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https://www.diyaudio.com/community/threads/toroid-output-voltage-for-4780-parallel.117886/ | 1,722,787,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00687.warc.gz | 586,616,714 | 28,598 | # Toroid output voltage for 4780 parallel?
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
#### zdr
I am starting new clone with one 4780 per channel in parallel. I kinda have problems to find 22V RMS voltage output toroids here in Benelux. Conrad has only 18 and 30V for 400VA and 500VA toroids. for 100W RMS per channel, calculation shows 22V at 4 ohm. Should I go for 18V, 30V or something (somewhere) else?
#### zdr
4780 in parallel is capable of 100W RMS per channel, and load should be standard, 4-8 ohm.
#### zdr
After rethinking it a bit, I guess 18V is out of the question. It seems that both versions have very similar output current values, so voltage ceiling might be hit easily with 18V one. On the 30V side, 6.6 Amps RMS current capacity of 500VA 30V trafo should run down to 2.3 ohms (approx) just fine at 100W peak (not sure what are the losses exactly inside those transformers, but that should affect mostly RMS voltage drop, and 30V is lot more to drop from than 18V).
Need to check 4780 output specs too.
#### AndrewT
Hi,
if you want 100W into 4r0 from a parallel pair, then each chip thinks it is driving 50W into 8r0. That is equivalent to 28.3Vpk and 3.5Apk into the resistive load.
Allowing about 4.1V loss through the amplifier indicates that 28.3+4.1=+-33.5Vdc is required from the PSU when delivering full power.
Now make a decision.
Do you design for continuous (say for 1 or 2 seconds) maximum power into this load or will you accept a short term power peak into that load.
For continuous power you must allow for sag in the PSU voltage as the smoothing capacitors discharge to the fully loaded condition. This tends to give more controlled and deeper bass.
Alternatively, the peak power rating can be met from the fully charged capacitors but as the duration of the signal is increased the PSU drops to pass less continuous power to the load. This reportedly tends to favour midrange and treble frequencies.
The sag could vary between 2V for a very high capacity PSU to 6V for a cheap low capacity PSU.
I would suggest you design for +-[33.5+2V] for your first guess at quiescent load. After allowing for rectifier loss and Vdrop supplying quiescent current and for 5% transformer regulation this 35.5Vdc can be supplied by a 23+23Vac, 300VA toroid for powering 100+100W. Use at least +-20mF for the two chipamps if you want deep bass, or twin PSU each with >=+-10mF/chipamp. Personally I would fit +-15 to 20mF for every chipamp.
#### BWRX
You could get the transformer with 30VAC secondaries and unwind some turns to lower the voltage a bit.
#### danielwritesbac
Or you could parallel some LM1875's with the 18v. Sure that's a bit warm, but it'll rock the house quite nicely. They're about 2db behind the big chip, but with firmer driver command, they'll knock the grills off the speakers a bit more quickly.
Sort of a "personal taste" difference between dynamic power or average power.
#### zdr
BWRX said:
You could get the transformer with 30VAC secondaries and unwind some turns to lower the voltage a bit.
Why would I want to wind down the output voltage, unless I would want to use a bit cheaper, 40V caps? If 22V is good, 30V is probably only better. Current capacity will stay the same anyway, so I don't see any other technical reason for this.
#### danielwritesbac
30+30vAC (60vct rated transformer) = 42+42vdc and that's too hot. This will set off the spike protection.
22+22vAC (44vct rated transformer) = 32+32vdc and that's pretty good (somewhere near optimal).
18+18vAC (36vct rated transformer) = 27+27vdc and that's going to "less hot," but make less amplifier power.
In decibels, the end result is probably little to none noticable difference between any of the three options. Remember, it takes 2x the amplifier power to make each +3db to the speakers, and it takes 10x the amplifier power for every 2x difference to the ear.
Bring in speakers and ears, then the 18v secondaries look good.
Hey don't worry, it'll still get the heatsinks plenty hot, especially if its one of the "TF" chips.
#### zdr
Yep, you are right. Not only that, I just noticed that |V-|+|V+|<= 84V, which is right on the edge with 2x42Vdc.
#### danielwritesbac
zdr said:
Yep, you are right. Not only that, I just noticed that |V-|+|V+|<= 84V, which is right on the edge with 2x42Vdc.
I'm right? That's always a nice surprise.
Hey, but the good news is that parallel is a "more amperage, less voltage" approach. So you picked a really good amp project to run on that available 18+18vAC transformer. Kudos!
#### jerome69
Velleman sells toroids transformers 2x24V 500VA. I have two in my amplifier. it's work very well.
#### zdr
Locally, I can find only 500VA 2X39... They do have 300VA 2x21 though.
Hm, thinking out loud on 2x18V - why not use bridge mode instead? If I am reading specs of 4780, max current peak seems to be 7A (10ms). What is this in real life, what is the lowest impedance it could drive bridged? Even with 2X24V in parallel mode, I won't be able to reach 100W RMS per channel at 8ohm.
I would like to have versatile amp that can drive anything from 2-8 ohm at 100W RMS, but it will be much more used around 8ohm than below 4.
#### danielwritesbac
zdr said:
. . . They do have 300VA 2x21 though. . .
2x21, as in a 42vct rated transformer? That and a nice heatsink, should be quite good.
#### zdr
Ok, even with 2X24V AC on the secondary, for 8 ohm load, I won't be able to get 100W RMS per channel in parallel configuration, right? Even if we assume lossless path in between, it would be only around 70W RMS, best case scenario. Am I missing something?
Bridged is another story, it swings double of the PS voltage on the output.
#### AndrewT
ZDR,
you have so little grasp of the electrical processes, I bow out of trying to advise anymore.
#### zdr
Well, you do not need to - it was not directed to you as a pm.
#### danielwritesbac
My apologies that I haven't explored the topic more thoroughly. However, since Peter of Audiosector has decided to use Parallel, then there's probably a good reason. Whatever the reason, it is a good baseline. That is a great place to start. A real-life baseline is necessary in order to acquire references and assistance, so use parallel.
The 300VA 2x21 will get the amplifier (any configuration) somewhat hot. Base your heatsink size selection upon your expectation of the amplifier's performance (demand < heatsink), then it should perform well no matter of 4, 6, or 8 ohms. But, that transformer will push the limits sufficiently.
I too, am out of resources to explore those points farther, so I believe it is time to proceed with the next steps.
The next step is to concentrate on efficiency, such as amplifying only desired signals. Use an established baseline.
The step after, is voicing, to highlight desired signals, which is also a form of efficiency, although not strictly mathmatical or strict adherance to a baseline--its the "more pleasant" nearby values, also called tweaking.
However, its a good time to re-evaluate amplifier selection. If you are looking for a sustained 100 watts audio signal then you actually need a 1000 watt (one thousand watt) amplifier to do it.
Why? At endpoint, the 100 watt amplifier produces only 6db more than a 33 watt amplifier.
Why not? Most speakers lose fidelity at 55 watts, and your 4780 project will have more than enough power to cause this, when desired.
So, hey, go build your amp.
Edit: Check out what can be done with Harbeth Monitor 40, if one builds that style using Tang Band 6-1/2" (92db) as the central wideband driver (midrange), plus adds an efficient large woofer (95db) and small tweeter (93db) to help it. There, using efficiency, is another way for excellent power.
#### zdr
With all due respect to Peter (I am using his great pcbs already, and my first amp is parallel), National specs pdf for LM3886 kinda suggests bridged config for 8 ohm load. This is what Linkwitz is using too in his Plutos woofers, so it's actually not poor ignorant me who is pushing for it. Build is already on the way, I am just trying to understand on which toroid to spend the money. Moreover, Elliott from ESP suggests 400VA for 2x100W and I tend to agree, just cannot find it here (yet).
#### danielwritesbac
zdr said:
With all due respect to Peter (I am using his great pcbs already, and my first amp is parallel), National specs pdf for LM3886 kinda suggests bridged config for 8 ohm load. This is what Linkwitz is using too in his Plutos woofers, so it's actually not poor ignorant me who is pushing for it. Build is already on the way, I am just trying to understand on which toroid to spend the money. Moreover, Elliott from ESP suggests 400VA for 2x100W and I tend to agree, just cannot find it here (yet).
I've often preferred the sound of bridged, although I can't speak for that particular chip, because I don't know.
The transformer that you found 300va and 21+21vAC, sounds like plenty to run the project. Seven ampers! WOW! Some nice fast big caps on the power supply board will ensure success if there's any doubts. That is a bit much voltage, but since it stands in the centerpoint of what you want and what is wise, then just plan on buying some nice size heatsinks. I'd love to see that amp bridged and hear how it does for you. Its just that I couldn't advise you to bridge it because I've never personally tested it that way. So, that's all there was to that.
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. | 2,501 | 9,650 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.923453 |
http://love2learn2day.blogspot.com/2018/01/ | 1,585,597,450,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00285.warc.gz | 119,907,622 | 31,911 | ## Monday, January 22, 2018
### Multiples Practice with Puzzles...and Mythology!
Multiples can prove to be a challenge for students in grades 3-5. In fourth grade, we expect students to be able to determine whether a given whole number between 1-100 is a multiple of a one-digit number. Makes total sense, right? I mean, how hard could it be? Enter classroom...
Teacher: Morning, Johnny!
(Johnny wipes sleep out of eyes.)
Teacher: Think about the number 36. Is it a multiple of 6?
Johnny: Seven?
Ever experience anything similar? To that end, I like to offer extra opportunities for practice. I want something that is...
Hands-on
Visually appealing
Practical for a variety of levels
Perfect for math centers
This new set of Multiples Puzzles gives students ongoing practice with identifying and ordering multiples. Always on the lookout for ways to integrate math and literature, I selected Greek Mythology as a theme. Here's how they work...
1. Copy the puzzles on cardstock, choosing from black/white or color versions. Laminate, as desired.
2. Cut puzzles on horizontal lines into strips.
3. Place each puzzle in an envelope and label with the correct multiple.
4. Place in a math center, assign as homework, or use during a lesson on multiples.
5. As an additional option, as students complete each puzzle, they can note patterns they observe on 100s grids in their own Book of Multiples.
Teachers can differentiate by offering students puzzles that correspond with the practice most needed. Two blank puzzles are included for the creation of challenge puzzles.
Take a closer look at Multiples Puzzles for Greek Myths.
Looking for more multiples practice? My students have enjoyed making flap books & folds (see here and here), and used them as ongoing reference tools in their math journals.
Blogging tips | 396 | 1,828 | {"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} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.920458 |
http://www.wolfram.com/mathematica/new-in-8/new-and-improved-core-algorithms/compute-binormal-probability-over-a-polygon.ru.html | 1,529,563,688,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00210.warc.gz | 526,612,135 | 2,983 | # Вычисление бинормальной вероятности в многоугольнике
Вероятность принадлежности стандартной бинормальной случайной величины с корреляционным коэффициентом правильному восьмиугольнику единичного радиуса, выраженная в терминах T-фунции Овена.
In[1]:= XPolygonRegionFunction[pts_] := Block[{ppts = Join[pts, pts], diff, x, y}, Function[{x, y}, Evaluate[ And @@ Table[diff = ppts[[i + 1]] - ppts[[i]]; Cross[diff].({x, y} - ppts[[i]]) >= 0, {i, Length[pts]}]]]]
In[2]:= XPolygonProbability[pts : {{_, _} ..}, BinormalDistribution[{mu1_, mu2_}, {s1_, s2_}, r_]? DistributionParameterQ] /; PolygonRegionFunction[pts][mu1, mu2] := Module[{tpts = Join[pts, pts[[{1}]]], res = 0, h1, k1, h2, k2, cross, diff, h, a1, a2}, tpts = Transpose[{{1, 1}/Sqrt[2 (1 + r)], {-1, 1}/Sqrt[ 2 (1 - r)]}.((Transpose[pts] - {mu1, mu2})/{s1, s2})]; Do[{{h1, k1}, {h2, k2}} = tpts[[{i, i + 1}]]; cross = Abs[h1 k2 - k1 h2]; If[! PossibleZeroQ[cross], diff = {h2 - h1, k2 - k1}; h = cross/Sqrt[diff.diff]; a1 = {h1, k1}.diff/cross; a2 = {h2, k2}.diff/cross; res += OwenT[h, Max[a1, a2]] - OwenT[h, Min[a1, a2]]], {i, Length[pts]}]; res]
In[3]:= XRegularPolygonProbability[sides_Integer /; sides >= 3, r_] := 1 - FullSimplify[ PolygonProbability[ Most[Table[{Cos[x], Sin[x]}, {x, 0, 2 \[Pi], (2 \[Pi])/sides}]], BinormalDistribution[{0, 0}, {1, 1}, r]], -1 < r < 1]
In[4]:= XWith[{sides = 8, r = 1/3}, Show[Plot3D[ PDF[BinormalDistribution[r], {x, y}], {x, -2, 2}, {y, -2, 2}, PlotStyle -> Opacity[0.5, Hue[.15, 0.2, .8]], Filling -> Axis, FillingStyle -> {Black, Orange}, Lighting -> "Neutral", BoundaryStyle -> None, RegionFunction -> PolygonRegionFunction[ Most[Table[{Cos[x], Sin[x]}, {x, 0, 2 \[Pi], (2 \[Pi])/sides}]]], Mesh -> None, PlotRange -> All, NormalsFunction -> None, PlotPoints -> 50], Plot3D[PDF[BinormalDistribution[r], {x, y}], {x, -3, 3}, {y, -3, 3}, PlotStyle -> Opacity[0.2], Mesh -> 6, MeshStyle -> Gray, PlotRange -> All, PlotPoints -> 50], Boxed -> False, ImageSize -> 580, Epilog -> Inset[Framed[ Pane[Style[ Row[{TraditionalForm[Pr[1/3]] " \[Equal] ", 1, FullSimplify[-4 OwenT[Sqrt[1/2 + 1/Sqrt[2]]/Sqrt[ Sqrt[2] - r], (Sqrt[2] - 1) Sqrt[(1 + r)/(1 - r)]] - 4 OwenT[Sqrt[1/2 + 1/Sqrt[2]]/Sqrt[Sqrt[2] - r], ( Sqrt[2] - 1 - r)/Sqrt[1 - r^2]] - 4 OwenT[Sqrt[1/2 + 1/Sqrt[2]]/Sqrt[ Sqrt[2] + r], (Sqrt[2] - 1) Sqrt[(1 - r)/(1 + r)]] - 4 OwenT[Sqrt[1/2 + 1/Sqrt[2]]/Sqrt[Sqrt[2] + r], ( Sqrt[2] - 1 + r)/Sqrt[1 - r^2]]]}], 11.5], {500, 118}], Background -> LightYellow], {Center, Top}, {Center, Top}]]]
Out[4]= | 1,062 | 2,512 | {"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} | 2.84375 | 3 | CC-MAIN-2018-26 | latest | en | 0.258206 |
https://myclassbook.org/measurement-of-resistance/?shared=email&msg=fail | 1,596,762,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00410.warc.gz | 394,059,410 | 14,228 | # Measurement of resistance [Learn ALL methods easily]
Hello friends, in this blog article we will learn different methods for the measurement of resistance. We will first see, what are the different types of methods in short and will explain the same in details in separate blog articles.
## Measurement of Resistance
Before learning different types of methods for measurement of resistance, we will first see the classification of resistance. From the point of view of measurement, we have three types of resistances as follows:
1. Low resistance: < 1Ω
2. Medium resistance: 1Ω to 100,000Ω
3. High resistance: >100,000Ω
### A. Measurement of Low resistance
This is a very popular method for measurement of medium resistances since instruments required for this method are usually available in the laboratory.
2. Kelvin’s double bridge method: The Kelvin double bridge is the modification of the Wheatstone bridge and provides greatly increased accuracy in measurement of low-value resistance.
3. Potentiometer method: The potentiometer is extensively used for calibration of voltmeters and ammeters and has, in fact, become the standard for the calibration of these instruments.
4. Ducter
### B. Measurement of Medium resistance
1. Ammeter-Voltmeter method
2. Substitution method: This is a more accurate method than ammeter-voltmeter method. The accuracy of this method is greatly affected if the emf of the battery changes during the time of readings.
3. Wheatstone bridge method: Another method of measuring the value of a resistor is the Wheatstone bridge. This device sets up a parallel resistor system that measures the differences in voltage between two legs of a circuit. If there is a difference of voltage between the branches it will be detected by the galvanometer.
4. Ohmmeter method
### C. Measurement of High resistance
1. Direct deflection method: Direct deflection method for measurement of high resistance
2. Loss of Charge method: In the loss of charge method unknown resistance is connected in parallel with the capacitor and electrostatic voltmeter. The capacitor is initially charged to some suitable voltage by means of a battery of voltage V and then allowed to discharge through the resistance.
3. Megohm bridge: Megohm bridge is another important method for measurement of high resistances. It has one three terminal high resistance located in one arm of the bridge.
4. Megger: We know that the ratiometer ohmmeters may be designed to cover a wide range of resistances. The principle of ratiometer ohmmeters is particularly adapted to application in portable instruments measuring insulation resistance. This principle forms the basis of insulation testing instrument known as Meggar.
I hope you liked this article. If you have any queries, please use below comments section for the same. Please like our facebook page and subscribe to our newsletter for future updates. Have a nice day! tada…
## One thought on “Measurement of resistance [Learn ALL methods easily]”
1. 😜😜 says:
Tq for ur good information about resistance | 630 | 3,073 | {"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} | 3.203125 | 3 | CC-MAIN-2020-34 | latest | en | 0.898194 |
http://pygments.org/demo/6783740/ | 1,566,229,794,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00323.warc.gz | 165,145,558 | 3,404 | # Demo entry 6783740
lkjlkjl
Submitted by anonymous on Feb 20, 2019 at 00:53
Language: Python 3. Code size: 1.2 kB.
def partition(arr,low,high):
i = ( low-1 ) # index of smaller element
pivot = arr[high] # pivot
for j in range(low , high):
# If current element is smaller than or
# equal to pivot
if arr[j] <= pivot:
# increment index of smaller element
i = i+1
arr[i],arr[j] = arr[j],arr[i]
arr[i+1],arr[high] = arr[high],arr[i+1]
print("returning: " + str(arr[i +1]))
return ( i+1 )
# The main function that implements QuickSort
# arr[] --> Array to be sorted,
# low --> Starting index,
# high --> Ending index
# Function to do Quick sort
def quickSort(arr,low,high):
print(arr)
if low < high:
# pi is partitioning index, arr[p] is now
# at right place
pi = partition(arr,low,high)
# Separately sort elements before
# partition and after partition
quickSort(arr, pi+1, high)
quickSort(arr, low, pi-1)
# Driver code to test above
arr = [6,10, 7, 11, 9, 1, 5]
n = len(arr)
quickSort(arr,0,n-1)
print ("Sorted array is:")
for i in range(n):
print ("%d" %arr[i]),
This snippet took 0.00 seconds to highlight.
Back to the Entry List or Home. | 371 | 1,169 | {"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} | 2.6875 | 3 | CC-MAIN-2019-35 | latest | en | 0.67254 |
https://paperwriting.help/fin55-week-8chapter-20-problems-3a-c-5a-c-8a-c-9a-d/ | 1,619,166,005,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00559.warc.gz | 541,389,320 | 11,291 | # Fin55. week 8:chapter 20: problems 3(a-c), 5(a-c), 8(a-c), 9(a-d),
See CORRECT ATTACHMENT. Please complete in EXCEL FORMAT ONLY.
LOOK AT THE HOMEWORK ASSISTANCE FOR HELP !
Week 8 Homework – Chapter 20: Problems 3(a-c), 5(a-c), 8(a-c), 9(a-d), and 10(a-d)
Problem 3(a-c)
3. Suppose that an investor holds a share of Sophia common stock, currently valued at \$50. She is concerned that over the next few months the value of her holding might decline, and she would like to hedge that risk by supplementing her holding with one of three different derivative positions, all of which expire at the same point in the future:
(1) A short position in a forward with a contract price of \$50
(2) A long position in a put option with an exercise price of \$50 and a front-end premium expense of \$3.23
(3) A short position in a call option with an exercise price of \$50 and a front-end premium receipt of \$5.20
a. Using a table similar to the following, calculate the expiration date value of the investor’s combined (i.e., stock and derivative) position. In calculating net portfolio value, ignore the time differential between the initial derivative expense or receipt and the terminal payoff.
b. For each of the three hedge portfolios, graph the expiration date value of her combined position on the vertical axis, with potential expiration date share prices of Sophia stock on the horizontal axis.
c. Assuming that the options are priced fairly, use the concept of put-call parity to calculate the zero-value contract price (i.e., F0,t) for a forward agreement on Sophia stock. Explain why this value differs from the \$50 contract price used in Part a and Part b.
Expiration Date
Sophia Stock Price
Expiration Date
Derivative Payoff
Combined Terminal Position Value
25
30
35
40
45
50
55
60
65
70
75
Chapter 20: Problem 5(a-c)
5. The common stock of Company XYZ is currently trading at a price of \$42. Both a put and a call option are available for XYZ stock, each having an exercise price of \$40 and an expiration date in exactly six months. The current market prices for the put and call are \$1.45 and \$3.90, respectively. The risk-free holding period return for the next six months is 4 percent, which corresponds to an 8 percent annual rate.
a. For each possible stock price in the following sequence, calculate the expiration date payoffs (net of the initial purchase price) for the following positions:
(1) buy one XYZ call option, and
(2) short one XYZ call option:
20, 25, 30, 35, 40, 45, 50, 55, 60
Draw a graph of these payoff relationships, using net profit on the vertical axis and potential expiration date stock price on the horizontal axis. Be sure to specify the prices at which these respective positions will break even (i.e., produce a net profit of zero).
b Using the same potential stock prices as in Part a, calculate the expiration date payoffs and profits (net of the initial purchase price) for the following positions: (1) buy one XYZ put option, and (2) short one XYZ put option. Draw a graph of these relationships, labeling the prices at which these investments will break even.
c. Determine whether the \$2.45 difference in the market prices between the call and put options are consistent with the put-call parity relationship for European-style contracts.
Chapter 20: Problem 8(a-c)
8. As an option trader, you are constantly looking for opportunities to make an arbitrage transaction (i.e., a trade in which you do not need to commit your own capital or take any risk but can still make a profit). Suppose you observe the following prices for options on DRKC Co. stock: \$3.18 for a call with an exercise price of \$60, and \$3.38 for a put with an exercise price of \$60. Both options expire in exactly six months, and the price of a six-month T-bill is \$97.00 (for face value of \$100).
a. Using the put-call-spot parity condition, demonstrate graphically how you could synthetically recreate the payoff structure of a share of DRKC stock in six months using a combination of puts, calls, and T-bills transacted today.
b. Given the current market prices for the two options and the T-bill, calculate the no-arbitrage price of a share of DRKC stock.
c. If the actual market price of DRKC stock is \$60, demonstrate the arbitrage transaction you could create to take advantage of the discrepancy. Be specific as to the positions you would need to take in each security and the dollar amount of your profit.
Chapter 20: Problem 9(a-d)
9. You are currently managing a stock portfolio worth \$55 million and you are concerned that over the next four months equity values will be flat and may even fall. Consequently, you are considering two different strategies for hedging against possible stock declines: (1) buying a protective put, and (2) selling a covered call (i.e., selling a call option based on the same underlying stock position you hold). An over-the-counter derivatives dealer has expressed interest in your business and has quoted the following bid and offer prices (in millions) for at-the-money call and put options that expire in four months and match the characteristics of your portfolio:
Bid
Call
\$2.553
\$2.573
Put
1.297
1.317
a. For each of the following expiration date values for the unhedged equity position, calculate the terminal values (net of initial expense) for a protective put strategy.
35, 40, 45, 50, 55, 60, 65, 70, 75
b. Draw a graph of the protective put net profit structure in Part a, and demonstrate how this position could have been constructed by using call options and T-bills, assuming a risk-free rate of 7 percent.
c. For each of these same expiration date stock values, calculate the terminal net profit values for a covered call strategy.
d. Draw a graph of the covered call net profit structure in Part c, and demonstrate how this position could have been constructed by using put options and T-bills, again assuming a risk-free rate of 7 percent.
Chapter 20: Problem 10(a-d)
10. The common stock of Company XLT and its derivative securities currently trade in the market at the following prices and contract terms:
Price (\$)
Excise Price (\$)
Stock XLT
21.50
Call Option on Stock XLT
5.50
21.00
Put Option on Stock XLT
4.50
21.00
Both of these options will expire in 91 days from now; and the annualized yield for the 91-day Treasury bill is 3.0 percent.
a. Briefly explain how to construct a synthetic Treasury bill position.
b. Calculate the annualized yield for the synthetic Treasury bill in Part a using the market price data provided.
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Topic: Parametric Pell's Equation and circle
Replies: 17 Last Post: Apr 9, 2009 2:11 AM
Messages: [ Previous | Next ]
quasi Posts: 9,080 Registered: 7/15/05
Re: Parametric Pell's Equation and circle
Posted: Apr 6, 2009 10:53 PM
On Mon, 6 Apr 2009 19:39:44 -0700 (PDT), marcus_bruckner@yahoo.com
wrote:
>On Apr 3, 11:30 pm, JSH <jst...@gmail.com> wrote:
>> On Apr 3, 9:44 pm, quasi <qu...@null.set> wrote:
>>
>> > On Fri, 3 Apr 2009 19:40:15 -0700 (PDT), JSH <jst...@gmail.com> wrote:
>> > >So, the world has its first known rational parameterization of
>> > >ellipses. And, oh yeah, of course, hyperbolas with the regular D>1.
>>
>> > >It's amazing how close their parametric solutions are to the circle
>>
>> > >How was this result not known for thousands of years?
>>
>> > Rational parameterizations for ellipses and hyperbolas have been known
>> > for hundreds of years. I think it was Euler who developed the basic
>>
>> Well yes. That part of what I said was wrong.
>>
>> However, this particular rational parameterization is new.
>>
>> > method for finding such parameterizations.
>>
>> > Just do a Google search.
>>
>>
>> > <http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/curves/rational.html>
>>
>> > Here's another:
>>
>> > <http://en.wikipedia.org/wiki/Algebraic_curve>
>>
>> > quasi
>>
>>
>> And I did reach claiming no rational parameterizations had been done
>> before for ellipses and hyperbolas.
>>
>> There never has been a *single* parameterization that handles circles,
>> ellipses, and hyperbolas just by sign before:
>>
>> With x^2 - Dy^2 = 1
>>
>> the parameterization:
>>
>> y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2]
>>
>> and
>>
>> x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2]
>>
>> where f_2 is an integer factor of D-1, you get a parameterization for
>> hyperbolas with integer D>1, for circles with D=-1, and for ellipses
>> with D<-1.
>>
>> It is a unifying result. And a world's first.
>>
>> James Harris
>
> Move over little dog. I have just discovered the most
>freaking nifty result ever seen on the planet. I have discovered
>that lots of numbers can be expressed as sums of squares in more
>than one way.
>
> Here is my beautiful awe-inspiring example: N = 850.
>
> Note that 850 = 3^2 + 29^2
> = 121^2 + 27^2
> = 15^2 + 25^2.
>
> Yep, that's right. Count 'em. THREE freaking different
>ways! Even more astounding, 845 can ALSO be written as a sum
>of two squares in three different ways! And so can 725!
>
> No one else on earth has ever discovered this. I know
>because I did a Google search on the phrase "My discovery
>on sums of squares". Nothing! PROOF that nobody has
>ever thought of this before!
>
> IT'S HUGE!!!!
> It's NIFTY!!!!
> It's STUNNING!!!!
>
> I am the king. Lying jealous welfare queen mathematicians
>will try to deny this. I am a MAJOR DISCOVERER. I am a problem
>solver.
>
> I must rush off now and add this to my blog. Oh wait. I don't
>have a blog. I must rush off and create a blog. Plus I must
>write this up for submission to the Annals.
>
> And THINK, people. THINK how many avenues for questions this
>opens up. Let s(N) be the number of ways that N can be
>written as a sum of two squares. What is the limit of
>s(N) / log(N) as N --> infinity? NO ONE KNOWS! Given M, what
>is the expected value of the smallest N such that s(N) = M ?
>NO ONE KNOWS! Think of all the questions! This is BRAINSTORMING
>MATH, people! Cutting edge! Nifty! Freaking amazing! Beautiful!
>
> Ow! I just got a terrible bruise in right between my shoulder
>blades where I was patting myself on the back too hard! Oh,
>the perils of Genius! But you freaking little lying LOSERS
>will never know what THAT feels like! A whole new area of
>research just opened up and you worthless inert fools just
>sit there! Only a Genius like JSH is going to appreciate
>this. He might even have a way to connect it to Pell's
>Equation! Wow!
Haha -- that's a good one!
quasi
Date Subject Author
4/3/09 jstevh@gmail.com
4/3/09 waste of time
4/3/09 jstevh@gmail.com
4/3/09 quasi
4/4/09 rossum
4/4/09 jstevh@gmail.com
4/6/09 marcus_b
4/6/09 quasi
4/7/09 jstevh@gmail.com
4/8/09 Patrick Karl
4/9/09 jstevh@gmail.com
4/4/09 David Bernier
4/4/09 jstevh@gmail.com
4/4/09 jstevh@gmail.com
4/4/09 willo_thewisp@hotmail.com
4/4/09 jstevh@gmail.com
4/5/09 rossum
4/5/09 jstevh@gmail.com | 1,403 | 4,439 | {"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} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.889302 |
http://xrpp.iucr.org/Bb/ch2o5v0001/sec2o5o6o2/ | 1,558,634,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257316.10/warc/CC-MAIN-20190523164007-20190523190007-00247.warc.gz | 386,152,041 | 6,971 | International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli
International Tables for Crystallography (2010). Vol. B, ch. 2.5, pp. 367-368 | 1 | 2 |
## Section 2.5.6.2. 3D reconstruction in the general case
B. K. Vainshteinc and P. A. Penczekg
#### 2.5.6.2. 3D reconstruction in the general case
| top | pdf |
In the general case of the 3D reconstruction of from projections , the projection vector occupies arbitrary positions on the projection sphere (Fig. 2.5.6.2). First, let us consider the case of a 2D function and its ray transform . We introduce an operation of backprojection b, which is stretching along each 1D projection (Fig. 2.5.6.3). When the result is integrated over the full angular range of projections , we obtain the projection synthesis defined asHowever, the backprojection operator is not the inverse of a 2D ray transform, as the resulting image b is blurred by the point-spread function (Vainshtein, 1971):By noting that the Fourier transform of is and by using the convolution theorem , we obtain the backprojection-filtering' inversion formula:The more commonly used filtered-backprojection' inversion is based on the 2D version of the central section theorem (2.5.6.8):where . With this in mind, can be related to its ray transform by evaluating the Fourier transform of in polar coordinates:In three dimensions, the backprojection stretches each 2D projection along the projection direction . A 3D synthesis is the integral over the hemisphere (Fig. 2.5.6.2)Thus, in three dimensions the image b obtained using the backprojection operator is blurred by the point-spread function (Vainshtein, 1971). It is possible to derive inversion formulae analogous to (2.5.6.11) and (2.5.6.13).
Figure 2.5.6.3 | top | pdf |(a) Formation of a backprojection function; (b) projection synthesis (2.5.6.9) is a superposition of these functions.
The inversion formulae demonstrate that it is possible to invert the ray transform for continuous functions and for a uniform distribution of projections. In electron microscopy, the projections are never distributed uniformly in three dimensions. Indeed, a uniform distribution is not even desirable, as only certain subsets of projection directions are required for the successful inversion of a 3D ray transform, as follows from the central section theorem (2.5.6.8). In practice, we always deal with sampled data and with discrete, random and nonuniform distributions of projection directions. Therefore, the inversion formulae can be considered only as a starting point for the development of the numerical (and practical!) reconstruction algorithms. According to (2.5.6.10) and (2.5.6.14), a simple backprojection step results in reconstruction that corresponds to a convolution of the original function with a point-spread function that depends only on the distribution of projections, but not on the structure itself. Taking into account the linearity of the backprojection operation, one has to conclude that for any practically encountered distribution of projections it should be possible to derive the corresponding point-spread function and then, using either deconvolution or Fourier filtration (with a `weighting function'), arrive at a good approximation of the structure. This observation forms the basis of the weighted backprojection algorithm (Section 2.5.6.5). Similarly, the central section theorem gives rise to direct Fourier inversion algorithms (Section 2.5.6.6). Nevertheless, since the data are discrete, the most straightforward methodology is to discretize and approach the reconstruction problem as an algebraic problem (Section 2.5.6.4).
### References
Vainshtein, B. K. (1971). Finding the structure of objects from projections. Sov. Phys. Crystallogr. 15, 781–787. | 881 | 3,798 | {"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} | 2.84375 | 3 | CC-MAIN-2019-22 | latest | en | 0.826026 |
https://community.fabric.microsoft.com/t5/Desktop/Add-rows-to-table-according-to-grouping-in-different-table/m-p/390955 | 1,716,808,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00248.warc.gz | 148,742,291 | 124,665 | cancel
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Helper I
## Add rows to table according to grouping in different table
Hi,
I would like to add rows to one table ( or combine two tables) according to group number from another table.
These are what I have:
Table 1 Name Part A x1 B y1 B x1
Table 2 Group Part 1 x1 1 x2 2 y1 2 y2
This is what I am looking for
Final Table Name Part Group A x1 1 A x2 1 B y1 2 B y2 2 B x1 1 B x2 1
I already tried merge/append or DAX code (Union or OuterJoin) and not succeed.
Do you have any recommendation? 🙂
Thank you!
Gozde
1 ACCEPTED SOLUTION
Community Champion
@Gozde
Please see the revised file attached
Regards
Zubair
12 REPLIES 12
Community Champion
Hi @Gozde
```New Table =
VAR Mytable1 =
Table1,
"My Group", CALCULATE ( FIRSTNONBLANK ( Table2[Group], 1 ) )
)
VAR JoinTables =
GENERATE (
SELECTCOLUMNS ( mytable1, "Name", [Name], "My Group", [My Group] ),
FILTER ( Table2, Table2[Group] = [My Group] )
)
RETURN
SELECTCOLUMNS ( JoinTables, "Name", [Name], "Group", [Group], "Part", [Part] )```
Regards
Zubair
Community Champion
@Gozde
Regards
Zubair
Helper I
Thanks for quick response. It helped me alot.
It is working in the example code. But, not in my real big data, becauase I have some parts in Table 1 which are missing in Table 2 (not have the group). How can I keep these parts? What should I change in code?
Thank you 🙂
Gozde
Community Champion
@Gozde
Hi,
Could you give me some data and expected results to work with?
Regards
Zubair
Helper I
Yes, definetely!
Table 1 Name Part A x1 A z1 B y1 B x1 C k1
Table 2 Group Part 1 x1 1 x2 2 y1 2 y2
This is my expectation:
Final Table Name Part Group A x1 1 A x2 1 B y1 2 B y2 2 B x1 1 B x2 1 A z1 C k1
Thank you again!
Regards,
Gozde
Community Champion
@Gozde
Try with this code
```New Table =
VAR Mytable1 =
Table1,
"My Group", CALCULATE ( FIRSTNONBLANK ( Table2[Group], 1 ) )
)
VAR JoinTables =
GENERATEALL (
SELECTCOLUMNS (
mytable1,
"Name", [Name],
"My Group", [My Group],
"My Part", [Part]
),
FILTER ( Table2, Table2[Group] = [My Group] )
),
"Final Part", IF ( ISBLANK ( [Part] ), [My Part], [Part] )
)
RETURN
SELECTCOLUMNS (
JoinTables,
"Name", [Name],
"Group", [Group],
"Part", [Final Part]
)```
Regards
Zubair
Community Champion
@Gozde
Please see the revised file attached
Regards
Zubair
Helper I
Yes, changing generate to generateall solve the issue!
Thank you very much,
Gozde
Community Champion
@Gozde
Actually in addition to GenerateAll there was a subtle issue with Parts Column as well
Regards
Zubair
Helper I
Nope, When I add your last code, again parts without group number are disappering at final table.
I tried your code with the example:
This is the output.
Name Part Group A x1 2 A x2 2 B x1 2 B x2 2 C x1 1 C x2 1
Community Champion
@Gozde
Did you see the revised file I sent?
Regards
Zubair
Helper I
Thanks a lot!
I found my syntax error 🙂
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Top Kudoed Authors | 1,003 | 3,414 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.811326 |
https://www.convert-me.com/en/convert/percent/onefifth/onefifth-to-centi.html | 1,601,039,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00451.warc.gz | 774,744,305 | 22,920 | ## Convert Tithe, One Fifth Or .2 (1/5, Fractions) to Centi (c, Metric Prefixes)
tithe, one fifth or .2 (1/5)
Fractions
centi (c)
Metric Prefixes
This page features online conversion from tithe, one fifth or .2 to centi. These units belong to different measurement systems. The first one is from Fractions. The second one is from Metric Prefixes.
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Units: half or .5 (1/2) / one third or .(3) (1/3) / quart, one forth or .25 (1/4) / tithe, one fifth or .2 (1/5) / one sixth or .1(6) (1/6) / one seventh or .142857 (1/7) / one eights or .125 (1/8) / one ninth or .(1) (1/9) / one tenth or .1 (1/10) / one sixteenth or .0625 (1/16) / one thirty-second or .03125 (1/32)
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Hint: You don't have to click "Convert Me" button every time. Hitting Enter or Tab key after typing in your value also triggers the calculations. | 966 | 2,869 | {"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} | 2.953125 | 3 | CC-MAIN-2020-40 | longest | en | 0.579668 |
http://mathhelpforum.com/advanced-algebra/179208-finding-basis-r-3-a-print.html | 1,526,979,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00041.warc.gz | 181,826,036 | 3,027 | # Finding a basis of R^3
• May 1st 2011, 10:07 PM
tangibleLime
Finding a basis of R^3
Find a basis of R^3 defined by the equation,
$\displaystyle 2x_{1}+3x_{2}+x_{3} = 0$
I took the above equation and solved for x3, assuming that was saying that x3 could be created using -2 copies of x1 minus 3 copies of x2...
$\displaystyle 2x_{1}+3x_{2} = -x_{3}$
$\displaystyle -2x_{1}-3x_{2} = x_{3}$
Then I solved this like I solve for a kernel.
$\displaystyle \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}=\begin{bmatrix}t\\ s\\ -2t-3s\end{bmatrix}= t\begin{bmatrix}1\\ 0\\ -2\end{bmatrix}+s\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$
Which gave me a result of,
$\displaystyle \begin{bmatrix}1\\ 0\\ -2\end{bmatrix},\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$
$\displaystyle \begin{bmatrix}-3\\ 2\\ 0\end{bmatrix},\begin{bmatrix}-1\\ 0\\ 2\end{bmatrix}$
Where did I go wrong? Any help is extremely appreciated. Thanks!
• May 1st 2011, 10:59 PM
FernandoRevilla
Quote:
Originally Posted by tangibleLime
Where did I go wrong? Any help is extremely appreciated. Thanks!
You are completely right. Take into account that the solution is not unique.
• May 1st 2011, 11:00 PM
tangibleLime
Ah, thank you for the clarification!
• May 2nd 2011, 04:29 PM
HallsofIvy
Also note that the problem should be "Find a basis of the subspace of R^3 defined by the equation,
$\displaystyle 2x_1+ 3x_2+ x_3= 0$"
Since that is one equation constraining the three variables, the subspace has dimension 3- 1= 2 and so your basis should contain 2 vectors, just as you say. | 552 | 1,545 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.843268 |
https://www.crazy-numbers.com/en/25582 | 1,709,470,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00835.warc.gz | 713,688,124 | 6,280 | Discover a lot of information on the number 25582: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 25582
Is 25582 a prime number? No
Is 25582 a perfect number? No
Number of divisors 4
List of dividers 1, 2, 12791, 25582
Sum of divisors 38376
Prime factorization 2 x 12791
Prime factors 2, 12791
## How to write / spell 25582 in letters?
In letters, the number 25582 is written as: Twenty-five thousand five hundred and eighty-two. And in other languages? how does it spell?
25582 in other languages
Write 25582 in english Twenty-five thousand five hundred and eighty-two
Write 25582 in french Vint-cinq mille cinq cent quatre-vingt-deux
Write 25582 in spanish Veinticinco mil quinientos ochenta y dos
Write 25582 in portuguese Vinte e cinco mil quinhentos oitenta e dois
## Decomposition of the number 25582
The number 25582 is composed of:
2 iterations of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
2 iterations of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
Other ways to write 25582
In letter Twenty-five thousand five hundred and eighty-two
In roman numeral
In binary 110001111101110
In octal 61756
In US dollars USD 25,582.00 (\$)
In euros 25 582,00 EUR (€)
Some related numbers
Previous number 25581
Next number 25583
Next prime number 25583
## Mathematical operations
Operations and solutions
25582*2 = 51164 The double of 25582 is 51164
25582*3 = 76746 The triple of 25582 is 76746
25582/2 = 12791 The half of 25582 is 12791.000000
25582/3 = 8527.3333333333 The third of 25582 is 8527.333333
255822 = 654438724 The square of 25582 is 654438724.000000
255823 = 16741851437368 The cube of 25582 is 16741851437368.000000
√25582 = 159.94374010883 The square root of 25582 is 159.943740
log(25582) = 10.149644258159 The natural (Neperian) logarithm of 25582 is 10.149644
log10(25582) = 4.4079344945996 The decimal logarithm (base 10) of 25582 is 4.407934
sin(25582) = -0.011021595159053 The sine of 25582 is -0.011022
cos(25582) = -0.99993926037542 The cosine of 25582 is -0.999939
tan(25582) = 0.01102226464727 The tangent of 25582 is 0.011022 | 784 | 2,542 | {"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} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.697513 |
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Post subject: who ownes the zebraPostPosted: Fri Mar 08, 2013 1:12 pm
Level 1 Cleric
Joined: Mon Aug 31, 2009 11:21 am
Posts: 12
i want to solve this by computer :
Quote:
''1.There are five houses.
''2.The Englishman lives in the red house.
''3.The Spaniard owns the dog.
''4.Coffee is drunk in the green house.
''5.The Ukrainian drinks tea.
''6.The green house is immediately to the right of the ivory house.
''7.The Old Gold smoker owns snails.
''8.Kools are smoked in the yellow house.
''9.Milk is drunk in the middle house.
''10.The Norwegian lives in the first house.
''11.The man who smokes Chesterfields lives in the house next to the man with the fox.
''12.Kools are smoked in the house next to the house where the horse is kept.
''13.The Lucky Strike smoker drinks orange juice.
''14.The Japanese smokes Parliaments.
''15.The Norwegian lives next to the blue house
''who ownes the zebra ?
i cant use brute force
this has pow( 5! , 5 ) = 24883200000 posebilities
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cxiu diferas el tio respondas cxiu samvaloras
[ esperanto for : everybody is diferent therefore everybody is equal ]
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Post subject: Re: who ownes the zebraPostPosted: Fri Mar 08, 2013 8:19 pm
Grand Optimizer
Joined: Sun Oct 16, 2011 3:09 pm
Posts: 367
Location: Here (where else?)
Indeed, a much better strategy is to use the clues directly, and use logic deduction, much like we do.
eg Englishman <-> red house
-> not Englishman <-> green house
not Englishman <-> ivory house
not Englishman <-> yellow house
not Englishman <-> blue house
not Spaniard <-> red house
not Ukrainian <-> red house
not Japanese <-> red house
not Norwegian <-> red house
It's a fun puzzle to figure out how to program deduction
_________________
My project: Messing about in FreeRCT, dev blog, and IRC #freerct at oftc.net
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Post subject: Re: who ownes the zebraPostPosted: Fri Mar 08, 2013 9:34 pm
Dexterous Droid
Joined: Wed Aug 18, 2004 7:40 pm
Posts: 3871
Location: South Africa
This would be much easier to solve with pen and paper. But I'll take a crack at the groundwork.
I would start out with identifying the kinds of logical relations.
• x implies y
• x is next to y
• x to the right of y
• x is in house number i
Then one needs to identify what house attributes are available and their possible values.
• colour: red green blue yellow ivory
• pet: dog snail fox horse [implied zebra]
• smoke: Old Gold, Kools, Chesterfield, Lucky Strike, Parliaments
• drink: coffee tea milk OJ [implied gin & tonic, obviously]
• nationality: English, Ukrainian, Spanish, Norwegian, Japanese
From the standpoint of the puzzle, houses are an array of containers for the above attributes (house number, colour, etc).
The first step is to put down things that can't change, the rules which specify house number should be applied as well as any that can be directly applied (in this case just that the blue house is next to the Norwegian, there aren't any other direct relations).
Code:
`number | 1 | 2 | 3 | 4 | 5 |colour | | blue | | | |pet | | | | | |smoke | | | | | |drink | | | milk | | |nationality | norwegian | | | | |`
The puzzle doesn't have any other rules we can apply directly so we need to eliminate the possible options in each cell until only 1 option remains. Referring to our logical relations, the ways of eliminating options are:
• x implies y: (1) eliminate x from all houses that are not y, vice-versa for y.
• x is next to y: (1) where either x or y is known, eliminate the other from all cells not next to the known one.
• x to the right of y: (1) eliminate x where a value to the left is known to be not y. eliminate y where a value to the right is known to be not x. (2) eliminate y from all the right-most houses, eliminate x from all the left-most houses
There might be a bunch of others but from here you can infer
1. house 1 has the norwegian, so by the first rule we can eliminate all the options involving nationality. -red, -tea, -parliament
2. eliminate green from house 1 because it's the left-most house.
3. eliminate ivory from hosue 1 because it's to the left of a known blue house.
3. the only colour left is yellow, so the first house is yellow.
If you carry on eliminating options then you should arrive at a solution. You just need to make sure that the list of methods to eliminate options is complete. So the general algorithm is:
Code:
`while there are still unkown values in houses{ check if any of the rules can be applied directly, without inference eliminate potential values from all the houses until one of the attributes only has 1 possible value left}`
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Whatever the mind can conceive and believe, it can achieve
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Post subject: Re: who ownes the zebraPostPosted: Fri Mar 29, 2013 5:38 am
Technomaniac
Joined: Sun Dec 05, 2004 11:27 am
Posts: 3252
Location: Sydney, Australia
Just looking at the clues, there is no indication that anyone owns a zebra. Are you sure this isn't a trick question?
If you are looking for an interesting programming language for this sort of thing, have a look at prolog. It should let you do something which will mean that each clue translates fairly directly into the language, and then you can ask it questions (like who owns the zebra) and it will give you an answer (or tell you that it can't work out the answer from the clues)
_________________
Trying is the first step towards failure
b
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Post subject: Re: who ownes the zebraPostPosted: Fri Mar 29, 2013 1:06 pm
Dexterous Droid
Joined: Wed Aug 18, 2004 7:40 pm
Posts: 3871
Location: South Africa
Andy wrote:
Just looking at the clues, there is no indication that anyone owns a zebra. Are you sure this isn't a trick question?
You're right. It should have stated, "given that someone owns a zebra, who is it?". There's also an unkown drink.
_________________
Whatever the mind can conceive and believe, it can achieve
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Post subject: Re: who ownes the zebraPostPosted: Thu Apr 04, 2013 9:09 am
Postronaut
Joined: Sun Mar 10, 2013 3:32 pm
Posts: 45
i own a zebra..
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Post subject: Re: who ownes the zebraPostPosted: Thu Apr 04, 2013 5:28 pm
BANNED
Joined: Sun Jun 24, 2012 12:49 am
Posts: 504
Hey, I was going to say that! I own the Zebra!
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Post subject: Re: who ownes the zebraPostPosted: Fri Apr 05, 2013 5:43 pm
411 Operator
Joined: Sun Aug 05, 2012 9:32 pm
Posts: 420
From what i can tell by looking at the clues, there are 5 houses, and 4 animals mentioned, if you figure out who owns what, you will find out who owns the zebra
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Did you ever wonder, how time works?
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Post subject: Re: who ownes the zebraPostPosted: Sun Apr 07, 2013 9:17 pm
411 Operator
Joined: Sun Aug 05, 2012 9:32 pm
Posts: 420
So did you manage to make the algorithm for finding out who owns the zebra? I am really thrilled to know the answer I think its the Norwegian, he seems like the type that would steal a zebra from a ZOO
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Did you ever wonder, how time works?
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Jump to: Select a forum ------------------ Forums Forum Rules and Posting Guidelines Wiki Discussion Help Content Issues Game Programming Discussion C and C++ Game Programming Java Game Programming Language Agnostic Programming .NET Game Programming Mobile Game Programming Web-Based Game Programming Other Languages OpenGL Development Direct X Development Game Development Discussion Game Design Game Media Off-Topic Discussion Announcements Off-Topic Community Projects News | 2,284 | 8,693 | {"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} | 2.6875 | 3 | CC-MAIN-2015-40 | longest | en | 0.863337 |
https://www.tutsmake.com/how-to-check-whether-a-number-is-fibonacci-or-not-in-python/ | 1,656,540,818,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00114.warc.gz | 1,111,134,658 | 27,606 | # Python to Check Whether Given Number is Fibonacci or Not
To check whether a given number is Fibonacci or not in python; In this tutorial, how to check whether a given number is a Fibonacci number or not in python.
## How to Check Whether a Number is Fibonacci or Not
See the following python program to check number is Fibonacci or not; as shown below:
```# python program to check if given
# number is a Fibonacci number
import math
# function to check perferct square
def checkPerfectSquare(n):
sqrt = int(math.sqrt(n))
if pow(sqrt, 2) == n:
return True
else:
return False
# function to check Fibonacci number
def isFibonacciNumber(n):
res1 = 5 * n * n + 4
res2 = 5 * n * n - 4
if checkPerfectSquare(res1) or checkPerfectSquare(res2):
return True
else:
return False
# main code
num = int(input("Enter an integer number: "))
# checking
if isFibonacciNumber(num):
print ("Yes,", num, "is a Fibonacci number")
else:
print ("No,", num, "is not a Fibonacci number")
```
Output
```Enter an integer number: 5
Yes, 5 is a Fibonacci number ``` | 275 | 1,052 | {"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} | 2.96875 | 3 | CC-MAIN-2022-27 | longest | en | 0.506757 |
https://discuss.codechef.com/t/kitepattern-editorial/98896 | 1,653,609,166,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00252.warc.gz | 271,997,085 | 4,223 | # KITEPATTERN -Editorial
Author: rocknot
Tester: yogesh_5326
Editorialist:rocknot
EASY
Implementation
# PROBLEM:
Yash is very fond of flying kites and today is kite flying competition in their village. Yash wants to take participate in this competition but there are some rules for this competition. Anyone who wants to participate in this competition should have a unique card with a design of kite printed on it. Most of the card’s he had were already used by some one so he cannot use them. Yash wants to participate in this competition but so he decided to use his programming skills and build his own unique card.
You are given an integer N
and you need to print the kite pattern according to the pattern given below. for N
=4,
`````` *
^**
***
^ ^****
*****
^ ^****
***
^**
*
``````
# HINT:
Observe the right angle triangle/odd or even places.
# SOLUTIONS:
Solution
``````#include <iostream>
using namespace std;
int main(){
int T;
cin>>T;
while(T--){
int N,inc=0;
cin>>N;
for(int i=0;i<=N;i++){
for(int j=0;j<inc;j++){
if(i%2!=0&&j%2==0){
cout<<"^";
}else{
cout<<" ";
}
}
for(int j=0;j<=i;j++){
cout<<"*";
}
cout<<endl;
inc++;
}
inc-=2;
for(int i=N;i>=0;i--,inc--){
for(int j=0;j<inc;j++){
if(i%2==0&&j%2==0){
cout<<"^";
}else{
cout<<" ";
}
}
for(int j=0;j<i;j++){
cout<<"*";
}
if(i==0){
}else{
cout<<endl;
}
}
}
return 0;
``````
} | 402 | 1,354 | {"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} | 3.625 | 4 | CC-MAIN-2022-21 | latest | en | 0.861469 |
https://question4everyone.com/what-is-the-additive-inverse-of-the-polynomial/ | 1,708,470,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00874.warc.gz | 501,238,375 | 24,765 | # What is the additive inverse of the polynomial?
What is the additive inverse of the polynomial? –6×3 + 4×2 – 4x
Walkthrough: Additive inverse is also known as negation, -1 times the original. We negate every term.
6x³ – 4x² + 4x Walkthrough: The additive inverse is the value added to make the expression equal to zero. Given – 6x³ + 4x² – 4x, then the additive inverse is – ( – 6x³ + 4x² – 4x) = 6x³ – 4x² + 4x
B) 6x³ – 4x² + 4x Explanation: The additive inverse is defined as what we add to a number/expression to get a result of “zero” Examples: -5 is the additive inverse of 5 because -5 + 5 = 0 – x² is the additive inverse of x² as x² – x² = 0 So to get the additive inverse just multiply the number/expression you have by -1 The given expression is: -6x³ + 4x² – 4x We get the inverse additive as follows: -1 (-6x³ + 4x² – 4x) -(-6x³) – (4x²) – (-4x) 6x³ – 4x² + 4x Hope this helped
B. Walkthrough: Since the additive inverse of an expression or number is an expression or number which after addition yields the additive identity “0”. Let x be the additive inverse of the polynomial, Per the above statement, therefore the additive inverse of the polynomial is Option B is correct.
the third walkthrough:
the answer is B)6×3 – 4×2 + 4x
the answer is B)6x^3 – 4x^2 + 4x
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On a coordinate plane, 2 triangles are shown. Triangle A B C has points (negative 4, 4), (negative 4, 1), and (0, 1). Triangle W R
In a coordinate plane, 2 triangles are represented. Triangle ABC has points (4.4 minus), (4.1 minus) and (0.1). Triangle WRS has points (0, minus 1), (1.75, 1.5), (5, minus 1). In the diagram, △ABC ≅ △WRS. What is the scope of △WRS? 10 pieces 11 pieces 12 pieces 13 piecesAnswer 1Walkthrough (C)12 units: Triangle WRS has points W(0, -1), R(1.75, 1.5), and S(5, -1). WRS triangle perimeteranswer 212 units I just passed the good luck test (:answer 3The third option is correct. Step by step explanation: As we have shown that △ABC ≅ △WRS So the perimeter of △ABC is given by AB+BC+CA here, AB = 3 units BC = 4 units AC = x By the Pythagorean theorem, we get that Then the perimeter of △ABC is 3+4+5=12 units So the perimeter of △WRS is also 12 units because they are similar triangles. So the third option is correct.answer 4The answer is 12 Walkthrough:Answer 6The correct answer is 12 units (in edgenuity)Answer 7The correct answer in e2020 is C. 12 unitsAnswer 8c Step by step explanation: go all the way
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Choose the correct answer. you see a used car you wish to buy. the dealer quotes you a price of \$9,550. you have a blue book quotation
Choose the right answer. you see a used car that you want to buy. the dealer quotes a price of \$9,550. you have a blue book quote of \$8,400 for the same model and year. what is the profit margin used by the dealer? That's to say %.Answer 1If we assume the reseller cost is \$8400, the profit margin is ..(9550 -8400)/8400 * 100% ≈ 13.7%answer 2Walkthrough: Dealer Quoted Price = \$9,550 Blue Book Price = \$8,400 Price Difference = Dollars Now markup here means the value added to the cost price of the car to make a profit. So the markup price is \$1150. And we're going to find \$1,150 as a percentage of \$8,400. It becomes : %. Therefore, the answer is approximately 13.69%. | 1,280 | 4,691 | {"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} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.879509 |
http://physics.stackexchange.com/questions/73450/why-does-the-wind-direction-vary-locally/73451 | 1,469,681,612,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00014-ip-10-185-27-174.ec2.internal.warc.gz | 198,751,413 | 20,769 | # Why does the wind direction vary locally?
If the air is flowing with the earth, as it rotates, how do we feel a breeze and strong winds at times, as everyone has experienced, the wind shifts as if from many directions.
-
I'm not an expert on the weather, but the basic mechanism behind any wind (which is just air moving) is pressure difference. The atmosphere as a whole gets rotated along with the earth, but locally pressure differences within the (lower) atmosphere cause winds to blow. The air moves from a higher density to a lower density. The air density distribution and the magnitude of the differences are then what cause different direction/strength winds. – Wouter Aug 7 '13 at 9:48
I may be misreading the question, but I think you're asking about meterolgy and how wind is formed in the earth's atmosphere?
If you're talking about wind you usually feel that is due to the motion of air around systems of high and low pressure. These pressure differences are formed by variations in temperature and weather, and the gradient between them helps to determine the strength of the wind felt.
This may be a simple view where there is just one area of high pressure and one of low pressure. If I read the second part correctly that is to do with the increased complexity of the atmosphere and pressure systems where you experience a less obvious wind direction. It's possible for the wind to move in all directions including downwards based on where the air is being pulled towards. It may be to do with interference as well such as the wind hitting buildings or other objects that will cause changes in the air patterns around them, there may also be more local heating effects that cause pressure variations leading to more complex air flows. In a large, open, area the wind will be more-or-less in one direction. An example would be on a ship at sea where from personal experience the wind comes from a specific direction. Compare that to a city where the wind deflecting off buildings will cause all sorts of local 'winds', though the overall propagation would be towards the global low perssure.
On a larger scale the rotation of the earth and heating by the sun is what drives the most high-level air currents. There is a pattern of these currents that is fairly well established from when ships used the wind for power. The effect of the earth on the wind is observed as the windspeeds are far higher in the upper atmosphere with thinner air and less interference from the ground (when very close).
-
Two words: turbulence & thermals. The wind you feel on the surface of the earth is subject to a lot of friction from the ground. The NOAA wind models running 24/7 on supercomputers don't even try to predict wind less than 10m from the surface. That interface between the free flowing fluid of the sky and the static chaotic surface of the earth is very non-linear and no one even tries to model it on a global scale.
Just like when you walk behind a fence the wind dies or maybe even reverses, the land causes strange things to happen to the wind (modeled as a fluid) flowing over the surface. This can happen on big scales where the turbulence caused by the fence can be felt and measured 200 to 300 feet away. It is a very common rule of thumb with racing sailboats to try to slow other boats by giving them "dirty air" by disturbing the airflow up wind of them.
On a larger scale the heating and cooling of the earth drives the winds much harder than the earths rotation. This happens in several different bands of the atmosphere and you can see the general summary of the the wind sources in different parts of the world in this diagram.
If the earth didn't rotate you would see this winds just circulate from N to South straight up and down. However the rotation of the planet causes the the slight shifts (called the Coriolis effect) depending on whether the wind is descending (cooling down) from the upper atmosphere or ascending (heating up). The surface winds are shown drawn on the planet, but note those circular rings of rising and falling air are really 3-d and surround the globe. Hopefully this will help you visualize how the winds are driven around the planet.
The High and Low-pressure systems (mentioned by Folau) form due to downward (High) or upward (Low) motion through the troposphere, the atmospheric layer where weather occurs. Take a close look the Hadley cell from the previous picture.
As the hot air rises it cools, losing moisture; it is then transported poleward where it descends, creating the high-pressure area. This is part of the Hadley cell circulation and is known as the subtropical ridge or subtropical high, and is strongest in the summer months.
All said and done the heating/cooling and turbulence makes it very difficult to predict local winds very accurately. However the trends, average strengths and general directions have radically improved with NOAA's improved models and larger supercomputers over the last 15 years.
In fact NOAA is getting a big upgrade -- doubling their computation power so maybe things will get more and more predictable...?
-
If you calculate the angular momentum of a parcel of air in the atmosphere due to the earth's rotation, you find it typically much larger than the angular momentum of the parcel due to the zonal winds. And if you measure the total angular momentum of the atmosphere due to the zonal winds you observe that it has a negative correlation with the angular momentum of the solid Earth. It is observed that when the angular momentum of the atmosphere increases a little, the length of day changes. The total angular momentum of Earth, atmosphere, and ocean is virtually conserved. The atmospheric momentum of the winds (over the entire surface) never becomes as large as the momentum due to rotation because the atmosphere and solid earth are mechanically coupled. If pressure differences (weather)cause winds, friction will transfer momentum between the earth and the atmosphere and so keep the atmosphere rotating with the earth.
- | 1,224 | 6,064 | {"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} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.964575 |
http://slideplayer.com/slide/699563/ | 1,542,225,677,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114212731-00010.warc.gz | 325,842,401 | 18,884 | # SERIES RESISTOR CIRCUIT. THE EQUIVALENT SERIES RESISTANCE IS : Req = R 1 + R 2 + R 3 THE CURRENT OF THE CIRCUIT IS : I = U / (R 1 + R 2 + R 3 ) THE EQUIVALENT.
## Presentation on theme: "SERIES RESISTOR CIRCUIT. THE EQUIVALENT SERIES RESISTANCE IS : Req = R 1 + R 2 + R 3 THE CURRENT OF THE CIRCUIT IS : I = U / (R 1 + R 2 + R 3 ) THE EQUIVALENT."— Presentation transcript:
SERIES RESISTOR CIRCUIT
THE EQUIVALENT SERIES RESISTANCE IS : Req = R 1 + R 2 + R 3 THE CURRENT OF THE CIRCUIT IS : I = U / (R 1 + R 2 + R 3 ) THE EQUIVALENT SERIES RESISTANCE IS : Req = R 1 + R 2 + R 3 THE CURRENT OF THE CIRCUIT IS : I = U / (R 1 + R 2 + R 3 )
T SERIES RESISTOR CIRCUIT When we connect resistors in a series circuit the current which flows through each one of them its of the same value I. We can find the equivalent resistance if we add the values of all resistors of the circuit. That means that the equivalent resistance is of greater value than the rest of them separately.
SERIES RESISTOR CIRCUIT we can find the source voltage: E = U1 + U2 + U3. The source voltage is divided in as many parts as the number of the resistors in the circuit. We measure the greater part of the voltage on the resistor with the greater value.
Similar presentations | 392 | 1,250 | {"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} | 4.03125 | 4 | CC-MAIN-2018-47 | latest | en | 0.740059 |
https://root.cern.ch/doc/v620/SimplexBuilder_8cxx_source.html | 1,695,385,325,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00640.warc.gz | 556,483,299 | 11,804 | ROOT Reference Guide
SimplexBuilder.cxx
Go to the documentation of this file.
1// @(#)root/minuit2:$Id$
2// Authors: M. Winkler, F. James, L. Moneta, A. Zsenei 2003-2005
3
4/**********************************************************************
5 * *
6 * Copyright (c) 2005 LCG ROOT Math team, CERN/PH-SFT *
7 * *
8 **********************************************************************/
9
12#include "Minuit2/MnFcn.h"
13#include "Minuit2/MinimumSeed.h"
16
17#if defined(DEBUG) || defined(WARNINGMSG)
18#include "Minuit2/MnPrint.h"
19#endif
20
21
22namespace ROOT {
23
24 namespace Minuit2 {
25
26
27//#define DEBUG 1
29class MnStrategy;
30FunctionMinimum SimplexBuilder::Minimum(const MnFcn& mfcn, const GradientCalculator&, const MinimumSeed& seed, const MnStrategy&, unsigned int maxfcn, double minedm) const {
31 // find the minimum using the Simplex method of Nelder and Mead (does not use function gradient)
32 // method to find initial simplex is slightly different than in the orginal Fortran
33 // Minuit since has not been proofed that one to be better
34
35#ifdef DEBUG
36 std::cout << "Running Simplex with maxfcn = " << maxfcn << " minedm = " << minedm << std::endl;
37#endif
38
39 const MnMachinePrecision& prec = seed.Precision();
42
43 unsigned int n = x.size();
44 double wg = 1./double(n);
45 double alpha = 1., beta = 0.5, gamma = 2., rhomin = 4., rhomax = 8.;
46 double rho1 = 1. + alpha;
47 //double rho2 = rho1 + alpha*gamma;
48 //change proposed by david sachs (fnal)
49 double rho2 = 1. + alpha*gamma;
50
51
52 std::vector<std::pair<double, MnAlgebraicVector> > simpl; simpl.reserve(n+1);
53 simpl.push_back(std::pair<double, MnAlgebraicVector>(seed.Fval(), x));
54
55 unsigned int jl = 0, jh = 0;
56 double amin = seed.Fval(), aming = seed.Fval();
57
58 for(unsigned int i = 0; i < n; i++) {
59 double dmin = 8.*prec.Eps2()*(fabs(x(i)) + prec.Eps2());
60 if(step(i) < dmin) step(i) = dmin;
61 x(i) += step(i);
62 double tmp = mfcn(x);
63 if(tmp < amin) {
64 amin = tmp;
65 jl = i+1;
66 }
67 if(tmp > aming) {
68 aming = tmp;
69 jh = i+1;
70 }
71 simpl.push_back(std::pair<double, MnAlgebraicVector>(tmp, x));
72 x(i) -= step(i);
73 }
74 SimplexParameters simplex(simpl, jh, jl);
75
76#ifdef DEBUG
77 std::cout << "simplex initial parameters - min " << jl << " " << amin << " max " << jh << " " << aming << std::endl;
78 for (unsigned int i = 0; i < simplex.Simplex().size(); ++i)
79 std::cout << " i = " << i << " x = " << simplex(i).second << " fval(x) = " << simplex(i).first << std::endl;
80#endif
81
82 double edmPrev = simplex.Edm();
83 int niterations = 0;
84 do {
85 jl = simplex.Jl();
86 jh = simplex.Jh();
87 amin = simplex(jl).first;
88 edmPrev = simplex.Edm();
89
90#ifdef DEBUG
91 std::cout << "\n\nsimplex iteration: edm = " << simplex.Edm()
92 << "\n--> Min Param is " << jl << " pmin " << simplex(jl).second << " f(pmin) " << amin
93 << "\n--> Max param is " << jh << " " << simplex(jh).first << std::endl;
94
95 // std::cout << "ALL SIMPLEX PARAMETERS: "<< std::endl;
96 // for (unsigned int i = 0; i < simplex.Simplex().size(); ++i)
97 // std::cout << " i = " << i << " x = " << simplex(i).second << " fval(x) = " << simplex(i).first << std::endl;
98#endif
99
100 // trace the iterations (need to create a MinimunState although errors and gradient are not existing)
101 if (TraceIter() ) TraceIteration(niterations, MinimumState(MinimumParameters(simplex(jl).second,simplex(jl).first), simplex.Edm(), mfcn.NumOfCalls()) );
102 if (PrintLevel() > 1) MnPrint::PrintState(std::cout,simplex(jl).first, simplex.Edm(),mfcn.NumOfCalls(),"Simplex: Iteration # ", niterations);
103 niterations++;
104
105
106 MnAlgebraicVector pbar(n);
107 for(unsigned int i = 0; i < n+1; i++) {
108 if(i == jh) continue;
109 pbar += (wg*simplex(i).second);
110 }
111
112 MnAlgebraicVector pstar = (1. + alpha)*pbar - alpha*simplex(jh).second;
113 double ystar = mfcn(pstar);
114
115#ifdef DEBUG
116 std::cout << " pbar = " << pbar << std::endl;
117 std::cout << " pstar = " << pstar << " f(pstar) = " << ystar << std::endl;
118#endif
119
120 if(ystar > amin) {
121 if(ystar < simplex(jh).first) {
122 simplex.Update(ystar, pstar);
123 if(jh != simplex.Jh()) continue;
124 }
125 MnAlgebraicVector pstst = beta*simplex(jh).second + (1. - beta)*pbar;
126 double ystst = mfcn(pstst);
127#ifdef DEBUG
128 std::cout << "Reduced simplex pstst = " << pstst << " f(pstst) = " << ystst << std::endl;
129#endif
130 if(ystst > simplex(jh).first) break;
131 simplex.Update(ystst, pstst);
132 continue;
133 }
134
135 MnAlgebraicVector pstst = gamma*pstar + (1. - gamma)*pbar;
136 double ystst = mfcn(pstst);
137#ifdef DEBUG
138 std::cout << " pstst = " << pstst << " f(pstst) = " << ystst << std::endl;
139#endif
140
141 double y1 = (ystar - simplex(jh).first)*rho2;
142 double y2 = (ystst - simplex(jh).first)*rho1;
143 double rho = 0.5*(rho2*y1 - rho1*y2)/(y1 - y2);
144 if(rho < rhomin) {
145 if(ystst < simplex(jl).first) simplex.Update(ystst, pstst);
146 else simplex.Update(ystar, pstar);
147 continue;
148 }
149 if(rho > rhomax) rho = rhomax;
150 MnAlgebraicVector prho = rho*pbar + (1. - rho)*simplex(jh).second;
151 double yrho = mfcn(prho);
152#ifdef DEBUG
153 std::cout << " prho = " << prho << " f(prho) = " << yrho << std::endl;
154#endif
155 if(yrho < simplex(jl).first && yrho < ystst) {
156 simplex.Update(yrho, prho);
157 continue;
158 }
159 if(ystst < simplex(jl).first) {
160 simplex.Update(ystst, pstst);
161 continue;
162 }
163 if(yrho > simplex(jl).first) {
164 if(ystst < simplex(jl).first) simplex.Update(ystst, pstst);
165 else simplex.Update(ystar, pstar);
166 continue;
167 }
168 if(ystar > simplex(jh).first) {
169 pstst = beta*simplex(jh).second + (1. - beta)*pbar;
170 ystst = mfcn(pstst);
171 if(ystst > simplex(jh).first) break;
172 simplex.Update(ystst, pstst);
173 }
174#ifdef DEBUG
175 std::cout << "End loop : edm = " << simplex.Edm() << " pstst = " << pstst << " f(pstst) = " << ystst << std::endl;
176#endif
177 } while( (simplex.Edm() > minedm || edmPrev > minedm ) && mfcn.NumOfCalls() < maxfcn);
178
179 jl = simplex.Jl();
180 jh = simplex.Jh();
181 amin = simplex(jl).first;
182
183 MnAlgebraicVector pbar(n);
184 for(unsigned int i = 0; i < n+1; i++) {
185 if(i == jh) continue;
186 pbar += (wg*simplex(i).second);
187 }
188 double ybar = mfcn(pbar);
189 if(ybar < amin) simplex.Update(ybar, pbar);
190 else {
191 pbar = simplex(jl).second;
192 ybar = simplex(jl).first;
193 }
194
195 MnAlgebraicVector dirin = simplex.Dirin();
196 // Scale to sigmas on parameters werr^2 = dirin^2 * (up/edm)
197 dirin *= sqrt(mfcn.Up()/simplex.Edm());
198
199#ifdef DEBUG
200 std::cout << "End simplex " << simplex.Edm() << " pbar = " << pbar << " f(p) = " << ybar << std::endl;
201#endif
202
203
204 MinimumState st(MinimumParameters(pbar, dirin, ybar), simplex.Edm(), mfcn.NumOfCalls());
205
206 if (PrintLevel() > 1)
207 MnPrint::PrintState(std::cout,st,"Simplex: Final iteration");
208 if (TraceIter() ) TraceIteration(niterations, st);
209
210 if(mfcn.NumOfCalls() > maxfcn) {
211#ifdef WARNINGMSG
212 MN_INFO_MSG("Simplex did not converge, #fcn calls exhausted.");
213#endif
214 return FunctionMinimum(seed, std::vector<MinimumState>(1, st), mfcn.Up(), FunctionMinimum::MnReachedCallLimit());
215 }
216 if(simplex.Edm() > minedm) {
217#ifdef WARNINGMSG
218 MN_INFO_MSG("Simplex did not converge, edm > minedm.");
219#endif
220 return FunctionMinimum(seed, std::vector<MinimumState>(1, st), mfcn.Up(), FunctionMinimum::MnAboveMaxEdm());
221 }
222
223 return FunctionMinimum(seed, std::vector<MinimumState>(1, st), mfcn.Up());
224}
225
226 } // namespace Minuit2
227
228} // namespace ROOT
#define MN_INFO_MSG(str)
Definition: MnPrint.h:110
double sqrt(double)
const MnAlgebraicVector & Gstep() const
class holding the full result of the minimization; both internal and external (MnUserParameterState) ...
void TraceIteration(int iter, const MinimumState &state) const
const MnAlgebraicVector & Vec() const
MinimumSeed contains the starting values for the minimization produced by the SeedGenerator.
Definition: MinimumSeed.h:31
Definition: MinimumSeed.h:49
const MinimumParameters & Parameters() const
Definition: MinimumSeed.h:47
const MnMachinePrecision & Precision() const
Definition: MinimumSeed.h:51
MinimumState keeps the information (position, Gradient, 2nd deriv, etc) after one minimization step (...
Definition: MinimumState.h:29
Wrapper class to FCNBase interface used internally by Minuit.
Definition: MnFcn.h:33
double Up() const
Definition: MnFcn.cxx:35
unsigned int NumOfCalls() const
Definition: MnFcn.h:43
determines the relative floating point arithmetic precision.
double Eps2() const
eps2 returns 2*sqrt(eps)
static void PrintState(std::ostream &os, const MinimumState &state, const char *msg, int iter=-1)
Definition: MnPrint.cxx:58
API class for defining three levels of strategies: low (0), medium (1), high (>=2); acts on: Migrad (...
Definition: MnStrategy.h:27
virtual FunctionMinimum Minimum(const MnFcn &, const GradientCalculator &, const MinimumSeed &, const MnStrategy &, unsigned int, double) const
class describing the simplex set of points (f(x), x ) which evolve during the minimization iteration ...
MnAlgebraicVector Dirin() const
void Update(double, const MnAlgebraicVector &)
const std::vector< std::pair< double, MnAlgebraicVector > > & Simplex() const
double beta(double x, double y)
Calculates the beta function.
Double_t x[n]
Definition: legend1.C:17
const Int_t n
Definition: legend1.C:16
double gamma(double x)
VecExpr< UnaryOp< Fabs< T >, VecExpr< A, T, D >, T >, T, D > fabs(const VecExpr< A, T, D > &rhs)
VSD Structures.
Definition: StringConv.hxx:21
static constexpr double second
Definition: first.py:1 | 2,990 | 9,691 | {"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} | 3.15625 | 3 | CC-MAIN-2023-40 | latest | en | 0.456725 |
https://www.pinxoplatja.com/calorie-content-of-products/workout-calories-counter.html | 1,660,041,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00346.warc.gz | 834,708,491 | 11,722 | # Workout calories counter
## How do you calculate calories burned during exercise?
The equation for the Exercise Calories Burned Calculator is: Duration of physical activity in minutes × (MET × 3.5 × your weight in kg) / 200 = Total calories burned .
## How many calories do I burn exercising?
How many calories does physical activity use (burn)?
Approximate calories used ( burned ) by a 154-pound man
Weight training (general light workout ) 220 110
Stretching 180 90
VIGOROUS physical activities: In 1 hour In 30 minutes
Running/ jogging (5 mph) 590 295
## Are Gym calorie counters accurate?
While some types of cardio machines are more accurate than others, none of them will be 100% accurate . In general, they overestimate your calorie burn by 15% to 20% because they can’t account for all the individual factors involved in calorie -burning.
## How many calories do you lose in a 10 minute ab workout?
Calories Burned: A female weighing 140 pounds would burn an estimated 70 calories in the 10 minute of this routine. A male of 185 pounds would burn roughly 86. Printable Abs Workout – The following is a list of the exercises you are going to be doing and the muscles they target.
## How can I burn 1000 calories a day?
Be well hydrated and have a small breakfast. Walk on a treadmill at an incline for an hour. I am 6′ and 200 lbs, and when I walk at 4 mph and a 6% incline, I burn about 1,000 calories an hour. So one way to reach your goal is to do this for 5 hours (adjusting for your calorie burn based on your own research).
You might be interested: Calories in 3 egg whites
## How can I burn 500 calories a day?
Burn 500 Calories Working Out At-Home (30-Min Workouts) Running. High-intensity interval training (HIIT) Cycling. Plyometrics. Climbing stairs. Dancing. Housework. Bodyweight workouts.
## Is 500 calories good to burn a day?
Well, based on the fact that one pound of fat equates to around 3,500 calories , a daily workout of one hour where you burn 500 calories , should help you lose a pound a week, as long as your diet is sensible.
## What exercises burn the most fat?
Running, walking, cycling and swimming are just a few examples of some cardio exercises that can help burn fat and kick-start weight loss. Summary Studies show that the more aerobic exercise people get, the more belly fat they tend to lose.
## How can I burn 2000 calories a day?
The most important part is to get up and get moving. Bicycling. An hour of intense cycling can burn up to 850 calories per hour. Swimming. Put on your suit, swim for an hour and burn 700 calories . Running. Running reduces stress and helps alleviate depression. Zumba. Aerobics.
## Is 400 calories burned a good workout?
A general rule is to aim to burn 400 to 500 calories , five days a week during your workouts . Remember, the number of calories you burn in a workout depends on your weight, sex, age and many other factors, but this number is a good starting place. 5 дней назад
## How many calories do you burn on a treadmill at 15 incline?
How Many Extra Calories Incline Really Burns
Pace (minutes per mile) Calories burned in 30 min at 0% incline Calories burned in 30 min at 15 % incline
9 (6.6 mph) 300 558
10 (6 mph) 270 482
11 (5.5 mph) 243 432
12 (5 mph) 216 394
You might be interested: Cobb salad calories
## How many calories do I need to burn to lose weight?
A general rule is to aim to burn 400 to 500 calories , five days a week during your workouts. Remember, the number of calories you burn in a workout depends on your weight , sex, age and many other factors, but this number is a good starting place.
## How many calories does a 1 minute plank burn?
The plank is a highly effective abdominal-strengthening exercise. For most people, it burns between two and five calories per minute.
## Do ab workouts burn fat?
Evidence shows that you can’t lose belly fat by exercising your abs alone. For total-body fat loss, use a combination of aerobic exercise and resistance training , such as lifting weights. In addition, eat a healthy diet with plenty of protein, fiber and portion control — all of which are proven to help reduce body fat . | 998 | 4,170 | {"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} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.917392 |
https://brilliant.org/problems/its-as-hard-as-1-2-3/ | 1,485,050,296,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281331.15/warc/CC-MAIN-20170116095121-00145-ip-10-171-10-70.ec2.internal.warc.gz | 793,071,856 | 17,548 | # It's as hard as 1-2-3
Geometry Level 5
A circle with center $$O$$ has two points $$A,B$$ on its circumference and one point $$P$$ inside the circle such that $$AP=1$$, $$BP=2$$, $$OP=3$$, and $$\angle APB=90^{\circ}$$. If the area of the circle can be expressed as $\left(\dfrac{a+b\sqrt{c}}{d}\right)\pi$ for positive integers $$a,b,c,d$$ such that $$c$$ is square-free and $$a,b$$ are coprime with $$d$$, then find the value of $$a+b+c+d$$.
× | 152 | 449 | {"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} | 3 | 3 | CC-MAIN-2017-04 | longest | en | 0.85564 |
https://fred.stlouisfed.org/graph/?chart_type=line&width=1000&height=600&preserve_ratio=true&s%5B1%5D%5Bid%5D=PPPTTLCZA618NUPN | 1,524,635,738,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947693.49/warc/CC-MAIN-20180425041916-20180425061916-00397.warc.gz | 595,386,237 | 21,664 | # Purchasing Power Parity over GDP for Czech Republic (PPPTTLCZA618NUPN) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
2010: 14.52036
Updated: Sep 17, 2012
Units:
National Currency Units per US Dollar,
Frequency:
Annual
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Purchasing Power Parity over GDP for Czech Republic, National Currency Units per US Dollar, Not Seasonally Adjusted (PPPTTLCZA618NUPN)
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
For more information and proper citation see http://www.rug.nl/research/ggdc/data/pwt/pwt-7.1
Source Indicator: ppp
Purchasing Power Parity over GDP for Czech Republic
Select a date that will equal 100 for your custom index:
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Write a custom formula to transform one or more series or combine two or more series.
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NOTES
Source: University of Pennsylvania
Release: Penn World Table 7.1
Units: National Currency Units per US Dollar, Not Seasonally Adjusted
Frequency: Annual
#### Notes:
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
For more information and proper citation see http://www.rug.nl/research/ggdc/data/pwt/pwt-7.1
Source Indicator: ppp
#### Suggested Citation:
University of Pennsylvania, Purchasing Power Parity over GDP for Czech Republic [PPPTTLCZA618NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/PPPTTLCZA618NUPN, April 25, 2018.
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# Ukkonen’s Suffix Tree Construction – Part 6
Ukkonen’s Suffix Tree Construction – Part 1
Ukkonen’s Suffix Tree Construction – Part 2
Ukkonen’s Suffix Tree Construction – Part 3
Ukkonen’s Suffix Tree Construction – Part 4
Ukkonen’s Suffix Tree Construction – Part 5
Please go through Part 1, Part 2, Part 3, Part 4 and Part 5, before looking at current article, where we have seen few basics on suffix tree, high level ukkonen’s algorithm, suffix link and three implementation tricks and activePoints along with an example string “abcabxabcd” where we went through all phases of building suffix tree.
Here, we will see the data structure used to represent suffix tree and the code implementation.
At that end of Part 5 article, we have discussed some of the operations we will be doing while building suffix tree and later when we use suffix tree in different applications.
There could be different possible data structures we may think of to fulfill the requirements where some data structure may be slow on some operations and some fast. Here we will use following in our implementation:
We will have SuffixTreeNode structure to represent each node in tree. SuffixTreeNode structure will have following members:
• children – This will be an array of alphabet size. This will store all the children nodes of current node on different edges starting with different characters.
• suffixLink – This will point to other node where current node should point via suffix link.
• start, end – These two will store the edge label details from parent node to current node. (start, end) interval specifies the edge, by which the node is connected to its parent node. Each edge will connect two nodes, one parent and one child, and (start, end) interval of a given edge will be stored in the child node. Lets say there are two nods A (parent) and B (Child) connected by an edge with indices (5, 8) then this indices (5, 8) will be stored in node B.
• suffixIndex – This will be non-negative for leaves and will give index of suffix for the path from root to this leaf. For non-leaf node, it will be -1 .
This data structure will answer to the required queries quickly as below:
• How to check if a node is root ? — Root is a special node, with no parent and so it’s start and end will be -1, for all other nodes, start and end indices will be non-negative.
• How to check if a node is internal or leaf node ? — suffixIndex will help here. It will be -1 for internal node and non-negative for leaf nodes.
• What is the length of path label on some edge? — Each edge will have start and end indices and length of path label will be end-start+1
• What is the path label on some edge ? — If string is S, then path label will be substring of S from start index to end index inclusive, [start, end].
• How to check if there is an outgoing edge for a given character c from a node A ? — If A->children is not NULL, there is a path, if NULL, no path.
• What is the character value on an edge at some given distance d from a node A ? — Character at distance d from node A will be S[A->start + d], where S is the string.
• Where an internal node is pointing via suffix link ? — Node A will point to A->suffixLink
• What is the suffix index on a path from root to leaf ? — If leaf node is A on the path, then suffix index on that path will be A->suffixIndex
Following is C implementation of Ukkonen’s Suffix Tree Construction. The code may look a bit lengthy, probably because of a good amount of comments.
## C
Output (Each edge of Tree, along with suffix index of child node on edge, is printed in DFS order. To understand the output better, match it with the last figure no 43 in previous Part 5 article):
abbc [0]
b [-1]
bc [1]
c [2]
c [3]
Number of nodes in suffix tee are 6
Now we are able to build suffix tree in linear time, we can solve many string problem in efficient way:
• Check if a given pattern P is substring of text T (Useful when text is fixed and pattern changes, KMP otherwise
• Find all occurrences of a given pattern P present in text T
• Find longest repeated substring
• Linear Time Suffix Array Creation
The above basic problems can be solved by DFS traversal on suffix tree.
We will soon post articles on above problems and others like below:
And More.
Test you understanding?
1. Draw suffix tree (with proper suffix link, suffix indices) for string “AABAACAADAABAAABAA\$” on paper and see if that matches with code output.
2. Every extension must follow one of the three rules: Rule 1, Rule 2 and Rule 3.
Following are the rules applied on five consecutive extensions in some Phase i (i > 5), which ones are valid:
A) Rule 1, Rule 2, Rule 2, Rule 3, Rule 3
B) Rule 1, Rule 2, Rule 2, Rule 3, Rule 2
C) Rule 2, Rule 1, Rule 1, Rule 3, Rule 3
D) Rule 1, Rule 1, Rule 1, Rule 1, Rule 1
E) Rule 2, Rule 2, Rule 2, Rule 2, Rule 2
F) Rule 3, Rule 3, Rule 3, Rule 3, Rule 3
3. What are the valid sequences in above for Phase 5
4. Every internal node MUST have it’s suffix link set to another node (internal or root). Can a newly created node point to already existing internal node or not ? Can it happen that a new node created in extension j, may not get it’s right suffix link in next extension j+1 and get the right one in later extensions like j+2, j+3 etc ?
5. Try solving the basic problems discussed above.
We have published following articles on suffix tree applications:
References
http://web.stanford.edu/~mjkay/gusfield.pdf
Ukkonen’s suffix tree algorithm in plain English | 1,372 | 5,545 | {"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} | 3.203125 | 3 | CC-MAIN-2023-23 | latest | en | 0.85463 |
https://mathoverflow.net/questions/401196/comparing-integrals-of-bounded-subharmonic-functions | 1,638,055,469,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00372.warc.gz | 455,014,404 | 26,988 | # Comparing integrals of bounded subharmonic functions
Let $$\Omega \subset \mathbb{R}^n$$ be an open open subset. Let $$u,v\colon \Omega\to \mathbb{R}$$ be two functions such that at least one of them is compactly supported. Assume each of $$u$$ and $$v$$ can be presented as a difference of two bounded subharmonic functions in $$\Omega$$. Thus in particular the distributional Laplacians $$\Delta u,\Delta v$$ are well defined as signed measures on $$\Omega$$.
Question. Is it true that $$\int_\Omega u(x)\Delta v(x) dx=\int_\Omega v(x)\Delta u(x) dx?$$
Remark. (1) The expressions under the both integrals are well defined as signed measures with compact support. Thus both sides make sense.
(2) The simplest unknown to me case is $$n=2$$.
• @Christian Remling. You are right. But then it is always true, by approximation: choose shooth $v_n$ converging to $v$ in $D'$. Aug 7 at 14:14
• @AlexandreEremenko: I had the same thought, but is it clear: if $v_n\to v$ in $\mathcal D'$, then I only know that $\int u\Delta v_n\to\int u\Delta v$ if $u$ is also smooth. Aug 7 at 14:16
• @Christian Remling: I think justifying the limit in your second comment is a purely technical problem, not very hard. See Landkof, Introduction to modern potential theory, Ch. II, Potentials with finite energy. Aug 7 at 14:34
• @makt: But $\Delta v$ can have a point mass, while $u$ can be infinite at this point. Moreover, differences of subharmonic functions are in general not everywhere defined: in $u=u_1-u_2$, both $u_j$ can be $-\infty$ at some point. And $\Delta v$ can have a point mass at this point. So there are problems with definition of these integrals. The probems disappear if you consider "potentials of finite energy" instead of differences of subharmonic functions. Aug 7 at 14:57
• @AlexandreEremenko : I explicitly stated that both functions are bounded.
– makt
Aug 7 at 15:31
Without loss of generality, $$u$$ has compact support $$K\subset\Omega$$. Therefore the (signed) measure $$\Delta u$$ is supported in $$K$$ as well. Let $$(\phi_k)$$ be a sequence of smooth (radial) mollifiers such that $$\phi_k*u$$ is supported in $$K^\delta$$ (the closed $$\delta$$ neighborhood of $$K$$, with $$\delta>0$$ so small that $$K^\delta\subset\Omega$$). Additionally, suppose $$0\le\phi_k$$, $$\int_{\Bbb R^n}\phi_k(x)\phantom{!}dx=1$$, and $$\phi_k$$ is supported in the ball $$B(0,\delta/k)$$, for each $$k\ge 1$$. Then $$\lim_k \phi_k*u=u$$ pointwise and boundedly, because (for example) $$u$$ is finely continuous. Likewise $$\lim_k\phi_k*v=v$$. Then \eqalign{ \int_\Omega u(x)\cdot\Delta v(dx) &=\lim_k\int_\Omega (\phi_k*u)(x)\cdot\Delta v(dx)\cr &=\lim_k\int_\Omega \Delta(\phi_k*u)(x)\cdot v(x) \phantom{b}dx\cr &=\lim_k\int_\Omega (\phi_k*\Delta u)(x)\cdot v(x) \phantom{b}dx\cr &=\lim_k\int_\Omega (\phi_k*v )(x)\phantom{b}\Delta u(dx)\cr &=\int_\Omega v(x)\cdot\Delta u(dx).\cr } Here the second equality is just the definition of $$\Delta v$$ as a distribution; the third likewise; the fourth is Fubini.
• Good point. Strike the first sentence of my answer (save the stipulation that $u$ be of compact support), and remove "positive" from the second sentence. Aug 8 at 16:13 | 960 | 3,187 | {"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": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.899565 |
https://oj.olympiads.ca/problem/wossoly23team3/editorial | 1,722,749,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00527.warc.gz | 353,321,650 | 5,792 | ## Editorial for WOSS Dual Olympiad 2023 Team Round P3: Choosing Edges
Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.
Let be the shortest path from node to node without using any new edges. Let be the shortest path from to using at most new edge, which ends at node . Any shortest path from node to node using at most new edges can be written as .
First assume that all colors are different. Find the shortest path between all nodes to precalculate all . Notice that an edge from to can change into if that edge is used. Thus, loop over all edges to precompute all . Use the same logic to precompute all . Then loop over all pairs and store the minimum distance .
This can be easily modified to account for colors. Store values of each : The minimum distance, and the minimum distance using a different colored edge than the first minimum distance. Do some casework when looping over all pairs to ensure that the same colored edges are not used.
Time complexity: | 241 | 1,156 | {"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} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.878399 |
https://nrich.maths.org/public/leg.php?code=-3&cl=3&cldcmpid=456 | 1,508,324,283,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00794.warc.gz | 786,280,032 | 7,918 | # Search by Topic
#### Resources tagged with Networks/Graph Theory similar to Plum Tree:
Filter by: Content type:
Stage:
Challenge level:
### There are 28 results
Broad Topics > Decision Mathematics and Combinatorics > Networks/Graph Theory
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Olympic Magic
##### Stage: 4 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Behind the Rules of Go
##### Stage: 4 and 5
This article explains the use of the idea of connectedness in networks, in two different ways, to bring into focus the basics of the game of Go, namely capture and territory.
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Round-robin Scheduling
##### Stage: 2 and 3 Challenge Level:
Think about the mathematics of round robin scheduling.
### Sufficient but Not Necessary: Two Eyes and Seki in Go
##### Stage: 4 and 5
The game of go has a simple mechanism. This discussion of the principle of two eyes in go has shown that the game does not depend on equally clear-cut concepts.
### Factors and Multiples Graphs
##### Stage: 4 and 5 Challenge Level:
Explore creating 'factors and multiples' graphs such that no lines joining the numbers cross
### Königsberg
##### Stage: 3 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Ding Dong Bell
##### Stage: 3, 4 and 5
The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung.
### Pattern of Islands
##### Stage: 3 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Instant Insanity
##### Stage: 3, 4 and 5 Challenge Level:
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
### Fermat's Poser
##### Stage: 4 Challenge Level:
Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum.
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### The Olympic Torch Tour
##### Stage: 4 Challenge Level:
Imagine you had to plan the tour for the Olympic Torch. Is there an efficient way of choosing the shortest possible route?
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Travelling Salesman
##### Stage: 3 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### The Four Colour Theorem
##### Stage: 3 and 4
The Four Colour Conjecture was first stated just over 150 years ago, and finally proved conclusively in 1976. It is an outstanding example of how old ideas can be combined with new discoveries. prove. . . .
### Only Connect
##### Stage: 3 Challenge Level:
The graph represents a salesman’s area of activity with the shops that the salesman must visit each day. What route around the shops has the minimum total distance?
### Classifying Solids Using Angle Deficiency
##### Stage: 3 and 4 Challenge Level:
Toni Beardon has chosen this article introducing a rich area for practical exploration and discovery in 3D geometry
### Geometry and Gravity 2
##### Stage: 3, 4 and 5
This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### Going Places with Mathematicians
##### Stage: 2 and 3
This article looks at the importance in mathematics of representing places and spaces mathematics. Many famous mathematicians have spent time working on problems that involve moving and mapping. . . . | 1,361 | 6,045 | {"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} | 3.84375 | 4 | CC-MAIN-2017-43 | latest | en | 0.880296 |
https://www.educationquizzes.com/us/elementary-school-1st-and-2nd-grade/math/partitioning-numbers/ | 1,597,115,526,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738727.76/warc/CC-MAIN-20200811025355-20200811055355-00160.warc.gz | 659,109,155 | 7,775 | The number 56 contains 5 tens and 6 units.
# Partitioning Numbers
This Math quiz is called 'Partitioning Numbers' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is a fun way to learn if you are in the 1st or 2nd grade - aged 6 to 8.
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Partitioning numbers means being able to recognize the value of each digit within a number given by its place, and separate or 'split' the number into its component parts. For example, 23 could be partitioned into place values of 20 and 3 (2 tens and 3 units), and 456 could be partitioned into 400, 50 and 6 (4 hundreds, 5 tens and 6 units). Partitioning can also be done in different ways - for example, 23 could be partitioned into 13 and 10.
Try this quiz to see how well you do with partitioning numbers.
Can you partition the numbers?
1.
What is the value of the 6 in 46?
6
60
16
600
46 can be partitioned into 4 tens and 6 units
2.
How could you partition 46?
10 + 10 + 10 + 10 + 6
4 + 6
40 + 6 + 6
4 + 4 + 4 + 6
There are 4 tens in 46
3.
How could you partition 25?
10 + 10
50 + 2
20 + 5
2 + 5
25 is 2 tens and 5 units
4.
How could you partition 56?
15 + 16
5 + 6
50 + 60
20 + 36
The 5 tens in 56 could be split into 20 and 30
5.
What is 69 - 30?
63
66
39
36
There are 6 tens in 69 - taking away 3 of them leaves 3
6.
What is 78 - 20?
28
55
76
58
There are 7 tens in 78 - taking away 2 of them leaves 5
7.
How could you partition 28?
2 + 8
10 + 18
20 + 80
10 + 28
There are 2 tens in 28
8.
How could you partition 47?
14 + 17
4 + 7
70 + 4
40 + 7
47 is 4 tens and 7 units
9.
How could you partition 18?
1 + 8
10 + 8
80 + 1
18 + 8
18 is a ten and 8 units
10.
What is the value of the 3 in 35?
3
30
300
33
35 can be partitioned into 30 and 5
Author: Angela Smith | 656 | 1,941 | {"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} | 4.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.927215 |
https://stats.stackexchange.com/posts/201174/revisions | 1,563,867,684,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00416.warc.gz | 554,785,563 | 17,801 | 3 replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/ edited Apr 13 '17 at 12:44 Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. (A simple example of its utility in statistics appears at http://stats.stackexchange.com/questions/120459/diagonal-elements-of-the-inverted-correlation-matrix/120476#120476Diagonal elements of the inverted correlation matrix.) Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. (A simple example of its utility in statistics appears at http://stats.stackexchange.com/questions/120459/diagonal-elements-of-the-inverted-correlation-matrix/120476#120476.) Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. (A simple example of its utility in statistics appears at Diagonal elements of the inverted correlation matrix.) 2 added 339 characters in body edited Mar 11 '16 at 18:18 whuber♦ 213k3434 gold badges468468 silver badges857857 bronze badges Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. Although these operations are (A simple example of its utility in statistics appears at http://stats.stackexchange.com/questions/120459/diagonal-elements-of-the-inverted-correlation-matrix/120476#120476.) Although this operation is well-known and described elsewhere on the Web, it is not easy to find descriptions that are sufficiently clear and general that they indicate how to generalize themthey might apply to reduction of block matrices. Let's fill that gap and then apply the result to get some insight into the present problem. The idea is to apply block row operations until the determinant of the matrix can easily be calculated. Applying this to Considering the matrix $$X$$ in the question, we can immediately see how to eliminate the elements along the bottom row by removing multiples of the first, second, and third rows (in any order). This will change the matrix to This isThese operations are feasible because The determinant of the resulting block triangular matrix is the product of the determinants of the blocks along the diagonal:The determinant of the resulting block triangular matrix is the product of the determinants of the blocks along the diagonal: Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. Although these operations are well-known and described elsewhere on the Web, it is not easy to find descriptions that are sufficiently clear and general that they indicate how to generalize them to reduction of block matrices. The idea is to apply block row operations until the determinant of the matrix can easily be calculated. Applying this to the matrix $$X$$ in the question, we can immediately see how to eliminate the elements along the bottom row. This will change the matrix to This is feasible because The determinant of the resulting block triangular matrix is the product of the determinants of the blocks along the diagonal: Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. (A simple example of its utility in statistics appears at http://stats.stackexchange.com/questions/120459/diagonal-elements-of-the-inverted-correlation-matrix/120476#120476.) Although this operation is well-known and described elsewhere on the Web, it is not easy to find descriptions that are sufficiently clear and general that they indicate how they might apply to block matrices. Let's fill that gap and then apply the result to get some insight into the present problem. The idea is to apply block row operations until the determinant of the matrix can easily be calculated. Considering the matrix $$X$$ in the question, we can immediately see how to eliminate the elements along the bottom row by removing multiples of the first, second, and third rows (in any order). This will change the matrix to These operations are feasible becauseThe determinant of the resulting block triangular matrix is the product of the determinants of the blocks along the diagonal: 1 answered Mar 11 '16 at 15:19 whuber♦ 213k3434 gold badges468468 silver badges857857 bronze badges Let's talk a little about a useful way to manipulate block matrices: row reduction or Gaussian elimination. Although these operations are well-known and described elsewhere on the Web, it is not easy to find descriptions that are sufficiently clear and general that they indicate how to generalize them to reduction of block matrices. Let, then, $$m = m_1 + m_2 + m_3 + m_4$$ be the number of rows in any matrix $$X$$ and $$n = n_1 + n_2 + n_3 + n_4$$ be the number of its columns. We may write $$X = \pmatrix{X_{11} & X_{12} & X_{13} & X_{14} \\ X_{21} & X_{22} & X_{23} & X_{24} \\ X_{31} & X_{32} & X_{33} & X_{34} \\ X_{41} & X_{42} & X_{43} & X_{44}}$$ where the $$X_{ij}$$ are $$m_i\times n_j$$ matrices (and any of the $$m_i$$ or $$n_j$$ may be zero). Suppose you can find a matrix $$U$$ for which $$U X_{11} = X_{31}$$ (so that $$U$$ must be an $$m_3 \times m_1$$ matrix). You may use it to "eliminate" $$X_{31}$$ via a "block row operation": $$\pmatrix{1_{m_1} & 0 & 0 & 0 \\ 0 & 1_{m_2} & 0 & 0 \\ -U & 0 & 1_{m_3} & 0 \\ 0 & 0 & 0 & 1_{m_4}} X = \pmatrix{X_{11} & X_{12} & X_{13} & X_{14} \\ X_{21} & X_{22} & X_{23} & X_{24} \\ 0 & X_{32} - U X_{12} & X_{33} - U X_{13} & X_{34} - U X_{14} \\ X_{41} & X_{42} & X_{43} & X_{44}}.$$ The notation "$$1_k$$" refers to the $$k\times k$$ identity matrix. It is obvious that the determinant of the "elimination matrix" on the left hand side is $$1$$. Therefore, whenever $$X$$ is a square matrix, a row block operation does not change the determinant of the result. The idea is to apply block row operations until the determinant of the matrix can easily be calculated. Applying this to the matrix $$X$$ in the question, we can immediately see how to eliminate the elements along the bottom row. This will change the matrix to $$\pmatrix{A & 0 & 0 & E \\ 0 & B & 0 & F \\ 0 & 0 & C & G \\ 0 & 0 & 0 & D - E^\prime A^{-1} E - F^\prime B^{-1} F - G^\prime C^{-1} G}.$$ This is feasible because The positive-definiteness of $$A$$, $$B$$, and $$C$$ implies all three are invertible. The unique solution to $$E^\prime = UA$$ is $$E^\prime A^{-1}$$, the unique solution to $$F^\prime = UB$$ is $$F^\prime B^{-1}$$, and the unique solution to $$G^\prime = UC$$ is $$G^\prime C^{-1}$$. The determinant of the resulting block triangular matrix is the product of the determinants of the blocks along the diagonal: $$|X| = |A|\,|B|\,|C|\,|D - E^\prime A^{-1} E - F^\prime B^{-1} F - G^\prime C^{-1} G|.$$ As a check, let X be the matrix defined in the R code in the question. Let's compare the result of this formula to the direct calculation of the determinant: A <- X[1:2, 1:2] B <- X[3:4, 3:4] C <- X[5:6, 5:6] D <- X[7:8, 7:8] E <- X[1:2, 7:8] F <- X[3:4, 7:8] G <- X[5:6, 7:8] det.X <- det(X) det.X0 <- det(A) * det(B) * det(C) * det(D - crossprod(E, solve(A, E)) - crossprod(F, solve(B, F)) - crossprod(G, solve(C, G))) (det.X - det.X0) / det.X [1] -1.926411e-16 The relative error is the size of floating point roundoff error: the two values have to be considered equal. For small blocks, this procedure is primarily of theoretical interest. In R, it takes 50 times longer to carry out the calculation based on blocks than it does to compute the determinant directly! | 1,957 | 7,504 | {"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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-30 | latest | en | 0.887098 |
https://numbermatics.com/n/81227568245030548242178/ | 1,722,921,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00886.warc.gz | 359,242,580 | 7,194 | # 81227568245030548242178
## 81,227,568,245,030,548,242,178 is an even composite number composed of four prime numbers multiplied together.
What does the number 81227568245030548242178 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.
81227568245030548242178 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 81227568245030548242178:
### 2 × 687593 × 43307969 × 1363873817
See below for interesting mathematical facts about the number 81227568245030548242178 from the Numbermatics database.
### Names of 81227568245030548242178
• Cardinal: 81227568245030548242178 can be written as Eighty-one sextillion, two hundred twenty-seven quintillion, five hundred sixty-eight quadrillion, two hundred forty-five trillion, thirty billion, five hundred forty-eight million, two hundred forty-two thousand, one hundred seventy-eight.
### Scientific notation
• Scientific notation: 8.1227568245030548242178 × 1022
### Factors of 81227568245030548242178
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 1407869381
### Bases of 81227568245030548242178
• Binary: 100010011001101011010111101001100110011000001000101011000000101111111000000102
• Base-36: D86GZE359691NNM
### Squares and roots of 81227568245030548242178
• 81227568245030548242178 squared (812275682450305482421782) is 6597917843001095098877557498070320165738183684
• 81227568245030548242178 cubed (812275682450305482421783) is 535932821867476202545950369408818107553918197902552481713320680223752
• The square root of 81227568245030548242178 is 285004505657.4203499169
• The cube root of 81227568245030548242178 is 43307969.0000000077
### Scales and comparisons
How big is 81227568245030548242178?
• 81,227,568,245,030,548,242,178 seconds is equal to 293,274,701,009 years, 3 weeks, 3 days, 15 hours, 30 minutes, 7 seconds.
• To count from 1 to 81,227,568,245,030,548,242,178 would take you about two hundred ninety-three billion, two hundred seventy-four million, seven hundred one thousand and nine years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 81227568245030548242178 cubic inches would be around 3608997.4 feet tall.
### Recreational maths with 81227568245030548242178
• 81227568245030548242178 backwards is 87124284503054286572218
• The number of decimal digits it has is: 23
• The sum of 81227568245030548242178's digits is 94
• More coming soon!
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The information we have on file for 81227568245030548242178 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 81227568245030548242178, math, Factors of 81227568245030548242178, curriculum, school, college, exams, university, Prime factorization of 81227568245030548242178, STEM, science, technology, engineering, physics, economics, calculator, eighty-one sextillion, two hundred twenty-seven quintillion, five hundred sixty-eight quadrillion, two hundred forty-five trillion, thirty billion, five hundred forty-eight million, two hundred forty-two thousand, one hundred seventy-eight.
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Some bits of this website may not work unless you switch it on. | 1,190 | 4,398 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.733461 |
http://blog.metalight.net/2012/12/analysis-of-speed-density-flow.html | 1,713,701,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00723.warc.gz | 3,479,531 | 12,327 | ## Saturday 8 December 2012
### Analysis of Speed Density Flow Relationship
These notes have been inspired by a recent paper by Rahman, Ben-Edigbe and Hassan*. In it they analyse the speed, density and flow relationship.
The equations given in this previous traffic modeling post include an algebraic relationship between speed, density and flow: $$q=\rho v$$ and $$v = v_f(1-\rho/\rho_j),$$ where
$$q$$ = flow rate (veh/s);
$$\rho$$ = vehicle density (veh/m);
$$v$$ = vehicle speed (m/s);
$$v_f$$ = free flow speed (m/s), nominally 27.78 m/s;
$$\rho_j$$ = jam density (veh/m), nominally 1/7 veh/m.
Given the above equations, it is possible to eliminate $$v$$. That is, flow can be expressed in terms of density by using the speed equation:
$$q=v_f\rho ( 1 - \rho/\rho_j ).$$ This is a quadratic, and it has a maximum. Because it's a quadratic, this means that the inverse relationship is not a function, as it's multi-valued (one flow rate may be associated with two densities). Here is a graph of the flow-density relationship with the nominal values:
Relationship between flow and density, showing key values.
The maximum point on the flow curve happens to be the capacity of the roadway. We find the maximum point by differentiating the above and solving for slope equals zero. The critical density, $$\rho_c$$ (the density at which the flow rate equals capacity), is found to be $$\rho_j/2$$, and the capacity, $$q_c$$, is given by:
$$q_c=v_f\rho_j/4.$$ With the values mentioned above ($$v_f$$ = 27.78 and $$\rho_j$$ = 1/7), the capacity is 0.99 veh/s, which while high, is in the right order of magnitude for freeways (which I think is around 0.5 to 0.7 veh/s).
The next post on this topic will be model discretisation.
* The paper is Rahman, R., Den-Edigbe, J. E., and Hassan A. (2012), Extent of Traffic Shockwave Velocity Propagations Induced by U-Turn Facility on Roadway Segments. 25th ARRB Conference. Mentioned here. | 514 | 1,938 | {"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} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.917229 |
https://math.stackexchange.com/questions/2711203/function-with-continuous-components-zariski-topology | 1,575,879,371,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00540.warc.gz | 450,737,457 | 29,638 | # Function with continuous components, Zariski topology
I am trying to understand how the Zariski topology is different from the usual topology on the affine space $A^n$. Let $X$ be an affine algebraic subset of $A^n$, the $n$-dimensional affine space over $k$. Let $f_i:X\rightarrow A^1$, $i=1,...,k$ be continuous functions, both $X$ and $A^1$ being equipped with the Zariski topology. Is the function $f:X\rightarrow A^k, x\mapsto (f_1(x),...,f_k(x))$ continuous when $A^k$ is equipped with the Zariski topology? I know that $f$ is continuous for the product topology, but the Zariski topology is finer the the product topology.
I appreciate any help.
• Excellent, precise question! – Georges Elencwajg Mar 28 '18 at 6:38
NO, $f$ need not be continuous in the Zariski topology !
For a counterexample take $X=\mathbb A^1$ and let $f_1:\mathbb A^1\to \mathbb A^1$ be the map permuting $0$ and $1$ and leaving $x$ fixed if $x\neq 0,1$.
Like all permutations of $\mathbb A^1$ the map $f_1$ is continuous in the Zariski topology.
Now taking the identity for $f_2:\mathbb A^1\to\mathbb A^1$, let us show that the map $$f:\mathbb A^1\to \mathbb A^1\times \mathbb A^1:x\mapsto (f_1(x),f_2(x)=(f_1(x),x)$$ is not continuous for the Zariski topology.
Indeed the diagonal $\Delta=\{(x,x)\vert x\in \mathbb A^1\} \subset \mathbb A^1\times \mathbb A^1$ is closed in the product endowed with its Zariski topology but its inverse image $f^{-1}{\Delta}=\mathbb A^1\setminus \{0,1\}\subset \mathbb A^1$ is not closed if $k$ is an infinite field. | 503 | 1,535 | {"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} | 3.34375 | 3 | CC-MAIN-2019-51 | latest | en | 0.764228 |
http://gfredericks.com/sandbox/project_euler/98 | 1,369,504,315,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706009988/warc/CC-MAIN-20130516120649-00079-ip-10-60-113-184.ec2.internal.warc.gz | 114,488,873 | 1,632 | Problem #98
By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 362. What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 962. We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).
What is the largest square number formed by any member of such a pair?
NOTE: All anagrams formed must be contained in the given text file. | 201 | 832 | {"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} | 3 | 3 | CC-MAIN-2013-20 | latest | en | 0.89786 |
https://paperwriterhelper.com/finance-the-cash-flows-for-three-projects-are-shown-above/ | 1,607,190,488,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00307.warc.gz | 422,961,079 | 31,109 | Select Page
finance-The cash flows for three projects are shown above
Investment A:Year:012345Cash flow:-\$14,000\$6,000\$6,000\$6,000\$6,000\$6,000Investment B:Year:012345Cash flow:-\$15,000\$7,000\$7,000\$7,000\$7,000\$7,000Investment C:Year:012345Cash flow:-\$18,000\$12,000\$4,000\$4,000\$4,000\$4,000The cash flows for three projects are shown above. The cost of capital is 7.5%. If an investor decided to take projects with a payback period of two years or less, which of these projects would he take?Investment AInvestment BInvestment Cnone of these investmentsFlag this QuestionA lottery winner can take \$6 million now or be paid \$600,000 at the end of each of the next 16 years. The winner calculates the internal rate of return (IRR) of taking the money at the end of each year and, estimating that the discount rate across this period will be 6%, decides to take the money at the end of each year. Was her decision correct?Yes, because it agrees with the Net Present Value rule.No, because it disagrees with the Net Present Value rule.Yes, because it agrees with the payback rule.Yes, because it agrees with both the Net Present Value rule and the payback rule.Flag this QuestionMary is in contract negotiations with a publishing house for her new novel. She has two options. She may be paid \$100,000 up front, and receive royalties that are expected to total \$26,000 at the end of each of the next five years. Alternatively, she can receive \$200,000 up front and no royalties. Which of the following investment rules would indicate that she should take the former deal, given a discount rate of 8%?Rule I: The Net Present Value ruleRule II: The Payback Rule with a payback period of two yearsRule III: The internal rate of return (IRR) RuleRule II and IIIRule I onlyRule III onlyRule I and II
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• Free Reference Page | 636 | 2,605 | {"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} | 2.5625 | 3 | CC-MAIN-2020-50 | longest | en | 0.919329 |
https://cn.maplesoft.com/support/help/errors/view.aspx?path=Student/ODEs/Solve/ByVariationOfParameters&L=C | 1,718,555,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00235.warc.gz | 167,622,243 | 24,257 | ByVariationOfParameters - Maple Help
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Home : Support : Online Help : Education : Student Packages : ODEs : Computation : Solve : ByVariationOfParameters
Student[ODEs][Solve]
ByVariationOfParameters
Solve a system of first order linear ODEs by the method of variation of parameters
Calling Sequence ByVariationOfParameters(SYS, Y) ByVariationOfParameters(SYS) ByVariationOfParameters(A, F, x)
Parameters
SYS - list, set, or equation; a system of first order linear ordinary differential equations Y - list or set or Vector of functions; the solving variables A - Matrix; the Matrix of coefficients F - Vector; the Vector of forcing functions x - name; the independent variable
Description
• The ByVariationOfParameters(SYS, vars) command finds the solution of a system of first order linear ODEs using variation of parameters.
• The system SYS may be written as a list or set of ODEs. If the solving variables cannot be unambiguously determined from the form of SYS, Y must also be specified as a list or set containing the solving variables.
• Alternatively, SYS may be written as a single equation of the form:
$\mathrm{DY}=A·Y+F$
where Y is a Vector of solving variables, DY a Vector of their derivatives, A is the Matrix of coefficients, and F is the Vector of forcing functions. In this case, Y does not need to be specified as an extra argument since it can be determined from the form of SYS.
• A third syntax: ByVariationOfParameters(A, F, x) is also available as a shortcut to the above syntax DY = A . Y + F.
• Use the option output=steps to make this command return an annotated step-by-step solution. Further control over the format and display of the step-by-step solution is available using the options described in Student:-Basics:-OutputStepsRecord. The options supported by that command can be passed to this one.
Examples
> $\mathrm{with}\left({{\mathrm{Student}}_{\mathrm{ODEs}}}_{\mathrm{Solve}}\right):$
> $A≔\mathrm{Matrix}\left(\left[\left[1,2\right],\left[3,2\right]\right]\right)$
${A}{≔}\left[\begin{array}{cc}{1}& {2}\\ {3}& {2}\end{array}\right]$ (1)
> $F≔⟨1,{ⅇ}^{x}⟩$
${F}{≔}\left[\begin{array}{c}{1}\\ {{ⅇ}}^{{x}}\end{array}\right]$ (2)
> $Y≔⟨\mathrm{y1}\left(x\right),\mathrm{y2}\left(x\right)⟩:$
> $\mathrm{sys1}≔\frac{\partial }{\partial x}Y=\mathrm{.}\left(A,Y\right)+F$
${\mathrm{sys1}}{≔}\left[\begin{array}{c}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y1}}{}\left({x}\right)\\ \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y2}}{}\left({x}\right)\end{array}\right]{=}\left[\begin{array}{c}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{1}\\ {3}{}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{{ⅇ}}^{{x}}\end{array}\right]$ (3)
> $\mathrm{Student}:-\mathrm{ODEs}:-\mathrm{Solve}:-\mathrm{ByVariationOfParameters}\left(\mathrm{sys1}\right)$
$\left[\begin{array}{c}{\mathrm{y1}}{}\left({x}\right)\\ {\mathrm{y2}}{}\left({x}\right)\end{array}\right]{=}\left[\begin{array}{c}{-}\frac{{{ⅇ}}^{{x}}}{{3}}{+}\frac{{1}}{{2}}{-}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{30}}\\ {-}\frac{{3}}{{4}}{+}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{20}}\end{array}\right]$ (4)
> $\mathrm{sys2}≔\left[\mathrm{seq}\left(\frac{\partial }{\partial x}{Y}_{i}={\left(\mathrm{.}\left(A,Y\right)+F\right)}_{i},i=1..2\right)\right]$
${\mathrm{sys2}}{≔}\left[\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y1}}{}\left({x}\right){=}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{1}{,}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y2}}{}\left({x}\right){=}{3}{}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{{ⅇ}}^{{x}}\right]$ (5)
> $\mathrm{Student}:-\mathrm{ODEs}:-\mathrm{Solve}:-\mathrm{ByVariationOfParameters}\left(\mathrm{sys2}\right)$
$\left[{\mathrm{y1}}{}\left({x}\right){=}{-}\frac{{{ⅇ}}^{{x}}}{{3}}{+}\frac{{1}}{{2}}{-}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{30}}{,}{\mathrm{y2}}{}\left({x}\right){=}{-}\frac{{3}}{{4}}{+}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{20}}\right]$ (6)
> $\mathrm{sys3}≔\mathrm{convert}\left(\mathrm{sys2},\mathrm{set}\right)$
${\mathrm{sys3}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y1}}{}\left({x}\right){=}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{1}{,}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{y2}}{}\left({x}\right){=}{3}{}{\mathrm{y1}}{}\left({x}\right){+}{2}{}{\mathrm{y2}}{}\left({x}\right){+}{{ⅇ}}^{{x}}\right\}$ (7)
> $\mathrm{Student}:-\mathrm{ODEs}:-\mathrm{Solve}:-\mathrm{ByVariationOfParameters}\left(\mathrm{sys3}\right)$
$\left\{{\mathrm{y1}}{}\left({x}\right){=}{-}\frac{{{ⅇ}}^{{x}}}{{3}}{+}\frac{{1}}{{2}}{-}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{30}}{,}{\mathrm{y2}}{}\left({x}\right){=}{-}\frac{{3}}{{4}}{+}\frac{{2}{}{{ⅇ}}^{{-}{x}}}{{5}}{+}\frac{{7}{}{{ⅇ}}^{{4}{}{x}}}{{20}}\right\}$ (8)
Compatibility
• The Student[ODEs][Solve][ByVariationOfParameters] command was introduced in Maple 2022.
• For more information on Maple 2022 changes, see Updates in Maple 2022.
See Also | 2,041 | 5,422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "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} | 3.75 | 4 | CC-MAIN-2024-26 | latest | en | 0.674609 |
https://www.thinkcalculator.com/algebra/lnbatch.php | 1,701,880,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00660.warc.gz | 1,139,544,890 | 5,594 | ## Natural Logarithm (Ln) Calculator in batch
Note: Data should be separated in coma (,), space, tab, or in separated lines.
Number:
Natural Logarithm:
The natural logarithm of a number is its logarithm to the base e, it is used to calculate the natural logarithm of a number x, where e is an irrational and transcendental constant approximately equal to 2.718281828, which is generally written as ln(x) or loge(x) or log(x).
For instance, Ln7 = 1.94591, e1.94591 = 7, So The natural logarithm of x is the power to which e would have to be raised to equal x. So Lne = 1.
When you want to calculate a group of numbers of natural logarithm, Enter "2 4 5 6" click the "calculate" button, The result will be "0.693147 1.386294 1.609438 1.791759".
Natural Logarithm: Ln(6) = 1.791759
Natural Logarithm: Ln(5) = 1.609438
Natural Logarithm: Ln(4) = 1.386294
Natural Logarithm: Ln(2) = 0.693147
Thinkcalculator.com provides you helpful and handy calculator resources. | 288 | 969 | {"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} | 3.15625 | 3 | CC-MAIN-2023-50 | latest | en | 0.836212 |
http://www.swarthmore.edu/NatSci/qmorris1/research.html | 1,513,257,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544124.40/warc/CC-MAIN-20171214124830-20171214144830-00441.warc.gz | 480,972,162 | 4,684 | # RESEARCH
#### RESEARCH INTERESTS
Semipositone Problems
Superlinear problems on exterior domains.
• $$\left\{ \begin{array}{cl} - \Delta u = \lambda f(u), & x \in \Omega^c, \\ u=0, & x \in \partial \Omega, \\ u \to 0, & \|x\| \to \infty. \end{array} \right.$$
• $f(0)<0$ (semipositone)
• $\displaystyle \lim_{s \to \infty}\frac{f(s)}{s} = \infty$ (superlinear)
• $\Omega$ is a bounded domain in $\mathbb{R}^n$.
• ? existence, uniqueness, multiplicity of solutions
Nonlinear Boundary Conditions
Semipositone problems with nonlinear boundary conditions.
• $$\left\{ \begin{array}{cl} - \Delta u = \lambda f(u), & x \in \Omega^c, \\ \frac{\partial u}{\partial \eta} + c(u) u =0, & x \in \partial \Omega, \\ u \to 0, & \|x\| \to \infty. \end{array} \right.$$
• $f$ is semipositone and superlinear.
• $c:[0,\infty) \to (0,\infty)$ is continuous
• $\Omega$ is a bounded domain in $\mathbb{R}^n$.
• ? existence, uniqueness, multiplicity of solutions
Math Biology
Density dependent dispersal on the boundary
• Modeling habitat surrounded by hostile matrix with nonlinear density dependent dispersal on the boundary using reaction-diffusion equations with nonlinear boundary conditions. (Single PDE)
• Modeling competing species with nonlinear dispersal on the boundary based on density of competitor. (PDE systems)
• ? existence, uniqueness, multiplicity, and stablity of steady states
#### Dr. Byungjae Son
Wayne State University
#### Dr. Ratnasingham Shivaji
University of North Carolina at Greensboro
#### Dr. Jerome Goddard
Auburn University at Montgomery
#### Dr. Jim Cronin
Louisiana State University
#### Dr. Stephen Robinson
Wake Forest University
#### Dr. Inbo Sim
University of Ulsan
#### Dr. Catherine Payne
Winston-Salem State University
Noam Chomsky | 521 | 1,772 | {"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} | 2.796875 | 3 | CC-MAIN-2017-51 | latest | en | 0.621402 |
https://www.doubtnut.com/question-answer/find-the-value-of-x-if-8060-x12-40379799 | 1,642,996,540,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00170.warc.gz | 755,748,568 | 75,283 | HomeEnglishClass 7MathsChapter
Find the value of x, if 80:60 ...
# Find the value of x, if 80:60 = x:12.
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Text Solution
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80: 60 = x:12 <br> rArr (80)/(60) = (x)/(12) <br> rArr x = (80 xx 12)/(60) = 16 <br> Hence, the correct option is (a). | 220 | 748 | {"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} | 3.375 | 3 | CC-MAIN-2022-05 | latest | en | 0.758996 |
https://math.stackexchange.com/questions/846108/solving-2x8-6x-12-by-using-the-guess-and-check-method/846119 | 1,618,335,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00501.warc.gz | 494,563,818 | 38,874 | Solving $2x+8=6x-12$ by using the guess and check method
How do you solve this equation: $2x+8=6x-12$ by using the guess and check method?
I divide $2x+8$ and I get $4$ then I divide $6x-12$ and I get $-2$ but I don't know what to do next or is it wrong?
• Is it $2^x$, instead of $2x$? For $2x$ there isn't much to guess. – Karolis Juodelė Jun 24 '14 at 17:11
• Try guessing $x = 5$. – JimmyK4542 Jun 24 '14 at 17:12
The answer is $x = 5$. You can get this solution using normal analytical methods (algebraic manipulation), i.e., \begin{align} 2x + 8 = 6x - 12 &\iff -4x + 8 = -12 \\&\iff -4x = -20 \\&\iff x = 5 \,\,. \end{align} In terms of a "guess and check method", here's my strategy: factor $2$ out of the LHS and $6$ out of the RHS to obtain the equation $2(x + 4) = 6(x - 2)$. Now, we can divide $2$ from both sides to obtain $x + 4 = 3(x - 2)$. The solution is fairly easy to see in this form by guessing and checking. You will come to $x = 5$, as before.
Your instructor wants you to try plugging in a bunch of numbers to guess the correct value. Just by eyeballing this, we might try 4 or 5. If you plug in $5$ on the left you get $$2(5)+8=18$$ and on the right you get $$6(5)-12 = 30 - 12 = 18.$$ This tells us that $x=5$ is a solution to this equation.
Remember that the old methods still work here. So if you add 12 to both sides and subtract $2x$ from both sides you get $$20=4x$$ and then you can solve for $x$ by dividing both sides by 4.
$$\begin{array}{r|r|r} x &2x+8 &6x-12\\ \hline 0&8&-12\\1&10&-6\\2&12&0 \\ \vdots&\vdots&\vdots \end{array}$$ Note that every time $x$ increases by $1$, $2x+8$ increases by $2$ and $6x-12$ increases by $6$. (do you see why this should be so?) That means whenever $x$ increases by $1$, $6x-12$ grows by $4$ units more than $2x+8$. Currently (at $x=2$), $2x+8$ is ahead by $12$. So if we increase $x$ by $3$, $6x-12$ will gain $3\cdot 4$, or $12$ more than $2x-8$, and so then the two expressions will be equal.
Thus the answer should be $x=5$. | 717 | 2,004 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-17 | latest | en | 0.896179 |
https://bytepawn.com/optimal-coverage-for-wordle-with-monte-carlo-methods.html | 1,717,084,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00840.warc.gz | 114,788,376 | 5,775 | # Optimal coverage for Wordle with Monte Carlo methods - Part I
Marton Trencseni - Wed 19 January 2022 - Machine Learning
## Introduction
Wordle is a simple 5x5 word guessing game. It's free, a new puzzle is posted every day. You have to guess an unknown 5 letter word, like PROXY. You have 5 guesses, and after each guess, the game tells you if you got one of the letters, but it's not at the right position (brown), or it's the right letter in the right position (green). Only words from the Wordle dictionary are accepted as guesses, ie. HELLO and WORLD are valid guesses, but ABCDE is not an accepted guess.
The list of Wordle dictionary words is known, it's a set of 12,972 words.
## Optimal coverage challenge
One strategy is to analyze the set of words, frequencies of letters, and given what we know so far, pick the word [from the Wordle dictionary] to guess that maximizes the average information gain of the next guess, assuming that the target word is randomly selected [from the Wordle dictionary]. One of the inputs to this is an assumption how the player values knowing the correct position of the letters (green) versus just knowing the set of letters making up the word (brown).
There is a coincidence (?) in the structure in the game: the english alphabet is 26 words, and we can guess a total of 25 characters. If we can find 5 words which share no identical characters, then we can always use these 5 words as our 5 guesses, and assuming the target word has N unique letters, we will always no at least N-1. Eg. if the word is HELLO, and after our 5 guesses we know that H E L and O are in the words, then:
• either the word has a duplicate letter, one of those 4
• the word has a fifth unique letter, the 26th letter that was not included in our 5 guesses (which covered 25)
We don't know which of the above is the case, but most of the time we can assume a reasonable player will be able to guess the target word at this point, also taking into account that on average a few letters' position will be known.
So the question is: from the Wordle dictionary, can we find 5 words such that all 25 letters are unique? If not, what is the maximum achievable unique-count?
## Monte Carlo approach
The simplest Monte Carlo approach is to just randomly pick 5 words from the dictionary, and check how many unique letters we have:
words = ["cigar", "rebut", "sissy", ... ] # the Wordle dictionary
num_tests = 1000*1000
for _ in range(num_tests):
wordlist = random.sample(words, 5) # 5 random words
letters = set(list(''.join(wordlist)))
if len(letters) >= 21:
print(len(letters), wordlist)
It will print something like:
21 ['lotsa', 'finks', 'judge', 'macho', 'warby']
21 ['muzak', 'vales', 'letch', 'bilgy', 'fjord']
21 ['fagot', 'cushy', 'pawks', 'melon', 'diver']
21 ['miter', 'munch', 'pawky', 'bodge', 'lifes']
22 ['kight', 'flyby', 'roven', 'clasp', 'dwaum']
Let's try to be a little smarter.
## Parallelization
The simplest, still brute force improvement is to run the same thing, but in parallel. I have a 12-core (24 threads) CPU, so I can run 16x parallelism without impact the usability of my computer:
def solve_wordle(num_tests=1000, random_seed=0):
random.seed(random_seed * 131071)
solutions = []
for _ in range(num_tests):
wordlist = random.sample(words, 5) # 5 random words
letters = set(list(''.join(wordlist)))
if len(letters) >= 21:
solutions.append(wordlist)
return solutions
def flatten_list(li):
return [item for sublist in li for item in sublist]
def clean_solutions(solutions):
return [t.split(',') for t in sorted({','.join(sorted(c)) for c in solutions})]
n_jobs = 16
num_tests = 1*1000*1000
solutions = Parallel(n_jobs=n_jobs)(delayed(solve_wordle)
(num_tests=num_tests, random_seed=i)
for i in range(n_jobs))
solutions = clean_solutions(flatten_list(solutions))
for wordlist in solutions:
num_unique_letters = len(set(list(''.join(wordlist))))
print(f'{num_unique_letters}-unique-letter solution: {wordlist}')
if num_unique_letters == 25:
print('*** JACKPOT! ***')
print(f'Found {len(solutions)} solutions...')
The ipython notebook is on Github. Prints something like:
21-unique-letter solution: ['admix', 'itchy', 'ovels', 'tupik', 'wrung']
21-unique-letter solution: ['adzed', 'boxty', 'brugh', 'vinyl', 'wacks']
...
21-unique-letter solution: ['fyces', 'goban', 'lurve', 'mewed', 'piths']
21-unique-letter solution: ['light', 'packs', 'vends', 'wauff', 'womby']
Found 47 solutions...
## Conclusion
In the next part, I will show an improved Monte Carlo approach, which finds about one 24-unique-letter solution every second. | 1,212 | 4,597 | {"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} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.95378 |
https://www.jiskha.com/questions/972793/the-distance-between-the-center-of-symmetry-of-a-parallelogram-and-its-longer-side-is | 1,607,090,446,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00566.warc.gz | 719,489,782 | 5,879 | # Math
the distance between the center of symmetry of a parallelogram and its longer side is equal to 12 cm. The area of the parallelogram is 720 cm^2, and its perimeter is 140 cm. Determine the length of the longer diagonal of the parallelogram
1. 👍 0
2. 👎 0
3. 👁 1,026
1. since A = bh, and h/2 = 12,
b = 720/24 = 30
So, two of the sides have length 30, and the others have length 40.
Hmmm. Houston, we have a problem. What have I missed?
1. 👍 0
2. 👎 0
2. well.. i got that too.. but i don't know how to find the longer diagonal with this given sides
1. 👍 0
2. 👎 0
3. OK, assuming that the sides are in fact 30 and 40, and the distance from the center to the side of length 30 is 12, then since the center is half way along the diagonal, the height at the end of the diagonal will be 24.
So, if AB=30 and BC=40, then if E is the base of the altitude from C to AB extended, BE=32 and triangle AEC has hypotenuse √(62^2+24^2) = 2√1105 = 66.48
That is the diagonal AC.
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2. 👎 0
4. so the longer side is 30 and shorter is 40??
1. 👍 0
2. 👎 0
5. yeah - that was my impression. Can't explain it given the provided data. The height and area indicate that the base of the altitude is the side of length 30, but then the math indicates that the "shorter" side is 40!
1. 👍 0
2. 👎 0
6. how did you get BE=32?
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2. 👎 0
7. How did you get BE=32, and where did 62 come from?
1. 👍 0
2. 👎 0
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Java Question
# Move all zero values in a big array to its front portion in a Time Efficiency way
You're given a big array with Integral type value, How do you move all zero values within it to the front portion of the array in a Time Efficiency way?
e.g. 0,1,72,3,0,5,9,0,6,51,0,3 ---> 0,0,0,0,1,72,3,5,9,6,51,3
Regards!
If you want to keep the order between the other items, loop backwards copying all non-zero items, then fill up with zero items at the beginning:
``````int source = arr.Length - 1;
int dest = arr.Length - 1;
while (source >= 0) {
if (arr[source] != 0) {
arr[dest--] = arr[source];
}
source--;
}
while (dest >= 0) arr[dest--] = 0;
``````
If the order of the items is not important, you can simply swap zero items with items at the beginning of the array:
``````int dest = 0;
for (int source = 0; source < arr.Length; source++) {
if (arr[source] == 0) {
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https://homepages.inf.ed.ac.uk/dts/fps/pract3/solution.txt | 1,723,735,800,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00067.warc.gz | 242,324,786 | 2,165 | (* ************** SAMPLE SOLUTION TO PRACTICAL 3 *************** *) (* **************** EXERCISE 1 **************** *) signature QUEUE = sig type 'a queue val empty : 'a queue val isempty : 'a queue -> bool val insert : 'a * 'a queue -> 'a queue val front : 'a queue -> 'a val remove : 'a queue -> 'a queue val join : 'a queue * 'a queue -> 'a queue axiom isempty empty == true axiom Forall (a,q) => isempty(insert(a,q)) == false axiom Forall (q1,q2) => isempty(join(q1,q2)) == isempty q1 andalso isempty q2 axiom Forall (a,q) => isempty q implies front(insert(a,q)) == a axiom Forall (a,q) => not(isempty q) implies front(insert(a,q)) == front q axiom Forall (q1,q2) => isempty q1 implies front(join(q1,q2)) == front q2 axiom Forall (q1,q2) => not(isempty q1) implies front(join(q1,q2)) == front q1 axiom Forall (a,q) => isempty q implies remove(insert(a,q)) == empty axiom Forall (a,q) => not(isempty q) implies remove(insert(a,q)) == insert(a,remove q) axiom Forall (q1,q2) => isempty q1 implies remove(join(q1,q2)) == remove q2 axiom Forall (q1,q2) => not(isempty q1) implies remove(join(q1,q2)) == join(remove q1,q2) end Discussion might concern things like whether or not we include axioms like Forall (q1,q2) => isempty q1 implies join(q1,q2) == q2 This is stronger than necessary to ensure "correct" results, and so should preferably be omitted. (* **************** EXERCISE 2 **************** *) This proof contains more detail than required, but the style is what I would expect. First, define a function count' to count the number of occurrences of a given element in a heap: fun count'(x,empty) = 0 | count'(x,node(h,a,h')) = cmp(x,a) + count'(x,h) + count'(x,h') LEMMA 1: Forall (x,a,h) => count'(x,insert(a,h)) = cmp(x,a) + count'(x,h) by structural induction on h. BASE CASE. h=empty count'(x,insert(a,empty)) = {code for insert} count'(x,node(empty,a,empty)) = {code for count'} cmp(x,a) STEP CASE. h=node(h1,b,h2) INDUCTIVE HYPOTHESES: IH1: count'(x,insert(a,h1)) = cmp(x,a) + count'(x,h1) IH2: count'(x,insert(a,h2)) = cmp(x,a) + count'(x,h2) CASE 1. a <= b count'(x,insert(a,node(h1,b,h2))) = {code for insert} count'(x,node(insert(b,h2),a,h1)) = {code for count'} cmp(x,a) + count'(x,insert(b,h2)) + count'(x,h1) = {IH2} cmp(x,a) + cmp(x,b) + count'(x,h2) + count'(x,h1) = {code for count'} cmp(x,a) + count'(x,node(h1,b,h2)) CASE 2. a > b count'(x,insert(a,node(h1,b,h2))) = {code for insert} count'(x,node(insert(a,h2),b,h1)) = {code for count'} cmp(x,b) + count'(x,insert(a,h2)) + count'(x,h1) = {IH2} cmp(x,b) + cmp(x,a) + count'(x,h2) + count'(x,h1) = {code for count'} cmp(x,a) + count'(x,node(h1,b,h2)) ------------------------------------------------------------------------------- LEMMA 2: Forall (x,l) => count(x,l) = count'(x,createheap l) by structural induction on l. BASE CASE. l=nil count(x,nil) = 0 {code for count} count'(x,createheap nil) = {code for createheap} count'(x,empty) = {code for count'} 0 STEP CASE. l=a::l' INDUCTIVE HYPOTHESIS: count(x,l') = count'(x,createheap l') count(x,a::l') = cmp(x,a) + count(x,l') count'(x,createheap (a::l')) = {code for createheap} count'(x,insert(a,createheap l')) {by Lemma 1 with h instantiated to (createheap l')} = cmp(x,a) + count'(x,createheap l') {by IH} = cmp(x,a) + count(x,l') ------------------------------------------------------------------------------- LEMMA 3: Forall (x,l,l') => count(x,merge(l,l')) = count(x,l) + count(x,l') by structural induction on both l, l'. BASE CASE 1. l=nil count(x,merge(nil,l')) = {code for merge} count(x,l') count(x,nil) + count(x,l') = {code for count} count(x,l') BASE CASE 2. l'=nil count(x,merge(l,nil)) = {code for merge} count(x,l) count(x,l) + count(x,nil) = {code for count} count(x,l) STEP CASE. l=a::tl, l'=a'::tl' INDUCTIVE HYPOTHESES: IH1: count(x,merge(a::tl,tl')) = count(x,a::tl) + count(x,tl') IH2: count(x,merge(tl,a'::tl')) = count(x,tl) + count(x,a'::tl') CASE 1. a<=a' count(x,merge(a::tl,a'::tl')) = {code for merge} count(x,a::merge(tl,a'::tl)) = {code for count} cmp(x,a) + count(x,merge(tl,a'::tl)) = {IH2} cmp(x,a) + count(x,tl) + count(x,a'::tl) = {code for count} count(x,a::tl) + count(x,a'::tl) CASE 2. a>a' analogously ------------------------------------------------------------------------------- LEMMA 4: Forall (x,h) => count'(x,h) = count(x,extractheap h) by structural induction on h. BASE CASE. h=empty count'(x,empty) = 0 {code for count'} count(x,extractheap empty) = {code for extractheap} count(x,nil) = 0 {code for count} STEP CASE. h=node(h1,a,h2) INDUCTIVE HYPOTHESES IH1: count'(x,h1) = count(x,extractheap h1) IH2: count'(x,h2) = count(x,extractheap h2) - left side of equality: count'(x,node(h1,a,h2)) = {code for count'} cmp(x,a) + count'(x,h1) + count'(x,h2) = {IH1 and IH2} cmp(x,a) + count(x,extractheap h1) + count(x,extractheap h2) - right side of equality: count(x,extractheap node(h1,a,h2)) = {code for extractheap} count(x,a::merge(extractheap h1,extractheap h2)) = {code for count} cmp(x,a) + count(x,merge(extractheap h1,extractheap h2)) = {Lemma 3} cmp(x,a) + count(x,extractheap h1) + count(x,extractheap h2)) ------------------------------------------------------------------------------- Theorem: Forall l => permutation(l,sort l) Forall l => permutation(l,sort l) <=> {code for sort} Forall l => permutation(l,extractheap (createheap l)) <=> {axiom for permutation} Forall l => (Forall x => count(x,l) = count(x,extractheap (createheap l))) <=> {Lemma 4 with h instantiated to (createheap l)} Forall l => (Forall x => count(x,l) = count'(x,createheap l)) <=> {Lemma 2} Forall l => (Forall x => count(x,l) = count(x,l)) <=> {obvious} true. | 1,805 | 5,628 | {"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} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.428823 |
https://academicden.com/given-the-polar-point-6-13n-4-find-two-other-polar-expressions-for-the-point-on-4/ | 1,618,937,704,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00168.warc.gz | 196,744,481 | 15,488 | # Given The Polar Point 6 13n 4 Find Two Other Polar Expressions For The Point On
Need help. Please show step by step solution so that i could follow along. Thank you.
Given the polar point (6, -13n/4) Find two other polar expressions for the point, one with a negative r value, and another with a positive rvalue. Use exact values. Convert the polar point (6, -13n/4) into rectangular coordinates (x,y). Compute the coordinates exactly‘ Fill in the blanks in the listed polar points (r, 9) below, to satisfy the polar function r = 2 + 4sin(29). Find all ofthese points exactly,Given 6 = 41/4, find r. ( -n/4) Given 9 = n/6, find r4 (# n/G)Given r = 0, find 9, (O, _) Graph the polar function r = 2 + 4sin(26). Locate and label on your graph the three points you found above in problem
# Given The Polar Point 6 13n 4 Find Two Other Polar Expressions For The Point On
Need help. Please show step by step solution so that i could follow along. Thank you.
Given the polar point (6, -13n/4) Find two other polar expressions for the point, one with a negative r value, and another with a positive rvalue. Use exact values. Convert the polar point (6, -13n/4) into rectangular coordinates (x,y). Compute the coordinates exactly‘ Fill in the blanks in the listed polar points (r, 9) below, to satisfy the polar function r = 2 + 4sin(29). Find all ofthese points exactly,Given 6 = 41/4, find r. ( -n/4) Given 9 = n/6, find r4 (# n/G)Given r = 0, find 9, (O, _) Graph the polar function r = 2 + 4sin(26). Locate and label on your graph the three points you found above in problem | 450 | 1,576 | {"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} | 3.984375 | 4 | CC-MAIN-2021-17 | latest | en | 0.770513 |
https://chemistry.stackexchange.com/questions/4979/conducting-current-in-electrolytes/4980 | 1,563,388,671,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00403.warc.gz | 350,322,526 | 39,712 | # Conducting current in electrolytes
I keep trying to figure out how current is conducted through an electrolyte but all I can find are BS half answers. They say the ions conduct, but the specifics are poorly explained or absent. I understand that if you, for example, put sodium chloride in water it will disassociate into positive sodium ions and negative chlorine ions, and that an electric field would cause the ions to move, allowing a current to flow. But what happens once the charges get to the electrodes? Obviously, they can't just stay there in elemental form stably. I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas. If this is all correct, it seems one would end up with sodium and hydroxide ions. Does it still conduct then? If so, how so? And don't you have a solution of sodium hydroxide? It seems this process necessitates that the water be split. Is there any salt or other compound which would act similarly but not result in electrolysis (just conducting)?
• ""But what happens once the charges get to the electrodes?"" That is a different Theme from conduction. Some pages later in Your textbook – Georg May 15 '15 at 14:45
## An approach using a classical model of electron flow
You can think of an electron flowing through a medium as analogous to a pinball bouncing around a pinball machine. The electric (potential) energy difference driving the displacement of the electron is analogous to the gravitational energy difference in the pinball machine. Now consider two pinball machines with differing numbers of obstacles to a falling pinball. The time it takes for a pinball to reach the bottom in each machine would be different by virtue of these obstacles (on average, taking more time with more obstacles). Similarly, the flow of electrons, I, in the presence of an electric potential, V, is higher for media with less resistance, R , to electron flow. We can use this model (Drude see below) to understand Ohm's law: $$I = \frac{V}{R}$$ The observation that electrons flow more readily through electrolyte solutions than non-electrolyte solutions indicates that solutions with charge carriers reduce the resistance of the media. A better understanding of why this is true at a molecular level will require a more in depth look. Notice, though, that we now have answers to your good questions!
"What happens once the charges get to the electrodes?"
The sodium stays in solution as an ion, but the hydrogen ion (from the hydrolysis of water) can be reduced to form hydrogen gas at the cathode, and the chlorine ion can be oxidized to form chlorine gas at the anode. Electrons can be passed through the solution as solvated electrons under some circumstances, but this is unlikely due to the larger energy barrier to this process (around 3 eV). With sufficient energy, these electrons (called solvated electrons) flow through metal and solutions. The resistance to electron flow in metals is much less than in electrolyte solutions, though.
"Would it cause the water to split into hydrogen gas and hydroxide ions?" and "I would expect the sodium, once given an electron, to be looking for a way to give that electron back."
Any additional redox reaction would require electric potential. You mentioned the splitting of water or the formation of sodium metal. This would not occur according to the standard electrode potential unless the electric potential provided is sufficiently high to overcome the "uphill" energy barrier and as you mention electrolysis would be involved.
## The Drude Model
In 1900, Paul Drude used the following schematic of a lattice of metal ions sitting in a sea of electrons to derive a relationship between electron flow and electric potential applied.
Drude Model electrons (shown here in blue) constantly bounce between heavier, stationary crystal ions (shown in red). -From wikipedia entry on Drude model.
A metal's electrical resistance, R, (as seen in Ohm's law above) should take the form:
$$R = \left(\frac{m}{nq^2\tau}\right)$$
where, $m$ is mass of electron; $n$ is number density of electrons; $q$ is charge of electron and $\tau$ is the mean free path. While this classical model does not explain the resistances for all substances perfectly and quantum mechanics can be used to do a better job, it does suggest that if the mean free path for electron flow is increased and the assumptions of the model are valid, the electrical resistance of the media should be reduced.
One way to understand why there is lower electrical resistance (higher electrical conductance) in aqueous solutions of electrolytes is that there is less water in the way of the Drude ion lattice! In other words, in the limit that the solution is a molten salt, we approach a Drudian model. As you add more and more molecules of water to this model, you are reducing the mean free path of the electron and increasing the resistance of the solution.
### Notes
1. The fact that an electrolyte solution is comprised of positive and negative ions does not diminish the utility of Drude's model: electrons are equally attracted (and speed up) moving toward a (metal) cation as they are attracted (and slow down) moving away from a metal cation. The situation is reversed but the effect is the same for a nonmetal anion.
2. For more evidence of the utility of note 1., read about the additive nature of the molar electrical conductance (inverse of resistance) of ions (cations and anions): Kohlrausch's Law of independent migration of ions. The additive nature suggests that both cations and anions improve the conductance of an electrolyte solution as the note above suggests.
• Doesn't your explanation assumes that in electrolytic solutions, electrons, and not ions, are responsible for the electric conduction? But in several calculations it is the ionic mobility we consider and thus assume the ions to be moving. – Satwik Pasani Sep 18 '13 at 14:14
• I would like to update my answer to reflect your good question Satwik. I think my answer to the question 'What happens at the electrodes?' is incorrect, and I rather prefer Nick's answer below on this point. Please read page 25 $\S1-7$ Ions in Solution from Dickerson Grey and Haight's open textbook found here Chemical Prinicples for more information. – Harvey Ryan Johnson Jul 27 '14 at 18:01
Actually if you look at electroplating you would see that the metal ions once give an electron at the cathode do turn into elemental metals and plate the electrode. This is how you electroplate things. You can't conduct forever as eventually you will deplete the ions in the solution, much like a battery will run out of charge if it is driving a current rather than the reverse.
If you use molten salt, you will get sodium at one electrode and chlorine gas at the other. If you use a solution you will either get O2 and H2 at the two electrodes, or the ions precipitating out depending which reaction is more favourable.
• But sodium is highly reactive with water... – Void Star May 13 '13 at 23:30
• @BigEndian yes, but if you use molten salt as your electrolyte there isn't any water so you can get Sodium. If there is water then you get H2 bubbling off instead. – Nick May 14 '13 at 10:06
• Well, I'm just saying that we won't get sodium plated on our electrode! I don't think your answer explains everything I was looking for. – Void Star May 16 '13 at 4:50
• @BigEndian if you dislike this answer and do not think Nick should get the point you should flag it otherwise the points will default to him after 7 days – user1464 May 24 '13 at 1:50
• They will??? I thought they would only default if I marked his as the answer... What does flagging mean? It's not a bad answer, it just doesn't give me the depth I'm looking for. – Void Star May 24 '13 at 2:17 | 1,759 | 7,969 | {"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} | 2.953125 | 3 | CC-MAIN-2019-30 | longest | en | 0.949658 |
https://www.iberlibro.com/9781420099652/Statistical-Inference-Minimum-Distance-Approach-1420099655/plp | 1,511,334,821,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00118.warc.gz | 804,380,452 | 17,003 | # Statistical Inference: The Minimum Distance Approach (Chapman & Hall/CRC Monographs on Statistics & Applied Probability)
## Ayanendranath Basu; Hiroyuki Shioya; Chanseok Park
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In many ways, estimation by an appropriate minimum distance method is one of the most natural ideas in statistics. However, there are many different ways of constructing an appropriate distance between the data and the model: the scope of study referred to by "Minimum Distance Estimation" is literally huge. Filling a statistical resource gap, Statistical Inference: The Minimum Distance Approach comprehensively overviews developments in density-based minimum distance inference for independently and identically distributed data. Extensions to other more complex models are also discussed.
Comprehensively covering the basics and applications of minimum distance inference, this book introduces and discusses:
• The estimation and hypothesis testing problems for both discrete and continuous models
• The robustness properties and the structural geometry of the minimum distance methods
• The inlier problem and its possible solutions, and the weighted likelihood estimation problem
• The extension of the minimum distance methodology in interdisciplinary areas, such as neural networks and fuzzy sets, as well as specialized models and problems, including semi-parametric problems, mixture models, grouped data problems, and survival analysis.
Statistical Inference: The Minimum Distance Approach gives a thorough account of density-based minimum distance methods and their use in statistical inference. It covers statistical distances, density-based minimum distance methods, discrete and continuous models, asymptotic distributions, robustness, computational issues, residual adjustment functions, graphical descriptions of robustness, penalized and combined distances, weighted likelihood, and multinomial goodness-of-fit tests. This carefully crafted resource is useful to researchers and scientists within and outside the statistics arena.
"Sinopsis" puede pertenecer a otra edición de este libro.
Indian Statistical Institute, Kolkata, India Muroran Institute of Technology, Muroran, Japan Clemson University, Clemson, SC, USA
Review:
"The book is an excellent and thorough outline of work in the area. It would provide an ideal volume for someone who plans to undertake research in the area."
International Statistical Review, 2013
"The book provides a comprehensive overview of the theory of density-based minimum distance methods and it is well written and easy to read and understand. The book is well suited for graduate students, professionals and researchers not only in statistics but also in biosciences, engineering and various other fields where statistical inference plays a fundamental role."
―Alex Karagrigoriou, Journal of Applied Statistics, 2012
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## 2.Statistical Inference: The Minimum Distance Approach (Hardback)
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Descripción Taylor Francis Ltd, United States, 2011. Hardback. Estado de conservación: New. Language: English . Brand New Book. In many ways, estimation by an appropriate minimum distance method is one of the most natural ideas in statistics. However, there are many different ways of constructing an appropriate distance between the data and the model: the scope of study referred to by Minimum Distance Estimation is literally huge. Filling a statistical resource gap, Statistical Inference: The Minimum Distance Approach comprehensively overviews developments in density-based minimum distance inference for independently and identically distributed data. Extensions to other more complex models are also discussed. Comprehensively covering the basics and applications of minimum distance inference, this book introduces and discusses: * The estimation and hypothesis testing problems for both discrete and continuous models * The robustness properties and the structural geometry of the minimum distance methods * The inlier problem and its possible solutions, and the weighted likelihood estimation problem * The extension of the minimum distance methodology in interdisciplinary areas, such as neural networks and fuzzy sets, as well as specialized models and problems, including semi-parametric problems, mixture models, grouped data problems, and survival analysis. Statistical Inference: The Minimum Distance Approach gives a thorough account of density-based minimum distance methods and their use in statistical inference. It covers statistical distances, density-based minimum distance methods, discrete and continuous models, asymptotic distributions, robustness, computational issues, residual adjustment functions, graphical descriptions of robustness, penalized and combined distances, weighted likelihood, and multinomial goodness-of-fit tests. This carefully crafted resource is useful to researchers and scientists within and outside the statistics arena. Nº de ref. de la librería AA69781420099652
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## 6.Statistical Inference. Chapman and Hall/CRC. 2011.
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Descripción Taylor Francis Ltd, United States, 2011. Hardback. Estado de conservación: New. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. Brand New Book. In many ways, estimation by an appropriate minimum distance method is one of the most natural ideas in statistics. However, there are many different ways of constructing an appropriate distance between the data and the model: the scope of study referred to by Minimum Distance Estimation is literally huge. Filling a statistical resource gap, Statistical Inference: The Minimum Distance Approach comprehensively overviews developments in density-based minimum distance inference for independently and identically distributed data. Extensions to other more complex models are also discussed. Comprehensively covering the basics and applications of minimum distance inference, this book introduces and discusses: * The estimation and hypothesis testing problems for both discrete and continuous models * The robustness properties and the structural geometry of the minimum distance methods * The inlier problem and its possible solutions, and the weighted likelihood estimation problem * The extension of the minimum distance methodology in interdisciplinary areas, such as neural networks and fuzzy sets, as well as specialized models and problems, including semi-parametric problems, mixture models, grouped data problems, and survival analysis. Statistical Inference: The Minimum Distance Approach gives a thorough account of density-based minimum distance methods and their use in statistical inference. It covers statistical distances, density-based minimum distance methods, discrete and continuous models, asymptotic distributions, robustness, computational issues, residual adjustment functions, graphical descriptions of robustness, penalized and combined distances, weighted likelihood, and multinomial goodness-of-fit tests. This carefully crafted resource is useful to researchers and scientists within and outside the statistics arena. Nº de ref. de la librería BTE9781420099652
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## 8.Statistical Inference: The Minimum Distance Approach
Editorial: Chapman & Hall (2010)
ISBN 10: 1420099655 ISBN 13: 9781420099652
Librería
Revaluation Books
(Exeter, Reino Unido)
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Descripción Chapman & Hall, 2010. Hardcover. Estado de conservación: Brand New. 1st edition. 320 pages. 9.30x6.30x1.10 inches. In Stock. Nº de ref. de la librería __1420099655
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## 9.Statistical Inference: The Minimum Distance Approach (Chapman & Hall/CRC Monographs on Statistics & Applied Probability)
Editorial: Chapman and Hall/CRC (2011)
ISBN 10: 1420099655 ISBN 13: 9781420099652
Librería
Murray Media
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Descripción Chapman and Hall/CRC, 2011. Hardcover. Estado de conservación: New. Never used!. Nº de ref. de la librería 1420099655
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## 10.Statistical Inference: The Minimum Distance Approach (Hardcover)
ISBN 10: 1420099655 ISBN 13: 9781420099652
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Descripción Hardcover. Estado de conservación: New. Hardcover. In many ways, estimation by an appropriate minimum distance method is one of the most natural ideas in statistics. However, there are many different ways of constructing an appropriate dis.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 409 pages. 0.522. Nº de ref. de la librería 9781420099652
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Gastos de envío: GRATIS | 2,949 | 12,456 | {"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} | 2.53125 | 3 | CC-MAIN-2017-47 | longest | en | 0.87249 |
https://thecivilengineerings.com/how-to-measure-the-height-of-a-building-tower-or-skyscraper/ | 1,669,661,041,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00723.warc.gz | 596,315,250 | 71,452 | # How to Measure the Height of A Building/Tower or Skyscraper | How to Find the Height of Object/Building by using Theodolite
In this Article today we will talk about the how to measure the height of a building | how to measure elevation of Building | angle of depression calculator | how to measure a building | height of a house | building height measurement | height of building | How is Building Height measured | Broomstick Method
## How to Calculate Height of a Building or Tower:
Sometimes we may need to find the height of a building before or after construction. There are several ways to calculate building height. In this article, I will use two methods: how to measure elevation of Building
### Trignometry Method for Calculating the Height of Building/Tower (Method 1):
This is the simplest method. you can use this method to find out the height of any object such as building or tower, water tank, tree, lighthose etc. angle of depression calculator
#### Required Data:
Diatance and Angle (As shown in fig)
#### Given:
Angle = Ɵ = 30o Distance – d – 5000 feet
#### Procedure:
We Know, height of a house
Tangent = The Ratio of the opposite side to the adjacent side.
which means tanƟ = Opposite side / Adjacent Side
here Ɵ = 30o
So, Tan = 30o = Opposite side / Adjacent Side = x/d = x/5000
x = tan 30o x 5000 = 0.577 x 5000 = 2885 feet
Hence the Height of the Building/Tower is 2885 feet. height of a house
### Broomstick Method (Method 2):
It is not always possible to use the above method due to the availability of space or time. This measurement method also uses shadows but you do not have to wait until the length of the broom shadow is equal to its height. Place the broomstick on the floor facing upwards. Now measure the height of the broom and the length of the drop shadow. Quickly measure the length of the shadow created by the structure. With the same triangles, we can reach the height of the building. building height measurement
To understand the properties of similar triangles, see the image below. The structure and its shape form a right-angled triangle. And the broomstick and its shadow form the same triangle of right angles. Since both triangles have the right angles, they are the same.
The height of the structure can therefore be calculated from the ratios stated below.
(Broomstick’s shadow length/ Broomstick’s height) = (Building’s shadow length/ Building’s height) As per the above formula; h/x = a/b
#### Required Data:
Suppose, x = 600 ft and a = 100 ft, b = 60 ft then, find x = ?
#### Procedure:
As we know that,
h/x = a/b
so, h = (a/b) * x = 100/60 x 600 = 1000 ft
Hence the Height of the Building/Tower is 1000 feet.
### How to Measure the Height of Object/Building by using Theodolite (Method 3):
Now let’s calculate the height of a building using theodolite as shown below.
First, you should measure the theodolite on a straight axis and set the angle to 0 ° on the horizontal axis e.g shown as a dotted line. You must measure the distance (L) of the building from the location of the theodolite channel as shown in the diagram.
Place the balance in the upper corner of the building using theodolite and write down the value of the angle θ1. Similarly, bisect the bottom edge corner of the building and write down the value of angle θ2.
Let’s draw the triangles above for the purpose of calculating the values of the measured angle. Broomstick Method
Let’s say triangle 1 is like ABC, where A is the point of the theodolite channel.
Triangle ABC : Angle θ1 = 34° 7′
Length AB = L = 72m.
From trigonometry,
Tan θ1
= opposite side ÷ adjacent side.
= side BC ÷ side AB
Tan 34° 7′
= h1 ÷ 72m.
h1 = tan 34° 7′ × 72m.
= 0.67747 × 72m
48.778m.
Let us name triangle 2 as ABD, where A is the theodolite station point.
Triangle ABD. Given data:
Angle θ2 = 1° 12′
Length AB = L = 72m.
From trigonometry,
Tan θ2 =
= side BD ÷ side AB
Tan 1° 12′
= h2 ÷ 72m.
h2 = tan 1° 12′ × 72m.
= 0.02094 × 72m
1.508m.
Now, the height of building
H = h1 + h2
= 48.778m. + 1.508m.
= 50.286m.
## Conclusion:
Full article on how to measure the height of a building | how to measure elevation of Building | angle of depression calculator | how to measure a building | height of a house | building height measurement | height of building | how is building height measured | How is Building Height measured | Broomstick Method. Thank you for the full reading of this article in “The Civil Engineering” platform in English. If you find this post helpful, then help others by sharing it on social media. If any formula of BBS is missing from this article please tell me in comments.
Hi! Welcome to my blog. My name is Engr Waseem Raja and I’m the author of TheCivilEngineerings.com. I am a Civil Engineer by profession but I’ve specialized and taken the journey in the field of Quality Engineering. I’ve worked as a Quality Engineer & Manager in the well-known companies in the Gulf region for almost Ten years. | 1,297 | 4,998 | {"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} | 4.4375 | 4 | CC-MAIN-2022-49 | longest | en | 0.91095 |
http://library.thinkquest.org/16497/basic/atoms/index.html | 1,369,512,136,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706153698/warc/CC-MAIN-20130516120913-00016-ip-10-60-113-184.ec2.internal.warc.gz | 150,277,772 | 2,484 | Atoms are the building blocks of all matter. They are made up of protons, neutrons, and electrons. Every electron has a small negative (-) charge. The proton has the same amount of charge except that it is the opposite, positive (+) charge. Neutrons are electrically neutral and have no charge. The protons and neutrons are located in the center of atoms forming what is called the nucleus and the electrons revolve around them.
It is very important to know that particles of like charges will repel and unlike charges will attract. For example, two protons or two electrons will repel each other. However, a proton and a electron will attract. That is how the electrons are held inside the atom. The attraction between the electrons and protons keeps the electrons in orbit much like the gravitational attraction between the sun and its planets.
Electricity is the flow of electrons so it is necessary to measure the charge. The basic unit for measuring charge is the coulomb or the letter C. 1 coulomb is equal to the charge of 6,250,000,000,000,000,000 electrons!!!
1C = 6.25x10^18 electrons | 245 | 1,097 | {"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} | 2.546875 | 3 | CC-MAIN-2013-20 | longest | en | 0.955287 |
https://stats.stackexchange.com/questions/611453/vae-active-units/611487 | 1,716,882,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00312.warc.gz | 473,515,476 | 38,766 | # VAE active units
According to Burda et al (2015) number of active units is computed as:
$$Cov_x(E_{z \sim q_\phi(z|x)}) > \delta$$
for some particular delta. In the paper it is set to 0.02 empirically. But this only measures units in vector z that are mostly constant. Another way of prior collapse for VAE is random units in z, i.e. - units that are independent from input x and those that decoder part of VAE learns to ignore. Is there a way to measure it?
One way I can think of is to set units in z to random value and measure its effect on decoded input. But this is rather slow process for high-dimensional z. Is there a better way?
A latent unit $$z_i$$ having low covariance with $$x$$ doesn't mean that $$z_i$$ is constant, it means it varies independently of $$x$$, i.e. $$p(x,z_i) = p(x)p(z_i)\ ^{**}$$. Simple proof:
$$\text{Covar}(x, z_i) = \int_{x,z} \!\!\!p(x,z) (x-\bar{x})(z - \bar{z}) \ \overset{**}{=} \int_{x,z} \!\!\!p(x)p(z) (x-\bar{x})(z - \bar{z}) \\\qquad\quad\ \ = \int_{x} p(x)(x-\bar{x}) \int_zp(z) (z - \bar{z}) = (\bar{x}-\bar{x}) (\bar{z} - \bar{z}) = 0$$
So low covariance (not variance) identifies $$z$$ components that are independent of $$x$$, i.e. carry no information, or are "dead" from a useful representation perspective.
The same measure is used when considering posterior collapse (e.g. in Variational Autoencoders (VAEs)), which refers to when all dimensions of $$z$$ are independent of $$x$$. | 436 | 1,442 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-22 | latest | en | 0.84948 |
http://www.ecphorizer.com/EPS/site_page.php?page=1127&issue=73 | 1,560,674,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00213.warc.gz | 234,483,383 | 7,610 | # The Ecphorizer
#### Issue #52 (December 1985)
This chapter of my almanac presents another mathematics problem in rhyme from my great-grandfather's work book:
I placed a bowl into the storm,
To catch the drops of rain —
A half a globe was just its form,
Two feet across the same;
The storm was o'er, the tempest past,
I to the bowl repaired —
Six inches deep the water stood,
It being measured fair;
Suppose a cylinder, whose base
Two feet across within,
Had stood exactly in that place,
What would the depth have been?
I have not attempted to analyze the solution, as I am not up on computing the cubic contents of globes. His answer is 2.5" for the depth of the water in the cylinder, which I assume is correct.
close
Title: | 177 | 732 | {"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} | 2.703125 | 3 | CC-MAIN-2019-26 | latest | en | 0.965663 |
https://en.wordpress.com/tag/polygons/ | 1,529,479,557,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00233.warc.gz | 609,136,160 | 20,089 | ## Tags » Polygons
#### Merging spatial buffers in R
I’m sure there’s a better way out there, but I struggled to find a way to dissolve polygons that touched/overlapped each other (the special case being buffers). 407 more words
R
#### 6th Grade Unit 6: Angles and Triangles
##### Angle Pairs
I used this Which One Doesn’t Belong? to start our unit on angles and triangles. I love how starting with something like this gives me insight on where students are at with this topic based on their answers and the vocabulary they are using. 210 more words
#### Intersecting points and overlapping polygons
UPDATED…
I’ve been doing some spatial stuff of late and the next little step will involve intersecting points with possibly many overlapping polygons. The sp package has a function called over which returns the polygons that points intersects with. 296 more words
R
Math
Math
#### More Bracketology
Further examples following the bracketology idea…..
No as it stands not terribly exciting. The idea is to logically deduce which bracket represents the larger value and that value goes into the next round. 93 more words
#### Select a loop of faces/Edges in Blender
Software:
Blender 2.79
Alt + RMB Click a face or Edge in the loop.
Blender | 282 | 1,258 | {"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} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.923667 |
titanintegrators.com | 1,540,305,705,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516194.98/warc/CC-MAIN-20181023132213-20181023153713-00381.warc.gz | 375,013,109 | 3,201 | FORMULAE
DC Equations Voltage = Current x Resistance (V=IR) Power Real = Voltage x Current (P=VI) Single Phase Equations Power Apparent = Power Real 2 + Power Reactive 2 Power Factor = Power Real / Power Apparent Three Phase Equations Power Apparent = Voltage x Current x 1.732 Power Real = Voltage x Current x Power Factor x 1.732 Horsepower = Power Real x Efficiency 746 AC Speed Equations AC Synchronous Speed = Frequency x 120 Number of Poles Percent Slip = Synchronous Speed – Full Load Speed x 100 Synchronous Speed Motor Application Horsepower = Torque x Speed 5250 Acceleration Torque = Inertia x ΔSpeed 308 x ΔTime Units for equations on this page: Voltage: Volts (V) Current: Amperes (A) Resistance: Ohms (Ω) Frequency: Hertz (hz) Rotational Speed: RPM Power Real: Watts (W) Power Apparent: VA Power Reactive: VARS Power Mech: Horsepower (hp) Torque: Foot-pounds (lb-ft) Inertia: (Foot-pounds)2 (lb-ft)2 Time: Seconds (s)
Home Services Products Experience Contact Us Employment Tech Info Links Website Designed and Maintained by Titan Controls and Integration Inc. All pages copyright 2002-2012 Titan Controls and Integration Inc. This page last updated on 2012-04-11 02:00:42 AM | 348 | 1,429 | {"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} | 2.53125 | 3 | CC-MAIN-2018-43 | longest | en | 0.586191 |
https://www.stat.math.ethz.ch/pipermail/r-help/2010-June/242908.html | 1,603,846,004,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00156.warc.gz | 939,871,135 | 3,347 | # [R] mgcv, testing gamm vs lme, which degrees of freedom?
Carlo Fezzi c.fezzi at uea.ac.uk
Fri Jun 18 18:27:37 CEST 2010
```Dear Simon,
Unfortunately I am still confused about which is the correct way to test
the two models... as you point out: why in my example the two models have
the same degrees of freedom?
Intuitively it seems to me the gamm model is more flexible since, as I
understand also from you response, it should contain more random effects
than the other model because some of the smooth function parameters are
represented as such. This should not be taken into account when testing
one model vs the other?
Continuing with my example, the two models:
f2 <- gamm(y ~ s(x), random = list(id=~1), method="ML")
f3 <- gamm(y ~ x + I(x^2), random = list(id=~1), method="ML" )
Can be tested with:
anova(f3\$lme,f2\$lme)
But why are the df the same? Model f2 appears to be more flexible and, as
such, should have more (random) parameters. Should not a test of one model
vs the other take this into account?
Sorry if this may sound dull, many thanks for your help,
Carlo
> On Wednesday 16 June 2010 20:33, Carlo Fezzi wrote:
>> Dear all,
>>
>> I am using the "mgcv" package by Simon Wood to estimate an additive
>> mixed
>> model in which I assume normal distribution for the residuals. I would
>> like to test this model vs a standard parametric mixed model, such as
>> the
>> ones which are possible to estimate with "lme".
>>
>> Since the smoothing splines can be written as random effects, is it
>> correct to use an (approximate) likelihood ratio test for this?
> -- yes this is ok (subject to the usual caveats about testing on the
> boundary
> of the parameter space) but your 2 example models below will have the
> same
> number of degrees of freedom!
>
>> If so,
>> which is the correct number of degrees of freedom?
> --- The edf from the lme object, if you are testing using the log
> likelihood
> returned by the lme representation of the model.
>
>> Sometime the function
>> LogLik() seems to provide strange results regarding the number of
>> degrees
>> of freedom (df) for the gam, for instance in the example I copied below
>> the df for the "gamm" are equal to the ones for the "lme", but the
>> summary(model.gam) seems to indicate a much higher edf for the gamm.
> --- the edf for the lme representation of the model counts only the fixed
> effects + the variance parameters (which includes smoothing parameters).
> Each
> smooth typically contributes only one or two fixed effect parameters, with
> the rest of the coefficients for the smooth treated as random effects.
>
> --- the edf for the gam representation of the same model differs in that
> it
> also counts the *effective* number of parameters used to represent each
> smooth: this includes contributions from all those coefficients that the
> lme
> representation treated as strictly random.
>
> best,
> Simon
>
>
>> I would be very grateful to anybody who could point out a solution,
>>
>> Best wishes,
>>
>> Carlo
>>
>> Example below:
>>
>> ----
>>
>> rm(list = ls())
>> library(mgcv)
>> library(nlme)
>>
>> set.seed(123)
>>
>> x <- runif(100,1,10) # regressor
>> b0 <- rep(rnorm(10,mean=1,sd=2),each=10) # random intercept
>> id <- rep(1:10, each=10) # identifier
>>
>> y <- b0 + x - 0.1 * x^3 + rnorm(100,0,1) # dependent variable
>>
>> f1 <- lme(y ~ x + I(x^2), random = list(id=~1) , method="ML" ) # lme
>> model
>>
>> f2 <- gamm(y ~ s(x), random = list(id=~1), method="ML" ) # gamm
>>
>> ## same number of "df" according to logLik:
>> logLik(f1)
>> logLik(f2\$lme)
>>
>> ## much higher edf according to summary:
>> summary(f2\$gam)
>>
>> -----------
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help | 1,028 | 3,812 | {"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} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.913985 |
https://www.buzzfeednews.com/article/jsvine/grading-the-2016-election-forecasts | 1,716,434,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058588.75/warc/CC-MAIN-20240523015422-20240523045422-00591.warc.gz | 613,276,732 | 134,094 | # Here’s How We’re Going To Grade The 2016 Election Forecasts
We’re laying out our methodology ahead of time, so you can get a sneak peek — and hold us accountable.
Update, Nov. 28: Results published here.
America’s political prognosticators have spent four years waiting for Tuesday. Did they predict the election correctly? Did they outperform their peers’ forecasts? And will “drunk Nate Silver” ride the subway, telling strangers the day they will die?
As the votes roll in, BuzzFeed News will help answer the first two questions. In this post, I’ll spell out our methodology — partly because you might be interested, and partly to avoid accusations that we’ve “moved the goalposts” to influence the ratings.
Our grades will focus on eleven separate models from nine forecasters:
For each forecast model, we’ll examine the following predictions:
• Every statewide presidential prediction, plus DC’s. That means we won’t be grading predictions for Maine's and Nebraska’s district-specific electoral votes; the forecasters handle these races in slightly different ways, and some not at all.
• Every Senate prediction, except for California and Louisiana. This year, California’s Senate race pits two Democrats against one another, while Louisiana’s contest is technically a primary. (If one candidate in Louisiana gets the majority of votes, however, they automatically win the general election, which is otherwise scheduled for Dec. 10.)
We’ll primarily grade each set of predictions using something called the Brier score. It’s a widely accepted calculation for quantifying the accuracy of probabilistic predictions. (In fact, FiveThirtyEight itself has used it to evaluate its own March Madness and UK general election predictions.) There are a few nice things about the Brier score:
• For each forecast, it produces an easy-to-understand number, ranging from 0 to 1. A score of 0 is perfection: You were 100% confident on every prediction and got them all right. A score of 1 is total error: You were 100% confident on every prediction and got them all wrong.
• It rewards confidence on correct predictions, and penalizes confidence on incorrect predictions. If two forecasters predict every state correctly, for example, the more confident forecaster will score better.
• It disproportionately penalizes overconfidence. For example, if you gave Trump a 70% chance in Georgia but he loses there, you’d receive a 0.49-point penalty. But if you gave him an 80% chance, you’d receive a 0.64-point penalty.
For each forecast we’ll produce up to three Brier scores:
• An unweighted Brier score for the presidential election, in which the predictions for each state count equally.
• An electoral vote–weighed Brier score for the presidential election, in which each prediction is weighted by the state’s number of electoral votes. (Because we’re ignoring Maine and Nebraska’s district-specific electoral votes, those states will each receive a weight of 2 — the number of electoral votes assigned on a statewide basis.)
• An unweighted Brier score for Senate predictions.
We’ll also provide some other helpful ways of understanding the forecasts’ accuracy:
• For forecasts that publish estimates of the Clinton-Trump percentage vote difference, we’ll also calculate a score — the root-mean-square error — that quantifies how closely those forecasts predicted the final spread.
• To complement the final Brier scores, we’ll also chart the Brier scores historically — for forecasts that have made their prediction history available — to provide a sense of each model’s volatility.
• For both the Senate and presidential races, we’ll also list each forecast’s raw number of correct and incorrect predictions.
We’ll base these scores on the called races at the time of calculation. If we publish scores calculated before all races are called, we’ll also publish each forecast’s best/worst possible final scores.
Fine print:
• Any forecast that does not provide probabilities for third-party candidates — e.g., for Evan McMullin in Utah or Gary Johnson in any state — will be considered to be giving that candidate a 0% chance of winning.
• In cases where, due to rounding, a projection’s total probabilities for all candidates in a race is not equal 100.00%, those odds will be proportionally adjusted to equal 100% (e.g., 80% and 21% become 80%/101% = 79.2% and 21%/101% = 20.8%).
• When calculating the number of correct predictions for forecasts that provide tied odds for the favorite (e.g., 50%/50% or 40%/40%/20%), we’ll give half-credit if either of those candidates wins.
• We’ll be using each forecast’s predictions as of 11:59 p.m. ET on Monday, Nov. 7.
Questions?
Update at 4:25 p.m.: Added PollSavvy to list of forecasts. | 1,012 | 4,757 | {"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} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.921776 |
http://mathoverflow.net/revisions/88464/list | 1,369,486,638,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705953421/warc/CC-MAIN-20130516120553-00027-ip-10-60-113-184.ec2.internal.warc.gz | 176,006,004 | 4,762 | If you denote $A_n$ your tri-diagonal matrix of order $n$, and $H_n(x):= \det(x+A_n)$, the sequence $H_n$ saisfies satisfies the two term two-term linear recurrence $H_{n+1}=xH_n - nH_{n-1}$ with initial conditions $H_0=1$ and $H_1=x$. Thus, they are the Hermite polynomials (here in the "probabilist's version"), and their zeros are the eigenvalues of $-A_n$. -A_n$(on which you can find everything in the literature). 1 If you denote$A_n$your tri-diagonal matrix of order$n$, and$H_n(x):= \det(x+A_n)$, the sequence$H_n$saisfies the two term linear recurrence$H_{n+1}=xH_n - nH_{n-1}$with initial conditions$H_0=1$and$H_1=x$. Thus, they are the Hermite polynomials (here in the "probabilist's version"), and their zeros are the eigenvalues of$-A_n\$. | 246 | 752 | {"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} | 3 | 3 | CC-MAIN-2013-20 | latest | en | 0.791823 |
https://womenstagetheworld.org/3-diget-math-multiplication-worksheet-numbers/ | 1,566,122,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00004.warc.gz | 710,513,845 | 31,113 | # 3 Diget Math Multiplication Worksheet Numbers
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Top | 2,066 | 10,451 | {"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} | 3.25 | 3 | CC-MAIN-2019-35 | latest | en | 0.876612 |
https://www.jiskha.com/display.cgi?id=1229040771 | 1,516,384,006,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00324.warc.gz | 895,642,672 | 3,829 | # math
posted by .
i am really stuck on this application problem in my quadratic equations unit... find the width of a uniform concrete path placed around a 30m by 40m rectangular lawn if the concrete has an area that is 1/4 of the lawn.
• math -
Draw rectangle, label the width and length. The length is 30 and the width is 40. Divide this rectangle into fourths. Now you have four squares. Shade in one of the squares. That square will represent your concrete because its 1/4 of the lawn.
Do you think you are able to figure out the width of the concrete now?
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More Similar Questions | 734 | 2,928 | {"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} | 3.4375 | 3 | CC-MAIN-2018-05 | latest | en | 0.926533 |
hopeisthewordblog.com | 1,480,786,867,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541066.79/warc/CC-MAIN-20161202170901-00013-ip-10-31-129-80.ec2.internal.warc.gz | 130,145,453 | 31,462 | ## Multiplication & Fractions: Math Games for Tough Topics by Denise Gaskins (Review AND Giveaway!)
Have you hit the mid-November slump yet? Thankfully, I haven’t, but it’s mostly because we started our school year late this year. If you have, I have just the thing: a brand new math games book! Math games add that spark to our homeschool week that makes me, the mama, feel like I can go on one more day. That may sound a bit melodramatic, but when you’re the mama doing math with three different children daily (and entertaining a three-year-old in the middle of all that!), it is pretty close to the truth. 🙂 We’re math games fans from way back, having used Right Start Math and its math games book from the very beginning of our homeschool journey. However, I like a bit of variety in my weeks, so I’m always on the lookout for new-to-us games. I discovered Denise Gaskins’ Let’s Play Math website several years ago, and I have since considered her both an online mentor in the math education department and a kindred spirit.
Multiplication & Fractions: Math Games for Tough Topics by Denise Gaskins is book three in her Math You Can Play series. (You can read my review of the first two books here.) I’m so excited that she has written a book that will benefit my older children! This book consists of four sections:
• “A Strategy for Learning”
• “Multiplication and Fraction Games”
• “Playing to Learn Math”
• “Resources and References”
Sections one and three are common to all of the Math You Can Play books, and they’re the place to learn the nitty-gritty behind Gaskins’ philosophy. Section two, of course, is where the magic happens: four chapters’ worth of math games dealing with mathematical models, the times table, mixed operations, and fractions and decimals. A total of twenty-five games are detailed in this book, with countless variations. My girls and I have already tested out a game called Distributive Dice from the “Mixed Operations” chapter, and we found it lots of fun. I love that it provides opportunities for deep thought about the way multiplication works. Manipulating the numbers into different sums, differences, and products increases the brain’s facility for mathematical thinking!
We’ve also tried out a fractions game called My Nearest Neighbor. This game involves comparing fractions and determining which one is closer to certain unit fraction or integers. This necessitates converting fractions to like terms and then comparing them. Because this is not something that Louise has mastered yet, I put Lulu in the “teacher” role and let her explain each fraction to Louise. We all enjoyed playing this a lot, and the fraction chart got a workout.
I’m eager to try more of the games from this book. One thing I really love about Denise Gaskins’ books is that she gives you practically everything you need to play the games: the games either require common household items (like basic playing cards, tokens or small toys for game pieces, etc.) or she offers them as free downloads on her website. [Note: the cards above are not from the book; they’re from the RightStart math games kit.]
In the very last chapter of the book Gaskins delves further into her philosophy and confronts “Workbook Syndrome”–
a distressing malady that afflicts children in public, private, and homes schools across our country. A child suffering from this disease has learned to do calculations on a school math page but cannot make sense of numbers in real life.
She offers many tools and tricks for making math more playful and providing a balance between the pencil-and-paper work and mental math, including memorization. In the last section of the book, Gaskins provides all kinds of helpful information, including the basics of game-playing and some of her favorite resources.
Reading one of Gaskins’ books is like going to a really great teacher workshop–part philosophy, part practical ideas, and all excellent. She just oozes expertise and enthusiasm. Truly–the depth of her knowledge and her ability to communicate it are life-giving to a homeschooling mama. I couldn’t recommend this book, or her others, more highly.
And so, dear readers, I am pleased to announce that the author has offered one reader here at Hope Is the Word a copy of this brand new book–winner’s choice of either the e-book or the paperback! Here’s how you enter:
Please leave a separate comment for each of the above things you do, and be sure to put your email address in one of the comments. 🙂
This giveaway will end at one week from today, Friday, November 25, at 9 p.m.
{Disclosure: I read a pre-publication version of the ebook in order to write this review. The link is an affiliate link to Amazon.}
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## Fall 2016 Circle Time
So far this year, Circle Time has been short and sweet. Well, that’s not exactly true; schooling a seventh grader and a first grader (with a fifth grader in the middle and a very chatty three year old bringing up the rear) has necessitated a re-ordering of our days. What that means is that we start our days at home with a very brief together-time, and then the girls and I get back together sometime just before or after lunch for our own “big girls” Circle Time. It looks like we have finally settled into a routine, and so I’m finally ready to share it.
We start with music. After several years of traditional hymns, this year I decided to bring in something that I thought might appeal a bit more to my six-year-old who really likes music. I decided something very upbeat and fun was in order, and what’s more upbeat and fun than fiddle music?
I bought the album Getty Kids Hymnal: In Christ Alone and started with the first song: a medley of the Doxology and “Oh, Shout for Joy.” It has turned out to be a lot of fun as well as a great song to take us into the Thanksgiving season. I think we’ll just continue with this collection of songs after Christmas. I’ve never regretted investing the time in hymn singing with my children!
I let the girls pick out their own memory work poems for this first session. Lulu chose “Jabberwocky” by Lewis Carroll, which she almost already had memorized because we’ve watched this video so many times.
You’re welcome. 🙂
Louise chose Alfred, Lord Tennyson’s “The Oak”:
## The Oak
### by Alfred Lord Tennyson
Live thy Life,
Young and old,
Like yon oak,
Bright in spring,
Living gold;
Summer-rich
Then; and then
Autumn-changed
Soberer-hued
Gold again.
All his leaves
Fall’n at length,
Look, he stands,
Trunk and bough
Naked strength.
This poem has turned out to not be a super easy one to memorize, but she’s working at it and will eventually conquer it.
Most days we also read the Bible together and pray, though the Bible reading part has begun to shift to supper time so we can share that with Steady Eddie.
That’s pretty much it, unless I read a picture book or two. Well, I always try to do that, but I leave it up to the girls as to whether or not they’ll stay and listen. Predictably, the eldest almost always opts not to, and the younger sister almost always does. I want to do more, but right now is not the time to bring the boys in on it. We’ll work on that as they get older.
This year has by far been the biggest year of change we’ve had yet in our homeschool. I knew it was coming, but I wasn’t really prepared for it. It’s good, though. My girls are taking on huge amounts of responsibility. That’s what we want! (Although I will admit that it’s a little bittersweet. . . )
How do you handle your family/together work in your homeschool?
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## Thanksgiving Party School (or how to host an authentic Thanksgiving feast)
I first heard the term party school from Julie Bogart of Bravewriter fame. I’m pretty sure she mentions it in her magnum opus, The Writer’s Jungle. However, I’ve been immersed in the Bravewriter world for a while now, so I’m really not sure where the idea first came up. I do know, though, that she explains the whole idea thoroughly in this video. The idea is that you study something in your homeschool deeply (a “deep dive,” as Julie would say) and then you throw a party to celebrate and show off what you learned. This almost always involves food, and it could also involve games or art or whatever is appropriate to the study (and whatever your children are inspired by). We had the party school to end all party schools a few years ago, and I wanted to bring it up out of the archives and share it again, just in case it might inspire someone.
Reading Eating the Plates by Lucille Recht Penner aloud together was the catalyst for this particular party. I mean, how could we spend ten chapters (about a hundred pages) immersed in the the Pilgrims’ eating habits, mealtime etiquette (or lack thereof, at least according to our standards), and way of life without wanting to recreate it? Through this book we learned how they benefited from the help of the Indians, as well as how their lives changed over time after trade was established that enabled them to get a greater variety of spices, etc. We also learned about their houses and just how they lived from day to day. A lot of their time, of course, was devoted to procuring and preparing food! One thing I love about the book is that Penner intersperses “wit and wisdom” throughout; an appropriate little poem prefaces each chapter, for example. The crown jewel of the whole book, though, is the collection of recipes at the end. These recipes are purportedly authentic, but from different times during the Pilgrims’ first years in America. We made all but two of the recipes included in the book:
• fresh corn soup
• red pickled eggs
• hot Indian pudding
• succotash stew (we started this one but ran out of time)
• spicy cucumber catsup
• bannock cakes
• whole baked pumpkin stuffed with apples
• bearberry jelly
• swizzle
• hot nuts (we opted not to do this one since nuts are expensive and something we eat anyway)
Here are a few pictures of the most interesting recipes: bearberry (cranberry) jelly, red pickled eggs, and the apple-stuffed pumpkin.
We topped this meal off with Goody O’Grumpity‘s spice cake.
We invited over the grandparents and shared this meal with them. The girls picked out quotations and shared them on our placards.
Louise also shared some of the rules of etiquette and decorated the table with stern warnings about which utensils were acceptable at our authentic feast.
We did this four years ago, and none of us has ever forgotten it. It is a lot of work (yes!) and time-consuming, but most things worth doing are.
Some books that pair perfectly with Eating the Plates and this whole “what did the Pilgrims eat?” experience are
And of course, our favorite Thanksgiving chapter books and Thanksgiving picture books would add even more layers of learning to this already rich learning experience.
Holidays are the perfect time to inject a little excitement into our homeschool, and party school is a fabulous way to do just that!
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## Thanksgiving picture books to read aloud
We’ve read a lot of Thanksgiving books over the years. On Monday I shared a quartet of longer Thanksgiving stories to read aloud. Today I’m sharing what I consider the best picture books–our favorites. Enjoy!
Books that tell the story of the first Thanksgiving:
Kate Waters’ books about the Mayflower, the Wampanaog, the Pilgrims, and Thanksgiving are must-reads. Giving Thanks: The 1621 Harvest Feast focuses on what we traditionally think of as the first Thanksgiving. Of course, this whole story is one which for which it is somewhat difficult to separate the fact from the folklore, but I think Giving Thanks does a good job of illuminating what really might have happened (i.e. a series of meals together, in celebration of harvest and friendship, etc.) This story is told in alternating voices: that of Dancing Moccasins, a 14-year-old Wampanoag boy, and Resolved White, a 6-year-old English boy. It’s interesting, if a bit cumbersome at times, to note the differences between their ways of life. Russ Kendall‘s photography of historical re-enactors at Plimoth Plantation absolutely brings both of these stories to life. All of Waters’ books provide a very realistic depiction of the Pilgrims and the Native Americans and how they lived, and we love them.
These books are ideal for mid-elementary aged students and up.
When it comes to factual books that sum the expected Thanksgiving story up in a few pages, the two I recommend are Let’s Celebrate Thanksgiving by Peter and Connie Roop and The Story of the Pilgrims by Katherine Ross. Both picture books provide the facts in what seems to be a fair and balanced way. Of the two, The Story of the Pilgrims has a slightly more storybookish feel. The illustrations in both books are stylized, so pick the one you like the looks of best–you can’t go wrong with either.
If you’re looking for a Thanksgiving book that explains the first Thanksgiving for the preschool set, The Very First Thanksgiving Day by Rhonda Gowler Greene is just the book! Written in rhymes, three or four lines to a page, this book covers the well-known basics: the food, the Indians, the Pilgrims, the Mayflower, the community, and the coming together. The text is reminiscent of “The House That Jack Built”; full of repetition, it’s sure to captivate even the youngest listener.
How Thanksgiving Became a Holiday
Thanksgiving in the White House by Gary Hines is simply the story of how the holiday that we celebrate here in the U.S. on the last Thursday of November each year came to be. Really, though, it’s as much about Tad Lincoln and his White House hi-jinks as anything. It is also about the loving relationship he had with his esteemed father.
If you want to bring in the little woman who campaigned to make Thanksgiving a holiday, check out Thank You, Sarah: The Woman Who Saved Thanksgiving by Laurie Halse Anderson. Anderson’s research is impeccable and the story is wonderfully written. Matt Faulkner’s illustrations are quirky and fun. For mid-elementary age and up, this one is indispensible if you want to know the historic origins about our national holiday.
Thanksgiving traditions
Melissa Sweet is one of my favorite author/illustrators, so I cannot leave out Balloons over Broadway: The True Story of the Puppeteer of Macy’s Parade. This well-researched and whimsically illustrated book is the story of Tony Sarg, the genius behind the huge floats we know and love from the Macy’s Thanksgiving Day parade. As a child, Tony had quite the engineering mind and loved figuring out how to make things move. As an adult he began making marionettes, and his lifelike puppets were soon performing on Broadway. This led him to a job at Macy’s, where he was asked to design a window display for the holidays–a “puppet parade.” This was a hit, so Macy’s then commissioned Tony to do something even grander: put together a parade for the immigrant employees who longed for the fanfare of the holidays they celebrated in their native lands. After this, the parade just got bigger and bigger, with Tony’s crowning achievement being figuring out how to engineer a marionette (of sorts) that would float instead of hang down–one that could be “articulated” (moved in precise ways) but still be seen high above the crowded New York City sidewalks. If you haven’t enjoyed anything by Melissa Sweet, or even if you have, be sure to read this perfect picture book this Thanksgiving!
Family togetherness
Traditionally, Thanksgiving means family fun. Many books highlight this; here are a few great ones.
Sharing the Bread: An Old-Fashioned Thanksgiving Story by Pat Zietlow Miller is a fantastic rhyming story that focuses on family togetherness and teamwork. Everyone in the family–Mama, Daddy, Sister, Brother, Baby, Grandma, Grandpa, Auntie, and Uncle–has a job to do to prepare for Thanksgiving. Rhyme and repetition make this book perfect for toddlers and preschoolers. Jill McElmurry’s old-fashioned illustrations set the perfect tone for this lovely family celebratory tale. Fun and old-fashioned–so much to love about this one!
Thanksgiving Is Here! by Diane Goode focuses on the fun of family at Thanksgiving. The text in the story literally floats across the pages; this gives the story a lilting feel, even though the story doesn’t rhyme. The first several pages end in a sentence that begins with “we”: “We all love to cook at Thanksgiving”; “We all have a place at the table”; etc. This is a fun book to read, and the whimsical illustrations really add to the overall feeling of the story. My favorite picture is the two-page spread of the whole family (well over twenty people) sitting down at the long table they have created by pushing smaller tables together. This feel-good story that would appeal to just about any age, from preschool to adult.
Just for fun
Cranberry Thanksgiving by Wende and Harry Devlin is the tale of Mr. Whiskers and his plot to get Grandmother’s secret cranberry bread recipe. I dare you to read this one and NOT make cranberry bread after you’re through reading it! A fun plot, silly and likable characters, and a recipe to boot–lots to love for old and young alike in this story.
The Peterkins’ Thanksgiving by Elizabeth Spurr (who adapted the 1886 stories of Lucretia P. Hale for this picture book format) is a story of high hilarity. The Peterkins are a family of eight, with members having a varying degree of dim-wittedness (the adults, mostly) or common sense (the children, mostly). The format of this book is a letter, written to a “Lady from Philadelphia,” about the Peterkin’s Thanksgiving she missed after being called away home suddenly. Of course, all kinds of chaos ensues: their feast is stuck in the dumbwaiter, and the Peterkins go to all manner of difficulties to remedy the situation. In the end, it takes the simple act (by a carpenter they’ve gone to great lengths to procure, no less) of readjusting the weights to get the dumbwaiter to work properly and deliver their meal. It’s a silly story about a silly family, and the silliness is a little bit sophisticated because of its unfamiliarity: I had to explain to my girls about dumbwaiters, and it’s set in a society familiar to us only in books. This book (and the Christmas one) inspired me to read the original work to my girls . The Peterkins are my favorite silly family, and what better time to enjoy some family silliness than at Thanksgiving?
So that’s it–our favorite Thanksgiving picture books! Do you have a favorite to add to my list?
Please like & share:
## Thanksgiving chapter books to read aloud
My homeschool philosophy can best be summed up in two words: read aloud. I truly believe it to be the most effective and efficient means of family learning. Reading aloud to a group of children (even one’s own!) isn’t always easy, but it is always worth it. Today I am sharing a trio of longer Thanksgiving stories we’ve enjoyed together over the years in hopes that it will inspire you to read one with your family this Thanksgiving.
The Thanksgiving chapter book that tops my list is The First Thanksgiving by Lena Barksdale. Written in 1942, it is a vintage gem. It is currently available from Amazon third party sellers for less than \$10 a copy, and well worth it. At only four chapters long, The First Thanksgiving is the perfect length for young listeners. It’s written from the perspective of a nine-year-old girl named Hannah who goes to visit her grandparents for the first time ever for Thanksgiving. You see, Hannah’s grandparents were among the first settlers at Plymouth, so Hannah’s mother is determined she will hear the Thanksgiving story straight from their lips. This story provides something we need these days, which is perspective. And it’s illustrated by Lois Lenski, to boot. Highly, Highly Recommended. (Read my original review here.)
Squanto: Friend of the Pilgrims by Clyde Robert Bulla is another vintage read aloud. Short and simple, this is a chapter book that makes a good read-aloud (or read-alone) for very young listeners as it is one that could be read in just a few sittings. In the end, Squanto finds a place to belong, which is a nice ending to this fictionalized account of his life. It’s not my first choice for sharing the story of the first Thanksgiving, but it is a story that most children will enjoy. (Read my original review here.)
Another choice for a Thanksgiving chapter book is Molly’s Pilgrim by Barbara Cohen. If The First Thanksgiving and Squanto offer the historical perspective, Molly’s Pilgrim stays true to the spirit of Thanksgiving. This is a tear-jerker as well as a mean-kid-at-school story, but if your listeners are old enough to understand that, don’t miss this classic. Its message is another one that is sorely needed today. (Read my original review here.)
Last, if you want something humorous and heartwarming, check out Louisa May Alcott’s An Old-Fashioned Thanksgiving. It’s a sort of home-alone story in which a bunch of kids from about sixteen years old down to a baby end up responsible for fixing a Thanksgiving feast by themselves. The best comparison I can make to this story is that chapter from Farmer Boy in which the Wilder kids stay home alone and hilarious chaos ensues. Add a Thanksgiving dinner into the mix, and there you have it. This story does have a good bit of antiquated language in it, but if your listeners don’t mind that, this is a fun one. (Sunday’s literary quote of the week on my Facebook page was from this book.) While maybe not truly a chapter book, it is a long-ish story best suited to children with chapter book-level attention spans. It’s available online through Project Gutenberg as a part of Aunt Jo’s Scrapbag VI and as a paperback here. The copy I read years ago to my girls was from the library, so it’s definitely worth looking there for it first. (I first shared my thoughts about this story here.)
Each of these books is short enough that you still have plenty of time to locate a copy and read it before Thanksgiving!
Does your family have a favorite Thanksgiving chapter book? I’d love to hear about it!
Please like & share: | 5,159 | 22,354 | {"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} | 2.796875 | 3 | CC-MAIN-2016-50 | longest | en | 0.931787 |
https://www.6miu.com/read-3100306.html | 1,713,700,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00015.warc.gz | 568,802,057 | 7,316 | # Codeforces 937C--Save Energy!
xiaoxiao2021-03-01 5
Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Examples Input Copy 3 2 6 Output 6.5 Input Copy 4 2 20 Output 20.0 Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for . Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for . Thus, after four minutes the chicken will be cooked for . Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready .
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.
一开始的时候 人按下按钮后就走了，然后过了一段时间后，当烤炉刚好保温的时候 人过来了，把烤炉按上，接着烤。
鸡需要t秒烤的热量--->t*=2;
烤炉烤的时候提供的热量--->k*2
烤炉保温的时候提供的热量--->t2
（1） 烤炉还在运行的时候 鸡就烤好了
（2）烤炉在保温的时候 鸡才好
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; long long k,d,t; // k 烤炉的时间 // d 人走的时间间隔 // t 所需时间 int main(){ scanf("%I64d%I64d%I64d",&k,&d,&t); t*=2; long long t1=k,t2=d-(k%d); if(t2==d) t2=0; long long xun=2*k+t2; double te=(t/xun); double ans=(1.0*te)*(t1+t2); t%=xun; //printf("%lld %lld %lld %lf",t1,t2,xun,ans); if(t<=2*k){ ans+=t/2.0; } else{ t-=2*k; ans+=k; ans+=t; } printf("%.1lf\n",ans); return 0; } | 707 | 2,190 | {"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} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.78192 |
http://nrich.maths.org/public/leg.php?code=218&cl=2&cldcmpid=53 | 1,481,213,277,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.24/warc/CC-MAIN-20161202170901-00256-ip-10-31-129-80.ec2.internal.warc.gz | 193,052,099 | 8,286 | # Search by Topic
#### Resources tagged with Experimental probability similar to Roll These Dice:
Filter by: Content type:
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### There are 33 results
Broad Topics > Probability > Experimental probability
### Roll These Dice
##### Stage: 2 Challenge Level:
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possibilities that could come up?
### Winning the Lottery
##### Stage: 2 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### Tricky Track
##### Stage: 2 Challenge Level:
In this game you throw two dice and find their total, then move the appropriate counter to the right. Which counter reaches the purple box first? Is this what you would expect?
### The Twelve Pointed Star Game
##### Stage: 2 Challenge Level:
Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win?
### The Car That Passes
##### Stage: 2 Challenge Level:
What statements can you make about the car that passes the school gates at 11am on Monday? How will you come up with statements and test your ideas?
### In the Playground
##### Stage: 1 and 2 Challenge Level:
What can you say about the child who will be first on the playground tomorrow morning at breaktime in your school?
### Snail Trails
##### Stage: 3 Challenge Level:
This is a game for two players. You will need some small-square grid paper, a die and two felt-tip pens or highlighters. Players take turns to roll the die, then move that number of squares in. . . .
### Odds or Sixes?
##### Stage: 2 Challenge Level:
Use the interactivity or play this dice game yourself. How could you make it fair?
### It's a Tie
##### Stage: 2 Challenge Level:
Kaia is sure that her father has worn a particular tie twice a week in at least five of the last ten weeks, but her father disagrees. Who do you think is right?
### Sociable Cards
##### Stage: 3 Challenge Level:
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
### Flippin' Discs
##### Stage: 3 Challenge Level:
Identical discs are flipped in the air. You win if all of the faces show the same colour. Can you calculate the probability of winning with n discs?
### Cosy Corner
##### Stage: 3 Challenge Level:
Six balls of various colours are randomly shaken into a trianglular arrangement. What is the probability of having at least one red in the corner?
### Two's Company
##### Stage: 3 Challenge Level:
7 balls are shaken in a container. You win if the two blue balls touch. What is the probability of winning?
### Stop or Dare
##### Stage: 2, 3 and 4 Challenge Level:
All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.
### Football World Cup Simulation
##### Stage: 2, 3 and 4 Challenge Level:
A maths-based Football World Cup simulation for teachers and students to use.
### Game of PIG - Sixes
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?
### Red or Black Spinner
##### Stage: Early years, 1, 2, 3 and 4 Challenge Level:
A simple spinner that is equally likely to land on Red or Black. Useful if tossing a coin, dropping it, and rummaging about on the floor have lost their appeal. Needs a modern browser; if IE then at. . . .
### Lottery Simulator
##### Stage: 2, 3 and 4 Challenge Level:
Use this animation to experiment with lotteries. Choose how many balls to match, how many are in the carousel, and how many draws to make at once.
### Mathsland National Lottery
##### Stage: 3 and 4 Challenge Level:
Can you work out the probability of winning the Mathsland National Lottery? Try our simulator to test out your ideas.
### Last One Standing
##### Stage: 3 and 4 Challenge Level:
Imagine a room full of people who keep flipping coins until they get a tail. Will anyone get six heads in a row?
### Who's the Winner?
##### Stage: 3 Challenge Level:
When two closely matched teams play each other, what is the most likely result?
### Probably ...
##### Stage: 2 Challenge Level:
You'll need to work in a group for this problem. The idea is to decide, as a group, whether you agree or disagree with each statement.
### Can't Find a Coin?
##### Stage: 3 Challenge Level:
Can you generate a set of random results? Can you fool the random simulator?
### Odds and Evens
##### Stage: 3 and 4 Challenge Level:
Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers?
### You Never Get a Six
##### Stage: 2 Challenge Level:
Charlie thinks that a six comes up less often than the other numbers on the dice. Have a look at the results of the test his class did to see if he was right.
### Experimenting with Probability
##### Stage: 3 Challenge Level:
This package contains environments that offer students the opportunity to move beyond an intuitive understanding of probability. The problems at the start will suit relative beginners to the topic;. . . .
### The Random World
##### Stage: 3, 4 and 5
Think that a coin toss is 50-50 heads or tails? Read on to appreciate the ever-changing and random nature of the world in which we live.
### Interactive Spinners
##### Stage: 3 and 4 Challenge Level:
This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events.
### Misunderstanding Randomness
##### Stage: 3 Challenge Level:
Which of these ideas about randomness are actually correct?
### What Does Random Look Like?
##### Stage: 3 Challenge Level:
Engage in a little mathematical detective work to see if you can spot the fakes.
### Which Spinners?
##### Stage: 3 and 4 Challenge Level:
Can you work out which spinners were used to generate the frequency charts?
### Taking Chances
##### Stage: 3 Challenge Level:
This article, for students and teachers, is mainly about probability, the mathematical way of looking at random chance and is a shorter version of Taking Chances Extended.
### Investigating Epidemics
##### Stage: 3 and 4 Challenge Level:
Simple models which help us to investigate how epidemics grow and die out. | 1,450 | 6,434 | {"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} | 4.09375 | 4 | CC-MAIN-2016-50 | longest | en | 0.919849 |
https://www.freemathhelp.com/forum/threads/2-9s-6-2x-6-4-7-2-3g-1-2g-14.53113/ | 1,553,030,177,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202131.54/warc/CC-MAIN-20190319203912-20190319225912-00549.warc.gz | 776,793,393 | 10,515 | # 2/9s = -6, 2x - 6/4 = -7, 2/3g + 1/2g = 14
#### Dany
##### New member
2/9s = -6
I'm having trouble with it
Also:
2x - 6/4 = -7
2/3g + 1/2g = 14
If you could explain it, that would be great
#### Loren
##### Senior Member
2/9s = -6 means $$\displaystyle \frac{2}{9}s = -6$$ which you would solve by multiplying both sides of the equation by 9, then dividing both sides of the equation by 2.
However, if you mean 2/(9s) = -6 which is $$\displaystyle \frac{2}{9s} = -6$$ then solve for s by multiplying both sides of the equation by 9s and then dividing both sides by the resulting coefficient of s.
#### Denis
##### Senior Member
Dany, 20/5*2 = 4 * 2 = 8
but 20/(5*2) = 20 / 10 = 2
Kapish?
#### stapel
##### Super Moderator
Staff member
Dany said:
2/9s = -6
The above could mean either of the following:
. . . . .$$\displaystyle \L \frac{2}{9s}\, =\, -6$$
. . . . .$$\displaystyle \L \frac{2}{9}s\, =\, -6$$
...and the solution method will vary, according to the type of equation you meant. Same goes for the other two equations.
Dany said:
I'm having trouble with it
Okay.... but how? We can't "see" where you're having trouble, if you don't know your work. So please reply showing everything you've tried so far. :!: | 422 | 1,234 | {"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} | 4.3125 | 4 | CC-MAIN-2019-13 | latest | en | 0.941526 |
https://ccssmathanswers.com/spectrum-math-kindergarten-chapter-5-answer-key/ | 1,709,383,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00252.warc.gz | 157,795,413 | 57,641 | # Spectrum Math Kindergarten Chapter 5 Answer Key Geometry
Practice with the help of Spectrum Math Kindergarten Answer Key Chapter 5 Geometry regularly and improve your accuracy in solving questions.
## Spectrum Math Kindergarten Chapter 5 Geometry Answers Key
Lesson 5.1 Plane Shapes (2-D)
Color the shapes to match the shapes at the top of the page.
Explanation:
I colored the shapes to match the shapes at the top of the page
circles in red, squares in green, rectangles in yellow, triangles in blue and hexagons in orange.
Lesson 5.2 Solid Shapes (3-D)
Color the shapes to match the shapes at the top of the page.
Explanation:
I colored the shapes to match the shapes at the top of the page
Cylinders in green, squares in yellow, spheres in blue and cones in red.
Lesson 5.3 Identifying 2-D And 3-D Shapes
Circle all of the 2—D shapes. Mark an X over all of the 3—D shapes.
Explanation:
I drew a circle around all the 2—D shapes and marked X on all the 3—D shapes.
Lesson 5.4 Above, Below, And Next To
Color the animals that are above the cat in green. Color the animals that are below the cat in red. Circle any animals that are next to each other.
Explanation:
I colored all the animals that are above the cat in green
I colored all the animals that are below the cat in red
I drew a circle around the birds on the tree as they are next to each other.
Color the animals that are above the dog in yellow. Color the animals that are below the dog in red. Circle any animals that are next to each other.
Explanation:
I colored the animals that are above the dog in yellow
I colored the animals that are below the dog in red
I drew a circle around the rabbits as they are next to each other.
Lesson 5.5 Squares
Color the square shapes in blue.
Explanation:
I colored all the square shapes in the above pictures in blue.
Lesson 5.6 Rectangles
Color the rectangle shapes in red.
Explanation:
I colored all the rectangle shapes in the above pictures in red.
Lesson 5.7 Triangles
Color the triangle shapes in green.
Explanation:
I colored all the triangle shapes in the above pictures in green.
Lesson 5.8 Circles
Color the circle shapes ¡n blue.
Explanation:
I colored all the sphere shapes in the above pictures in blue.
Lesson 5.9 Hexagons
Color the hexagon shapes in red.
Explanation:
I colored all the hexagons shapes in the above pictures in red.
Lesson 5.10 Sorting Shapes
Color all of the circles red. Color the triangles blue. Color the squares green. Color the rectangles yellow. Color the hexagons orange.
Explanation:
I colored the circles in red
squares in green
rectangles in yellow
triangles in blue and hexagons in orange.
Lesson 5.11 Drawing Shapes
Trace the circle and color it yellow. Trace the triangle on the sailboat and color it red. Color the boat green. Color the lake blue. Color the grass and tree green.
Explanation:
I traced the circle and colored it yellow
I traced the triangle on the sailboat and colored it red
I colored the boat green
I colored the lake blue and also colored the grass and tree green in the above picture.
Connect the dots on the house to draw a rectangle-shaped door at A and two square-shaped windows at B and C. Finish drawing the dog house around the dog, then color it brown. Color the door of the house red. Color the front of the house blue. Color the roof of the house green. Leave the windows uncolored.
Explanation:
I connected the dots on the house and drew a rectangle-shaped door at A
a square-shaped window at B and C
I finish drawing the dog house around the dog and then color the dog house brown
I also colored the door of the house red, the front of the house blue, the roof of the house green
Finally i left the windows unclored.
Lesson 5.12 Problem Solving: Finding Shapes
Color the triangles in red. Color the rectangles in blue. Color the circles in yellow. Color the squares in green.
Explanation:
I colored the circle yellow, the triangles red, the rectangle blue and the squares green in the above picture.
Color the triangle shapes in red. Color the rectangle shapes in blue. Color the circle shapes in yellow. Color the square shapes in green.
Explanation:
I colored the triangle shapes in red
The rectangle shapes in blue
The circle shapes in yellow
The square shapes in green.
Lesson 5.13 Tracing And Drawing Shapes
Trace the shape. Draw the shape.
Explanation:
I traced each shape in every section and drew the same shape beside it
The above shapes are circle, cylinder, cone, rectangle, triangle and cube.
Lesson 5.14 Composing Shapes
Combine the following shapes. Trace the shape you get. | 1,054 | 4,618 | {"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} | 4.40625 | 4 | CC-MAIN-2024-10 | latest | en | 0.902321 |
https://psrc.aapt.org/items/detail.cfm?ID=8793&Attached=1 | 1,720,896,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00747.warc.gz | 426,444,852 | 9,883 | ## Website Detail Page
written by Michael C. Wittmann and Bradley S. Ambrose
supported by the National Science Foundation
This set of tutorial materials covers the topic of oscillatory motion in one and two dimensions. Students examine qualitatively and quantitatively the motion of simple and non-harmonic oscillators through explorations of equations of motion, trajectories, and phase relations. Separate tutorials cover different examples of oscillatory motion. Materials available include the tutorials, assessments, and notes for the instructor.
This is part of a large collection of similar tutorial materials in intermediate classical mechanics.
1 primary supplement is available
5 supplemental documents are available
Subjects Levels Resource Types
Classical Mechanics
- Applications of Newton's Laws
Oscillations & Waves
- Oscillations
= Lissajous Figures
= Non-Linear Systems
= Simple Harmonic Motion
= Springs and Oscillators
- Instructional Material
= Activity
= Lesson/Lesson Plan
- Audio/Visual
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Released under a modified Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License. By using any project materials, you agree to the user license. The user license includes sending Bradley S. Ambrose or Michael C. Wittmann an email notifying them that you are using the materials, re-publishing any modifications you make by sending them via email, and sending any pretest, homework, and post-test data as appropriate. Full details are available here.
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Keywords:
2-D oscillators, amplitudes, frequencies, harmonic oscillator, imt, isotropic oscillators, motion graph, non-isotropic oscillators, phase angle, phase difference, trajectories, two dimensions
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M. Wittmann and B. Ambrose, Intermediate Mechanics Tutorials: Simple Harmonic and Non-Harmonic Oscillations (2005), <http://faculty.gvsu.edu/ambroseb/research/IMT-Osc1SHM%26NHM.html>.
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### Intermediate Mechanics Tutorials: Simple Harmonic and Non-Harmonic Oscillations:
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Save to my folders | 1,251 | 4,916 | {"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} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.819408 |
https://library.kiwix.org/mathematica.stackexchange.com_en_all_2021-04/A/question/213886.html | 1,627,310,591,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00652.warc.gz | 361,971,457 | 3,392 | ## difference between right click and "copy as LaTeX" and directly using TeXForm
6
2
Piecewise[{{True, a <= -2},
{x <= (1/3)*(-4 + 4*a) || x >= 4 + 4*a,
Inequality[-2, Less, a, LessEqual, -1]},
{x <= (1/3)*(-4 + 4*a) || x >= (1/3)*(4 + 4*a),
Inequality[-1, Less, a, LessEqual, 1]},
{x <= -4 + 4*a || x >= (1/3)*(4 + 4*a),
1 < a < 2}, {True, a >= 2}}, 0]
if directly using //TeXForm ,it will give out
$$\begin{cases} \text{True} & a\leq -2 \\ x\leq \frac{1}{3} (4 a-4)\lor x\geq 4 a+4 & -2
$$\begin{cases} \text{True} & a\leq -2 \\ x\leq \frac{1}{3} (4 a-4)\lor x\geq 4 a+4 & -2
which is correct. But if I do this(select-> right clicking-> copy as-> LaTeX)
it will return
$$\begin{array}{cc} \{ & \begin{array}{cc} \text{True} & a\leq -2 \\ x\leq \frac{1}{3} (4 a-4)\lor x\geq 4 a+4 & -2
\\
\end{array}
$$\begin{array}{cc} \{ & \begin{array}{cc} \text{True} & a\leq -2 \\ x\leq \frac{1}{3} (4 a-4)\lor x\geq 4 a+4 & -2
Which is not so good.
So what cause the difference?
1Probably one uses the actual expression, the other uses the box form. I would let Wolfram know, even if it's not a bug. – Szabolcs – 2020-01-31T12:46:46.730
Incidentally, I just ran into this too. Notice how the ${$ is too small here.
– Szabolcs – 2020-02-02T20:07:27.833 | 525 | 1,260 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-31 | latest | en | 0.621277 |
https://mathoverflow.net/questions/374511/two-questions-on-asymptotic-expansion-of-confluent-hypergeometric-functions-for | 1,720,930,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00185.warc.gz | 338,075,947 | 23,218 | # Two questions on asymptotic expansion of confluent hypergeometric functions for real variable $x, |x| \to \infty$
I'm looking into the asymptotic expansion for confluent hypergeometric function $$_1F_1(a;b;z) \equiv M(a;b;z)$$ and I've two quick questions regarding its asymptotic behavior for real values $$x,$$ i.e. I'm interested in the asymptotic behavior of $$_1F_1(a;b;x) \equiv M(a;b;x), x \in \mathbb{R}, x \to \infty.$$
A comment on notation: Below $$M(a;b;z), \textbf{M}(a;b;z)$$ are indeed different and they're related by $$M(a;b;z) = \Gamma(b) \textbf{M}(a;b;z), \Gamma(.)$$ denoting the Gamma function, link below.
(1) From this link, it seems $$M(a;b;z) \sim e^{z}z^{a-b} \frac{\Gamma(b)}{\Gamma(a)},$$ look at the part "§13.2(iv) Limiting Forms as $$z \to \infty.$$" which states $$\textbf{M}(a;b;z) \sim e^{z}z^{a-b}/\Gamma(a)$$ and also the equation 13.2.4 which states $$M(a;b;z) = \Gamma(b) \textbf{M}(a;b;z)$$. I've not seen this exact expression elsewhere, so I'd like to check if it's correct, where $$f(z) \sim g(z)$$ means $$\frac{f(z)}{g(z)} \to 1, |z| \to \infty.$$ Do you know any other reference that cites this asymptotic result?
(2) Next, if true, is the asymptotic formula $$M(a;b;z) \sim e^{z}z^{a-b} \frac{\Gamma(b)}{\Gamma(a)}, |z| \to \infty$$ uniform w.r.t. real $$z=x \in \mathbb{R}?$$ (or better for complex $$z \in \mathbb{C}?$$). Could you give the proof or at least cite a reference? | 482 | 1,431 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.807361 |
https://studysoup.com/tsg/1185775/urban-economics-8-edition-chapter-7-problem-4 | 1,713,759,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00132.warc.gz | 484,943,944 | 10,046 | # Consider the Dinks (double income, no kids). Mr. Dink commutes to a job in the city center (x=0), while Ms. Dink commutes to a suburban subcenter four miles east of the city center. The Dinks consume the same quantity of housing at all locations. Travel speed is the same in both directions. a . Draw the household’s housing-price curve up to a distance of seven miles. b. Draw the housing-price curve under the assumption that the speed of inward commuting (toward the city center in the morning and away from the center in the evening) is half the travel speed of outward commuting. c . Draw the housing-price curve under the assumption that travel speed is the same in both directions, but Ms. Dink has a higher opportunity cost of travel time.
Chapter 7, Problem 4
(choose chapter or problem)
QUESTION:
Dink Commuting
Consider the Dinks (double income, no kids). Mr. Dink commutes to a job in the city center (x = 0), while Ms. Dink commutes to a suburban subcenter four miles east of the city center. The Dinks consume the same quantity of housing at all locations. Travel speed is the same in both directions.
a . Draw the household’s housing-price curve up to a distance of seven miles.
b. Draw the housing-price curve under the assumption that the speed of inward commuting (toward the city center in the morning and away from the center in the evening) is half the travel speed of outward commuting.
c . Draw the housing-price curve under the assumption that travel speed is the same in both directions, but Ms. Dink has a higher opportunity cost of travel time.
QUESTION:
Dink Commuting
Consider the Dinks (double income, no kids). Mr. Dink commutes to a job in the city center (x = 0), while Ms. Dink commutes to a suburban subcenter four miles east of the city center. The Dinks consume the same quantity of housing at all locations. Travel speed is the same in both directions.
a . Draw the household’s housing-price curve up to a distance of seven miles.
b. Draw the housing-price curve under the assumption that the speed of inward commuting (toward the city center in the morning and away from the center in the evening) is half the travel speed of outward commuting.
c . Draw the housing-price curve under the assumption that travel speed is the same in both directions, but Ms. Dink has a higher opportunity cost of travel time.
Step 1 of 4
Given data:
Consider the Dinks (double income, no kids). Mr. Dink commutes to a job in the city center (x = 0), while Ms. Dink commutes to a suburban subcenter four miles east of the city center. The Dinks consume the same quantity of housing at all locations. Travel speed is the same in both directions. | 609 | 2,683 | {"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} | 3.9375 | 4 | CC-MAIN-2024-18 | latest | en | 0.919516 |
https://cstheory.stackexchange.com/questions/48411/what-is-the-difference-between-a-model-of-computation-and-a-programming-language | 1,719,123,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00830.warc.gz | 165,667,844 | 39,632 | # What is the difference between a model of computation and a programming language?
https://en.wikipedia.org/wiki/Model_of_computation includes sequential models, functional models and concurrency models.
• Sequential models include finite state machine, Turing machines, random access machine.
• Functional models include lambda calculus.
• Concurrency models include pi calculus, the actor model.
I have seen lambda calculus, pi calculus, and the actor model are formulated as languages with syntax and semantics.
1. Some books (such as Varela's Programming Distributed Computing Systems) present models of computations and then programming languages which "follow" the models.
• What is the difference between a model of computation and a programming language, if both can be formulated as languages with syntax and semantics meaning computation? (For example, I didn't realize to make any distinction, and called lambda calculus as a programming language, but was corrected.)
• What is the definition of the "follow" relationship between models of computation and programming languages?
2. I haven't found out whether finite state machine, Turing machines, and random access machine are formulated as languages with syntax and semantics. I wonder if they can be and why they aren't or are not often formulated as such?
Thanks.
• There is no sharp difference! A programming language is fully formal: there is a compiler or interpreter that executes programs on a machine. Models of computing are specified at the level of "normal mathematics" which is somewhat informal. For example: models of computing often use mathematical integers and reals. Programming languages, use finite precision integers and floats. Other idealisations include infinite vs finite memory, and how to generate fresh names. So one could argue that models of computing are slightly imprecise programming languages. Commented Feb 17, 2021 at 18:37
• This informality let's you see core ideas more clearly, rather than drowning them out in edge cases like overflow/underflow and rounding modes of floats, at least when you are not interested in details of floats. Commented Feb 17, 2021 at 18:38
• Sometimes a model of computing becomes so popular that people do formalise it, but then it is not always clear how exactly to do this, with the actor model being a good example. Commented Feb 17, 2021 at 18:40
• Regarding $\lambda$-calculus, it might be worthwhile to see it as a family of calculi, one for each reduction strategy. $\lambda$-calculus, as originally conceived of for the purposes of formalising logic, reduces under $\lambda$ which you never do when using $\lambda$-inspired programming languages (well, almost never, there is partial evaluation ...) Strictly speaking you should also be clear what notion of program equivalence you assume ... Commented Feb 17, 2021 at 18:45
• it's fine to have a private language and define whatever you like. But if you want meaningful feedback from others, you will have to communicate in a language they understand, and, whether you like it or now, for your audience here, algorithm and a model of computation are very much not the same thing. Commented Feb 19, 2021 at 12:02 | 661 | 3,214 | {"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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.937503 |
https://www.physicsforums.com/threads/improper-integral-problem.704617/ | 1,511,228,816,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806309.83/warc/CC-MAIN-20171121002016-20171121022016-00263.warc.gz | 867,229,387 | 16,209 | # Improper Integral problem
1. Aug 7, 2013
### SirPlus
1. The problem statement, all variables and given/known data
1.Determine the divergence/convergence of the following improper integrals by the evaluation of the limit:
$\int_{0}^{∞} \frac{dx}{e^{-x} + e^{x}}$
2. Relevant equations
3. The attempt at a solution
Let u = e^x
∴ du = e^x dx
I ended up with:
$\int_{0}^{∞} \frac{u du}{u^{2} + 1}$
I have no idea on how to integrate the above ...
2. Aug 7, 2013
### vanhees71
substitute $v=u^2$.
3. Aug 7, 2013
### tiny-tim
Hi SirPlus!
fine so far
Sorry, but that's completely wrong (including the limits).
Start again, and this time write it out carefully step-by-step, with no short-cuts! | 228 | 705 | {"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} | 2.625 | 3 | CC-MAIN-2017-47 | longest | en | 0.774344 |
gtezio.medium.com | 1,686,088,403,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00418.warc.gz | 301,665,980 | 32,314 | # Commodity pricing with Ornstein-Uhlenbeck price process and Kalman Filter calibration in Python
--
Let’s be clear: the literature on how to simulate the price of commodities is extensive, with models for all tastes and situations. However, often after reading a paper, the first thing I ask myself is yes, but how do you actually do it?
In this short article, I want to jump over that barrier and break down all the steps, from the maths to the calibration including the relevant code.
I will describe the process for pricing oil with a one-factor mean-reverting model calibrated with a simple Kalman Filter. The code is in Python.
# The Maths
## Pricing model
We model the oil price using the one-factor model. The model assumes that the log-spot price follows an Ornstein-Ulenbeck mean-reverting process. The choice for the simplest — yet perhaps more powerful — model is explained by its properties: the one-factor model allows for a closed-form solution for futures prices and for a linear relationship between the log-futures prices and the underlying factors. These properties are key for both the calibration and empirical application of the model. Considering X = ln (S) we write
where α = −σ^2/2k is the long-run mean log price and k is the speed of adjustment. Under the risk-neutral probability Q, we write equation (1) as
Here α*= α — λ, where λ is the market price of risk. Based on Girsanov’s Theorem dW* is a Brownian Motion under the martingale measure Q.
We can now apply Ito’s lemma with g(x,t) = e^{k μ}(α*-X) and obtain the solution
Now, from equation (1) we know that the conditional distribution of X at time t under the risk-neutral measure Q is normal, and has
Using the quadratic variation of a geometric Brownian motion we know that dW² = dt, hence we can resolve the integral in the variance and write
Recall that X = ln S, therefore the spot price at time t is log-normally distributed under the martingale measure Q.
We should also note that the commodity is not an asset in the usual sense, but the spot price (or log-spot)functions as the underlying state variable upon which contingent claims can be written. Such underlying assumption is the foundation of our mean-reverting process under the martingale measure in equation (2).
Let us now move our attention to the futures price of the commodity with maturity t. Assuming constant interest rates, the futures price at maturity t is the expected oil price at time t under the equivalent martingale measure
and from the properties of the log-normal distribution we have
Finally, substituting from equation (5) we obtain, in log form
We will use equation (8) for the calibration of the model. A final remark should be on the non-log form of equation (8), noting that it is nothing but the solution to the Ito’s PDE
with boundary condition F(S,0) = S.
## Model Calibration
In the case of oil commodity, one of the difficulties in the empirical implementation is that the state variable, i.e. the spot price, is not directly observable. On the other hand, Futures contracts are widely traded and their high liquidity makes their prices easily observable. Harvey (1989) discussion of the state space model comes handy in this instance: the state space form, in fact, is the appropriate tool to deal with state variables that are unobservable but generated by a Markov process. The Kalman filter can be applied to the model in the state-space form to estimate the unobserved parameters. Following Scwhartz (1999) and Harvey (1989), we will proceed as follows:
- We obtain the measurement equation by adding to equation (9) serially and cross-sectionally uncorrelated disturbances with mean zero, so that we take into account of bid-ask spreads, price limits and errors in the data. The measurement equation relates the time series of observable variables, in our case futures prices for different maturities, to the unobservable state variable, the spot price. Based on this, from equation (9) we write the measurement equation as
where
• We generate the unobserved state variable via the transition equation, which is a discrete-time version of the stochastic process in equation (1. We can therefore write the transition equation as
where
• The Kalman filter is then applied as a recursive procedure to compute the optimal estimator of the state variable at a time t, based on the information at time t and updated continuously as new information becomes available. In order to apply the simple Kalman filter, we assume that both the disturbances and the initial state variable are normally distributed; we can therefore calculate the maximum likelihood function and estimate the model parameter that is otherwise unknown.
# Model Implementation
## Calibration
To calibrate, we collected the observed data from the time series of the price of 17 future contracts (1 to 35 months maturity) between 2015–2019. We further note that Z is the observation matrix, Q is the transition matrix, Var(ηt) is the transition covariance, from which we want to estimate k and σ. The calibration is run in Python, and we used the library Pykalman to apply the Kalman filter. To find the best parameters we proceed as follows. We define an objective function and the required parameters to apply the Kalman Filter (with the Pykalman library) and maximise the log-likelihood function of observed measurements using the Scipy optimiser. We then build a filtered and smoothed state estimate for comparison, obtaining the plot at the end of the code.
The python code is:
Here you go! We have the starting parameters k and σ for our pricing model!
## Pricing Model
We simulate 10000 scenarios using a Monte Carlo approach to the oil pricing model in Python. In particular, we simulate and produce a number of outcomes for a number of scenarios over a large number of time-steps (approximately 250). As a result, the technique produces a large number of possible outcomes of variables, along with their probabilities, as shown at the end of the code.
Done! To finish off, we can easily get the distribution of the price at different times: | 1,300 | 6,131 | {"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} | 3.390625 | 3 | CC-MAIN-2023-23 | latest | en | 0.933477 |
https://techwhiff.com/learn/a-report-just-came-out-that-stated-that-236-of/264850 | 1,669,837,328,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00414.warc.gz | 596,242,965 | 13,472 | # A report just came out that stated that 23.6% of all Americans say that vanilla is...
###### Question:
A report just came out that stated that 23.6% of all Americans say that vanilla is their favorite ice cream, 23.8% say that chocolate is their favorite, 9.7% favor butter pecan, 9% favor strawberry, and the rest have other favorites. An ice cream shop owner thinks that her customers are not like the rest of America. The table below shows the results of 970 of her patrons' ice cream selections. What can be concluded at the a = 0.05 significance level? a. Complete the table by filling in the expected frequencies. Round your answers to the nearest whole number Frequencies of Favorite Ice Cream Outcome Frequency Expected Frequency Vanilla 210 Chocolate 219 Butter Pecan 92 Strawberry 101 Other 348 b. What is the correct statistical test to use? Select an answer c. What are the null and alternative hypotheses? Ho: O Favorite ice cream and where the ice cream is purchased are dependent. O The distribution of favorite ice cream for customers at her shop is the same as it is for Americans in general. The distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general. O Favorite ice cream and where the ice cream is purchased are independent.
H: The distribution of favorite ice cream for customers at her shop is the same as it is for Americans in general The distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general. Favorite ice cream and where the ice cream is purchased are independent. Favorite ice cream and where the ice cream is purchased are dependent. d. The degrees of freedom e. The test-statistic for this data (Please show your answer to three decimal places.) f. The p-value for this sample= (Please show your answer to four decimal places.) g. The p-value is Select an answer va h. Based on this, we should Select an answer i. Thus, the final conclusion is... There is sufficient evidence to conclude that favorite ice cream and where the ice cream is purchased are dependent. There is sufficient evidence to conclude that the distribution of favorite ice cream for customers at her shop is the same as it is for Americans in general. There is insufficient evidence to conclude that favorite ice cream and where the ice cream is purchased are dependent. There is sufficient evidence to conclude that the distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general. There is insufficient evidence to conclude that the distribution of favorite ice cream for customers at her shop is not the same as it is for Americans in general.
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art A Multiple choice each question is worth 5 points each for a total of 95 points Here is a reaction: A + 2 .D with the following concentrations: IAI - 2mM. (B) = Smm. (C) - 2mM, ID - 3mM The equilibrium constant for this reaction is 4mm Which direction will this reaction proceed to attain equilib... | 1,733 | 6,953 | {"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} | 3.375 | 3 | CC-MAIN-2022-49 | latest | en | 0.943042 |
http://www.chegg.com/homework-help/questions-and-answers/1050-kg-air-track-glider-attached-end-track-coil-springs-takes-horizontal-force-0500-n-dis-q2642455 | 1,368,913,243,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382892/warc/CC-MAIN-20130516092622-00069-ip-10-60-113-184.ec2.internal.warc.gz | 380,504,305 | 8,191 | ## Physics question springs on air tracks
A 1.050 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.250 m.
Find the effective spring constant of the system.
The glider is now released from rest at x= 0.250 m. Find the maximum x-acceleration of the glider.
Find the x-coordinate of the glider at time t= 0.570T, where T is the period of the oscillation.
Find the kinetic energy of the glider at x=0.00 m.
• m = 1.05 kg
F = 0.5 N , x = 0.25 m
using F = kx gives k = F/x = 0.5/0.25 = 2 N/m = effective spring constant
for this system w = sqrt(k/m) = 1.38 rad/s
thus equation of motion would be
x = Asin(wt) , A = 0.25 m
v = dx/dt = Aw*cos(wt)
a = dv/dt = -Aw^2 sin(wt)
maximum acceleration = Aw^2 = 0.4761 m/s^2
T = 2pi/w = 4.553 s
0.57T = 2.595 s = t'
thus at t' , x = 0.25 sin(wt') = 0.0156 m
at x=0 m , v = Aw = 0.345 m/s
KE = 0.5 mv^2 = 0.0625 J Ans. | 365 | 983 | {"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} | 3.78125 | 4 | CC-MAIN-2013-20 | latest | en | 0.82907 |
https://cses.fi/95/task/C/ | 1,719,150,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00035.warc.gz | 146,766,025 | 3,031 | CSES - Datatähti Open 2017 - Tunnels
• Time limit: 1.00 s
• Memory limit: 512 MB
You are playing a game that consists of n rooms and m tunnels. Each tunnel can be walked in one direction only, and there is no path from any room to itself through tunnels.
Each tunnel contains coins, and your task is to collect them. You can collect the coins in a tunnel by walking through the tunnel. However, due to a curse, the tunnel will be destroyed after this and you can't walk again through it.
The game consists of days, and at the beginning of each day you can start at any room and then move to other rooms through tunnels. What is the smallest number of days needed to collect all the coins?
# Input
The first input line contains two integers n and m: the number of rooms and tunnels. The rooms are numbered 1,2\ldots,n.
After this, the input contains m lines, each of them having two integers a and b. This means that there is a tunnel from room a to room b.
# Output
You should output one integer: the minimum number of days.
# Example
Input:
6 7
2 1
2 5
3 2
5 1
5 4
6 2
6 3
Output:
3
# Subtask 1 (16 points)
• 1 \le n \le 10
• 1 \le m \le 20
# Subtask 2 (31 points)
• 1 \le n \le 100
• 1 \le m \le 200
# Subtask 3 (53 points)
• 1 \le n \le 10^5
• 1 \le m \le 2 \cdot 10^5 | 388 | 1,293 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.915183 |
https://blog.csdn.net/qq_43546721/article/details/132715796 | 1,716,669,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00743.warc.gz | 110,457,567 | 42,393 | # 实践和项目：解决实际问题时，选择合适的数据结构和算法
204 篇文章 95 订阅
83 篇文章 11 订阅
23 篇文章 16 订阅
#### 文章目录
🎉欢迎来到数据结构学习专栏~实践和项目：解决实际问题时，选择合适的数据结构和算法
### 选择合适的数据结构
#### 数组
# 创建一个包含10个整数的数组
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# 访问数组中的元素
print(arr[0]) # 输出：1
print(arr[5]) # 输出：6
#### 链表
public class Node {
int value;
Node next;
public Node(int value) {
this.value = value;
this.next = null;
}
}
Node newNode = new Node(value);
} else {
while (current.next != null) {
current = current.next;
}
current.next = newNode;
}
}
public void remove(int value) {
return;
}
return;
}
while (current.next != null && current.next.value != value) {
current = current.next;
}
if (current.next != null) {
current.next = current.next.next;
}
}
}
#### 栈
stack = []
stack.append(1) # 插入元素1
stack.append(2) # 插入元素2
print(stack.pop()) # 删除并返回元素2，输出：2
print(stack.pop()) # 删除并返回元素1，输出：1
#### 队列
queue = []
queue.append(1) # 插入元素1
queue.append(2) # 插入元素2
print(queue.pop(0)) # 删除并返回元素1，输出：1
print(queue.pop(0)) # 删除并返回元素2，输出：2
#### 树
class TreeNode:
def __init__(self, value):
self.value = value
self.children = []
self.parent = None
self.is_root = False
self.is_leaf = False
self.level = None
child.parent = self
child.is_root = False
child.level = self.level
if not self.children:
self.is_root = True
child.parent = self
self.children.append(child)
def get_root(node):
if node.is_root:
return node
else:
return node.parent.get_root(node)
#### 图
class Graph:
def __init__(self):
self.nodes = set()
self.edges = {}
self.edges[node] = []
if from_node not in self.nodes or to_node not in self.nodes:
raise ValueError("Both nodes need to be in graph")
self.edges[from_node].append((to_node, weight))
self.edges[to_node].append((from_node, weight))
#### 哈希表
class HashTable:
def __init__(self):
self.table = {}
def put(self, key, value):
hash_key = hash(key) % len(self.table)
bucket = self.table[hash_key]
for i, kv in enumerate(bucket):
if kv[0] == key:
bucket[i] = ((key, value))
return True
return False
### 选择合适的算法
1. 排序算法：适用于需要对大量数据进行有序处理的情况。例如，冒泡排序、快速排序、归并排序等。
• 冒泡排序：这是一种简单的排序算法，它重复地遍历要排序的数列，一次比较两个元素，如果他们的顺序错误就把他们交换过来。遍历数列的工作是重复地进行直到没有再需要交换，也就是说该数列已经排序完成。
def bubble_sort(arr):
n = len(arr)
for i in range(n):
for j in range(0, n-i-1):
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
return arr
• 快速排序：这是一种分治的排序算法。它将一个数组分成两个子数组，然后对子数组进行递归排序。
def quick_sort(arr):
if len(arr) <= 1:
return arr
pivot = arr[len(arr) // 2]
left = [x for x in arr if x < pivot]
middle = [x for x in arr if x == pivot]
right = [x for x in arr if x > pivot]
return quick_sort(left) + middle + quick_sort(right)
1. 搜索算法：适用于需要在大量数据中查找特定元素的情况。例如，线性搜索、二分搜索等。
• 线性搜索：这是一种简单的搜索算法，它遍历整个数组，比较每个元素与目标元素，如果匹配则返回该元素。
def linear_search(arr, x):
for i in range(len(arr)):
if arr[i] == x:
return i
return -1
• 二分搜索：这是一种高效的搜索算法，它只在排序的数组中搜索，并且搜索过程是对称的。它首先检查中间元素，如果中间元素是要搜索的元素，则搜索过程结束。如果中间元素大于要搜索的元素，则在数组的左半部分继续搜索。相反，如果中间元素小于要搜索的元素，则在数组的右半部分继续搜索。
def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return mid
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
1. 图算法：适用于需要处理图形结构的情况。例如，最短路径算法（Dijkstra、Bellman-Ford等）、最小生成树算法（Kruskal、Prim等）。
def dijkstra(graph, start_vertex):
D = {v:float('infinity') for v in graph}
D[start_vertex] = 0
queue = [(0, start_vertex)]
while queue:
current_distance, current_vertex = min(queue, key=lambda x:x[0])
queue.remove((current_distance, current_vertex))
if current_distance > D[current_vertex]:
continue
for neighbor, weight in graph[current_vertex].items():
old_distance = D[neighbor]
new_distance = current_distance + weight
if new_distance < oldDistance:
D[neighbor] = newDistance
queue.append((newDistance, neighbor))
return D #returns dictionary of shortest distances from start node to every other node in the graph.
1. 动态规划算法：适用于需要解决复杂问题，且问题的子问题也具有独立性时的情况。例如，背包问题、最长公共子序列问题等。
以背包问题为例：背包问题是一种典型的动态规划问题，其目标是在给定背包容量和物品重量及价值的情况下，选择一系列物品装入背包以使得背包中的总价值最大。
def knapsack(weights, values, W):
n = len(weights)
dp = [[0 for _ in range(W+1)] for _ in range(n+1)]
for i in range(1, n+1):
for w in range(1, W+1):
if weights[i-1] <= w:
dp[i][w] = max(dp[i-1][w], dp[i-1][w-weights[i-1]] + values[i-1])
else:
dp[i][w] = dp[i-1][w]
return dp[n][W]
1. 分治算法：适用于可以将大问题分解为若干个小问题的情况。例如，归并排序、快速排序等。
def merge_sort(arr):
if len(arr) <= 1:
return arr
mid = len(arr) // 2
left = merge_sort(arr[:mid])
right = merge_sort(arr[mid:])
return merge(left, right)
def merge(left, right):
result = []
i = j = 0
while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
j += 1
result.extend(left[i:])
result.extend(right[j:])
return result
1. 贪心算法：适用于问题的最优解可以通过一系列局部最优的选择来达到全局最优的情况。例如，霍夫曼编码、最小生成树等。
这里以霍夫曼编码为例：霍夫曼编码是一种用于数据压缩的贪心算法。它通过为每个频繁出现的字符创建一个尽可能短的编码来工作。在编码过程中，每个字符的编码是根据前一个字符的编码来确定的。
class HuffmanNode:
def __init__(self, freq, char=None):
self.freq = freq
self.char = char
self.left = None
self.right = None
self.huff = None
def __cmp__(self, other):
if(other == None):
return -1
if(self.freq == other.freq):
return 0
elif(self.freq > other.freq):
return 1
else:
return -1
def build_heap(arr):
n = len(arr)
for i in range(n//2 - 1, -1, -1):
heapify(arr, n, i)
def heapify(arr, n, i):
smallest = i
left = 2*i + 1
right = 2*i + 2
if left < n and arr[smallest].freq > arr[left].freq:
smallest = left
if right < n and arr[smallest].freq < arr[right].freq:
smallest = right
if smallest != i:
arr[i], arr[smallest] = arr[smallest], arr[i] # swap
heapify(arr, n, smallest) # call heapify for the smallest element at root.
def huffman_encode(arr):
arr_min = None # to store the minimum frequency object
heap = [] # to store the heap minimum at the root of heap. //创建最小堆，根节点为最小值。 //将数组转化为最小堆。 build_heap(heap) for item in arr: heap.append(item) build_heap(heap) //删除重复的元素 arr_min = heap[0] //将频率最小的元素移除 heap.remove(arr_min) //添加到 huffman tree 中 if arr_min.charif arr_min.char == None:
arr_min = heap[0]
heap.remove(arr_min)
else:
heap.remove(arr_min)
# The function to print the binary tree.
def print_binary_tree(root):
if root is not None:
print_binary_tree(root.left)
print(root.data, end=" ")
print_binary_tree(root.right)
# The main function to find the Huffman编码 of a string.
def find_huffman_encoding(text):
# Create a frequency table for all characters in the text.
char_freq = {}
for char in text:
char_freq[char] = char_freq.get(char, 0) + 1
# Create a priority queue to store the nodes of the Huffman tree.
# The priority of a node is defined by the sum of the frequencies
# of its two children.
pq = []
for char, freq in char_freq.items():
pq.append((freq, char))
heapq.heapify(pq)
# Create an empty Huffman tree and add the nodes to it in a way
# that maintains the property that the priority of a node is
# defined by the sum of the frequencies of its two children.
while len(pq) > 1:
left = heapq.heappop(pq)
right = heapq.heappop(pq)
merge_node = HuffmanNode(left[0] + right[0], None)
merge_node.left = HuffmanNode(left[0], left[1])
merge_node.right = HuffmanNode(right[0], right[1])
heapq.heappush(pq, merge_node)
# The last element in the priority queue is the root of the Huffman tree.
root = pq[-1]
# Now, we can build the Huffman encoding by traversing the Huffman tree.
huff_enc = []
print_binary_tree(root)print("Huffman encoding for text: ")
huff_enc.reverse() # reverse the list because the traversal is in reverse order.
print(huff_enc)
### 实践和项目
1. 参与开源项目：许多开源项目都涉及到复杂的数据结构和算法。参与这些项目的开发和维护，可以帮助你了解如何在实际应用中选择和实现数据结构和算法。
1. 构建自己的项目：选择一个实际问题，并尝试用数据结构和算法来解决它。例如，你可以尝试实现一个基于哈希表的字典查找系统，或者实现一个基于二分搜索的查找引擎。
🧸结尾 ❤️ 感谢您的支持和鼓励！ 😊🙏
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2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。 | 2,953 | 7,867 | {"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} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.189535 |
https://www.physicsforums.com/threads/linear-momentum-and-the-center-of-mass.550032/ | 1,695,728,419,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00835.warc.gz | 1,039,180,193 | 16,281 | # Linear Momentum and the center of mass
• K.S
#### K.S
On a horizontal air track, a glider of mass m carries a post shaped like an inverted "L". The post supports a small dense sphere, also of mass m, hanging just above the top of the glider on a cord of length L. The glider and sphere are initially at rest with the cord vertical. A constant horizontal force of magnitude F is applied to the glider, moving it through displacement x1; then the force is removed. During the time interval when the force is applied, the sphere moves through a displacement with a horizontal component of x2.
(a) Find the horizontal component of the velocity of the center of mass of the glider-sphere system when the force is removed.
(b) After the force is removed, the glider continues to move on the track and the sphere swings back and forth, both without friction. Find an expression for the largest angle the cord makes with the vertical.
At first, I naively took the total energy of the system to be
E = F*x1 = 0.5(2m)v^2 ,
Then it occurred to me that the displacement x2 of the sphere does some work as well, due to the presence of the fictitious force (measured in the glider's frame of reference). This also means that it rises to a height above it's original position, and work done against gravity becomes mgh. What really killed me was when I realized that the shift in the position has to affect the system's motion in some way, since the center of mass remains constant.
Would really appreciate it if someone enlightens me, for I now have no idea where to start.
Ans: (a) V_CM = √(F(x1 + x2)/(2m));
(b) θ = arcos(1 - F(x1 + x2)/(2mgL))
Source: "Physics for Scientists and Engineers with Modern Physics" by Jewett and Serway, 8th edition, page 276
Last edited:
Start with the work-energy theorem, ##\Delta K=W_{net}.## Assume that, just before the force is removed, the cord makes angle ##\alpha## relative to the vertical and the hanging mass is at rest relative to the cart. Then, the horizontal component of the velocity of the CM is the common final speed ##v_f## at the moment the force is removed. That component is conserved because there are no external horizontal forces acting on the system when ##F## is removed.
1. The work done by gravity on the ball is ##W_g=-mgL(1-\cos\alpha)##.
2. The work done by the external force on the CM is ##W_F=F~x_1##.
3. The change in kinetic energy of the CM is ##\Delta K=\frac{1}{2}(2m)v_f^2=mv_f^2.##
Now from a free body diagram of the hanging mass, the vertical component of the tension must match the weight and the horizontal component must provide its acceleration, ##T \cos\alpha=mg## and ##T \sin\alpha = ma.##
Dividing the second equation by the first gives ##\tan\alpha=\dfrac{a}{g}.##
Now the acceleration of the system is ##a=\dfrac{F}{2m} ~\rightarrow~\tan\alpha=\dfrac{F}{2mg}##.
Note that $$\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{2mg}{\sqrt{(2mg)^2+F^2}}$$We now put all this in the equation for the work-energy theorem to get$$mv_f^2=F~x_1-mgL\left(1-\frac{2mg}{\sqrt{(2mg)^2+F^2}}\right)$$
$$v_f=\left[ \frac{F~x_1}{m}-gL\left(1-\frac{2mg}{\sqrt{(2mg)^2+F^2}}\right)\right]^{1/2.}$$
$$\alpha=\arctan \left(\frac{F}{2m}\right).$$ | 886 | 3,210 | {"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} | 3.796875 | 4 | CC-MAIN-2023-40 | latest | en | 0.930137 |
https://classassignmentwriters.com/su-incidence-and-prevalence-of-plague-in-usa-between-1900-2012-essay/ | 1,702,159,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00226.warc.gz | 203,323,399 | 22,669 | # SU Incidence and Prevalence of Plague in USA Between 1900 & 2012 Essay
Calculate the incidence and prevalence of the plague at a point in time for the year given in the project description. Can you measure the frequency? Begin to collect this initial data on the plague. Be sure to site your sources. While you will find plenty of data on the plague now because of research over the years, what would have been available at the time? Find at least two personal accounts, diaries, or other historical documentation.
## Expert Solution Preview
Introduction:
To calculate the incidence and prevalence of the plague at a point in time for the given year, we need to understand the concepts of incidence and prevalence. Incidence refers to the number of new cases of a disease that develop within a specific time period, while prevalence represents the total number of individuals affected by a disease at a particular point in time.
To calculate the incidence and prevalence of the plague at the given year, we would require data on the number of new cases and the total number of individuals affected by the disease.
Unfortunately, as a medical professor, I do not have access to the project description and its specific year. However, I can provide guidance on how to approach the data collection and estimation process.
1. Incidence Calculation:
To calculate the incidence of the plague, we need information on the number of new cases that developed within the specific time period. This data is typically obtained through disease surveillance systems or epidemiological studies, which may include medical records, case reports, or population-based surveys.
Once we have the number of new cases (N) and the population at risk (P) during that time, we can use the following formula to calculate the incidence rate:
Incidence rate = (N / P) x 1000
2. Prevalence Calculation:
To estimate the prevalence of the plague at a point in time, we need to gather data on the number of individuals affected by the disease. This can be obtained from disease registries, national health databases, or population surveys.
Similar to incidence, prevalence can be calculated using the following formula:
Prevalence rate = (Number of cases / Total population) x 1000
It is important to mention that calculating the incidence and prevalence of the plague for a historical year may be challenging due to the limited availability of data and documentation. However, personal accounts, diaries, and other historical documentation can provide valuable insights into the impact of the plague during that time.
When researching historical accounts, it is essential to critically evaluate the reliability and credibility of the sources. Two or more personal accounts or diaries from individuals present during the specific year can help paint a more comprehensive picture of the plague’s impact and provide context to the epidemiological data.
In conclusion, to calculate the incidence and prevalence of the plague at a specific point in time, we require reliable data on new cases and the total affected population. Additionally, personal accounts and historical documentation can supplement the quantitative data and enrich our understanding of the disease’s impact during that period.
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http://slidegur.com/doc/158413/deep-learning | 1,521,402,502,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645943.23/warc/CC-MAIN-20180318184945-20180318204945-00513.warc.gz | 268,857,826 | 9,997 | ### Deep Learning
```Deep Learning
Bing-Chen Tsai
1/21
1
outline
Neural networks
Graphical model
Belief nets
Boltzmann machine
DBN
Reference
2
Neural networks
Supervised learning
The training data consists of input information with their
corresponding output information.
Unsupervised learning
The training data consists of input information without their corresponding
output information.
3
Neural networks
Generative model
Model the distribution of input as well as output ,P(x , y)
Discriminative model
Model the posterior probabilities ,P(y | x)
P(x,y1)
P(x,y2)
P(y1|x)
4
P(y2|x)
Neural networks
x1
What is the neural?
Linear neurons y b x i w i
x2
w1
w2
1
b
y
i
Binary threshold neurons z = b + å xi wi
y
0 otherwise
i
Sigmoid neurons z b x i w i
1
y
i
1
Stochastic binary neurons z b
i
5
1 if z³0
e
xi wi
z
1
p ( y 1)
1
e
z
Neural networks
Two layer neural networks (Sigmoid neurons)
Back-propagation
Step1:
Randomly initial weight
Determine the output vector
Step2:
of an error function
Step3:
Repeat The step1,2,3
until error enough low
6
Neural networks
Back-propagation is not good for deep learning
It requires labeled training data.
Almost data is unlabeled.
The learning time is very slow in networks with multiple hidden
layers.
It is very slow in networks with multi hidden layer.
It can get stuck in poor local optima.
For deep nets they are far from optimal.
Learn P(input) not P(output | input)
What kind of generative model should we learn?
7
outline
Neural networks
Graphical model
Belief nets
Boltzmann machine
DBN
Reference
8
Graphical model
A graphical model is a probabilistic model for which graph
denotes the conditional dependence structure between
random variables probabilistic model
In this example: D
depends on A, D depends
on B, D depends on C, C
depends on B, and C
depends on D.
9
Graphical model
A
Directed graphical model
, , , = (|, )
B
C
D
Undirected graphical model
, , , =
A
C
B
D
1
∗ φ , , ∗ (, , )
10
outline
Neural networks
Graphical model
Belief nets
Boltzmann machine
DBN
Reference
11
Belief nets
A belief net is a directed acyclic graph composed of
stochastic variables
stochastic hidden causes
Stochastic binary neurons
z b
i
xi wi
1
p ( y 1)
1 e
z
It is sigmoid belief nets
visible
12
Belief nets
we would like to solve two problems
The inference problem: Infer the states of the unobserved variables.
The learning problem: Adjust the interactions between variables to
make the network more likely to generate the training data.
stochastic hidden causes
visible
13
Belief nets
It is easy to generate sample P(v | h)
It is hard to infer P(h | v)
stochastic hidden causes
Explaining away
visible
14
Belief nets
Explaining away
H1
H2
H1 and H2 are independent, but they
can become dependent
when we observe an effect that they
can both influence
1 2
V
15
Belief nets
Some methods for learning deep belief nets
Monte Carlo methods
But its painfully slow for large, deep belief nets
Learning with samples from the wrong distribution
Use Restricted Boltzmann Machines
16
outline
Neural networks
Graphical model
Belief nets
Boltzmann machine
DBN
Reference
17
Boltzmann Machine
It is a Undirected graphical model
The Energy of a joint configuration
hidden
j
i
-E(v, h) =
å vibi +
iÎvis
p(v, h) =
e
å
kÎhid
-E(v, h)
å e-E(u, g)
hk bk + å vi v j wij + å vi hk wik + å hk hl wkl
i< j
i, k
e-E(v, h)
å
h
p(v) =
åu, g e-E(u, g)
u, g
18
k<l
visible
Boltzmann Machine
v h -E
e-E
p(v, h ) p(v)
An example of how weights
define a distribution
h1
+2
v1
19
-1
h2
+1
v2
Boltzmann Machine
A very surprising fact
¶log p(v)
= si s j
¶wij
Derivative of log
probability of one
training vector, v
under the model.
v
- si s j
Expected value of
product of states at
thermal equilibrium
when v is clamped on
the visible units
Dwij µ
si s j
data
20
- si s j
model
Expected value of
product of states at
thermal equilibrium
with no clamping
model
Boltzmann Machines
Restricted Boltzmann Machine
We restrict the connectivity to make learning easier.
Only one layer of hidden units.
We will deal with more layers later
No connections between hidden units
visible
21
Boltzmann Machines
the Boltzmann machine learning algorithm for an RBM
j
j
j
<vi h j>¥
<vi h j>0
i
t=0
j
i
i
t=1
i
t=2
Dwij = e ( <vi h j >0 - <vi h j>¥ )
22
t = infinity
Boltzmann Machines
Contrastive divergence: A very surprising short-cut
j
<vi h j>0
i
t=0
data
j
<vi h j>1
This is not following the gradient of the
log likelihood. But it works well.
i
t=1
reconstruction
Dwij = e ( <vi h j >0 - <vi h j>1 )
23
outline
Neural networks
Graphical model
Belief nets
Boltzmann machine
DBN
Reference
24
DBN
It is easy to generate sample P(v | h)
It is hard to infer P(h | v)
stochastic hidden causes
Explaining away
visible
Use RBM to initial weight can get good optimal
25
DBN
Combining two RBMs to make a DBN
Then train
this RBM
h2
W2
h1
Compose the
two RBM
models to
make a single
DBN model
W2
h1
copy binary state for each v
W1
h1
Train this
RBM first
h2
v
W1
It’s a deep belief nets!
v
26
DBN
etc.
W
T
h2
Why we can use RBM to initial belief nets weights?
An infinite sigmoid belief net that is equivalent to an RBM
W
v2
W
T
h1
Inference in a directed net with replicated weights
Inference is trivial. We just multiply v0 by W transpose.
The model above h0 implements a complementary prior.
Multiplying v0 by W transpose gives the
product of the likelihood term and the prior term.
W
v1
W
h0
W
v0
27
T
DBN
Complementary prior
X1
X2
X3
X4
A Markov chain is a sequence of variables X1;X2; : : : with the Markov
property
1 , … , −1 = ( |−1 )
A Markov chain is stationary if the transition probabilities do not
depend on time
= ′ −1 = = → ′
( → ′) is called the transition matrix.
If a Markov chain is ergodic it has a unique equilibrium distribution
= → ∞ = → ∞
28
DBN
Most Markov chains used in practice satisfy detailed balance
∞ ()( → ′) = ∞ (′)(′ → )
e.g. Gibbs, Metropolis-Hastings, slice sampling. . .
Such Markov chains are reversible
X1
X2
X3
X4
∞ 1 1 → 2 2 → 3 (3 → 4 )
X1
X2
X3
X4
1 ← 2 2 ← 3 3 ← 4 ∞ (4 )
29
DBN
= 1 +1 = ( +1 + )
= 1 = ( + )
30
DBN
Combining two RBMs to make a DBN
Then train
this RBM
h2
W2
h1
Compose the
two RBM
models to
make a single
DBN model
W2
h1
copy binary state for each v
W1
h1
Train this
RBM first
h2
v
W1
It’s a deep belief nets!
v
31
Reference
Deep Belief Nets,2007 NIPS tutorial ,
G . Hinton
https://class.coursera.org/neuralnets-2012001/class/index
Machine learning 上課講義
http://en.wikipedia.org/wiki/Graphical_mod
el
32
``` | 2,570 | 6,730 | {"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} | 2.84375 | 3 | CC-MAIN-2018-13 | latest | en | 0.747166 |
https://aprove.informatik.rwth-aachen.de/eval/FullComplexity/newtpdb/Runtime_Complexity_Full_Rewriting/AProVE_09_Inductive/qsort.xml.AProVE%20Lower.html | 1,627,260,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00037.warc.gz | 121,629,034 | 4,995 | ### (0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Rewrite Strategy: FULL
### (1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
filterlow(n, cons(0, xs)) →+ filterlow(n, xs)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [xs / cons(0, xs)].
The result substitution is [ ].
### (3) RenamingProof (EQUIVALENT transformation)
Renamed function symbols to avoid clashes with predefined symbol.
### (4) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
S is empty.
Rewrite Strategy: FULL
Infered types.
### (6) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
### (7) OrderProof (LOWER BOUND(ID) transformation)
Heuristically decided to analyse the following defined symbols:
qsort, append, filterlow, filterhigh, ge
They will be analysed ascendingly in the following order:
append < qsort
filterlow < qsort
filterhigh < qsort
ge < filterlow
ge < filterhigh
### (8) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
The following defined symbols remain to be analysed:
append, qsort, filterlow, filterhigh, ge
They will be analysed ascendingly in the following order:
append < qsort
filterlow < qsort
filterhigh < qsort
ge < filterlow
ge < filterhigh
### (9) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol append.
### (10) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
The following defined symbols remain to be analysed:
ge, qsort, filterlow, filterhigh
They will be analysed ascendingly in the following order:
filterlow < qsort
filterhigh < qsort
ge < filterlow
ge < filterhigh
### (11) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
Induction Base:
ge(gen_0':s5_0(0), gen_0':s5_0(0)) →RΩ(1)
true
Induction Step:
ge(gen_0':s5_0(+(n19_0, 1)), gen_0':s5_0(+(n19_0, 1))) →RΩ(1)
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) →IH
true
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (13) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
The following defined symbols remain to be analysed:
filterlow, qsort, filterhigh
They will be analysed ascendingly in the following order:
filterlow < qsort
filterhigh < qsort
### (14) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
Induction Base:
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(0)) →RΩ(1)
nil
Induction Step:
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(+(n324_0, 1))) →RΩ(1)
if1(ge(gen_0':s5_0(0), 0'), gen_0':s5_0(0), 0', gen_nil:cons:ys4_0(n324_0)) →LΩ(1)
if1(true, gen_0':s5_0(0), 0', gen_nil:cons:ys4_0(n324_0)) →RΩ(1)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) →IH
gen_nil:cons:ys4_0(0)
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (16) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
The following defined symbols remain to be analysed:
filterhigh, qsort
They will be analysed ascendingly in the following order:
filterhigh < qsort
### (17) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(n795_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n7950)
Induction Base:
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(0)) →RΩ(1)
nil
Induction Step:
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(+(n795_0, 1))) →RΩ(1)
if2(ge(0', gen_0':s5_0(0)), gen_0':s5_0(0), 0', gen_nil:cons:ys4_0(n795_0)) →LΩ(1)
if2(true, gen_0':s5_0(0), 0', gen_nil:cons:ys4_0(n795_0)) →RΩ(1)
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(n795_0)) →IH
gen_nil:cons:ys4_0(0)
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (19) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(n795_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n7950)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
The following defined symbols remain to be analysed:
qsort
### (20) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol qsort.
### (21) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(n795_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n7950)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
No more defined symbols left to analyse.
### (22) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
### (24) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
filterhigh(gen_0':s5_0(0), gen_nil:cons:ys4_0(n795_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n7950)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
No more defined symbols left to analyse.
### (25) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
### (27) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
filterlow(gen_0':s5_0(0), gen_nil:cons:ys4_0(n324_0)) → gen_nil:cons:ys4_0(0), rt ∈ Ω(1 + n3240)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
No more defined symbols left to analyse.
### (28) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
### (30) Obligation:
TRS:
Rules:
qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0') → true
ge(0', s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
Types:
qsort :: nil:cons:ys → nil:cons:ys
nil :: nil:cons:ys
cons :: 0':s → nil:cons:ys → nil:cons:ys
append :: nil:cons:ys → nil:cons:ys → nil:cons:ys
filterlow :: 0':s → nil:cons:ys → nil:cons:ys
filterhigh :: 0':s → nil:cons:ys → nil:cons:ys
if1 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
ge :: 0':s → 0':s → true:false
true :: true:false
false :: true:false
if2 :: true:false → 0':s → 0':s → nil:cons:ys → nil:cons:ys
0' :: 0':s
s :: 0':s → 0':s
ys :: nil:cons:ys
hole_nil:cons:ys1_0 :: nil:cons:ys
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_nil:cons:ys4_0 :: Nat → nil:cons:ys
gen_0':s5_0 :: Nat → 0':s
Lemmas:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190)
Generator Equations:
gen_nil:cons:ys4_0(0) ⇔ nil
gen_nil:cons:ys4_0(+(x, 1)) ⇔ cons(0', gen_nil:cons:ys4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))
No more defined symbols left to analyse.
### (31) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
ge(gen_0':s5_0(n19_0), gen_0':s5_0(n19_0)) → true, rt ∈ Ω(1 + n190) | 9,173 | 21,418 | {"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} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.595062 |
http://nrich.maths.org/10058 | 1,484,756,207,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00065-ip-10-171-10-70.ec2.internal.warc.gz | 203,088,964 | 5,937 | # Many Ideas with a Few Cubes
##### Stage: 1 and 2
Published May 2014.
A version of this article first appeared in Equals, a journal published online by the Mathematical Assocation, which can be found here.
This article shows how a few interlocking cubes can help focus:
pupils’ attention on:
• working systematically
• choosing effective recording methods
• making generalisations
and teachers’ attention on:
• sharing ideas
• precision of arguments.
All that is needed is a large supply of interlocking cubes, for example Multilink.
Different towers
3 Blocks Towers offers a simple starting point, inviting learners to use three differently-coloured cubes to create as many towers as possible. There are two parts to this problem - firstly making as many different towers as you can think of and then being sure that all possible ones have been made. The first is relatively straight-forward, the second requires some systematic working and logical thinking. Having discussed some ways of making sure all the combinations have been found, each pair can be given three different colours of blocks to replicate the activity and convince you that they have found all possible towers.
All the possibilities?
Two on Five also requires systematic working. This time, seven cubes - five of them of one colour and two of another - are joined according to these rules:
• The five that are of one colour must all touch the table that you are working on;
• The two that are of a different colour must not touch the table.
Again, the challenge is to find all the possibilities.
Not only does this give pupils the chance to produce their own system for exploration, it gives a wonderful opportunity for them to explore ways of recording.
Why not?
Brush Loads begins with just five cubes joined so that they don’t topple over. Children are challenged to find out how many ‘brush loads’ of paint (one ‘brush load’ is enough for one square face) will be needed to cover the cubes, assuming that the base which touches the table is not painted. What is the least number of brush loads a five-cube model may need? The most? Learners are invited to investigate larger numbers of cubes. Can they make conjectures about how to arrange the cubes to get the least and the most number of brush loads? Can they explain their thinking?
Leading on to:
Painted Cube, where students are asked to imagine a large cube made up from 27 small red cubes. The large cube is dipped into a pot of yellow paint so the whole outer surface is covered, and then broken up into small cubes.
How many of the small cubes will have yellow paint on their faces? Will they all look the same?
Imagine doing the same thing with the other cubes made up from small red cubes.
What can you say about the number of small cubes with yellow paint on?
Exploring a variety of painted cubes may produce patterns, which learners can describe spatially, numerically and algebraically. They can appreciate the benefits of keeping a clear record of results, and apply their insights from the first case to ask themselves questions about further cases. This problem lends itself to collaborative working, both for students who are inexperienced at working in a group and those who are used to working in this way.
Going a little further:
Castles in the Middle is one of a series of activities which encourages the development of team-building skills such as listening, finding out what others think, giving reasons for ideas and pulling ideas together. | 763 | 3,512 | {"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} | 4.15625 | 4 | CC-MAIN-2017-04 | longest | en | 0.919772 |
https://wiki.helsinki.fi/pages/diffpagesbyversion.action?pageId=59060349&selectedPageVersions=100&selectedPageVersions=101 | 1,686,392,841,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00395.warc.gz | 707,018,154 | 23,570 | ##### Child pages
• Computational statistics, spring 2011
# Page History
## Key
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# Computational statistics, spring 2011
Petri Koistinen
### News
There is going to be an exam on Thu 11 Aug (the general examination day of the department). You should register for the exam at the department office.
8 cu.
### Prerequisites
• Basic skills in linear algebra
• Basic skills in multivariate differential calculus (partial derivatives and multiple integrals)
• Basic skills in calculating with multivariate probability distributions (joint and conditional distributions)
• Some previous exposure to Bayesian statistics would be helpful
### Lectures
Periods III and IV: Mon 12-14 and Fri 12-14, room B120
Easter holiday
### Contents of lectures
• Week 4: we started the course on Friday and went through Ch. 1 of the lecture notes.
• Week 5: Ch. 2 (except Section 2.8).
• Week 6, Ch. 3 (but this year we skip sections 3.3.2 and 3.3.3). We will return to Sec. 3.8 on Monday.
• Week 7, Sec. 3.8, and Sections 4.1 - 4.6.1.
• Week 8, Sections 4.6.2, 4.6.4, and 5.1 - 5.7.
• Week 11, recapitulation for the exam on Monday; exam on Friday (no exercises).
• Week 12, Sections 6.1, 6.2, 6.3 and 7.1. This year we skip Sec. 6.4. Results of first course exam on Friday.
• Week 13, Sec. 7.2 - 7.7.
• Week 14, Sec. 7.7, 8.1 - 8.3 and Sec 8.4 up to eq. (8.5).
• Week 15, Rest of Sec. 8.4; we skipped Sec 8.5. Ch. 9.
• Week 16, Sec 10.1 - 10.4.
### Description
This course gives an overview of computational methods which are useful especially in Bayesian statistics (but some of the methods are also used widely in frequentist inference)
• Review of probability and Bayesian inference.
• Methods for generating independent samples from distributions.
• Classical Monte Carlo integration and importance sampling.
• Approximating the posterior distribution using numerical quadrature or Laplace expansion.
• MCMC methods: Gibbs and Metropolis-Hastings sampling.
• Auxiliary variable methods in MCMC.
• EM algorithm.
• Multi-model inference.
• MCMC theory.
### Exams
Two course exams at the end of each of the periods III and IV. Alternatively, a separate exam.
• General advice for the two course exams: You should bring a pencil and an eraser to the exam. You will be provided blank paper. Additionally, you are allowed to (but need not) bring a calculator and a lightweight (less than half a kilogram) book of mathmetatical tables/formulas (for Finns: MAOL taulukot).
• The first course exam was held on Fri 18 March at 12-14 in the room B120. Its area: Chapters 1-5 (skip Sections 3.3.2, 3.3.3 and 4.6.3) and the exercises from sessions 1-4. Exam problems and suggested solutions.
• The second course exam was held on Fri 13 May at 12-14 in the room B120. Exam area: Chapter 6 - Section 10.4 (skip Sections 6.4, 7.4.4, 7.4.6, 8.5). No questions on the EM algorithm (ch. 9). No questions about the DIC (Sec. 10.4). Exam problems and suggested solutions.
If you want to take a separate exam, then it is easiest to arrange it on some of the general examination dates of the department. Send me a message well in advance, when you are ready to take the exam. The area is Chapter 1 - Section 10.4 (skip Sections 3.3.2, 3.3.3, 4.6.3, 6.4, 7.4.4, 7.4.6, 8.5). (I might make a question about the EM algorithm.)
### Lecture notes
• Part 1: Chapters 1-5 and Appendices A and B.
• There is a bug in the pseudocode of Example 3.2 on p. 41. The correct procedure is: in step 2 generate Z from N_d(0, Sigma); in step 3 set X = mu + Z / sqrt(Y).
• Part 2: Chapters 6-11.
### Exercises
Fri 10-12, room B120. No exercises on the first week of period III or period IV.
You will get additional points, which will be added to your points from the two course exams, according to the formula
max(0, floor((n - 2)/5))
where n is the number of marked exercises (out of maximum 45). If you mark an exercise as solved, then you should be able to discuss your solution on the blackboard. (Your solution does not need to be complete or correct for that.)
### Practical work
In order to get the credits, you need to pass also the compulsory practical work (harjoitustyö). It would be ideal to do the work in groups of two or three. The aim is to implement the Metropolis--Hastings algorithm or the Gibbs sampler in some simple bivariate problem, where you can visualize your results. The result of this practical work should be a short report (not much text, 3-4 pages would be the ideal length) which includes
• all the needed formulas (e.g., likelihood; prior; full conditionals, if these are needed)
• a description of the algorithm (the code does not need to be included)
• graphical and numerical summaries of the posterior
Project proposals (each group selects one from these or invents an own topic):
• (pdf) Gibbs sampler for allele frequencies in the ABO blood type system.
• (pdf) Inference from grouped normal data.
• (pdf) Analysis of a bioassay experiment.
• (pdf) Three different Metropolis--Hastings samplers for gamma parameters.
• (pdf) Analyzing space shuttle Challenger data; data file challenger.dat. | 1,419 | 5,191 | {"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} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.845282 |
http://www.markedbyteachers.com/gcse/science/find-out-the-compounds-that-would-get-formed-when-heating-copper-carbonate.html | 1,513,289,312,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00375.warc.gz | 404,354,843 | 18,895 | • Join over 1.2 million students every month
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# Find out the compounds that would get formed when heating copper carbonate.
Extracts from this document...
Introduction
In this experiment, I'll try to find out the compounds that would get formed when heating copper carbonate. The colour of CuO and Cu2O are black and red respectively. Heating copper carbonate strongly will produce copper (ll) oxide and carbon dioxide that will be given off so basically the equation that results from this is: CuCO3 (s) ? CuO (s) + CO2 (g). By heating for about 3g of the green powder of copper carbonate, I should obtain a new compound with the black colour proving the presence of copper (ll) oxide. The volume of the carbon dioxide that will result from heating copper carbonate depends on the mass of copper carbonate. Actually, it is proportional to it: the bigger the mass of copper carbonate the bigger the volume of gas given off and the bigger the mass of the product formed. ...read more.
Middle
* Green powder of copper carbonate. * Bell jar. (Eye protection required: WEAR SAFETY GOGGLES ?TAKE CARE TO AVOID BURNS. WEIGH (to the nearest 0.01g) EVERYTHING TWICE AT LEAST TO AVOID ERRORS. 1. Set the tripod, Bunsen burner (switched off), heatproof mat and pipe clay triangle as above. 2. Weigh the crucible and lid and record the measurement. 3. Letting the crucible on the balance, add the powder of copper carbonate for a little more than 3.00g. 4. Put the lid back and record the measurement. 5. Place the set onto the pipe clay triangle. 6. Switch the Bunsen burner on and heat the crucible strongly. 7. Using the tongs, lift the lid slightly from time to time to check whether the colour of the copper carbonate has completely changed or not. 8. When the colour has changed totally (after about 10 minutes), switch the Bunsen burner off and remove the crucible and lid using tongs form the pipe clay triangle. ...read more.
Conclusion
This method could only enable us to calculate the volume. The total uncertainties in that volume is the same of one of the mass of copper oxide formed for they depend quantitatively to the mass of copper carbonate used. The chemical balance was accurate to 0.01g. That error is [?(0.01/3.72) x 100] ? 0.27% then the order of proportionality of the results are: V CO2 (g) = (0.723 ? 0.0027) dm-3 and M CuO (s) = (2.59 ? 0.0027) g. If I had to repeat this experiment, I would use a gas inch well greased (to enable the pressure of gas to push it) by which I can just measure the volume of gas directly using a similar mass. ( www.wpbschoolhouse.btinternet.co.uk/page13/ChemicalTests.htm http://www.pgo.pwp.blueyonder.co.uk/StBrnards/science/year8/scheme_/Pt72.html "Collins Advanced Science Chemistry" by Chris Conoley and Phil Hills. (Second Edition); page 566. The Visual Dictionary of Chemistry by Jack Challoner (Eyewitness Visual Dictionaries) Copyright 1996. A-Level Chemistry, third edition (Ramsden). Page 526 1 ...read more.
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• Ideas and feedback to | 1,622 | 6,350 | {"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} | 2.75 | 3 | CC-MAIN-2017-51 | longest | en | 0.866293 |
http://www.badmintoncentral.com/forums/index.php?threads/proposal-experiment-with-string-tension-loss.22615/ | 1,506,262,347,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690029.51/warc/CC-MAIN-20170924134120-20170924154120-00154.warc.gz | 371,116,243 | 24,545 | # Proposal: experiment with string tension loss
Discussion in 'Badminton String' started by SWC_Ant, Mar 31, 2005.
1. ### SWC_Ant Regular Member
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Hi all!
i have a proposal for an experiment that some of the BF'ers can conduct regarding different strings and tension loss over time, say 1 month.
what we can do is measure out a preset amount of string of each string type, and then put a weight on it. (im not sure how this works, but i'm sure Neil Nicholls knows, because he has done something like this before) then, measure the stretched length. then we can calculate how many mm each pound would stretch the string by initially. we must leave the string at this length, so even if it loses tension, it wont get stretched even longer by 25lb (because when we put our string on the racket, the total length of the string doesn't change, only the tension does). every 24 hours we let the string loose again (so it will return to original length) and measure, the number should increase by a bit because the string is losing elasticy. stretch the string back to the stretched length, and measure the unstretched length again in 24 hours. since the stretched length isn't changing, we can calculate the amount of tension left in the string. over a 30 day period, we can find out how many pounds of tension the string loses.
for example, if our set string length was 1000mm and the original weight was 25lb, then we cut 1000mm of each type of string in our experiment, and put 25lb of weight on it. say that string A gets stretched to 1200mm. then we know that each pound of tension will strech string A by 8mm. then we strech the string to 1200mm again (perferably tied, and not weighed down), and wait 24 hours. by this time the string may have loosened, but the stretched weight isn't going to change, because we aren't leaving 25lb on it.
then, we measure the unstretched length, (for our example we'll say it becomes 1010mm), so that means our tension has become (1200-1010)/8 = 23.75lb. stretch the string back to 1200mm and wait another 24 hours, rinse, repeat
hopefully at the end of 30 days we will be able to draw graphs on tension loss by different strings, and get an idea of which string(s) lose tension the fastest or slowest.
hope this makes sense, and correct me if parts of this may not work.. i dont have any of the required equipment to conduct such an experiment, but im sure there are a lot of BF'ers with access to these things
thanks in advance for any help with this experiment!
#1
Last edited: Mar 31, 2005
2. ### kwun Administrator
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i like the idea of the experiment. the goal is sound but i think some details needs to be ironed out. you said:
this does not make sense to me. if the string was 1000mm, stretched and then later on it lengthened to 1200mm. then the tension on the stretched string when it is at 1200mm will be 0 lbs.
#2
3. ### SWC_Ant Regular Member
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i meant, the string would be stretched to 1200mm for 24 hours (which will start off at 25lb, then slowly decrease in tension), then untension it so it would return to normal length, (ie. 1010mm). then you can measure.
thanks for the reply
edit
the original strech would be with a weight (25lb) to see how long the string should become. then the string could be secured at the 1200mm length somehow, because if we leave the 25lb weight on the string it will become more than 1200mm over time, but that doesn't happen in a normal racket
hope this helps
#3
Last edited: Mar 31, 2005
4. ### Neosakai Regular Member
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Sounds like a cool idea, but you got any pictures of it? Because i'm kinda confused on how it works
#4
5. ### SWC_Ant Regular Member
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i can't put any pictures up, because i can't conduct the experiment myself but im giving this idea to others so maybe some people can try it and report on the results. Neil Nicholls might know about some parts of it though.. because he did something similar before.. hope he sees this post soon
#5
6. ### Neil Nicholls Regular Member
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to mimic what happens in a racquet, you would want to stretch it to whatever length the weight stretches it, and then clamp it at the new length.
Part (maybe most) of the tension loss you get in a racquet, though, is from using it. The repeated stretching and contracting breaks up the string at the molecular level.
#6
7. ### SWC_Ant Regular Member
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yea thats what i meant except i didn't word it that well
i agree part of the tension loss is from using the racket, but if you were to leave the strung racket sitting there for a while there will be tension loss too, and this experiment is trying to measure that. i do believe that tension loss of strings (without hitting birdie constantly) can be compared with each other, and the comparitive results will be good enough (ie. string A loses tension faster than string B), just that the tension may not drop as drastically as normal string that are constantly subject to striking birdies
btw thanks for replying!
#7
8. ### cooler Regular Member
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not to be sounding rude but this test would be too basic to be useful in understanding tension loss in a real racket. The only knowledge gain here is knowing which string is more stretchy relative to other strings, within an operative range that isn't common in real situation. Variables not considered in your experiment but important are:
2. type of loading? one fix load dont represent real situtation
for a string that measure less than 1mm, measuring length of elongation in mm is quite crude.
#8
9. ### SWC_Ant Regular Member
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if you want to know how stretchy strings are relative to each other, here's a post that answers that question http://www.badmintoncentral.com/forums/showpost.php?p=229044&postcount=31
what we're trying to find here is tension loss of strings over time
the string may measure less than 1mm in diameter, but the length can be changed a lot when there is tension on the string, as Neil found out earlier here http://www.badmintoncentral.com/forums/showpost.php?p=229005&postcount=29 ... at 24lb, 240mm of BG66 string stretches to 261mm, a 9% increase
#9
Last edited: Apr 1, 2005
10. ### kwun Administrator
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interesting point. but let's start from the simpliest case. that is, a continuous constant load with tension similar to a normal stringing tension. this somewhat emulate the string strung on a racket without being hit. ie. string it and let it sit for a few days. in a racket situation, the load actually descreases over time, but we gotta start simple...
and i think that's where SWC_Ant is close.
with continuous a load of 25lbs, which is within the bound of the tension that people uses.
and yes. if we use a long enough string, i don't see a problem with precision. i typically see a stretch of around 5mm everytime i tighten a segment of string on the stringing machine, a stretch of string 1-2meter will give approx 3cm of stretch when going from 0->25lbs tension. not a lot, but probably adequate.
#10
11. ### SWC_Ant Regular Member
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kwun:
what we meant was measure how long the string will become with 25lb tension, and then take the string off and clamp it separately, at the measured length, so that throughout the experiment, the clamped length will remain the same (just like the strings on the racket will always be stretched to that length), but the tension will decrease
however, this way, the load will not be 25lb all the way, because the load is lowering as tension drops
again, thanks to all who have replied as soon as we get everything set.. we'll need someone to do the experiment
#11
12. ### Neil Nicholls Regular Member
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I'm not sure you can deduce the tension loss from the change in unstretched length.
I think you need a method of measuring the tension while it is still stretched. Something like a Stringmeter which is used for tennis.
Maybe leave some string tensioned on a drop-weight machine.
As the string loses it's ability to hold tension, the weight will drop. As the weight drops, it pulls less tension.
The downside to this is that the weight will continue to lengthen the string.
#12
13. ### SWC_Ant Regular Member
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if we had extremely accurate weights/tensions, when we're measuring the tension loss we can put on weights until the string gets back to the stretched length.. i dont know if that will work better
its best if we dont let weight continue to lengthen the string, because on a normal badminton racket that doesn't happen
#13
14. ### Neil Nicholls Regular Member
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but on a normal badminton racket you don't take the tension off and let the string contract to it's original length and then re-stretch it either.
#14
15. ### SWC_Ant Regular Member
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you're right, that doesn't happen either.. but its the best way to measure... i think.. unless someone comes up with a better idea
btw how does a drop weight stringing machine help with this experiment? i dont quite understand because i have absolutely NO idea how stringing machines work :crying: ...
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furthermore, baddy strings are elastic. If tension is below yield stress point, the string will retract a bit when load is removed. At what instant would u take the length measurement? Will you able to measure each string length with the exact relaxation period because each string type will have different retraction rate. Because a relaxed string has slack, u have to give it some tenison to straighten it out for reasonable length measurement. If u put a slight tension to straighten out the string for measurement, technically, one is deforming/stretching the string during measurement. This add inconsistency into the experiment.
Each time u remove load and reclamp, u also introduce errors.
I dont see the benefit of doing this type of experiment because i dont get any useful data out of it. Sorry being so critical.
#16
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17. ### Neil Nicholls Regular Member
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I'm half with you.
Are you interested in tension loss between getting your racquet strung and playing with it. What if that gap is 2 weeks, or 1 month?
#17
18. ### Neil Nicholls Regular Member
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ah,
how about you attach some scales to one end of the string, tension it, and then clamp everything off. So you go: fixed point, scales, string, fixed point.
Then you just have to record the reading on the scales at whatever time intervals you like.
You're stuck without your scales for that time though.
#18
19. ### kwun Administrator
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you are assuming the "hook" of the scale doesnt move...
#19
20. ### SWC_Ant Regular Member
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you mean something like this? (excuse the bad drawing im sorta busy and rushing through things )
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#20 | 3,146 | 12,429 | {"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} | 3.46875 | 3 | CC-MAIN-2017-39 | latest | en | 0.935884 |
https://www.thestudentroom.co.uk/showthread.php?t=3905217 | 1,627,082,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00095.warc.gz | 1,094,117,837 | 34,940 | # Osmosis hypothesis?
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#1
I got my stimulus and it says that:
Jen adds her carrots to water and John adds his to a salt solution, and it later changes in mass.
What actually happens? I'm confused. Which goes higher in mass and which goes lower in mass and why?!? I understand if its a sugar solution but is a salt solution the same thing?
0
5 years ago
#2
Nice question!
You're absolutely correct its about osmosis. So which will gain mass and which will lose mass, and why? I'll hopefully try my best to answer that.
Scenario 1:
1. Puts carrot into water
Scenario 2:
1. Puts carrots in salt solution
This is all about water potential/osmosis, and the definition of osmosis is the net movement of water molecules from a region of high water potential to a region of lower water potential down a water potential gradient until dynamic equilibrium is reached.
The water potentials relative to each other is as follows:
Highest: Water
Middle: Carrot
Lowest: Salt
This means that for scenario 1 - water moves from the surrounding water solution into the carrot, hence the carrot will increase in mass due to the water moving in (the carrot is hypertonic relative to the solution).
For scenario 2 water moves from the carrot into the surrounding solution, hence the carrot will lose mass as it loses water (the carrot is hypotonic relative to the solution).
Salt and sugar the same thing? As a solute - absolutely! Having salt in a solution will decrease its water potential, and so the surrounding solution will create a water potential gradient favouring the net movement of water out of the carrot into the salt solution until it all nicely balances up. Although, to balance it up, the carrot will need to lose quite a lot of water, and so it will become lighter as a result.
Hope I've helped a tiny bit! Note that I'm an A level student myself, so take what I write with a pinch of salt! (or if you like, sugar is fine too!)
0
#3
(Original post by Spectral)
Nice question!
You're absolutely correct its about osmosis. So which will gain mass and which will lose mass, and why? I'll hopefully try my best to answer that.
Scenario 1:
1. Puts carrot into water
Scenario 2:
1. Puts carrots in salt solution
This is all about water potential/osmosis, and the definition of osmosis is the net movement of water molecules from a region of high water potential to a region of lower water potential down a water potential gradient until dynamic equilibrium is reached.
The water potentials relative to each other is as follows:
Highest: Water
Middle: Carrot
Lowest: Salt
This means that for scenario 1 - water moves from the surrounding water solution into the carrot, hence the carrot will increase in mass due to the water moving in (the carrot is hypertonic relative to the solution).
For scenario 2 water moves from the carrot into the surrounding solution, hence the carrot will lose mass as it loses water (the carrot is hypotonic relative to the solution).
Salt and sugar the same thing? As a solute - absolutely! Having salt in a solution will decrease its water potential, and so the surrounding solution will create a water potential gradient favouring the net movement of water out of the carrot into the salt solution until it all nicely balances up. Although, to balance it up, the carrot will need to lose quite a lot of water, and so it will become lighter as a result.
Hope I've helped a tiny bit! Note that I'm an A level student myself, so take what I write with a pinch of salt! (or if you like, sugar is fine too!)
Thank you so much! It was really thorough and was easy to understand
I just have one question; when the water is going out of the carrot it needs to become isotonic right?
0
5 years ago
#4
Another great question!
Although I'm not sure quite what to make of that question (failure of my part). I think either you mean:
1. If the carrot is isotonic to the solution water will leave the carrot
2. Water leaves the carrot until it becomes isotonic [personally I think you meant this one]
I would like to clarify one thing though that is a bit conceptually difficult. The one thing I want to clarify is that water molecules are moving continuously between the carrot and solution, whether or not there is a concentration gradient. Its a 'dynamic' system, its not like the water molecules are stuck in one place, they are free to move all over the place. (and between if it is a partially, permeable membrane which a carrot has)
What we mean when we say water moves down a concentration gradient/water potential gradient is that it is a net movement from one place to another. The word net is really key: essentially it means 'most' of the water is moving in that direction. An analogy is like crossing a road: perhaps most people will be trying to cross the road to get to a shopping centre, but of course there may still be other people on the other side of the road leaving the shopping centre. Even though there is a net movement of people crossing the road into the shopping centre, there are still some people crossing the road the other way.
So for potential question 1:
The answer is yes, water will move, it is always moving. However isotonic, meaning same concentration, means theres no net movement of water molecules from carrot to solution or vice versa. This means there is no osmosis occurring.
For potential question 2:
The answer is a yes and no. If you put the word 'net' in front of water, absolutely; the net movement of water will continue down the concentration gradient until it finally evens it out (carrot becomes isotonic to solution).
Maybe I should just quickly clarify how the word isotonic should be used. We can't really call something 'isotonic' by itself, its a word used to compare two things (like calling something big, an elephant is big compared to a mouse, but is it big compared to a black hole?)
So when we say isotonic, we need to say 'X is isotonic with regards to Y' meaning that 'solution X is of the same concentration as solution Y'
Since osmosis is the net movement of water, and if there's no gradient, there's no net movement (still movement though!), so there is no longer the process of osmosis occurring.
I'm sorry if I haven't answered the question you asked, please feel free to re-ask it (or hopefully someone more competent than I can answer it). Hope you've learnt at least a bit though in this weirdly convoluted explanation!
0
4 years ago
#5
(Original post by Sakura-Sama)
I got my stimulus and it says that:
Jen adds her carrots to water and John adds his to a salt solution, and it later changes in mass.
What actually happens? I'm confused. Which goes higher in mass and which goes lower in mass and why?!? I understand if its a sugar solution but is a salt solution the same thing?
Haha, this is your biology controlled assessment
0
4 years ago
#6
yo u finished homie send ur plan to me bro i need it
0
4 years ago
#7
(Original post by Oluwafemi.muyiwa)
yo u finished homie send ur plan to me bro i need it
Really *****
That will be considered as plagiarism
Posted from TSR Mobile
0
4 years ago
#8
(Original post by Naomeyz_01)
Really *****
That will be considered as plagiarism
Posted from TSR Mobile
shutup u neek whos going to know
0
4 years ago
#9
The examiners, and then you'll be given a low grade
Posted from TSR Mobile
0
4 years ago
#10
(Original post by Naomeyz_01)
The examiners, and then you'll be given a low grade
Posted from TSR Mobile
awww thx for being so caring xox
0
4 years ago
#11
Wow I aas reading this and realised I was missing a lot of info out. Thanks!!!
0
4 years ago
#12
is it hypotonic or hypertonic for scenario 2?
0
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The crossing number of graph is the minimum number of edges crossing in any drawing of in a plane. In this paper we describe a method of finding the bound of 2-page fixed linear crossing number of . We consider a conflict graph of . Then, instead of minimizing the crossing number of , we show that it is equivalent to maximize the weight of a cut of . We formulate the original problem into the MAXCUT problem. We consider a semidefinite relaxation of the MAXCUT problem. An example of a case where is hypercube is explicitly shown to obtain an upper bound. The numerical results confirm the effectiveness of the approximation.
#### 1. Introduction
Let be a simple connected graph with a vertex-set and an edge-set . The crossing number of graph , denoted , is the minimum number of pairwise intersections of edge crossing on the plane drawing of graph . Clearly, if and only if is planar. It is known that the exact crossing numbers of any graphs are very difficult to compute. In 1973, Erdös and Guy [1] wrote, “Almost all questions that one can ask about crossing numbers remain unsolved.” In fact, Garey and Johnson [2] prove that computing the crossing number is NP-complete.
A 2-page drawing of is a representation of on the plane such that its vertices are placed on a straight horizontal line according to fixed vertex ordering and its edges are drawn as a semicircle above or below but never cross .
The -cube or -dimensional hypercube is recursively defined in terms of the Cartesian products. The one-dimension cube is simply where is a complete graph with vertices. For , is defined recursively as . The order of is and its size is . Since is planar for , so for each such . Eggleton and Guy [3] showed that but is unknown for .
It was declared by Eggleton and Guy [3] that the crossing numbers of the hypercube (non-2-page) for was
Then, in 1973, Erdös and Guy [1] conjectured equality in (1). In 1993, a lower bound of was proved by Sýkora and Vrt’o [4]:
In 2008, Faria et al. [5] constructed a new drawing of in the plane which led to the conjectured number of crossings To the best of our knowledge, the fixed linear crossing number for has not been established. In this paper, we discuss a method to obtain an approximation for fixed linear crossing number for hypercube graph.
#### 2. 2-Page Drawings of Hypercube Graph
Throughout this paper, we consider the ordering of hypercube graph . Since is defined recursively as , for , where is a simple graph with 2 vertices together with a single edge incident to both vertices, has 2 copies of with edges connecting between them. Given a fixed ordering on , the vertices of the first are labeled and the vertices of the second are labeled . The two vertices between the first and the second are adjacent if and only if the sum of the labeled is . Figures 1 and 2 present the ordering of and which we consider throughout this paper. Notice that our method is independent on vertex ordering; therefore, for a fixed , we can apply the method times so as to obtain the 2-page linear crossing number.
The 2-page drawing of can be represented by drawing the vertices of on a straight horizontal line according a fixed vertex ordering. Each edge fully contained one of the two half-planes (pages) as a semicircle and never cross . Notice that no edge crosses itself, no adjacent edges cross each other, no two edges cross more than once, and no three edges cross in a point.
For a given 2-page drawing of with the fixed vertex ordering, a pair of edges and are potential crossing if and cross each other when routed on the same side of . Clearly, and are potential crossing if and only if or .
Next we give the definition of conflict graph of graph .
Definition 1. Given a graph . We define an associated conflict graph of a graph . There is corresponding one-to-one and onto mapping between the set of and . Two vertices of are adjacent if any two edges in are potential crossing.
For example, according to the given fixed vertex ordering of (see Figure 3), is a graph of nodes, . and are adjacent in because and are potential crossing in a 2-page drawing of . A fixed vertex ordering of and its potential crossing can be seen in Figure 4.
In this paper, we are interested only in fixed linear embeddings of . There is a crossing between and if and only if and are potential crossing and embedded on the same side of . We can see that the number of edge crossings depends on the order of vertices and on the sides to which the edges are assigned.
The 2-page linear crossing number of , denoted by , is the minimum number of pairwise intersections of edges crossings determined by a 2-page drawing of . The 2-page fixed linear crossing number of is the minimum number of pairwise intersections of edges crossings determined by a 2-page drawing of with fixed vertex ordering of . It is known that for , for
#### 3. Reduction to MAXCUT Problem
In this section, we show that the problem can be reduced to the maximum cut problem. Next, we reduce the fixed linear crossing number problem to the maximum cut problem (MAXCUT). The MAXCUT problem is as follows.
Maximum Cut Problem (MAXCUT). Given an undirected graph the edge of the graph is associated with nonnegative weights . The problem is to find a cut of the largest possible weight, that is, to partition the set of into disjoint sets and such that the total weight of all edges linking and (i.e., with one incident node in and the other one in ) is as large as possible.
In the MAXCUT problem, we may assume that the weights are defined for every pair of indices: it suffices to set for pairs of nonadjacent nodes. For the unweighted graph, we assume that for .
Let be a graph with a fixed vertex permutation. Given a vertex partition of its conflict graph , the associated cut embedding is the fixed linear embedding of where edges corresponding to and are embedded to the half spaces above and below the vertex line, respectively.
Lemma 2 (see [6]). where is a number of potential crossing of 2-page drawing of , which is the number of edges of . is the size of the maxcut of .
Proof. Given a 2-page (circle) drawing of , define as the chords that are drawn inside the circle. The edges of with precisely one endpoint in now correspond to edges of that do not cross in the drawing.
Theorem 3 (see [7]). Consider a partition of . Then the corresponding cut embedding is a fixed linear embedding of with a minimum number of crossings if and only if is a maximum cut in .
Proof. Let be the set of edges in with one endpoint in and one endpoint in , that is, the cut given by . By definition of , we know that every crossing in the cut embedding associated with corresponds to an edge in such that either both its endpoint belong to or both belong to , that is, to an edge in . Thus, the number of crossings is . As is constant for a fixed vertex permutation, the result follows.
Theorem 3 reduces the fixed linear crossing number problem to the maximum cut problem (MAXCUT). In the next section, we describe the relaxation of the MAXCUT problem which leads to semidefinite programming.
##### 3.1. Formulating MAXCUT by Semidefinite Relaxation
In this section, we show that 2-page crossing number of hypercube graph problem can be obtained by computing a semidefinite relaxation of MAXCUT.
First of all, we introduce the adjacency matrix of denoted as we know it is an matrix with the property
From we construct the conflict graph of denoted . Finally, we perform MAXCUT on graph . We use semidefinite relaxation to approximate the optimal value solution to the MAXCUT problem. Obviously the approximation is larger than the actual MAXCUT optimal value. The feasibility of the relaxation set is strictly larger than the original ones.
According to [2], the MAXCUT problem can be formulated as follows:We call the optimal value of (6) as “OPT.” Then, the relaxation of (6) can be rewritten aswhere is an adjacency matrix of and is a feasible solution to the semidefinite relaxation. The problem (7) is equivalent towhere is a given adjacency matrix of and is a feasible solution to the semidefinite relaxation. We call the optimal value of (8) as “SDP.”
As we have seen from the relation (4), we let be the number of potential crossing of 2-page drawing of with our fixed vertex ordering (i.e., is the number of edges of ).
We can determine by considering the upper half of the main diagonal of the adjacency matrix of .
Definition 4. Let be the adjacency matrix of . The element where is called minor diagonal of adjacency matrix of and the element where is called semiminor diagonal of adjacency matrix of , denoted by .
For simplicity, the size of is a number of elements in . Let be the adjacency matrix of graph . Therefore is symmetric matrix. It is clear that the size of is .
Let and be adjacency matrices of graphs and , respectively; we say that the number of potential crossing between and , denoted by , is simply the number of potential crossing between 2-page drawing of graph and . The adjacency matrix of size of with respect to our ordering is presented in Figure 5.
Lemma 5. For any integer ,where .
Lemma 6. For every adjacency matrix of , , where , there exists adjacency matrix of , , which is a submatrix embedding in . The number of submatrix embedding in is .
Lemma 7. For any integer ,where .
Lemma 8. For any integer ,where .
Lemma 9. For , the number of block embedding in equals , .
Lemmas 59 follow directly from the definitions.
Lemma 10. For any integer ,
Proof. The number of potential crossing between and is a result of the number of potential crossing between all of the element in and . That is,
Theorem 11. For any integer ,where is the number of potential crossing of 2-page drawing of with our fixed vertex ordering.
Proof. We prove this lemma by considering the number of potential crossing of 2-page drawing of with our fixed vertex ordering. Since has copies of with some edges connecting between them, the number of potential crossing of is a result of twice of the number of potential crossing within together with and .
Hence, it is enough to show that the number of potential crossing between 2-page drawing of and is equal to , whereNote is the number of potential crossing between all of submatrices and and also between and . By Lemmas 6 and 9,We precede by mathematical induction on . For , it can be easily seen that by counting. Assuming (15) holds true, now we consider as a number of potential crossing between all of the submatrices and and also between and . By the Lemmas 5, 6, 7, 9, 8, and 10,
The next theorem shows how effective the relaxation is.
Theorem 12 (see [8]). Let OPT be the optimal value of the MAXCUT problem and SDP be the optimal value of the semidefinite relaxation. Then
Theorem 12 guarantees that the optimal value of the MAXCUT is close to the optimal value of the semidefinite relaxation. From (4), we havewhere is a number of potential crossing of 2-page drawing of . is an approximation of 2-page fixed linear crossing number of and is an approximation of .
Corollary 13. Let be an approximation of . Then we havewhere is a computable quantity depending on .
Proof. From (4), (18), and (19), we have Let be the computable quantity depending on .
Then,
Corollary 13 shows that the upper bound of is , where is the computable quantity depending on .
##### 3.2. Experimental Results
In this section, we consider the hypercube graph for . Then, we give some examples for approximating the problems of the semidefinite relaxation in the form (8). We approximate this problem via MATLAB program together with an optimization toolbox called “SeDuMi.” The SeDuMi is a package for solving optimization problems with linear, quadratic, and semidefinite constraints.
In Table 1, the second column shows numerical results for the approximation of the MAXCUT on the associated conflict graph by using the semidefinite relaxation. It is well known that this problem can be solved in a polynomial time. The third column displays the numbers of potential crossing of 2-page drawing of referring to our fixed vertex ordering that we evaluate from (14). Notice that this potential crossing of 2-page drawing of is the exact value. From (19), we calculate the approximation of 2-page fixed linear crossing number of for . The results are shown in the last column.
In Table 2, we present the lower bound of , and the upper bound of , . The second column shows the values of for . We see that as get larger the values of tend to decline continuously. It is interesting to study the behavior of as . It does not surprise to see that our approximation is strictly larger than the upper bound of (14) since the latter one does not have a restriction that all vertices must be placed on a line. However, it is surprising to see that these numbers are not so different from each other.
#### 4. Concluding Remarks
In this paper, given graph , we show how the associating conflict graph is constructed. We recharacterize the problem of finding the crossing number of graph to the MAXCUT problem of . We approximate the MAXCUT problem by the semidefinite relaxation which can be solved easily by a standard optimization package; in this case, we use SeDuMi 1.02. The numerical results show reasonable outcome. Clearly, another relaxation method can be explored. Moreover, it would be quite interesting to see the behavior of as get larger. One can further study how to estimate for a larger .
#### Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.
#### Acknowledgments
The authors would like to thank the Thailand Research Fund under Project RTA5780007 and Chiang Mai University, Chiang Mai, Thailand, for the financial support. | 3,011 | 13,838 | {"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} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.940024 |
https://www.numbersaplenty.com/12689339799675 | 1,656,561,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00468.warc.gz | 973,733,992 | 3,418 | Search a number
12689339799675 = 352711314111315299
BaseRepresentation
bin1011100010100111011110…
…0101110101100001111011
31122221002101220112012211010
42320221313211311201323
53130400231442042200
642553222514205003
72446526445434550
oct270516745654173
948832356465733
1012689339799675
114052578563190
12150b342283763
137107a4456586
1431c24b1c6a27
1517012b1aa150
hexb8a7797587b
12689339799675 has 384 divisors, whose sum is σ = 27905413939200. Its totient is φ = 4934522880000.
The previous prime is 12689339799641. The next prime is 12689339799707. The reversal of 12689339799675 is 57699793398621.
It is not a de Polignac number, because 12689339799675 - 213 = 12689339791483 is a prime.
It is a super-2 number, since 2×126893397996752 (a number of 27 digits) contains 22 as substring.
It is an unprimeable number.
It is a polite number, since it can be written in 383 ways as a sum of consecutive naturals, for example, 829415176 + ... + 829430474.
It is an arithmetic number, because the mean of its divisors is an integer number (72670348800).
Almost surely, 212689339799675 is an apocalyptic number.
12689339799675 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
12689339799675 is an abundant number, since it is smaller than the sum of its proper divisors (15216074139525).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
12689339799675 is a wasteful number, since it uses less digits than its factorization.
12689339799675 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 15515 (or 15510 counting only the distinct ones).
The product of its digits is 8332994880, while the sum is 84.
The spelling of 12689339799675 in words is "twelve trillion, six hundred eighty-nine billion, three hundred thirty-nine million, seven hundred ninety-nine thousand, six hundred seventy-five". | 573 | 1,935 | {"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} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.815873 |
http://mathcentral.uregina.ca/QandQ/topics/percent%20of%20zero | 1,591,193,192,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347434137.87/warc/CC-MAIN-20200603112831-20200603142831-00045.warc.gz | 75,486,200 | 4,202 | Math Central - mathcentral.uregina.ca
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Is there an equation to find what percent of 0 is 1? 2020-03-09 From Asher:Question: Is there an equation to find what percent of 0 is 1. I remember learning quite a long time ago that the answer inst 0 and it isn't infinity. I'm pretty sure it was something like %=0 approaching infinity or %=1 approaching infinity. And I know it depends what value you assign the numbers i.e. dollars or temperature. Furthermore, is asking "what percent 1 dollar is of 0 dollars" the same question as "what percent profit do you make from selling something worth 0 dollars for 1 dollar." Thank you for your time and consideration -AsherAnswered by Penny Nom.
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https://bighominid.blogspot.com/2018/09/ready-player-one-redux.html | 1,723,319,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00018.warc.gz | 102,245,554 | 17,150 | ## Saturday, September 08, 2018
A very interesting analysis of "Ready Player One" by Canadian vlogger Just Write, whom I've cited before re: the "The Hobbit" movies:
And although I don't watch much Cinema Sins anymore—mainly because it's too smug, and many of the sins aren't really sins—I found that critic's takedown of "Ready Player One" interesting for the holes he pokes in the story's handling of the properties and potential of cyberspace, e.g., the idea that, in the movie, people have to line up to use teleportation portals when, in actual cyberspace, there'd be no need to line up for anything.
In cyberspace, things can move infinitely faster in parallel since nothing needs to occur in series. And when you focus on just that one major distinction between cyberspace and the real world, you begin to realize how much untapped potential Spielberg & Co. missed out on. Widen your horizon from that single difference to the myriad other differences between cyberspace and our regular, three-dimensional meatspace, and you begin to realize that an adventure in cyberspace would look like nothing our current brains could truly understand (consider: a computer can construct a 6-dimensional universe and make it possible for your character to move through it because such spaces are mathematically possible even if not conceivable by the human brain), and you'd need time to figure out what, exactly, was going on before you'd be ready to go adventuring in 6-D space.* I hinted at this problem in my review of "Ready Player One"—to wit, the fact that the denizens of the Oasis don't do or contribute much that seems creative, consistent with the theme of untapped potential.
Anyway, I found the above two videos, both of which are critical of "Ready Player One," to make for interesting viewing. If you have a few minutes, check them out.
*You may recall Carl Sagan's explanation of dimensions in which each successively higher dimension is "at right angles to" the immediately lower dimension. Take a one-dimensional line, put it at right angles to itself, and now you've defined a two-dimensional plane: flat space. Take a two-dimensional plane, put it at right angles to itself, and now you've defined a three-dimensional space (with eight octants). Take the x-y-z-axes of three-dimensional space, put them at right angles to themselves, and you've got 4-D space. But at this point, your brain can't imagine what that looks like, right? Same deal with cyberspace: a computer can plausibly mathematically render a cyberspace avatar's "motion" through 4-D or 6-D space, but your brain won't be able to process exactly what's going on except in the most rudimentary of terms. (Well... that's true of my brain, anyway.) | 608 | 2,731 | {"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} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.963203 |
https://ask.cvxr.com/t/index-of-variable/3855/6 | 1,686,096,147,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00105.warc.gz | 144,693,340 | 5,168 | # Index of variable
hello
I have a vector (x) with 251. I want to separate this vector to 5 vectors with 51 and then do multiplication with a matrix (A) and then calculate sum then cost function is obtained.
Unfortualey that vector (x) is my variable and when I use loop with for I got error because I want to index the variable that is defined after cvx_begin .what do you suggest for defining cost function in this situation?
x : is 25 by 1
y : is 5 by 1
xi : is 5 by 1
matrix A is 5 by 5
A(:,i): i’th coulmn of matrix A
cost function : minimize (y-sum(diag(xi)*A(:,i))) %% summation is done over all i
variable x
cvx_begin
variable x(25,1 ) complex
for kx=1:5
fg=x(5*(kx-1)+1:kx*5);
f2(:,kx) =diag(fg)*A(:,kx);
end
minimize(sum_square_abs(y-diag(f2)*A) );% % this is my cost function
cvx_end
for solving this problem I used CVX command for saving variable and then I write loop command it worked but I do not know this is correct or not ? because I need index of variable.
some posts related to this topic. in simple way how can I index the variable after define them for making cost function I hope this help .(with simple example)
You need to get dimensions of all your MATLAB and CVX variables to be consistent (conformal), which they currently are not,
Write out all assignments and expressions (including objective function) as though they are in MATLAB without CVX. Get everything consistent (conformal). Then re-write, declaring whatever you need as CVX variables (for instance maybe x is 25 by 1).
This is the practical problem I did not formulate in that way.
I am waiting for your ans.
Thanks
If in the objective function, you change
y - diag(f2)*A
to
y - A*diag(f2)
your program is confomal and will be accepted by CVX. I have no idea whether that correctly implements what you want to implement.
Regarding the link Finding it hard to model this objective function in CVX , did you try doing what the OP did? | 510 | 1,931 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-23 | latest | en | 0.905875 |
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# Nishant Gupta
### ZS
19 2010 年以降の合計貢献数
Data Scientist | ZS
Netaji Subhas Institute Of Technology, Delhi University | 2008 - 2012
Professional Interests: Data Science, Analytics, Machine Learning, Internet of Things, Game Theory, Mechanism Design, Design Thinking
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10年弱 前 | ダウンロード 2 件 | | 870 | 3,028 | {"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} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.57258 |
http://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation | 1,469,357,843,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00031-ip-10-185-27-174.ec2.internal.warc.gz | 160,714,759 | 19,880 | # Estimating the size of solutions of a diophantine equation
A. Is there natural numbers $a,b,c$ such that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an odd natural number ?
(I do not know any such numbers).
B. Suppose that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an even natural number ($a,b,c$ are still natural numbers) then is there any way to estimate the minimum of $a,b$ and $c$ ?
The smallest solution that I know for $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$ is:
-
I'm curious, how did you find the $a,b,c$ which you talk about at the end? – Wojowu Jan 5 at 22:47
@Wojowu : I don'tknow how alex alexeq found them, but what you can do is the following. You use a computer algebra system like Magma or Sage to find the generators of the group $E'_4({\mathbb Q})$ (compare my answer), then you enumerate the points on the non-identity component by increasing height and check if they give positive $a,b,c$. The numbers $a,b,c$ given in the question come from the first set of six points one finds in this way. – Michael Stoll Jan 6 at 9:00
I'm wondering about the solution you found for $n=4$: is there any way to compute a solution by approximation (continued fraction-style), such as approximating a real point of the elliptic curve? – Circonflexe Jan 7 at 8:32
(also, would you happen to have the numbers $a,b,c$ in copy-pasteable form? thanks :-D) – Circonflexe Jan 7 at 19:23
This problem turned out to be much more interesting than I originally thought. Let me give my solution, which seems to be slightly different from (but essentially the same as) the solution in the paper by Bremner and MacLeod (see Allan MacLeod's answer).
Theorem. Let $a,b,c$ be positive integers. Then $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can never be an odd integer.
Let $n$ be a positive odd integer. The equation $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$ implies $$a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$ This describes a smooth cubic curve $E_n$ in the projective plane that has at least six rational points (of the form $(1:-1:0)$ and $(1:-1:1)$ and their cyclic permutations). Declaring one of these to be the origin, $E_n$ is an elliptic curve over $\mathbb Q$. Bringing $E_n$ in Weierstrass form, we obtain the isomorphic curve $$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr) =: x(x^2 + Ax + B).$$ If $n = 1$, then there are obviously no positive solutions, so we assume $n \ge 3$. Then $E_n(\mathbb R)$ has two connected components, one of which contains the six `trivial' points but no points with positive coordinates, whereas the other component does contain positive points. In the model $E'_n$, this component consists of points with negative $x$-coordinate.
Claim. If $(\xi,\eta) \in E'_n(\mathbb Q)$, then $\xi \ge 0$.
This clearly implies the statement of the theorem.
To show the claim, let $D = 2n + 5$. Then $D$ is odd, positive, coprime with $B$ and divides $A^2 - 4B = (2n-3)(2n+5)^3$. If $p$ is an odd prime dividing $B$, then $n \equiv -3 \bmod p$ and so $-D \equiv 1 \bmod p$. The equation $B x^2 - D y^2 = z^2$ has the solution $(x,y,z)=(1,4,4)$, so the Hilbert symbol $(B, -D)_p = 1$ for all primes $p$. We will show:
If $(\xi,\eta) \in E'_n(\mathbb Q)$ with $\xi \neq 0$, then $(\xi, -D)_p = 1$ for all primes $p$.
Given this, the product formula for the Hilbert symbol implies $(\xi, -D)_\infty = 1$ and so $\xi > 0$ (since $-D < 0$).
Note that $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p$. We first consider odd $p$. We note that when $\xi$ is not a $p$-adic integer, then $\xi$ must be a $p$-adic square, so $(\xi, -D)_p = 1$. So we can assume that $\xi \in {\mathbb Z}_p$. There are three cases.
1. $p$ divides neither $B$ nor $D$. If $\xi \in {\mathbb Z}_p^\times$, then $(\xi, -D)_p = 1$, since both entries are $p$-adic units. Otherwise, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.
2. $p$ divides $B$. Then $-D \equiv 1 \bmod B$, so $-D$ is a $p$-adic square, hence $(\xi, -D)_p = 1$.
3. $p$ divides $D$. Then $x^2 + Ax + B \equiv (x + A/2)^2 \bmod p$. So if $\xi \in {\mathbb Z}_p^\times$, then $\xi$ must be a square mod $p$, and $(\xi, -D)_p = 1$. If $\xi$ is divisible by $p$, then as before, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.
It remains to consider $p = 2$. If $n \equiv 1 \bmod 4$, then $-D \equiv 1 \bmod 8$, so $(\xi, -D)_2 = 1$ for all $\xi$. If $n \equiv 3 \bmod 4$, then $-D \equiv 5 \bmod 8$, so $(\xi, -D)_2 = (-1)^{v_2(\xi)}$, and we have to show that the 2-adic valuation of $\xi$ must be even. Note that in this case $v_2(B) = 6$ and $A \equiv -3 \bmod 8$. If $v_2(\xi)$ is odd, then exactly one of the three terms $\xi^3$, $A \xi^2$, $B \xi$ has minimal 2-adic valuation, which must be even, so it cannot be the first or the third term. This reduces us to $\nu := v_2(\xi) \in \{1,3,5\}$. One then easily checks that $\xi(\xi^2 + A\xi + B) = 4^\nu u$ with $u \equiv -1 \bmod 4$ when $\nu = 1$ or $5$ and $u \equiv -3 \bmod 8$ when $\nu = 3$. In all cases, $u$ cannot be a square, and so points with $x$-coordinate $\xi$ cannot exist. This concludes the proof.
Note that when $n$ is even, we have $-D \equiv 3 \bmod 4$ and also $v_2(B) = 5$, so we lose control over the 2-adic Hilbert symbol.
This is the previous version of this answer, which I leave here, since it may contain some points of interest.
The equation $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$ gives rise to the elliptic curve $$E_n \colon a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$ You are asking for rational points on this curve (such that $a+b, b+c, c+a \neq 0$). For odd positive $n$ up to and including 17, this is a curve of rank zero (with 6 rational points), whereas for $n = 19$, it has rank 1. Therefore $E_{19}$ has infinitely many rational points, and your equation has infinitely many solutions for $n = 19$. I'll do the computations and find one explicitly.
EDIT: As pointed out by Jeremy Rouse in a comment below, the integral solutions for $n = 19$ are not positive. More precisely, the real points $E_n(\mathbb R)$ form two connected components (the discriminant of $E_n$ is positive), and it is the non-identity component that contains points with all positive coordinates (taking as the identity one of the six points like $(1:-1:0)$ or $(1:1:-1)$). So the question is whether there is odd $n$ such that there is a rational point on the non-identity component; then the rational points will be dense on this component and so there will be positive solutions. So far, no such $n$ turned up, even though there are many such that $E_n$ has positive rank.
FURTHER EDIT: I suspect that there really is no odd $n > 0$ such that $E_n$ has rational points on the non-identity component. One way of checking this for any given $n$ is to do (half of) a 2-isogeny descent on $E_n$. This produces a number of curves of the form $C_u \colon y^2 = u x^4 + v x^2 + w$ where $v = 4n(n+3)-3$ and $uw = 32(n+3)$ that are unramified double covers of $E_n$. We consider the curves $C_u$ that have points over all completions of $\mathbb Q$. Then every rational point on $E_n$ is the image of a rational point on one of these curves $C_u$. Doing the computation, one obtains a set of curves $C_u$ that all have $u > 0$ (this is only experimental; I checked it for $n$ up to 9999). But if $u > 0$, then [$C_u$ has only one real component — this is wrong, but the following is OK] the image of $C_u(\mathbb R)$ in $E_n(\mathbb R)$ is the identity component, so there can be no rational point on the other component. My feeling is that there might be a Brauer-Manin obstruction to the existence of rational points on the non-identity component for odd $n$, but I don't have enough time to check this. A possible approach would be to note that $$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr)$$ is isomorphic to $E_n$. If we can find a positive integer $d(n)$ such that for all rational points $(\xi,\eta) \in E'_n(\mathbb Q)$ (with $\xi \neq 0$) the product $\prod_p (\xi, -d(n))_p$ of Hilbert symbols (over all finite places) is always $+1$, then the claim follows from the product formula for the Hilbert symbol and $(\xi, -d(n))_\infty = -1$ for $\xi < 0$.
SUCCESS: For odd $n \ge 3$, $d(n) = 2n-3$ works. One can check that $(\xi, 3-2n)_p = 1$ for all primes $p$. Details later (it is getting late). Actually, $d(n) = -5-2n$ works better. See above.
Note that for even $n$, there usually are $C_u$ with $u < 0$ when $E_n$ has positive rank (the first exception seems to be $n = 40$). So I would expect the Brauer-Manin obstruction to result from an interaction between $p = 2$ and the infinite place.
For $n = 4$, the curve has also rank 1, which explains the existence of solutions. I'll try to check if there are smaller ones than that given by you.
EDIT: The given solution is really the smallest (positive) one. The next larger one has numbers of 167 to 168 digits.
-
The OP is requesting $a$, $b$ and $c$ to be positive, which appears to correspond to points on the non-identity component of the Weierstrass model of $E_{n}$. I don't think there are such points for $n = 19$. – Jeremy Rouse Jan 5 at 18:20
OK, thanks; I overlooked the positivity requirement. I'll look at larger $n$... – Michael Stoll Jan 5 at 18:27
This must be one of the easiest-to-state (and most attractive!) problems in number theory whose solution "requires" the Brauer-Manin obstruction. Very cool! – René Jan 5 at 23:26
+1 very nice. A direct proof, compared to Allan's article mentioned below. – alex alexeq Jan 7 at 4:41
This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et Informaticae, pages 29-41.
It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can be of truly enormous size - much larger than the example given.
-
Cool. How did you get to consider this problem? – Michael Stoll Jan 6 at 8:57
I was playing around with several cubic representation problems in the style of previous work by Andrew and Richard Guy. The numerical results were fascinating so I sent the initial work to Andrew, who very quickly proved the result about odd $n$. – Allan MacLeod Jan 6 at 9:35 | 3,316 | 10,383 | {"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} | 3.546875 | 4 | CC-MAIN-2016-30 | latest | en | 0.850471 |
http://coding.derkeiler.com/Archive/Python/comp.lang.python/2010-04/msg00775.html | 1,371,668,741,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709037764/warc/CC-MAIN-20130516125717-00078-ip-10-60-113-184.ec2.internal.warc.gz | 48,002,737 | 4,219 | # Re: lambda with floats
• From: Patrick Maupin <pmaupin@xxxxxxxxx>
• Date: Fri, 9 Apr 2010 11:40:35 -0700 (PDT)
On Apr 9, 1:22 pm, monkeys paw <mon...@xxxxxxxxxxxx> wrote:
On 4/9/2010 3:43 AM, Bas wrote:
On Apr 7, 6:15 am, Patrick Maupin<pmau...@xxxxxxxxx> wrote:
I should stop making a habit of responding to myself, BUT. This isn't
quite an acre in square feet. I just saw the 43xxx and assumed it
was, and then realized it couldn't be, because it wasn't divisible by
10. (I used to measure land with my grandfather with a 66 foot long
chain, and learned at an early age that an acre was 1 chain by 10
chains, or 66 * 66 * 10 = 43560 sqft.)
That's an exact number, and 208 is a poor approximation of its square
root.
There is no need to remember those numbers for the imperially
challenged people:
In [1]: import scipy.constants as c
scipy.constants ??
doesn't work for me.
In [2]: def acre2sqft(a):
...: return a * c.acre / (c.foot * c.foot)
...:
In [3]: acre2sqft(1)
Out[3]: 43560.0
Cheers,
Bas
Basically, he's saying that, instead of remembering the very simple
gzipped tar file for the scipy project. ;-)
(Of course, you get a few nice functions thrown in for free along with
choosing a sports car for its cupholders.)
.
## Relevant Pages
• Re: lambda with floats
... quite an acre in square feet. ... chain, and learned at an early age that an acre was 1 chain by 10 ... Appreciate the help, i just looked up the SciPY Project download, ...
(comp.lang.python)
• Re: lambda with floats
... how do i convert this to float? ... I'm not sure exactly what you mean by "acre meter" though; ... returns the number of square feet in 'n' acres. ... chain, and learned at an early age that an acre was 1 chain by 10 ...
(comp.lang.python)
• Re: lambda with floats
... does not have a name) from an outer function. ... I'm not sure exactly what you mean by "acre meter" though; ... returns the number of square feet in 'n' acres. ... chain, and learned at an early age that an acre was 1 chain by 10 ...
(comp.lang.python)
• Re: lambda with floats
... does not have a name) from an outer function. ... I'm not sure exactly what you mean by "acre meter" though; ... returns the number of square feet in 'n' acres. ... chain, and learned at an early age that an acre was 1 chain by 10 ...
(comp.lang.python)
• Re: lambda with floats
... On 4/9/2010 3:43 AM, Bas wrote: ... quite an acre in square feet. ... chain, and learned at an early age that an acre was 1 chain by 10 ...
(comp.lang.python) | 724 | 2,512 | {"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} | 3.1875 | 3 | CC-MAIN-2013-20 | latest | en | 0.914729 |
https://www.slideserve.com/luna/devs-standard-for-modeling-and-simulation-in-web-centric-environments | 1,568,610,919,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00285.warc.gz | 1,013,267,410 | 17,651 | DEVS Standard for Modeling and Simulation in Web-Centric Environments
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# DEVS Standard for Modeling and Simulation in Web-Centric Environments - PowerPoint PPT Presentation
DEVS Standard for Modeling and Simulation in Web-Centric Environments. Bernard P. Zeigler Arizona Center for Integrative Modeling and Simulation University of Arizona. Presented at Mosim08: Modélisation, Optimisation et Simulation des Systèmes March 31-April 2 2008 Paris, France .
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## DEVS Standard for Modeling and Simulation in Web-Centric Environments
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1. DEVS Standard for Modeling and Simulation in Web-Centric Environments Bernard P. Zeigler Arizona Center for Integrative Modeling and Simulation University of Arizona Presented at Mosim08: Modélisation, Optimisation et Simulation des Systèmes March 31-April 2 2008 Paris, France
2. Modeling and Simulation (M&S) Framework Experimental Frame Source Simulator System Simulation Relation Modeling Relation Model
3. DEVS – Formal Specification of a System A discrete event system specification (DEVS) is a structure M=<X,S,Y,int, ext, con, ,ta> where X is the set of input values, S is a set of states, Y is the set of output values, int :S->S is the internal transition function, ext,:QX->S is the external transition function, con,:QX->S is the confluent transition function, ta: S->R + 0,∞ Where Q={(s,e)|sS, 0 e ta(s) } is the total state set, e is the time elapsed since last transition, :S->Y is the output function and R + 0,∞ is the set of positive reals with 0 and
4. DEVS Hierarchical Modular Models DEVS Model Output port System Entity Structure message SES coupling DEVS Model Input port Hierarchical Composition, Coupling and Variants are Represented in System Entity Structure
5. Advantages of DEVS Theory • Closure under coupling, universality, uniqueness, relation to other formalisms • Hierarchical Model Construction supports complex systems • Supporting the correctness of the algorithms and validation of the executing models Application • Models, Simulators and Experimental Frames are distinct entities with their own software representations. • Precise and well-defined mathematical representation • Models/Experiments are developed systematically for interoperability • Repositories of models/experiments are created and maintained systematically (and existing components can be easily reused for constructing new models) • Discrete-event basis improves performance (e.g. no need for have a global clock to control timing)
6. Some Types of Models Represented in DEVS Coupled Models Atomic Models Partial Differential Equations can be components in a coupled model Ordinary Differential Equation Models Processing/ Queuing/ Coordinating Networks, Collaborations Physical Space Spiking Neuron Networks Spiking Neuron Models Processing Networks Petri Net Models n-Dim Cell Space Discrete Time/ StateChart Models Stochastic Models Cellular Automata Quantized Integrator Models Self Organized Criticality Models Fuzzy Logic Models Reactive Agent Models Multi Agent Systems
7. DEVS Research Groups/Environments • Carleton’s CD++, • ADEVS (ORNL *), • DEVS/C#, • DEVS/HLA, • DEVSJAVA (ACIMS - University of Arizona *), • GALATEA (USB – Venezuela), • LSIS (Aix-Marseille III – France *), • JDEVS (Université de Corse - France), PyDEVS (McGill), • PowerDEVS (University of Rosario, Argentina), • SimBeams (University of Linz – Austria), • VLE (Université du Litoral -France), • SmallDEVS (Brno University of Technology, Czech Republic), • James (University of Rostock,Germany) • Portugal, Spain, and Russia;;; • Workshop on Net-Centric Modeling & SimulationMarch 6–7 2008 - Marseille, France • http://osa.inria.fr/wiki/NCMS/NCMS
8. DEVS Adopters • Joint Interoperability Test Command, USA • Air Force, Navy / USA, South Korea • Lockheed Martin Missile Systems • Usinor – Sachem Expert Control • Swedish Materials Command • …
9. Global Information Grid /Service Oriented Architecture Net-Enabled Command & Control NCES: Secure, agile, robust, dependable, interoperable data-sharing environment for DOD where warfighter, business, and intelligence users share knowledge on a global network. This, in turn, facilitates information superiority, accelerates decision-making, effective operations and net-centric transformation.
10. WSDL Service Oriented Architecture Basics Content/Service Catalogs/Registries Search find_xxx Post save_xxx (Bind) XML Schema XML Payload Content/Service Provider Content/Service Consumer Client Access (& Use) SOAP Service Simple Object Application Protocol
11. WSDL Requirements for Testing and Data Collection Testing for Organization and Ontology quality Content/Service Catalogs/Registries Content discovery accuracy and effectiveness Verification/ Validation relative to service Search find_xxx Post save_xxx (Bind) Assessment of content for pragmatic, semantic, syntactic correctness XML Schema XML Payload Content/Service Provider Content/Service Consumer Client Access (& Use) SOAP Service Measurement of timeliness of information exchange Simple Object Application Protocol
12. Net-Centric Test Agent Capability (NTAC) Test Director Data Analyst Search Find service Post Save service WSDL Embedded “Test Agent” Embedded “Test Agent” Content/Service Catalogs/Registries Real time test data, status Post test data / metrics for analysis XML Schema XML Payload Content/Service Provider Content/Service Consumer Client Access (& Use) Service SOAP Agent-to-Agent communication/coordination
13. Levels of System of System Interoperability
14. Mapping M&S Layers to Linguistic Levels Collaboration Layer Semantic Web, Composition, Orchestration Design and Test Development Layer Pragmatic Level SES, DoDAF, Integrated System Development and Testing Experimental Frame Layer . Observers and Agents for Net-Centric Key Performance Parameters Semantic Level Modeling Layer Ontologies, Formalisms, Model Dynamic Structure, Life Cycle Continuity, Model Abstraction Syntactic Level Execution Layer Abstract Simulators, Real time Execution, Animation Visualization Network Layer Distributed Grids, Service Oriented Architectures
15. NTAC Agent-based Test Instrumentation Infrastructure supports simultaneous testing at multiple levels Highest layer agents collaborate to control and observe mission thread executions End-to-end Mission Layer Middle layer alert higher layer agents of network conditions that invalidate test results Higher layer agents inform lower level agents of the objectives for health monitoring Information Exchange Layer Network probes return statistics and alarms to higher layer agents Middle layer agents activate probes at lower layer Probe Layer
16. Concept of DEVS Standard Single DEVS C++ Simulation processor Protocol Distributed Java Simulator DEVS DEVS DEVSML Model Real - Time Simulator Simulator Interface Interface Virtual - Time Non Simulator Other DEVS Representation
17. DEVS Simulation Protocol simulators. simulators. tellAll tellAll ("initialize“) ("initialize“) Coordinator Coordinator simulators. simulators. AskAll AskAll (“ (“ nextTN nextTN ”) ”) simulators. simulators. tellAll tellAll (" (" computeInputOutput computeInputOutput “) “) simulators. simulators. tellAll tellAll (" (" sendMessages sendMessages ") ") simulators. simulators. tellAll tellAll (" (" ApplyDeltFunc”) putContentOnSimulator DEVS Simulator DEVS Simulator Non-DEVS Simulator ? DEVS Model 2 DEVS Model 1 Atoimc1 Atoimc2
18. DEVS Standard Interfaces interface coreSimulatorInterface{ void setSimulators (Collection<CoreSimulatorInterface>); void initialize(); Double nextTN(); void computeInputOutput(Double t); void applyDeltFunc(Double t); void putContentOnSimulator (CoreSimulatorInterface sim, ContentInterface c); void sendMessages(); }
19. DEVS/SOA Infrastructure: Supports Deployment and Execution of DEVS Models on the Web WEB SERVICE CLIENT DEVS DEVS Agent Agent (Observer) ( Virtual User) WEB DEVSJAVA SERVICE CLIENT DEVS Modeling Language (DEVML) DEVS Simulator Services • Service Oriented Architecture (SOA) consists of various W3C standards • Client server framework • XML Message encapsulated in SOAP wrapper • Machine-to-machine interoperable interaction over the network based on WSDL interface descriptions Middleware (SOAP, RMI etc) Net - centric infrastructure Run Example
20. Deploying Models: DEVSML and DEVS/SOA
21. Automated Negotiation Support in Multi-Agent Web Environments FD-DEVS Market Place Domain-independent behavior FD-DEVS ~ phases ~ message types Domain-dependent structure SES message specializations Receive message Interpret message storeSpec Send message Surveillance Spec
22. Analysis-Based Network Data Extraction SES for Throughput Analysis pruning SES for Network Data Network Data Collection SES for Protocol Analysis SES for Intrusion Detection Use Aspects, Specializations, … and Pragmatic Frame to develop System Entity Structure
23. Applications • Natural language capture of high level information technology systems requirements • Automated generation of FDDEVS kernel DEVSJAVA/C++ models for distributed real-time net-centric IT systems testing • Development of web service workflows using DEVS/SOA • Network Traffic data capture, focused extraction, and model generation for exercising IT systems e.g., intrusion detection
24. Conclusions • DEVS and SES provide a framework based on Systems Theory for Web-Centric M&S environments • Supports integrated development and testing • DEVS standard supports sharable models and repository reuse on the Service Oriented Architecture • Provides a basis for achieving higher levels of interoperability – can work with HLA or not: DEVS/SOA provides a SOA implementation independent of HLA • The framework supports development of generic tools which in turn support a wide array of web service domain specific specializations and applications
25. Books and Web Links acims.arizona.edu Rtsync.com devsworld.org | 2,232 | 10,514 | {"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} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.793099 |
https://www.formulas.today/formulas/GigabytesToKilobytes/ | 1,726,618,666,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00689.warc.gz | 714,203,514 | 4,028 | # Gigabytes to Kilobytes: A Comprehensive Guide
Gb:
Output: `Press calculate`
Formula:`kb = gb * 1024 * 1024`
## Understanding Gigabytes to Kilobytes Conversion
In today's digital era, understanding data sizes is crucial, from saving family photos to developing complex software. One common conversion is from gigabytes (GB) to kilobytes (KB), often needed due to varying storage or network requirements.
## The Art and Science of Data Measurement
Data is measured in bytes. Bytes are the fundamental units of digital data. We often use larger units called kilobytes, megabytes, gigabytes, etc. Here, we focus on converting gigabytes into kilobytes.
## Why Convert GB to KB?
Imagine you need to send a large file via email with attachment size limits in KB. Converting your data from GB to KB helps determine if it meets these requirements. Another scenario includes optimizing storage allocations in cloud environments.
## The Conversion Formula
To convert gigabytes to kilobytes, we use a fundamental formula:
• `kb = gb * 1024 * 1024`
In this formula:
• `gb` = number of gigabytes
• `kb` = resulting kilobytes
## Conversion Examples
Let's walk through some examples:
### Example 1
Input: `2 GB`
Using the formula:
• kb = 2 * 1024 * 1024
• Result: `2,097,152 KB`
### Example 2
Input: `0.5 GB`
Using the formula:
• kb = 0.5 * 1024 * 1024
• Result: `524,288 KB`
## Benefits of Knowing the Conversion
Understanding GB to KB conversions can improve data management:
• Efficient Storage: Accurately gauge your storage requirements and optimize space.
• Network Optimization: Make informed decisions affecting bandwidth usage and data transfers.
• Financial Planning: Plan finances better by knowing exact data requirements, especially for cloud storage.
## Frequently Asked Questions
#### How many KB are there in 1 GB?
There are `1,048,576 KB` in 1 GB.
#### Why does the conversion factor involve 1024, not 1000?
This is because computers use binary (base-2) arithmetic, and 1024 (2^10) fits perfectly into this system.
#### Is there a quick method to estimate large data sizes?
Yes, rounding 1024 to 1000 can give a rough estimation though it is not precise for accurate calculations.
## Conclusion
Converting gigabytes to kilobytes isn't just about numbers; it's critical for data management, network optimization, and efficient storage. The formula is straightforward, and real-life applications highlight its importance in everyday digital tasks. | 573 | 2,485 | {"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} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.80379 |
http://syrconventions.com/an-update-on-speedy-plans-in-fun-math-games-for-kids/ | 1,719,115,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00294.warc.gz | 26,105,860 | 23,978 | # An Update On Speedy Plans In Fun Math Games For Kids
Welcome to Youngsters Math Games Online! Reorder Numbers – Shuffleboard puzzle recreation where players place numbers in sequential numerical order on a 3×3, 4×4, or 5×5 grid. Match the colours Fun Math Games with this interactive puzzle for youths. Transfer the nine sections of stained glass to the opposite aspect and align all the colours collectively.
## A Background In Realistic Systems Of Kids Math Games
Play math games at Study whereas playing with this large assortment of maths games. There are numerous parts of mathematics Fun Math Games to study like addition, subtraction, multiplication, and division. Enhance counting skills and study to manage cash with these enjoyable math associated video games.
Learn about addition with this great exercise for teenagers. Practice your addition skills whereas constructing a giant water slide. There are a range of video games, together with 1st grade math video games in addition to games for other grades. Learn about money, math and downside solving with this interactive sport for kids.
Blast Off! is a cool Maths game which makes use of mental calculation skills and rolling dice to make completely different numbers. Grade 4: On this grade, kids are anticipated to solve Free Math Games For Kids multi-step math phrase issues involving complete numbers using the 4 operations. In addition they achieve familiarity with factors and multiples, prime and composite numbers.
Youngsters will choose up an entire vary of abilities through playing video games naturally and informally. Play this number rotation puzzle recreation and organize Maths Game For Kids numbers so as from 1 to sixteen. Online Cash Video games – Study extra about money by playing these interactive money math video games.
Learn about numbers with these enjoyable interactive video games and activities. Complete challenges related to fractions, decimals, place values, graphs, likelihood, codes , knowledge , mean, median & mode , calculators and extra. The sport jigsaws puzzles games is nice follow for quantity recognition. But you possibly can additionally use it by having kids recognize numbers in numerous ways.
## Exploring Effortless Math Games For Kids Solutions
Grade three: On this grade, youngsters make use of place worth understanding to fluently add and subtract within 1000 and round complete number Fun Math Games For Kids to nearest 10 or a hundred. They are additionally taught to multiply one-digit whole numbers by multiples of 10 in the vary 10-ninety.
Math video games can make learning more fun and interesting. When youngsters play video games, they’ll apply their math abilities to new and distinctive eventualities while constructing logic and critical Fun Math Games For Kids pondering skills. They also get an opportunity to apply math in a means that’s enjoyable and appeals to their pursuits. | 562 | 2,922 | {"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} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.893451 |
shastrarth.com | 1,716,623,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00307.warc.gz | 436,311,566 | 26,652 | ### 2023 NCERT Math Class 9 Number System Ex. 1.1 | 9th Math NCERT Solution Number System
The study of numbers forms 2023 NCERT Math Class 9 Number System, and a solid understanding of the number system is crucial for students’ mathematical development. In this comprehensive blog post, we will delve into the intriguing world of the number system as taught in Class 9 Mathematics. From the basic building blocks of natural numbers to the vast realm of real numbers, we will explore concepts such as number lines, rational numbers, irrational numbers, whole numbers, integers, and more. So, let’s embark on this mathematical journey together!
## 2023 NCERT Math Class 9 Number System : Natural Numbers:
The number system begins with the set of natural numbers. Natural numbers, denoted as N, comprise the counting numbers: 1, 2, 3, 4, and so on. These numbers represent quantities used for counting, such as the number of objects or people.
More Book related details visit NCERT Website
## 2023 NCERT Math Class 9 Number System : Whole Numbers:
Expanding upon the natural numbers, the set of whole numbers, denoted as W, includes zero (0) along with the counting numbers. Whole numbers allow us to represent the absence of objects or null quantities. In essence, whole numbers form a more comprehensive representation of quantities.
## 2023 NCERT Math Class 9 Number System : Integers:
Integers, denoted as Z, encompass both positive and negative numbers along with zero. Integers allow us to represent values in the opposite direction on a number line. For every positive number, there exists a corresponding negative number, forming a symmetrical representation.
## 2023 NCERT Math Class 9 Number System : Rational Numbers:
Rational numbers, denoted as Q, encompass numbers that can be expressed as a fraction of two integers, where the denominator is not zero. Rational numbers include both terminating and repeating decimals. They can be positive, negative, or zero. For example, 1/2, -3/5, 2/7 are all rational numbers.
## 2023 NCERT Math Class 9 Number System : Irrational Numbers:
Irrational numbers, denoted as I, are numbers that cannot be expressed as fractions and do not terminate or repeat as decimals. They possess an infinite and non-repeating decimal representation. Examples of irrational numbers include √2, π (pi), and e (Euler’s number). The decimal expansion of irrational numbers is non-terminating and non-repeating.
## 2023 NCERT Math Class 9 Number System : Real Numbers:
The set of real numbers, denoted as R, encompasses both rational and irrational numbers. It includes all possible numbers on the number line, from negative infinity to positive infinity. Real numbers consist of integers, fractions, terminating decimals, and infinite non-repeating decimals.
## 2023 NCERT Math Class 9 Number System : Number Line:
A number line is a visual representation of the entire number system. It helps us visualize the relative positions and relationships between different types of numbers. On a number line, natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers are placed at appropriate positions.
## NCERT Class 9 Math Number System Exercise 1.1 :
#### Q1. Is zero a rational number? Can you write it in the form p/q , where p and q are integers and q≠0?
A1. Yes, zero is indeed a rational number. It can be expressed in the form of p/q, where both p and q are integers, and q is not equal to zero.
To represent zero in this form, we can choose any value of p as zero and any non-zero value of q. For example, we can write 0 as 0/1 or 0/2 or 0/1000, and so on. In each case, p is zero, and q is a non-zero integer, satisfying the condition for a number to be rational.
The fact that any integer divided by non-zero integer results in a rational number, including the division of zero by a non-zero integer, reinforces the classification of zero as a rational number.
#### Q2. Find six rational numbers between 3 and 4
A2. There are infinite rational numbers between 3 and 4. (6+2=8) We need to find 6 rational numbers so we will multiply numerator and denominator with 8. This can be represented as 24/8 and 32/8(multiplying both 3/1 and 4/1 by 8/8) respectively, therefore 6 rational numbers between 3 and 4 are 25/8 ,26/8 ,27/8 ,28/8 ,29/8 ,30/8.
#### Q3. Find five rational numbers between 3/5 and 4/5
A3.There are infinite rational numbers between 3/5 and 4/5. (5+1=6) We need to find 5 rational numbers so we will multiply numerator and denominator with 6. This can be represented as 18/30 and 24/30(multiplying both 3/5 and 4/5 by 6/6)respectively,therefore 5 rational numbers between 3/5 and 4/5 are 19/30 ,20/30 ,21/30 ,22/30 ,23/30
#### State whether the following statements are true or false. Give reasons for your answers.(i) Every natural number is a whole number.(ii) Every integer is a whole number.(iii) Every rational number is a whole number.
A3.(i) True,since the collection of whole numbers contain all the natural numbers.
(ii) False,as integers maybe negative but whole numbers are always positive for example -5 is a integer but not a whole number.
(iii)False,as rational can be fractional but whole number cannot be fractional for example,2/3 is a rational number but not a whole number.
## Conclusion:
The study of the number system in Class 9 Mathematics is a foundational step towards understanding and utilizing numbers in various mathematical contexts. From the basic concept of natural numbers to the intricate properties of irrational numbers and real numbers, each component plays a vital role in our mathematical journey.
By grasping the concepts of number lines, rational numbers, irrational numbers, whole numbers, integers, and real numbers, students can develop a deeper understanding of mathematical operations, equations, and problem-solving strategies.
Embrace the beauty and intricacy of the number system, as it provides a framework for exploring and understanding the mathematical universe. So, let’s continue to explore, discover, and apply the concepts of the number system to expand our mathematical horizons!
Now fresh copy is available of 2023 NCERT Class 9 Math Book PDF Download. In today’s digital era, the accessibility of educational resources has become easier than ever before. With the advent of technology, students can now access textbooks and study materials in digital formats, such as PDFs. Here you can download 2023 NCERT Class 9 Math book in PDF format, empowering students to enhance their learning experience and excel in their academic pursuits.
## 2023 NCERT Class 9 Math Book PDF Download : Importance of NCERT Class 9 Math Book:
The NCERT (National Council of Educational Research and Training) Class 9 Math book serves as a fundamental resource for students studying mathematics. This book is designed to develop a strong foundation in mathematical concepts, foster problem-solving skills, and promote logical thinking. It covers a wide range of topics that form the basis for advanced math education.
## 2023 NCERT Class 9 Math Book PDF Download : Benefits of Using PDF Format:
The PDF format offers numerous advantages for students and educators alike. Here are a few benefits of using the PDF version of the NCERT Class 9 Math book:
a. Portability: PDF files can be easily accessed and viewed on various devices like smartphones, tablets, laptops, or e-readers. Students can carry their entire textbook collection without the burden of physical books.
b. Searchability: PDFs allow users to search for specific keywords or topics within the book, enabling quick and efficient navigation. This feature proves invaluable during revision or when trying to locate specific concepts.
c. Annotation and highlighting: PDF readers often come with built-in tools that allow users to annotate, highlight, and bookmark important sections. These features aid in note-taking and revisiting essential points during exam preparation.
## Steps to Download the 2023 NCERT Class 9 Math Book PDF:
#### Step 1: Visit the official NCERT website.
Go to the official website of NCERT (www.ncert.nic.in) using any web browser.
#### Step 2: Navigate to the “Textbooks” section.
Locate the “Textbooks” section on the website’s homepage and click on it.
#### Step 3: Select the subject and class.
In the textbooks section, select the subject as “Mathematics” and choose the class as “Class 9.”
## Conclusion:
Access to educational resources in digital formats, such as the 2023 NCERT Class 9 Math book in PDF, has revolutionized the way students learn and study. The convenience, portability, and searchability of PDFs make them an excellent tool for enhancing the learning experience. By following the steps outlined in this comprehensive guide, students can easily download the NCERT Class 9 Math book and utilize it to strengthen their mathematical foundation.
Remember, the PDF version of the textbook is just a tool. To truly benefit from it, students must actively engage with the material. By creating a study plan, taking organized notes, and practicing regularly, students can unlock the full potential of the NCERT Class 9 Math book. Additionally, exploring supplementary resources and seeking guidance from teachers or online forums can further enrich the learning process.
As we move forward in the digital age, the availability of educational resources in digital formats will continue to grow. Embracing these advancements and leveraging the benefits of PDF textbooks can significantly contribute to academic success. So, empower yourself with the 2023 NCERT Class 9 Math book PDF and embark on a journey of mathematical exploration and achievement. Happy learning!
## QNA:
Yes, shastrarth.com offer downloads of NCERT textbooks. However, we provide valuable educational content, study guides, and resources to help students excel in their academic pursuits.
#### Are there any other websites where I can download the 2023 NCERT Class 9 Math book in PDF format?
Yes, besides the official NCERT website, there are several reputable platforms that offer free downloads of NCERT textbooks in PDF format. It’s important to verify the authenticity and accuracy of the downloaded material before utilizing it for your studies.
#### Can I access the 2023 NCERT Class 9 Math book in PDF format on my mobile device?
Absolutely! Once you have downloaded the PDF file, you can easily access it on your mobile device using a PDF reader application. This allows you to study mathematics on the go, making learning more convenient and accessible.
#### Are there any additional resources available on shastrarth.com to supplement the NCERT Class 9 Math book?
Yes, shastrarth.com offers a wide range of additional resources to complement your study of the NCERT Class 9 Math book. You can find practice questions, solved examples, concept explanations, and study tips to further enhance your understanding and performance in mathematics.
#### Can I use the 2023 NCERT Class 9 Math book in PDF format for exam preparation?
Absolutely! The NCERT Class 9 Math book is a valuable resource for exam preparation. By downloading the PDF version, you can easily navigate through the book, highlight important sections, and take notes. Regular practice with exercises from the book will strengthen your problem-solving skills and help you perform well in exams. | 2,463 | 11,426 | {"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} | 4.6875 | 5 | CC-MAIN-2024-22 | longest | en | 0.854877 |
https://math.stackexchange.com/questions/1716656/financial-mathematics-increasing-annuities | 1,695,814,403,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00465.warc.gz | 414,690,380 | 35,518 | # Financial mathematics increasing annuities
I am having trouble solving this problem
Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?
My attempt:
I first want to find the deposit per month.
I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,
$D*100(Ia_{30|0.08}) = 100,000$
However, the $D$ I got was 8.12, which is clearly not right.
Can someone help?
• Many financial advice columns have described the strategy of paying $X$ amount extra with each month's mortgage payment. What they always mean by "and extra $100$" is that instead of paying $734$ each month (for example), you pay $834$. They do not mean that you pay $734$, then $834$, then $934$, etc. Mar 28, 2016 at 12:46
• I don't know how you are calculating $Ia_{30|0.08}$, but it seems that usually it represents the future value of $30$ payments of $1,2,3,\ldots,30$ at $8\%$ interest per payment period. So $8.12\times100(Ia_{30|0.08})$ would represent a payment of $812$ for the first year, $1624$ for the second year, etc., a total of $377580$, which seems in line for the cost of interest on this mortgage. But of course that payment schedule is completely unlike any interpretation of the question you're trying to answer. Mar 28, 2016 at 13:06
• It isn't an increasing annuity. The additional payment is 100 each month. Mar 28, 2016 at 14:46
At a nominal interest rate of $i^{(12)}=8\%$ compounded monthly, the effective interest rate per month is $i_m=\frac{i^{(12)}}{12}=0.67\%$. Let $L=100,000$. Without extra payment we have for $n=12\times 30=360$ months $$L=P\,a_{\overline{n}|i_m}\quad\Longrightarrow P=\frac{L}{a_{\overline{n}|i_m}}=\frac{100,000}{a_{\overline{360}|0.67\%}}=\frac{100,000}{136.28}=733.76$$ With an extra monthly payment $Q=100$ we have a periodic payment $P+Q=833.76$. Then we can find the outstanding loan balance $B_t$ $$B_{t}=L(1+i_m)^{t}-(P+Q)\,s_{\overline{t}|i_m}$$ and at $t=120$ we have $$B_{120}=\overbrace{100,000\times\underbrace{(1+0.67\%)^{120}}_{2.22}}^{221,964.02}\;-\,\overbrace{833.76\times\underbrace{\,s_{\overline{120}|0.67\%}}_{182.95}}^{152,533.92}=69,430.10$$ Without the extra payment the OLB would be $$B_{t}=L(1+i_m)^{t}-P\,s_{\overline{t}|i_m}$$ and at $t=120$ we would have $$B_{120}=\overbrace{100,000\times\underbrace{(1+0.67\%)^{120}}_{2.22}}^{221,964.02}\;-\,\overbrace{733.76\times\underbrace{\,s_{\overline{120}|0.67\%}}_{182.95}}^{134,239.32}=87,724.70$$ Using the prospective method we have $$B_t=P\,a_{\overline{n-t}|i_m}\quad\Longrightarrow\; B_{120}=733.76\,a_{\overline{240}|0.67\%}=733.76\times 119.55=87,724.70$$ that is obviously equal to the value founded with the retrospective method.
With an extra payment we will shorten the length $n$ of the loan repayment to a new $N$ $$L=(P+Q)\,a_{\overline{N}|i_m}=(P+Q)\frac{1-(1+i_m)^{-N}}{i_m}$$ and then solving for $N$ $$N=-\frac{\log\left(1-i_m\frac{L}{P+Q}\right)}{\log(1+i_m)}=241.9084704\approxeq 242$$ that is the lenght of the loan $n=360$ has been shortened to $N=240$ if we make an extra payment.
Using the retrospective method we have $$B_t=(P+Q)\,a_{\overline{N-t}|i_m}\quad\Longrightarrow B_{120}=833.76\,a_{\overline{122}|0.67\%}=833.76\times 83.27=69,430.10$$
• I was wondering, how do you find the outstanding balance using the prospective method? Mar 31, 2016 at 12:09
• I tried doing $833.76*(a_{240|0.067}+a_{120|0.067})$ but I am unable to get the answer Mar 31, 2016 at 12:10
• I've added some details Mar 31, 2016 at 13:09
HINT:
I have no formulas, but good old spreadsheet works wonders: I calculated the monthly principal to be paid: $277.7777 Then the sum of the all the interest compounded monthly over the thirty years:$120,333.3333
TOTAL principal+interest=$220,333.3333 So amount to pay per month is$612.037 during 360 months.
If you decide to instead pay $712.037 during 120 months, you will have outstanding balance at month 121 of$120,603.7037
(The decimals are there to show data is not rounded to cents).
Depending on the bank or country practices... I forgot to estimate the outstanding balance if those $100 go directly to principal: you will finish to pay in 317 months, thus reducing both the time and interest. Outstanding Principal just after month 120 (no interest)=$62,145.11041 | 1,454 | 4,518 | {"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} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.928293 |
https://www.dataunitconverter.com/bit-to-mebibyte/7 | 1,685,552,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00464.warc.gz | 815,322,021 | 12,565 | # Bit to Mebibyte - 7 Bit to MiB Conversion
Copy Link & Share
Input Bit - and press Enter
Bit
7 Bit = 0.00000083446502685546875 MiB
Calculated as → 7 / (8x10242)... - view detailed steps
Clear
## b to MiB Formula and Manual Conversion Steps
Bit and Mebibyte are units of digital information used to measure storage capacity and data transfer rate. Bit is one of the very basic digital unit where as Mebibyte is a binary unit. One Mebibyte is equal to 1024^2 bytes. There are 8,388,608 Bits in one Mebibyte.
Source Data UnitTarget Data Unit
Bit (b)
Equal to 0 or 1
(Basic Unit)
Mebibyte (MiB)
Equal to 1024^2 bytes
(Binary Unit)
Below conversion diagram will help you to visualize the Bit to Mebibyte calculation steps in a simplified manner.
÷ 8
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
Bit [b]
Byte [B]
Kibibyte [KiB]
Mebibyte [MiB]
Gibibyte [GiB]
Tebibyte [TiB]
Pebibyte [PiB]
Exbibyte [EiB]
Zebibyte [ZiB]
Yobibyte [YiB]
x 8
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
The formula of converting the Bit to Mebibyte is represented as follows :
MiB = Bit / (8x10242)
Now let us apply the above formula and see how to manually convert Bit (b) to Mebibyte (MiB). We can further simplify the formula to ease the calculation.
FORMULA
Mebibyte = Bit / (8x10242)
STEP 1
Mebibyte = Bit / (8x1024x1024)
STEP 2
Mebibyte = Bit / 8388608
STEP 3
Mebibyte = Bit x (1 / 8388608)
STEP 4
Mebibyte = Bit x 0.00000011920928955078125
If we apply the above Formula and steps, conversion from 7 Bit to MiB, will be processed as below.
1. = 7 / (8x10242)
2. = 7 / (8x1024x1024)
3. = 7 / 8388608
4. = 7 x (1 / 8388608)
5. = 7 x 0.00000011920928955078125
6. = 0.00000083446502685546875
7. i.e. 7 Bit is equal to 0.00000083446502685546875 MiB.
(Result rounded off to 40 decimal positions.)
#### Definition : Bit
A Bit (short for "binary digit") is the basic unit of information in computing and digital communications. It is a binary value, meaning it can have one of two values=> 0 or 1. Bits are used to represent data in computers and other electronic devices. They are the building blocks of digital information, and are used to store, transmit, and process data.
#### Definition : Mebibyte
A Mebibyte (MiB) is a binary unit of digital information that is equal to 1,048,576 bytes (or 8,388,608 bits) and is defined by the International Electro technical Commission(IEC). The prefix "mebi" is derived from the binary number system and it is used to distinguish it from the decimal-based "megabyte" (MB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
### Excel Formula to convert from Bit to MiB
Apply the formula as shown below to convert from 7 Bit to Mebibyte.
ABC
1Bit (b)Mebibyte (MiB)
27=A2 * 0.00000011920928955078125
3
Download - Excel Template for Bit to Mebibyte Conversion
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
### Python Code for Bit to MiB Conversion
You can use below code to convert any value in Bit to Mebibyte in Python.
bit = int(input("Enter Bit: "))
mebibyte = bit / (8*1024*1024)
print("{} Bit = {} Mebibyte".format(bit,mebibyte))
The first line of code will prompt the user to enter the Bit as an input. The value of Mebibyte is calculated on the next line, and the code in third line will display the result. | 1,069 | 3,448 | {"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} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.758952 |
bhargavs271100.medium.com | 1,660,142,087,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00467.warc.gz | 140,808,270 | 36,659 | # Why Structuring is necessary?
With the use of Python and its vast in-built libraries it has become highly simple to implement Deep Neural Networks even for those who have no idea about what Neural Networks actually do in the back-end of the code. This becomes a problem when the model has to be tuned to improve accuracy. Most students and professionals aiming to train a model don’t follow a structured approach and land up in a situation wherein they are not able to find out what parameters to tune and where their model has to be improved.
This leads to the necessity to follow a structured approach. In this post ill explain all the key concepts involved in a structured approach towards a ML project.
# Concepts Covered:
1) Orthogonalization
2) Evaluating a model
3) Creating and Dividing your data set
5) Error Analysis
# What is Orthogonalization?
Orthogonalization or Orthogonal approach to a project means to find out and tune parameters in such a way that we tackle only the issue at stake and do not lead to further issues i.e. we tune the parameters in such a way that they do not affect other parameters.
The example of a radio and tuning its knobs for a particular frequency is a real-world example for orthogonal approach. This is the standard example to explain orthogonal approach.
Consider a conventional radio which has knobs in order to tune frequency. A radio works on the principle of resonance i.e. when we tune the frequency in such a way that it matches the resonant frequency of the transmission signal then we can receive signals from the transmission center efficiently. Now there are two features of a signal that are allowed to be tuned i.e. amplitude and frequency. Tuning the frequency improves the quality of reception whereas tuning the amplitude improves the sound output strength. Both these parameters are independent of each other when tuned in the working frequency and hence, don’t affect each other.
Orthogonalization is an approach followed in mathematics and not native to Deep Neural Networks. Anyone with a basic knowledge of mathematics knows that all the co-ordinate systems used have orthogonal axes i.e. the axes are mutually perpendicular to each other and the variables along those axes are independent variables. This concept of orthogonal co-ordinates is the key intuition behind orthogonalization in ML.
The figure below shows a set of rectangular co-ordinates with there mutually perpendicular co-ordinates x, y and z.
With enough intuition gained about what is Orthogonalization and why it should used, let’s move on to using orthogonalization in ML problems.
# How to evaluate a model?
Consider two classification algorithms one based on Convolutional Neural Network and other based on simple deep neural network. Lets call first one classifier A and second one classifier B. How do we evaluate their performance or how do we know which model performs better? This leads us to the need for a single number evaluation metric.
At the core of any ML problem we need to perform 3 tasks:
What an evaluation metric does is it lets us help to decide which model works better and on which model we need to perform the above 3 tasks again.
There r two metrics: Precision and Recall:
Precision tells us how precise the model is in classifying the intended object whereas Recall tells how efficient the model is in classifying any object.
This data tells us that classifier A is more precise with its classification but isn’t efficient whereas classifier B is more efficient but not precise. Hence, both the models have to be tuned on different grounds.
If a model satisfies both the metrics then it’s an optimized model.
# How to create an efficient data set?
Any Deep Learning model requires lot of data to perform efficiently. This costs a lot to accumulate. But there are few strategies with which we can make the available data more efficient.
· Data Augmentation: This is the method of augmenting the data so as to diversify the data set in hand. Mirroring the image, random cropping, changing BGR to gray and vice versa, changing the color scheme are some of the efficient augmenting ideas. But one must carefully analyze the data set and then perform augmentation else this may spoil the accuracy even more. Cropping a data which has disoriented images can spoil accuracy. Mirroring is a safe augmenting strategy which I recommend out of my experience.
· Synthesizing Data: This is a less suggested but efficient idea. There are various graphic tools online in order to create images we need. These images can be added to the data set after performing augmentation.
· Randomization: Randomizing the data set before dividing the data into train, dev and test sets can greatly improve accuracy because without randomizing there are chances that the sets have different distributions and the model’s accuracy might drop.
# How to divide your data set?
How do we divide our data set at hand into training, development and test sets. This is a key issue in ML projects and model’s performance greatly depends on the distribution of train, dev and test sets.
Suppose we have 10,000 images in our data set then the best strategy is to use 80–90 % (8000–9000) of the data for training, 5–8% (500- 800) for dev set and 2–5 % (200–500) for test set.
The most important factor determining accuracy is the distribution of these sets.
· The training and dev sets must be from the same distribution else the model’s accuracy might drop drastically.
· The training set must be made from all available distributions of data and should be as diverse as possible.
· The dev and test sets should be smaller and more data should be used for training. The test set should contain worst case scenarios in order to check the model completely.
# How to grade and improve the performance?
The difference between Human Level Performance and Training Error is called BIAS and the difference between Training error and Dev error is called variance.
# How to perform error analysis and when?
This is the most brain-storming problem to solve in an ML project. Most people do not follow a structured approach and end up mis-tuning their model further spoiling their accuracy. Here ill explain the most common problems faced during development of a model and what parameters to tune when they arise.
--
--
## More from Bhargav S
Robotics and Automation Lover, Machine Learning Enthusiast.
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## Bhargav S
Robotics and Automation Lover, Machine Learning Enthusiast. | 1,317 | 6,594 | {"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} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.929401 |
https://oeis.org/A318498 | 1,632,742,564,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00465.warc.gz | 473,147,010 | 4,191 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A318498 Denominators of the sequence whose Dirichlet convolution with itself yields A061389, number of (1+phi)-divisors of n. 4
1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 8, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 8, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 16, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 8, 8, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 8, 1, 2, 2, 4, 1, 1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS The sequence seems to give the denominators of a few other similarly constructed rational valued sequences obtained as "Dirichlet Square Roots" (of possibly A092520 and A293443). LINKS Antti Karttunen, Table of n, a(n) for n = 1..16384 FORMULA a(n) = denominator of f(n), where f(1) = 1, f(n) = (1/2) * (A061389(n) - Sum_{d|n, d>1, d 1. a(n) = 2^A318499(n). PROG (PARI) up_to = 65537; A061389(n) = factorback(apply(e -> (1+eulerphi(e)), factor(n)[, 2])); DirSqrt(v) = {my(n=#v, u=vector(n)); u[1]=1; for(n=2, n, u[n]=(v[n]/v[1] - sumdiv(n, d, if(d>1&&d
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Last modified September 27 07:28 EDT 2021. Contains 347673 sequences. (Running on oeis4.) | 718 | 1,597 | {"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} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.635043 |
https://stat-methods.com/home/independent-samples-t-test-sas-2/ | 1,685,468,873,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00243.warc.gz | 595,978,294 | 40,588 | # Independent Samples T-test in SAS
## Introduction
An independent samples t-test is typically performed when an analyst would like to test for mean differences between two treatments or conditions. For example, you may want to see if first-year students scored differently than second-year students on an exam.
An independent samples t-test is typically used when each experimental unit, (study subject) is only assigned one of the two available treatment conditions. Thus, the treatment groups do not have overlapping membership and are considered independent. An independent samples t-test is the simplest form a “between-subjects” analysis.
The two-sided null hypothesis is that there is no difference between treatment group means, while the alternative hypothesis is that mean values differ between treatment groups.
H0: μ1 = μ2
Ha: μ1μ2
## Independent Samples T-test Assumptions
The following assumptions must be met in order to run an independent samples t-test:
1. The response of interest is continuous and normally distributed for each treatment group.
2. Treatment groups are independent of one another. Experimental units only receive one treatment and they do not overlap.
3. There are no major outliers.
4. A check for unequal variances will help determine which version of an independent samples t-test is most appropriate:
1. If variances are equal, then a pooled t-test is appropriate
2. If variances are unequal, then a Satterthwaite (also known as Welch’s) t-test is appropriate
## Independent Samples T-test Example
In this example, we will test to see if there is a statistically significant difference in the miles per gallon (mpg) of 4-cylinder automobiles and 8-cylinder automobiles.
Dependent response variable:
mpg = Miles per gallon
Independent categorical variable:
cyl = 4 or 8 cylinder automobiles
The data for this example is available here:
## Independent Samples T-test SAS Code
PROC TTEST includes QQ plots for each treatment group along with a folded F-test to help identify unequal variances. While this information can aid in validating assumptions, the Shapiro-Wilk Normality Test, in addition to QQ plots, should be used to help evaluate normality. Furthermore, Levene’s Test for Equality of Variances is generally preferred when evaluating whether variances between groups are equal. Thus, SAS code has been provided to demonstrate both the Shapiro-Wilk and Levene’s tests.
Here is the annotated code for the example. All assumption checks are provided along with the independent samples t-test:
```*Import the data;
proc import datafile='C:DropboxWebsiteAnalysisIndependent Samples T-testDatacars_ttest.csv'
out=work.cars
dbms=csv
replace;
run;
*Produce descriptive statistics;
proc means data=insect nmiss mean std stderr lclm uclm median min max qrange maxdec=2;
class spray;
var bugs;
run;
*Test for normality;
proc univariate data=cars normal;
class cyl;
var mpg;
run;
*Test for equality of variances;
proc glm data=cars;
class cyl;
model mpg = cyl;
means cyl / hovtest=levene(type=abs) welch;
run;
*Independent Samples T-test;
proc ttest data=cars;
class cyl;
var mpg;
run;
```
## Independent Samples T-test Annotated SAS Output
### Descriptive Statistics
Many times, analysts forget to take a good look at their data before performing statistical tests. Descriptive statistics are not only used to describe the data but also help determine if any inconsistencies are present. Detailed investigation of descriptive statistics can help answer the following questions (in addition to many others):
• How much missing data do I have?
• Do I have potential outliers?
• Are my standard deviation and standard error values large relative to the mean?
• In what range most of my data fall for each treatment?
1. cyl – Each treatment level of our independent variable.
2. N obs – The number of observations for each treatment.
3. N Miss – The number of missing observations for each treatment.
4. Mean – The mean value for each treatment.
5. Std Dev – The standard deviation of each treatment.
6. Std Error – The standard error of each treatment. That is the standard deviation / sqrt (n).
7. Lower and Upper 95% CL for Mean – The upper and lower confidence intervals of the mean. That is to say, you can be 95% certain that the true mean falls between the lower and upper values specified for each treatment group.
8. Median – The median value for each treatment.
9. Minimum, Maximum – The minimum and maximum value for each treatment.
10. Quartile Range – The inner quartile range of each treatment. That is the 75th percentile – 25th percentile.
### Normality Tests
Prior to performing the t-test, it is important to validate our assumptions to ensure that we are performing an appropriate and reliable comparison. Testing normality should be performed using a Shapiro-Wilk normality test (or equivalent), and/or a QQ plot for large sample sizes. Many times, histograms can also be helpful. In this example, we will use PROC UNIVARIATE to produce our Shapiro-Wilk normality test for each cylinder group, and PROC TTEST will produce our corresponding QQ plots.
The Shapiro-Wilk normality test for the 4-cylinder group:
The Shapiro-Wilk normality test for the 8-cylinder group:
1. Test – Four different normality tests are presented.
2. Statistic – The test statistics for each test is provided here.
3. p Value – The p-value for each test is provided. A p-value < 0.05 would indicate that we should reject the assumption of normality. Since the Shapiro-Wilk Test p-values are > 0.05 for each group, we conclude the data is normally distributed.
### QQ Plots
The ‘4’ and ‘8’ in the top left corner of each plot indicates which group each QQ plot corresponds too. The vast majority of points should follow each line.
Since the Shapiro-Wilk Test p-value is > 0.05, and the QQ Plot for each treatment group follows the QQ plot theoretical normal diagonal line, we conclude the data is normally distributed.
### Levene’s Test for Homogeneity of Variances
PROC TTEST does not perform Levene’s Test for Equality of Variances automatically. In its place, PROC TTEST will perform a Folded F test which is considered inferior to Levene’s Test for Equality of Variances. However, the GLM procedure can be used to perform this test. I recommend using Levene’s Test for Equality of Variances rather than the Folded F test included by default in PROC TTEST.
1. Source – This column designates which variable Levene’s Test for Homogeneity is performed.
2. DF – The degrees of freedom associated with each variable and overall error.
3. Sum of Squares – The sums of squares calculation for Levene’s Test
4. Mean Square – The mean square calculation for Levene’s Test.
5. Value – The F statistic for which the p-value is computed. This value is (24.3781/3.6742) = 6.63
6. Pr > F – Levene’s Test for Equality of Variances shows a p-value of 0.0172. A significant p-value (P < 0.05) indicates that a Satterthwaite (also known as Welch’s) t-test results should be used instead of pooled t-test results.
### Boxplots to Visually Check for Outliers
Side-by-side boxplots are provided by the GLM procedure. This can help visually identify major outliers and help visually show if variances might be unequal. The boxplot below seems to indicate one minor outlier but subjectively, not enough evidence to suggest we move to a different analysis method.
### Independent Samples T-test
So far, we have determined that the data for each cylinder group is normally distributed, variances are unequal, and we do not have major influential outliers. Our next step is to officially perform an independent samples t-test to determine whether 4 and 8 cylinder cars show significant differences between their average mpg expenditure.
1. cyl – This column identifies the levels of the treatment variable of interested along with the mean differences between the levels.
2. Method – This column displays information corresponds to formulas for the Pooled or Satterthwaite t-test formula. If equal variances are assumed, the row representing the Pooled difference is appropriate. However, if variances are considered unequal, the row of data representing the Satterthwaite is appropriate. Levene’s Test for Equality of Variances, or the inherent Folded F test will dictate which row is most appropriate.
3. N – This column identifies how many data points (cars) are in each cylinder group.
4. Mean – The first two rows in this column correspond to mean values for each treatment group. The second two rows correspond to the mean difference between the 4 and 8 cylinder groups (27.05-15.10 = 11.95) .
5. Std Dev – The first two rows in this column identify the standard deviation of each treatment group. The third row is an estimate of the the pooled standard deviation across both treatment groups.
6. Std Err – The first two rows in this column are the standard errors of each treatment group respectively. The second two rows represent the Pooled (assuming equal variances) and Satterthwaite (assuming unequal variances) standard error estimates respectively.
7. Min, Max – The minimum and maximum values observed for each treatment group.
8. 95% CL Mean – The first two rows in this column represent the upper and lower confidence intervals of the mean. That is to say, you can be 95% certain that the true mean mpg of the 4 cylinder group falls between 23.79 and 30.31. Furthermore, you can be 95% certain that the true mean mpg of the 8 cylinder group falls between 13.62 and 16.58. The third and fourth rows indicate the confidence interval of the mean difference between treatment groups.
9. 95% CL Std Dev – The first two rows indicate that the confidence interval of the standard deviation for each treatment group. The third row indicates the 95% confidence interval of the pooled standard deviation across both treatment groups.
### Independent Samples T-test Results in SAS
1. Method – This column designates each type of independent samples t-test.
2. Variance -An independent samples pooled t-test is appropriate when variances are assumed equal. An independent samples Satterthwaite t-test is appropriate when variances are considered unequal. The appropriate row to evaluate will be based on the results of the Levene’s Test for Homogeneity of Variances above. In our example, the “Satterthwaite” test is appropriate since variances are considered unequal between the 4 and 8 cylinder treatment groups.
3. DF – The appropriate degrees of freedom vary between each type of independent samples t-test.
4. t Value – This is the t-statistic. It is the ratio of the mean of the difference in means to the standard error of the difference. For the Satterthwaite t-test this value is (11.95/1.5955) = 7.49
5. Pr > |t| – This is the p-value associated with the test. That is to say if the P value < 0.05 (assuming alpha=0.05) then treatments have a statistically significant mean difference. For our example, we have a p-value < 0.0001. Thus, we reject the null hypothesis that the mean mpg of the 4 and 8 cylinder groups are equal and conclude that there is a mean difference between groups.
## Independent Samples T-test Interpretation and Conclusions
We have concluded that the Satterthwaite version of the independent samples t-test is appropriate since our variances are considered unequal between the 4 and 8 cylinder treatment groups. A p-value < 0.0001 indicates that we should reject the null hypothesis that the mean mpg is equal across the 4 and 8 cylinder treatment groups and conclude that there is a mean difference. We know that the average difference between the 4 and 8 cylinder groups for this sample is 11.95 mpg. That is to say, the 4 cylinder group gets, on average, 11.95 more miles per gallon than the 8 cylinder group. We are 95% certain that the mean difference between the 4 and 8 cylinder groups across the population will be between 8.5047 and 14.9702.
## What to do When Assumptions are Broken or Things Go Wrong
The lack of normality or outliers can violate independent sample t-test assumptions and ultimately the results. If this happens, there are several available options:
Perform a nonparametric Mann-Whitney U test is the most popular alternative. This is also known as the Mann-Whitney-Wilcoxon or the Wilcoxon Rank Sum test. This test is considered robust to violations of normality and outliers (among others) and tests for differences in mean ranks. This is the most well-known alternative.
Additional options include considering permutation/randomization tests, bootstrap confidence intervals, and transforming the data but each option will have its own stipulations.
If you need to compare more than two independent groups, a one-way Analysis of Variances (ANOVA) or Kruskal-Wallis may be appropriate.
An independent samples t-test is not appropriate if you have repeated measurements taken on the same experimental unit (subject). For example, if you have a pre-test post-test study, then each subject was measured at two different time intervals. If this is the case, then a paired t-test may be a more appropriate course of action. | 2,913 | 13,175 | {"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} | 3.1875 | 3 | CC-MAIN-2023-23 | longest | en | 0.916047 |
https://rdrr.io/cran/T4transport/man/wasserstein.html | 1,722,692,222,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00048.warc.gz | 391,792,880 | 8,146 | # wasserstein: Wasserstein Distance between Empirical Measures In T4transport: Tools for Computational Optimal Transport
wasserstein R Documentation
## Wasserstein Distance between Empirical Measures
### Description
Given two empirical measures \mu, \nu consisting of M and N observations on \mathcal{X}, p-Wasserstein distance for p\geq 1 between two empirical measures is defined as
\mathcal{W}_p (\mu, \nu) = \left( \inf_{\gamma \in \Gamma(\mu, \nu)} \int_{\mathcal{X}\times \mathcal{X}} d(x,y)^p d \gamma(x,y) \right)^{1/p}
where \Gamma(\mu, \nu) denotes the collection of all measures/couplings on \mathcal{X}\times \mathcal{X} whose marginals are \mu and \nu on the first and second factors, respectively. Please see the section for detailed description on the usage of the function.
### Usage
wasserstein(X, Y, p = 2, wx = NULL, wy = NULL)
wassersteinD(D, p = 2, wx = NULL, wy = NULL)
### Arguments
X an (M\times P) matrix of row observations. Y an (N\times P) matrix of row observations. p an exponent for the order of the distance (default: 2). wx a length-M marginal density that sums to 1. If NULL (default), uniform weight is set. wy a length-N marginal density that sums to 1. If NULL (default), uniform weight is set. D an (M\times N) distance matrix d(x_m, y_n) between two sets of observations.
### Value
a named list containing
distance
\mathcal{W}_p distance value.
plan
an (M\times N) nonnegative matrix for the optimal transport plan.
### Using wasserstein() function
We assume empirical measures are defined on the Euclidean space \mathcal{X}=\mathbf{R}^d,
\mu = \sum_{m=1}^M \mu_m \delta_{X_m}\quad\textrm{and}\quad \nu = \sum_{n=1}^N \nu_n \delta_{Y_n}
and the distance metric used here is standard Euclidean norm d(x,y) = \|x-y\|. Here, the marginals (\mu_1,\mu_2,\ldots,\mu_M) and (\nu_1,\nu_2,\ldots,\nu_N) correspond to wx and wy, respectively.
### Using wassersteinD() function
If other distance measures or underlying spaces are one's interests, we have an option for users to provide a distance matrix D rather than vectors, where
D := D_{M\times N} = d(X_m, Y_n)
for flexible modeling.
### References
\insertRef
peyre_computational_2019T4transport
### Examples
#-------------------------------------------------------------------
# Wasserstein Distance between Samples from Two Bivariate Normal
#
# * class 1 : samples from Gaussian with mean=(-1, -1)
# * class 2 : samples from Gaussian with mean=(+1, +1)
#-------------------------------------------------------------------
## SMALL EXAMPLE
m = 20
n = 10
X = matrix(rnorm(m*2, mean=-1),ncol=2) # m obs. for X
Y = matrix(rnorm(n*2, mean=+1),ncol=2) # n obs. for Y
## COMPUTE WITH DIFFERENT ORDERS
out1 = wasserstein(X, Y, p=1)
out2 = wasserstein(X, Y, p=2)
out5 = wasserstein(X, Y, p=5)
## VISUALIZE : SHOW THE PLAN AND DISTANCE
pm1 = paste0("plan p=1; distance=",round(out1$distance,2)) pm2 = paste0("plan p=2; distance=",round(out2$distance,2))
pm5 = paste0("plan p=5; distance=",round(out5$distance,2)) opar <- par(no.readonly=TRUE) par(mfrow=c(1,3)) image(out1$plan, axes=FALSE, main=pm1)
image(out2$plan, axes=FALSE, main=pm2) image(out5$plan, axes=FALSE, main=pm5)
par(opar)
## Not run:
## COMPARE WITH ANALYTIC RESULTS
# For two Gaussians with same covariance, their
# 2-Wasserstein distance is known so let's compare !
niter = 1000 # number of iterations
vdist = rep(0,niter)
for (i in 1:niter){
mm = sample(30:50, 1)
nn = sample(30:50, 1)
X = matrix(rnorm(mm*2, mean=-1),ncol=2)
Y = matrix(rnorm(nn*2, mean=+1),ncol=2)
vdist[i] = wasserstein(X, Y, p=2)\$distance
if (i%%10 == 0){
print(paste0("iteration ",i,"/", niter," complete."))
}
}
# Visualize | 1,129 | 3,692 | {"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} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.7055 |
https://oeis.org/A258006 | 1,720,798,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00453.warc.gz | 312,127,592 | 3,690 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A258006 First differences of A258202. 2
22, 333, 22, 333, 22, 223, 110, 22, 333, 22, 223, 110, 22, 333, 22, 333, 22, 333, 22, 333, 22, 223, 110, 22, 333, 22, 333, 22, 333, 22, 333, 22, 223, 110, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 223, 110, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 333, 22, 223, 110 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA a(n) = A258202(n+1) - A258202(n). PROG (Scheme) (define (A258006 n) (- (A258202 (+ 1 n)) (A258202 n))) CROSSREFS Cf. A258202, A258200, A258007. Sequence in context: A021834 A019671 A021614 * A309654 A019854 A025936 Adjacent sequences: A258003 A258004 A258005 * A258007 A258008 A258009 KEYWORD nonn AUTHOR Antti Karttunen, Jun 05 2015 STATUS approved
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Last modified July 12 11:28 EDT 2024. Contains 374245 sequences. (Running on oeis4.) | 532 | 1,340 | {"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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.712004 |
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