3 + a^2 \, X^2 + a$ $\alpha^{11}$ $a^2 \, Z^3 + Z + 1$ $X^4 + X^3 + a \, X^2 + a^2$ $\alpha^{12}$ $a \, Z^3 + a \, Z^2 + 1$ $X^4 + a \, X^3 + a \, X^2 + a \, X + a$ $\alpha^{13}$ $a \, Z^2 + a$ $X^4 + a \, X + 1$ $\alpha^{14}$ $a \, Z^3 + a \, Z^2 + a^2$ $X^4 + a \, X^3 + a^2 \, X^2 + a \, X + a^2$
Type Ⅱ $[72, 36, 12]$ self-dual codes who are binary images of $[36, 18]_4$ self-dual $\theta$-cyclic codes
Coefficients of $g$ $\alpha$ $\left[ a^2, 0, a^2, a^2, 1, 1, a^2, 1, a^2, 0, 1, a^2, 1, a^2, a^2, 1, 1, 0, 1 \right]$ -2820 $\left[ 1, a, a, a, a^2, a^2, a, 0, 0, 0, 0, 0, a^2, a, a, a^2, a^2, a^2, 1 \right]$ -3204 $\left[ 1, a^2, a^2, 1, 1, a^2, 1, a, 0, 0, 0, a^2, 1, a, 1, 1, a, a, 1 \right]$ -3276 $\left[ a^2, 1, 1, 0, a, 1, 1, a^2, 0, 0, 0, 1, a^2, a^2, a, 0, a^2, a^2, 1 \right]$ -3312 $\left[ a^2, a^2, 1, a, a^2, 0, 0, 0, a, 0, a, 0, 0, 0, 1, a, a^2, 1, 1 \right]$ -3336 $\left[ a^2, a, a, 0, 1, a, a, a^2, a^2, 0, 1, 1, a, a, a^2, 0, a, a, 1 \right]$ -3372 $\left[ a^2, 0, 0, a^2, 0, a, a, a^2, a, a, a, 1, a, a, 0, 1, 0, 0, 1 \right]$ -3408 $\left[ 1, 1, a, a^2, 1, 1, 1, a, 0, 1, 0, a^2, 1, 1, 1, a, a^2, 1, 1 \right]$ -3420 $\left[ a, a, 1, a, a^2, a^2, 1, a^2, a^2, a^2, a^2, a^2, a, a^2, a^2, 1, a, 1, 1 \right]$ -3456 $\left[ a, a^2, 1, a, 0, a, 0, a^2, a, a^2, 1, a^2, 0, 1, 0, 1, a, a^2, 1 \right]$ -3504 $\left[ 1, a, a^2, a, a^2, 1, 1, a, a, 0, a^2, a^2, 1, 1, a, a^2, a, a^2, 1 \right]$ -3540 $\left[ a, 1, a^2, a^2, a, 0, a, 0, 0, 0, 0, 0, 1, 0, 1, a^2, a^2, a, 1 \right]$ -3564 $\left[ 1, 0, 0, a, 1, 1, a^2, a, 0, 1, 0, a^2, a, 1, 1, a^2, 0, 0, 1 \right]$ -3576 $\left[ 1, 1, a^2, a^2, 1, a^2, a, a^2, a, 0, a^2, a, a^2, a, 1, a, a, 1, 1 \right]$ -3600 $\left[ 1, 0, 0, 0, 1, 1, 1, 0, a, 1, a^2, 0, 1, 1, 1, 0, 0, 0, 1 \right]$ -3612 $\left[ 1, 0, 0, 0, 1, a^2, 0, 1, a^2, 0, a, 1, 0, a, 1, 0, 0, 0, 1 \right]$ -3636 $\left[ a, a^2, a^2, a^2, 1, 1, a^2, 0, a, 0, 1, 0, a^2, a, a, a^2, a^2, a^2, 1 \right]$ -3660 $\left[ 1, 0, 0, a, 0, a, a, 1, 1, 1, 1, 1, a^2, a^2, 0, a^2, 0, 0, 1 \right]$ -3696 $\left[ a, 0, 0, a, a, 1, a^2, a^2, a, a^2, 1, a^2, a^2, a, 1, 1, 0, 0, 1 \right]$ -3732 $\left[ a, 0, a, a, 1, a^2, 0, a^2, 0, 0, 0, a^2, 0, a^2, a, 1, 1, 0, 1 \right]$ -3744 $\left[ a^2, a, 1, 1, a^2, a^2, 1, 0, a^2, a, 1, 0, a^2, 1, 1, a^2, a^2, a, 1 \right]$ -3768 $\left[ 1, 1, a^2, 0, a, 0, a, 1, 0, 1, 0, 1, a^2, 0, a^2, 0, a, 1, 1 \right]$ -3816 $\left[ 1, a^2, a^2, a, 0, a^2, a, a, a, 1, a^2, a^2, a^2, a, 0, a^2, a, a, 1 \right]$ -3828 $\left[ 1, a, a, 1, 0, a^2, 0, a^2, 0, 0, 0, a, 0, a, 0, 1, a^2, a^2, 1 \right]$ -3924
Coefficients of $g$ $\alpha$ $\left[ a^2, 0, a^2, a^2, 1, 1, a^2, 1, a^2, 0, 1, a^2, 1, a^2, a^2, 1, 1, 0, 1 \right]$ -2820 $\left[ 1, a, a, a, a^2, a^2, a, 0, 0, 0, 0, 0, a^2, a, a, a^2, a^2, a^2, 1 \right]$ -3204 $\left[ 1, a^2, a^2, 1, 1, a^2, 1, a, 0, 0, 0, a^2, 1, a, 1, 1, a, a, 1 \right]$ -3276 $\left[ a^2, 1, 1, 0, a, 1, 1, a^2, 0, 0, 0, 1, a^2, a^2, a, 0, a^2, a^2, 1 \right]$ -3312 $\left[ a^2, a^2, 1, a, a^2, 0, 0, 0, a, 0, a, 0, 0, 0, 1, a, a^2, 1, 1 \right]$ -3336 $\left[ a^2, a, a, 0, 1, a, a, a^2, a^2, 0, 1, 1, a, a, a^2, 0, a, a, 1 \right]$ -3372 $\left[ a^2, 0, 0, a^2, 0, a, a, a^2, a, a, a, 1, a, a, 0, 1, 0, 0, 1 \right]$ -3408 $\left[ 1, 1, a, a^2, 1, 1, 1, a, 0, 1, 0, a^2, 1, 1, 1, a, a^2, 1, 1 \right]$ -3420 $\left[ a, a, 1, a, a^2, a^2, 1, a^2, a^2, a^2, a^2, a^2, a, a^2, a^2, 1, a, 1, 1 \right]$ -3456 $\left[ a, a^2, 1, a, 0, a, 0, a^2, a, a^2, 1, a^2, 0, 1, 0, 1, a, a^2, 1 \right]$ -3504 $\left[ 1, a, a^2, a, a^2, 1, 1, a, a, 0, a^2, a^2, 1, 1, a, a^2, a, a^2, 1 \right]$ -3540 $\left[ a, 1, a^2, a^2, a, 0, a, 0, 0, 0, 0, 0, 1, 0, 1, a^2, a^2, a, 1 \right]$ -3564 $\left[ 1, 0, 0, a, 1, 1, a^2, a, 0, 1, 0, a^2, a, 1, 1, a^2, 0, 0, 1 \right]$ -3576 $\left[ 1, 1, a^2, a^2, 1, a^2, a, a^2, a, 0, a^2, a, a^2, a, 1, a, a, 1, 1 \right]$ -3600 $\left[ 1, 0, 0, 0, 1, 1, 1, 0, a, 1, a^2, 0, 1, 1, 1, 0, 0, 0, 1 \right]$ -3612 $\left[ 1, 0, 0, 0, 1, a^2, 0, 1, a^2, 0, a, 1, 0, a, 1, 0, 0, 0, 1 \right]$ -3636 $\left[ a, a^2, a^2, a^2, 1, 1, a^2, 0, a, 0, 1, 0, a^2, a, a, a^2, a^2, a^2, 1 \right]$ -3660 $\left[ 1, 0, 0, a, 0, a, a, 1, 1, 1, 1, 1, a^2, a^2, 0, a^2, 0, 0, 1 \right]$ -3696 $\left[ a, 0, 0, a, a, 1, a^2, a^2, a, a^2, 1, a^2, a^2, a, 1, 1, 0, 0, 1 \right]$ -3732 $\left[ a, 0, a, a, 1, a^2, 0, a^2, 0, 0, 0, a^2, 0, a^2, a, 1, 1, 0, 1 \right]$ -3744 $\left[ a^2, a, 1, 1, a^2, a^2, 1, 0, a^2, a, 1, 0, a^2, 1, 1, a^2, a^2, a, 1 \right]$ -3768 $\left[ 1, 1, a^2, 0, a, 0, a, 1, 0, 1, 0, 1, a^2, 0, a^2, 0, a, 1, 1 \right]$ -3816 $\left[ 1, a^2, a^2, a, 0, a^2, a, a, a, 1, a^2, a^2, a^2, a, 0, a^2, a, a, 1 \right]$ -3828 $\left[ 1, a, a, 1, 0, a^2, 0, a^2, 0, 0, 0, a, 0, a, 0, 1, a^2, a^2, 1 \right]$ -3924
Type Ⅰ $[72, 36, 12]$ self-dual codes who are binary images of $[36, 18]_4$ self-dual $\theta$-cyclic codes.
Coefficients of $g$ $\beta$ $\gamma$ $\left[ a^2, a, 1, 1, a^2, a^2, 1, 1, a, a, a, a^2, a^2, 1, 1, a^2, a^2, a, 1 \right]$ 201 0 $\left[ 1, 0, 0, a, 0, a^2, a, 0, a, 0, a^2, 0, a^2, a, 0, a^2, 0, 0, 1 \right]$ 237 0 $\left[ 1, a^2, a^2, a^2, a, 1, a^2, a, a, 1, a^2, a^2, a, 1, a^2, a, a, a, 1 \right]$ 249 0 $\left[ 1, 0, 1, a, a, 0, 1, 1, a^2, 1, a, 1, 1, 0, a^2, a^2, 1, 0, 1 \right]$ 273 0 $\left[ a, 1, 1, 1, a^2, a, 1, a^2, 1, a^2, a, a^2, a, 1, a^2, a, a, a, 1 \right]$ 273 36 $\left[ a^2, a^2, 1, 0, 1, 0, 0, a^2, 1, 0, a^2, 1, 0, 0, a^2, 0, a^2, 1, 1 \right]$ 309 0 $\left[ 1, 1, a, 1, a^2, a^2, a^2, 0, a, 1, a^2, 0, a, a, a, 1, a^2, 1, 1 \right]$ 345 0 $\left[ a^2, a, 1, a^2, 0, 0, 1, 0, a^2, a, 1, 0, a^2, 0, 0, 1, a^2, a, 1 \right]$ 381 0 $\left[ 1, a, a, a^2, 0, a, a, 1, 1, 1, 1, 1, a^2, a^2, 0, a, a^2, a^2, 1 \right]$ 393 36 $\left[ a, a, a^2, a, 1, a, 0, a, a^2, 0, a^2, 1, 0, 1, a, 1, a^2, 1, 1 \right]$ 489 36
Coefficients of $g$ $\beta$ $\gamma$ $\left[ a^2, a, 1, 1, a^2, a^2, 1, 1, a, a, a, a^2, a^2, 1, 1, a^2, a^2, a, 1 \right]$ 201 0 $\left[ 1, 0, 0, a, 0, a^2, a, 0, a, 0, a^2, 0, a^2, a, 0, a^2, 0, 0, 1 \right]$ 237 0 $\left[ 1, a^2, a^2, a^2, a, 1, a^2, a, a, 1, a^2, a^2, a, 1, a^2, a, a, a, 1 \right]$ 249 0 $\left[ 1, 0, 1, a, a, 0, 1, 1, a^2, 1, a, 1, 1, 0, a^2, a^2, 1, 0, 1 \right]$ 273 0 $\left[ a, 1, 1, 1, a^2, a, 1, a^2, 1, a^2, a, a^2, a, 1, a^2, a, a, a, 1 \right]$ 273 36 $\left[ a^2, a^2, 1, 0, 1, 0, 0, a^2, 1, 0, a^2, 1, 0, 0, a^2, 0, a^2, 1, 1 \right]$ 309 0 $\left[ 1, 1, a, 1, a^2, a^2, a^2, 0, a, 1, a^2, 0, a, a, a, 1, a^2, 1, 1 \right]$ 345 0 $\left[ a^2, a, 1, a^2, 0, 0, 1, 0, a^2, a, 1, 0, a^2, 0, 0, 1, a^2, a, 1 \right]$ 381 0 $\left[ 1, a, a, a^2, 0, a, a, 1, 1, 1, 1, 1, a^2, a^2, 0, a, a^2, a^2, 1 \right]$ 393 36 $\left[ a, a, a^2, a, 1, a, 0, a, a^2, 0, a^2, 1, 0, 1, a, 1, a^2, 1, 1 \right]$ 489 36
Type Ⅰ $[72, 36, 12]$ self-dual codes who are binary images of $[36, 18]_4$ self-dual extended $\theta$-cyclic codes.
Coefficients of $g$ $v$ $\beta$ $\delta$ $\left[ 1, 1, 0, 0, a, 0, a, 1, 1, 1, a^2, 0, a^2, 0, 0, 1, 1 \right]$ $\left[0, 1, 1, 0\right]$ 221 0 $\left[ 1, a^2, 1, 1, a, a^2, a^2, a^2, 0, a, a, a, a^2, 1, 1, a, 1 \right]$ $\left[0, 1, 1, 0\right]$ 323 0 $\left[ a, 1, a, 1, 0, 0, a, a^2, 0, a^2, 1, 0, 0, a, 1, a, 1 \right]$ $\left[0, a^2, a, 0\right]$ 238 0 $\left[ a, a, 0, a, a^2, a, a, 0, 0, 0, 1, 1, a^2, 1, 0, 1, 1 \right]$ $\left[0, a^2, a, 0\right]$ 391 0 $\left[ a, a, 0, 1, 0, 0, a, 0, a^2, 0, 1, 0, 0, a, 0, 1, 1 \right]$ $\left[0, a^2, a, 0\right]$ 289 0 $\left[ a^2, 1, 0, a, 0, 1, a^2, a, a, a, 1, a^2, 0, a, 0, a^2, 1 \right]$ $\left[0, a, a^2, 0\right]$ 102 0 $\left[ a, 0, 1, a^2, 0, a, 0, a^2, 0, a^2, 0, 1, 0, a^2, a, 0, 1 \right]$ $\left[0, a^2, a, 0\right]$ 255 0 $\left[ a, a^2, a, 0, 0, 1, a, 1, 0, a, 1, a, 0, 0, 1, a^2, 1 \right]$ $\left[0, a^2, a, 0\right]$ 153 0
Coefficients of $g$ $v$ $\beta$ $\delta$ $\left[ 1, 1, 0, 0, a, 0, a, 1, 1, 1, a^2, 0, a^2, 0, 0, 1, 1 \right]$ $\left[0, 1, 1, 0\right]$ 221 0 $\left[ 1, a^2, 1, 1, a, a^2, a^2, a^2, 0, a, a, a, a^2, 1, 1, a, 1 \right]$ $\left[0, 1, 1, 0\right]$ 323 0 $\left[ a, 1, a, 1, 0, 0, a, a^2, 0, a^2, 1, 0, 0, a, 1, a, 1 \right]$ $\left[0, a^2, a, 0\right]$ 238 0 $\left[ a, a, 0, a, a^2, a, a, 0, 0, 0, 1, 1, a^2, 1, 0, 1, 1 \right]$ $\left[0, a^2, a, 0\right]$ 391 0 $\left[ a, a, 0, 1, 0, 0, a, 0, a^2, 0, 1, 0, 0, a, 0, 1, 1 \right]$ $\left[0, a^2, a, 0\right]$ 289 0 $\left[ a^2, 1, 0, a, 0, 1, a^2, a, a, a, 1, a^2, 0, a, 0, a^2, 1 \right]$ $\left[0, a, a^2, 0\right]$ 102 0 $\left[ a, 0, 1, a^2, 0, a, 0, a^2, 0, a^2, 0, 1, 0, a^2, a, 0, 1 \right]$ $\left[0, a^2, a, 0\right]$ 255 0 $\left[ a, a^2, a, 0, 0, 1, a, 1, 0, a, 1, a, 0, 0, 1, a^2, 1 \right]$ $\left[0, a^2, a, 0\right]$ 153 0
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http://ecoechoinvasives.blogspot.com/2017/ | ## Peanut Gallery Independent Praise of Drew Hempel's research
You have the best mental focus of any student I've met. Just don't let it go to your head! Qigong Master Jim Nance 2015
Your mind is very powerful from years of meditation. Just focus on the lower tan t'ien. Qigong Master Chunyi Lin 2015
My second source for the superiority of full lotus is the teaching of qigong master Chunyi Lin, founder of Spring Forest Qigong. According to qigong researcher (and all around wild and crazy guy) Drew Hempel, Chunyi Lin said that 20 minutes in full lotus is worth 4 hours of any other meditation practice. That sounds like an exaggeration, but Drew has had some pretty far out results from his practice. Chris, Tai Chi in San Diego, 2011
2/2/11: Long time Steamshovel correpsondent Drew Hemphill has had his research cited extensively in a new article by Dan Eden (Gary Vey) who had exposed James Hurtak's ties to Project HAARP and the NSA: Kenn Thomas
I think Drew Hempel offers us some clues about chirality and harmonics.
1) There is a self-sustaining cosmic resonance (hidden coherence) - what he calls alchemical resonance?
2) Natural resonance cycles change over time (based on precession of the equinox?) where there is a switch from carbon to silicone based life?
3) The resonance change between Saturn and Jupiter will have a major effect upon Earth?
4) Aligning the ions of plasma with the ions of matter creates complimentary opposites of energy and mass?
5) Positronium (electron-position coupling) creates mirror matter that passes through black holes?
The Secret of Pre-Established Harmony and the Image of Time
gritzle70 on The Next Level forum in 2008
According to the polymath and University of Minnesota alum Drew Hempel, the vagus nerve climax can be achieved by male practitioners of meditation and Qigong. According to Hempel, these practices can ionize neurotransmitters in the gut — our body's largest repository of serotonin — which can then travel through the vagus nerve and cross the normally prudish blood-brain barrier, thus giving neck-pulsing, orgasmic-like sensations written about in Daoist texts. Qigong and meditation, Hempel said, are the opposite of dopamine obsession: they focus on present awareness and a lack of anticipation for future events.
David Logan, Op-ed, Daily Nebraskan, 2011
Drew Hempel has caught my interest for some time as many of his articles on the ‘Natural Resonance Revolution’, so closely matches much of our groups ( and my own ) experiences and thoughts on the nature of the phenomena we’ve been exposed to via the frequencies I’ve created for our use. A lot of what Drew speaks about is congruent to where we are at and I think it really serves all of us to reflect upon his writings especially in relation to what we have been journeying through in the real world.
Donald Adams, aka Dr. Starz, computer frequency "sonification" healer
The final phase though of the Exopolitical NWO exoreligion roll out is the Borg-like assimilation philosophies of the Transhuman Movement
and their matrix upload plans so well outlined by Drew Hempel.
UFO Culture blog 2009
Drew Hempel went that far. He's a scientist who has plumbed the depths of these training techniques.
He kind of takes...at one point he called it Pre-Socratic African Taoist Yoga training. Which is I guess a pretty bare one-liner about his research but you know qigong is something that will become more culturally significant in the way that Tai Chi is now.Philosophy professor Dave Leech podcast discussion with David Metcalfe and Dr. Tim Brigham
I will tell you how to do this. I followed the information which was posted in online forums and articles by someone called 'drew hempel' as he studied something called 'spring forest qigong' and 'taoist yoga'. I followed the same training he did and it's very powerful. He studied SFQ with qigong master chun yi lin (http://www.bornahealer.com/)...
- there is a lot of **** on the net because it seems like no one has a clue how to do real energy work, drew hempel was the only person I found who made any sense to me, so I followed what he did and it worked....
Dreambot, 2011 Spiritualforums post
Drew Hempel's third eye is connected like only a handful of people on this planet's are. He's able to see things in a way very few can. If you're interested in metaphysics and the such, he's worth a read.
"Riskfactor" posting on a forum in 2011.
The disgustingly typical common dismissal of yours of Drew's worth is telling. Can't handle anything too far beyond your comfort zone so you don't at all engage, as Drew would say, with his information. Much of which is simply discerning the fundamental structure linking revered traditional scriptures and cuttingedge science and it's explication in human life on Earth.
Then of course you project by claiming he is not engaging well... you're right, but you are still denying your hypocrisy, as you are too scared to admit that the reason you say you don't appreciate any of his writing is probably not because it is all false but because it is occult to your limited mind.
You can't argue with the facts and are too dumb to handle his points so you resort to smear tactics.
Who cares if some nut spams the net. It is only disruptive to the weak of mind. Fact is Drew has an "original" conclusion that deserves global discussion and you censor it like a welltrained shill, which I assume you are not.
Drew's idea. 2:3 is Yang, 3:4 is Yin... ie 666 is Yang. But he might repeat it like the example below, at which 99% of readers will throw their hands up in incomprehension and only folk like me, a stubborn OneTruthLover, bother to really get what he is saying. To say his is crazy is stupidly missing the point, plus to which he seems saner than most of the zombies who pretend humanity. He may a mixed up lake but at least he is not a puddle of puke.
Brace yourselves,
"I've posted that in fact musicologist Ernest McClain had already gleaned this from Plato -- see his Pythagorean Plato -- and that in fact while 5/4 was the Greek Miracle as the cube root of two leading to amplitude as classical mathematics for 3 dimensional form
the Tai Chi Perfect 5th-4th as 2:3 and 3:4 Yang and Yin (my discovery) is the QUANTUM secret of 2-dimensional energy intensity from frequency, not classical amplitude."
Needless to say he condenses much meaning in few words so you gotta read more his of his spiel to get it. His earliest piece on the greek Miracle is best. Cloud Tiger, a Tai Chi practitioner in the UK, 2009
Hempel, proficient in a wide sprectrum of intellectual pursuits including math and music --- and far from being deluded in any sense of the word as brought forth in the above quote by Hui Neng --- is also an eastern spiritual renaissance master as well. Just as much too, in the aforementioned areas and others far afield, as a search of his name will show, or did before he just simply disappeared into cyber space or infinity, he was one of the most prolific contributors on the net. His writings and books lean heavily toward a variety of eastern spiritual aspects, with a strong personal emphasis on qigong and meditation executed almost exclusively through the use of the full lotus position.
Additionally, even though Drew has resurfaced and available for interaction to all comers, I have however, NOT changed any of the above content of the original page because in my opinion it still remains highly relevant, showing all kinds of insight into who and where Hempel comes from, from any number of angles. Good to hear you are back. The internet wasn't the same without you.
The Wanderling, advaita meditation online site page on Drew Hempel
First we have to address the root of this possible problem.
the main root of all of this is the conceptual axiom that the parts of a system should be symmetrical if it desires to be in HARMONY. This in math is called COMMUTATIVE RING, or AxB=BxA.
From this principle (the commutative ring) we created a body of rules and paradigms in science, and even in religion, that separates our perception from nature itself (our game of denying being part of it) For example, as someone on this forum talked about maxwell´s quaternions in here. To work with an asymmetrical principle (most of things in nature) with our math that has as a main basic principle the commutative ring we first need to CHANGE what we gonna study, I mean, our science takes the information in nature which is assymétrical and DIVERGENT and converts it to an symmetrical and CONVERGENTE group of data to work with.
This is, as a principle, exactly what we did with the musical scales, as drew demonstrated.
In ancient music we got the asymmetrical intervals and, mainly since old non-pre-socratic Greece, we develop the obsession and the desire to “correct” it´s ugliness (natural asymmetry), so we transformed ancient music into symmetrical intervals. Buy by doing so we generate a comma, an extra group of data that can´t be used and should be discarded and watched over to avoid its return, because if this comma returns to the scale, it will create a hell of a mess to the fragile artificial symmetry we applied to it. So, conceptualy, the kind of technology we create, it prefers to change nature into artificial symmetry instead of flowing with it.
So it´s a matter of the right tool for the right job. And our Math has no asymmetrical tool for the job. IT´s all about the commutative ring, so even when working with asymmetrical systems (like nature) it has to convert it into symmetrical to process it. IT´s simple, drew is saying that western math and music take what is divergent (spiraling out) by nature and converge it before giving any kind of result. It´s a fact. Check, I don´t know, Wikipedia?, to get informations about the commutative ring and the non-commutative ring.
Now, I´m not 100% with drew on the consequences of “raping” nature with symmetry. I do know drew for the last 10yrs. Or so. I´m very good at math and good enough in music. I´m so-so in physics, i´ve studied electrical engineer for two yrs before change my mind and go to journalism school and drop it too to work as a photographer and illustrator and now as a computer graphics animator and advertising film director. So I have some background to understand a great chunk of drew hypothesis.
We try to solve all with symmetry, as the case with the music intervals, therefore, as in the case of music intervals, we create commas in science and all the range of activities that rely on science.
So this is the radical part of this use of musical model to explain things: if we try to understand nature (that is asymmetrical and divergent) with concepts and tools that are symmetrical and convergent) we gonna create dissonances in return and enter in a cycle of solving a dissonance with more symmetry to generate even more dissonance in the next octave ´till we # everything up! (we are already doing so! Our tech gives us marvelous stuff at the same time it acidify the oceans ending up #ing up with animals that depends on ultrasound for communication to reproduction. Our tech gives us marvelous tools as computer hi-res screens, but in return any cycle above 30hz is with time stressful to the human cells, etc, etc.)
non-commutative is so alien to our understand of thing that, as an example, if you go to Mr. Wolfram Math repository site you gonna find only ONE EMPTY page about it!!non-commutative is so alien to our understand of thing that, as an example, if you go to Mr. Wolfram Math repository site you gonna find only ONE EMPTY page about it!!
Now let´s go to the possible solutions for this.
1.full-lotus as a solution. Drew said full-lotus can transform the excess of electrochemical information in the lower body into electromagnectic information in the upper body. Some angry scientific trolls started to make fun of it.
for the materialistic people the thing is in the physiology.
siting in full-lotus will put a great chunk of pressure into your SACRUM! The sacrum is empty and pressure there will start to create small levels of ultra sound that will TRANSDUCE the energy from electrochemical into electromagnetic (remember, SOUND is PRESSURE WAVES, and it´s been prove that ultrasound can transduce energy from one state to other… use uncle google, he is your friend!! )… so this is one of the main reasons why full-lotus is a great tool.
It took me one year of daily chikung to sit in full-lotus with ease, before my one year of spring forest qigong I could barely sit in freestyle (like an north American indian) for more than a couple of minutes without felling a hell of pain in my back and legs!
After i´ve learned to sit in full-lotus, it´s a #ing bliss, full-lotus and half-lotus all day long. I even cut the legs from all my tables in my studio so I can sit in full-lotus and half-lotus anternated all day!!
So it´s one solution, but it´s a personal one.
2. Music create patterns in our brain.
It´s known that music create patterns in our brain. So if we just listen to music made with symmetrical and proportional intervals, some kind of patterns will form… with non-western music (mostly live, since hard to find speakers in home and computers that can produce ultrasound! ) so with non-western music we can generate other kind of patterns, since the intervals are asymmetrical. Drew have posted links to papers demonstrated, among otherthings, how ultrasound can stimulate the brain, the center of the brain, so on.
Also, Pineal Gland is a place made for sonoluminescence production. See? Pineal gland has fluid and cavitation in it! So with ultrasound stimulus in it we can ionize it a lot, we can produce tons of melatonie, and also we can generate little bubbles of light! By my personal experience, seeing light while practicing small universe (the oldest meditation that exist, it´s part of spring forest chikung and also is a “secret” inside the kriya yoga school) is relaxing, insightful and healing.
So, changing the king of music we listen to is a way of changing the patterns in our brain and so maybe it will reflect in the patterns we create in the world too.
3.creating a science where natural resonance ratios are important.
Well, this third item i´ll say nothing, since it´s what i´m working for the past 5 yrs. But a simple example of using natural resonance (non-comutative) musical ratios in technology is The Sonic Bloom, it´s a kit with audio from birds generating the fifth and forth in a repeated way so the intensity will affect the plants cells to expand more, taking more nutrients. It´s amazing. I use it to feed my home grown food (i´m a raw vegetarian) as my wife garden.
And this is just the tip of the iceberg…
Now those open minded, the daily practice of small universe from spring forest qigong, or the listening of music without symmetrical intervals, or the use of tech like the sonic bloom are just a few of the examples of the changes we can make personally or collectively to a better and more harmonious world.
And those so full of the “right” paradigm, well, you can keep projecting your anger on drew, it´s funny to read your assumptions about my crazy druid friend!
KALLISTI and have a nice day and a lovely life!
Guto Novo.
Correspondent for 15 years in Brazil
Seriously drew is next level, beyond even Dobbs, Ebert, Heidegger, Mcluhan etc.
Another UK practitioner-philosopher
"The universe and I came into being together; I and everything therein are One."
"If then all things are One, what room is there for speech? On the other hand, since I can say the word 'one' how can speech not exist? If it does exist, we have One and speech -- two; and two and one -- three(14) from which point onwards even the best mathematicians will fail to reach (the ultimate); how much more then should ordinary people fail?">"
- Chuang Tzu, 300 BCE
"Sound is the bridge between God and the soul" SHRI DHYANYOGI
a “universal scaling system”, ... this discrete scaling manifests itself in acoustic systems, as is well known in western classical music, where the two scalings correspond, respectively, to passing to the octave (frequency ratio of 2) and transposition (the perfect fifth is the frequency ratio 3/2), with the approximate value log(3)/ log(2) ∼ 19/12 responsible for the difference between the “circulating temperament” of the Well Tempered
Clavier and the “equal temperament” of XIX century music. It is precisely the irrationality of log(3)/ log(2) which is responsible for the noncommutative [complementary opposites as yin/yang] nature of the quotient corresponding to the three places {2, 3,∞}. -
Math professor (Fields medal) Alain Connes on quantum music as noncommutative time-frequency origin of reality from infinite spiral of perfect fifths!
Our brain is an incredible ....perceives things in momentum space of the photons we receive and manufactures a mental picture. Which is geometric. But what I am telling you is that I think ...that the fundamental thing is spectral [frequency]....And somehow in order to think we have to do an enormous Fourier Transform...on geometry. By talking about the "music of shapes" is really a fourier transform of shape and the fact that we have to do it in reverse. Alain Connes, 2012
Stanford Quantum physicist Eddie Oshins:
This representation only works for the (more fundamental) 1/2-integral representations (i.e. spinors/turns/quaternions) but also lets one build the vector and tensor representations. The converse does not hold....this property of "noncommutivity" in itself might be valuable in some way.
My claim, and original idea, has been that this is circumnavigating a T'ai Chi (Yin/Yang) symbol! More recently (Oshins, 1993b) I have suggested that this proximate technique can be used to realize Wing Chun kung-fu's "bong sau/tan sau" movement out of the Kauffman/Oshins "quaternionic arm" discussed and referenced below in end note 5.
I believe that this may be a way to get mind to code the relative relationship of part of oneself with respect to the rest of oneself (self-referential motion) and can explain the concepts of being "centered"/"one"/"integrated"/"extended"/"whole" etc. which one strives for in meditation.
Oshins, E. (1993). Oshins, E. (1993). A test for classical psychospinors. http://www.quantumpsychology.com/pdf/Test-ClassicalPsychospinors.pdf In Abdullah, F. (Ed.) Conservation and Invariance. Cambridge, UK: Alternative Natural Philosophy Association, London England.
Manfred Euler's 2013 description of acoustic STM - Scanning Tunnelling Microscopy builds on his previous connections between the quantum acoustic realms.
Binaural hearing is the acoustic analog of the interferometer or double-slit experiments. The two ears can be regarded as an acoustic interferometer, which recovers the phase difference of signals between the two ears by binaural correlation.
Near-field imaging with sound waves compelling demonstrates the inadequacy of pictorial realism and promotes more abstract views of the reality displayed.
A comparison of sound and matter waves clarifies that these [noncommutative] limitations exist in principle.
2016: de Broglie clocks as synchronization: a tangible model of how mass emerges.
matter waves are locally in phase with the particle clocks (de Broglie's Law of Phase Harmony).
The clock runs forever so it's self-sustaining (consciousness-energy). It resonates with the quantum vacuum. The harmonic beats create dynamic energy.
So then you have a "phase particle" that can be faster than the speed of light - superluminal - and a "mass particle" that is slower than the speed of light as the "group wave" of the "phase wave." The beats of the phase wave then are "in resonance" with the quantum vacuum - and so create mass from the massless field, explaining the Higgs mechanism.
"Universal coherence" - a "mind boggling outlook."
Phase harmony in de Broglie theory relates a local periodic phenomenon (the 'particle clock') to a periodic propagating field in such a way that relativistic invariance is satisfied. If a similar phenomenon in the cell is relevant it should couple the global oscillation pattern locally with periodic (mechanic, electric, biochemical ???) processes.
Coherence as consciousness.
"Ghost Tones"
Manfred Euler is a Professor Emeritus of Physics at the University of Kiel.
It's actually a Klein Bottle so you can't see the 5th dimension.
The red is a logarithmic singularity - but that is just classical physics.
So the foundation of reality is quantum - which is noncommutative phase as the 5th dimension.
So light as a photon is a point but as a wave it is nonlocal - but this means it is in 2 places at the same time - as the 5th dimension that is noncommutative.
People think that doesn't make sense - how can it be in two places at the same time? Actually basic music theory explains this.
So for example the Perfect Fifth is C to F as subharmonic 2/3 while the Perfect Fifth is C to G overtone harmonic as 3/2.
So C = 2 while F=3=G at the same time. That is noncommutative phase. It is also called "Fourier Uncertainty" or "time-frequency uncertainty" - and that is the true foundation of reality.
So for example de Broglie - studying relativity - realized Einstein had a problem - as energy goes to the speed of light then time as wavelength slows down, gets bigger. That means as frequency goes higher then time also gets bigger. That violates a fundamental property since Pythagoras that frequency is inverse to time as wavelength (from music theory).
So de Broglie realized since quantum physics is real there has to be a phase wave that is faster than the speed of light. So you have a mass wave that is the "group phase" - and then you have a particle phase that is called the "internal clock" or a 2nd clock that is the phase wave of the particle - also called the "pilot wave" - and this comes back from the future. So there is a Harvard physicist Andrew Strominger who works with Stephen Hawking - and says that the future already exists and reality is a holograph. So the 4D universe is actually a projection of this 5th dimension of noncommutative phase that is time-like.
This is how precognition is real.
It was Olivier Costa de Beauregard who emphasized that precognition was part of de Broglie's pilot wave model and in 1956 de Broglie said to Costa de Beauregard that there was "an incompatibility with our conventional notions of space and time." Nevertheless scientists have continued this parapsychology research based on de Broglie: pdf AY Temkin 2011
Peter Kingsley notes, "In terms not only of formal and structural analogies but also of historical contacts, there can be no separating the Thracian Orpheus [of Pythagorean equivalence] from central-Asiatic shamanic tradition." This connection with Taoism is also made explicit by the motif of the Pythagorean master Empedocles who, "dies a miraculous death by vanishing into thin air but who leaves a tell-tale item.... A more classic Taoist concept is that of achieving the divine state either by fashioning a spirit-body...thoughtfully leaving a pile of discarded garments...."
Ancient Philosophy, Mystery and Magic, Oxford Ph.D. thesis, 1995, p. 236.
## Wednesday, December 13, 2017
### Natalia Shakhova and The Methane Bomb that may blow next summer
The Methane Bomb.
This vid gives Natalia Shakhova talking the background of why the ESAS (East Siberian Arctic Shelf) has so much methane built up, ready to blow as a huge bomb
Here Nick Breeze posts a long interview with Natalia Shakhova
Nick Breeze told me that climate scientists are not concerned about the methane since the mainstream models think this release won't happen for another couple hundred years. Why? Because their models don't take into account many of the positive feedbacks going on in the arctic.
Nick Breeze then said how even if the Methane Bomb goes off he thinks it won't be more than 5 billion tons a year.
The upper limit of the methane bomb is 50 billion tons but even if it was just 5 billion tons in one year that would have a huge affect on global warming.
Because the real effect of methane is 100 times worse than CO2, not just 30 times worse as is claimed.
We currently emit supposedly just 30 billion tons of CO2 a year - so 5 billion tons of methane would be at least another 150 billion tons - and so global warming would increase another 1 degree C.
That is the BEST case scenario!! At that heat level we could not farm at huge scale since that 2 degree C. is a global average with the interior of the continents being much warmer.
Jennifer Francis interview this spring
Keeling Lecture on Arctic - this spring: GROUND ZERO for global warming
China's Air Apocalypse Smog is from Arctic melting
The Polar Vortex IS the Jet Stream - extreme weather from Arctic ice melting
A new report by the World Meteorological Organization concluded that carbon dioxide increased in the atmosphere at record speed last year and has reached a level not seen in more than 3 million years. At that time, the average atmospheric temperature was 3 degrees Celsius warmer than today, which melted glaciers in Greenland and the Antarctic and pushed sea levels at least 30 feet higher than they are now.
### Treasure Bottle Chi, Master Nan, Huai-Jin and Wim Hof
Thanks again to Serious Buddhist who donated to my full lotus meditation long distance healing experiments! This is a great translation he did of Master Nan, Huai-jin teaching full lotus meditation.
Notice that Master Nan, Huai-jin clearly emphasizes the key secret of meditation is that the yin and yang channels are interacting with the legs crossed and various breathing exercises and hand mudras.
This video is one of the clearest, best and most direct - of real meditation - that I have seen. Master Nan, Huai-jin was considered a living Buddha and had memorized the teachings in both practice and written script.
So then he goes into the Treasure Bottle Chi exercise which builds up to holding the breath. Master Nan, Huai-jin teaches that the breath should only be held after exhale and this is also exactly what Wim Hof teaches. Wim Hof studied Tibetan tummo and Master Nan, Huai-jin also studied Tibetan tummo.
And the final exercise Master Nan, Huai-jin shows is a standing exercise that he recommended males do every night before sleeping in order to maintain celibacy. The key for me, as I have detailed, is for the legs shaking to an extreme sympathetic nervous system response that triggers an opposite extreme of the parasympathetic nervous system.
It is very refreshing to see Master Nan, Huai-jin so clearly detail the secrets of real meditation training as based on complementary opposites energy of the yin and yang channels and to state how we lose our energy with our eyes open but through meditation we build up the energy by circulating it as yin-yang. He understood, just like connecting the negative to positive channels in a battery, the energy then is activated for flowing.
So in the end he talks about the abilities that can also be achieved as "fruits" of the cultivation but the beginning is focused on the goal of meditation as stillness or the Emptiness and how the great mystery of the Universe is whether it is stillness or eternal motion.
This is really a mystery about the spiritual ego and how the source of light, our spiritual ego, is actually the qi or prana energy as intelligence that is Emptiness - and eternal motion achieved when the light goes into stillness and our ego is gone. So it is an inherent mystery of reality and yet can be experienced through meditation.
### Calling B.S. on the Church Queering Crap
The blog did not post comment nbr. 1 but DID post comment nbr. 2. HAHA. Censorship! Hilarious.
http://queering-the-church.blogspot.com/2010/08/some-gods-of-homosexual-love.html
Comment Nbr. 1:
So you think raping of boys is a good thing? I have a relative who is Mayan and he was raped as a boy and now he has physically and sexually harassed me because he can not control his ejaculation addiction. This is twisted and some academics have called this out. Why do you think Mayan civilization collapsed? And why do you think Central America is the most violent region on Earth? Male ejaculation addiction.
The original human culture, the San Bushmen, did not have any homosexuality, nor did they know what it was. Nor did their close relatives the Pygmy culture - all humans originate from the San Bushmen culture who lived sustainably without rape and without warfare. Why? Because all males were required a puberty to train in celibacy "staying power" of N/om energy that is from the female psychic energy of the vagus nerve. I have the secret of this training - I was raised Christian and I recommend people learn to be real Christians. Ejaculation spikes cortisol as a sympathetic nerve reaction and so you get ejaculation addiction as a positive feedback of dopamine and cortisol - with the dopamine causing violence. This is why anger and lust are so closely connected since they are both controlled by the hypothalamus. Robert Sapolsky goes into this science.
There is nothing "saintly" about ejaculation - saints are from celibacy and ionization of neurohormones creating healing energy. Because the West does not have this training therefore the Church is rife with ejaculation addiction male priests that have abused other humans. It is disgusting and evil.
Comment Nbr 2. The Mayan God "Chin" originated from the Aztecs. Before the Aztecs spread pederasty rape into Mayan culture, the Mayan culture was based on complementary opposites of male and female sacred energy, as detailed in Martin Prechtel's book "Secrets of the Talking Jaguar" and his several other books. He trained as a village Mayan shaman that preserves the original Mayan culture before it was corrupted by pederasty rape.
## Echo Timing, Consciousness and Acoustic light from Phonon Ether
It is pointed out that the mystery of how biological systems measure their lengths vanishes away if one premises that they have discovered a way to generate linear waves analogous to compressional sound. These can be used to detect length at either large or small scales using echo timing and fringe counting.
The engineering principle is shown to be very general and functionally the same as that used by hearing organs.
Self-amplification is secret of life-consciousness - Nobel physicist Robert Betts Laughlin at Stanford
an ubiquitous superfluid which fills the universe. Here we analyze light propagation through this “dark superfluid” (which also dark matter would be a hydrodynamic manifestation of) by considering a photon-phonon analogy, where photon is a quasi-particle which acoustically propagates through
### Hydrodynamics of the dark superfluid: II. photon-phonon analogy - Hal
by M Fedi - 2017 pdf
From the point of view of a fluid approach, the de Broglie-Bohm's pilot wave could perhaps be explained as an ether wave, a special fluid...
"The modern concept of the vacuum of space, confirmed every day by experiments, is a relativistic ether." Noble physicist Robert Betts Laughlin
the Planck constant refers to mass circulating along a closed loop in a given time....this kinetic energy adds energy to the system while the turn is performed, so we understand the meaning of [change in frequency energy x change in time aka time-frequency uncertainty]....a mass circulation in a quantum vortex....a continuous hydrodynamic fluctuation.
a photon is actually a special spin-1 phonon propagating through dark energy....Light could be nothing more than "the sound of dark energy" and c the speed of sound through dark energy. A sound that we perceive through our eyes....producing a transversal wave due to spin.
...Dark energy could not be dark at all but it could be perceived as the most luminous thing we know. Light itself.
Marco Fedi
### The interwebs as a "deer in the headlights" entropy biophoton interaction of spiritual ecology healing
Thanks for that amazing ice shelf report - and so I posted Kent A. Peacock, philosophy of science prof stating that ice shelf dynamics were not even included in IPCC since they stated it was too unknown - and yet there can definitely be abrupt increase in water levels from huge calving of ice shelves - in just a matter of weeks or months. So yes we are in a very amazingly fragile situation - and the fact is the truth is not a popularity contest. Guy Mcpherson was an award winning professor and yet he is dismissed as a doomsday cult hero and when he posts it only reaches maybe tens of thousands of people at most - and there are almost 8 billion people on the planet. So do the math.
The fact is that modern human consciousness is a psychic energy blockage - we got cut off from the truth when our left brain dominance made the mistake that infinity (Mother Nature) could be "contained" by geometry (physical reality) using symbolic language (analytic geometry), and so by using right-hand dominant technology we could find happiness, prosperity, etc. This is actually just the Pyre in Empire - the great funeral Pyre of Mother Earth.
Music lets us know the truth of reality as all humans rely on music from natural harmonics and the original human culture did live sustainable without warfare and without rape - just as the bonobo primates are closest cousins have the females control the land and the females control the males. It is only female human primates living in Nature in a group that synchronize with the lunar cycle at 29.5 days and this is the pineal gland psychic energy that also governs the water cycles on Earth for life to evolve. Water is a macroquantum molecule and so is negentropic and enables life to evolve via proton-proton gradient cycling - delocalized quantum coherent negentropic energy from the relativistic mass of light, the hidden spin momentum energy of light.
So for example it is proven that mushrooms and trees communicate via biophotons and quantum jitters are the source of the supposed "random mutations" of ecological evolution. In fact quantum biology shows that these quantum jitters are quantum coherent and then amplify up into the macro-level - so that there is actually a protoconsciousness of the universe that is inherently nonlocal - when light is turned around as self-aware consciousness, at zero time, there is infinite frequency that is actually energy from the future, as precognition. So what we perceive as the 4D universe is actually a left-brain projection via right-hand technology that is the inverse of ecological destruction of Mother Earth, that is a 5D reality of reverse time or negentropic evolution. This is what the Egyptians called Nuit as the secret of alchemy and now we call nothingness but it is from the lunar psychic energy that controls water on Earth.
So we can not even expect science to give us answers - because the technology we rely on is just making the problem worse - and this has been inherent to the mathematics of WEstern science, the symmetric math is a lie from the wrong music theory. So the interwebs is a "deer in the headlights" phenomenon and since "sex sells" then what goes viral is T & A ejaculation addiction material and so the truth remains hidden and then censored. There has to be a conscious sublimation of neurohormone emotional energy by the male human primates to then access the truth of reality via the heart-pineal gland connection and so resonate with the inherent negentropic information that is superluminal. As Guy McPherson says we all will die but what he does not mention is that we all have coherent biophoton spirits inside us and they leave our bodies after death. But these spirits actually exist in a 5D universe that is formless awareness as negentropic energy - self-healing "qi" or "prana' or what the original humans called N/om energy.
So the problem right now is actually ghost pollution as when modern humans die they turn into ghosts with emotional blockages - we can not see this from satellite readings as it is a macroquantum coherent light with emotional blockages and then us alive people suck up the emotional blockages of ghosts. So we can say that actually left-brain dominance is a kind of mass mind control possession by ghosts - and the original human culture was constantly healing this energy blockage through trance dance healing that recreated the original creation - the dream time of reality - as Dr. Bradford Keeney details. So for example when the original humans killed a bird for food - they do not eat the bird until the next day. Why? Because they need to wait for the bird's spirit-ghost to clear its emotional blockages - or else the spirit-ghost of the bird will tell the other birds that humans are killing the birds.
So that is a much more sophisticated "real time" satellite reading of reality - a macroquantum coherent biophoton energy reading of ecology that then guides the interactions with using the "resources" of nature, from the truth of the whole universe. Our technological toys have been fun but remain out of touch with the formless awareness consciousness truth of reality - and so physicists using satellites are out of touch with the accelerating entropy on Earth that is inherent to the supposed increasing entropy of the Universe as a whole. The dark energy accelerating the expansion of the Universe is actually the same dark energy that caused the Big Bang and is also the same Cosmic Mother energy that is noncommutative time-frequency quantum nonlocality. So we can access this energy when we meditate and "turn the light around" - as two thirds of our brain's energy is used for vision. It is proven that we emit biophoton energy when our eyes are open and so using the internet we are imprinting our spirit energy into the technology. This is how energy blockages can also be read through the internet and also healing done through the internet.
So yes we are now dependent on our technology to give us the supposed "real time" increasing disaster of life on Earth but in fact this left-brain dominant means of accessing information is the past to a future of increasing entropy, whereas right-brain dominant meditation enables the reverse process to occur, whereby virtual photons from the future are captured and then reverse the entropy of the past, to increase the healing of the present as increased negentropic energy from the future - the quantum force momentum energy that is not conserved.
Funny I was gonna eat tons of garlic to try to defend myself from sexual-physical harassment by my in-laws visiting this xmas. I used to eat bulbs of garlic - like a bulb a day of organic garlic when I lived off dumpster-diving food for almost 10 years - and so my skin turned permanently yellow from all the sulfur. haha. I am not longer a white person. haha. Just like Ellen Page got orange skin in her arm pits and palms from drinking too much carrot juice. But special forces actually get hydrogen sulfide injections because the hydrogen doubles your strength and endurance as it directly powers the mitochondria, that are symbiotic bacteria living in each of our cells. When we take a crap then the bacterial levels of our bodies are reduced by half but if we are full of crap then there are more bacterial cells in our body than actual human cells - and so we literally are ecological bags of bacteria walking around - and there are hundreds of viruses in the air at all times that we suck in. So our immune system is directly controlled by the neurohormone levels of adrenaline via the sympathetic nervous system maxing out to create a reverse extreme of the parasympathetic nervous system that then stores up potassium charge in our neurons. We can then literally irradiate and fry any external bacteria and viruses from the increased potassium charge created by increased serotonin levels via increased ACTH of the pituitary-hypothalamus connection to the adrenal glands - doing deep breathing exercises, as Wim Hof discovered. So eating garlic will kill the bad bacteria but supposedly we have evolved eating garlic so long that our good bacteria can resist the extreme antioxidants of the garlic causing the burning sensation and internal tingling. We can achieve this same internal tingling by "earthing" - with barefoot on Earth and legs bent so legs shaking pushing the sympathetic nervous system to extreme - and so this internal tingling is from lightning hitting earth as the Schumann resonance - with legs shaking 7 to 9 beats per second as with trance drum-dancing. So then this clears out the shit as an opposite parasympathetic rebound that flushes the colon with serotonin - and the internal tingling is then the increased negative ions that neutralize the free radicals from eating sugar or from bad bacteria and viruses.
Sure REXXON was gonna strike a multi-billion oil deal with Putin and so there is now a global alliance of ejaculation addiction males - so for example Kissinger sits on China's national oil petroleum board of directors - and these dudes all have Post-apolcaypse plans of living underground - just like Cheney, with his home state Wyoming as a "national disaster zone" - this is official government policy - and he planned Continuity of Government as FEMA is a literal underground government with above ground fascism - and so after his 9/11 plan kicked to still the last oil-natural gas around the world - what did Cheney do? He literally lived underground for 3 months. haha. These dudes are insane - as GEE points out - in 2026 there might some elites left in bunkers but would you really want to be with those psychopaths? No thanks.
## Monday, December 11, 2017
### Nonlocal Phonon Spin Sound Ether: Kent A. Peacock philosophy of science
So there are two very amazing articles by philosophy of science professor Kent A. Peacock and both mention the case of the nonlocally entangled phonon vibrations - created between two diamonds. Because we typically think of a vibration needing a medium to "vibrate" - and the definitive cause is sound - which then changes based on the medium. But the question is as the speed of light is proven to be invariant for relativity of spacetime then how can one interpret the potential of an inherent quantum nonlocal medium? In other words there should be no "special" frame of reference (i.e. God) and yet if a nonlocal medium ether exists then why isn't there this inherent real of "no motion" (i.e. a standing wave)?
This is a very fascinating discussion of philosophy of science - and so first of all let's link his two papers.
### Would superluminal influences violate the principle of relativity?
KA Peacock - arXiv preprint arXiv:1301.0307, 2013 - arxiv.org
Abstract: It continues to be alleged that superluminal influences of any sort would be
inconsistent with special relativity for the following three reasons:(i) they would imply the existence of a distinguished'frame;(ii) they would allow the detection of absolute motion; and
### Happiest Thoughts: Great Thought Experiments of Modern Physics
KA Peacock - 2016 - philsci-archive.pitt.edu
This is a review of those key thought experiments in physics from the late 19th century
onward that seem to have played a particular role in the process of the discovery or
advancement of theory. Among others the paper discusses Maxwell's demon, several of
Einstein's thought experiments in relativity, Heisenberg's microscope, the Einstein-
Schrödinger cat, and the EPR (Einstein-Podolsky-Rosen) thought experiment.
And so who is Kent A. Peacock? He doesn't seem to be referenced much.
### So then who is John Norton? http://www.pitt.edu/~jdnorton/jdnorton.html a fellow philosophy of science professor, of course.
Our urge to oversimplify has led to many myths about what powered Einstein's discoveries. Naive thinking? Capricious rule-breaking? Operational thinking? I correct some myths and try to give a more accurate picture of how Einstein made two discoveries: special relativity and the light quantum.
So he is writing on the same topic. I'll check it out.
O.K. so Peacock states that in terms of light then even zero and infinity are relative to each other.
So it's not that nonlocality does not exist but rather when the light is turned around, there is no individual sense of I, but instead the light is so bright there is an Emptiness of the ego: Time goes to zero and then reverse time as the ether phonon energy is created.
And so there is no "absolute" time but rather two sets of time based on relativity and quantum physics - as de Broglie also realized - one from the future that is superluminal - and measured relative to the past of the observer who is now, in the present, just pure light relative to him(her) self ego.
And so since there is no longer any "absolute simultaneity" as Einstein had to ponder (he read Hume and Mach for his own philosophy of science research) what this means is that noncommutative time-frequency is also the source as eternal motion of consciousness - formless awareness - always in two places at the same time (relative to phase).
The tricky part to realize is that frequency is noncommutative to time and so Dr. Hameroff emphasizes that music as consciousness is actually "anharmonic" as in Indian music. So if 1 is the I-thought as C, the root tonic note, then the octave frequency is 2 also as the same note or pitch, C but 3 is the Perfect Fifth harmonic as G yet at the same time 3 is the Perfect Fifth harmonic as F, only as an undertone subharmonic. So this means G=3=F at the same time - this is called noncommutative phase that is the secret of nonlocal entanglement. Quantum physics is based on a linear operator so it inherently has noncommutative time-frequency so that if you change the order of observation or measurement then the value also changes and so reality is inherently participatory. So this means that relativity then collapses the quantum nonlocality into a subjective sense of I-thought whereas the objective reduction of the waveform is due to the protoconsciousness of the universe that is inherently nonlocal as infinite noncommutative time-frequency resonance. So we can listen faster than time-frequency uncertainty because listening is nonlinear and our brains are quantum coherent at the microsecond pulse or wavelength that has ultrasound frequency - between the right and left ear phase for stereophonic 4D spacetime reality.
But this ultrasound then resonates the brain as a whole, the microtubules, as quantum non-local consciousness. So when we visualize - we "turn the light around" - the biophotons of consciousness and so at zero time, light should be a standing wavelength with the nodes at rest but due to relativity this can only be measured externally - such that light has no rest mass but light does have relativistic mass. So light has then a hidden momentum energy that is not conserved - this is the spin force or the phonon nonlocal ether of light due to noncommutative spacetime. Dr. Stuart Hameroff calls it a noncommutatve scalar field as per Sir Roger Penrose's research into asymmetric time as the origin of the universe. So we can then say the protoconsciousness of the Universe is listening to us all the time and if we turn the light around, as in meditation, since we are the light as our ego consciousness, self, awareness, when the light is turned back to its source, it then captures virtual photons from the future. This is how precognition is also real along with free will, as Dr. Stuart Hameroff details in one of his academic articles.
### A biophysical approach to cancer dynamics: Quantum chaos and ...
www.sciencedirect.com/science/article/pii/S0303264716301903
Hence, eCBs and serotonin-like endopsychedelics (e.g. DMT) may interconnect the mental/cognitive state of a conscious organism to its inflammatory-immune pathways resulting in cancer. Tumors actively interact with their microenvironment to maintain malignant phenotypes and one of such signaling cascades is the ...
So when I discovered this paper - the title - the abstract was so wild -
Quantum tunneling in mutagenesis, vacuum energy field dynamics, and cytoskeletal networks in tumor morphogenesis have revealed the applicability for description of cancer dynamics, which is discussed with a brief account of endogenous hallucinogens, bioelectromagnetism and water fluctuations.
Wait - the corrections are accessible as pdf! Oh it's his move to Concordia since McGill medical got discredited - but he is not listed in the Concordia University directory. oops.
I wondered who is this person? Turns out using Excel at age 4? it is this science kid genius (honorary medical dr. at age 17) named: background
his parents, who are Tamil, came here from Sri Lanka.”My parents motivated us to get into the sciences,” he says, “and to make a contribution and an impact on society.”
When he was at Coronation elementary school, Uthamacumaran worked on a robotics program. He travelled abroad, to Japan in 2005 and Germany in ’06, to participate in science competitions.
For the German competition, the students created Scibot, a robot Uthamacumaran says was “as tall as me at the age of 10.” For the exhibit, Uthamacumaran dressed as a mad scientist and danced to James Brown’s I Feel Good while the robot replicated his movements.
Cancer research is more complex and involves less dancing. Uthamacumaran’s study of the disease has led him to a holistic approach that emphasizes the steps one can take to reduce the risk of contracting the disease.
“Dietary options, nutrition, daily activities and the radiation sources you’re surrounded by, like television and music players,” Uthamacumaran
said. He said everything we do in our modern industrialized society impacts the occurrence of cancer.
He believes anyone can achieve whatever they want. When you put your mind to it you can do it, he said the sky extends forever. There is no sky.”"
# abicumaran uthamacumaran
### Canada's 17 year old Abicumaran Uthamacumaran Fights for the Cure ...
Oct 15, 2011 - Attending McGill University is 17-year-old Abicumaran (Abi) a recent graduate of Marymound High School who hopes to find a cure for cancer. It appears to be his life mission. Abi took an interest in medicine at an early age. At 5 he said he was reading Darwin and the Origins of life. Abi's parents are .
Problem is the paper is behind fire wall - so I'm trying to coax out tidbits using search engines.
So we can see the bioelectromagnetics from resonating the cytoskeleton - piezoelectric energy via the microtubules.
### 'Quantum jitters' could form basis of evolution, cancer - Phys.org
https://phys.org › Chemistry › Biochemistry
Mar 11, 2015 - The study, which appears March 12 journal Nature, indicates that these jitters appear at about the same frequency that the DNA copying machinery makes mistakes, which might make them the basis of random genetic changes that drive evolution and diseases like cancer.
And so the quantum origin of evolution is now being confirmed! And what was considered random jitters can now be resonated as quantum coherence via the de Broglie Law of Phase Harmony, as translated into qigong alchemy training.
### DNA typos to blame for most cancer mutations : Nature News ...
www.nature.com/news/dna-typos-to-blame-for-most-cancer-mutations-1.21696
Mar 23, 2017 - Many of the genetic mutations in tumour cells such as these are created by DNA-replication errors. Nearly two-thirds of the mutations that drive cancers are caused by errors that occur when cells copy DNA, mathematical models suggest. The findings, published in Science on 23 March, are the latest ...
Epigenetics driving evolution and cancer?
plasticity of an organism’s characteristics, or phenotype, foreshadows its evolution. In essence, you can start with an epigenetic variant — think calloused hands — and later that particular trait can become permanently fixed in the genes.
Famously, West-Eberhard said, “Genes are followers, not leaders, in evolution.” Now that same idea is invading the theory of cancer. It seems that cancer cells, too, can first begin to change through temporary epigenetic changes, instead of by means of mutations in the DNA.
### DNA and Phonons by Southern Jameson West - YouTube
Apr 11, 2011 - Uploaded by shouwei86
Qubit Lab 1,947 views · 5:08. 'Quantum jitters' may drive DNA mutations {Duke University Research ...
But you can see he goes from quantum spin and DNA phonons to then doing political analysis.
Hopes to cure cancer, talk by abicumaran uthamacumaran 2013
So we'll see if he gets into DMT in that talk. Nope but this is what he is talking about:
In conclusion, the shRNA- Crk resulted in the significant reduction of cell adhesion and
migration and lead to dramatic cytoskeletal rearrangements. Variations in levels of Cas and FAK protein binding interactions were also identified. Further, understanding these protein interactions will help us derive safer anti-cancer treatments and identify potential anti-metastatic therapeutic targets.
If not we can check out another talk. 2013
Wow - so the eyes absorb the microwaves from cell phones! Amazing.
Tryptophan (TRP) is essential for many physiological processes, and its metabolism changes in some diseases such as infection and cancer. The most studied aspects of TRP metabolism are the kynurenine and serotonin pathways. A minor metabolic route, tryptamine and N,N-dimethyltryptamine (DMT) biosynthesis, has received far less attention, probably because of the very low amounts of these compounds detected only in some tissues, which has led them to be collectively considered as trace amines.
# abicumaran uthamacumaran is describing some kind of miracle healing of not just cancer but also anti-aging! He discovered some kind of protein transcription treatment - is it DMT-based?
Wow - Bohmian mechanics? Wish I cud access the paper. So I emailed Dr. Uthabicumaran to see if he could send me a copy of his paper. In the mean time here is the list of the references he uses.
## Saturday, December 9, 2017
### Life Rhythm as a Symphony of Oscillatory Patterns: Electromagnetic Energy and Sound Vibration Modulates Gene Expression for Biological Signaling and Healing
What? Yep. Just discovered this gem. Have not even read it yet!
Dr. Roger Jahnke has a great talk on his channel on qigong and he mentions the vagus nerve and biophotons
These strategies may represent a new tool to allow selective tuning of cell/tissue/organ homeostasis, paving the way for the use of sound physics and music for optimization as a cell therapy in regenerative medicine. Also, it is interesting to note that parallels have been drawn between traditional Chinese medicine (TCM) and sonocytology, with the suggestion that nanotechnology may shed light upon the acupuncture meridian system and contribute to the modernization of TCM techniques, including those that traditionally use musical tones for preventive treatment of disease.204 Cumulatively, the results found using AFM techniques support the hypothesis that cell decisions are not restricted to biochemical effectors, but can be orchestrated though nanomechanical regulators, and exposure to specific acoustic-range vibrational modes or “music” may represent a worthy field of investigation for “informative reprogramming” of cellular behavior. If such nanomechanical bioinformation can be identified, then nanomechanical/EMF patterns orchestrating stem cell commitment and differentiation might be retained and stored as an informative “nanomechanical signature” of the “sounds” or “music” emitted and functionally received by cells and organs. Such sounds might communicate the informational memory of the biofield and be used to enhance regulation of a variety of processes including differentiation, stem cell reprogramming, and the maintenance and manipulation of homeostasis.
## BIOINFORMATION: TOWARDS A NEW LANGUAGE OF VIBRATION
In this article, we have presented a portion of the large and extremely rapidly growing body of evidence suggesting the existence of a vibrational bioinformation regulation system operating across the subcellular, cellular and organismic levels. To reiterate, it has been clearly demonstrated that the microtubular cytoskeleton has functional electronic properties beyond the stabilization of cellular shape, that endogenous EMFs generated by the intracellular network of micro-tubules, centrosomes and chromosomes play a fundamental role in regulating the dynamics of mitosis and meiosis,139 and that endogenous EMFs in the nuclear DNA-containing chromatin also play a key role in chromosome packing during the mitotic cell-cycle phase.205
http://www.laszloinstitute.com/people/david-muehsam/
Dear Dr. David Muehsam: I did my first year of college at Hampshire where I took quantum mechanics from Professor Herbert Bernstein, with my focus on music-based philosophy, in 1990. I ended up finishing a master's degree in "sound-current nondualism and radical ecology" in 2000 at University of Minnesota. I did intensive qigong yoga meditation with a Chinese spiritual healer, Chunyi Lin, of http://springforestqigong.com. His "external qi" heaing was corroborated by a "randomized controlled" study healing chronic pain, not treatable by Western medicine for 5 years, peer-reviewed published, led by Dr. Ann Vincent of the Mayo Clinic.
So in regards to your Life Rhythm article, I just wanted to point out that the secret is noncommutative phase a macroquantum resonance. Eddie Oshins, the quantum physicist at Stanford also figured this out. He was tasked to study quantum logic and quantum psychology but he also taught Wing Chun. He realized that Daoist alchemy neigong is based on noncommutative phase.
So this is very simple yet very radical. Math professor Alain Connes also realized this insight, that the empirically true music theory is noncommutative and therefore a "Universal scaling system" that he calls the quantum music of the spheres. But Connes is biased toward Western music and so although he argues the foundation of reality is noncommutative time-frequency, as a "triple spectral" - he uses Western music as his metaphor.
So it was math professor Luigi Borzacchini who realized that music theory was a "negative judgment paradox" at the origin of Western math, as a "secret of the sect" that he called "really astonishing" and "shocking." He said this created a "pre-established deep disharmony" that guides Western science.
O.K. so basically there is a trick in music theory that people just don't think about - and it is noncommutative math. Archtyas used the equation arithmetic mean x harmonic mean = geometric mean squared. So therefore he could only use 3/2 as the ratio for the Perfect Fifth, as an overtone, for C to G. So if C is 1 as the root tonic, and the octave C is 2, then G is 3. But equally valid is 2/3 as the Perfect Fifth with geometric value C to F. But that had to be covered up and hidden.
So I discovered that the secret of Daoism is based on these complementary opposites of Perfect Fifths/Octaves/Perfect Fourths. It is also the secret origin of the "three gunas" the oldest philosophy of India. And the original human culture, the San Bushmen, relied on music based on the octave/Perfect Fifth/Fourth harmonics. Neuroscience has proven that these harmonics have the strongest amplitude resonance in the brain.
But what WEsterners don't realize, that Easterners did realize, is that it is an infinite spiral of fifths, not as just some modular math, but because it is noncommutative. Anyway Eddie Oshins realized that the hand movements of Neigong were based on the quantum logic of noncommutative phase.
As I said I realized this as well - so for example the left hand is yang, for males, and lower body yin, while the right hand is yin and upper body yang. So that is the secret to turning the body into a macroquantum harmonic oscillator - if you face the palms of the hands in those positions it creates a natural noncommutative resonance.
So the secret of OM in India is also based on this infinite spiral of fifths as noncommutative phase that is non-local consciousness. De Broglie also figured this out with his Law of Phase Harmony. So when we "turn the light around" in meditation this means time goes to zero, but there is a hidden momentum that is the phase wave from the future, as negentropic energy, the relativistic mass of light from the noncommutative spacetime ether.
My blog has more details: http://ecoechoinvasives.blogspot.com
thanks,
drew hempel.
P.s. any comments, questions, criticisms would be appreciated. I spread the research for free so if you are interesting in sharing that would be great.
Were you supervising Division 1 projects with Music Professor Margo McKay-Simmons (sp.?) at Hampshire College. When I saw that you studied with Yusef Lateef - suddenly your voice sounded familiar to me and then I realized you looked familiar. I composed "Troll Dance" as a fugue on Moog synthesizer - recorded on a 4-trac that was out of sync. So the fugue decomposed itself in the recording, in contrast to the actual composition. You two (if I remember correctly) thought it was pretty funny. haha.
drew hempel
https://www.pdf-archive.com/2017/04/10/idiot-s-guide-to-taoist-alchemy/
That is the training overview based on the noncommutative phase music model.
This goes back to the original human culture - have you read Dr. Victor Grauer's book, "Sounding the Depths" on the original human culture music?
thanks again,
interactive Heart biofeedback as sound-visual art - Dr. David Muehsam in Italy pdf
David is a specialist in the biophysical transduction mechanisms and therapeutic applications of electromagnetic fields (EMFs). He has practiced meditation and yoga since the early 1980’s and has taught extensively around the world, leading workshops and yoga teacher-trainings in the US, EU, Asia and S. America. Dr. Muehsam holds a degree in Physics from Hampshire College, Amherst, USA, and PhD in Neurophysiology from the University of Bologna. His work in bioelectromagnetics began in 1991 at the Bioelectrochemistry Laboratory at Mt. Sinai Medical Center, NY, USA, with emphasis on EMF biophysical interaction mechanisms and therapeutics. Dr. Muehsam has authored extensively on biophysical mechanisms of EMF bioeffects, EMF therapeutics, mathematical modeling of EMF bioeffects, alternative and complementary medicine and has conducted studies on Pranayama and Qigong practices. Dr. Muehsam is the Director of Technology Development for the Consciousness and Healing Initiative, is on the Scientific Advisory Board of the Global Coherence Project, organizations dedicated to the study of consciousness, mind/body interactions and healing, and is Senior Biophysicist for Rio Grande Neurosciences. Dr. Muehsam is also an internationally recognized yoga/meditation teacher. He has studied extensively in the Iyengar and Ashtanga schools, and his practice and teaching is deeply influenced by the traditions of Satyananda Saraswati and the Thai Theravada lineage of Ajahn Chah. Dr. Muehsam has received written authorization from K. Pattabhi Jois to teach Ashtanga Vinyasa Yoga, holds Certificate IV in Teacher Training and Assessment from the Australian Government, Yoga Alliance 500-Hour certification, completed yoga teacher training with Richard Freeman, and founded the Fana Yoga teacher training school, internationally certifying yoga instructors. Dr. Muehsam has extensive international teaching experience, leading workshops and teacher-trainings, in recent years in Australia, Chile, China, Germany, India, Indonesia, Ireland, Italy, Japan, Malaysia, Singapore, Spain, Switzerland, and the United States. Dr. Muehsam is also an accomplished musician, performing on flute and saxophone since childhood. As a child and young adult he was exposed to the musical influences of New York City and Western Massachusetts, and he began to perform professionally as a teenager. His primary musical mentor is Yusef Lateef, with whom he maintained a close relationship for 15 years while living in the USA, and he has also studied with Ray Copeland, Jimmy Giuffre, George Garzone, Barry Harris, John Castellano, and more recently, Garry Dial, Antonio Hart and Allaudin Mathieu. In India, David studied bansuri flute with Ravi Shankar Mishra and All India Radio Artist Sri Upadhyaya. David has worked with his own groups, and performed with Emery Smith, Chris Daily, the Fred Clayton International Rhythm Connection, African Chic, the Baton Rouge Symphony Orchestra, and others. His experiences living and working in a variety of cultures have given him a deep interest in World Music, the use of music for healing, the potential benefits of yoga/meditation for reducing stress, improving emotional and mental well being, and how this could lead to a better understanding of health and healing.
http://vidartscience.org/
### How Music Theory is the secret to the original Sustainable spiritual human culture
http://www.quantumpsychology.com/back.html If you scroll down on Eddie Oshins' papers - he was the "in house" quantum physicist as Stanford particle accelerator - working on quantum psychology. So he gets into the Boolean logic paradoxes as quantum logic. I think you might be interested. I independently made the same discovery as him as he taught Wing Chun and so he realized the secret of the nonwestern harmonics is noncommutative math. So I had this realization in high school as my piano teacher also taught music theory in the evening. I pondered - why, if C is 1, then the Perfect fifth as C to G has to be 3/2 and can not be 2/3 whereas the Perfect Fifth, as C to F can not be 2/3 but instead has to be changed to a Perfect Fourth as 4/3, C to F.
Then her husband was her music professor at the University of Minnesota and so he also taught private music training. So I studied orchestration, ear training, composition from him. So he showed me a cool trick - find a piano and then silently press down a higher harmonic - like a G note. Keep it held down, then strike the lower C strongly and you hear the G note. Then try it with a different note that is not a close harmonic and it doesn't work. O.K. so as I was then taught equal-tempered tuning based on logarithms - as a "compromise" - I discovered the book "Science and Music" by Sir James Jean, the quantum physicist. He emphasizes that the empirical truth of music is an infinite spiral of fifths. So since I began training classical piano since age 5 - and was into music deeply from age 3, as I sat quietly through my mom's piano lessons - listening - and so it's now proven that if you study an instrument strongly before age 7 then you get a much larger corpus callosum.
So I had this insight that Pythagoras must have been correct and I was against the logarithmic equal-tempered tuning. My private teacher, the former music professor, did not like hearing this. I actually had to keep my insight to myself but my piano teacher did get me a book on spiritual insights of music composers and her husband got me a book called the "Creative Process" about how dreams informed both science and art. And then after my senior piano recital - wherein I memorized Bach's Italian Concerto in F, a Mozart Sonata, a Brahms piece and I performed John Cage prepared piano and my own compositions - so then my music professor gave me a copy of Gregory Bateson's book "Mind and Nature: A necessary Unity" and also a copy of the Tao Te Ching. Then when I took quantum mechanics my first year of college, the professor emphasized that when people learn classical physics in high school then they have the wrong foundation for science. And his teacher's assistant was reading the same book that I was reading - Mind and Nature. haha.
O.K. so my blog has all my research details http://ecoechoinvasives.blogspot.com - where I quote Alain Connes - and so I played all types of music but after the first year of college, I could no longer play Western music much. I did end up playing free jazz improv with a street flute player - but I kept pursuing my vision. In 1996 I had written a manifesto called The Fundamental Force - and I had an image of Bloch's Wall as an attempt to describe my insight into music theory. Mainly the essay was a critique of Slavoj Zizek and so I sent him a copy. He replied it looked very interesting and he would read it and get back to me. So I never heard back but his next book in 1997, the Plague of Fantasies, was largely a "strawman argument" against my critique of him! haha. He was slamming ecofeminism and paranormal psychology and he also emphasized Platonic music, etc.
So when you talk of this self-organizing self - the key is what Schrodinger stated in his book, "What is Life?" and he said it has to be an "aperiodic" - or what Stuart Hameroff with Roger Penrose call "anharmonic." And so Hameroff - with whom I've corresponded - he very much emphasizes that music is the closest model to what consciousness is. And he also clearly states not Western music. His research was corroborated by this Indian scientist who has a lab on Japan - I have also corresponded with him as well. So we had quite a lengthy correspondence - and so he admitted that no one can know the original phase of the quantum Bloch sphere. But essentially he did the nano-research that tested the electrical conductance of microtubules - and he discovered that ultrasound resonates microtubules at 1000 times greater amplitude than any other frequency. Also the microtubules have to be quantum coherent when they resonate. So this was amazing corroboration of Penrose and Hameroff. So Hameroff actually tested ultrasound against his skull and reported a great bliss that lasted for a while. But from my own research I had already deduced this and in music theory it's called the Hypersonic Effect.
Actually the CIA mind control scientist Dr. Andrija Puharich figured this out - and he was a real pioneer and he was part of the Actual Matrix Plan that I discovered - soon after my master's thesis on radical ecology and sound-current nondualism. So anyway it's not just ultrasound - the Mozart Effect is real and from the slow middle movements of Baroque and early classical - as Mozart was strongly influenced by Bach (He was close friend's with Bach's son). And so when I had memorized Bach's Italian Concerto, my piano teacher was impressed at how fast I memorized the piece, and so she got me into this fancy chamber music camp up on Lake Superior, Madeline Island. But at the time I was already in a punk rock hardcore band and playing jazz, etc. So I just caused trouble mainly but my roommate was a composer who aced his SAT and so he got a full ride scholarship to Harvard. I did go on to study music composition at Smith College and UMASS - and Hampshire - but I only stayed out east for a year.
UW-Madison accepted me into their music composition program but what happened was hilarious! I had focused in a self-design degree on what I called global blues music and I was openly critical of the avant-garde serial music of the composition professor - at his newly built house on the prairie. So he said his music was just "ahead of the times" and so no one could appreciate it. haha. So then I was getting As and Bs in my music orchestration class but I was also protesting all the time, near the music building, on the Mall - and one of the professors must have seen me. I protested against NAFTA and I worked for the UW-Greens. So then he called me into his office and said my transposition of a Schoenberg piano sonata into a string quartet was Communist because I was writing "equal play for equal pay." I was so shocked I started to tear up right in front of him after he demanded that I redo all my past assignments, even though I had already gotten good grades on them. So I just dropped out - but those professors bragged how they could get any of their music students into the avant-garde computer music programming graduate degrees at Princeton, etc. haha.
So since then it's been proven that the brain neurons have the strongest amplitude resonance from the Perfect Fifth and Octave harmonics. And then finally I realized that the infinite spiral of fifths is from noncommutative math with the Perfect Fourth. Also I discovered this was the origin of Daoism and the "three gunas" of India - the oldest philosophy of India. Then if you study the original human culture - it is completely focused on music trance dance training with the emphasis on the octave, Perfect Fifth and Perfect Fourth as this infinite alchemical resonance energy.
So the Mozart Effect increases the alpha brain waves and serotonin levels - there is a nonlinear feedback between the ultrasound and ELF harmonics. It's been proven that the qigong energy healers have strong ELF waves as a standing resonance with Earth's Schumann resonance - as a theta-alpha brain wave that also resonates the heart energy. Actually this is based on quantum coherent biophotons. As I finished my master's degree doing intensive meditation with a Chinese qigong master - then I saw ghosts and I did healing and I smelled cancer - and this is exactly what the original human culture had all males train to do from 100,000 years ago. It was the secret of our sustainable culture.
As Gregory Bateson said it's based on "difference of the difference" as orders of logical type - but chaos math is all based on logistic equations that are symmetric math. So the origin of quantum non-locality is noncommutative time-frequency energy. For example Heisenberg's uncertainty actually originates from noncommutative time-frequency. So music was before math and before language even - music is right-brain dominant and Western music is wrong. Sonoluminescence is real - as is sonofusion - from infinite time-frequency resonance as noncommutative phase.
So Alain Connes - his problem is that he practices Chopin at home and so he is biased towards Western music. He thinks the future will be quantum consciousness whereby people here multiple spacetimes in their head just as in orchestration. I have studied orchestration - so I know the brain twister of playing multiple clefs - transposing in real time - on the piano - as the whole orchestra. I did that training as private study while I was in high school. But Connes is completely projecting his Western bias onto what music is. But Connes is correct about noncommutative time-frequency from music theory. He calls this the quantum music of the sphere as a universal scaling system.
So reality is based on triple spectrals - as frequency - and then spacetime geometry is created from pure frequency that is noncommutative to time. Actually de Broglie figured all this out with his Law of Phase Harmony - as quantum relativity and essentially it is because Einstein's relativity went against the based Pythagorean principle that frequency is inverse to time. This cover up started with Archytas and Plato - as math professor Luigi Borzacchini calls it - a "negative judgment paradox" that is from music theory and was concealed as a "secret of the sect" that is "really astonishing" and "shocking." So I do not believe in math - I tested in the 92% for the ACT in math - in high school but I disagreed with the Pythagorean Theorem due to my music theory training. And so Alain Connes admits that set theory paradox between real and discrete numbers is only solved by noncommutative geometry with time-frequency as the origin of reality.
So he is trying to "solve" the problem - but Roger Penrose argues that this noncommutative time-frequency creates ego awareness from gravity collapsing the non-locality as phase (the 5th dimension) - so essentially our left-brain awareness is essentially the same as gravity (or logarithmic symmetric math) with right-hand technology as inherent to left-brain dominance. For example does Roger Penrose even discuss the ecological crisis? Of course not yet he admits the secret of the Big Bang is that time is dramatically "asymmetric" and that is the secret origin of the Big Bang. Not only is time asymmetric - but negentropy is based on reverse time. Stuart Hameroff has a great paper on free will and that, he says, also explains precognition. I know for a fact that precognition can happen as visions that come true years into the future! I had this happen - I even wrote the dream down. And in now way was it a "statistical possibility."
So as - who was it - well I recommend Eddie Oshins - he studied statistics enough but realized that the secret was noncommutative phase as macroquantum resonance. Eddie Oshins was a quantum physicist at Stanford and he worked closely with math professor Louis Kauffmann with whom I've corresponded. But Oshins taught Wing Chun and Neigong and he realized the secret of the nonwestern neigong (alchemy) is noncommutative macroquantum harmonics - that he called "quantum psychology." He was pissed at how that term got co-opted for the New Age analysis.
So basically I was against math since I knew - I mean I took statistic in college but didn't really try at it. Also for example for chemistry in high school - I figured out the curve and just guessed on all the multiple choice since I knew I cud beat the curve. haha. And then I took quantum physics my first year of college at Hampshire College. Professor Herbert Bernstein emphasized that everyone should take quantum physics first - and he taught quantum entanglement. So Lee Smolins also took his first quantum physics from Herbert Bernstein at Hampshire. And so Lee Smolins has developed a whole theory about the evolution of life in the Universe - it is quite interesting - and all based on the idea that time is very real along with entropy and so time can not be just a symmetric outside parameter. But again how does that effect the ecological crisis? For example math professor Joe Mazur wrote a book called Euclid in the Rainforest - and so I mailed him - saying hey - I studied conservation biology and sustainable development in Costa Rica and (where the book took place) I don't think Western math is "saving" the rainforest as he seemed to imply - quite the contrary.
So I told him my music-math analysis. His brother, as you may know, is a big number theory mathematician - Barry. So anyway Joe Mazur encouraged me to dig deeper and I discovered this quote from David Fowler saying that he thought music theory would explain the connection of continued proportions to irrational numbers. So then I discovered Luigi Borzacchini's research on a math forum. I had this dream of a math-music equation and I mailed it, scribbled down to Borzacchini. He said my math was good but I didn't have any historical proof that it had existed. Anyway Borzacchini says that this music cover up of math then has created a "pre-established deep disharmony" that secretly guides the evolution of science. By the Way - David F. Noble was fired from M.I.T. and the American Historical Association backed up his claim, in court, that he was fired for political reasons. Why? You should read his books, "America by Design" and "Forces of Production" - all his books are amazing. His last book was a follow up to the Religion of Technology.
Let me give you another example. I read the book "Evolving the Alien" by math professor Ian Stewart. He's a chaos mathematician. So then he wrote - well if aliens exist they must be solitary like hermits. I said well qigong masters can already travel out into space and to do so they have to be hermits. Now Ian Stewart also wrote a book on the history of math - as it is all based on symmetry. The beauty of truth is symmetry is what he writes. I completely disagree. The secret of nonwestern music is noncommutative time-frequency. Or what in Daoism is called complementary opposites.
See I made that mistake in my master's thesis - and this physicist Charles Madden wanted to published my thesis as a book - back in 2000. But he kept asking me questions and not understanding. Finally he admitted actually his wife had wanted him to publish my master's thesis. So then I read Charles Madden's book on Music and Fractals and I discovered the error in my master's thesis. Madden points out the yin-yang or Tai Chi symbol can not be a fractal since it is not symmetric! And then of course Steve STrogatz makes the same point about the truth of reality. I corresponded with Strogatz as well - he calls Riemanns Hypothesis a "conspiracy between atom and arithmetic, nature and number" - but what is it really but noncommutative time-frequency! And anyway - Strogatz points out that fractals are not REAL in Nature - because due to their symmetric math, Fractals are a computerized Platonic ideal - they only exist in computer reality. But try to tell that to the New Age scene! haha.
Freemasonry, as David F. Noble, traces back to the 9th Century A.D., as the secret origin of modern science, is based on Platonic philosophy which actually was the origin of symmetric math as irrational numbers and yet secretly from the wrong music theory! haha. So then Ian Stewart wrote about quasi telepathy in electric fish - paddle fish and sharks - and I told him that I had finished my master's degree training with a Daoist qigong master and so I know that telepathy is really. I achieved telepathy! He said he did not believe in natural telepathy but that it could be possible with quantum computing. Then soon after he wrote a letter published in the journal Nature - as a warning about Matrix-style mind control whereby the quantum computing is using our minds as telepathy. We think it's telepathy but actually our minds are being controlled, etc.
I had written this expose already as the Actual Matrix Plan and it's also based on music theory - i discovered this soon after my master's thesis. So the problem is that Ian Stewart is trying to project his symmetric math - but he had an article in the New Scientist questioning if quantum physics is really based on randomness - he admitted that no one really knows. Quantum biology shows otherwise! In fact, if you read the award-winning Life on the Edge quantum biology book - 2016 - it details how the foundation of reality is quantum coherence - not randomness. It's because the Copenhagen interpretation is wrong. Position and Momentum already assumes symmetric math and also a continuum whereas it's actually derived from noncommutative time-frequency energy resonance.
I corresponded with Basil J. Hiley about this - the quantum physics professor. So when time is zero there there is a novel quantum force - a "self-force" that is quantum non-locality or noncommutative phase as infinite frequency. In fact this is reverse time energy as the relativistic mass of light. So when we "turn light around" as in visualization during meditation - then we are going to zero time but in fact there is a hidden momentum that is superluminal, as de Broglie realized - this is the non-local consciousness that is formless awareness and creates negentropic quantum coherence healing energy - called "Yuan qi" or prana or N/om by the original human culture. So de Broglie realized that frequency is to time as momentum is to wavelength. So essentially the superluminal phase creates a momentum energy that reverses the entropy of wavelength as mass.
## Friday, December 8, 2017
### Male Ejaculation Addiction as Nuclear Armageddon: Psychic Energy Vampires
Only a nuke can clean this country. Fucking dirty people.
r u from pak or india???..then i will answer ur dislike..
Dirty and clean , right and wrong ,good or bad is the perception of our own , the positive thinking can change this world then the destruction. Love is the better way.
Fayedd They do not stone women and fuck camels and goats
Sayed do you fuck goats and camels?
I just found this choice exchange on a video about a holy man who lives in Nepal.
It sums up the dilemma of modern males controlled by their Kundabuffer. One male offers a nuke and the other male countries that such males threatening nukes like to ejaculate into goats and camels while stoning women.
Yes - this is the great dilemma of our time. The males with the nukes are motivated by ejaculation addiction and yet act holier-than-thou about people they use their left-brain dominance to dismiss as subhumans (i.e. liberal Democrats or "fucking dirty" or deadbeat poor scum, etc.).
Shaolin Iron Testicles is based on celibacy and qi energy training to suck the balls back into the body.
No one is say in such an environment. If you try to complain about "unwanted touching" or non-consensual physical contact" as several females have done against the supposedly progressive stalwart, comedian promoter of the genocidal attack on Iraq, Al Franken - or even if one "physical assault" at least construed as grabbing someone's back without their consent - as Garrison Keillor did - "rubbing" is what he called it. O.K. so the victims in those cases are attacked by the Liberal Elite technofeminists.
Who are these so-called "anonymous" females complaining about being grabbed or touched without consent? asks the liberal elite technofeminists! How dare these females - who were frozen into "involuntary immobility" voice their complaint and therefore "hurt" the males and be "mean" to the males.
Even worse is if a male complains about another male doing it to him. haha. We live in strange times of high-tech "sophistication" ruled over by modern chimps who can not control their lower bodies and so seek out physical contact without consent to suck off people's energy for ejaculation. These are energy vampires. If they don't get their fix then they feel "victimized" and if that doesn't get what they want then they turn violent. | 2017-12-14 04:29:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.329702228307724, "perplexity": 3023.622888439863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00424.warc.gz"} |
https://www.transtutors.com/questions/9-building-occupancy-and-security-records-the-computer-department-s-share-of-the-de-1384738.htm | # (9) Building occupancy and security—records the computer department’s share of the depreciation,...
(9) Building occupancy and security—records the computer department’s share of the depreciation, maintenance, heat, and security costs of the building; the firm allocates these costs to the department on the basis of square feet occupied.
## Related Questions in Managerial Accounting - Others
• ### Refer to Exercise 7-8 . Required 1. Calculate the allocation ratios to be used under the reciprocal...
January 06, 2016
Refer to Exercise 7-8. Required 1. Calculate the allocation ratios to be used under the reciprocal method for S1 and S2. Carry your calculations out to four digits. (S1 will have three allocation ratios—one each for S2, Cutting, and Sewing. S2 will have three allocation ratios—one
• ### ACCT 2127 TAKEHOME EXAM QUESTIONSVERSION 2 Course Code: ACCT2127Course Description: Accounting for...
October 15, 2018
on the last digit in your student number. Ignore any letter at the end of this number.• If the last digit in your student number is 0, 2, 4, 6, 8 – then you must do Takehome Exam version 1.• If the last digit in your student number is 1, 3, 5, 7, 9 – then you must do Takehome Exam version...
• ### Acct
(Solved) November 23, 2013
a plant asset was purchased on january 1 for $60,000 with an estimated salvage value of$12,000 at the end of its useful life. The current year's depreciation expense is $4,000 calculated on the straight-line basis and the balance of accumulated depreciation account at the end of the year is$20
The useful life is calculated through the formula for depreciation expense using the straight line method. (Cost - Salvage value)/ Useful life = Depreciation expense (60,000 - 12,000) /...
• ### Relevant costs for decision making
(Solved) November 20, 2013
• ### Weighted Average Method, Physical Flow, Equivalent Units, Unit Costs, Cost Assignment Mimasca Inc....
October 17, 2018
Weighted Average Method, Physical Flow, Equivalent Units, Unit Costs, Cost Assignment Mimasca Inc. manufactures various holiday masks. Each mask is shaped from a piece of rubber in the molding department. The masks are then transferred to the finishing department, where they are painted and have e | 2018-11-20 00:15:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1785045713186264, "perplexity": 5066.852449573446}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746171.27/warc/CC-MAIN-20181119233342-20181120015342-00109.warc.gz"} |
http://blog.dancecentral.info/2010/07/standard-ballroom-technique-foot.html | ## Tuesday, July 27, 2010
### Standard Ballroom Technique: Foot Alignment
Step relative to where you are now, not where you will be. For example, in the Waltz Natural Spin Turn the man's first step is forward, and then after placing his right foot he turns 1/4 to his right.
Some might be tempted to turn the body and hips before the step, so that it goes DW against LOD, or the foot is placed turned out. This would take the gentleman right into his partner, rather than stepping past her. | 2021-04-21 01:16:43 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8366707563400269, "perplexity": 3501.7306687461687}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00133.warc.gz"} |
https://wilkelab.org/cowplot/reference/rectangle_key_glyph.html | These functions create customizable legend key glyphs, such as filled rectangles or circles.
rectangle_key_glyph(colour = NA, fill = fill, alpha = alpha,
size = size, linetype = linetype, padding = unit(c(0, 0, 0, 0),
"pt"), color)
circle_key_glyph(colour = NA, fill = fill, alpha = alpha,
size = size, linetype = linetype, padding = unit(c(0, 0, 0, 0),
"pt"), color)
## Arguments
colour, color Unquoted name of the aesthetic to use for the outline color, usually colour, color, or fill. Can also be a color constant, e.g. "red". Unquoted name of the aesthetic to use for the fill color, usually colour, color, or fill. Can also be a color constant, e.g. "red". Unquoted name of the aesthetic to use for alpha, usually alpha. Can also be a numerical constant, e.g. 0.5. Unquoted name of the aesthetic to use for the line thickness of the outline, usually size. Can also be a numerical constant, e.g. 0.5. Unquoted name of the aesthetic to use for the line type of the outline, usually linetype. Can also be a constant, e.g. 2. Unit vector with four elements specifying the top, right, bottom, and left padding from the edges of the legend key to the edges of the key glyph.
## Examples
library(ggplot2)
set.seed(1233)
df <- data.frame(
x = sample(letters[1:2], 10, TRUE),
y = rnorm(10)
)
ggplot(df, aes(x, y, color = x)) +
geom_boxplot(
key_glyph = rectangle_key_glyph(fill = color, padding = margin(3, 3, 3, 3))
)
ggplot(df, aes(x, y, color = x)) +
geom_boxplot(
key_glyph = circle_key_glyph(
fill = color,
color = "black", linetype = 3, size = 0.3,
padding = margin(2, 2, 2, 2)
)
) | 2019-12-10 13:13:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.366886168718338, "perplexity": 9486.260723082141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00116.warc.gz"} |
https://www.electropedia.org/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/d8b5a72bf4ba22d0c1257f320050f59d?OpenDocument | IEVref: 521-01-05 ID: Language: en Status: Standard Term: Maxwell-Boltzmann velocity-distribution law Synonym1: Synonym2: Synonym3: Symbol: Definition: algebraic equation giving the number dN of particles of a non-quantized system, the components of velocity of which are comprised in the intervals (u, u + du), (v, v + dv), (w, w + dw) respectively: $\text{d}N=A\cdot \text{exp}\left[\frac{-m\left({u}^{2}+{v}^{2}+{w}^{2}\right)}{2kT}\right]\text{d}\text{\hspace{0.17em}}u\cdot \text{d}\text{\hspace{0.17em}}v\cdot \text{d}\text{\hspace{0.17em}}w$ where $A=N{\left[\frac{m}{\left(2\pi \cdot kT\right)}\right]}^{3/2}$ N is the total number of particles m is the mass of a particle T is the thermodynamic temperature k is the Boltzmann constant NOTE – dN/N represents the probability that a particle has its components of velocity within the intervals considered. Publication date: 2002-05 Source: Replaces: Internal notes: 2017-06-02: Cleanup - Remove Attached Image 521-01-051.gif 2017-06-02: Cleanup - Remove Attached Image 521-01-052.gif CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5: | 2021-05-08 22:11:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2886919677257538, "perplexity": 9008.69245952951}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00395.warc.gz"} |
https://www.semanticscholar.org/paper/A-Philosophical-Treatise-of-Universal-Induction-Rathmanner-Hutter/ab850f1f49e40de524e6c4803678a10ebbbcb0b6 | # A Philosophical Treatise of Universal Induction
@article{Rathmanner2011APT,
title={A Philosophical Treatise of Universal Induction},
author={Samuel Rathmanner and Marcus Hutter},
journal={Entropy},
year={2011},
volume={13},
pages={1076-1136}
}
• Published 28 May 2011
• Philosophy
• Entropy
Understanding inductive reasoning is a problem that has engaged mankind for thousands of years. This problem is relevant to a wide range of fields and is integral to the philosophy of science. It has been tackled by many great minds ranging from philosophers to scientists to mathematicians, and more recently computer scientists. In this article we argue the case for Solomonoff Induction, a formal inductive framework which combines algorithmic information theory with the Bayesian framework…
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Towards common-sense reasoning via conditional simulation: legacies of Turing in Artificial Intelligence
• Computer Science
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• 2014
This work describes a computational formalism centered around a probabilistic Turing machine called QUERY, which captures the operation of Probabilistic conditioning via conditional simulation and demonstrates how the QUERY abstraction can be used to cast common-sense reasoning as probabilism inference in a statistical model of observations and the uncertain structure of the world that generated that experience.
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Probabilities on Sentences in an Expressive Logic
• Philosophy, Computer Science
J. Appl. Log.
• 2013
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• Psychology
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• 2013
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## References
SHOWING 1-10 OF 70 REFERENCES
An Introduction to Kolmogorov Complexity and Its Applications
• Computer Science
Texts and Monographs in Computer Science
• 1993
The book presents a thorough treatment of the central ideas and their applications of Kolmogorov complexity with a wide range of illustrative applications, and will be ideal for advanced undergraduate students, graduate students, and researchers in computer science, mathematics, cognitive sciences, philosophy, artificial intelligence, statistics, and physics.
Dutch Bookies and Money Pumps
C ONCLUSIVE arguments in philosophy are rare, so any such argument we find we prize. If it is not only conclusive but clever, all the better for it. The Dutch book arguments of the theory of
Universal Intelligence: A Definition of Machine Intelligence
• Computer Science
Minds and Machines
• 2007
A number of well known informal definitions of human intelligence are taken, and mathematically formalised to produce a general measure of intelligence for arbitrary machines that formally captures the concept of machine intelligence in the broadest reasonable sense.
Contemporary debates in philosophy of science
Notes on Contributors.Preface.Introduction: What is the Philosophy of Science?.Part I: Do Thought Experiments Transcend Empiricism?.1. Why Thought Experiments Transcend Empiricism: James Robert Brown
A Computer Scientist's View of Life, the Universe, and Everything
Basic concepts of Kolmogorov complexity theory are applied to the set of possible universes, and perceived and true randomness, life, generalization, and learning in a given universe are discussed.
The Need for Biases in Learning Generalizations
The notion of bias in generalization problems is defined, and it is shown that biases are necessary for the inductive leap.
Complexity-based induction systems: Comparisons and convergence theorems
Levin has shown that if tilde{P}'_{M}(x) is an unnormalized form of this measure, and P( x) is any computable probability measure on strings, x, then \tilde{M}'_M}\geqCP (x) where C is a constant independent of x .
Algorithmic information theory
This article is a brief guide to the field of algorithmic information theory, its underlying philosophy, the major subfields, applications, history, and a map of the field are presented.
No free lunch theorems for optimization
• Computer Science
IEEE Trans. Evol. Comput.
• 1997
A framework is developed to explore the connection between effective optimization algorithms and the problems they are solving. A number of "no free lunch" (NFL) theorems are presented which
Universal Artificial Intellegence - Sequential Decisions Based on Algorithmic Probability
• Marcus Hutter
• Education
Texts in Theoretical Computer Science. An EATCS Series
• 2005
Reading a book as this universal artificial intelligence sequential decisions based on algorithmic probability and other references can enrich your life quality. | 2022-08-15 22:21:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4422801434993744, "perplexity": 2695.6991868677655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572212.96/warc/CC-MAIN-20220815205848-20220815235848-00508.warc.gz"} |
https://www.electro-tech-online.com/threads/esd-damage-to-mclr-pin-of-pic18f26k20.151149/ | ESD damage to MCLR pin of PIC18F26K20?
Status
Not open for further replies.
Flyback
Well-Known Member
Hello,
We have some offline, linear regulator based LED drivers which are dimmable with their PIC18F26K20. (we are using ICSP with Pickit3 to program it)
Some of the PCBs work, most don’t, and some work the first time they are powered up but never thereafter. By “non-working”, I mean they light up but don’t respond to DALI dimming commands sent to the PIC18F26K20.
We have had 1000 of the PCBs made and assembled, and then realised that we forgot to add a 10k pullup resistor from the MCLR pin to Vdd. (MCLR reset is disabled, so its an input pin).
We then had an external modification applied to the PCBs which involves wiring a 10k resistor from MCLR pin to Vdd, and a 10n capacitor from MCLR pin to Vss. I undid this modification on one of the few working boards and then found that stopped it working. –But when I re-did this modification the PCB still did not work.
I am assuming that we have violated the dreaded MCLR pin here… The MCLR pin has no ESD protection diodes due to its use in ICSP, and so am I right in saying that the MCLR pin is supersensitive to ESD, and dies very easily?
(To make matters worse our supply capacitor next to the micro is just a 4n7, 0402 capacitor. However, there are two 10u, 0805 ceramic capacitors about 8mm away from the PIC.)
Do you think we are wasting our time trying to modify these boards? Is the MCLR pin so ultra-sensitive to ESD that everywhere from the PCB assembly house to our factory it is going to get its MCLR pin killed because we forgot to add the 10k resistor from MCLR to Vdd?
PIC18F26K20 datasheet...
Last edited:
cowboybob
Well-Known Member
From page 53 of the Data Sheet:
Did you include R1 on the PCB? Without it, sure looks like bad things can happen...
Pommie
Well-Known Member
If MCLR is disabled in config then no resistor or capacitor are needed. Sounds more like a software bug - an uninitialised variable - a badly timed interrupt - even some SFRs are in a random state at powerup. Double check your code.
Mike.
be80be
Well-Known Member
Like Mike said there more then likely something wrong in the code. If you set mclr as input it shouldn't be a problem. I've used this chip it's a pain in the butt to set but after I got it right it worked fine.
Code:
FOSC(FOSC) = [LP, XT, HS, RC, EC, ECIO6, HSPLL, RCIO6, INTIO67, INTIO7],
FCMEN(FCMEN) = [OFF, ON],
IESO(IESO) = [OFF, ON],
PWRT(PWRT) = [ON, OFF],
BOREN(BOREN) = [OFF, ON, NOSLP, SBORDIS],
BORV(BORV) = [30, 27, 22, 18],
WDTEN(WDTEN) = [OFF, ON],
WDTPS(WDTPS) = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768],
CCP2MX(CCP2MX) = [PORTBE, PORTC],
LPT1OSC(LPT1OSC) = [OFF, ON],
HFOFST(HFOFST) = [OFF, ON],
MCLRE(MCLRE) = [OFF, ON], /first place to look
STVREN(STVREN) = [OFF, ON],
LVP(LVP) = [OFF, ON],
XINST(XINST) = [OFF, ON],
DEBUG(DEBUG) = [ON, OFF],
CP0(CP0) = [ON, OFF],
CP1(CP1) = [ON, OFF],
CP2(CP2) = [ON, OFF],
CP3(CP3) = [ON, OFF],
CPB(CPB) = [ON, OFF],
CPD(CPD) = [ON, OFF],
WRT0(WRT0) = [ON, OFF],
WRT1(WRT1) = [ON, OFF],
WRT2(WRT2) = [ON, OFF],
WRT3(WRT3) = [ON, OFF],
WRTC(WRTC) = [ON, OFF],
WRTB(WRTB) = [ON, OFF],
WRTD(WRTD) = [ON, OFF],
EBTR0(EBTR0) = [ON, OFF],
EBTR1(EBTR1) = [ON, OFF],
EBTR2(EBTR2) = [ON, OFF],
EBTR3(EBTR3) = [ON, OFF],
EBTRB(EBTRB) = [ON, OFF]
Last edited:
Flyback
Well-Known Member
Thanks, it seems a coincidence that i undid the RC mod on the MCLR pin of the working unit, then re-did it, and it then stopped working. -Kind of made me think that my fiddling with the MCLR pin killed it.
BTW we dont have resistor R1 in Cowboybobs kindly given post. We'd have to add it in manually because as discussed we have nothing connected to the MCLR piin on the PCB.....other than a blind track going off to a via used in ISCP which we do with "spring pins" into the five programming vias.
Pommie
Well-Known Member
If your using it for ICSP then you shouldn't have a capacitor. Use a resistor in series with a diode to stop Vpp (12V) from getting to Vdd(5V). Or once debugged turn it off and leave it floating. As I said earlier, triple check your code.
Mike.
be80be
Well-Known Member
I had these on a pcb that jon made call the tap I thought it was the chip is bad it was configure setting. It worked one time and then wouldn't power on. When powered.
If you hook it to your programmer and it tells you there a 18f26k20 hooked to it mclr is still good.
JonSea
Well-Known Member
You may want to review Commandments For Using PICs. This is a collection of problems seen in various forums that may cause weird or intermittent operation. If /MCLR actually is disabled in the code, it may be a red herring and some of tbese common problems the fault.
I'll put my bet on insufficient bypass caps (since we've seen that you've worked hard to save a penny no matter what the cost) or that you've failed to tie all of the micro's ground connetions together.
The linked paper is by yours truly.
MrDEB
Well-Known Member
This is a lesson to be learned, build a breadboard and get it working properly THEN order ONLY the minimum of 10 boards.
Get them working to verify Everything IS CORRECT, THEN order the quantity needed.
Sorry to hear that you may have to scrap 1000 boards.
I as well as most every member here has never had boards made without some issue. I have 10 boards that are now used as drink coasters.
be80be
Well-Known Member
From reading the post. Is probably not that bad check the post part of the board that shows the program pads and mlcr. I had one it's took me a week ot figure out it would run one time if you had the pickit3 hooked up and dead as door nail under it's own power it was the config that was wrong
I think it ran with the pickit3 cause it was pulling the mlcr pin up to vdd
JonSea
Well-Known Member
It is ironic that MrDEB was one of the inspirations in writing Commandments For Using PICs.
One thing MrDEB has never grasped - before you start changing things (or in this case, throw out a batch of boards), understand the cause of the fault. It's extremely premature to condemn a batch of boards to the scrap heap. The /MCLR connection (or lack thereof) will not cause the problems seen unless there is a software error. A hardware fix should not be required as described for the /MCLR pin.
Likely are other hardware issues i described in the document above that have resulted in intermitent behavior.
MrDEB
Well-Known Member
if you read the first post, Flyback mentioned that the pcboards wee made without the 10K resistor.
We have had 1000 of the PCBs made and assembled, and then realised that we forgot to add a 10k pullup resistor from the MCLR pin to Vdd. (MCLR reset is disabled, so its an input pin).
Jonseas Commandments For Using PICs is great reading.
As for condemning the batch of boards Flyback stated that the 10k resistor is lacking but maybe I missed something but never heard of using a 10n cap between MCLR and VCC.
If he had built one board using a minimum of 10 boards and discovered the issue weather it is code or ?? he could save lots of before having 1000 boards made.
It could still be the code. My suggestion, using Commandments For Using PICs, get an led to blink and take it one step at a time. This has helped me when testing out different code sections.
JonSea
Well-Known Member
MrDEB,
/MCLR is disabled. Therefore, a pullup resistor is not required. You have done this in your past "designs" despite it creating complications.
When /MCLR is enabled (probably the more normal case), a 10k pullup resistor is required to keep the chip out of reset.
JonSea
Well-Known Member
MrDEB,
Flytrap is also in a totally different situation than you. These boards are not hobby boards for some party trick. These are part of a commercial product. Furthermore, he didn't design the boards or create the code. He is trying to troubleshoot someone else's work in order to move the product along.
In a production environment, you don't get the take a guess at what the proper board might be, change the board randomly as you try to form a proper IF/THAN statement and try to stumble on a solution that sort of does what you want for Lord knows why and pat yourself on the back for being so smart.
be80be
Well-Known Member
If your using mplab X Post the this file if you can. If your guy did it like this the file is mcc.c
That's made by mcc it show's the configure setting. And the pin_manager.c
They just show how the configure bit's are set and the chip pins are set.
Status
Not open for further replies. | 2021-05-15 11:09:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2258691042661667, "perplexity": 4411.379631938145}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00112.warc.gz"} |
https://vroomlab.wordpress.com/2020/12/ | # That first sip of coffee in the morning
It is a good idea to enjoy a cup of coffee before starting a busy day.
Suppose the coffee fresh out of the pot with temperature $\alpha^{\circ} C$ is too hot, we can immediately add cream to reduce the temperature by $\delta^{\circ} C$ instantly, then wait for the coffee to cool down naturally to $\omega^{\circ} C$ before sipping it comfortably. We can also wait until the temperature of the coffee drops to $(\omega+\delta)^{\circ} C$ first, then add the cream to further reduce it instantly to $\omega^{\circ} C$.
Typically, $\alpha = 90, \omega = 75$, and $\delta = 5$.
If we are in a hurry and want to wait the shortest possible time, should the cream be added right after the coffee is made, or should we wait for a while before adding the cream?
The heat flow from the hot water to the surrounding air obeys Newton’s cooling and heating law, described by the following ordinary differential equation:
$\frac{d}{dt}\theta(t) = k (E-\theta(t))$
where $\theta(t)$, a function of time $t$, is the temperature of the water, $E$ is the temperature of its surroundings, and $k>0$ is a constant depends on the heat transfer mechanism, the contact are with the surroundings, and the thermal properties of the water.
Fig. 1 a place where Newton’s law breaks down
Under normal circumstances, we have
$E \ll \omega < \omega+\delta < \alpha-\delta < \alpha\quad\quad\quad(1)$
Based on Newton’s law, the mathematical model of coffee cooling is:
$\begin{cases} \frac{d}{dt}\theta(t) = k (E-\theta(t)) \\ \theta(0)= \theta_0\end{cases}\quad\quad\quad(2)$
Fig. 2
Solving (2), an initial-value problem (see Fig. 2) gives
$\theta(t) = E + e^{-kt}(\theta_0 - E).$
Therefore,
$t = \frac{\log\left(\frac{E-\theta_0}{E-\theta(t)}\right)}{k}=\frac{\log\left(\frac{\theta_0-E}{\theta(t)-E}\right)}{k}.\quad\quad\quad(3)$
If cream is added immediately (see Fig. 3),
Fig. 3 : cream first
by (3),
$t_1=\frac{\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)}{k}.$
Otherwise (see Fig. 4),
Fig. 4: cream last
$t_2=\frac{\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)}{k}.$
And so,
$t_1-t_2 = \frac{\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)}{k}- \frac{\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)}{k}=\frac{1}{k}\left(\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)-\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)\right)\quad(4)$
Fig. 5
Since
$\frac{(\alpha-\delta)-E}{\omega-E}- \frac{\alpha-E}{(\omega+\delta)-E}=\frac{(\alpha-\delta-E)(\omega+\delta-E)-(\omega-E)(\alpha-E)}{(\omega-E)(\omega+\delta-E)}=\frac{-\delta(\omega+\delta-\alpha)}{(\omega-E)(\omega+\delta-E)}\overset{(1)}{>0}$
implies
$\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)-\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)>0,\quad\quad\quad(5)$
from (4) , we see that
$t_1-t_2 > 0;$
i.e.,
$t_1 > t_2$
Hence,
If we are in a hurry and want to wait the shortest possible time, we should wait for a while before adding the cream!
Exercise-1 Solve (2) without using a CAS.
Exercise-2 Show that $\frac{\alpha-\delta-E}{\omega-E}\cdot\frac{\omega+\delta-E}{\alpha-E} >1.$
# An ODE to Thanksgiving
A turkey is taken from the refrigerator at ${\theta_0}^{\circ} C$ and placed in an oven preheated to $E^{\circ} C$ and kept at that temperature; after $t_1$ minutes the internal temperature of the turkey has risen to ${\theta_1}^{\circ} C$. The fowl is ready to be taken out when its internal temperature reaches ${\theta_2}^{\circ} C$.
Typically, ${\theta_0} = 2, E=200, t_1=30, \theta_1=16, \theta_2=88$.
Determine the cooking time required.
According to Newton’s law of heating and cooling (see “Convective heat transfer“), the rate of heat gain or loss of an object is directly proportional to the difference in the temperatures between the object and its surroundings. This law is best described by the following ODE (Ordinary Differential Equation):
$\frac{d}{dt}{\theta(t)} = k\cdot(E-\theta(t)),\quad\quad\quad(1)$
where $\theta(t), E$ are the temperatures of the object and its surroundings respectively. $k > 0$ is the constant of proportionality.
Fig. 1
We see that (1) has a critical point $\theta^* = E$. Fig. 1 illustrates the fact that depending on its initial temperature, an object either heats up or cools down, trending towards $E$ in both cases.
We formulate the problem as a system of differential-algebraic equations:
$\begin{cases} \frac{d}{dt}{\theta(t)} = k\cdot(E-\theta(t)) \\ \theta(0)=\theta_0\\ \theta(t_1)=\theta_1 \\ \theta(\boxed{t_2}) = \theta_2\end{cases}(2)$
To find the required cooking time, we solve (2) for $t_2$ (see Fig. 2).
Fig. 2
Using Omega CAS Explorer, the typical cooking time is found to be approximately $4$ hours $(3.88 \cdots\approx 4)$
Luise Lange of Woodrow Wilson Junior College once wrote (see ” A Century of Calculus, Part I”, p. 50):
“In many calculus texts problems are formulated too one-sidedly in terms of particular, numerical data rather than in general terms. While pedagogically it may be wise to begin a new type of problem with some numerical examples, it is only the general formulation, and the interpretation of the answer in general terms, which can give insight into the functional relation between the given and the derived data.”
I agree with her wholeheartedly! On encountering a mathematical modeling problem stated with numerical values, I prefer to re-state it using symbols first. Then solve the problem symbolically. The numerical values are substituted for the symbols at the very end.
This post is a case in point, as the problem is re-formulated from page 1005 of Jan Gullberg’s “Mathematics From the Birth of Numbers”:
Exercise-1 Solving (2) without using a CAS.
Exercise-2 Given $\theta_0< \theta_1 < E$, show that
$k = \frac{t\log(\frac{E-\theta_0}{E-\theta_1})}{t_1} > 0.$
Exercise-3 Given $\theta_0 < \theta_1< E$, verify that $\lim\limits_{t\rightarrow \infty} \theta(t) = E$ from
$\theta(t) = -Ee^{ -\frac{t\log(\frac{E-\theta_0}{E-\theta_1})}{t_1} } + \theta_0 e^{-\frac{t\log(\frac{E-\theta_0}{E-\theta_1})}{t_1}} + E.$
Exercise-4 A slice is cut from a loaf of bread fresh from the oven at $180^{\circ} C$ and placed in a room with a constant temperature of $20^{\circ} C$. After 1 minute, the temperature of the slice is $140^{\circ} C$. When has the slice of bread cooled to $32^{\circ} C$?
# Eye Of The Tiger
Evaluate
$\int \frac{x^2e^x}{(x+2)^2}\;dx$
is an exercise accompanied my previous post “Integration by Parts Done Right“. It is a special case of
$\int \frac{x^2e^x}{(x+\boxed{a})^2}\;dx$.
For various $a$, Omega CAS Explorer gives
$\int \frac{x^2}{(x+a)^2}e^x\;dx$
$= \int \left(\frac{x}{x+a}\right)^2 e^x\;dx$
$= \int \left(\frac{x+a-a}{x+a}\right)^2 e^x\;dx$
$=\int \frac{(x+a)^2-2a(x+a)+a^2}{(x+a)^2}e^x\;dx$
$= \int \left(1-\frac{2a}{x+a}+\frac{a^2}{(x+a)^2}\right)e^x\;dx$
$= \int e^x-\frac{2a}{x+a}e^x + \frac{a^2}{(x+a)^2}e^x\;dx$
$= \int e^x\;dx-\int \frac{2a}{x+a}e^x\;dx + \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-2a\int \frac{e^x}{x+a}\;dx + \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-2a\int \frac{1}{x+a}(e^x)'\;dx + \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-2a\left(\frac{1}{x+a}e^x-\int (\frac{1}{x+a})' e^x\;dx\right) + \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-2a\left(\frac{1}{x+a}e^x-\int\frac{-1}{(x+a)^2}e^x\;dx\right)+ \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-2a\left(\frac{1}{x+a}e^x+\int\frac{1}{(x+a)^2}e^x\;dx\right)+ \int \frac{a^2}{(x+a)^2}e^x\;dx$
$= e^x-\frac{2a}{x+a}e^x-2a\int\frac{e^x}{(x+a)^2}\;dx + a^2\int \frac{e^x}{(x+a)^2}\;dx$
$= e^x-\frac{2a}{x+a}e^x-\left(2a-a^2\right)\int\frac{1}{(x+a)^2}e^x\;dx$
$\overset{2a-a^2=0}{=} \frac{x-a}{x+a}e^x$
$= \begin{cases} e^x, a=0\\ \frac{x-2}{x+2}e^x, a=2\end{cases}$ | 2023-03-21 21:31:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 67, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5448670983314514, "perplexity": 655.5096827854152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00032.warc.gz"} |