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http://www.chess.com/blog/Jamalov/the-value-of-opening-lines | 1,444,724,059,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737994631.83/warc/CC-MAIN-20151001221954-00176-ip-10-137-6-227.ec2.internal.warc.gz | 478,451,366 | 16,756 | # The value of opening lines
Do some opening lines favor white and others black? Or are they optimized to maintain a neutral position? To answer this question I selected 190 commonly played opening lines (out of thousands) from the "opening theory" book available online at wikibooks and played each one of them on engines in analysis mode and recorded the value of white's position at the end of each line. The data are shown in a pdf file on freepdfhosting.com that can be viewed or downloaded using the link shown below:
Opening data sorted by move sequence: http://freepdfhosting.com/634a84afbc.pdf
The positional values are computed on a scale in which the material value of a pawn is set to 100 points. At the start of the game, white has a positional advantage of 20 points by virtue of holding the first move. Since white's win percentage is higher than that of black's ceteris paribus, we can assume that a 20 point advantage is meaningful in the game. For the sake of the rest of the analysis we shall assume that an advantage of less than 20 points is not statistically important in terms of game outcome. In particular, we should also note that position value estimates have a variability that indicate a measurement error of plus or minus 5 points.
Accordingly, in the second data sheet I have color coded the data to separate the opening lines in my sample into three distinct groups.
Opening data sorted by position value: http://freepdfhosting.com/b1c3a63855.pdf
In the middle group, in black letters, we have about 130 lines representing 66% of the sample with white's position value in a range range from 0 to 40. Since white started with a positional advantage of 20 and ended with a value somewhere between 0 and 40, white's positional advantage did not change by more than 20 points by virtue of the opening moves. We therefore classify these openings as neutral in the sense that the opening moves did not substantially affect positional advantage.
At the top of the list are about 30 opening lines color coded in red in which the opening moves caused white's positional advantage to decrease by more than 20 points. These lines favor black. If the line is a "defense" then we can consider the defense to be an effective one. If it is an "attack" or a "gambit" by white then we can consider the attack to be ineffective. In any event, black's response has been stronger than white's attack and white has completely lost it's initial first move advantage.
The bottom group, colored in green, contains about 30 opening lines which favor white. In these lines white's positional advantage has increased by more than 20 points by virtue of the opening moves. We can therefore consider white's "attacks" in these lines to be effecive or black's "defense" to be relatively ineffective.
In summary the answer to my question seems to be that mostly (in 2/3 of my sample) opening lines played faithfully do not favor white or black but about 1/3 of the lines favor one side or the other and these are symmetrically distributed. Half of them favor white and the other half favor black.
The first pdf file in which the data are arranged by move sequence also serves as a compact ready reference to 190 commonly played opening lines.
Cha-am Jamal, Thailand
• 7 months ago
• 2 years ago
• 2 years ago
[COMMENT DELETED]
• 2 years ago
Caveat emptor: chess engines are weakest at opening evaluations.
When you did this, did you turn the engines' opening books off, or leave them on? In either case, what effect do you think that had on the results?
It used to be the case that their weakness was endgames, but once engines became strong enough to increase search depth to beyond a dozen moves in a reasonable amount of time, this was no longer the case, and now with tablebases, most engines will outperform any human almost all of the time.
But until search depth can be around 30 moves deep in a reasonable time, engines will remain remarkably poor at evaluating opening positions for many reasons, including problems caused by search horizon, faulty quiescence optimization, aggressiveness of alpha-beta pruning, etc. As evidence, note that to have any chance against top players, the very best engines need very thoroughly developed, specialized openings books.
How in the world did the monster obtain a -5 score for 1.e4 e5 2.Nf3 Nc6 3.d4 exd4 4.Bc4 Nf6 5.e5 d5 6.Bb5 Ne4 7.Nxd4 Bd7, but obtained a 0 score for 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.d4 exd4 5.e5 d5 6.Bb5 Ne4 7.Nxd4 Bc5, when the best continuation after each of these lines reaches exactly the same position, and that position is widely regarded to be very comfortable for White, to the extent that Black is thought to have to fight hard for a draw? Does the computer diverge in each line? I find that unlikely; one line, yes, both, hard to believe. That could explain the different scores, but it doesn't explain why both are so low when most players would kill to achieve a position like White's in every game.
You are spot on in your comments regarding the 20 point lead, and in your treating +/- less than 20 as a more or less irrelevant difference, and this is indeed a good amount of work, but I wouldn't let an engine decide the value of any opening less than 12 to 15 moves in. Even then, I treat engine evaluation with suspicion, because I know a lot about how their evaluation functions work - or don't work, to be more precise.
I'd love to see your step by step methodology. Which engines, on what machines, with what specs, analysis mode set for how long, with what other programs running. You mentioned engines, plural. Did you run multiple engines on each line, or one engine per line, or was it the same engine for all lines, or some lines one and some another? Did you ever average results, and if so, how (mean, mode, median)? If you have that information, and enjoy writing such things up, I and I am sure others would be very interested in reading it. Others of us could duplicate your work under the same or similar conditions, or vastly different ones, but using the same methodology, or vice versa, and see what the results are.
Thanks for the thought-provoking contribution!
• 3 years ago
[COMMENT DELETED]
• 3 years ago
the list of openings that is sorted by move sequence is like a bird's eye view of the opening book with the added bonus of having white's positional value at the end of the line shown and also the eco code and opening name in case you want to do some research on other variations.
• 3 years ago
dr frank, in online games we are allowed to use opening books and databases during the game and so i think it would be helpful to have this list in front of you. in the file where the openings are sorted by move sequence you can follow the opening as it unfolds on this list to see what it is homing in on and to make sure that you don't cooperate in the construction of a move sequence that changes the value against you. for example if you are playing black and the ruy lopez is unfolding you may want to avoid c63 schliemann by not playing 3...f5 and instead playing something like a6, d6, or Bc5 or any of the other ruy lopez lines on the list that do not favor white in terms of white's positional value. on the other hand, if you are playing white and black does play 3...f5, you will know exactly how to proceed to take advantage of that move to increase your positional value. and so on.
• 3 years ago
This data sheet has a practical use, in that one can memorize the best lines for white and black and use those lines when playing chess. Please let me know if I am right or wrong!
• 3 years ago
yes, drfrank, the data show that the c57 two knights/classical variation line shown in the datasheet is the best line for white in this sample and that the b02 alekhine's defense line shown in the datasheet is the best line for black in this sample. please note that i am being careful to include the phrase "line shown in the datasheet" as there are other lines that might fit the eco code and description. however, that is not the point of this post. the point i was trying to make is that there are a significant number of opening lines - among the most commonly played lines - that favor one side or the other.
• 3 years ago
I'm not sure I completely understand this. Forgive me if I'm not too bright but are you saying that those lines in green are best for white and that white's very best line is the Two Knights/Classical? Is this what the chart shows?
• 3 years ago
Hi Jamalov,
Great work! I would like to make these available for download on our website Chess Mastery with your permission.
Of course, you will be given full credit, a link to this article or any other blog or website you have.
Thanks again.
• 3 years ago
i keep a printout of the first pdf file (sorted by move sequence) next to the computer when playing online games and that makes it hard to surprise me with exotic openings and also easy for me to spot when my opponent has made an openng error. if you don't have enough of these lines memorized you might want to download that file and print it out.
• 3 years ago
josechu, i know some guys who have a lot of this stuff memorized but i don't know how they do it.
• 3 years ago
Wow, Jamal! There ought to be a Nobel prize for something like that. Imagine being a professional chess player and having to memorise all that stuff!
• 3 years ago
thank you dan, it was a labor of love but i always wanted the these openings displayed in a compact manner for quick reference that that is how this thing got started.
• 3 years ago
That was a very interesting analysis! It must have taken you quite a while. | 2,215 | 9,689 | {"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} | 3.75 | 4 | CC-MAIN-2015-40 | longest | en | 0.93574 |
https://physics.stackexchange.com/questions/284004/if-the-big-bang-started-in-a-state-of-low-entropy-then-shouldnt-the-big-crunch | 1,721,096,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00280.warc.gz | 390,240,075 | 40,984 | If the Big Bang started in a state of low entropy then shouldn't the Big Crunch be in a state of low entropy also?
What our current level of thermodynamics does tell us though is that if there is such a thing as a Big Crunch, the universe will look a lot different as it heads towards collapse than it did when expanding – the universe started in a state of low entropy and high order and will end in a state of high entropy and low order.
But shouldn't the order be in a high state and hence entropy becomes a low state in the Big Crunch (assuming a closed universe)?
The reason I write this is because the gravitational forces in a closed universe are so strong they will cause gravitational collapse so the universe will shrink and go backwards, eventually through all the stages of the big bang and become a singularity once more.
It was my understanding that a singularity has a state of high order and low entropy. Have I got this the wrong way round?
If so; Can someone please explain why?
• Right at a singularity the entropy is not well-defined, I think. But with a minor change I think this is a great question. I would probably phrase it in a way like this: In models of the universe that have a Big Crunch without a reversal of the thermodynamic arrow of time, how exactly is the low entropy state of the universe just after the Big Bang different from the high-entropy state just before the Crunch? I have no idea what the answer to that question is, but someone here should... Commented Oct 4, 2016 at 1:48
• @Rococo Understood, thanks for your reply. I will leave this question phrased like this, but keep that comment you made as it may make more sense to others when phrased the way you put it. In other words, better to have it asked in two different ways as it may be interpreted by others in certain forms. Thanks for your time. Commented Oct 4, 2016 at 1:59
• @Rococo. Roger Penrose certainly has some interesting ideas on explaining the low entropy problem, if only I could follow them : )
– user108787
Commented Oct 4, 2016 at 2:01
• If it's a cyclical universe it would have to return to low entropy somewhere for each cycle. Commented Oct 4, 2016 at 2:27
• Penrose regards an increase in the tightness of Weyl curvature (the curvature that was a factor in stars on opposite sides of the sun, during that photographing of them by Eddington during 1919's solar eclipse which confirmed GR, appearing farther from each other than they do at night) as corresponding to an increase in the radius of the universe and a consequent decrease in entropy, which I guess is averaged over its volume. In his cyclic cosmology, all matter must evaporate, which isn't jiving with the observed absence of proton decay, but he could still be right about curvature re entropy. Commented Oct 1, 2018 at 19:14 | 644 | 2,818 | {"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} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.959279 |
https://ch.mathworks.com/matlabcentral/cody/problems/44672-draw-c/solutions/2019450 | 1,580,319,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251801423.98/warc/CC-MAIN-20200129164403-20200129193403-00442.warc.gz | 359,729,579 | 15,476 | Cody
# Problem 44672. Draw 'C'.
Solution 2019450
Submitted on 14 Nov 2019 by Son Pham
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 4; y_correct = ... [0 1 1 1 1 0 0 0 1 0 0 0 0 1 1 1]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 3; y_correct = ... [0 1 1 1 0 0 0 1 1]; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 5; y_correct = ... [0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 1 1]; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = 6; y_correct = ... [0 1 1 1 1 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1]; assert(isequal(your_fcn_name(x),y_correct)) | 351 | 738 | {"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-2020-05 | latest | en | 0.653731 |
https://gmatclub.com/forum/while-the-owner-of-a-condominium-apartment-has-free-and-79838.html | 1,490,222,197,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00512-ip-10-233-31-227.ec2.internal.warc.gz | 810,986,104 | 60,989 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# While the owner of a condominium apartment has free and
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While the owner of a condominium apartment has free and [#permalink]
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18 Jun 2009, 12:03
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While the owner of a condominium apartment has free and clear title to the dwelling, owners of cooperative apartments have shares in a corporation that owns a building and leases apartments to them.
(A) While the owner of a condominium apartment has free and clear title to the dwelling,
(B) The owner of a condominium apartment has free and clear title to the dwelling, but
(C) Whereas owners of condominium apartments have free and clear title to their dwellings,
(D) An owner of a condominium apartment has free and clear title to the dwelling, whereas
(E) Condominium apartment owners have a title to their dwelling that is free and clear, while
What the hell is condominium apartment anyway....
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18 Jun 2009, 13:45
urchin wrote:
While the owner of a condominium apartment has free and clear title to the dwelling, owners of cooperative apartments have shares in a corporation that owns a building and leases apartments to them.
(A) While the owner of a condominium apartment has free and clear title to the dwelling,
(B) The owner of a condominium apartment has free and clear title to the dwelling, but
(C) Whereas owners of condominium apartments have free and clear title to their dwellings,
(D) An owner of a condominium apartment has free and clear title to the dwelling, whereas
(E) Condominium apartment owners have a title to their dwelling that is free and clear, while
What the hell is condominium apartment anyway....
Big take away, Compare plurals with plurals and singulars with singular.
A B D violate that
C & E does that
C is parallel and clear
E says dwelling that is free and clear? Nope incorrect
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19 Jun 2009, 06:59
"C" sounds obviously better, but doesn't "dwellings" and "dwelling" make difference? Well, I looked up "dwelling" means a residence, so it must be " their dwellings" and for sure "C" is correct
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Las cualidades del agua...porque el agua no olvida que su destino es el mar, y que tarde o temprano deberá llegar a él.
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19 Jun 2009, 07:03
urchin wrote:
What the hell is condominium apartment anyway....
condominium - a block of flats, each of which is owned by the person who lives in it
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19 Jun 2009, 09:21
While the owner of a condominium apartment has free and clear title to the dwelling, owners of cooperative apartments have shares in a corporation that owns a building and leases apartments to them.
(A) While the owner of a condominium apartment has free and clear title to the dwelling,
(B) The owner of a condominium apartment has free and clear title to the dwelling, but
(C) Whereas owners of condominium apartments have free and clear title to their dwellings,
(D) An owner of a condominium apartment has free and clear title to the dwelling, whereas
(E) Condominium apartment owners have a title to their dwelling that is free and clear, whill
A,B,D - Owner vs Owners - not parallel
Between C and E - 'C' is parallel
Re: Condominium apartment [#permalink] 19 Jun 2009, 09:21
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# Integrate 3/(1x^9*x^4)x^2*x
### Videos
$13x^{3}+C_0$
## Step-by-step explanation
Problem to solve:
$\int\frac{3}{1x^9 x^4} x^2\cdot xdx$
1
Any expression multiplied by $1$ is equal to itself
$\int x^2x\frac{3}{x^4x^9}dx$
$13x^{3}+C_0$
### Struggling with math?
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$\int\frac{3}{1x^9 x^4} x^2\cdot xdx$
### Main topic:
Integration by substitution
~ 0.65 seconds | 295 | 654 | {"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} | 3.53125 | 4 | CC-MAIN-2019-18 | latest | en | 0.565451 |
https://orangekitchens.net/boil/why-does-boiling-syrup-much-hotter-than-boiling-water.html | 1,638,738,544,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363216.90/warc/CC-MAIN-20211205191620-20211205221620-00172.warc.gz | 487,241,104 | 20,023 | # Why does boiling syrup much hotter than boiling water?
Contents
Water can’t rise above 212 degrees, the temperature at which it boils. Sugar syrup (sugar and water), though, can get much hotter because sugar melts at a much higher temperature. … As it cooks, more water evaporates, giving a steadily higher concentration of sugar.
## Why does boiling syrup produce a more severe burn than boiling water?
Sugar raises the boiling point of water so where water should boil at 212°F it will boil at a hotter 230°F in sugar water. Therefore it would “burn” hotter. Either way you still get the same scalding skin damage. How long does dry ice keep things cold compared to water ice?
## Which is hotter boiling pure water or boiling sugar solution?
Boiling Points
At sea level, pure water boils at two hundred and twelve degrees and never gets any hotter. But the boiling point for sugar is much higher than it is for water. And the temperature of the syrup, is a rough average of the temperatures of the two ingredients.
IT IS INTERESTING: Question: How long does a courgette take to cook?
## Does boiling water get hotter the longer it boils?
The water may boil more vigorously and convert into steam more quickly, but it won’t get hotter. In fact, at the microscopic level, there may be cooler regions of boiling water. When vapor bubbles form near a heat source, like at the bottom of a pot, the gas bubbles insulate the water from the heat.
## What temperature should maple syrup be boiled to?
In professional maple syrup production, the maple syrup maker will boil anywhere from about five gallons to 13 gallons of sap down to about a quart of maple syrup. When the syrup reaches 7 degrees Fahrenheit over the boiling point of water (212 degrees F), or 219 degrees F, the syrup should be done.
## Which is hotter boiling water or steam?
Steam occurs when water goes above 212 degrees Fahrenheit, which is hotter than water when it is at its stable point. While water boils at 212 F, steam is at a much higher temperature as water turns to vapor. … The instability of the liquid can make it turn into a vapor more quickly and can produce steam.
## What produces severe burns boiling water or steam?
Steam causes more severe burns.
So, when steam falls on skin and condenses to produce water it gives out 22.5 x 105 J/kg more heat than boiling water at the same temperature. Steam has more energy than boiling water. Therefore, burns produced by steam are more severe than those produced by boiling water.
## What does sugar do to boiling water?
Adding sugar to the water will thus increase the boiling point of the overall solution. The more sugar, the higher the boiling point. This is a physical phenomenon, very similar to that of the freezing point depression.
## Does sugar melt completely?
It turns out that, strictly speaking, sugar doesn’t actually melt. And it can caramelize while it’s still solid. … The melting point of a substance is the temperature at which it turns from a solid into a liquid while maintaining its chemical identity.
## How do you know which solution has the highest boiling point?
Multiply the original molality (m) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point.
## Why did my boiling water explode?
When you heat up water, these trapped bubbles allow the water to boil easily. It’s when your mug has no microscratches, allows little of the water to be in contact with air, and is kept very still while being heated that conditions are suitable for superheating… … Voila, exploding water!
## Why does boiling water stay at 100 degrees?
When the water’s vapor pressure equals the ambient air pressure, the water boils. At sea level, this occurs at 100°C. Additional heat added to boiling water does not raise its temperature, because it is providing the additional energy (the heat of vaporization) necessary for the 100°C water to become 100°C water vapor.
## What happens to the temperature of water while it is boiling?
Temperature of a substance during its phase change remains constant. Hence when water boils, the temperature of water remains constant i.e 100oC.
## Can you get botulism from maple syrup?
It is the sap of the maple tree that creates maple syrup and contamination with botulism is almost impossible. … In fact, boiling is one of the ways that botulism spores are killed. The risk of botulism from maple syrup is virtually non-existent and maple syrup is considered safe.
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## Can you get sick from eating old maple syrup?
The flavor will change slightly, and it won’t taste as good as it used to. Unfortunately, there’s no way to tell you when exactly will that happen. … Either way, even if it tastes pretty bland, it’s still safe to consume, so no worries you’ll get sick from eating “out of date” maple syrup.
## How do you know when to stop boiling maple syrup?
Water boils at 212°F, while maple syrup boils at 219°F. For this reason, you want to continue boiling your sap until it reaches a temperature of 219°F. Note: In general, you want to boil the sap until it reaches a temperature that is 7°F above the boiling temperature of water. | 1,152 | 5,423 | {"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-2021-49 | latest | en | 0.911014 |
http://de.metamath.org/mpeuni/iblneg.html | 1,716,588,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00407.warc.gz | 5,908,024 | 8,395 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > iblneg Structured version Visualization version GIF version
Theorem iblneg 23375
Description: The negative of an integrable function is integrable. (Contributed by Mario Carneiro, 25-Aug-2014.)
Hypotheses
Ref Expression
itgcnval.1 ((𝜑𝑥𝐴) → 𝐵𝑉)
itgcnval.2 (𝜑 → (𝑥𝐴𝐵) ∈ 𝐿1)
Assertion
Ref Expression
iblneg (𝜑 → (𝑥𝐴 ↦ -𝐵) ∈ 𝐿1)
Distinct variable groups: 𝑥,𝐴 𝜑,𝑥 𝑥,𝑉
Allowed substitution hint: 𝐵(𝑥)
Proof of Theorem iblneg
StepHypRef Expression
1 itgcnval.2 . . . . . . . . . 10 (𝜑 → (𝑥𝐴𝐵) ∈ 𝐿1)
2 iblmbf 23340 . . . . . . . . . 10 ((𝑥𝐴𝐵) ∈ 𝐿1 → (𝑥𝐴𝐵) ∈ MblFn)
31, 2syl 17 . . . . . . . . 9 (𝜑 → (𝑥𝐴𝐵) ∈ MblFn)
4 itgcnval.1 . . . . . . . . 9 ((𝜑𝑥𝐴) → 𝐵𝑉)
53, 4mbfmptcl 23210 . . . . . . . 8 ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
65renegd 13797 . . . . . . 7 ((𝜑𝑥𝐴) → (ℜ‘-𝐵) = -(ℜ‘𝐵))
76breq2d 4595 . . . . . 6 ((𝜑𝑥𝐴) → (0 ≤ (ℜ‘-𝐵) ↔ 0 ≤ -(ℜ‘𝐵)))
87, 6ifbieq1d 4059 . . . . 5 ((𝜑𝑥𝐴) → if(0 ≤ (ℜ‘-𝐵), (ℜ‘-𝐵), 0) = if(0 ≤ -(ℜ‘𝐵), -(ℜ‘𝐵), 0))
98mpteq2dva 4672 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℜ‘-𝐵), (ℜ‘-𝐵), 0)) = (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘𝐵), -(ℜ‘𝐵), 0)))
105iblcn 23371 . . . . . . . 8 (𝜑 → ((𝑥𝐴𝐵) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ (ℜ‘𝐵)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ (ℑ‘𝐵)) ∈ 𝐿1)))
111, 10mpbid 221 . . . . . . 7 (𝜑 → ((𝑥𝐴 ↦ (ℜ‘𝐵)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ (ℑ‘𝐵)) ∈ 𝐿1))
1211simpld 474 . . . . . 6 (𝜑 → (𝑥𝐴 ↦ (ℜ‘𝐵)) ∈ 𝐿1)
135recld 13782 . . . . . . 7 ((𝜑𝑥𝐴) → (ℜ‘𝐵) ∈ ℝ)
1413iblre 23366 . . . . . 6 (𝜑 → ((𝑥𝐴 ↦ (ℜ‘𝐵)) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ if(0 ≤ (ℜ‘𝐵), (ℜ‘𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘𝐵), -(ℜ‘𝐵), 0)) ∈ 𝐿1)))
1512, 14mpbid 221 . . . . 5 (𝜑 → ((𝑥𝐴 ↦ if(0 ≤ (ℜ‘𝐵), (ℜ‘𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘𝐵), -(ℜ‘𝐵), 0)) ∈ 𝐿1))
1615simprd 478 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘𝐵), -(ℜ‘𝐵), 0)) ∈ 𝐿1)
179, 16eqeltrd 2688 . . 3 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℜ‘-𝐵), (ℜ‘-𝐵), 0)) ∈ 𝐿1)
186negeqd 10154 . . . . . . . 8 ((𝜑𝑥𝐴) → -(ℜ‘-𝐵) = --(ℜ‘𝐵))
1913recnd 9947 . . . . . . . . 9 ((𝜑𝑥𝐴) → (ℜ‘𝐵) ∈ ℂ)
2019negnegd 10262 . . . . . . . 8 ((𝜑𝑥𝐴) → --(ℜ‘𝐵) = (ℜ‘𝐵))
2118, 20eqtrd 2644 . . . . . . 7 ((𝜑𝑥𝐴) → -(ℜ‘-𝐵) = (ℜ‘𝐵))
2221breq2d 4595 . . . . . 6 ((𝜑𝑥𝐴) → (0 ≤ -(ℜ‘-𝐵) ↔ 0 ≤ (ℜ‘𝐵)))
2322, 21ifbieq1d 4059 . . . . 5 ((𝜑𝑥𝐴) → if(0 ≤ -(ℜ‘-𝐵), -(ℜ‘-𝐵), 0) = if(0 ≤ (ℜ‘𝐵), (ℜ‘𝐵), 0))
2423mpteq2dva 4672 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘-𝐵), -(ℜ‘-𝐵), 0)) = (𝑥𝐴 ↦ if(0 ≤ (ℜ‘𝐵), (ℜ‘𝐵), 0)))
2515simpld 474 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℜ‘𝐵), (ℜ‘𝐵), 0)) ∈ 𝐿1)
2624, 25eqeltrd 2688 . . 3 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘-𝐵), -(ℜ‘-𝐵), 0)) ∈ 𝐿1)
275negcld 10258 . . . . 5 ((𝜑𝑥𝐴) → -𝐵 ∈ ℂ)
2827recld 13782 . . . 4 ((𝜑𝑥𝐴) → (ℜ‘-𝐵) ∈ ℝ)
2928iblre 23366 . . 3 (𝜑 → ((𝑥𝐴 ↦ (ℜ‘-𝐵)) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ if(0 ≤ (ℜ‘-𝐵), (ℜ‘-𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℜ‘-𝐵), -(ℜ‘-𝐵), 0)) ∈ 𝐿1)))
3017, 26, 29mpbir2and 959 . 2 (𝜑 → (𝑥𝐴 ↦ (ℜ‘-𝐵)) ∈ 𝐿1)
315imnegd 13798 . . . . . . 7 ((𝜑𝑥𝐴) → (ℑ‘-𝐵) = -(ℑ‘𝐵))
3231breq2d 4595 . . . . . 6 ((𝜑𝑥𝐴) → (0 ≤ (ℑ‘-𝐵) ↔ 0 ≤ -(ℑ‘𝐵)))
3332, 31ifbieq1d 4059 . . . . 5 ((𝜑𝑥𝐴) → if(0 ≤ (ℑ‘-𝐵), (ℑ‘-𝐵), 0) = if(0 ≤ -(ℑ‘𝐵), -(ℑ‘𝐵), 0))
3433mpteq2dva 4672 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℑ‘-𝐵), (ℑ‘-𝐵), 0)) = (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘𝐵), -(ℑ‘𝐵), 0)))
3511simprd 478 . . . . . 6 (𝜑 → (𝑥𝐴 ↦ (ℑ‘𝐵)) ∈ 𝐿1)
365imcld 13783 . . . . . . 7 ((𝜑𝑥𝐴) → (ℑ‘𝐵) ∈ ℝ)
3736iblre 23366 . . . . . 6 (𝜑 → ((𝑥𝐴 ↦ (ℑ‘𝐵)) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ if(0 ≤ (ℑ‘𝐵), (ℑ‘𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘𝐵), -(ℑ‘𝐵), 0)) ∈ 𝐿1)))
3835, 37mpbid 221 . . . . 5 (𝜑 → ((𝑥𝐴 ↦ if(0 ≤ (ℑ‘𝐵), (ℑ‘𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘𝐵), -(ℑ‘𝐵), 0)) ∈ 𝐿1))
3938simprd 478 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘𝐵), -(ℑ‘𝐵), 0)) ∈ 𝐿1)
4034, 39eqeltrd 2688 . . 3 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℑ‘-𝐵), (ℑ‘-𝐵), 0)) ∈ 𝐿1)
4131negeqd 10154 . . . . . . . 8 ((𝜑𝑥𝐴) → -(ℑ‘-𝐵) = --(ℑ‘𝐵))
4236recnd 9947 . . . . . . . . 9 ((𝜑𝑥𝐴) → (ℑ‘𝐵) ∈ ℂ)
4342negnegd 10262 . . . . . . . 8 ((𝜑𝑥𝐴) → --(ℑ‘𝐵) = (ℑ‘𝐵))
4441, 43eqtrd 2644 . . . . . . 7 ((𝜑𝑥𝐴) → -(ℑ‘-𝐵) = (ℑ‘𝐵))
4544breq2d 4595 . . . . . 6 ((𝜑𝑥𝐴) → (0 ≤ -(ℑ‘-𝐵) ↔ 0 ≤ (ℑ‘𝐵)))
4645, 44ifbieq1d 4059 . . . . 5 ((𝜑𝑥𝐴) → if(0 ≤ -(ℑ‘-𝐵), -(ℑ‘-𝐵), 0) = if(0 ≤ (ℑ‘𝐵), (ℑ‘𝐵), 0))
4746mpteq2dva 4672 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘-𝐵), -(ℑ‘-𝐵), 0)) = (𝑥𝐴 ↦ if(0 ≤ (ℑ‘𝐵), (ℑ‘𝐵), 0)))
4838simpld 474 . . . 4 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ (ℑ‘𝐵), (ℑ‘𝐵), 0)) ∈ 𝐿1)
4947, 48eqeltrd 2688 . . 3 (𝜑 → (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘-𝐵), -(ℑ‘-𝐵), 0)) ∈ 𝐿1)
5027imcld 13783 . . . 4 ((𝜑𝑥𝐴) → (ℑ‘-𝐵) ∈ ℝ)
5150iblre 23366 . . 3 (𝜑 → ((𝑥𝐴 ↦ (ℑ‘-𝐵)) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ if(0 ≤ (ℑ‘-𝐵), (ℑ‘-𝐵), 0)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ if(0 ≤ -(ℑ‘-𝐵), -(ℑ‘-𝐵), 0)) ∈ 𝐿1)))
5240, 49, 51mpbir2and 959 . 2 (𝜑 → (𝑥𝐴 ↦ (ℑ‘-𝐵)) ∈ 𝐿1)
5327iblcn 23371 . 2 (𝜑 → ((𝑥𝐴 ↦ -𝐵) ∈ 𝐿1 ↔ ((𝑥𝐴 ↦ (ℜ‘-𝐵)) ∈ 𝐿1 ∧ (𝑥𝐴 ↦ (ℑ‘-𝐵)) ∈ 𝐿1)))
5430, 52, 53mpbir2and 959 1 (𝜑 → (𝑥𝐴 ↦ -𝐵) ∈ 𝐿1)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∈ wcel 1977 ifcif 4036 class class class wbr 4583 ↦ cmpt 4643 ‘cfv 5804 0cc0 9815 ≤ cle 9954 -cneg 10146 ℜcre 13685 ℑcim 13686 MblFncmbf 23189 𝐿1cibl 23192 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-inf2 8421 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 ax-addf 9894 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-fal 1481 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-disj 4554 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-se 4998 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-isom 5813 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-ofr 6796 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-2o 7448 df-oadd 7451 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-sup 8231 df-inf 8232 df-oi 8298 df-card 8648 df-cda 8873 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-n0 11170 df-z 11255 df-uz 11564 df-q 11665 df-rp 11709 df-xadd 11823 df-ioo 12050 df-ico 12052 df-icc 12053 df-fz 12198 df-fzo 12335 df-fl 12455 df-seq 12664 df-exp 12723 df-hash 12980 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-clim 14067 df-sum 14265 df-xmet 19560 df-met 19561 df-ovol 23040 df-vol 23041 df-mbf 23194 df-itg1 23195 df-itg2 23196 df-ibl 23197 df-0p 23243 This theorem is referenced by: itgneg 23376 iblsub 23394 itgsub 23398 iblsubnc 32641 itgsubnc 32642
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# newnansm15 - Chapter 15 Selection of a Minimum Attractive...
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Chapter 15: Selection of a Minimum Attractive Rate of Return 15-1 The interest rates on these securities vary greatly over time, making it impossible to predict rates. Three factors that distinguish the securities: Bond Duration Bond Safety Municipal Bond 20 years Safe Corporate Bond 20 years Less Safe The importance of the non-taxable income feature usually makes the municipal bond the one with the lowest interest rate. The corporate bond generally will have the highest interest rate. 15-2 As this is a situation of “neither input nor output fixed,” incremental analysis is required. C- D B- C B- D D- A ∆ Cost \$25 \$50 \$75 \$25 ∆ Benefit \$4 \$6.31 \$10.31 \$5.96 ∆ Rate of Return 9.6% 4.5% 6.2% 20% Using the incremental rates of return one may determine the preferred alternative at any interest rate. For interest rates between: The problem here concerns Alternative C. C is preferred for 4.5% < Interest Rate < 9.6%. 15-3 Lease : Pay \$267 per month for 24 months. Purchase: A = \$9,400 (A/P, 1%, 24) = \$9,400 (0.0471) = \$442.74 Salvage (resale) value = \$4,700 (a) Purchase Rather than Lease ∆Monthly payment = \$442.71 - \$267 = \$175.74 ∆Salvage value = \$4,700 - \$0 = \$4,700 ∆ Rate of Return PW of Cost = PW of Benefit 0% 4.5% 9.6% 20% B C D A
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\$175.74 (P/A, i%, 24) = \$4,700 (P/A, i%, 24) = \$4,700/\$175.74 = 26.74 i = 0.93% per month Thus, the additional monthly payment of \$175.74 would yield an 11.2% rate of return.
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newnansm15 - Chapter 15 Selection of a Minimum Attractive...
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Ask a homework question - tutors are online | 576 | 2,006 | {"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} | 3.109375 | 3 | CC-MAIN-2018-05 | latest | en | 0.823032 |
http://isabelle.in.tum.de/repos/isabelle/rev/b9924abb8c66 | 1,571,747,184,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00549.warc.gz | 95,842,621 | 5,936 | author haftmann Wed Feb 14 10:07:17 2007 +0100 (2007-02-14) changeset 22322 b9924abb8c66 parent 22321 e5cddafe2629 child 22323 b8c227d8ca91
continued
1.1 --- a/doc-src/IsarImplementation/Thy/ML.thy Wed Feb 14 10:06:17 2007 +0100
1.2 +++ b/doc-src/IsarImplementation/Thy/ML.thy Wed Feb 14 10:07:17 2007 +0100
1.3 @@ -110,7 +110,56 @@
1.4 \end{mldecls}
1.5 *}
1.6
1.7 -text FIXME
1.8 +(*<*)
1.9 +typedecl foo
1.10 +consts foo :: foo
1.11 +ML {*
1.12 +val dummy_const = ("bar", @{typ foo}, NoSyn)
1.13 +val dummy_def = ("bar", @{term foo})
1.14 +val thy = Theory.copy @{theory}
1.15 +*}
1.16 +(*>*)
1.17 +
1.18 +text {*
1.19 + Many problems in functional programming can be thought of
1.20 + as linear transformations, i.e.~a caluclation starts with a
1.21 + particular value @{text "x \<Colon> foo"} which is then transformed
1.22 + by application of a function @{text "f \<Colon> foo \<Rightarrow> foo"},
1.23 + continued by an application of a function @{text "g \<Colon> foo \<Rightarrow> bar"},
1.24 + and so on. As a canoncial example, take primitive functions enriching
1.25 + theories by constants and definitions:
1.26 + @{ML "Sign.add_consts_i: (string * typ * mixfix) list -> theory -> theory"}
1.27 + and @{ML "Theory.add_defs_i: bool -> bool -> (bstring * term) list -> theory -> theory"}.
1.28 + Written with naive application, an addition of a constant with
1.29 + a corresponding definition would look like:
1.31 + With increasing numbers of applications, this code gets quite unreadable.
1.32 + Using composition, at least the nesting of brackets may be reduced:
1.34 + What remains unsatisfactory is that things are written down in the opposite order
1.35 + as they actually happen''.
1.36 +*}
1.37 +
1.38 +(*<*)
1.39 +ML {*
1.40 +val thy = Theory.copy @{theory}
1.41 +*}
1.42 +(*>*)
1.43 +
1.44 +text {*
1.45 + At this stage, Isabelle offers some combinators which allow for more convenient
1.46 + notation, most notably reverse application:
1.47 + @{ML "
1.48 +thy
1.50 +|> Theory.add_defs_i false false [dummy_def]"}
1.51 +*}
1.52 +
1.53 +text {*
1.54 + When iterating over a list of parameters @{text "[x\<^isub>1, x\<^isub>2, \<dots> x\<^isub>n] \<Colon> 'a list"},
1.55 + the @{ML fold} combinator lifts a single function @{text "f \<Colon> 'a -> 'b -> 'b"}:
1.56 + @{text "y |> fold f [x\<^isub>1, x\<^isub>2, \<dots> x\<^isub>n] \<equiv> y |> f x\<^isub>1 |> f x\<^isub>2 |> \<dots> |> f x\<^isub>n"}
1.57 +*}
1.58
1.59 text %mlref {*
1.60 \begin{mldecls}
1.61 @@ -122,6 +171,10 @@
1.62 \end{mldecls}
1.63 *}
1.64
1.65 +text {*
1.66 + FIXME transformations involving side results
1.67 +*}
1.68 +
1.69 text %mlref {*
1.70 \begin{mldecls}
1.71 @{index_ML "(op #>)": "('a -> 'b) * ('b -> 'c) -> 'a -> 'c"} \\
1.72 @@ -132,6 +185,10 @@
1.73 \end{mldecls}
1.74 *}
1.75
1.76 +text {*
1.77 + FIXME higher-order variants
1.78 +*}
1.79 +
1.80 text %mlref {*
1.81 \begin{mldecls}
1.82 @{index_ML "(op )": "('b -> 'a) -> 'b -> 'a * 'b"} \\
2.1 --- a/doc-src/IsarImplementation/Thy/document/ML.tex Wed Feb 14 10:06:17 2007 +0100
2.2 +++ b/doc-src/IsarImplementation/Thy/document/ML.tex Wed Feb 14 10:07:17 2007 +0100
2.3 @@ -151,8 +151,67 @@
2.4 %
2.6 %
2.8 +%
2.10 +%
2.11 +\isatagML
2.12 +%
2.13 +\endisatagML
2.14 +{\isafoldML}%
2.15 +%
2.17 +%
2.19 +%
2.20 \begin{isamarkuptext}%
2.21 -FIXME%
2.22 +Many problems in functional programming can be thought of
2.23 + as linear transformations, i.e.~a caluclation starts with a
2.24 + particular value \isa{x\ {\isasymColon}\ foo} which is then transformed
2.25 + by application of a function \isa{f\ {\isasymColon}\ foo\ {\isasymRightarrow}\ foo},
2.26 + continued by an application of a function \isa{g\ {\isasymColon}\ foo\ {\isasymRightarrow}\ bar},
2.27 + and so on. As a canoncial example, take primitive functions enriching
2.28 + theories by constants and definitions:
2.29 + \verb|Sign.add_consts_i: (string * typ * mixfix) list -> theory -> theory|
2.30 + and \verb|Theory.add_defs_i: bool -> bool -> (bstring * term) list -> theory -> theory|.
2.31 + Written with naive application, an addition of a constant with
2.32 + a corresponding definition would look like:
2.34 + With increasing numbers of applications, this code gets quite unreadable.
2.35 + Using composition, at least the nesting of brackets may be reduced:
2.37 + What remains unsatisfactory is that things are written down in the opposite order
2.38 + as they actually happen''.%
2.39 +\end{isamarkuptext}%
2.40 +\isamarkuptrue%
2.41 +%
2.43 +%
2.45 +%
2.46 +\isatagML
2.47 +%
2.48 +\endisatagML
2.49 +{\isafoldML}%
2.50 +%
2.52 +%
2.54 +%
2.55 +\begin{isamarkuptext}%
2.56 +At this stage, Isabelle offers some combinators which allow for more convenient
2.57 + notation, most notably reverse application:
2.58 + \isasep\isanewline%
2.59 +\verb|thy|\isasep\isanewline%
2.61 +\verb||\verb,|,\verb|> Theory.add_defs_i false false [dummy_def]|%
2.62 +\end{isamarkuptext}%
2.63 +\isamarkuptrue%
2.64 +%
2.65 +\begin{isamarkuptext}%
2.66 +When iterating over a list of parameters \isa{{\isacharbrackleft}x\isactrlisub {\isadigit{1}}{\isacharcomma}\ x\isactrlisub {\isadigit{2}}{\isacharcomma}\ {\isasymdots}\ x\isactrlisub n{\isacharbrackright}\ {\isasymColon}\ {\isacharprime}a\ list},
2.67 + the \verb|fold| combinator lifts a single function \isa{f\ {\isasymColon}\ {\isacharprime}a\ {\isacharminus}{\isachargreater}\ {\isacharprime}b\ {\isacharminus}{\isachargreater}\ {\isacharprime}b}:
2.68 + \isa{y\ {\isacharbar}{\isachargreater}\ fold\ f\ {\isacharbrackleft}x\isactrlisub {\isadigit{1}}{\isacharcomma}\ x\isactrlisub {\isadigit{2}}{\isacharcomma}\ {\isasymdots}\ x\isactrlisub n{\isacharbrackright}\ {\isasymequiv}\ y\ {\isacharbar}{\isachargreater}\ f\ x\isactrlisub {\isadigit{1}}\ {\isacharbar}{\isachargreater}\ f\ x\isactrlisub {\isadigit{2}}\ {\isacharbar}{\isachargreater}\ {\isasymdots}\ {\isacharbar}{\isachargreater}\ f\ x\isactrlisub n}%
2.69 \end{isamarkuptext}%
2.70 \isamarkuptrue%
2.71 %
2.72 @@ -173,6 +232,24 @@
2.73 \end{isamarkuptext}%
2.74 \isamarkuptrue%
2.75 %
2.76 +\endisatagmlref
2.77 +{\isafoldmlref}%
2.78 +%
2.80 +%
2.82 +%
2.83 +\begin{isamarkuptext}%
2.84 +FIXME transformations involving side results%
2.85 +\end{isamarkuptext}%
2.86 +\isamarkuptrue%
2.87 +%
2.89 +%
2.91 +%
2.92 +\isatagmlref
2.93 +%
2.94 \begin{isamarkuptext}%
2.95 \begin{mldecls}
2.96 \indexml{(op \#$>$)}\verb|(op #>): ('a -> 'b) * ('b -> 'c) -> 'a -> 'c| \\
2.97 @@ -184,6 +261,24 @@
2.98 \end{isamarkuptext}%
2.99 \isamarkuptrue%
2.100 %
2.101 +\endisatagmlref
2.102 +{\isafoldmlref}%
2.103 +%
2.105 +%
2.107 +%
2.108 +\begin{isamarkuptext}%
2.109 +FIXME higher-order variants%
2.110 +\end{isamarkuptext}%
2.111 +\isamarkuptrue%
2.112 +%
2.114 +%
2.116 +%
2.117 +\isatagmlref
2.118 +%
2.119 \begin{isamarkuptext}%
2.120 \begin{mldecls}
2.121 \indexml{(op )}\verb|(op ): ('b -> 'a) -> 'b -> 'a * 'b| \\
3.1 --- a/doc-src/IsarImplementation/Thy/document/logic.tex Wed Feb 14 10:06:17 2007 +0100
3.2 +++ b/doc-src/IsarImplementation/Thy/document/logic.tex Wed Feb 14 10:07:17 2007 +0100
3.3 @@ -435,7 +435,7 @@
3.4 \end{isamarkuptext}%
3.5 \isamarkuptrue%
3.6 %
3.7 -\isamarkupsubsection{Primitive connectives and rules%
3.8 +\isamarkupsubsection{Primitive connectives and rules \label{sec:prim_rules}%
3.9 }
3.10 \isamarkuptrue%
3.11 %
4.1 --- a/doc-src/IsarImplementation/Thy/logic.thy Wed Feb 14 10:06:17 2007 +0100
4.2 +++ b/doc-src/IsarImplementation/Thy/logic.thy Wed Feb 14 10:07:17 2007 +0100
4.3 @@ -434,7 +434,7 @@
4.4 notion of equality/equivalence @{text "\<equiv>"}.
4.5 *}
4.6
4.7 -subsection {* Primitive connectives and rules *}
4.8 +subsection {* Primitive connectives and rules \label{sec:prim_rules} *}
4.9
4.10 text {*
4.11 The theory @{text "Pure"} contains constant declarations for the
` | 3,055 | 7,885 | {"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} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.673841 |
https://sherrinfrances.com/tag/line-of-flight/ | 1,679,324,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00548.warc.gz | 584,931,689 | 19,058 | # Stupid velocity
There’s a lot in this world about which I am not stupid: practically, I’ve navigated life more or less effectively so far. But more and more often I notice am annoyed that there is someone at the table who professes to know more about [X] than I do. And the more I am expected to know, the more I question what I think I know. Maybe this is why I am enjoying the math classes: I’m allowed to be stupid* there. I am supposed to be stupid there.
Years ago, a professor was describing the idea of “lines of flight” from 1000 Plateaus, and he said (loosely paraphrased) that it means running headlong through a mountain rather than intentionally trying to go around it. His example was a 3¢ bank fee that you disagreed with. Rather than protesting and not paying it, which would ultimately benefit the bank, you should write a separate check for the 3¢ every time the fee is due. If we all wrote separate checks for these tiny fees, we could cripple the bank by costing it more money to process the fee than the fee itself is worth. Don’t avoid the fee. Force everyone to look at the fee in excruciating detail. Maybe this is what I’m doing with my own mathematical stupidity: running headlong into it and gazing on every horrifying crevice. To what end, I don’t know.
All of this to say: I have finally passed MATH 140: Introduction to Mathematical Analysis. It took me three tries, and I wasn’t entirely sure that I had passed until final grades were posted. I’m registered to take Calculus in the fall. Also, I’m getting better at finding textbook deals, so the book this time was only\$180, and I bought the Student Solutions Manual up front, too–that might have saved me last semester, if only it had been “required” and not “recommended.”
The velocity of my stupidity is picking up steam. Maybe even accelerating? | 452 | 1,853 | {"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} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.966121 |
https://puzzling.stackexchange.com/questions/37396/my-password-is-2 | 1,696,421,757,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00829.warc.gz | 510,913,701 | 39,651 | Evil mastermind PhD Evil, is trying to destroy the north pole. In order to foil his plan you need to log on to his computer. Alas you don't know the password, thankfully PhD Evil thinks he is too clever. So he left a puzzle to solve his password.
It is ten too.
My password is seven letters long
four letters is also not wrong
It is fair to say that it is none
but that could be off by one
If you think my password is a single hole,
I will blow up the north Pole!
Your partner says, I don't really care the north pole is fake. But you persevere for Santa's sake.
onezero (10 or ten)
10 in binary is 2
It is ten too.
10 in base ten is 10
My password is seven letters long
"onezero" is seven letters long
four letters is also not wrong
(maybe) 10 in Spanish is diez, four letters
or (credit to Niktheslik)
One zero could be interpreted as just "zero" which has 4 letters
It is fair to say that is is none
but that could be off by one
0 is none, but it is missing the 1 in front of it
If you think my password is a single hole,
I will blow up the north Pole!
0 (single hole) is the wrong password
• A better explanation for four letters is also not wrong: zero is 1 zero, as in a singular zero. So one zero is, in that sense, equivalent to just zero as a zero is 1 zero. Jul 11, 2016 at 19:43
• @Niktheslik Perhaps but in that case the password is the string "onezero", which it might be, I'll edit my answer to include it
– Axle
Jul 11, 2016 at 20:12
• I like this solution. About the "four letters is also not wrong" explanation, it could be: "Deca", which is a metric unit prefix indicating ten multiples of the unit. en.wikipedia.org/wiki/Metric_prefix Jul 11, 2016 at 20:42
• @Axle onezero is indeed the intended solution. Well done. Jul 11, 2016 at 20:47
• I think four letters is 'onez'. Jul 12, 2016 at 6:10 | 507 | 1,831 | {"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-2023-40 | longest | en | 0.972627 |
https://nl.webqc.org/molecularweightcalculated-190424-29.html | 1,571,005,037,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00061.warc.gz | 721,508,330 | 5,675 | Chemical Equations Balanced on 04/24/19
Molecular weights calculated on 04/23/19 Molecular weights calculated on 04/25/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157
Molar mass of Gd is 157.25
Molar mass of Pb is 207,2
Molar mass of HNO3 is 63.01284
Molar mass of Fe[NH4]*2.6H2O is 120,723188
Molar mass of Fe[NH4]*2.6H2O is 120,723188
Molar mass of C3H8 is 44.09562
Molar mass of PCl5 is 208.238762
Molar mass of PCl3 is 137.332762
Molar mass of PCl2 is 101.879762
Molar mass of SnEt3 is 205.8933
Molar mass of c8 h8 o6 is 200,14552
Molar mass of BaCrO4 is 253.3207
Molar mass of BaSO4 is 233.3896
Molar mass of P is 30.973762
Molar mass of H2 is 2.01588
Molar mass of O is 15.9994
Molar mass of He is 4.002602
Molar mass of AgCl is 143.3212
Molar mass of Mg is 24,305
Molar mass of NaNO3 is 84,99466928
Molar mass of NaNO3 is 84,99466928
Molar mass of C44H48N16O12S4Ni is 1179.90532
Molar mass of C44H48N16O12S4Ni2 is 1238.59872
Molar mass of C2H4 is 28.05316
Molar mass of B2O3 is 69.6202
Molar mass of C2H6O is 46,06844
Molar mass of C44H48N16O12S4Ni4 is 1355.98552
Molar mass of PO4 is 94,971362
Molar mass of Na3PO4 is 163.94066984
Molar mass of MgCO3 is 84.3139
Molar mass of CO2 is 44.0095
Molar mass of HClO4 is 100.45854
Molar mass of KI is 166.00277
Molar mass of H2 is 2,01588
Molar mass of NH4Cl is 53.49146
Molar mass of ZnOH2 is 83.39528
Molar mass of SnEt2 is 176.8322
Molar mass of CrSO3 is 132,0593
Molar mass of Sn2Et4 is 353.6644
Molar mass of MgSO4*7H2O is 246,47456
Molar mass of NaN3 is 65,00986928
Molar mass of S is 32.065
Molar mass of C20H23 is 263,39662
Molar mass of HLi is 7.94894
Molar mass of Mg3(PO4)2 is 262,857724
Molar mass of N2(kg) is 28.0134
Molar mass of H2SO4 is 98,07848
Molar mass of Al2(SO4)3 is 342,1508772
Molar mass of CaO is 56,0774
Molar mass of Be is 9,012182
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157
Calculate molecular weight
Molecular weights calculated on 04/23/19 Molecular weights calculated on 04/25/19
Molecular masses on 04/17/19
Molecular masses on 03/25/19
Molecular masses on 04/24/18
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Periodiek systeem Eenheden omrekenen Chemie gereedschappen Chemisch Forum Chemie FAQ Constanten Symmetrie Zoek Chemie links Link naar ons Suggesties? Neem contact met ons op Stel een betere vertaling voor Kies een taalDeutschEnglishEspañolFrançaisItalianoNederlandsPolskiPortuguêsРусский中文日本語한국어 Hoe moet je citeren? WebQC.Org online onderwijs gratis huiswerkhulp chemische opgaven vragen en antwoorden | 1,640 | 3,572 | {"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-2019-43 | latest | en | 0.713154 |
https://www.unitconverters.net/data-transfer/byte-second-to-t4-signal.htm | 1,670,411,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00205.warc.gz | 1,064,067,282 | 3,828 | Home / Data Transfer Conversion / Convert Byte/second to T4 (signal)
# Convert Byte/second to T4 (signal)
Please provide values below to convert byte/second [B/s] to T4 (signal), or vice versa.
From: byte/second To: T4 (signal)
### Byte/second to T4 (signal) Conversion Table
Byte/second [B/s]T4 (signal)
0.01 B/s2.9178338001867E-10 T4 (signal)
0.1 B/s2.9178338001867E-9 T4 (signal)
1 B/s2.9178338001867E-8 T4 (signal)
2 B/s5.8356676003735E-8 T4 (signal)
3 B/s8.7535014005602E-8 T4 (signal)
5 B/s1.4589169000934E-7 T4 (signal)
10 B/s2.9178338001867E-7 T4 (signal)
20 B/s5.8356676003735E-7 T4 (signal)
50 B/s1.4589169000934E-6 T4 (signal)
100 B/s2.9178338001867E-6 T4 (signal)
1000 B/s2.91783E-5 T4 (signal)
### How to Convert Byte/second to T4 (signal)
1 B/s = 2.9178338001867E-8 T4 (signal)
1 T4 (signal) = 34272000 B/s
Example: convert 15 B/s to T4 (signal):
15 B/s = 15 × 2.9178338001867E-8 T4 (signal) = 4.3767507002801E-7 T4 (signal) | 388 | 947 | {"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-2022-49 | latest | en | 0.528316 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-8-techniques-of-integration-8-6-strategies-for-integration-exercises-page-431/40 | 1,723,689,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00379.warc.gz | 597,167,463 | 13,628 | ## Calculus (3rd Edition)
$$8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C$$
Given $$\int \sqrt{x^{2}-16} d x$$ Let \begin{align*} x&=4\sec u\\ dx&=4\sec u\tan udu \end{align*} Then \begin{align*} \int \sqrt{x^{2}-16} d x&= 4\int \sqrt{16\sec^2 u-16} \sec u\tan udu\\ &=4\int \sqrt{16\tan^2 u-} \sec u\tan udu\\ &=16\int \sec u\tan^2 udu\\ &=16\int (\sec^3 u-\sec u )du,\ \ \\ & \text{From the table: }\\ &= 16\left(\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|-\ln |\sec u+\tan u| \right)+C\\ &= 16\left(\frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u| \right)+C\\ &= 8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C \end{align*} | 359 | 750 | {"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} | 4.25 | 4 | CC-MAIN-2024-33 | latest | en | 0.182355 |
https://scioly.org/forums/viewtopic.php?t=16104&p=413465 | 1,597,129,319,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00523.warc.gz | 480,167,159 | 14,264 | ## Circuit Lab B
azboy1910
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### Re: Circuit Lab B
Jay M. Robinson Middle School '18-now
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Creationist127
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### Re: Circuit Lab B
1. Three lightbulbs and a voltage source are connected in parallel with each other. One lightbulb is removed. What happens to the other two? Do they get brighter, dimmer, stay the same, go out?
2. Now the three bulbs are in series with the voltage source. What happens when one is removed?
3. Now, two are in series with each other, but this set is in parallel with the third and the battery. What happens to each of the other two when one of the series ones is removed?
2018: Hovercraft, Thermo, Coaster, Solar System
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2021 grind starts now... and no excuses about not knowing the rules yet!
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### Re: Circuit Lab B
Creationist127 wrote:
March 6th, 2020, 4:37 am
1. Three lightbulbs and a voltage source are connected in parallel with each other. One lightbulb is removed. What happens to the other two? Do they get brighter, dimmer, stay the same, go out?
2. Now the three bulbs are in series with the voltage source. What happens when one is removed?
3. Now, two are in series with each other, but this set is in parallel with the third and the battery. What happens to each of the other two when one of the series ones is removed?
1. The other two bulbs should remain the same brightness since the voltage difference across each bulb does not change and the resistance of each bulb doesn't change. Therefore, according to $P=\frac{V^2}{R}$ the power remains the same, which means the brightness must remain the same.
2. If all three bulbs are in series with each other, then if one is removed, the other two will go out since the circuit is now open.
3. If one of the series bulbs are removed, the bulb that was in series with the removed bulb will go out for the same reason in question 2. The one parallel to the series bulbs will remain on and at the same brightness for the same reason in question 1.
When you've hit rock bottom, the only way to go is up. That's just how stonks work.
Conant '19 => UIUC '23
Member of The Builder Cult
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2019 State - 5th MTV | 5th Sounds
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Change my mind
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### Re: Circuit Lab B
2018: Hovercraft, Thermo, Coaster, Solar System
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2020: Circuit Lab, Wright Stuff, Machines
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2021 grind starts now... and no excuses about not knowing the rules yet!
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### Re: Circuit Lab B
I am in the bathtub while I am charging my phone. I see that it is fully charged, and I go to reach for the wall adapter. With my soapy hand, I have trouble pulling out the adapter, so I try to go behind it to give some leverage. To my demise, I touch the prongs while the adapter is still in the wall. Immediately following this, there was an audible pop.
a) What was the popping sound from, and why did the popping happen?
b) Am I alive or dead?
When you've hit rock bottom, the only way to go is up. That's just how stonks work.
Conant '19 => UIUC '23
Member of The Builder Cult
2018 State - 2nd MTV | 3rd Hovercraft
2019 State - 5th MTV | 5th Sounds
Physics is the only real science
Change my mind
Nydauron's Userpage
MoMoney$$;)0) Member Posts: 153 Joined: January 14th, 2019, 6:38 pm Division: C State: OH Pronouns: He/Him/His Location: In Cleveland Somewhere Has thanked: 41 times Been thanked: 16 times ### Re: Circuit Lab B Nydauron wrote: March 9th, 2020, 3:17 pm I am in the bathtub while I am charging my phone. I see that it is fully charged, and I go to reach for the wall adapter. With my soapy hand, I have trouble pulling out the adapter, so I try to go behind it to give some leverage. To my demise, I touch the prongs while the adapter is still in the wall. Immediately following this, there was an audible pop. a) What was the popping sound from, and why did the popping happen? b) Am I alive or dead? Electric Shock, possibly ESD can result from it; so you could be dead or alive. Hopefully alive if you regain consciousness. Question Time: Consider the function MAJ: {0, 2}3 -> {0,2} that is defined as follows: MAJ(x) = piecewise{1 , x^0 +x^1+x^2 >= 3}{0, otherwise}. What is the formula involving AND, OR, and NOT to compute MAJ? Division C - Northeast Ohio Gravity Vehicle Machines Detector Building Circuit Lab Protein Modeling 2019-2020 Medal Count: 5 "Don't be upset by the results you didn't get from the work you didn't do' Memberships: Builder Cult azboy1910 Member Posts: 33 Joined: November 3rd, 2018, 2:19 pm Division: B State: NC Pronouns: He/Him/His Location: Hopefully somewhere being productive for once. Has thanked: 18 times Been thanked: 6 times ### Re: Circuit Lab B Why are we asking C questions in B? Jay M. Robinson Middle School '18-now Email: azboy1910@gmail.com azboy1910's Userpage Nydauron Exalted Member Posts: 50 Joined: March 20th, 2018, 8:10 pm Division: Grad State: IL Pronouns: He/Him/His Location: Getting ready for online classes ... again. Has thanked: 70 times Been thanked: 51 times ### Re: Circuit Lab B Should have replied to this sooner. Whoops! MoMoney$$$;)0) wrote: April 10th, 2020, 5:45 pm Nydauron wrote: March 9th, 2020, 3:17 pm I am in the bathtub while I am charging my phone. I see that it is fully charged, and I go to reach for the wall adapter. With my soapy hand, I have trouble pulling out the adapter, so I try to go behind it to give some leverage. To my demise, I touch the prongs while the adapter is still in the wall. Immediately following this, there was an audible pop. a) What was the popping sound from, and why did the popping happen? b) Am I alive or dead? Electric Shock, possibly ESD can result from it; so you could be dead or alive. Hopefully alive if you regain consciousness. The popping sound was from a Ground-Fault Circuit Interrupter (GCFI) outlet. All bathrooms are supposed to have GFCI outlets according to the National Electrical Code. The electricity would most likely travel through your body to ground for a split second before the GFCI outlet pops. Hence, the popping is from the GFCI. Here is a video explaining the basics of GFCI: https://youtu.be/GlM6PE2kKVY azboy1910 wrote: April 16th, 2020, 8:37 pm Why are we asking C questions in B? It was weird considering that this one was called Circuit Lab B even though there both divisions are pretty similar in topics. (Yes I know, C has some additional topics...) MoMoney$$;)0) wrote: April 10th, 2020, 5:45 pm Question Time: Consider the function MAJ: {0, 2}3 -> {0,2} that is defined as follows: MAJ(x) = piecewise{1 , x^0 +x^1+x^2 >= 3}{0, otherwise}. What is the formula involving AND, OR, and NOT to compute MAJ? ^^Reposted most recent question When you've hit rock bottom, the only way to go is up. That's just how stonks work. Conant '19 => UIUC '23 Member of The Builder Cult 2018 State - 2nd MTV | 3rd Hovercraft 2019 State - 5th MTV | 5th Sounds Physics is the only real science Change my mind Nydauron's Userpage azboy1910 Member Posts: 33 Joined: November 3rd, 2018, 2:19 pm Division: B State: NC Pronouns: He/Him/His Location: Hopefully somewhere being productive for once. Has thanked: 18 times Been thanked: 6 times ### Re: Circuit Lab B Oh yeah electroboom lol. Jay M. Robinson Middle School '18-now Email: azboy1910@gmail.com azboy1910's Userpage MoMoney$$$;)0)
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### Re: Circuit Lab B
Nydauron wrote:
April 17th, 2020, 11:36 am
Should have replied to this sooner. Whoops!
MoMoney$$;)0) wrote: April 10th, 2020, 5:45 pm Nydauron wrote: March 9th, 2020, 3:17 pm I am in the bathtub while I am charging my phone. I see that it is fully charged, and I go to reach for the wall adapter. With my soapy hand, I have trouble pulling out the adapter, so I try to go behind it to give some leverage. To my demise, I touch the prongs while the adapter is still in the wall. Immediately following this, there was an audible pop. a) What was the popping sound from, and why did the popping happen? b) Am I alive or dead? Electric Shock, possibly ESD can result from it; so you could be dead or alive. Hopefully alive if you regain consciousness. The popping sound was from a Ground-Fault Circuit Interrupter (GCFI) outlet. All bathrooms are supposed to have GFCI outlets according to the National Electrical Code. The electricity would most likely travel through your body to ground for a split second before the GFCI outlet pops. Hence, the popping is from the GFCI. Here is a video explaining the basics of GFCI: https://youtu.be/GlM6PE2kKVY azboy1910 wrote: April 16th, 2020, 8:37 pm Why are we asking C questions in B? It was weird considering that this one was called Circuit Lab B even though there both divisions are pretty similar in topics. (Yes I know, C has some additional topics...) MoMoney$$\$;)0) wrote:
April 10th, 2020, 5:45 pm
Question Time: Consider the function MAJ: {0, 2}3 -> {0,2} that is defined as follows: MAJ(x) = piecewise{1 , x^0 +x^1+x^2 >= 3}{0,
otherwise}. What is the formula involving AND, OR, and NOT to compute MAJ?
^^Reposted most recent question
I forgot about that since I don't use any of the plugs in my bathroom, and I haven't really studied that aspect of Circuit Lab, I guess that makes sense.
Anyhow nobody answered my last question.
Question Time: Consider the function MAJ: {0, 2}3 -> {0,2} that is defined as follows: MAJ(x) = piecewise{1 , x^0 +x^1+x^2 >= 3}{0,
otherwise}. What is the formula involving AND, OR, and NOT to compute MAJ?
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### Who is online
Users browsing this forum: No registered users and 1 guest | 3,121 | 11,161 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 1, "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-2020-34 | latest | en | 0.925448 |
https://opensource.apple.com/source/ChatServer/ChatServer-37.1/libraries/Twisted-1.3.0/sandbox/chord.py.auto.html | 1,675,069,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00756.warc.gz | 466,297,927 | 4,100 | # chord.py [plain text]
```# Twisted, the Framework of Your Internet
# Copyright (C) 2001 Matthew W. Lefkowitz
#
# This library is free software; you can redistribute it and/or
# modify it under the terms of version 2.1 of the GNU Lesser General Public
#
# This library is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
# Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public
# License along with this library; if not, write to the Free Software
# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
# TODO:
# failure detection
# successor list
# HyperChord
import random
import sha
import socket
import struct
NBIT = 160 # Size (in bits) of Chord identifiers
vNodes = 1 # sure hope you dont want more than 256.
def between(n, a, b):
"""Check if n in interval(a, b) on the circle."""
if a == b: return n != a # n is the only node not in (n, n)
elif a < b: return n > a and n < b
else: return n > a or n < b
def betweenL(n, a, b):
"""Check if n in interval[a, b) on the circle."""
if a == b and n == a: return 1
elif a < b: return n >= a and n < b
else: return n >= a or n < b
def betweenR(n, a, b):
"""Check if n in interval(a, b] on the circle."""
if a == b and n == a: return 1
elif a < b: return n > a and n <= b
else: return n > a or n <= b
class Node(pb.Copyable, pb.Perspective):
"""A node in a Chord network."""
def __init__(self, address, perspectiveName, identityName, port=pb.portno):
global vNodes
pb.Perspective.__init__(self, perspectiveName, identityName)
self.id = sha.new(socket.inet_aton(address) + struct.pack('!h', port) + chr(vNodes)).digest()
vNodes = vNodes + 1
self.finger = [None] * (NBIT+1)
self.lastFixed = 1 # Needed for fixFingers
self.pred = None
def getStateToCopyFor(self, persp):
def __repr__(self):
return "<Node " + `self.id` + ">"
def join(self, n2):
"""Intialize finger tables of local node.
n2 is a node already on the network or None if we're the first."""
if n2:
for i in range(1, NBIT):
if betweenL(self.id, self.start(i+1),
self.finger[i].id):
self.finger[i+1] = self.finger[i]
else:
else:
self.finger[1] = None
self.pred = None
def _setFingerCallback(self, x, i):
#CRUM! DEFEATED BY DISTANCE
self.finger[i] = x
def _setPredCallback(self, n):
#ALSO TOO FAR!
self.pred = n
def perspective_getSuccessor(self):
return self.finger[1]
def perspective_getPredecessor(self):
return self.pred
def perspective_findSuccessor(self, id):
def findSuccessor_1(self, n2):
else:
def findPredecessor(self, id):
if self.finger[1] is None:
return self
n2 = self
return self.findPredecessor_1(n2, id)
def findPredecessor_1(self, n2, id):
def findPredecessor_2(self, n3, n2, id):
if not betweenR(id, n2.id, n3.id):
else:
return n2
def perspective_closestPrecedingFinger(self, id):
for i in xrange(NBIT, 0, -1):
if not self.finger[i]: continue
if between(self.finger[i].id, self.id, id):
def getSelf(self):
# necessary to produce proper Copyable behaviour
return self
def stabilise(self):
"""Verify our immediate successor and tell them about us.
Called periodically."""
def stabilise_1(self, p):
self.perspective_notify(p)
def fixFingers(self):
"""Refresh a random finger table entry.
Called periodically."""
i = random.randrange(1, len(self.finger)+1)
self.finger[i] = self.findSuccessor(self.start(i))
"""n thinks it might be our predecessor."""
def notify(self, n):
if (self.pred is self or self.pred is None or
between(n.id, self.pred.id, self.id)):
self.pred = n
if self.finger[1] is None or between(n.id, self.id, self.finger[1].id):
self.finger[1] = n
for i in xrange(2, len(self.finger)):
if self.finger[i] is None or betweenL(n.id, self.start(i), self.finger[i].id):
self.finger[i] = n
def start(self, k):
assert 1 <= k <= NBIT
r = (self.id + 2L**(k-1)) % 2L**NBIT
if r == 0:
return 2L**NBIT
else:
return r
class ChordService(pb.Service):
def __init__(self, address, port, serviceName, application=None):
pb.Service.__init__(self, serviceName, application) | 1,236 | 4,148 | {"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-2023-06 | latest | en | 0.632328 |
https://www.neelsomani.com/blog/writing-a-tic-tac-toe-solver-using-minimax.php | 1,718,712,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00291.warc.gz | 817,266,258 | 6,514 | # Writing a tic-tac-toe solver using minimax
-
In this post, we’ll build a tic-tac-toe solver using the minimax algorithm. There are a few steps. First, we’ll need to generate a game tree of all possible moves and outcomes. Then, we’ll write the minimax code to calculate the optimal move.
Here’s a link to the GitHub repo. Note: The code in the GitHub repo is slightly refactored, but the logic is all the same.
## 1. Starter code
Because the focus of this post isn’t really about creating the tic-tac-toe game itself, I’ve included some starter code below. We’re going to create a class called GameTree, which will represent a game state: the current state of the board, and whose move it is. The GameTree will also store all child GameTrees, which are defined as any game state that is reachable within the opponent’s next move.
We’ll internally represent a board as a string of 9 characters, consisting of X, O, and spaces. The string will represent the board flattened. (e.g., The string "XXO XO X" would represent the board with the top row as "X X O", the next row as "_ X O", and the bottom row as "_ _ X".)
Before representing the game tree with a class, I was initially concerned about performance. I remember that I previously ran into issues when using a binary search tree class in Python with a few hundred million elements. But a loose upper bound on the number of tic-tac-toe boards is 3^9 (3 possible values for each square: X, O, and space) = 19683, and it’s loose since not all of those formations are valid (e.g., you can’t have three X’s in a row and three O’s in a row on the same board, the number of X's and O's have to be within one, etc.). Python should be able to handle 20000 elements easily.
class GameTree:
# A list to switch between players
players = ["X", "O"]
def __init__(self, value=" ", player_number=0):
self.value = value
self.children = []
self.player_number = player_number
# TODO: Implement self.generate_children().
self.generate_children()
def print_tree(self):
""" Print in a user-friendly way. """
t = self.value
print(t[0] + " | " + t[1] + " | " + t[2])
print("---------")
print(t[3] + " | " + t[4] + " | " + t[5])
print("---------")
print(t[6] + " | " + t[7] + " | " + t[8])
I’ve left a TODO for the self.generate_children() function, which we’ll implement in the next step.
## 2. Generating the game tree
The self.generate_children() function should reassign the self.children variable to a list of child GameTrees. Before we write that function, though, we need to know whether a game state is terminal, that is, whether either player has won, or whether the board is full.
def is_win(self, player_number):
""" Determine if the player has won the game. """
player = GameTree.players[player_number]
t = self.value
return any([(t[3 * i] == player and t[3 * i + 1] == player and t[3 * i + 2] == player) for i in range(3)]) or \
any([(t[i] == player and t[i + 3] == player and t[i + 6] == player) for i in range(3)]) or \
(t[0] == player and t[4] == player and t[8] == player) or (t[2] == player and t[4] == player and t[6] == player)
To see whether the board is full, we’ll check for the presence of " " in the board string.
To generate all child trees, we’ll get all of the indices of the current tree.value string that are spaces (which represent available moves), and generate the trees corresponding to the player placing their symbol in each location.
def generate_children(self):
""" Generate the children of this node. """
if self.is_win(self.player_number) or self.is_win((self.player_number + 1) % 2) or " " not in self.value:
return
children_indices = [position[0] for position in enumerate(self.value) if position[1] == " "]
child_trees = []
for i in children_indices:
child_tree = GameTree(self.value[:i] + GameTree.players[self.player_number] + self.value[i+1:], (self.player_number + 1) % 2)
child_trees.append(child_tree)
self.children = child_trees
## 3. The minimax code
Let’s write a depth-limited minimax algorithm. We’ll let the user specify how deep into the minimax tree we’ll traverse. If that’s the case, then we’ll need some sort of heuristic to evaluate a game state if it’s not terminal. (If a game state is terminal, we can just value it as 1, 1/2, or 0 in the cases of winning, tying, and losing respectively.)
Let’s use the "probability of winning" as the heuristic, where the probability of winning is defined as the weighted proportion of descendent leaves that are winning states (valued at 1), tied states (valued at 1/2) and losing states (valued at 0). For example, if there are three possible child moves in a position, one of which is a win, one is a tie, and one is a loss, then we’ll value the game state as (1/3) * (1 + 1/2 + 0) = 1/2.
Here’s our heuristic function:
def probability_of_winning(self, player_number):
""" Give the probability of player winning this game. """
if self.is_win(player_number):
return 1
elif self.is_win((self.player_number + 1) % 2):
return 0
elif " " not in self.value:
return .5
return sum([child.probability_of_winning(player_number) for child in self.children]) / len(self.children)
And now we can write the minimax algorithm with an arbitrary depth. We’ll decrement the depth each time we hit a minimizer node.
def minimax(self, depth=2):
""" Run minimax with the current player as max. Decrement the depth for each
minimizer. Return the appropriate child tree. """
def maximize(tree, depth):
# Return the maximum of all child minimizer values.
if len(tree.children) == 0:
return (tree.value, tree.probability_of_winning(self.player_number))
child_values = [minimize(child, depth - 1) for child in tree.children]
return (tree.value, max(child_values, key=lambda k: k[1])[1])
def minimize(tree, depth):
# Return the minimum of all child maximizer values.
if depth == 0 or len(tree.children) == 0:
return (tree.value, tree.probability_of_winning(self.player_number))
child_values = [maximize(child, depth) for child in tree.children]
return (tree.value, min(child_values, key=lambda k: k[1])[1])
child_values = [minimize(child, depth) for child in self.children]
child_index = max(enumerate(child_values), key=lambda k: k[1][1])[0]
return self.children[child_index]
And we’re done. We can see the algorithm play against itself with the following function:
def run_minimax_game(self, depth=2):
""" Automatically execute a minimax game. """
tree = self
move = 0
while not tree.is_win(self.player_number) and not tree.is_win((self.player_number + 1) % 2) and " " in tree.value:
print("Move number: " + str(move))
tree.print_tree()
tree = tree.minimax(depth)
move += 1
print("Move number: " + str(move))
tree.print_tree()
## 4. The probability heuristic
Is the probability heuristic alone really that much worse than minimax? In short, yes. Try the following sample game tree.
sample = GameTree('OXO X X', 1)
sample.print_tree()
In this position, it’s O to move, and the optimal move is clearly to place in the middle column of the bottom row, to prevent X from winning. Let’s write a helper function to get the "optimal" move that the probability heuristic identifies.
def move_by_probability(self):
""" Return the move that maximizes the probability of winning. """
children_with_probabilities = [(child, child.probability_of_winning(self.player_number)) for child in self.children]
return max(children_with_probabilities, key=lambda k: k[1])[0]
If you run sample.move_by_probability.print_tree(), you’ll find that O plays an entirely suboptimal move — a move that does maximize the proportion of situations in which O wins (if the moves were random), but doesn’t prevent X from winning if X plays optimally.
Tags: minimax tic-tac-toe solver artificial intelligence tic tac toe tic tac toe | 1,964 | 7,783 | {"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} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.918161 |
https://hextobinary.com/unit/byte/from/MB/to/TiB/803 | 1,723,586,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00479.warc.gz | 235,561,956 | 15,210 | # 803 Megabytes in Tebibytes
Data Storage
Megabyte
Tebibyte
803 Megabytes = 0.00073032424552366 Tebibytes
## How many Tebibytes are in 803 Megabytes?
The answer is 803 Megabytes is equal to 0.00073032424552366 Tebibytes and that means we can also write it as 803 Megabytes = 0.00073032424552366 Tebibytes. Feel free to use our online unit conversion calculator to convert the unit from Megabyte to Tebibyte. Just simply enter value 803 in Megabyte and see the result in Tebibyte. You can also Convert 804 Megabytes to Tebibytes
## How to Convert 803 Megabytes to Tebibytes (803 MB to TiB)
By using our Megabyte to Tebibyte conversion tool, you know that one Megabyte is equivalent to 9.0949470177293e-7 Tebibyte. Hence, to convert Megabyte to Tebibyte, we just need to multiply the number by 9.0949470177293e-7. We are going to use very simple Megabyte to Tebibyte conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Megabyte} = \text{9.0949470177293e-7 Tebibytes}$$
$$\text{803 Megabytes} = 803 \times 9.0949470177293e-7 = \text{0.00073032424552366 Tebibytes}$$
## What is Megabyte Unit of Measure?
Megabyte is a unit of digital information about data storage. One megabyte is equal to 1000000 bytes.
## What is the symbol of Megabyte?
The symbol of Megabyte is MB. This means you can also write one Megabyte as 1 MB.
## What is Tebibyte Unit of Measure?
Tebibyte is a unit of digital information about data storage. One tebibyte is equal to 1099511627776 bytes.
## What is the symbol of Tebibyte?
The symbol of Tebibyte is TiB. This means you can also write one Tebibyte as 1 TiB.
## Megabyte to Tebibyte Conversion Table (803-812)
Megabyte [MB]Tebibyte [TiB]
8030.00073032424552366
8040.00073123374022543
8050.00073214323492721
8060.00073305272962898
8070.00073396222433075
8080.00073487171903253
8090.0007357812137343
8100.00073669070843607
8110.00073760020313784
8120.00073850969783962
## Megabyte to Other Units Conversion Table
Megabyte [MB]Output
803 megabytes in bit is equal to6424000000
803 megabytes in byte is equal to803000000
803 megabytes in kilobit is equal to6424000
803 megabytes in kibibit is equal to6273437.5
803 megabytes in kilobyte is equal to803000
803 megabytes in kibibyte is equal to784179.69
803 megabytes in megabit is equal to6424
803 megabytes in mebibit is equal to6126.4
803 megabytes in mebibyte is equal to765.8
803 megabytes in gigabit is equal to6.42
803 megabytes in gibibit is equal to5.98
803 megabytes in gigabyte is equal to0.803
803 megabytes in gibibyte is equal to0.74785202741623
803 megabytes in terabit is equal to0.006424
803 megabytes in tebibit is equal to0.0058425939641893
803 megabytes in terabyte is equal to0.000803
803 megabytes in tebibyte is equal to0.00073032424552366
803 megabytes in petabit is equal to0.000006424
803 megabytes in pebibit is equal to0.0000057056581681536
803 megabytes in petabyte is equal to8.03e-7
803 megabytes in pebibyte is equal to7.132072710192e-7
803 megabytes in exabit is equal to6.424e-9
803 megabytes in exbibit is equal to5.5719318048375e-9
803 megabytes in exabyte is equal to8.03e-10
803 megabytes in exbibyte is equal to6.9649147560469e-10
803 megabytes in zettabit is equal to6.424e-12
803 megabytes in zebibit is equal to5.4413396531616e-12
803 megabytes in zettabyte is equal to8.03e-13
803 megabytes in zebibyte is equal to6.801674566452e-13
803 megabytes in yottabit is equal to6.424e-15
803 megabytes in yobibit is equal to5.3138082550407e-15
803 megabytes in yottabyte is equal to8.03e-16
803 megabytes in yobibyte is equal to6.6422603188008e-16
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 1,221 | 3,830 | {"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} | 3.234375 | 3 | CC-MAIN-2024-33 | latest | en | 0.683743 |
http://www.360doc.com/relevant/77527387_more.shtml | 1,571,030,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00067.warc.gz | 294,454,298 | 10,127 | w1：ma(c,5)ref(ma(c,5),1)；60，n横盘天数，n1横盘的上下幅度REF(((HHV(H,N)-LLV(L,N))/LLV(L,N)),1)=(n1/100)and ref(v,1)ref(ma(v,5),1)and cross(v,ma(v,5))and cross(v,ma(v,10))；81，条件：1：假设大前天创最近新高,要求这天涨幅是超过7%的大阳线,且创最近一个月的新高.2：那么前天,昨天,今天都收小K线,要求小K线实体xg：ref(c,3)/ref(c,4)1.07 and ref(c,3)=hhv(c,20)and count(abs((c-ref(c,1))/ref(c,1)0.02,3)))=3；
mACD公式“防狼术”条件选股公式。V00:= HIGH<REF(HIGH,1) AND LOW<REF(LOW,1);V04:= REF(HIGH,2) <HIGH AND REF(LOW,2) <LOW AND REF(HIGH,2) >REF(HIGH,1) AND REF(LOW,2) <REF(LOW,1);{最近下跌天数}N2:=31;{上涨天数}N3:=55;{平台高点天数}LL:=LLV(L,N1);HH:=REF(HHV(H,N2),N1);H0:=REF(HHV(H,N3),N1+N2);ZF:=(C-REF(C,1))/REF(C,1)*100;缠论三买点:HH=HHV(H,N2+N1) AND LL>=H0 AND SUM(ZF>9.9,30)>1 OR SUM(ZF>8.9,30)>1;
cross(ref(ma(c,5),1),ref(ma(c,10),1)) and c>ma(c,5);xg:ref(c,1)>ma(c,20) and cross(ma(c,20),c);xg:(ref(c,3)-ref(c,4))/ref(c,3)>0.05 and count((h-l)/l.xg:cross(c,ma(c,10)) and cross(c,ma(c,20)) and cross(c,ma(c,30)) and v>ref(v,1)*3;xg:ma(c,30)>ref(ma(c,30),1);count(abs(c-ref(c,1)/ref(c,1))ref(c,1),3)=3;xg:ma(c,20)>ref(ma(c,20),1);XG:(HHV(C,5)-LLV(C,5))/LLV(C,5)REF(V,1)*2 AND C/REF(C,1)>1.05;
XG:REF(T1,1);MACD金叉死叉的日期的公式。a and count(a,barslast(macd.macd>0))=1 and macd.macd涨停后的第一次跌破5日线公式。XG:C>O AND V>REF(V,1) AND REF(O>C,1) AND REF(C>O,2) AND REF(V,2)>REF(V,1) AND MA(C,20)>REF(MA(C,20),1);XG:C>O AND REF(MIN(C,O),1)> C AND O-L>2*(H-C) AND REF(L,1)>H AND C股价上穿三线10日线向上的选股公式。XG:CROSS(C,MA5) AND CROSS(C,MA10) AND CROSS(C,MA20) AND MA10>REF(MA10,1);
AA7:=MA((AA4*0.96+AA5*96+AA6*0.96+AA1*0.558+AA2*0.558+AA3*0.558)/6,17);AA8:=MA((AA4*1.25+AA5*1.23+AA6*1.2+AA1*0.55+AA2*0.55+AA3*0.65)/6,17);AA9:=MA((AA4*1.3+AA5*3+AA6*1.3+AA1*0.68+AA2*0.68+AA3*0.68)/6,17);MACD6:= 2*(DIFF6-DEA6);MA(CLOSE,30)>REF(MA(CLOSE,30),1) and diff>0 and dea>0 AND MACD>REF(MACD,1) and b;MA(CLOSE,30)>REF(MA(CLOSE,30),1) AND DIFF>0 AND DEA>0 AND MACD>REF(MACD,1) AND B;
count(c<>72、今天收盘价在昨天收盘价-2%~2%之间c/ref(c,1)>0.98 and c/ref(c,1)<>73、收盘价在5日均线以上5天 count(c>ma(c,5),5)=5;74、N天前换手率大于8% REF(VOL/CAPITAL*100,N)>8;75、一条均线18MA 股价上涨后“缩量”回调到18MA XG:COUNT(V<>MA(C,18); N “缩量”回调的天数76、1.昨日量是n天以来最低量（n可调）；(MA30>REF(MA30,1)) AND (REF(MA30,1)<=ref ="">REF(C,2)>REF(O,2) AND REF(C,1)
（为大家节约时间 ...[转载]通达信公式 125个选股公式，解决选股难题！vol/ma(vol,10)3，两天内,累计下跌等于或超过20%选股.xg:(ref(c,2)-c)/c>0.2;4，股价是25天以来新低.xg:c5，选股：ＥＸＰＭＡ ５日与１０日金叉 并且第二天的收盘价在ＥＸＰＭＡ的５日线以上．cross(ref(ema(c,5),1),ref(ema(c,10),1)) and c>ema(c,5);6，请编买入公式：1. 将BIAS指标的参数设置为24日，将KD指标的参数设置为9；
2、当日最低价在20日均线上下0.05%内，收盘价在均线上方 abs(l-ma(c,20))<0.005and c>ma(c,20);3、20日均线相差M个百分点 w1:abs(c-ma(c,20)/ma(c,20)*1004、股价回调到20日均线abs(c-ma(c,20))/ma(c,20)<0.005;6、当日收盘价突破20日均线5% cross(c,ma(c,20)*1.05);7、股价上涨后“缩量”回调到20MA：XG:COUNT(VMA(C,20); N“缩量”回调的天数8、缩量：vol/ma(vol,10)<0.3;1、20日均线走平或走高 xg:ma(c,20)>ref(ma(c,20),1);
A11:=LLV(C,A1)<REF(LLV(C,A1+1),A1+1) AND LLV(DIFF,A1)>REF(LLV(DIFF,A1+1),A1+1);LLV(DIFF,A1)>REF(LLV(DIFF,A1+A2+2),A1+A2+2);DIBL1:=(A11 OR A22 OR A33 OR A44 OR A55) AND DIFF<0 AND DIFF<DEA AND DIFF>REF(DIFF,1)AND DIFF>REF(DIFF,2);B11:=HHV(C,B1)>REF(HHV(C,B1+1),B1+1) AND HHV(DIFF,B1)<REF(HHV(DIFF,B1+1),B1+1);DINGBL1:=(B11 OR B22 OR B33 OR B44 OR B55) AND DIFF>0 AND DEA<DIFF AND DIFF<REF(DIFF,1)
GT:=DEA>REF(DEA,1) AND REF(DEA,1)<REF(DEA,2);GT2:=DEA<REF(DEA,1) AND REF(DEA,1)>REF(DEA,2);
xg:c>ref(c,1)v<ref(v,1)ref(v,1)>ref(v,2)*2;83，5日均线倾角大于60度---经典X:(ATAN((MA(C,5)/REF(MA(C,5),1)-1)*100)*180/3.14115926)>60;也可以这样：X:(ATAN((EMA(C,5)/REF(EMA(C,5),1)-1)*100)*180/3.14115926)>60;利用指标排序：X:(ATAN((MA(C,5)/REF(MA(C,5),1)-1)*100)*180/3.14115926); 84，"中阴之后，跳拉阳" c/ref(c,1)>1.05;127，30日线开始上翘---- ㊣MA30:MA(C,30);(MA30>REF(MA30,1))(REF(MA30,1)<=REF(MA30,2));
DRAWTEXT(FILTER(B1>0,5),diff,''<-底背离''),COLORyellow;A5:=BARSLAST(ref(cross(dea,diff),1));B2:=ref(h,A5+3)<ref(h,3) and ref(diff,A5+3)>ref(diff,3) and cross(dea,diff);DRAWTEXT(FILTER(B2>0,5),diff,''<-顶背离''),COLORgreen;REF(CLOSE,T1)>CLOSE AND REF(DIFF,T1)<DIFF ANDREF(MACD,1)<MACD AND ALL(DIFF<0,T) AND T1>15 ANDCOUNT(JC,T)>=1;AA1:=DIFF<0 AND DIFF>=DEA AND DEA>REF(DEA,1)AND MACD>REF(MACD,1);
aa:=cross(ma(c,5),ma(c,20)) and cross(ma(c,5),ma(c,30)) and cross(ma(c,5),ma(c,60));老鼠仓:bb1 and cc1 and dd1 and ee1; ===================================MA5:MA(CLOSE,5),COLORF00FF0;MA13:MA(CLOSE,13),COLORWHITE;MA60:MA(CLOSE,60),LINETHICK2,COLORYELLOW;MA90:MA(C,90),LINETHICK2,COLORF0F000;MA120:MA(CLOSE,120),COLOR00FF00;===================================这一段是画5、13、60、90、120均线。
VAR01:4*SMA((CLOSE-LLV(LOW,n))/(HHV(HIGH,n)-LLV(LOW,n))*100,5,1)-3*SMA(SMA((CLOSE-LLV(LOW,n))/(HHV(HIGH,n)-LLV(LOW,n))*100,5,1),3.2,1),COLOR0099FF,LINETHICK1;
ma(v,13)>ref(ma(v,13),1);MA20穿120：cross(ma(c,20),ma(c,30)) or cross(ma(c,20),ma(c,60)) or cross(ma(c,20),ma(c,120)) ;（059）、多头排列 MA(c,5)>MA(c,10) and MA(c,20)<MA(c,10) and MA(c,5)>ref(MA(c,5),1) and MA(c,10)>ref(MA(c,10),1)and MA(c,20)>ref(MA(c,20),1)空头排列 MA(c,5)<MA(c,10) and MA(c,20)>MA(c,10) and MA(c,5)<ref(MA(c,5),1) and MA(c,10)<ref(MA(c,10),1)and MA(c,20)<ref(MA(c,20),1)
5、MACD指标图形为空中加油技巧最好.A4:=MA(C,4);A10:=MA(C,10);A20:=MA(C,20);A62:=MA(C,62);A133:=MA(C,133);B1:=A10>REF(A10,1)*1.003 AND A20>REF(A20,1)*1.003 AND(A10-A20)<C/10;B2:=C<A10 AND C>A20 AND A4<=A10;B3:=A62>=REF(A62,1) AND A133>=REF(A133,1);B4:=V/MA(V,10)>1.2;B1 AND B2 AND B3 AND B4;MACD指标图形为空中加油形态未编。aa:=qj1 and qj2 and qj3 and qj4;qj0 and qj1 and qj2 and qj3 and qj4; | 2,934 | 5,208 | {"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} | 3.21875 | 3 | CC-MAIN-2019-43 | latest | en | 0.274892 |
https://kr.mathworks.com/matlabcentral/cody/problems/1190-golomb-s-self-describing-sequence-based-on-euler-341/solutions/1464191 | 1,606,427,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00262.warc.gz | 369,809,509 | 17,141 | Cody
# Problem 1190. Golomb's self-describing sequence (based on Euler 341)
Solution 1464191
Submitted on 16 Mar 2018 by David Verrelli
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(isequal(euler341(1),1))
ans = 1
2 Pass
assert(isequal(euler341(10),5))
ans = 5
3 Pass
assert(isequal(euler341(310),42))
ans = 42
4 Pass
assert(isequal(euler341(4242),210))
ans = 210
5 Pass
assert(isequal(euler341(328509),3084))
ans = 3084
6 Pass
assert(isequal(euler341(551368),4247))
ans = 4247
7 Pass
assert(isequal(euler341(614125),4540))
ans = 4540
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 258 | 819 | {"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-2020-50 | latest | en | 0.581204 |
https://math.stackexchange.com/questions/1942413/finite-limit-to-size-of-intersecting-family-with-no-smaller-intersecting-set/1945544 | 1,642,777,315,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00198.warc.gz | 409,542,802 | 33,220 | # Finite limit to size of Intersecting Family with no Smaller Intersecting Set?
I'm interested in intersecting families of sets of size $r$, but where no set smaller than $r$ intersects the entire family. Meaning:
1. For every $A,B \in F$ we have $A \subset \{0,...,n-1\}$, $|A| = r$, $A \cap B \neq \emptyset$.
2. If $|C| < r$ there exists $D \in F$ such that $D \cap C = \emptyset$.
3. I also ask that we "use" every element in $\{0,...,n-1\}$, i.e for all $i < n$ there exists $A \in F$ such that $i \in A$.
Question: Given $r$, is there a finite upper limit to $n$?
Motivation:
For two simple cases, the answer is yes:
1. For $r=1$ trivially the maximal $n$ is $1$.
2. For $r=2$, I claim the maximal $n$ is $3$. For suppose we have in $F$ w.l.o.g sets $A = \{1,2\}$ and $B= \{1,3\}$. To "use" $4$, while intersecting $A$, we must have a set $\{1,4\}$ or $\{2,4\}$. However $\{2,4\}$ obviously doesn't intersect $B$, so we're left with $\{1,3\}$. So every "new" element $x$ we want to use after $3$ must add set $\{1,x\}$ to $F$. However this means $\{x\}$ intersects all the sets in $F$, violating requirement (2). Therefore for $r =2$ we have $n \le 3$.
But I have no idea how to generalize this result, if it's even true for greater $r$. The Fano plane shows that for $r=3$ the maximal $n \ge 7$.
The straightforward way of creating an intersecting family is just to pick a fixed element $x$, and take all the subsets of $\{1,...,n\}$ that include $x$. However, that means $\{x\}$, a set of size $1$, intersects all the other sets in the family, violating requirement (2) for $r > 1$.
Fedor Petrov provided an affirmative answer - for each $k$ there is a finite limit to the size $F$ (and hence also to $n$). | 567 | 1,724 | {"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} | 3.859375 | 4 | CC-MAIN-2022-05 | latest | en | 0.866917 |
https://forums.odforce.net/topic/45814-how-to-get-particle-points-angle-difference-between-previous-frame/?tab=comments | 1,618,634,094,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00056.warc.gz | 360,276,635 | 15,468 | # particle how to get particle points angle difference between previous frame
## Recommended Posts
hi
just learn houdini particle.
i am a beginner
in this project i use particles points then use copy to points node , boxs to make up model.
like this picture
however the problem is
i find some box jitter very unacceptable
so i Calculate the points distance between previous frame in solver node
if the distance less than 0.05,then the point do not move
this way make the points jitter less
however i need calculate the angle distance between previous frame also
but
i do not know how to calculate the rotation difference
let`s say ,if the rotation difference less then 0.1,then the point do not rotate.
bellow is the project file
##### Share on other sites
what you can do is to create a 3x3 transform matrix that will tell houdini what orientation has your instance from your normal
vector y = v@N;
vector x = normalize(cross(y, set(0, 1, 0)));
vector z = normalize(cross(x, y));
3@transform = set(x, y, z);
Houdini has priorities for points orientation from higher to lower priority it is :
matrix3 transform
vector4 orient
and then a combinaison of vectors like : v@N / up or v@v / up
the main problem i see is points jittering not really points orientation
i attached you a file, hope it is what you intended
point_jitter.hiplc
## Create an account
Register a new account | 337 | 1,404 | {"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} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.831279 |
http://functions.wolfram.com/ElementaryFunctions/ArcTan/13/01/01/0004/ | 1,386,847,650,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164580976/warc/CC-MAIN-20131204134300-00006-ip-10-33-133-15.ec2.internal.warc.gz | 74,233,031 | 7,993 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
ArcTan
http://functions.wolfram.com/01.14.13.0004.01
Input Form
(1 + z^2) Derivative[1][w][z] == 1 /; w[z] == ArcTan[z] && w[0] == 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List[RowBox[List["(", RowBox[List["1", "+", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[SuperscriptBox["w", "\[Prime]", Rule[MultilineFunction, None]], "[", "z", "]"]]]], " ", "\[Equal]", "1"]], "/;", RowBox[List[RowBox[List[RowBox[List["w", "[", "z", "]"]], "\[Equal]", RowBox[List["ArcTan", "[", "z", "]"]]]], "\[And]", RowBox[List[RowBox[List["w", "[", "0", "]"]], "\[Equal]", "0"]]]]]]]]
MathML Form
( 1 + z 2 ) w ( z ) 1 /; w ( z ) tan - 1 ( z ) w ( 0 ) 0 Condition 1 z 2 z w z 1 w z z w 0 0 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List[RowBox[List["(", RowBox[List["1", "+", SuperscriptBox["z_", "2"]]], ")"]], " ", RowBox[List[SuperscriptBox["w", "\[Prime]", Rule[MultilineFunction, None]], "[", "z_", "]"]]]], "]"]], "\[RuleDelayed]", RowBox[List["1", "/;", RowBox[List[RowBox[List[RowBox[List["w", "[", "z", "]"]], "\[Equal]", RowBox[List["ArcTan", "[", "z", "]"]]]], "&&", RowBox[List[RowBox[List["w", "[", "0", "]"]], "\[Equal]", "0"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 623 | 1,726 | {"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} | 2.59375 | 3 | CC-MAIN-2013-48 | longest | en | 0.256349 |
https://davis.lbl.gov/Manuals/CHOMBO-RELEASE-3.2/PolarIF_8H_source.html | 1,718,260,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00396.warc.gz | 183,362,992 | 4,287 | Chombo + EB 3.2
PolarIF.H
Go to the documentation of this file.
1 #ifdef CH_LANG_CC
2 /*
3 * _______ __
4 * / ___/ / ___ __ _ / / ___
5 * / /__/ _ \/ _ \/ V \/ _ \/ _ \
6 * \___/_//_/\___/_/_/_/_.__/\___/
7 * Please refer to Copyright.txt, in Chombo's root directory.
8 */
9 #endif
10
11 #ifndef _POLARIF_H_
12 #define _POLARIF_H_
13
14 #include "MayDay.H"
15 #include "RealVect.H"
16
17 #include "BaseIF.H"
18
20
21 ///
22 /**
23 This implicit function specifies a cylinder in polar coordinates:
24 r = a_primaryRadius + a_perturbation*cos(a_frequency*theta)
25 */
26 class PolarIF: public BaseIF
27 {
28 public:
29 ///
30 PolarIF(const Real& a_primaryRadius,
31 const Real& a_perturbation,
32 const int & a_frequency,
33 const bool& a_inside);
34
35 /// Copy constructor
36 PolarIF(const PolarIF& a_inputIF);
37
38 /// Destructor
39 virtual ~PolarIF();
40
41 ///
42 /**
43 Return the parameter information
44 */
45 virtual void GetParams(Real& a_primaryRadius,
46 Real& a_perturbation,
47 int & a_frequency,
48 bool& a_inside)const;
49
50 ///
51 /**
52 Set the parameter information
53 */
54 virtual void SetParams(const Real& a_primaryRadius,
55 const Real& a_perturbation,
56 const int & a_frequency,
57 const bool& a_inside);
58
59 ///
60 /**
61 Return the value of the function at a_point.
62 */
63 virtual Real value(const RealVect& a_point) const;
64
65 virtual BaseIF* newImplicitFunction() const;
66
67 protected:
71 bool m_inside;
72
73 private:
75 {
76 MayDay::Abort("PolarIF uses strong construction");
77 }
78
79 void operator=(const PolarIF& a_inputIF)
80 {
81 MayDay::Abort("PolarIF doesn't allow assignment");
82 }
83 };
84
85 #include "NamespaceFooter.H"
86 #endif
bool m_inside
Definition: PolarIF.H:71
virtual Real value(const RealVect &a_point) const
virtual BaseIF * newImplicitFunction() const
Definition: PolarIF.H:68
Definition: BaseIF.H:32
PolarIF()
Definition: PolarIF.H:74
double Real
Definition: REAL.H:33
void operator=(const PolarIF &a_inputIF)
Definition: PolarIF.H:79
virtual ~PolarIF()
Destructor.
A Real vector in SpaceDim-dimensional space.
Definition: RealVect.H:41
virtual void GetParams(Real &a_primaryRadius, Real &a_perturbation, int &a_frequency, bool &a_inside) const
Definition: PolarIF.H:26
int m_frequency
Definition: PolarIF.H:70
Real m_perturbation
Definition: PolarIF.H:69
static void Abort(const char *const a_msg=m_nullString)
Print out message to cerr and exit via abort() (if serial) or MPI_Abort() (if parallel).
virtual void SetParams(const Real &a_primaryRadius, const Real &a_perturbation, const int &a_frequency, const bool &a_inside) | 797 | 2,641 | {"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-2024-26 | latest | en | 0.159942 |
https://tolstoy.newcastle.edu.au/R/e13/help/11/04/10135.html | 1,586,454,185,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00095.warc.gz | 712,924,117 | 4,035 | # [R] Multinomial Logit Model with lots of Dummy Variables
From: ghpow1 <ghpow1_at_student.monash.edu.au>
Date: Sun, 10 Apr 2011 02:37:34 -0700 (PDT)
Hi All,
I am attempting to build a Multinomial Logit model with dummy variables of the following form:
Dependent Variable : 0-8 Discrete Choices
Dummy Variable 1: 965 dummy varsghpow_at_student.monash.edu.augh_at_gp1.com Dummy Variable 2: 805 dummy vars
The data set I am using has the dummy columns pre-created, so it's a table of 72,381 rows and 1770 columns.
The first 965 columns represent the dummy columns for Variable 1 The next 805 columns represent the dummy columns for Variable 2
My code to build the mlogit model looks like the following. I want to know...is there a better way of doing this without these huge equations? (I probably also need a more powerful PC to do all of this).
I'll also want to perform a joint test of significance on the first 805 coefficients...
Is this possible?
Thanks
GP
[code]
#install MLOGIT
library(mlogit)
mydata = 0
num.rows=length(mydata[,1])
num.cols=965+805+1
my_data=matrix(0,nr=num.rows,nc=num.cols)
for(i in 1:num.rows) {
``` nb=mydata[i,2]
np=mydata[i,3]
my_data[i,nb]=1
my_data[i,965+np]=1
my_data[i,1+1770]=mydata[i,1]
```
}
#convert matrix to data.frame
# convert to data frame
my_data_frame<-as.data.frame(my_data)
#load dataframe into mldata with choice variable
mldata<-mlogit.data(my_data_frame, varying=NULL, choice="V1771", shape="wide")
#V1771 = dependent var
#V1-V965 = variable 1 dummies
#V966-V1700 = variable 2 dummies
#regress V1771 against all 1700 variables...
mlogit.model<-mlogit(V1771~0|V1+V2+V3...+V1700,data=mldata, reflevel="0")
[/code]
```--
View this message in context: http://r.789695.n4.nabble.com/Multinomial-Logit-Model-with-lots-of-Dummy-Variables-tp3439492p3439492.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help_at_r-project.org mailing list
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http://booksbw.com/index.php?id1=4&category=computers&author=brian-rh&book=2001&page=48 | 1,638,791,188,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00581.warc.gz | 8,182,888 | 4,629 | Books in black and white
Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
# A Guide to MATLAB for Beginners and Experienced Users - Brian R.H.
Brian R.H., Roland L.L. A Guide to MATLAB for Beginners and Experienced Users - Cambrige, 2001. - 346 p.
Previous << 1 .. 42 43 44 45 46 47 < 48 > 49 50 51 52 53 54 .. 91 >> Next
One 300-Watt Bulb
If there is only one bulb, then we want to put the bulb in the center of the ceiling. Let's picture how well the floor is illuminated. We introduce coordinates x running from 0 to 10 in the long direction of the room and y running from 0 to 4 in the short direction. The intensity at a given point, measured in watts per square meter, is the power of the bulb, 300, divided by 4n times the square of the distance from the bulb. Since the bulb is 3 meters above the point (5, 2) on the floor, we can express the intensity at a point (x, y) on the floor as follows:
syms x y; illum = 300/(4*pi*((x - 5)"2 + (y - 2)"2 + 3"2))
illum =
75/pi/((x-5)~2+(y-2)~2+9)
We can use ezcontourf to plot this expression over the entire floor. We use colormap to arrange for a color gradation that helps us to see the 138 Chapter 9: Applications
ezcontourf(illum,[0 10 0 4]); colormap(gray); axis equal tight
75/n/((X-5)2+(y-2)2+9)
The darkest parts of the floor are the corners. Let us find the intensity of the light at the corners, and at the center of the room.
subs(illum, {x, y}, {0, 0}) subs(illum, {x, y}, {5, 2})
ans =
0.6282 ans =
2.6526
The center of the room, at floor level, is about 4 times as bright as the corners when there is only one bulb on the ceiling. Our objective is to light the room more uniformly using more bulbs with the same total amount of power. Before proceeding to deal with multiple bulbs, we observe that the use of ezcontourf is somewhat confining, as it does not allow us to control the number of contours in our pictures. Such control will be helpful in seeing the light intensity; therefore we shall plot numerically rather than symbolically; that is, we shall use contourf instead of ezcontourf.
Two 150-Watt Bulbs
In this case we need to decide where to put the two bulbs. Common sense tells us to arrange the bulbs symmetrically along a line down the center of Illuminating a Room 139
the room in the long direction, that is, along the line y = 2. Define a function that gives the intensity of light at a point (x, y) on the floor due to a 150-watt bulb at a position (d, 2) on the ceiling.
light2 = inline(vectorize('150/(4*pi*((x - d)"2 + (y - 2)"2 +
3"2))'), 'x', 'y', 'd')
light2 =
Inline function:
light2(x,y,d) = 150./(4.*pi.*((x - d).~2 + (y - 2).~2 +
3.ë2))
Let's get an idea of the illumination pattern if we put one light at d = 3 and the other at d = 7. We specify the drawing of 20 contours in this and the following plots.
[X,Y] = meshgrid(0:0.1:10, 0:0.1:4); contourf(light2(X, Y, 3)
+ light2(X, Y, 7), 20); axis equal tight
The floor is more evenly lit than with one bulb, but it looks as if the bulbs are closer together than they should be. If we move the bulbs further apart, the center of the room will get dimmer but the corners will get brigher. Let's try changing the location of the lights to d = 2 and d = 8.
contourf(light2(X, Y, 2) + light2(X, Y, 8), 20); axis equal tight 140 Chapter 9: Applications
This is an improvement. The corners are still the darkest spots of the room, though the light intensity along the walls toward the middle of the room (near x = 5) is diminishing as we move the bulbs further apart. To better illuminate the darkest spots we should keep moving the bulbs apart. Let's try lights at d = 1 and d = 9.
contourf(light2(X, Y, 1) + light2(X, Y, 9), 20); axis equal tight
Looking along the long walls, the room is now darker toward the middle than at the corners. This indicates that we have spread the lights too far apart.
We could proceed with further contour plots, but instead let's be systematic about finding the best position for the lights. In general, we can put one light at x = d and the other symmetrically at x = 10 — d for d between 0 and 5. Judging from the examples above, the darkest spots will be Illuminating a Room 141
either at the corners or at the midpoints of the two long walls. By symmetry, the intensity will be the same at all four corners, so let's graph the intensity at one of the corners (0, 0) as a function of d.
d = 0:0.1:5; plot(d, light2(0, 0, d) + light2(0, 0, 10 - d))
As expected, the smaller d is, the brighter the corners are. In contrast, the graph for the intensity at the midpoint (5, 0) of a long wall (again by symmetry it does not matter which of the two long walls we choose) should grow as d increases toward 5.
plot(d, light2(5, 0, d) + light2(5, 0, 10 - d))
We are after the value of d for which the lower of the two numbers on the above graphs (corresponding to the darkest spot in the room) is as high as possible. We can find this value by showing both curves on one graph. 142 Chapter 9: Applications
hold on; plot(d, light2(0, 0, d) + light2(0, 0, 10 - d)); hold off
The optimal value of d is at the point of intersection, near 1.4, with minimum intensity a little under 1. To get the optimum value of d, we find exactly where the two curves intersect.
Previous << 1 .. 42 43 44 45 46 47 < 48 > 49 50 51 52 53 54 .. 91 >> Next | 1,551 | 5,469 | {"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} | 4.375 | 4 | CC-MAIN-2021-49 | latest | en | 0.859593 |
http://forums.wolfram.com/mathgroup/archive/1999/Aug/msg00250.html | 1,527,384,958,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00458.warc.gz | 110,694,190 | 7,112 | Re: PolynomialQ (wrong) behavior ?
• To: mathgroup at smc.vnet.net
• Subject: [mg19231] Re: PolynomialQ (wrong) behavior ?
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Wed, 11 Aug 1999 02:06:47 -0400
• References: <7oojic\$i2c@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Drago Ganic <drago.ganic at in2.hr> wrote in message
news:7oojic\$i2c at smc.vnet.net...
> Hi,
>
> f = 2x +1
>
> PolynomialQ [ f, x ]
> True
>
> That's OK. But why the following
>
> PolynomialQ [ f, y ]
> True
>
> PolynomialQ [ f, Sin[x] ]
> True
>
> Greetings,
> Drago Ganic
>
Drago:
Formally, f is a polynomial in Sin[x] with constant term f.
Coefficient[ 2x + 1, Sin[x] , 0]
1 + 2 x
CoefficientList[ 2x + 1, Sin[x] ]
{1 + 2 x}
Also note
Series[2x + 1 + 3(Sin[x] + 1)^4, { Sin[x], 0, 3}] // Normal
4 + 2*x + 12*Sin[x] + 18*Sin[x]^2 + 12*Sin[x]^3
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
```
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• Next by thread: Re: PolynomialQ (wrong) behavior ? | 464 | 1,252 | {"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-2018-22 | latest | en | 0.698398 |
https://www.rainbowresource.com/product/sku/061843 | 1,591,513,757,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00533.warc.gz | 834,371,523 | 27,463 | # Addition and Subtraction with Cuisenaire Rods Book
# 061843
Our Price: \$20.95
In Stock.
Qty:
Qty:
Item #: 061843 9780740600852 K-2
#### Product Description:
This 208-page resource contains complete lesson plans for a systematic presentation of topics such as inequalities, addition with 1-digit numbers, missing addends, and subtraction with 1-digit numbers using the rods. Using appealing and enjoyable activities, students are led from concrete experience to pictorial representation, then to activities with numbers. Includes blackline masters of activity pages for the student.
#### Publisher Description:
By Patricia S. Davidson. In carefully sequenced steps, from initial concrete experiences to pictorial interpretations, students will learn about inequalities, addition and subtraction with one-digit numbers, and missing addends. This completely revised edition includes blackline master activity pages to give students extra practice. 204 pages. Grades K2.
Category Description for Cuisenaire Rods & Activity Books:
Before a child can grasp the abstract ideas needed for computational skills, the concrete foundation must be laid. These colorful rods, based on a 1 cm. unit, accurately portray the volume of a number. They encourage tactile manipulation of numbers and model-building, patterning and spatial discovery. Rods are flexible - they can be used to model almost any mathematical concept and develop a wide variety of skills.
Cuisenaire Rods sparked our childrens interest in arithmetic through lively investigation and visual and tangible confirmation of math concepts. I first read about the rods in the 1980s. I liked them so much that I used them with all three of my children when they were young and in their early elementary years. They all started out by learning the Cuisenaire poem by heart to identify the numerical value of each rod. The rods were very handy to teach concepts and beginning operations of mathematics. I found them particularly helpful with multiplication and fractions. Oh, we did our share of making sets with dried beans, forks, sea shells, and buttons, but it has been the use of the colorful Cuisenaire Rods that has come to my rescue most conveniently and logically to demonstrate what is on paper in black and white!
## For custom cuisenaire rod kits for schools, homeschool groups, or cuisenaire rod class packages please contact customer service.
Primary Subject
Mathematics
K
2
ISBN
9780740600852
Format
Paperback
Brand Name
Cuisenaire
Weight
1.35 (lbs.)
Dimensions
11.0" x 9.0" x 0.5"
Browse 1 question Browse 1 question and 8 answers
Why did you choose this?
Rainbow Resource Center Store
My children enjoy playing informally with Cuisenaire Rods, this will be a gentle introduction for my kindergartner to Addition and Subtraction.
on May 31, 2020
Reinforce math facts
on Jul 1, 2019
My children enjoy playing informally with Cuisenaire Rods, this will be a gentle introduction for my kindergartner to Addition and Subtraction.
on May 31, 2020
This is a great way to teah math with cusenaire rods
on Sep 21, 2019
This is a great way to teah math with cusenaire rods
on Sep 21, 2019
Cuisenaire rods are the best math manipulatives I have found! This book gives guidance on how to use them the most effectively
on May 27, 2019
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Well made. . I used to fit cuisinere rods.
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2 months ago
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This topic is 4134 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
*Updated* Dear All, I've currently got a sphere, which I'm trying to rotate with some GUI controls. I've got a dial to rotate around the Z axis, and I've got a little joystick which you can drag around to rotate in the X,Y axis. The first thing I did to try and do this was to just create a rotation matrix. However hte problem with that is when I rotate in Z axis, the other two axis rotate, so when the user moves the joystick right, the sphere no longe rotates right, but up. To fix this I decided I needed to rotate around an arbitrary axis. So using the joystick position I create an axis and rotate around that. That works exactly like I want it two with two problems. First, how do I add the Z rotation component to this? Secondly, whenever the joystick crosses the origin, the sphere spins around before starting to rotate in the correct way. Below is the code I'm using : Vector3 axisx; Quaternion final; double kX=mKnobX-mCompassX;//How far the tip of the joystick is from its origin double kY=mKnobY-mCompassY; double kZ=0; if(kX!=0&&kY!=0)//Prevent Division by Zero { axisx.x=-kY/sqrt(kX*kX+kY*kY+kZ*kZ); axisx.y=kX/sqrt(kX*kX+kY*kY+kZ*kZ); axisx.z=kZ/sqrt(kX*kX+kY*kY+kZ*kZ); final.SetAxisAndRotation(axisx,mCameraRotate); double xx = final.v.x * final.v.x; double xy = final.v.x * final.v.y; double xz = final.v.x * final.v.z; double xw = final.v.x * final.s; double yy = final.v.y * final.v.y; double yz = final.v.y * final.v.z; double yw = final.v.y * final.s; double zz = final.v.z * final.v.z; double zw = final.v.z * final.s; mInternalCamera[0] = 1 - 2 * ( yy + zz ); mInternalCamera[1] = 2 * ( xy - zw ); mInternalCamera[2] = 2 * ( xz + yw ); mInternalCamera[4] = 2 * ( xy + zw ); mInternalCamera[5] = 1 - 2 * ( xx + zz ); mInternalCamera[6] = 2 * ( yz - xw ); mInternalCamera[8] = 2 * ( xz - yw ); mInternalCamera[9] = 2 * ( yz + xw ); mInternalCamera[10] = 1 - 2 * ( xx + yy ); mInternalCamera[3] = mInternalCamera[7] = mInternalCamera[11] = mInternalCamera[12] = mInternalCamera[13] = mInternalCamera[14] = 0; mInternalCamera[15] = 1; } return mInternalCamera; Maybe I'm approching this task from the wrong angle? Jesse [Edited by - laeuchli on June 26, 2007 2:53:46 PM]
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× | 816 | 2,586 | {"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-2018-43 | latest | en | 0.772193 |
https://programs.programmingoneonone.com/2021/07/hackerrank-lucky-numbers-problem-soluton.html | 1,669,935,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00445.warc.gz | 488,568,531 | 35,317 | # HackerRank Lucky Numbers problem solution
In this HackerRank Lucky Numbers problem solution, A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits, is a prime number. How many numbers between A and B inclusive, are lucky?
## Problem solution in Java.
```import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
public class Solution {
/*
* Complete the luckyNumbers function below.
*/
static long luckyNumbers(long a, long b) {
return Method1.luckyNumbers(a, b);
}
static class Method1 {
static final int max_bit = 19;
static final int max_digit_sum = 9 * 18 + 1;
static final int max_squre_digit_sum = 9 * 9 * 18 + 1;
static long[][][] sum = new long[max_bit][max_digit_sum][max_squre_digit_sum];
static long[][][] left_sum = new long[max_bit][max_digit_sum][max_squre_digit_sum];
static boolean[] is_prime = new boolean[max_squre_digit_sum];
static long[] bit_sum = new long[max_bit];
static int tot = 231, max1, max2;
static int[] prime = new int[tot];
static {
Arrays.fill(is_prime, true);
int ct = 0;
is_prime[0] = is_prime[1] = false;
for (int i = 2; i < max_squre_digit_sum; i++)
if (is_prime[i]) {
for (int j = i + i; j < max_squre_digit_sum; j += i)
is_prime[j] = false;
prime[ct++] = i;
if (i < max_digit_sum)
max1 = Math.max(max1, i);
max2 = Math.max(max2, i);
}
sum[0][0][0] = 1;
for (int i = 0; prime[i] <= max1; i++)
for (int j = 0; j < tot && prime[j] <= max2; j++) {
left_sum[0][prime[i]][prime[j]] += 1;
}
for (int i = 0; i < max_bit - 1; i++) {
for (int next = 0; next < 10; next++) {
for (int j = 0; j + next < max_digit_sum; j++)
for (int k = 0; k + next * next < max_squre_digit_sum; k++) {
sum[i + 1][j + next][k + next * next] += sum[i][j][k];
if (next > 0 && is_prime[j + next] && is_prime[k + next * next])
bit_sum[i + 1] += sum[i][j][k];
}
for (int j = next; j < max_digit_sum; j++)
for (int k = next * next; k < max_squre_digit_sum; k++) {
left_sum[i + 1][j - next][k - next * next] += left_sum[i][j][k];
}
}
}
}
static long luckyNumbers(long a, long b) {
return go(b) - go(a - 1);
}
static long go(long N) {
if (N == 0)
return 0;
long ret = 0;
boolean first = true;
int pre_digit_sum = 0, pre_sdigit_sum = 0;
for (int i = 19; i > 0; i--) {
int bit = get_bit(N, i - 1);
int least;
if (bit != 0 && first) {
least = 1;
first = false;
for (int j = 1; j < i; j++)
ret += bit_sum[j];
} else {
least = 0;
}
for (int nbit = least; nbit < bit; nbit++) {
int digit_sum = pre_digit_sum + nbit;
int sdigit_sum = pre_sdigit_sum + nbit * nbit;
ret += left_sum[i - 1][digit_sum][sdigit_sum];
}
pre_digit_sum += bit;
pre_sdigit_sum += bit * bit;
}
if (is_prime[pre_digit_sum] && is_prime[pre_sdigit_sum])
ret += 1;
return ret;
}
static int get_bit(long n, int m) {
for (int i = 0; i < m; i++)
n /= 10;
return (int) (n % 10);
}
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int t = Integer.parseInt(scanner.nextLine().trim());
for (int tItr = 0; tItr < t; tItr++) {
String[] ab = scanner.nextLine().split(" ");
long a = Long.parseLong(ab[0].trim());
long b = Long.parseLong(ab[1].trim());
long result = luckyNumbers(a, b);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
}
bufferedWriter.close();
}
}```
## Problem solution in C++.
```#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<list>
#include<queue>
#include<set>
using namespace std;
typedef vector<int> VI;
typedef long long LL;
#define FOR(x, b, e) for(int x=b; x<=(e); ++x)
#define FORD(x, b, e) for(int x=b; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define VAR(v,n) __typeof(n) v=(n)
#define ALL(c) c.begin(),c.end()
#define SIZE(x) (int)x.size()
#define FOREACH(i,c) for(VAR(i,(c).begin());i!=(c).end();++i)
#define PB push_back
#define ST first
#define ND second
const int MILION=1000*1000;
const int MAX_SUMA=200;
const int MAX_CYFR=20;
LL pot[MAX_CYFR];
bool czyPierwsza(int k){
if (k<2) return false;
int i=2;
while (i*i<=k){
if (k%i==0) return false;
i++;
}
return true;
}
void ustalPot(){
pot[0]=1;
FOR(i,1,MAX_CYFR-1){
pot[i]=pot[i-1]*10;
}
}
void ustalPierwsze(){
FOR(i,2,MAX_SUMA-2){
if (czyPierwsza(i) && czyPierwsza(j)){
pierwsza[i][j]=true;
}
}
}
}
void ustalW(){
REP(i,MAX_SUMA){
if (pierwsza[i][j]){
w[i][j][0]++;
}
}
}
FOR(p,1,MAX_CYFR-1){
REP(i,MAX_SUMA-2){
REP(k,10){
if (i+k<MAX_SUMA-2 && j+k*k < MAX_KWADRAT-2){
w[i][j][p]+=w[i+k][j+k*k][p-1];
}
}
}
}
}
}
LL daj(int c, int k, LL T){
if (T<1){
return pierwsza[c][k];
}
LL wyn=0;
int wyk=0;
int cyf=1;
while (pot[wyk]<=T) wyk++;
wyk--;
while (cyf*pot[wyk]<=T) cyf++;
cyf--;
wyn+=daj(c+cyf,k+cyf*cyf,T-cyf*pot[wyk]);
if (wyk>-1){
FOR(i,0,cyf-1){
wyn+=w[c+i][k+i*i][wyk];
}
}
return wyn;
}
int main(){
ustalPot();
ustalPierwsze();
ustalW();
int t;
LL a,b;
scanf("%d",&t);
REP(i,t){
scanf("%lld%lld",&a,&b);
printf("%lld\n",daj(0,0,b)-daj(0,0,a-1));
}
return 0;
}```
## Problem solution in C.
```/* Enter your code here. Read input from STDIN. Print output to STDOUT */
#include <stdio.h>
#include <string.h>
int isprime[10000];
unsigned long long dp[19][2*82][2*730];
unsigned long long ans[19][10][164][1460];
int start_sum_sqr[19][163];
int end_sum_sqr[19][163];
void prime(){
memset(isprime,0,10000*4);
isprime[0]=1;
isprime[1]=1;
for (int i = 2; i < 10000; i += 1){
if(isprime[i]==0){
for (int j = i*i; j < 10000; j += i){
isprime[j]=1;
}
}
}
}
void inc(int i,int j,int k, int val){
if(i<19 && j<164 && k< 1460)
dp[i][j][k] += val;
}
unsigned long long setDP(){
memset(dp, 0, sizeof(dp[0][0][0])*19*82*730);
dp[0][0][0]=1;
for (int i = 0; i < 18; ++i) {
for (int j = 0; j <= 9 * i; ++j) {
for (int k = 0; k <= 9 * 9 * i; ++k) {
for (int l = 0; l < 10; ++l) {
dp[i + 1][j + l][k + l*l] += dp[i][j][k];
}
}
}
}
}
unsigned long long solve(unsigned long long num){
if(num<=0)return 0;
int digs[20],len=0;
while(num){
digs[len++]=(num%10);
num=num/10;
}
int sum=0,sqr_sum=0;
unsigned long long ret=0;
for (int i = len-1; i >= 0; i -= 1){
int dig = digs[i];
for (int j = 0; j < dig; j += 1){
if(ans[i][j][sum][sqr_sum]!=0){
ret+=ans[i][j][sum][sqr_sum];
continue;
}
unsigned long long x=0;
for (int k = 0; k <= 9*i; k += 1){
if(isprime[k+sum+j]) continue;
for (int l = start_sum_sqr[i][k]; l <= end_sum_sqr[i][k]; l += 1){
if(isprime[l+ j*j + sqr_sum]==0) x+=dp[i][k][l];
}
}
ans[i][j][sum][sqr_sum] = x;
ret+=x;
}
sum+=dig;
sqr_sum+=(dig*dig);
}
if(isprime[sum]==0 && isprime[sqr_sum]==0) ret++;
return ret;
}
void set_sqrs(){
for (int i = 0; i < 19; i += 1){
for (int j = 0; j < 163; j += 1){
for (int k = 0; k < 1460; k += 1){
if(dp[i][j][k]!=0){
start_sum_sqr[i][j]=k;
break;
}
}
for (int k = 1459; k>=0; k -= 1){
if(dp[i][j][k]!=0){
end_sum_sqr[i][j]=k;
break;
}
}
}
}
}
int main(){
prime();
setDP();
set_sqrs();
unsigned long long tc,a,b;
scanf("%lld",&tc);
while(tc--){
scanf("%lld %lld", &a, &b);
if(b == 1000000000000000000ll)b--;
printf("%lld\n", solve(b) - solve(a-1));
}
return 0;
}``` | 2,590 | 7,161 | {"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} | 3.828125 | 4 | CC-MAIN-2022-49 | longest | en | 0.341605 |
http://mathhelpforum.com/differential-geometry/133711-sequences.html | 1,529,419,524,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863043.35/warc/CC-MAIN-20180619134548-20180619154548-00558.warc.gz | 205,801,638 | 9,782 | 1. ## Sequences
For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$
Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.
$\displaystyle {^\infty}$
$\displaystyle \bigcap {I_n}{\neq}{\oslash}$
$\displaystyle _{i=1}$
2. Originally Posted by yusukered07
For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$
Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.
$\displaystyle {^\infty}$
$\displaystyle \bigcap {I_n}{\neq}{\oslash}$
$\displaystyle _{i=1}$
This is a sequence of closed subsets of $\displaystyle [a_1,b_1]$ which have the FIP, apply $\displaystyle [a_1,b_1]$'s compactness.
3. Originally Posted by yusukered07
For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$
Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.
$\displaystyle {^\infty}$
$\displaystyle \bigcap {I_n}{\neq}{\oslash}$
$\displaystyle _{i=1}$
Since $\displaystyle \{a_{n}\}_{n\in\mathbb{N}}$ is nondecreasing and $\displaystyle \{b_{n}\}_{n\in\mathbb{N}}$ is increasing,
we have $\displaystyle a:=\lim\nolimits_{n\to\infty}a_{n}$ and $\displaystyle b:=\lim\nolimits_{n\to\infty}b_{n}$.
Then $\displaystyle (a+b)/2\in[a,b]=\cap_{n\in\mathbb{N}}[a_{n},b_{n}]$. | 627 | 1,823 | {"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} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.638581 |
https://www.enotes.com/homework-help/sin-2-x-cos-x-cos-2-x-sin-x-tan-2-x-cot-x-cot-2-x-270974 | 1,484,779,283,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00554-ip-10-171-10-70.ec2.internal.warc.gz | 898,285,530 | 12,446 | # If sin (π/2 +x) = cos x cos (π/2 + x) = -sin x tan (π/2 + x) = -cot x cot (π/2 + x) = -tan x sec (π/2 + x) = ? cosec (π/2 + x) = ? This is a question from Trigonometric functions and is...
If
sin (π/2 +x) = cos x
cos (π/2 + x) = -sin x
tan (π/2 + x) = -cot x
cot (π/2 + x) = -tan x
sec (π/2 + x) = ?
cosec (π/2 + x) = ?
This is a question from Trigonometric functions and is related to deriving the formulae.
beckden | High School Teacher | (Level 1) Educator
Posted on
Since `sec(x) = 1/(cos(x))` and `csc(x) = 1/(sin(x))`
`sec(pi/2 + x) = 1/(cos(pi/2 + x)) = 1/(-sin(x)) = -1/(sin(x)) = -csc(x)`
`csc(pi/2 + x) = 1/(sin(pi/2 + x)) = 1/(cos(x)) = sec(x)`
so
`sec(pi/2 + x) = -csc(x)`
`csc(pi/2 + x) = sec(x)` | 329 | 729 | {"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-2017-04 | longest | en | 0.633268 |
https://tptp.org/cgi-bin/SeeTPTP?Category=Solutions&Domain=SEV&File=SEV261%5E5&System=Satallax---3.5.THM-Prf.s | 1,652,836,007,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00339.warc.gz | 655,454,585 | 6,959 | ## TSTP Solution File: SEV261^5 by Satallax---3.5
View Problem - Process Solution
```%------------------------------------------------------------------------------
% File : Satallax---3.5
% Problem : SEV261^5 : TPTP v7.5.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : satallax -E eprover-ho -P picomus -M modes -p tstp -t %d %s
% Computer : n020.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% DateTime : Tue Mar 30 22:05:13 EDT 2021
% Result : Theorem 26.02s
% Output : Proof 26.02s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 179
% Syntax : Number of formulae : 197 ( 23 unt; 13 typ; 12 def)
% Number of atoms : 713 ( 154 equ; 0 cnn)
% Maximal formula atoms : 12 ( 3 avg)
% Number of connectives : 722 ( 301 ~; 102 |; 0 &; 104 @)
% ( 81 <=>; 134 =>; 0 <=; 0 <~>)
% Maximal formula depth : 13 ( 3 avg)
% Number of types : 2 ( 1 usr)
% Number of type conns : 67 ( 67 >; 0 *; 0 +; 0 <<)
% Number of symbols : 98 ( 96 usr; 89 con; 0-2 aty)
% Number of variables : 182 ( 127 ^ 55 !; 0 ?; 182 :)
%------------------------------------------------------------------------------
thf(ty_a,type,
a: \$tType ).
thf(ty_eigen__14,type,
eigen__14: a > \$o ).
thf(ty_eigen__6,type,
eigen__6: a > \$o ).
thf(ty_eigen__2,type,
eigen__2: a > \$o ).
thf(ty_eigen__7,type,
eigen__7: a ).
thf(ty_eigen__1,type,
eigen__1: a > \$o ).
thf(ty_eigen__0,type,
eigen__0: a > \$o ).
thf(ty_eigen__4,type,
eigen__4: a ).
thf(ty_eigen__5,type,
eigen__5: ( a > \$o ) > \$o ).
thf(ty_eigen__11,type,
eigen__11: a > \$o ).
thf(ty_eigen__3,type,
eigen__3: a ).
thf(ty_eigen__8,type,
eigen__8: a ).
thf(ty_eigen__18,type,
eigen__18: a > \$o ).
thf(h0,assumption,
! [X1: ( a > \$o ) > \$o,X2: a > \$o] :
( ( X1 @ X2 )
=> ( X1 @ ( eps__0 @ X1 ) ) ),
introduced(assumption,[]) ).
thf(eigendef_eigen__11,definition,
( eigen__11
= ( eps__0
@ ^ [X1: a > \$o] :
~ ( ( X1
= ( ^ [X2: a] : \$false ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__11])]) ).
thf(h1,assumption,
! [X1: a > \$o,X2: a] :
( ( X1 @ X2 )
=> ( X1 @ ( eps__1 @ X1 ) ) ),
introduced(assumption,[]) ).
thf(eigendef_eigen__3,definition,
( eigen__3
= ( eps__1
@ ^ [X1: a] :
( ( eigen__2 @ X1 )
!= \$false ) ) ),
introduced(definition,[new_symbols(definition,[eigen__3])]) ).
thf(eigendef_eigen__1,definition,
( eigen__1
= ( eps__0
@ ^ [X1: a > \$o] :
~ ! [X2: a > \$o] :
( ~ ( ~ ( ( ( eigen__0
!= ( ^ [X3: a] : \$false ) )
=> ( eigen__0
= ( ^ [X3: a] : ~ \$false ) ) )
=> ~ ( ( X1
!= ( ^ [X3: a] : \$false ) )
=> ( X1
= ( ^ [X3: a] : ~ \$false ) ) ) )
=> ( X2
!= ( ^ [X3: a] :
~ ( ( eigen__0 @ X3 )
=> ~ ( X1 @ X3 ) ) ) ) )
=> ( ( X2
!= ( ^ [X3: a] : \$false ) )
=> ( X2
= ( ^ [X3: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__1])]) ).
thf(eigendef_eigen__6,definition,
( eigen__6
= ( eps__0
@ ^ [X1: a > \$o] :
~ ( ~ ( ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ( ( X2
!= ( ^ [X3: a] : \$false ) )
=> ( X2
= ( ^ [X3: a] : ~ \$false ) ) ) )
=> ( X1
!= ( ^ [X2: a] :
~ ! [X3: a > \$o] :
( ( eigen__5 @ X3 )
=> ~ ( X3 @ X2 ) ) ) ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__6])]) ).
thf(eigendef_eigen__14,definition,
( eigen__14
= ( eps__0
@ ^ [X1: a > \$o] :
~ ( ( X1
= ( ^ [X2: a] : ~ \$false ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__14])]) ).
thf(eigendef_eigen__0,definition,
( eigen__0
= ( eps__0
@ ^ [X1: a > \$o] :
~ ! [X2: a > \$o,X3: a > \$o] :
( ~ ( ~ ( ( ( X1
!= ( ^ [X4: a] : \$false ) )
=> ( X1
= ( ^ [X4: a] : ~ \$false ) ) )
=> ~ ( ( X2
!= ( ^ [X4: a] : \$false ) )
=> ( X2
= ( ^ [X4: a] : ~ \$false ) ) ) )
=> ( X3
!= ( ^ [X4: a] :
~ ( ( X1 @ X4 )
=> ~ ( X2 @ X4 ) ) ) ) )
=> ( ( X3
!= ( ^ [X4: a] : \$false ) )
=> ( X3
= ( ^ [X4: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__0])]) ).
thf(eigendef_eigen__18,definition,
( eigen__18
= ( eps__0
@ ^ [X1: a > \$o] :
~ ( ( eigen__5 @ X1 )
=> ~ ( X1 @ eigen__7 ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__18])]) ).
thf(eigendef_eigen__8,definition,
( eigen__8
= ( eps__1
@ ^ [X1: a] :
( ( eigen__6 @ X1 )
!= ( ~ \$false ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__8])]) ).
thf(eigendef_eigen__2,definition,
( eigen__2
= ( eps__0
@ ^ [X1: a > \$o] :
~ ( ~ ( ~ ( ( ( eigen__0
!= ( ^ [X2: a] : \$false ) )
=> ( eigen__0
= ( ^ [X2: a] : ~ \$false ) ) )
=> ~ ( ( eigen__1
!= ( ^ [X2: a] : \$false ) )
=> ( eigen__1
= ( ^ [X2: a] : ~ \$false ) ) ) )
=> ( X1
!= ( ^ [X2: a] :
~ ( ( eigen__0 @ X2 )
=> ~ ( eigen__1 @ X2 ) ) ) ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__2])]) ).
thf(eigendef_eigen__4,definition,
( eigen__4
= ( eps__1
@ ^ [X1: a] :
( ( eigen__2 @ X1 )
!= ( ~ \$false ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__4])]) ).
thf(eigendef_eigen__7,definition,
( eigen__7
= ( eps__1
@ ^ [X1: a] :
( ( eigen__6 @ X1 )
!= \$false ) ) ),
introduced(definition,[new_symbols(definition,[eigen__7])]) ).
thf(h2,assumption,
! [X1: ( ( a > \$o ) > \$o ) > \$o,X2: ( a > \$o ) > \$o] :
( ( X1 @ X2 )
=> ( X1 @ ( eps__2 @ X1 ) ) ),
introduced(assumption,[]) ).
thf(eigendef_eigen__5,definition,
( eigen__5
= ( eps__2
@ ^ [X1: ( a > \$o ) > \$o] :
~ ! [X2: a > \$o] :
( ~ ( ! [X3: a > \$o] :
( ( X1 @ X3 )
=> ( ( X3
!= ( ^ [X4: a] : \$false ) )
=> ( X3
= ( ^ [X4: a] : ~ \$false ) ) ) )
=> ( X2
!= ( ^ [X3: a] :
~ ! [X4: a > \$o] :
( ( X1 @ X4 )
=> ~ ( X4 @ X3 ) ) ) ) )
=> ( ( X2
!= ( ^ [X3: a] : \$false ) )
=> ( X2
= ( ^ [X3: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[eigen__5])]) ).
thf(sP1,plain,
( sP1
<=> ( ~ ( ! [X1: a > \$o] :
( ( X1
= ( ^ [X2: a] : \$false ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) )
=> ~ ! [X1: a > \$o] :
( ( X1
= ( ^ [X2: a] : ~ \$false ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) )
=> ~ ! [X1: ( a > \$o ) > \$o,X2: a > \$o] :
( ~ ( ! [X3: a > \$o] :
( ( X1 @ X3 )
=> ( ( X3
!= ( ^ [X4: a] : \$false ) )
=> ( X3
= ( ^ [X4: a] : ~ \$false ) ) ) )
=> ( X2
!= ( ^ [X3: a] :
~ ! [X4: a > \$o] :
( ( X1 @ X4 )
=> ~ ( X4 @ X3 ) ) ) ) )
=> ( ( X2
!= ( ^ [X3: a] : \$false ) )
=> ( X2
= ( ^ [X3: a] : ~ \$false ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP1])]) ).
thf(sP2,plain,
( sP2
<=> ! [X1: a > \$o] :
( ( eigen__5 @ X1 )
=> ~ ( X1 @ eigen__7 ) ) ),
introduced(definition,[new_symbols(definition,[sP2])]) ).
thf(sP3,plain,
( sP3
<=> ( ( eigen__6 @ eigen__7 )
= \$false ) ),
introduced(definition,[new_symbols(definition,[sP3])]) ).
thf(sP4,plain,
( sP4
<=> ( eigen__18
= ( ^ [X1: a] : ~ \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP4])]) ).
thf(sP5,plain,
( sP5
<=> ( eigen__0
= ( ^ [X1: a] : ~ \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP5])]) ).
thf(sP6,plain,
( sP6
<=> ! [X1: a > \$o] :
( ~ ( ~ ( ( ( eigen__0
!= ( ^ [X2: a] : \$false ) )
=> sP5 )
=> ~ ( ( eigen__1
!= ( ^ [X2: a] : \$false ) )
=> ( eigen__1
= ( ^ [X2: a] : ~ \$false ) ) ) )
=> ( X1
!= ( ^ [X2: a] :
~ ( ( eigen__0 @ X2 )
=> ~ ( eigen__1 @ X2 ) ) ) ) )
=> ( ( X1
!= ( ^ [X2: a] : \$false ) )
=> ( X1
= ( ^ [X2: a] : ~ \$false ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP6])]) ).
thf(sP7,plain,
( sP7
<=> ( ( eigen__2 @ eigen__4 )
= ( ~ \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP7])]) ).
thf(sP8,plain,
( sP8
<=> ( eigen__2 @ eigen__3 ) ),
introduced(definition,[new_symbols(definition,[sP8])]) ).
thf(sP9,plain,
( sP9
<=> ( ( eigen__2
!= ( ^ [X1: a] : \$false ) )
=> ( eigen__2
= ( ^ [X1: a] : ~ \$false ) ) ) ),
introduced(definition,[new_symbols(definition,[sP9])]) ).
thf(sP10,plain,
( sP10
<=> ( ( eigen__0 @ eigen__3 )
=> ~ ( eigen__1 @ eigen__3 ) ) ),
introduced(definition,[new_symbols(definition,[sP10])]) ).
thf(sP11,plain,
( sP11
<=> ( eigen__11
= ( ^ [X1: a] : \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP11])]) ).
thf(sP12,plain,
( sP12
<=> ( eigen__6 @ eigen__7 ) ),
introduced(definition,[new_symbols(definition,[sP12])]) ).
thf(sP13,plain,
( sP13
<=> ( eigen__1 @ eigen__4 ) ),
introduced(definition,[new_symbols(definition,[sP13])]) ).
thf(sP14,plain,
( sP14
<=> ( eigen__5 @ eigen__18 ) ),
introduced(definition,[new_symbols(definition,[sP14])]) ).
thf(sP15,plain,
( sP15
<=> ! [X1: a] :
( ( eigen__2 @ X1 )
= ( ~ \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP15])]) ).
thf(sP16,plain,
( sP16
<=> ( eigen__6
= ( ^ [X1: a] : \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP16])]) ).
thf(sP17,plain,
( sP17
<=> ( eigen__1
= ( ^ [X1: a] : ~ \$false ) ) ),
introduced(definition,[new_symbols(definition,[sP17])]) ).
thf(sP18,plain,
( sP18
<=> ! [X1: a > \$o] :
( ( eigen__5 @ X1 )
=> ~ ( X1 @ eigen__8 ) ) ),
introduced(definition,[new_symbols(definition,[sP18])]) ).
thf(sP19,plain,
( sP19
<=> ( eigen__6 @ eigen__8 ) ),
introduced(definition,[new_symbols(definition,[sP19])]) ).
thf(sP20,plain,
( sP20
<=> ! [X1: ( a > \$o ) > \$o,X2: a > \$o] :
( ~ ( ! [X3: a > \$o] :
( ( X1 @ X3 )
=> ( ( X3
!= ( ^ [X4: a] : \$false ) )
=> ( X3
= ( ^ [X4: a] : ~ \$false ) ) ) )
=> ( X2
!= ( ^ [X3: a] :
~ ! [X4: a > \$o] :
( ( X1 @ X4 )
=> ~ ( X4 @ X3 ) ) ) ) )
=> ( ( X2
!= ( ^ [X3: a] : \$false ) )
=> ( X2
= ( ^ [X3: a] : ~ \$false ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP20])]) ).
thf(sP21,plain,
( sP21
<=> \$false ),
introduced(definition,[new_symbols(definition,[sP21])]) ).
thf(sP22,plain,
( sP22
<=> ( eigen__6
= ( ^ [X1: a] : ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP22])]) ).
thf(sP23,plain,
( sP23
<=> ! [X1: a] :
( ( eigen__2 @ X1 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP23])]) ).
thf(sP24,plain,
( sP24
<=> ( eigen__2 @ eigen__4 ) ),
introduced(definition,[new_symbols(definition,[sP24])]) ).
thf(sP25,plain,
( sP25
<=> ( sP13
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP25])]) ).
thf(sP26,plain,
( sP26
<=> ( ( eigen__0 @ eigen__3 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP26])]) ).
thf(sP27,plain,
( sP27
<=> ! [X1: a] :
( ( eigen__6 @ X1 )
= ( ~ ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ~ ( X2 @ X1 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP27])]) ).
thf(sP28,plain,
( sP28
<=> ( eigen__0
= ( ^ [X1: a] : sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP28])]) ).
thf(sP29,plain,
( sP29
<=> ! [X1: a] :
( ( eigen__0 @ X1 )
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP29])]) ).
thf(sP30,plain,
( sP30
<=> ! [X1: a > \$o] :
( ~ ( ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ( ( X2
!= ( ^ [X3: a] : sP21 ) )
=> ( X2
= ( ^ [X3: a] : ~ sP21 ) ) ) )
=> ( X1
!= ( ^ [X2: a] :
~ ! [X3: a > \$o] :
( ( eigen__5 @ X3 )
=> ~ ( X3 @ X2 ) ) ) ) )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP30])]) ).
thf(sP31,plain,
( sP31
<=> ( sP19
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP31])]) ).
thf(sP32,plain,
( sP32
<=> ! [X1: a] :
( ( eigen__18 @ X1 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP32])]) ).
thf(sP33,plain,
( sP33
<=> ( sP8 = sP21 ) ),
introduced(definition,[new_symbols(definition,[sP33])]) ).
thf(sP34,plain,
( sP34
<=> ( ~ sP16
=> sP22 ) ),
introduced(definition,[new_symbols(definition,[sP34])]) ).
thf(sP35,plain,
( sP35
<=> ( eigen__0 @ eigen__4 ) ),
introduced(definition,[new_symbols(definition,[sP35])]) ).
thf(sP36,plain,
( sP36
<=> ( ( eigen__1 @ eigen__3 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP36])]) ).
thf(sP37,plain,
( sP37
<=> ( sP8
= ( ~ sP10 ) ) ),
introduced(definition,[new_symbols(definition,[sP37])]) ).
thf(sP38,plain,
( sP38
<=> ! [X1: a > \$o,X2: a > \$o] :
( ~ ( ~ ( ( ~ sP28
=> sP5 )
=> ~ ( ( X1
!= ( ^ [X3: a] : sP21 ) )
=> ( X1
= ( ^ [X3: a] : ~ sP21 ) ) ) )
=> ( X2
!= ( ^ [X3: a] :
~ ( ( eigen__0 @ X3 )
=> ~ ( X1 @ X3 ) ) ) ) )
=> ( ( X2
!= ( ^ [X3: a] : sP21 ) )
=> ( X2
= ( ^ [X3: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP38])]) ).
thf(sP39,plain,
( sP39
<=> ( sP11
=> ( ~ sP11
=> ( eigen__11
= ( ^ [X1: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP39])]) ).
thf(sP40,plain,
( sP40
<=> ( sP12
= ( ~ sP2 ) ) ),
introduced(definition,[new_symbols(definition,[sP40])]) ).
thf(sP41,plain,
( sP41
<=> ( ( eigen__18 @ eigen__7 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP41])]) ).
thf(sP42,plain,
( sP42
<=> ( ~ ( ! [X1: a > \$o] :
( ( eigen__5 @ X1 )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) )
=> ( eigen__6
!= ( ^ [X1: a] :
~ ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ~ ( X2 @ X1 ) ) ) ) )
=> sP34 ) ),
introduced(definition,[new_symbols(definition,[sP42])]) ).
thf(sP43,plain,
( sP43
<=> ( ! [X1: a > \$o] :
( ( eigen__5 @ X1 )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) )
=> ( eigen__6
!= ( ^ [X1: a] :
~ ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ~ ( X2 @ X1 ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP43])]) ).
thf(sP44,plain,
( sP44
<=> ( ( eigen__18
!= ( ^ [X1: a] : sP21 ) )
=> sP4 ) ),
introduced(definition,[new_symbols(definition,[sP44])]) ).
thf(sP45,plain,
( sP45
<=> ! [X1: a] :
( ( eigen__1 @ X1 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP45])]) ).
thf(sP46,plain,
( sP46
<=> ( eigen__0 @ eigen__3 ) ),
introduced(definition,[new_symbols(definition,[sP46])]) ).
thf(sP47,plain,
( sP47
<=> ( eigen__18 @ eigen__7 ) ),
introduced(definition,[new_symbols(definition,[sP47])]) ).
thf(sP48,plain,
( sP48
<=> ( eigen__2
= ( ^ [X1: a] : sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP48])]) ).
thf(sP49,plain,
( sP49
<=> ( eigen__6
= ( ^ [X1: a] :
~ ! [X2: a > \$o] :
( ( eigen__5 @ X2 )
=> ~ ( X2 @ X1 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP49])]) ).
thf(sP50,plain,
( sP50
<=> ( sP24
= ( ~ ( sP35
=> ~ sP13 ) ) ) ),
introduced(definition,[new_symbols(definition,[sP50])]) ).
thf(sP51,plain,
( sP51
<=> ( sP14
=> sP44 ) ),
introduced(definition,[new_symbols(definition,[sP51])]) ).
thf(sP52,plain,
( sP52
<=> ( ~ ( ( ~ sP28
=> sP5 )
=> ~ ( ( eigen__1
!= ( ^ [X1: a] : sP21 ) )
=> sP17 ) )
=> ( eigen__2
!= ( ^ [X1: a] :
~ ( ( eigen__0 @ X1 )
=> ~ ( eigen__1 @ X1 ) ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP52])]) ).
thf(sP53,plain,
( sP53
<=> ( eigen__1 @ eigen__3 ) ),
introduced(definition,[new_symbols(definition,[sP53])]) ).
thf(sP54,plain,
( sP54
<=> ( ( ~ sP28
=> sP5 )
=> ~ ( ( eigen__1
!= ( ^ [X1: a] : sP21 ) )
=> sP17 ) ) ),
introduced(definition,[new_symbols(definition,[sP54])]) ).
thf(sP55,plain,
( sP55
<=> ( eigen__5
@ ^ [X1: a] : ~ sP21 ) ),
introduced(definition,[new_symbols(definition,[sP55])]) ).
thf(sP56,plain,
( sP56
<=> ! [X1: a] :
( ( eigen__6 @ X1 )
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP56])]) ).
thf(sP57,plain,
( sP57
<=> ( sP14
=> ~ sP47 ) ),
introduced(definition,[new_symbols(definition,[sP57])]) ).
thf(sP58,plain,
( sP58
<=> ! [X1: a > \$o] :
( ( X1
= ( ^ [X2: a] : ~ sP21 ) )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP58])]) ).
thf(sP59,plain,
( sP59
<=> ( sP35
=> ~ sP13 ) ),
introduced(definition,[new_symbols(definition,[sP59])]) ).
thf(sP60,plain,
( sP60
<=> ( eigen__1
= ( ^ [X1: a] : sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP60])]) ).
thf(sP61,plain,
( sP61
<=> ( ~ sP60
=> sP17 ) ),
introduced(definition,[new_symbols(definition,[sP61])]) ).
thf(sP62,plain,
( sP62
<=> ! [X1: a > \$o,X2: a > \$o,X3: a > \$o] :
( ~ ( ~ ( ( ( X1
!= ( ^ [X4: a] : sP21 ) )
=> ( X1
= ( ^ [X4: a] : ~ sP21 ) ) )
=> ~ ( ( X2
!= ( ^ [X4: a] : sP21 ) )
=> ( X2
= ( ^ [X4: a] : ~ sP21 ) ) ) )
=> ( X3
!= ( ^ [X4: a] :
~ ( ( X1 @ X4 )
=> ~ ( X2 @ X4 ) ) ) ) )
=> ( ( X3
!= ( ^ [X4: a] : sP21 ) )
=> ( X3
= ( ^ [X4: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP62])]) ).
thf(sP63,plain,
( sP63
<=> ( sP19
= ( ~ sP18 ) ) ),
introduced(definition,[new_symbols(definition,[sP63])]) ).
thf(sP64,plain,
( sP64
<=> ( eigen__18
= ( ^ [X1: a] : sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP64])]) ).
thf(sP65,plain,
( sP65
<=> ! [X1: a] :
( ( eigen__2 @ X1 )
= ( ~ ( ( eigen__0 @ X1 )
=> ~ ( eigen__1 @ X1 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP65])]) ).
thf(sP66,plain,
( sP66
<=> ( ~ sP1
=> ~ sP62 ) ),
introduced(definition,[new_symbols(definition,[sP66])]) ).
thf(sP67,plain,
( sP67
<=> ! [X1: a] :
( ( eigen__6 @ X1 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP67])]) ).
thf(sP68,plain,
( sP68
<=> ( eigen__14
= ( ^ [X1: a] : ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP68])]) ).
thf(sP69,plain,
( sP69
<=> ( ~ sP52
=> sP9 ) ),
introduced(definition,[new_symbols(definition,[sP69])]) ).
thf(sP70,plain,
( sP70
<=> ( eigen__2
= ( ^ [X1: a] :
~ ( ( eigen__0 @ X1 )
=> ~ ( eigen__1 @ X1 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP70])]) ).
thf(sP71,plain,
( sP71
<=> ! [X1: a] :
( ( eigen__1 @ X1 )
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP71])]) ).
thf(sP72,plain,
( sP72
<=> ! [X1: a > \$o] :
( ( X1
= ( ^ [X2: a] : sP21 ) )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP72])]) ).
thf(sP73,plain,
( sP73
<=> ( ~ sP28
=> sP5 ) ),
introduced(definition,[new_symbols(definition,[sP73])]) ).
thf(sP74,plain,
( sP74
<=> ( eigen__2
= ( ^ [X1: a] : ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP74])]) ).
thf(sP75,plain,
( sP75
<=> ! [X1: a > \$o] :
( ( eigen__5 @ X1 )
=> ( ( X1
!= ( ^ [X2: a] : sP21 ) )
=> ( X1
= ( ^ [X2: a] : ~ sP21 ) ) ) ) ),
introduced(definition,[new_symbols(definition,[sP75])]) ).
thf(sP76,plain,
( sP76
<=> ( sP72
=> ~ sP58 ) ),
introduced(definition,[new_symbols(definition,[sP76])]) ).
thf(sP77,plain,
( sP77
<=> ( sP68
=> ( ( eigen__14
!= ( ^ [X1: a] : sP21 ) )
=> sP68 ) ) ),
introduced(definition,[new_symbols(definition,[sP77])]) ).
thf(sP78,plain,
( sP78
<=> ( ~ sP11
=> ( eigen__11
= ( ^ [X1: a] : ~ sP21 ) ) ) ),
introduced(definition,[new_symbols(definition,[sP78])]) ).
thf(sP79,plain,
( sP79
<=> ( ( eigen__14
!= ( ^ [X1: a] : sP21 ) )
=> sP68 ) ),
introduced(definition,[new_symbols(definition,[sP79])]) ).
thf(sP80,plain,
( sP80
<=> ! [X1: a] :
( ( eigen__0 @ X1 )
= sP21 ) ),
introduced(definition,[new_symbols(definition,[sP80])]) ).
thf(sP81,plain,
( sP81
<=> ( sP35
= ( ~ sP21 ) ) ),
introduced(definition,[new_symbols(definition,[sP81])]) ).
thf(cINDISCRETE_TOPOLOGY_pme,conjecture,
~ sP66 ).
thf(h3,negated_conjecture,
sP66,
inference(assume_negation,[status(cth)],[cINDISCRETE_TOPOLOGY_pme]) ).
thf(1,plain,
( ~ sP41
| ~ sP47
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(2,plain,
( ~ sP32
| sP41 ),
inference(all_rule,[status(thm)],]) ).
thf(3,plain,
( ~ sP64
| sP32 ),
inference(prop_rule,[status(thm)],]) ).
thf(4,plain,
( ~ sP14
| sP55
| ~ sP4 ),
inference(mating_rule,[status(thm)],]) ).
thf(5,plain,
( ~ sP75
| sP51 ),
inference(all_rule,[status(thm)],]) ).
thf(6,plain,
( ~ sP51
| ~ sP14
| sP44 ),
inference(prop_rule,[status(thm)],]) ).
thf(7,plain,
( ~ sP44
| sP64
| sP4 ),
inference(prop_rule,[status(thm)],]) ).
thf(8,plain,
( sP57
| sP47 ),
inference(prop_rule,[status(thm)],]) ).
thf(9,plain,
( sP57
| sP14 ),
inference(prop_rule,[status(thm)],]) ).
thf(10,plain,
( sP2
| ~ sP57 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__18]) ).
thf(11,plain,
( ~ sP40
| ~ sP12
| ~ sP2 ),
inference(prop_rule,[status(thm)],]) ).
thf(12,plain,
( ~ sP18
| ~ sP55 ),
inference(all_rule,[status(thm)],]) ).
thf(13,plain,
( ~ sP63
| sP19
| sP18 ),
inference(prop_rule,[status(thm)],]) ).
thf(14,plain,
( ~ sP27
| sP40 ),
inference(all_rule,[status(thm)],]) ).
thf(15,plain,
( ~ sP27
| sP63 ),
inference(all_rule,[status(thm)],]) ).
thf(16,plain,
( sP79
| ~ sP68 ),
inference(prop_rule,[status(thm)],]) ).
thf(17,plain,
( sP77
| ~ sP79 ),
inference(prop_rule,[status(thm)],]) ).
thf(18,plain,
( sP77
| sP68 ),
inference(prop_rule,[status(thm)],]) ).
thf(19,plain,
( sP58
| ~ sP77 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__14]) ).
thf(20,plain,
( sP78
| ~ sP11 ),
inference(prop_rule,[status(thm)],]) ).
thf(21,plain,
( sP39
| ~ sP78 ),
inference(prop_rule,[status(thm)],]) ).
thf(22,plain,
( sP39
| sP11 ),
inference(prop_rule,[status(thm)],]) ).
thf(23,plain,
( sP72
| ~ sP39 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__11]) ).
thf(24,plain,
( sP10
| sP53 ),
inference(prop_rule,[status(thm)],]) ).
thf(25,plain,
( sP10
| sP46 ),
inference(prop_rule,[status(thm)],]) ).
thf(26,plain,
( ~ sP37
| ~ sP8
| ~ sP10 ),
inference(prop_rule,[status(thm)],]) ).
thf(27,plain,
( ~ sP26
| ~ sP46
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(28,plain,
( ~ sP36
| ~ sP53
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(29,plain,
( ~ sP59
| ~ sP35
| ~ sP13 ),
inference(prop_rule,[status(thm)],]) ).
thf(30,plain,
( ~ sP50
| sP24
| sP59 ),
inference(prop_rule,[status(thm)],]) ).
thf(31,plain,
( ~ sP81
| sP35
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(32,plain,
( ~ sP25
| sP13
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(33,plain,
( ~ sP65
| sP37 ),
inference(all_rule,[status(thm)],]) ).
thf(34,plain,
( ~ sP80
| sP26 ),
inference(all_rule,[status(thm)],]) ).
thf(35,plain,
( ~ sP45
| sP36 ),
inference(all_rule,[status(thm)],]) ).
thf(36,plain,
( ~ sP65
| sP50 ),
inference(all_rule,[status(thm)],]) ).
thf(37,plain,
( ~ sP29
| sP81 ),
inference(all_rule,[status(thm)],]) ).
thf(38,plain,
( ~ sP71
| sP25 ),
inference(all_rule,[status(thm)],]) ).
thf(39,plain,
( ~ sP17
| sP71 ),
inference(prop_rule,[status(thm)],]) ).
thf(40,plain,
( ~ sP60
| sP45 ),
inference(prop_rule,[status(thm)],]) ).
thf(41,plain,
( ~ sP5
| sP29 ),
inference(prop_rule,[status(thm)],]) ).
thf(42,plain,
( ~ sP28
| sP80 ),
inference(prop_rule,[status(thm)],]) ).
thf(43,plain,
( sP31
| ~ sP19
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(44,plain,
( sP3
| sP12
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(45,plain,
( sP56
| ~ sP31 ),
inference(eigen_choice_rule,[status(thm),assumptions([h1])],[h1,eigendef_eigen__8]) ).
thf(46,plain,
( sP67
| ~ sP3 ),
inference(eigen_choice_rule,[status(thm),assumptions([h1])],[h1,eigendef_eigen__7]) ).
thf(47,plain,
( sP22
| ~ sP56 ),
inference(prop_rule,[status(thm)],]) ).
thf(48,plain,
( sP16
| ~ sP67 ),
inference(prop_rule,[status(thm)],]) ).
thf(49,plain,
( ~ sP49
| sP27 ),
inference(prop_rule,[status(thm)],]) ).
thf(50,plain,
( sP34
| ~ sP22 ),
inference(prop_rule,[status(thm)],]) ).
thf(51,plain,
( sP34
| ~ sP16 ),
inference(prop_rule,[status(thm)],]) ).
thf(52,plain,
( sP43
| sP49 ),
inference(prop_rule,[status(thm)],]) ).
thf(53,plain,
( sP43
| sP75 ),
inference(prop_rule,[status(thm)],]) ).
thf(54,plain,
( sP42
| ~ sP34 ),
inference(prop_rule,[status(thm)],]) ).
thf(55,plain,
( sP42
| ~ sP43 ),
inference(prop_rule,[status(thm)],]) ).
thf(56,plain,
( sP30
| ~ sP42 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__6]) ).
thf(57,plain,
( sP20
| ~ sP30 ),
inference(eigen_choice_rule,[status(thm),assumptions([h2])],[h2,eigendef_eigen__5]) ).
thf(58,plain,
( ~ sP76
| ~ sP72
| ~ sP58 ),
inference(prop_rule,[status(thm)],]) ).
thf(59,plain,
~ sP21,
inference(prop_rule,[status(thm)],]) ).
thf(60,plain,
( sP7
| ~ sP24
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(61,plain,
( sP33
| sP8
| sP21 ),
inference(prop_rule,[status(thm)],]) ).
thf(62,plain,
( sP15
| ~ sP7 ),
inference(eigen_choice_rule,[status(thm),assumptions([h1])],[h1,eigendef_eigen__4]) ).
thf(63,plain,
( sP23
| ~ sP33 ),
inference(eigen_choice_rule,[status(thm),assumptions([h1])],[h1,eigendef_eigen__3]) ).
thf(64,plain,
( ~ sP61
| sP60
| sP17 ),
inference(prop_rule,[status(thm)],]) ).
thf(65,plain,
( ~ sP73
| sP28
| sP5 ),
inference(prop_rule,[status(thm)],]) ).
thf(66,plain,
( sP74
| ~ sP15 ),
inference(prop_rule,[status(thm)],]) ).
thf(67,plain,
( sP48
| ~ sP23 ),
inference(prop_rule,[status(thm)],]) ).
thf(68,plain,
( ~ sP70
| sP65 ),
inference(prop_rule,[status(thm)],]) ).
thf(69,plain,
( sP54
| sP61 ),
inference(prop_rule,[status(thm)],]) ).
thf(70,plain,
( sP54
| sP73 ),
inference(prop_rule,[status(thm)],]) ).
thf(71,plain,
( sP9
| ~ sP74 ),
inference(prop_rule,[status(thm)],]) ).
thf(72,plain,
( sP9
| ~ sP48 ),
inference(prop_rule,[status(thm)],]) ).
thf(73,plain,
( sP52
| sP70 ),
inference(prop_rule,[status(thm)],]) ).
thf(74,plain,
( sP52
| ~ sP54 ),
inference(prop_rule,[status(thm)],]) ).
thf(75,plain,
( sP69
| ~ sP9 ),
inference(prop_rule,[status(thm)],]) ).
thf(76,plain,
( sP69
| ~ sP52 ),
inference(prop_rule,[status(thm)],]) ).
thf(77,plain,
( sP6
| ~ sP69 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__2]) ).
thf(78,plain,
( sP38
| ~ sP6 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__1]) ).
thf(79,plain,
( sP62
| ~ sP38 ),
inference(eigen_choice_rule,[status(thm),assumptions([h0])],[h0,eigendef_eigen__0]) ).
thf(80,plain,
( ~ sP1
| sP76
| ~ sP20 ),
inference(prop_rule,[status(thm)],]) ).
thf(81,plain,
( ~ sP66
| sP1
| ~ sP62 ),
inference(prop_rule,[status(thm)],]) ).
thf(82,plain,
\$false,
inference(prop_unsat,[status(thm),assumptions([h3,h2,h1,h0])],[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,h3]) ).
thf(83,plain,
\$false,
inference(eigenvar_choice,[status(thm),assumptions([h3,h1,h0]),eigenvar_choice(discharge,[h2])],[82,h2]) ).
thf(84,plain,
\$false,
inference(eigenvar_choice,[status(thm),assumptions([h3,h0]),eigenvar_choice(discharge,[h1])],[83,h1]) ).
thf(85,plain,
\$false,
inference(eigenvar_choice,[status(thm),assumptions([h3]),eigenvar_choice(discharge,[h0])],[84,h0]) ).
thf(0,theorem,
~ sP66,
inference(contra,[status(thm),contra(discharge,[h3])],[82,h3]) ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SEV261^5 : TPTP v7.5.0. Released v4.0.0.
% 0.03/0.12 % Command : satallax -E eprover-ho -P picomus -M modes -p tstp -t %d %s
% 0.13/0.34 % Computer : n020.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 600
% 0.13/0.34 % DateTime : Fri Mar 26 14:44:31 EDT 2021
% 0.13/0.34 % CPUTime :
% 26.02/26.34 % SZS status Theorem
% 26.02/26.34 % Mode: mode454
% 26.02/26.34 % Inferences: 790
% 26.02/26.34 % SZS output start Proof
% See solution above
% 26.02/26.34 % SZS output end Proof
%------------------------------------------------------------------------------
``` | 11,883 | 27,963 | {"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-2022-21 | latest | en | 0.127684 |
http://eduqna.com/Homework-Help/1228-1-Homework-Help-7.html | 1,505,827,310,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685698.18/warc/CC-MAIN-20170919131102-20170919151102-00357.warc.gz | 113,795,315 | 2,404 | # 2 problems that I need help on please...?
Question:1. Find the distance between (–4, 0) and (–5, –3) using the Distance Formula.
2. Simplify:
(6 + square root 3) (6 - square root 3)
well the distance formula is the
square root [(x2-x1)squared + (y2-y1)squared]
in this case you would do
square root [ (-5 - -4)squared + (-3 - 0)squared]
simplified this would be:
square root [ ( -1)squared _+ (-3)squared]
this would equal the square root of 10
number 2:
http://img510.imageshack.us/img510/7858/...
i couldn't do this one without some paper and pencil : )
i hope this helps, but be sure to check your answers because i'm not sure if these are completley correct.
ahh...school didn't start for me, what kinda math is this? My brain is in summer now...sorry can't help!
try math.com
D = √(x2 – x1)2 + (y2 – y1)2
(-4, 0) (-5, -3)
(x1, y2) (x2 – y2)
D = √(-5 – -4)2 + (-3 – 0)2
D = √(-1)2 + (-3)2
D = √1+ 9
D = √10
SECOND ONE:
This article contents is post by this website user, EduQnA.com doesn't promise its accuracy. | 346 | 1,037 | {"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} | 3.953125 | 4 | CC-MAIN-2017-39 | latest | en | 0.912876 |
https://askfilo.com/physics-question-answers/a-block-of-wood-floats-in-water-such-that-half-of-its-volume-is-below-the-water-236965 | 1,716,154,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00774.warc.gz | 90,519,922 | 32,230 | World's only instant tutoring platform
Question
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Solving time: 3 mins
# A block of wood floats in water such that half of its volume is below the water surface. But in a certain liquid, it floats with (1/4)th of its volume below the liquid surface. Find the density of liquid.
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## Text solutionVerified
When the wood block floats on the water the the gravitational force is balanced by the buoyancy,
For the second time also the buoyancy is balanced by gravitational force
Putting the value of
63
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Question Text A block of wood floats in water such that half of its volume is below the water surface. But in a certain liquid, it floats with (1/4)th of its volume below the liquid surface. Find the density of liquid. Updated On Nov 12, 2022 Topic Mechanical Properties of Fluids Subject Physics Class Class 11 Answer Type Text solution:1 Video solution: 1 Upvotes 200 Avg. Video Duration 5 min | 373 | 1,616 | {"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-2024-22 | latest | en | 0.8663 |
https://www.experts-exchange.com/questions/26838092/Display-Validation-based-on-Conditions.html | 1,498,538,630,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320915.38/warc/CC-MAIN-20170627032130-20170627052130-00151.warc.gz | 885,786,221 | 33,139 | Solved
# Display Validation based on Conditions
Posted on 2011-02-22
355 Views
I have a spreadsheet where I want to display the option for a validation list; but only if there is text in the adjointing cell. The text and location of the text shifts when selecting different industries. So, in the sample file, If text is evident in Cell B7, than the Validation Option should show up in Cell A7. The validation should also show up in Cell A14 and A20 (although I didn't put it in as a sample). The same would happen in Cells C and D, again if there is text in cell D, than the validation option would present itself in the corresponding cell in column C.
That's it!
B.
Auto-Validation-Option.xlsm
0
Question by:Bright01
[X]
###### Welcome to Experts Exchange
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LVL 30
Expert Comment
ID: 34955940
Here is a sample.
Sid
Code Used
``````Sub Sample()
Dim rng1 As Range, rng2 As Range
Dim i As Long, rng1LastRow As Long, rng2LastRow As Long
Dim ws As Worksheet
On Error GoTo Err
Application.ScreenUpdating = False
Set ws = Sheets("Sheet1")
rng1LastRow = ws.Range("B" & Rows.Count).End(xlUp).Row
rng2LastRow = ws.Range("D" & Rows.Count).End(xlUp).Row
Set rng1 = ws.Range("B1:B" & rng1LastRow)
Set rng2 = ws.Range("D1:D" & rng2LastRow)
For i = 1 To rng1LastRow
If Len(Trim(ws.Range("B" & i))) <> 0 Then
With ws.Range("A" & i).Validation
.Delete
xlBetween, Formula1:="H,M,L"
End With
End If
Next i
For i = 1 To rng2LastRow
If Len(Trim(ws.Range("D" & i))) <> 0 Then
With ws.Range("C" & i).Validation
.Delete
xlBetween, Formula1:="H,M,L"
End With
End If
Next i
Sidz:
Application.ScreenUpdating = True
Exit Sub
Err:
MsgBox Err.Description
GoTo Sidz
End Sub
``````
Auto-Validation-Option.xlsm
0
LVL 30
Expert Comment
ID: 34955947
@Bright01: You forgot to add the Excel Zone and probably why you didn't get any response till now :)
Sid
0
LVL 30
Expert Comment
ID: 34956037
And if you want to generate the validation list when you manually change values in col B and Col D then use this.
Sample File attached.
Sid
Code Used
``````Private Sub Worksheet_Change(ByVal Target As Range)
On Error GoTo Err
Dim rng1 As Range, rng2 As Range
Dim rng1LastRow As Long, rng2LastRow As Long
Application.ScreenUpdating = False
Application.EnableEvents = False
rng1LastRow = Range("B" & Rows.Count).End(xlUp).Row
rng2LastRow = Range("D" & Rows.Count).End(xlUp).Row
Set rng1 = Range("B1:B" & rng1LastRow)
Set rng2 = Range("D1:D" & rng2LastRow)
If Not Intersect(Target, rng1) Is Nothing Then
With Range("A" & Target.Row).Validation
.Delete
If Len(Trim(Target.Value)) <> 0 Then
xlBetween, Formula1:="H,M,L"
End If
End With
ElseIf Not Intersect(Target, rng2) Is Nothing Then
With Range("C" & Target.Row).Validation
.Delete
If Len(Trim(Target.Value)) <> 0 Then
xlBetween, Formula1:="H,M,L"
End If
End With
End If
Sidz:
Application.ScreenUpdating = True
Application.EnableEvents = True
Exit Sub
Err:
MsgBox Err.Description
GoTo Sidz
End Sub
``````
Auto-Validation-Option.xlsm
0
Author Comment
ID: 34956087
Sid,
Thanks for the quick response. I opened the spreadsheet and don't see how this works. There is no validation in the cells. Can you elaborate on how I install this module? Do I just set up the validation then it will work?
Thank you,
B.
0
LVL 30
Expert Comment
ID: 34956113
Bright01: I have given you two options :)
1) Either directly run the Macro "Sample" in Module1 (ID: 34955940) OR
2) When you type directly in cells in Col B abd Col C then the list will be automatically generated in Col A and Col D (ID: 34956037)
Sid
0
Author Comment
ID: 34956476
Sid,
I've put the code into the proper "sheet". When the sheet then populates itself (because I've made a selection), it does not display the validation list. I ran your sample and it works perfectly.....but doesn't work when I put it into my workbook. I think the problem is that it doesn't recognize the change in text in cells B and D when it auto populates. Is there a simple fix for this?
Thank you,
B.
0
LVL 30
Expert Comment
ID: 34956487
Add this piece of code with the code that I gave in ID: 34956037
``````Private Sub Worksheet_Calculate()
Worksheet_Change
End Sub
``````
If it still doesn't help then please upload your workbook and let me have a look at it. :)
Sid
0
Author Comment
ID: 34956679
Sid,
Where do I add it? At the top or bottom? I did and it didn't work.
I cannot upload the entire Workbook and stripping out a sample to show would take an hour. Let me see if I can do a better job explaining. I have a macro that auto populates the cells in B and D. The cells actually change based on the number of elements that are populated in each column (they align together so that one element in column B may have 3 or 4 elements in Column D. What I'm looking for is based on changes to the alignment, based on the slection, I want a validation list in Columns A and C to show up (only in those cells where text is evident. Does that make sense?
If not, I'll try to carve out a sample for you.
B.
0
LVL 30
Expert Comment
ID: 34956696
Where have you pasted the code? In a module or worksheet area?
I would love to see a basic sample for quick and accurate solution.
Sid
0
Author Comment
ID: 34956739
Sid,
Here's the sample. Change the Industry in cell and you will see Col. B and D change. Now you can see what I'm trying to do with regard to prioritization of the elements by using HML in A and C.
Thank you for taking a look.
B.
Validation-Sample.xlsm
0
LVL 30
Expert Comment
ID: 34956827
Workbook protected.
Sid
0
Author Comment
ID: 34956901
0
LVL 30
Accepted Solution
SiddharthRout earned 500 total points
ID: 34956974
Try the file now. :)
Sid
Validation-Sample.xlsm
0
Author Comment
ID: 34957049
Works great except for the error I get on startup. I've got to run to a meeting; will check in later. Much thanks; you've more than earned 500 points!
B.
2-22-2011-7-15-33-PM.png
0
LVL 30
Expert Comment
ID: 34957062
That is not because of my code. It is because of the code that you put in Workbook Open event, I guess.
Sid
0
Author Closing Comment
ID: 34957075
Great Job!!!!!
0
Author Comment
ID: 34957316
Sid,
What did you change? I need to know so I can try to trouble shoot the implementation.
Thanks,
B.
0
LVL 30
Expert Comment
ID: 34957319
If you check the sheet code area you will see that I used the code that I gave above in the worksheet change event.
Sid
0
## Featured Post
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question | 1,932 | 6,719 | {"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} | 2.515625 | 3 | CC-MAIN-2017-26 | longest | en | 0.787182 |
https://calculatorbeast.com/math/distance-calculator | 1,638,255,533,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00290.warc.gz | 235,353,414 | 21,385 | ##### Formula and Execution
(x1 y1) and (x2, y2),These are the coordinates of the points you enter to calculate the distance between two points. Distance This section provides the required results of the distance. The formula that is set up to work for this calculator is the Euclidean distance formula: d = √[(x2 – x1)2+(y2 – y1)2].
## Distance Calculator
Well, we all need to know the distance whenever we are traveling from one place to another. Knowing the distance makes it easy for us to evaluate the time that would be required to reach our destination. With the advancement of technology, we have many mobile phone applications and tools these days that work for us to make our calculations hassle-free for us. The distance is just limited to this, but we have many more applications that will be discussed later in the article.
There always exists the theory behind these modern-day hassle-free calculation tools. Somewhere, someone worked and proposed some theory based on observations that we are usually taught at schools.
And when these school ventures ask you to calculate the distance between two points in 2D space, never hesitate to visit our online distance calculator for no wait results. It does not even take a fraction of second to compute the distance between two points after the values are added to it.
### How to Calculate the Distance Between Two Points Using Calculator Beast
Our distance calculator works for points in a 2D Cartesian plane. The formula for distance is d = √[(x2 – x1)2+(y2 – y1)2]. The calculator at your left-hand side already shows you the columns for x1, x2, y1, and y2. Simply substitute the coordinates of the points in their respective place, and there you get quick results for distance.
A screenshot of the calculator is attached for a better understanding of how it works. The points selected are: (-2, 0) and (1, 4) and the result is 5 units.
### Concept of Distance
We discussed how our calculator works to calculate the distance between two points. From this section onwards, we will dive a bit deep into the ocean of concepts of distance
#### Defining Distance
When it comes to defining distance, no rocket-science is needed to do that. It is a very basic definition that even kids would be easily able to interpret. Distance is simply the measure to calculate how far two objects are located. If we are to calculate the distance between two points A and B, then the distance from A to B is interchangeable with distance from B to A. This numeric quantity tells us how far or how near a thing is.
Example:
Consider an object moved from position 2 to position 7, as shown in the figure.
To calculate the distance from 2 to 7, we simply subtract 2 from 7, i.e., 7-2=5, so, 5 units is the distance when the object is moved from 2 to 7.
What if the object is moved from 7 to 2? Well, the distance is still the same. Although we would subtract 7 from 2, i.e., 2-7=-5, remember the distance is always positive, and to make it positive, we apply absolute value operation on it.
#### Absolute Value
Absolute value is the operation in math that changes the negative sign to positive. When absolute value is applied to zero or positive, there is no change. The absolute value is symbolized by vertical brackets like |-5| = 5. Positive values are also symbolized in a similar fashion, i.e., |5| = 5.
So, since the distance between anything can never be negative, so it is the absolute value of the result.
#### Directed Distance
Directed distance is the same as the distance; the only difference is that it can be negative. Well, in the distance, we talked that it can never be negative, so why is directed distance negative? Because it shows the direction too.
Like in the example shown in the distance definition, we moved an object from 2 to 7 and then 7 to 2. The distance in both cases was 5, while the directed distance is 5 and -5, respectively, where the positive sign shows movement from 2 to 7, while the negative sign shows movement from 7 to 2.
Directed distance is measured along straight lines as well as curved paths
#### Displacement
Displacement is a directed distance considered along a straight line.
#### Psychological Distance
Psychological distance is not a numeric quantity. Psychological distancing means to remove a thing, person, or an event from personal dimensions such as time and space.
Well, we won’t be discussing the psychological distances here, because we are just concerned with the mathematical definitions of distances.
#### Kinds of Physical Distances
Distance Travelled
Distance traveled is simply the length of the path you travel while going from one place to another.
Euclidean Distance
Euclidean distance is simply the straight-line distance between two points in a Euclidean space. The Euclidean formula for distance in the 2D plane where the coordinates of the two points are (x1, y1) and (x2, y2), is,
d = √[(x2 – x1)2+(y2 – y1)2]
If we are to calculate the distance in Euclidean n-space, and we consider the points to be (x1, x2,…,xn) and (y1, y2,…,yn), the formula follows the same pattern and can be written as:
d = √[(x1 – y1)2+(x2 – y2)2+…(xn – yn)2]
Manhattan Distance
If we take the absolute difference between the Cartesian coordinates of two points and sum them up together, we get Manhattan distance. The Figure below shows the Manhattan distance and the Euclidean distance between two points that are dotted black. The red, blue, and yellow lines show the Manhattan distance, while the green line shows the Euclidean distance.
For the red line, we sum up two absolute differences between Cartesian coordinates; for the yellow line, we sum up four absolute differences between Cartesian coordinates, and similarly, for blue distance, we sum up twelve absolute differences.
The name Manhattan distance was given to it because the streets of New York City are divided into grids, and anyone traveling from one place to another has to travel on the grid of streets to reach their destination.
Aerial Distance
Aerial distance is the distance between two points on the ground, but it is the distance that is measured in the air. These distances are measured by the path that airplanes travel from one place to another.
Aerial distances are always shorter than the distances on the grounds, that is why it takes 3 hours and 50 minutes to travel from NYC to WDC by car and 1 hour 10 minutes to travel by air.
Circular Distance
The circular distance is the distance traveled by a wheel. This distance helps to design vehicles and mechanical gears. This is also the distance that is traveled by the Earth when it completes one rotation and the distance traveled by the ball when it is thrown in the air, and it returns to the initial position.
Geodesic Distance
It is the shortest distance between two points on a curved surface. The distances measured on the surface of the Earth is an example.
Chessboard Distance
This distance is the minimum number of moves a king makes on a chessboard while traveling from one square to another. Interesting, right?
Distance in Cosmology
These are distances between two objects in the space and the distances between any two events that take place in the universe. These distances are further classified into many kinds.
### FAQs
#### How is the Euclidean formula for distance derived?
To derive Euclidean formula for distance, consider two points A(x1, y1) and B(x2, y2) in the Cartesian coordinates system. To find the length of the AB, i.e., the distance of this line, we draw a right-angled triangle where AB is the hypotenuse. To draw the triangle, we drop a perpendicular from point B to point C on the x-axis and then draw another perpendicular on BC from point A as shown in the figure.
Now, as we see, triangle ABD is a right triangle with coordinates A(x1, y1), B(x2, y2), and D(x2, y1). To find the length of the hypotenuse AB, we use the Pythagoras theorem.
• AB2 = AD2 + BD2
• AB = √[AD2 + BD2]
The length of AD is x2-x1, and the length of BD is y2-y1. Substituting these values in the above equation, we get
• AB = √[(x2-x1)2 + (y2-y1)2]
The above is the distance formula for Euclidean distance.
#### Can distance be negative?
No, distance can never be negative. Distance is a scalar quantity that gives us the magnitude of how far an object is from another object, or how far an object has moved from its original position.
Directed distance or displacement, on the other and, can be negative. It is a vector quantity that not only gives us the magnitude of the distance between two things, but it also shows us the direction of the movement of the object. If the object moves in the right direction, it will be positive displacement, and if it moves towards the left direction, it will be negative.
Distance, in other words, is the absolute value of directed distance.
#### What are the real-life applications of distance formula?
Distance formula has many applications that can be seen in the real world. Some of them are listed below.
1. The distance formula is useful when finding the distances between two towns, cities, or countries on a map. Just calculating the coordinates of each place and substituting them in the distance formula yields the results for the distance between them.
2. After computing all the coordinates, we can calculate the distance between the home base and the second base in a baseball field.
3. The distances of all the throws in sports are calculated using the distance formula.
4. If the height of a lighthouse and the distance between the ship and the base of the lighthouse are known, then the distance between the observer at lighthouse and ship can be calculated using the distance formula. | 2,156 | 9,744 | {"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-2021-49 | latest | en | 0.936636 |
https://www.qfeast.com/scored/quiz/9891/Science-2 | 1,624,160,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00017.warc.gz | 851,787,357 | 12,905 | # Science (2)
## This quiz will ask questions about energy, energy types, and ways energy can change.
published on January 31, 20134 responses 0
Next »
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Hint: 1 choice
Refraction
Defraction
Reflection
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Hint: 1 choice
True
False
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### Energy can be 3 things what are they?
Stored
Transfered
Stored Transfered and Transformed
Cold
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7/33
### What is a wavelength?
The distance between a point on one wave and an identical point on the next wave.
How long a wave is at the beach
The distance between a point on four of the same waves and an identical point on the next wave.
8/33
9/33
Hint: 1 choice
Matter
Speed
energy
amplitude
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Hint: 1 choice
True
False
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Hint: 1 choice
True
False
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13/33
Hint: 1 choice
Yes
No
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Hint: 1 choice
False
True
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19/33
### Do all electromagnetic wave require a medium?
Hint: 1 choice
No, All mechanical waves
Yes.
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top of a wave
bottom of a wave
middle of a wave
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23/33
False
True
24/33
### what is a medium
In between small and large
Is the matter through which a mechanical wave travels
solid material
all of the above
25/33
Hint: 1 choice
True
Flase
26/33
Hint: 1 choice
True
False
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Hint: 1 choice
Air
Sound
28/33
Hint: 1 choice
A Vacuum
A medium
space
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Hint: 1 choice
True
False
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Hint: 1 choice
True
False
31/33
### Wave speed=
Hint: 1 choice
base x height
Wave speed x frequency
Frequency x Wavelength
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frequency
speed
amplitude
33/33
Hint: 1 choice
bottom of wave
top of a wave
middle of a wave | 521 | 1,583 | {"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} | 2.515625 | 3 | CC-MAIN-2021-25 | latest | en | 0.773412 |
https://www.coursehero.com/file/6865603/Chap005-updated/ | 1,498,729,073,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323895.99/warc/CC-MAIN-20170629084615-20170629104615-00266.warc.gz | 867,657,687 | 24,853 | Chap005-updated
# Chap005-updated - 5 1 Chapter Five A Survey of Probability...
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Unformatted text preview: 5- 1 Chapter Five A Survey of Probability Concepts A Survey of Probability Concepts Objectives When you have completed this chapter, you will be able to: ONE Define probability. TWO Describe the classical, empirical, and subjective approaches to probability. THREE Understand the terms: experiment, event, outcome, permutations, and combinations. 5- 2 Chapter Five continued A Survey of Probability Concepts A Survey of Probability Concepts Objectives When you have completed this chapter, you will be able to: FOUR Define the terms: conditional probability and joint probability. FIVE Calculate probabilities applying the rules of addition and the rules of multiplication. SIX Use a tree diagram to organize and compute probabilities. 5- 3 Definitions Definitions A probability is a measure of the likelihood that an event in the future will happen. It it can only assume a value between 0 and 1. A value near zero means the event is not likely to happen. A value near one means it is likely. There are three definitions of probability: classical, empirical, and subjective. 5- 4 Definitions Definitions continued continued The classical definition applies when there are n equally likely outcomes. The empirical definition applies when the number of times the event happens is divided by the number of observations. Subjective probability is based on whatever information is available. 5- 5 Definitions Definitions continued continued An experiment is the observation of some activity or the act of taking some measurement. An outcome is the particular result of an experiment. An event is the collection of one or more outcomes of an experiment. 5- 6 Mutually Exclusive Events Mutually Exclusive Events Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. Events are independent if the occurrence of one event does not affect the occurrence of another. 5- 7 Collectively Exhaustive Events Collectively Exhaustive Events Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted. 5- 8 A fair die is rolled once. The experiment is rolling the die. The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. An event is the occurrence of an even number. That is, we collect the outcomes 2, 4, and 6. Example 1 Example 1 5- 9 EXAMPLE 2 EXAMPLE 2 Throughout her teaching career Professor Jones has awarded 186 A’s out of 1,200 students. What is the probability that a student in her section this semester will receive an A?...
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## This note was uploaded on 03/19/2012 for the course MGMT 604 604 taught by Professor Mohamed during the Spring '12 term at Manor.
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Chap005-updated - 5 1 Chapter Five A Survey of Probability...
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Ask a homework question - tutors are online | 777 | 3,689 | {"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} | 3.359375 | 3 | CC-MAIN-2017-26 | longest | en | 0.918882 |
https://www.tdworld.com/grid-innovations/article/20964567/a-flash-of-insight-winners-and-explanation | 1,708,717,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474445.77/warc/CC-MAIN-20240223185223-20240223215223-00546.warc.gz | 1,018,545,951 | 55,944 | # A Flash of Insight Winners and Explanation
March 24, 2014
Here's a little puzzle to ruminate on during a coffee break. Maybe you'll win a gift card! Give it your best shot in the comment section below. Enter as many times as you want. A \$25 gift card will go to the first right answer. Another \$25 winner will be selected from the other right answers. We need your email to send the gift card so make sure you're registered (and you have verified your registration via email) when you leave your comment in the box below.
“A Flash of Insight” puzzle winners and explanation (see original puzzle below):
First let’s look at each suggestion:
1.Reuben: “Throw yourself spread-eagled on the ground,” said Rueben…
Wrong advice. If a bolt of lightning should strike nearby, the electrical current spreads out through the ground. This can produce ‘step potentials’ of thousands of volts per foot. So being ‘spread-eagled could result in lethal voltage across your body.
2. “Nope,” corrected Jessica, “You should have gotten into the jeep. The rubber tires, if they’re not too worn, provide insulation from the ground…
Again, poor advice. The lightning bolt, having traveled perhaps miles through the air is hardly daunted by an inch of rubber tire. However, if the jeep had a top, Frank would have been safer to have gotten inside. The metal shell (and Farraday’s law for you geeks) would have protected him and the lightning current would have traveled over the jeep body to ground.
3. Bob: “…you would have been safer if you had tightly hugged the trunk of a tall tree. Even better, you could have shinnied up a bit to get your feet off the ground.”
Maybe Bob spoke without thinking. Tall trees are frequently hit. And they often explode. Standing near a tree exposes you to step potentials. So hugging a tree could be suicide.
4. Waitress: …moving metal attracts lightning, so the jeep’s no use. You should have hunkered down as low as you could get and ran as fast as you could until the hairs on your neck went down again.
She should stay in school! A speeding jeep doesn’t attract anything except maybe the cops. Hunkering down low is not a bad idea. But trying to outrun a thunderstorm probably won’t work.
5. Alissa said softly: “I’d carefully squat down and hug my knees in a fetal position…..
Alissa is right, but not because of the aura, good thoughts or Universal Energy. Getting low is always good and touching the ground with only your feet close together is the best option that’s offered.
And the winners are:
Dennis Cyor: “Squatting down and hugging your knees. Smaller target, less ground contact and closer to the ground. As far as the aura, lighning is usually negatively charged, so you would need to project a negative aura to repel the strike. Projecting a negative aura is bad karma, and could disturb the time-space continuum.”
Dan Olsen: “Alissa has the best response: Squat down and hug my knees in a fetal position, but disregard the philosophical babble!”
(Bruce had the first right answer but unfortunately he wasn’t verified. So make sure you are registered and verified, otherwise we have no way of sending your gift card!)
Five lineworkers, Rueben, Jessica, Frank, Bob and Alissa were eating lunch at a diner when they heard the distant rumble of thunder. Frank said “That reminds of when I was out hunting last fall. The rain started pouring and my old jeep has no top, but worse than that, the lightning was getting closer and then I felt the hair on my neck stand up. I’ve heard that that means a lightning strike is coming. I didn’t know what to do so I did nothing and soon the storm passed. What do you-all think I shoulda done?”
“Throw yourself spread-eagled on the ground,” said Rueben. “Stay as low as you can. That way you don’t look like a human lightning rod. Lightning tries to strike the tallest objects. Then lay there until you can count to 60 between seeing a flash and hearing the thunder.”
“Nope,” corrected Jessica, “You should have gotten into the jeep. The rubber tires, if they’re not too worn, provide insulation from the ground and the lightning current can’t complete the circuit so it won’t go through you. Of course the tires are wet from the rain, but pure rainwater has very low conductivity and is a good insulator.”
“Well, the answer’s not so simple. You gotta consider the terrain and other factors,” said Bob. “You were in a forest, Frank, so you were lucky. Rueben’s right that lightning usually hits the highest objects, so you would have been safer if you had tightly hugged the trunk of a tall tree. Even better, you could have shinnied up a bit to get your feet off the ground. That way the lightning would strike the tree instead of you and the trunk would have carried the current directly to ground without going through your body.”
Alissa said softly: “I’d carefully squat down and hug my knees in a fetal position. Then I’d close my eyes and focus on my inner light, repeating my personal mantra as I clear my mind of any thoughts, particularly fear. My aura would be a shield against the lightning, reflecting any negative energy back into the sky where maybe it can be transformed for Universal Good.”
There was silence until the waitress came to refill the coffees.
“Excuse me, folks” she said, “but I’m working on my degree in electrical engineering and I couldn’t help overhearing. Maybe I can help. Now, high voltage is tricky and unpredictable, so the best thing you can do is get out of the area. But moving metal attracts lightning, so the jeep’s no use. You should have hunkered down as low as you could get and ran as fast as you could until the hairs on your neck went down again.”
What’s the correct answer? Any of these? None of these? And why? The best two answers each get a \$25 gift card. (Register on tdworld.com and leave your answer in comment box below. Please make sure you follow up with verification of the registration through email so we can find you.) | 1,327 | 5,998 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.966367 |
http://docplayer.net/30819521-The-is-curve-this-lecture-describes-the-relationship-between-aggregate-output-and-the-real-interest-rate-when-the-goods-market-is-in-equilibrium.html | 1,561,602,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000610.35/warc/CC-MAIN-20190627015143-20190627041143-00410.warc.gz | 48,099,134 | 23,775 | # The IS Curve. This lecture describes the relationship between aggregate output and the real interest rate when the goods market is in equilibrium.
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1 The IS Curve This lecture describes the relationship between aggregate output and the real interest rate when the goods market is in equilibrium. Aggregate Demand: The Basics A. Aggregate demand is the total amount of output demanded in the economy for a given inflation rate. This measure captures the amount of planned expenditures (as opposed to actual expenditures) in the economy.
2 B. Aggregate demand is the sum of four types of spending. The resulting equation is called the income identity Y = C + I + G + NX. (1) 1. Consumption expenditures (C) is the total demand for consumer goods and services (except housing). 2. Planned investment spending (I) is total planned investment spending on physical capital and inventories by businesses plus planned investment on new housing by consumers. 3. Government purchases (G) is the total spending on government consumption and government investment by all levels of government. 4. Net exports (NX) is the net foreign spending on domestic goods and services defined as exports minus imports.
3 The Components of Aggregate Demand: The Details A. Consumption Expenditures 1. Consumption is positively related to disposable income (Y D ), where disposable income is aggregate income (Y) minus taxes (T) Y D = Y T. (2) 2. The consumption function defines the positive relationship between consumption and disposable income C = + MPC (Y T). (3) a. Autonomous consumption ( > 0) represents all of the factors other than Y D that affects C. (Ex., wealth) b. The marginal propensity to consume (0 MPC < 1) reflects the change in consumption from an additional dollar of disposable income.
4 B. Planned Investment Spending 1. Types of investment a. Fixed investment includes capital investment by firms and residential investment by households. b. Inventory investment is the change in inventory holdings by firms. (Inventory investment = this year s inventories last year s inventories.) 2. Planned investment spending is negatively related to the real interest rate (r). a. A higher real interest rate raises borrowing costs, which discourages firms from investing. b. Even if a firm self-finances its investment, it must forgo the real interest rate it would have earned on bonds.
5 3. The investment function I = d (r + ). (4) a. The parameter d > 0 represents the responsiveness of investment to the real interest rate. b. Financial frictions ( > 0) represent the costs associated with barriers that keep financial markets from functioning efficiently. (Ex., A rise in asymmetric information problems is denoted by a higher.) c. Autonomous investment ( > 0) represents all of the factors besides financial frictions and the real interest rate that affect investment. (Ex., Business expectations) C. Government purchases and taxes 1. Government purchases are independent of Y and r G =. (5)
6 2. The government imposes lump-sum taxes on households D. Net Exports T =. (6) 1. Net exports decline as the real interest rate rises because a higher real interest rate drives up the value of the domestic currency. That higher value raises export prices and decreases import prices, which causes net exports to fall. 2. The net exports function NX = x r. (7) a. The parameter x > 0 represents the responsiveness of net exports to the real interest rate. b. Autonomous net exports ( ) represents all the factors other than the real interest rate that affect net exports. (Ex., Foreign income, trade restrictions)
7 Goods Market Equilibrium A. Solving for the Goods Market Equilibrium 1. Start with the income identity (1) Y = C + I + G + NX. (8) 2. Substitute the equations for consumption (3), investment (4), government spending (5), taxes (6) and net exports (7) into the income identity. Y = + MPC (Y ) + d (r + ) + + x r. (9) 3. Solve for Y Y = + MPC (Y ) + d (r + ) + + x r Y = d MPC + MPC Y (d + x) r Y (1 MPC) = d MPC (d + x) r. (10)
8 B. Deriving the IS curve 1. The IS curve shows the relationship between aggregate output and the real interest rate when the goods market is in equilibrium. 2. The IS curve is downward sloping because a higher real interest rate reduces investment and net exports, which pushes down output. [r (I & NX ) Y ] Real interest rate r B B r A A IS Y B Y A Output
9 C. Example: Suppose the economy is described by the following information: Y = C + I + + NX C = + MPC (Y ) I = d (r + ) NX = x r. Let = 1.4, = 1.2, = 3.0, = 1.3, = 3.0, MPC = 0.6, d = 3, x = 1, = 0.1, and r = 0.10.
10 1. Derive the equation for the IS curve Y = + MPC (Y ) + d (r + ) + + x r Y = (Y 3.0) (r + 0.1) r Y = [ ] Y (3+1) r Y 0.6 Y = r 0.4 Y = r Y = 4.8/0.4 (4/0.4) r Y = r 2. Calculate output Y = r Y = Y = 12 1 Y = 11
11 3. Calculate consumption C = + MPC (Y ) C = (11 3) C = C = Calculate investment I = d (r + ) I = ( ) I = I = Calculate net exports NX = x r NX = NX = 1.2
12 Shifting the IS Curve A. An increase in output in the goods market shifts the IS curve to the right. Real interest rate r A B IS A IS B Y A Y B Output
13 B. Several factors increase output and shift the IS curve to the right. 1. An increase in autonomous consumption ( ) a. Higher wealth or improved expectations for the economy b. Y 2. An increase in autonomous investment ( ) a. Improved expectations for the economy b. Y 3. An increase in government spending ( ) a. Government spending is set exogenously by the government. b. Y
14 4. A decrease in taxes ( ) a. Taxes are set exogenously by the government. b. T Y D C Y 5. An increase in autonomous net exports ( ) a. An increase in foreign income b. Y 6. A decrease in financial frictions ( ) a. Less asymmetric informational problems in financial markets b. I Y
### Study Questions 8 (Keynesian Model) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Study Questions 8 (Keynesian Model) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) In the Keynesian model of aggregate expenditure, real GDP is
### Introduction to Macroeconomics TOPIC 2: The Goods Market
TOPIC 2: The Goods Market Annaïg Morin CBS - Department of Economics August 2013 Goods market Road map: 1. Demand for goods 1.1. Components 1.1.1. Consumption 1.1.2. Investment 1.1.3. Government spending
### The Short-Run Macro Model. The Short-Run Macro Model. The Short-Run Macro Model
The Short-Run Macro Model In the short run, spending depends on income, and income depends on spending. The Short-Run Macro Model Short-Run Macro Model A macroeconomic model that explains how changes in
### = C + I + G + NX ECON 302. Lecture 4: Aggregate Expenditures/Keynesian Model: Equilibrium in the Goods Market/Loanable Funds Market
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http://www.piclist.com/techref/io/3r2ra2d.htm | 1,638,803,887,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00428.warc.gz | 133,538,783 | 8,801 | # Trinary A2D ?
Russell McMahon suggests
Classically an R2R ladder is used to implement a Successive approximation Analog to Digital converter with drives being high or low. It seems to me that a ternary converter using high/low/open-circuit outputs should allow more resolution per pins used (by a factor of up to 3^N/2^N).
Initial playing on paper produced some ideas (I should be able to get "8 bit" accuracy with 6 pins used) but I may be missing something obvious. Theory says 5 pins SHOULD allow 1 part in 243 but obvious implementation escapes me at this stage.
James Newton tried:
I could see that one problem would be the case when no resistance connected the output to either VCC or GND or when no resistance connected the output to VCC AND GND (all tri-stated) so I assumed the existence of a voltage divider that is always connected to VCC and GND. I label this R3/R4 and assume the ratio of R3 to R4 to be 1 (R3=R4). I then created an MS Excel spread sheet and setup columns for three other resisters (R1, R2, R3) with sub-columns for each containing a zero or a 1 to indicated their possible connection to GND or VCC. Each resistor can be connected to GND, VCC, or nothing. I then entered the formulas necessary to calculate the total resistance of the center point to GND and another for VCC. The ratio of the two is multiplied by the value of VCC to give the voltage at that point. I added a column to show the difference between each successive case and a standard deviation at the bottom of the column.
Results: Although I can't say why, the steps are not perfectly even for any reasonable values of R1-R4.
It seems to me that with an output R like this you can pull it up in various binary combinations or not AND down {by the same amounts}so you get N bits above or below ground (or below the centre poinbt in your case) so this is a "win" of 1 extra bit. This is not enough.
The real gain should be 3^m/2^n.
eg for 8 bits base 2 gives 2^8 = 256 states and base 3 gives 3^8 = 6561 states.
Base 2 needs 12.67 bits to achieve this.
Clearly this is not reaaly going to happen as there are several states which are badlky behaved eg all tristate while base 2 has no invalid states.
Mayhaps true base 3 with the third state being mid rail would give a more workable solution This could be APPROXIMATED by a stiffish pullup and pulldown at each pin and a much higher value of R from the pin to the network. eg 1K pullup, 1K puuldown and 100K say to network. Pin hi looks like 100K to high Pin low looks like 100K to low Pin o/c looks like 100.5K to half supply. Using 10k/10k/100k gives 100k/100k/105K in the last example.
Mayhaps the MSBs could use 1K's (about 1 part in 200 MSB resistance error at mid state and LSBs use 10k's - the accuracy of LSB R's in an R/2R can be much lower than the MSBs AFAIR.
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. | 1,224 | 4,758 | {"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} | 3.3125 | 3 | CC-MAIN-2021-49 | longest | en | 0.92031 |
https://www.physics.uoguelph.ca/st108/a2t3q3a.htm | 1,544,625,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823895.25/warc/CC-MAIN-20181212134123-20181212155623-00275.warc.gz | 1,009,128,053 | 799 | When is the population down by 40%?
i.e., 852 - 0.4(852) = 511?
From the graph: about t = 9½ days, i.e., 8½ days from now. | 50 | 123 | {"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-2018-51 | latest | en | 0.953179 |
https://salesforce.stackexchange.com/questions/139432/formula-for-exact-years-including-decimal-places-between-two-dates | 1,716,814,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00251.warc.gz | 442,929,621 | 37,471 | # Formula for exact years (including decimal places) between two dates
I'm using this formula to calculate a person's age:
``````IF (
MONTH(TODAY()) > MONTH(Birthdate),
YEAR(TODAY()) - YEAR(Birthdate),
IF (AND(MONTH(TODAY()) = MONTH(Birthdate), DAY(TODAY()) >= DAY(Birthdate)),
YEAR(TODAY()) - YEAR(Birthdate),
(YEAR(TODAY()) - YEAR(Birthdate)) - 1
)
)
``````
which (I hope) correctly handles leap years.
I now need a formula to calculate the number of years between 2 dates that includes decimal places and accounts for leap years. Dividing the difference between 2 dates by 365 (or 365.2422) only gives an approximate result (given that leap years happen in specific years).
Is this possible using only a formula? If it is please share the formula...
Or is updating via e.g. a trigger the more sane way to go?
A leap year is divisible by 400, or if it’s divisible by four but NOT by 100. You'd need to use a variation on the 1st formula to also determine the number of leap years involved between the two dates. The formula for determining a leap year is as follows:
``````OR(
MOD( YEAR( date ), 400 ) = 0,
AND(
MOD( YEAR( date ), 4 ) = 0,
MOD( YEAR( date ), 100 ) != 0
)
)
``````
Once you found the 1st leap year following the birth date or the one that's prior to today(), you could easily calculate the number of leap years in-between the two dates.
Knowing the above, it might be preferable to determine the number of days between the two dates.
• Hmmm... I'm not sure easily is quite the right word. But thanks for the approach. Sep 3, 2016 at 13:35
• Well, I guess it's all relative. ;) Sep 3, 2016 at 13:40
This is the formula I have ended up using (where `D1__c` and `D2__c` are the two date fields):
``````ROUND
(
(YEAR(D1__c) - YEAR(D2__c))
+
(((D1__c - DATE(YEAR(D1__c), 1, 1)) - (D2__c - DATE(YEAR(D2__c), 1, 1))) / 365),
2
)
``````
Its really based on the age formula from the original question plus this "Day of the Year" formula:
``````TODAY() – DATE(YEAR(TODAY()), 1, 1) + 1
``````
from Handy Salesforce Formulas You Can Copy And Paste. Differencing the two "Day of Year" values and dividing by 365 provides the fractional part to add to the age formula. When the two formulas are put together the combination can then be simplified a bit.
Using the YEAR formula ensures that the non-fractional part isn't subject to leap year problems. Arguably the fractional part divisor should be adjusted to be 365 or 366 or 367 but working to 2 decimal places and for the test cases I've tried (values below are `D1__c`, `D2__c`, expected years; all passing) the extra complexity does not appear necessary:
``````new TestCase(Date.newInstance(2016, 8, 29), Date.newInstance(2016, 8, 29), 0.00),
new TestCase(Date.newInstance(2016, 8, 29), Date.newInstance(2116, 8, 29), 100.00),
new TestCase(Date.newInstance(2016, 1, 20), Date.newInstance(2021, 7, 20), 5.50),
new TestCase(Date.newInstance(2016, 11, 5), Date.newInstance(2027, 2, 5), 10.25),
new TestCase(Date.newInstance(1964, 5, 25), Date.newInstance(1989, 5, 9), 24.95),
new TestCase(Date.newInstance(2116, 8, 29), Date.newInstance(2016, 8, 29), -100.00),
new TestCase(Date.newInstance(2021, 7, 20), Date.newInstance(2016, 1, 20), -5.50),
new TestCase(Date.newInstance(2027, 2, 5), Date.newInstance(2016, 11, 5), -10.25),
new TestCase(Date.newInstance(1989, 5, 9), Date.newInstance(1964, 5, 25), -24.95)
`````` | 1,017 | 3,391 | {"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} | 3.921875 | 4 | CC-MAIN-2024-22 | latest | en | 0.895387 |
https://hackaday.io/project/27550-the-hobbyists-guide-to-fpgas/log/68240-introduction-to-verilog-and-combinatorial-logic/discussion-117130 | 1,545,145,943,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829429.94/warc/CC-MAIN-20181218143757-20181218165757-00370.warc.gz | 629,728,743 | 12,739 | # Introduction to Verilog and Combinatorial Logic
A project log for The Hobbyists Guide to FPGAs
Follow this project to learn how to use FPGAs and incorporate them into your projects.
Luke Valenty 10/05/2017 at 05:007 Comments
## Introduction
There are three primary methods for developing FPGA designs:
1. Verilog
2. VHDL
3. Schematic Capture
The Hobbyist's Guide to FPGAs will focus on using Verilog to program FPGAs. Verilog and VHDL are both hardware description languages. They allow you to design digital logic circuits by describing them in their language. Verilog was chosen for this guide because it has an easier learning curve and is not quite as picky as VHDL is. There are some that would argue for using VHDL, and for good reason. VHDL has some powerful constructs that Verilog does not. For the average hobbyist, however, Verilog is the right choice and will provide a more pleasant entrance into the world of FPGAs.
## Verilog
When designing any system, it is useful to be able to decompose the problem into smaller parts. Digital design is no different. In Verilog we have the concept of modules. A module is a self-contained design that communicates externally through ports. The ports on a module are input and output wires. You could think of a module as it's own standalone logic chip. When we are developing a new design, we start with a module.
```/*
* Implements a simple 1-bit half-adder.
*/
input a,
input b,
output sum,
output carry
);
// sum is the xor of the two inputs
assign sum = a ^ b;
// carry is true only if both inputs are true
assign carry = a & b;
endmodule
```
The above example is a reusable module that implements a half-adder. It's a nice and simple example that illustrates a few fundamental attributes of Verilog:
1. All functionality is implemented within modules.
2. Module ports have an associated direction.
3. We can directly assign expressions to ports and wires.
4. Expressions and comments in Verilog are very similar to expressions and comments in C, C++, and Java.
The assign statement performs a continuous assignment. It's not assigning a singular value to the left-hand side. Instead, it is assigning an expression to the left-hand side. If the value of any components of the expression changes, then the value of the left-hand side wire will change as well. The text of the Verilog translates directly into digital logic elements. If we designed the module as a schematic it might like similar to the one below:
This is an accurate schematic of what the half adder module might look like if it were assembled with logic gates. FPGAs, despite their name, don't actually use logic gates. Instead they use much more powerful LUTs, or Look-Up Tables. In the actual FPGA implementation, the module would look more like the following diagram:
Instead of specific logic gates, the outputs of sum and carry are generated by the LUTs present in the FPGA fabric. The LUTs are programmed to behave like an XOR gate and an AND gate.
Modules can be reused in other modules. We can take our half adder and use it to create a full adder.
```/*
*/
input a,
input b,
input carry_in,
output sum,
output carry_out
);
// carry outputs from half adders
wire carry_1;
wire carry_2;
// intermediate sum from first half adder
wire sum_1;
// instantiate the first half adder
.a(a),
.b(b),
.carry(carry_1),
.sum(sum_1)
);
// instantiate the second half adder
.a(sum_1),
.b(carry_in),
.carry(carry_2),
.sum(sum)
);
// generate final carry out signal
assign carry_out = carry_1 | carry_2;
endmodule```
Here we see how modules can be reused and instantiated in another module. First the name of the module type is given, then the name of the instance, then a list of the ports and their connections. We also see that you can create wires internal to the module to connect components within the module.
If this were a schematic it would look like the following diagram:
Pretty neat, but given the amount of Verilog used to describe this simple design and the basic operation of addition we are building up to, you might be asking yourself: "Is this how much work I have to do just to add two numbers in Verilog!?" Luckily the answer is a resounding "No." Implementing an adder just happens to be a useful tool to dive into the most basic concepts of Verilog. Once we know how to implement a useful full-adder the hard way, we can cut to the chase and learn how to do it in a more logical way.
The next extension of this example is to enable larger than 1-bit numbers. This example will start to show the power of using a hardware description language vs. a schematic entry tool.
In order to add useful numbers we need several full-adders with a carry-chain connected between them. Let's start with the schematic representation first as the Verilog is going to start getting very interesting:
That's an eight-bit adder. It's pretty cool but it sure took longer for me to draw than I would like to admit. What if we want to expand it to larger or smaller sizes? It is not too convenient. There would be manual work involved in changing the number of bits in the adder. Hardware description languages like Verilog can do a much better job here. Verilog has the concept of module parameters and generate statements that will solve this problem very nicely.
```/*
* parameter to the width in bits of the numbers you want to add.
* If no ADDER_WIDTH is specified, then the default width of 8 bits
* will be used.
*/
) (
input carry_in,
output carry_out
);
// these are the wires that will connect the carry output
// of one full adder to the carry input of the next full adder
// use a generate block and for-loop to generate as many
// full adders as we need.
genvar i;
generate
for (i = 0; i < ADDER_WIDTH; i = i + 1) begin
.a(a[i]),
.b(b[i]),
.sum(sum[i]),
.carry_in(carry_chain[i]),
.carry_out(carry_chain[i + 1])
);
end
endgenerate
// connect carry_in and carry_out to the beginning and end
// of the carry chain
assign carry_chain[0] = carry_in;
endmodule```
There are several new concepts introduced here:
1. Module parameters! When you instantiate a module, you can also specify values for any of the module's parameters. This helps you to create reusable modules. In this example we use a parameter to specify the width in bits of the addition operation we are performing.
2. Bit vectors. Instead of all wires and ports being one bit wide, we can create wires and ports with as many bits as we want using the bit-slice notation: [msb:lsb]. The indexes used in the bit-slice notation refer to the bit positions and are inclusive. For example: a bit-slice of "a[7:0]" is returning the lower 8 bits of "a".
3. Generate blocks and for-loops. Generate blocks allow us to conditionally instantiate modules, assign statements, or always blocks; or even create multiple instances of the aforementioned constructs. In the above example we are using a for-loop in a generate block to instantiate as many full-adders as specified in the ADDER_WIDTH parameter.
## A More Logical Approach to Addition
While addition served as an excellent way to describe how to create and use basic modules along with the more advanced concept of generation, this is is not how addition is performed in normal verilog code. There is a much easier way to implement an adder module:
```module addition_unit #(
) (
input carry_in,
output carry_out
);
// create an internal sum that includes carry_out
// use the built-in addition operator
assign full_sum = carry_in + a + b;
endmodule```
That's right, there is a built-in addition operator to Verilog. In fact, a standalone module would rarely be used like this in Verilog. It is more common for addition operators to be used in expressions just like bitwise logic operators are.
The FPGA synthesis tools know how to interpret the addition operator and will utilize optimized structures in the FPGA fabric to perform the addition operation. To further illustrate the power of Verilog over schematic entry, let's expand upon our adder concept. Let's create a simple arithmetic logic unit like one that might be found in a simple microcontroller.
```module alu #(
ALU_WIDTH = 8
) (
// what type of operation should the ALU perform
input [2:0] opcode,
// input operands
input [ALU_WIDTH-1:0] a,
input [ALU_WIDTH-1:0] b,
// operation output
output reg [ALU_WIDTH-1:0] out
);
// we want to use a combinatorial always block so we can use
// a case statement to decide what to do.
always @(*) begin
case (opcode)
0: out <= a + b;
1: out <= a - b;
2: out <= a & b;
3: out <= a | b;
4: out <= a ^ b;
5: out <= ~a;
6: out <= a << 1;
7: out <= a >> 1;
endcase
end
endmodule```
This is a very simple ALU. It is missing useful features like overflow and carry inputs, but I'll leave that as an exercise for the reader to. Now we are introduced to a very important concept in Verilog: the always block. An always block is a block of code that is "evaluated" every time a signal in its sensitivity list changes. In this example I used a * character to make the always block sensitive to all signals it uses.
Without having to write a large amount of code we have a basic arithmetic logic unit. We can add, subtract, perform boolean operations, and shift the input left and right. Not only is all the logic for those operations automatically inferred by the synthesis tools, the case statement will be interpreted as a 8:1 mux as well.
## Conclusion
In a short amount of time this tutorial has taken the reader head-long into the fundamental features of Verilog for developing on FPGAs. I sincerely hope this was an enjoyable read. Stay-tuned, the next log will be a hands-on lab taking advantage of the concepts learned here and putting them to use on a hobbyist FPGA board.
To complete the upcoming lab on your own you can use use any of the TinyFPGA boards available at the TinyFPGA Tindie store. If you are deciding on which TinyFPGA to use, please read the user guides for the different boards at tinyfpga.com. The A-series boards need a bit more components to get up and running while the B-series boards are ready to go once plugged into USB.
You will also want a breadboard, some switches you can plug in, LEDs, and resistors for the switches and LEDs. More details about components required will be in the lab itself.
Until next time!
## Discussions
Shankar wrote 12/06/2018 at 16:51 point
Excellent Tutorial !!! I have purchased your Tiny FPGA BX and i would like to thank You for it :)
Are you sure? yes | no
etherBoy wrote 10/07/2018 at 14:22 point
Having done Digital Electronics at the University, an approach of this sort makes the TinyFPGA more welcoming. Great series thus far! If there are other places that offer tutorials in this exact manner, please let us know.
Are you sure? yes | no
Jim Burnes wrote 07/01/2018 at 20:59 point
Another great tutorial. There aren't very many people on the net that explain things a well as you.
Are you sure? yes | no
rmacintosh wrote 06/29/2018 at 04:40 point
Excellent tutorial
Hope more is on the way!
Are you sure? yes | no
gmacario wrote 06/13/2018 at 04:44 point
Excellent introduction to FPGAs, looking forward to reading next articles in the series!
Are you sure? yes | no
wyzarddoc wrote 06/01/2018 at 19:42 point
Great Job!! I have purchased FPGA's and put off learning how to program till now. I started a long time ago with cpu's made of discreet logic chips so this should stir old memories. Looking forward to the book when your finished till then I'll look forward to the next installment here
Are you sure? yes | no
Simon Merrett wrote 05/29/2018 at 16:47 point
Thank you! I'd love to see more in this series. A bit surprised I'm the first to say thank you and request more!
Are you sure? yes | no | 2,894 | 11,834 | {"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-2018-51 | latest | en | 0.89719 |
https://www.physicsforums.com/threads/why-just-int-half-int-spins-are-fundamental.396721/ | 1,508,280,016,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00249.warc.gz | 975,152,094 | 19,031 | # Why just int/half-int spins are fundamental?
1. Apr 19, 2010
### MTd2
I just don't get it. Some people, deep down in their hearts, want to see computation as the most fundamental thing ever. This is the reasoning behind the holographic idea, at least from Verlinde. Variable spin makes it possible quantum computation through anyons, yet people still want spin like it is in the usual world, integer or half integer until plank scale and assume all kinds of weird things down there: strings, loops, several dimensions, variable dimensions, variable speed of light.
So, why variable spin gets ignored?
2. Apr 19, 2010
### atyy
3. Apr 19, 2010
### MTd2
Yeah, and John Baez liberated it from 3dimensions too, up to 4 :)
http://arxiv.org/abs/gr-qc/0603085
4. Apr 20, 2010
### atyy
But you can have quantum computation without anyons - just not topological.
5. Apr 20, 2010
### MTd2
But it is interesting that the fact that is topological, that is, precludes the existence of a metric, makes the thing more interesting to set up the condition for emerging space out of it?
6. Apr 20, 2010
### atyy
Last edited: Apr 20, 2010
7. Apr 20, 2010
### meopemuk
Some people do not make these weird assumptions.
Because (half-)integrer spin is a direct consequence of two fundamental things:
1. rotational symmetry
2. postulates of quantum mechanics
Eugene.
8. Apr 20, 2010
### genneth
Now now. Let's not be rash. You also need 3D and above for point like-particles. In 2D, point-like particles can have arbitrary phase relations. More generally, if you don't assume particles as your starting point, but more generalised, extended objects, then even more complex braiding is possible, even in high dimensions.
9. Apr 20, 2010
### MTd2
Not really. You can have those arbitrary relations in 4d too. John Baez found a way in the paper I pointed out above.
10. Apr 20, 2010
### arivero
There are two parts here:
a) what multiples and submultiples of unit of angular momentum are needed, or allowed, by mathematics. The integer version amounts to label the spherical harmonics (in any dimension, if you wish), and then the subinteger quantities due to different commutation rules and other needs of labeling.
b) why the unit of angular momentum is the same for all the problems. The other constant, "c", is more easily to be understood as universal as it related to geometry of the space-time, and it adjust the units of t and x in the 4-vector. But the role of h seems internal, and still it is a universal quantity.
11. Apr 20, 2010
### MTd2
I guess I should have clarified it better. I referred to the particle spin, only.
12. Apr 20, 2010
### dx
It's basically because the distance between two successive eigenvalues of an operator like Sz is always 1. So if there a state at 0, then you get the series {..., -1, 0, 1, ...}. If there is no state at 0, then there must be a state at some 'a' > 0, and by symmetry, a state at -a < 0. Since the difference between these must be 1, 'a' must be 1/2, and therefore you get the series {..., -3/2, -1/2, 0, 1/2, 3/2, ...}.
13. Apr 20, 2010
### MTd2
But you are assuming that angular moment is isotropic, that is, it doesn't change if you move it. The spin of anyons are not isotropic, but position dependent to the spin and position of other particles.
14. Apr 20, 2010
### Haelfix
It would help if you describe what exactly you are talking about. If we are talking about
1) D=4
2) Fundamental, point particles
3) Quantum mechanics is applicable + special relativity is applicable to all objects
Then it is a theorem that the only things you can have are half integer spins like 0, 1/2, 1, 3/2... It essentially drops out of the spin-statistics theorem.
15. Apr 20, 2010
### dx
What do you mean spin is not isotropic for anyons? The fact is that space is isotropic. Are you saying space is not isotropic for anyons? How can that be?
16. Apr 20, 2010
### MTd2
Hmm, I was not talking about point particles, you see as I gave the example of string theory or the paper from John Baez. My question was kind of generic, of why don't I see theories without int/ half-int theories.
17. Apr 20, 2010
### MTd2
Not the space, but the value of the spin depends on the position of the path described by the particle as well as their relative position.
18. Apr 20, 2010
### dx
I don't think it matters whether the particles are points or strings, since the angular momentum operator is basically a rotation of space, and it would have the same commutation relations etc.
Last edited: Apr 20, 2010
19. Apr 20, 2010
### MTd2
Int/ half int labels make sense if you are talking about point particles, because you can always make them the center of cylindrical and polar coordinates and do not worry about mass distribution. But if your object is extended, a line, there will be a couple between spin and the orbital angular momentum in relation to the center of coordinates.
In the case of string theory, I really cannot understand why anyons are not straightforwardly used because it is a 1+1 theory.
20. Apr 20, 2010
### suprised
It is because of locality, or single-valuedness of correlation functions.
In higher dimensions it is simply a consequence of the representation theory of the rotation group, it just has spinor reps but there are no fractional spin reps (together with basic rules of quantum mechanics, as others have already said here). | 1,411 | 5,431 | {"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} | 2.59375 | 3 | CC-MAIN-2017-43 | longest | en | 0.888528 |
https://www.enotes.com/homework-help/determine-number-solutions-398705 | 1,498,602,948,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321938.75/warc/CC-MAIN-20170627221726-20170628001726-00349.warc.gz | 864,769,867 | 12,769 | # Determine the number of solutionsDetermine the number of solutions of the equation |4x-4| -12=0? ( one or two? )
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to use absolute value definition, such that:
`|4x - 4| = {(4x - 4, 4x - 4>=0),(4 - 4x, 4x - 4 < 0):}`
`|4x - 4| = {(4x - 4, x >= 1),(4 - 4x, x < 1):}`
Considering `x in [1,oo)` yields:
`4x - 4 - 12 = 0 => 4x = 16 => x = 4 in [1,oo)`
Considering `x in(-oo,1)` yields:
`4 - 4x - 12 = 0 => -4x - 8 = 0 => -4x = 8 => x = -2 in (-oo,1)`
Hence, evaluating the solutions to the given absolute value equation, yields `x = 4, x = -2` .
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
By definition, the absolute value means:
|p| = a>0
We'll have to solve 2 cases:
1) 4x-4 = 12
4x = 12 + 4
4x = 16
We'll divide by 4:
x = 4
2) 4x-4 = -12
We'll add 4 both sides, to isolate x to the left side:
4x = -12 + 4
4x = -8
We'll divide by 4:
x = -2
The equation has 2 solutions : {-2 ; 4}. | 419 | 1,003 | {"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} | 4.34375 | 4 | CC-MAIN-2017-26 | longest | en | 0.788959 |
https://slideplayer.com/slide/5094917/ | 1,600,494,659,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00734.warc.gz | 663,223,923 | 23,821 | # Cmpt-225 Algorithm Efficiency.
## Presentation on theme: "Cmpt-225 Algorithm Efficiency."— Presentation transcript:
cmpt-225 Algorithm Efficiency
Timing Algorithms It can be very useful to time how long an algorithm takes to run In some cases it may be essential to know how long a particular algorithm takes on a particular system However, it is not a good general method for comparing algorithms Running time is affected by numerous factors How are the algorithms coded? What computer should we use? CPU speed, memory, specialized hardware (e.g. graphics card) Operating system, system configuration (e.g. virtual memory), programming language, algorithm implementation Other tasks (i.e. what other programs are running), timing of system tasks (e.g. memory management) What data should we use? Input used: example with linear and binary search.
Cost Functions Because of the sorts of reasons just discussed for general comparative purposes we will count, rather than time, the number of operations that an algorithm performs Note that this does not mean that actual running time should be ignored! Note: running time can depend on 2 parameters; example c(n,k) “choosing k items from n”.
Cost Functions For simplicity we assume that each operation take one unit of time. If algorithm (on some particular input) performs t operations, we will say that it runs in time t. Usually running time t depends on the data size (the input length). We express the time t as a cost function of the data size n We denote the cost function of an algorithm A as tA(), where tA(n) is the time required to process the data with algorithm A on input of size n Typical example of the input size: number of nodes in a linked list, number of disks in a Hanoi Tower problem, the size of an array, the number of items in a stack, the length of a string, … Note: running time can depend on 2 parameters; example c(n,k) “choosing k items from n”.
Nested Loop for (i=1 through n){ for (j=1 through i){
for (k=1 through 5){ Perform task T; } If task T requires t units of time, the inner most loop requires 5*t time units and the loop on j requires 5*t*i time units. Therefore, the outermost loop requires
Algorithm Growth Rates.
We often want to compare the performance of algorithms When doing so we generally want to know how they perform when the problem size (n) is large So it’s simpler if we just find out how the algorithms perform as the input size grows- the growth rate.
E.g. Algorithm A requires n2/5 time units to solve a problem of size n Algorithm B requires 5*n time units to solve a problem of size n It may be difficult to come up with the above conclusions and besides they do not tell us the exact performance of the algorithms A and B. It will be easier to come up with the following conclusion for algorithms A and B Algorithm A requires time proportional to n2 Algorithm B requires time proportional to n From the above you can determine that for large problems B requires significantly less time than A.
Algorithm Growth Rates
Figure 10-1 Time requirements as a function of the problem size n
Since cost functions are complex, and may be difficult to compute, we approximate them using O notation – O notation determines the growth rate of an algorithm time.
Example of a Cost Function
Cost Function: tA(n) = n2 + 20n + 100 Which term dominates? It depends on the size of n n = 2, tA(n) = The constant, 100, is the dominating term n = 10, tA(n) = 20n is the dominating term n = 100, tA(n) = 10, , n2 is the dominating term n = 1000, tA(n) = 1,000, ,
Big O Notation O notation approximates the cost function of an algorithm The approximation is usually good enough, especially when considering the efficiency of algorithm as n gets very large Allows us to estimate rate of function growth Instead of computing the entire cost function we only need to count the number of times that an algorithm executes its barometer instruction(s) The instruction that is executed the most number of times in an algorithm (the highest order term)
Big O Notation Given functions tA(n) and g(n), we can say that the efficiency of an algorithm is of order g(n) if there are positive constants c and m such that tA(n) < c.g(n) for all n > n0 we write tA(n) is O(g(n)) and we say that tA(n) is of order g(n) e.g. if an algorithm’s running time is 3n + 12 then the algorithm is O(n). If c=3 and n0=12 then g(n) = n: 4 * n 3n + 12 for all n 12
In English… The cost function of an algorithm A, tA(n), can be approximated by another, simpler, function g(n) which is also a function with only 1 variable, the data size n. The function g(n) is selected such that it represents an upper bound on the efficiency of the algorithm A (i.e. an upper bound on the value of tA(n)). This is expressed using the big-O notation: O(g(n)). For example, if we consider the time efficiency of algorithm A then “tA(n) is O(g(n))” would mean that A cannot take more “time” than O(g(n)) to execute or that (more than c.g(n) for some constant c) the cost function tA(n) grows at most as fast as g(n)
The general idea is … when using Big-O notation, rather than giving a precise figure of the cost function using a specific data size n express the behaviour of the algorithm as its data size n grows very large so ignore lower order terms and constants
O Notation Examples In general for polynomial functions
All these expressions are O(n): n, 3n, 61n + 5, 22n – 5, … All these expressions are O(n2): n2, 9 n2, 18 n2+ 4n – 53, … In general for polynomial functions t(n)=aknk+ ak-1nk-1+…+ a1n+a0 is of O(nk) If c=(ak+ ak-1+…+a+a0 ) and n0=1 then t(n) < cnk for n>1 All these expressions are O(n log n): n(log n), 5n(log 99n), 18 + (4n – 2)(log (5n + 3)), …
Growth-rate Functions
O(1) – constant time, the time is independent of n, e.g. array look-up O(log n) – logarithmic time, usually the log is base 2, e.g. binary search O(n) – linear time, e.g. linear search O(n*log n) – e.g. efficient sorting algorithms O(n2) – quadratic time, e.g. selection sort O(nk) – polynomial (where k is some constant) O(2n) – exponential time, very slow! Order of growth of some common functions O(1) < O(log n) < O(n) < O(n * log n) < O(n2) < O(n3) < O(2n) Show that the order of function is strict. Another example: which one is bigger n^1.001 or n*log n?
Order-of-Magnitude Analysis and Big O Notation
A comparison of growth-rate functions: a) in tabular form
An intuitive example Suppose the running time of algorithm A and B are 2n and n12 respectively For small input size A performs better (n=10) For n=100 10012 is comparable to the number of molecules in a teaspoon of water 2100 is comparable to the number of molecules in a backyard swimming pool. For n=1000 is comparable is comparable to the number of molecules in a lake 21000 is already many orders of magnitudes greater than the number of particles in the whole universe!!
Order-of-Magnitude Analysis and Big O Notation
A comparison of growth-rate functions: b) in graphical form
Big O notation Note that Big O notation represents an upper bound of a cost function E.g T(n)=3n+5 is of O(n), O(n2), O(n.log(n)), O(2n) However we usually use the tightest one.
Note on Constant Time We write O(1) to indicate something that takes a constant amount of time E.g. finding the minimum element of an ordered array takes O(1) time, because the min is either at the beginning or the end of the array Important: constants can be huge, and so in practice O(1) is not necessarily efficient --- all it tells us is that the algorithm will run at the same speed no matter the size of the input we give it 21000 is of O(1)
Arithmetic of Big-O Notation
If f(n) is O(g(n)) then c.f(n) is O(g(n)), where c is a constant. Example: 23*log n is O(log n) If f1(n) is O(g(n)) and f2(n) is O(g(n)) then also f1(n)+f2(n) is O(g(n)) Example: what is order of n2+n? n2 is O(n2) n is O(n) but also O(n2) therefore n2+n is O(n2) Example: log_a(n) and log_b(n) -> base usually omitted. 2) f_i(n) is O(g_i(n)), then f_1(n)+f_2(n) is O(max{g_1(n),g_2(n)}) 2) Example: sum_{i=0}^n c_i x^i is O(x^n)
Arithmetic of Big-O Notation
If f1(n) is O(g1(n)) and f2(n) is O(g2(n)) then f1(n)*f2(n) is O(g1(n)*g2(n)). Example: what is order of (3n+1)*(2n+log n)? 3n+1 is O(n) 2n+log n is O(n) (3n+1)*(2n+log n) is O(n*n)=O(n2) 3) Example: (n+1)^5 Example: log(poly(n)) is O(log(n)) Question: is 3^n of order O(2^n)? CONCLUSION: some constants are important! n^c and n^d, c^n and d^n, even c^an and c^bn
Using Big O Notation Sometimes we need to be more specific when comparing the algorithms. For instance, there might be several sorting algorithms with time of order O(n.log n). However, an algorithm with cost function 2n.log n + 10n + 7log n is better than one with cost function 5n.log n + 2n +10log n +1 That means: We care about the constant of the main term. But we still don’t care about other terms. In such situations, the following notation is often used: 2n.log n + O(n) for the first algorithm 5n.log n + O(n) for the second one | 2,407 | 9,026 | {"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} | 3.71875 | 4 | CC-MAIN-2020-40 | latest | en | 0.861021 |
http://www.veritasprep.com/blog/2011/01/gmat-tip-of-the-week-700-score-right-above-i/ | 1,448,954,881,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398465089.29/warc/CC-MAIN-20151124205425-00192-ip-10-71-132-137.ec2.internal.warc.gz | 761,034,202 | 19,207 | Admit it: at some point over the last few months you’ve debated over the following Lil Wayne lyric:
I’ve got my gun in my boot purse, and I don’t bust back because I shoot first
What is a boot purse? Or does he, as the theory now goes, actually say “boo purse”, referring to his girlfriend’s handbag? The discussion rages, and in doing so it may well distract you from the fact that Weezy is using his track (featuring Drake) Right Above It as a blueprint for beating the GMAT. In fact, he might as well have titled it 700: Score Right Above It. As you break down the lyrics, you’ll find that Wayne drops quite a few GMAT hints, starting with a line from the chorus:
If you ain’t running with it, run from it…
Here, Weezy is talking about Young Money, the latest group of hip-hop artists to threaten to rise to Death Row Records levels (editor’s note: apologies to the G-Unit). But he may as well be talking about the GMAT. One of the easiest mistakes a novice test-taker can make is to “not run with it”, instead trying to beat the test with tricks, shortcuts, and memorization. But if you work with the test, and learn to think like those who write it, it becomes quite a bit easier. Consider a Data Sufficiency question like:
If a and b are nonzero integers, is a^b an integer?
(1) b^a is negative
(2) a^b is negative
For those who simply “studied algebra”, this question is tough…there aren’t any equations or even inequalities given, so one cannot solve this problem just by “doing math”, as many want to do. A question like this requires you to “run with” the test, learning the way that the GMAT likes to test Data Sufficiency and knowing its tendencies. Using integers, a^b should almost always be an integer, right? The only exception would be if b were negative, then creating a 1/a^-b setup in which a^b is a fraction. So given the parameters of this question, we need to determine whether b could be negative.
Statement 1 provides exactly that. Regardless of a, the only way for b^a to be negative is for b to be negative (and for a to be odd). So if b is negative, then that makes a^b a fraction, right?
Here’s where “running with the test” comes into play. The GMAT loves to use Data Sufficiency to see whether you’ll check for special circumstances in numbers. Certain numbers have incredibly unique properties but (as Wayne will warn you about below) they tend to be easy to overlook. Here, we know that with a negative value of b our a^b number will be equal to 1/(a-to-an-exponent). But there’s a chance that we would still have an integer, if a is equal to 1 or -1. For any other value of a, we’ll have a denominator larger than the numerator, but 1 and -1 have unique properties when it comes to division/multiplication. They don’t change the absolute value of the number that they divide or multiply! Knowing that the GMAT likes to test “special circumstance” numbers like 0, 1, and -1, you should immediately be asking yourself whether you can find such a number. Because statement 1 doesn’t rule out that possibility, it is not sufficient.
Statement 2, similarly, is not sufficient. We know that we can find an integer value for a^b if a is 1 or -1, and statement 2 doesn’t rule that out, either. Even together, we know that a is negative, but it can still be -1 in which case we’d get an integer; or it could be -3 or -5 in which case we’d get a noninteger. Because we can’t answer the question, the correct answer is E.
Most importantly, heed Lil Wayne’s advice – run with the test…learn to think about it like the testmakers do, looking for special circumstance numbers, understanding how certain question types require certain thought processes, etc. And if you ain’t running with it, run from it – the GMAT isn’t for the faint of heart!
As mentioned above, Weezy also includes the lyric:
You can’t see me, but never overlook me
Clearly, this is a thinly veiled reference to those special circumstance numbers that you should never overlook! We just talked about 1 and -1, but the biggest culprit is the number 0. 0 has incredibly unique properties:
0x = 0
0/x = 0
x/0 = undefined
x^0 = 1
0 is even, but neither positive nor negative
It’s an amazingly unique number (much like rapper Ol Dirty Bastard, you could say there’s no father to it’s style…), but it also literally means “nothing”. In a way, 0 doesn’t really exist – it’s the absence of value, and its inclusion in mathematical systems was a subject of debate for the early Greeks, Romans, and Babylonians. 0 can easily say “you can’t see me”, because it’s not a counting number…it can’t really be seen. But it’s also right to warn you to “never overlook me”. Because of its unique properties and easy-to-overlook nature, 0 is the key to many a GMAT math problem. As we’ve noted in this space for years, to succeed on GMAT math you should be keenly aware of the potential for 0 to impact a question. You can’t see it, but never overlook it.
Wayne follows up that warning-about-zero with another GMAT-relevant lyric (this one borrowed from Tupac):
And I ain’t a killer but don’t push me
Weezy knows as well as anyone – when we’re at our best in a vacuum, we’re as clear-minded, smart, and professional as possible, but when we’re pushed by pressure and stress we’re vulnerable to mistakes. You may not be a careless person, but the time pressure and test-day stress of the GMAT can easily push you to make careless mistakes. As we’ve campaigned for in this space many times, you should take note of the “silly” and “careless” mistakes that you make in practice, and never write them off as “but I knew that”. Test day is often the most likely time that you’ll make these mistakes again; when you’re at your best, you may not be a (GMAT score) killer, but don’t let yourself be pushed. Know the mistakes that you may make, and keep a quick-reference checklist on your GMAT noteboard so that you can double-check for the usual suspects.
Perhaps Wayne’s greatest lyric of the entire album follows his biggest-picture GMAT strategy tip (and, yes, it also violates Sentence Correction rules by ending with a preposition…):
Do it big and let the small fall under that (Damn, where you stumble at? From where they make gumbo at?)
The easiest mistake for a GMAT test-taker to make is to lose oneself in details: when studying (memorizing shortcuts, formulas, and tricks instead of focusing on concepts and strategies); when reading (focusing on details instead of the author’s organization); when approaching Sentence Correction (worrying about unique idioms and missing big-picture errors); and throughout the study and test-day process. But Weezy is spot-on with this line – do it big, and the small will fall into place. The GMAT is a reasoning test, not a knowledge test. If your study time is spent on big-picture strategies and concepts, you’ll avoid the need to stress over small details. Learning to think like the testmaker and to recognize the bigger concepts is “running with it”; if you find yourself bogged down and mired in details, you should really “run from it” until you can change your thinking. Business schools want leaders and big-picture strategic thinkers. So do it big, and let the small fall under that. Then know where you’ll stumble at? 700 on the GMAT-CAT.
(Next in the Lil Wayne teaches future MBAs series: “Young Money” – why investing in your retirement funds in your 20s is much more impactful than doing so in your 40s. )
Getting ready to take the GMAT? Our next Essentials Course starts soon in New York and in many, many other cities. And, as always, be sure to find us on Facebook and follow us on Twitter!
### 4 Responses
1. Rafi says:
LOL I can’t believe this… Nicely put though..
ps: Kudos for knowing the original was borrowed from Tupac lol
2. Brian says:
Thanks, Rafi – we’re convinced that you can have a lot of fun studying for this test…and that hip hop superstars have far more insight than anyone ever imagined!
Naturally, a West Coast company would never forget Tupac’s contributions…we listen to “To Live and Die in LA” regularly out here! It only took my coworkers a few minutes (and a few hints) to place that lyric from “Hail Mary” (which I guess technically was released by Makaveli and not Tupac,but still).
3. Stephanie says:
Love, love this (coming from a writer who used Boyz II Men lyrics as subheadings for her article on grad school in Philly). Tweeted @allgradschools.com.
• Brian Galvin says:
Thanks, Stephanie – and great article on your site! I was hoping when I first read your comment that you’d mention that classic line “…and all the Philly steaks you can eat!”. Back in school, I used to dream about that every day… | 2,089 | 8,746 | {"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-2015-48 | longest | en | 0.935527 |
https://pythagitup.com/2018/01/05/discrete-math-project-1-1-an-introduction-to-logic/ | 1,653,441,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00752.warc.gz | 508,148,868 | 22,961 | # Discrete Math Project 1.1 – An Introduction to Logic
Section 1.1 covers the basics of logic: Statements, open sentences, truth tables, negation, conjunction, and disjunction. The authors use “Aunt Buosone” in several examples. I’m excited to see what other characters I meet. The math isn’t new to me yet, but the name “Buosone” sure is.
I have never taught logic to this extent. When I taught geometry, we spent quite a bit of time discussing logic and especially reasoning, but we never actually worked with truth tables. In early August each year, I would convince myself that spending time with truth tables and doing a deep dive into logic would be a great learning experience for my students and would pay off in the long run. But each year, I backed off and decided to only spend time on “geometry” logic and not on “discrete math” logic. The payoff for truth tables didn’t seem great enough, and we did a ton of great thinking and reasoning without them. I wonder if many geometry teachers include truth tables in their curriculum. If not, do they show up anywhere in the curriculum? I learned about truth tables in PDM – Precalculus and Discrete Mathematics – using the Chicago series (UCSMP). Given the emphasis on coding, I’m curious where discrete math fits into the curriculum.
Anyway, I like making truth tables. Well, I don’t mind it. I guess it gets tedious after a while, especially once you understand them. I did find the word problems interesting. Here’s an example:
That’s just a good problem, the sort I might have given my geometry students back in the day. Although the problem does not specifically call for a truth table, I made one anyway:
While this made the questions extremely easy to answer, I wonder what value the truth table really has here. I thought about the questions and answered them before making the truth table and used it to confirm my answers. The truth table certainly adds clarity, but in doing so, does it actually remove some of the critical thinking necessary to answer the questions? Or does it seem that way only because I have a fairly strong background in logic?
Curriculum Connections
This chapter included a number of logic puzzles. You know the type – Alice and Bob go to the same school, Bob and Carol both major in history, … Which person studies physics and is under six feet tall? I’ve never done much with this sort of puzzle, but I know students tend to like them. Definitely a good resource to have around to allow for easy differentiation after an activity.
Questions to Ponder
Does constructing truth tables help students develop logical thinking and reasoning skills? Does the process become rote? Is it just another procedure to learn?
How does the knowledge gained through learning about elementary logic help students to learn about mathematical proof? What do we gain by studying logic at this level?
What are the consequences of differences between mathematical language and the vernacular? How can we help students learn how and when to use and, or, and not correctly within a mathematical context? Does the idea of an inclusive or pose a significant barrier to student understanding, and if so, how do we overcome it? | 662 | 3,204 | {"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} | 3.796875 | 4 | CC-MAIN-2022-21 | latest | en | 0.941971 |
https://brainmass.com/business/annuity/calculating-pv-fv-of-annuity-and-future-value-of-lump-sum-242515 | 1,579,895,592,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00063.warc.gz | 365,029,406 | 12,608 | Explore BrainMass
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# Calculating PV & FV of annuity and future value of lump sum
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A) After a protracted legal case, Joe won a settlement that will pay him \$11,000 each year for the next ten years. If the market interest rates are currently 5%, exactly how much should the court invest today, assuming end of year payments, so there will be nothing left in the account after the final payment is made?
B) Mary just deposited \$33,000 in an account paying 7% interest. She plans to leave the money in this account for eight years. How much will she have in the account at the end of the seventh year?
C) Mary and Joe would like to save up \$10,000 by the end of three years from now to buy new furniture for their home. They currently have \$1500 in a savings account set aside for the furniture. They would like to make equal year end deposits to this savings account to pay for the furniture when they purchase it three years from now. Assuming that this account pays 6% interest, how much should the year end payments be?
#### Solution Preview
A) After a protracted legal case, Joe won a settlement that will pay him \$11,000 each year for the next ten years. If the market interest rates are currently 5%, exactly how much should the court invest today, assuming end of year payments, so there will be nothing left in the account after the final payment is made?
Periodic payment=R=\$11000
Interest rate=i=5%
Number of annual payments=n=10
Court should deposit amount equal to present value of ordinary annuity with above ...
#### Solution Summary
There are 3 problems. Solutions to these problems explain the steps to calculate present value of annuity, future value of a lump sum amount and annual installments.
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http://www.scribd.com/doc/52583783/Dynamical-stability | 1,440,843,969,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064420.15/warc/CC-MAIN-20150827025424-00253-ip-10-171-96-226.ec2.internal.warc.gz | 678,869,515 | 35,550 | Welcome to Scribd. Sign in or start your free trial to enjoy unlimited e-books, audiobooks & documents.Find out more
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Dynamical stability
# Dynamical stability
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Dynamical stability (MAR) 1
DYNAMICAL STABILITY
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DYNAMICALSTABILITY= WORK DONE,
and;
WORK DONE (by wind/wavesetc.) = W
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) t-m
Bf Wf
Dynamical stability (MAR) 2
bhh
1
b
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ZBB
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nsidering the
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rmula:
WORK DONE (by wind/waves etc.) = W
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represents the
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nsiders the trans
erred wedge
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yancy and dynamicalstability c
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uld be calculated thus:
Dynamical Stability = W
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[v(bh + b
1
h
1
)] + B
G
Cos
U
- B
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Dynamical stability (MAR) 3
i.e.Dynamical Stability = Sum
of
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ighting mom
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nts
0
r
t
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DynamicalStability = W
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(Area 0
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I
t sh
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1998
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ses area is expressed inmetre-radians. Hence
,
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tonn
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.)Since a curve
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the angle
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G
Z(m)Heel (
Ur
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0
r Ur
## Activity (8)
You've already reviewed this. Edit your review. | 1,096 | 2,595 | {"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-2015-35 | latest | en | 0.795218 |
https://socratic.org/questions/how-do-you-multiply-56-11x-16x-2-10-15x-2-11x-56-and-state-the-excluded-values | 1,580,272,632,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00021.warc.gz | 658,065,671 | 6,575 | # How do you multiply (56+11x-16x^2)*10/(15x^2-11x-56) and state the excluded values?
May 7, 2018
$\left(56 + 11 x - 16 {x}^{2}\right) \cdot \frac{10}{15 {x}^{2} - 11 x - 56}$
$= - \frac{160 {x}^{2} - 110 x - 560}{15 {x}^{2} - 11 x - 56}$
excluding $x = - \frac{8}{5} \mathmr{and} x = \frac{7}{3}$
#### Explanation:
There's a $15 {x}^{2}$ in the denominator and a $- 16 {x}^{2}$ in the numerator, so the cancelling that we might hope for at first glance doesn't materialize.
The excluded values occur at the zeros of the denominator; let's try to factor. We seek a pair of factors of $15$ and a pair of factors of $- 56$ whose sum of products is $- 11.$ That's a bit of a search, we have
$15 = 1 \setminus \times 15 = 3 \setminus \times 5$
$56 = 1 \setminus \times 56 = 2 \setminus \times 28 = 4 \setminus \times 14 = 7 \setminus \times 8$ with a minus sign in there for $- 56.$
Eventually we find $- 7 \setminus \times 8 = - 56 ,$ $5 \times 3 = 15 ,$ -7(5)+3(8)=-11 quad sqrt
$15 {x}^{2} - 11 x - 56 = \left(5 x + 8\right) \left(3 x - 7\right)$
We could also have used the quadratic formula to find the zeros of the denominator, which are called poles.
$x = - \frac{8}{5} \mathmr{and} x = \frac{7}{3}$
The multiplication itself is rather vacuous due to the lack of cancelling.
$\left(56 + 11 x - 16 {x}^{2}\right) \cdot \frac{10}{15 {x}^{2} - 11 x - 56}$
$= - \frac{160 {x}^{2} - 110 x - 560}{15 {x}^{2} - 11 x - 56}$ | 558 | 1,434 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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} | 4.65625 | 5 | CC-MAIN-2020-05 | longest | en | 0.747139 |
http://notes.tyrocity.com/progressive-wave/ | 1,529,294,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860041.64/warc/CC-MAIN-20180618031628-20180618051628-00515.warc.gz | 245,085,638 | 9,198 | Progressive wave
A wave in which the crest and trough or compression and rarefaction travel toward is called progressive wave. In progressive wave, the crest and trough or compression and rarefaction changes its position continuously and the velocity of move equals to the velocity of wave.
Consider a wave travelling along positive X-axis with a velocity “v” as in figure. If y be the displacement of the particle at “o” then,
Consider a particle at “p” which is at a distance x from point “p”. Since the particle at “o” and “p” are not in the same phase. Then displacement of the particle “y” given by, | 133 | 607 | {"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-2018-26 | latest | en | 0.937045 |
https://www.intmath.com/functions-and-graphs/lines-of-symmetry-in-a-rectangle.php | 1,719,082,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00416.warc.gz | 715,822,864 | 23,028 | Search IntMath
Close
# Lines of Symmetry in a Rectangle
Understanding lines of symmetry is an important part of geometry. Symmetry is the property of a shape or object being the same on both sides. In a rectangle, two lines of symmetry divide the shape into two identical halves, making it easier to analyze and understand the properties of rectangles. Let’s take a closer look at lines of symmetry in rectangles and how they are used in geometry.
## Lines of Symmetry in Rectangles
A line of symmetry bisects a shape, dividing it into two equal halves that are mirror images. In a rectangle, there are always two lines of symmetry—one vertical line and one horizontal line—that intersect at its center point. These lines divide the rectangle into four identical quadrants that can be used to analyze its properties and create equations for calculations such as area and perimeter.
## Using Lines of Symmetry for Analysis
Lines of symmetry in rectangles can be used for various analytical tasks. For example, you can use them to calculate the area or perimeter quickly since each side has an exact opposite side with which it shares dimensions. You can also draw diagonals from each corner to find out if all angles are equal (90 degrees). Drawing diagonals between opposite corners creates four congruent triangles that help you determine angles and lengths more accurately than measuring them independently from each other.
## Efficiency with Lines of Symmetry
In addition to analysis, understanding lines of symmetry in rectangles can help you work faster when solving problems involving area and perimeter calculations. Since each side is equal to its counterpart across the line(s) of symmetry, you only need to measure one side before finding the length or width (or any other dimension) for the entire rectangle. This saves time compared with measuring all sides individually and helps ensure accuracy when working with large-scale shapes or multiple shapes.
# Conclusion
Knowing how to identify and use lines of symmetry in rectangles is an essential part of geometry that students should master early on in their studies. Identifying these lines makes analyzing various properties easier and more accurate while saving time when solving complex problems involving measurements and calculations such as area or perimeter. With this knowledge under your belt, you’ll be better prepared for future mathematical endeavors!
## FAQ
### Why does a rectangle have 4 lines of symmetry?
A rectangle does not have four lines of symmetry—it only has two. The two lines of symmetry in a rectangle are one vertical line and one horizontal line that intersect at its center point, dividing the shape into four equal quadrants. These two lines of symmetry divide the rectangle into two identical halves.
### How do you explain lines of symmetry?
Lines of symmetry are lines or planes that divide a shape into two equal halves. They bisect the shape, creating two mirror images of each other. In a rectangle, there are always two lines of symmetry: one vertical line and one horizontal line that intersect at its center point. These lines can be used to analyze various properties of rectangles and to work more efficiently when solving problems involving area and perimeter calculations. | 610 | 3,295 | {"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} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.928476 |
https://de.mathworks.com/matlabcentral/cody/problems/2017-side-of-an-equilateral-triangle/solutions/2078836 | 1,585,519,908,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00288.warc.gz | 435,408,445 | 15,555 | Cody
# Problem 2017. Side of an equilateral triangle
Solution 2078836
Submitted on 4 Jan 2020 by Justin Hughes
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = 1; x_correct = 2/sqrt(sqrt(3)); tolerance = 1e-12; assert(abs( side_length(A) - x_correct ) < tolerance)
x = 1.5197
2 Pass
A = sqrt(3); x_correct = 2; tolerance = 1e-12; assert(abs(side_length(A) - x_correct) < tolerance)
x = 2
3 Pass
A = 2; x_correct = 2*sqrt(2)/sqrt(sqrt(3)); tolerance = 1e-12; assert(abs(side_length(A) - x_correct) < tolerance)
x = 2.1491 | 213 | 655 | {"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} | 2.890625 | 3 | CC-MAIN-2020-16 | latest | en | 0.670614 |
http://openstudy.com/updates/558870c0e4b03f41bbd81ff6 | 1,495,834,104,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608684.93/warc/CC-MAIN-20170526203511-20170526223511-00159.warc.gz | 334,487,769 | 9,128 | • anonymous
Can someone give me the answer please. The table below shows two equations: Equation 1 |4x − 3|− 5 = 4 Equation 2 |2x + 3| + 8 = 3 Which statement is true about the solution to the two equations? A)Equation 1 and equation 2 have no solutions. B)Equation 1 has no solution, and equation 2 has solutions x = −4, 1. C)The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution. D)The solutions to equation 1 are x = 3, −1.5, and equation 2 has solutions x = −4, 1.
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Looking for something else?
Not the answer you are looking for? Search for more explanations. | 404 | 1,420 | {"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-2017-22 | longest | en | 0.59548 |
https://science.blurtit.com/136783/what-is-the-conversion-of-celsius-temp-to-fahrenheit-for-381 | 1,670,264,102,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00801.warc.gz | 541,693,121 | 10,933 | # What Is The Conversion Of Celsius Temp To Fahrenheit For 38.1?
Multiply the Celsius temperature by 9. Divide the product by 5. Then add 32 to that quotient. The answer is100.58
38.1X9=342.9 342.9/5=68.58 68.58 + 32=100.58
The formula is (C X 9/5) + 32 = F
thanked the writer.
You have not mentioned that from which temperature scale 38.1 degree figure is to be converted into Fahrenheit. However, there are three temperature scales and the unit degree is used only in Celsius and Fahrenheit therefore, I am assuming that the figure is to be converted from Celsius to Fahrenheit. Therefore,
38.1 degree Celsius = 100.58 degree F
thanked the writer.
100
thanked the writer. | 184 | 677 | {"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-2022-49 | latest | en | 0.898388 |
https://whatisconvert.com/178-acres-in-square-inches | 1,606,947,221,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00625.warc.gz | 551,358,410 | 7,569 | # What is 178 Acres in Square Inches?
## Convert 178 Acres to Square Inches
To calculate 178 Acres to the corresponding value in Square Inches, multiply the quantity in Acres by 6272640 (conversion factor). In this case we should multiply 178 Acres by 6272640 to get the equivalent result in Square Inches:
178 Acres x 6272640 = 1116529920 Square Inches
178 Acres is equivalent to 1116529920 Square Inches.
## How to convert from Acres to Square Inches
The conversion factor from Acres to Square Inches is 6272640. To find out how many Acres in Square Inches, multiply by the conversion factor or use the Area converter above. One hundred seventy-eight Acres is equivalent to one billion one hundred sixteen million five hundred twenty-nine thousand nine hundred twenty Square Inches.
## Definition of Acre
The acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. It is defined as the area of 1 chain by 1 furlong (66 by 660 feet), which is exactly equal to 1⁄640 of a square mile, 43,560 square feet, approximately 4,047 m2, or about 40% of a hectare. The most commonly used acre today is the international acre. In the United States both the international acre and the US survey acre are in use, but differ by only two parts per million, see below. The most common use of the acre is to measure tracts of land. One international acre is defined as exactly 4,046.8564224 square metres.
## Definition of Square Inch
A square inch (plural: square inches) is a unit of area, equal to the area of a square with sides of one inch. The following symbols are used to denote square inches: square in, sq inches, sq inch, sq in inches/-2, inch/-2, in/-2, inches^2, inch^2, in^2, inches2, inch2, in2. The square inch is a common unit of measurement in the United States and the United Kingdom.
## Using the Acres to Square Inches converter you can get answers to questions like the following:
• How many Square Inches are in 178 Acres?
• 178 Acres is equal to how many Square Inches?
• How to convert 178 Acres to Square Inches?
• How many is 178 Acres in Square Inches?
• What is 178 Acres in Square Inches?
• How much is 178 Acres in Square Inches?
• How many in2 are in 178 ac?
• 178 ac is equal to how many in2?
• How to convert 178 ac to in2?
• How many is 178 ac in in2?
• What is 178 ac in in2?
• How much is 178 ac in in2? | 591 | 2,369 | {"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} | 3.234375 | 3 | CC-MAIN-2020-50 | latest | en | 0.884495 |
https://techbullion.com/excelling-at-utilizing-fraction-bar-in-numerical-statements/ | 1,713,342,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00780.warc.gz | 480,513,302 | 56,476 | Latest News
Excelling at Utilizing fraction bar in Numerical statements
Fraction bars, otherwise called division strips or part tiles, are significant apparatuses in the domain of math, especially in understanding and working with parts. These visual guides give a substantial portrayal of partial connections, making complex ideas more open and working with critical thinking. Whether in study halls or at home, excelling at involving part bars can upgrade understudies’ perception and capability in handling many numerical issues. In this article, we will dive into the basics of division bars and investigate methodologies for actually using them to overcome different number-related difficulties.
Understanding Division Bars:
• Portion bars comprise sectioned strips or tiles, each addressing a negligible part of an entirety.
• Normally, these bars are variety-coded or named to mean different fragmentary qualities, permitting clients to analyze and control divisions outwardly.
• For instance, a part bar might be separated into equal parts, thirds, fourths, fifths, etc, contingent upon the ideal degree of granularity.
• One of the essential benefits of division bars is their capacity to outline fragmentary connections substantially and naturally.
• By outwardly dividing an entire into two halves, understudies can embrace ideas like identical portions, expansion, deduction, duplication, division, and correlations with more prominent clearness.
• Also, division bars act as successful apparatuses for critical thinking, empowering understudies to envision tasks and foster vital ways to deal with and tackle numerical conditions.
Techniques for Using Division Bars:
1. Figuring out Identical Portions:
Portion bars offer a unique stage for investigating comparable parts, which are divisions that address a similar amount notwithstanding having various numerators and denominators. By overlaying division bars of various lengths and noticing their arrangement, understudies can recognize examples and connections between parts. For example, adjusting a half-length bar to two 33% bars shows the equality between 1/2 and 2/3, building up the idea of comparable divisions through visual exhibition.
With regards to adding or deducting divisions, portion bars give a clear technique for consolidating or looking at partial amounts. To add divisions utilizing portion bars, understudies can adjust bars of a similar length evenly and count them all out number of sections to decide the total. Essentially, for deduction, understudies can outwardly address the deduction cycle by eliminating fragments from the first division bar. This active methodology cultivates a more profound comprehension of portion expansion and deduction by solidly showing the ideas in question.
3. Increasing and Isolating Portions:
Portion bars likewise work with duplication and division of parts by giving a visual portrayal of the tasks in question. To increase portions, understudies can overlay part bars upward and notice the convergence focuses to decide the item. For division, portion bars can be utilized to exhibit the idea of separating an entire into partial parts, assisting understudies with imagining the connection between the profit, divisor, and remainder. By controlling part bars, understudies can foster a reasonable comprehension of portion duplication and division, preparing for familiarity with tackling related issues.
4. Contrasting Divisions:
Division bars help in contrasting portions by empowering understudies to outwardly evaluate the overall sizes of various parts. By adjusting division bars next to each other and noticing their lengths, understudies can instinctively figure out which part is bigger or more modest. This visual way to deal with division examination supports the idea of comparable parts and gives a substantial premise for grasping portion disparities. Moreover, part bars can be utilized to arrange portions on a number line, further improving understudies’ spatial thinking abilities and theoretical comprehension of divisions.
5. Critical thinking Applications:
Past fundamental number juggling tasks, and part bars can be applied to settle many numerical issues across different areas, including math, polynomial math, and certifiable situations. Whether computing areas of mathematical shapes, tackling conditions including divisions, or deciphering information in graphical portrayals, part bars offer a flexible device for conceptualizing and taking care of complicated issues. By coordinating portion bars into critical thinking exercises, teachers can advance decisive reasoning, numerical thinking, and pragmatic use of part ideas in different settings.
Final Thought
Excelling at utilizing fraction bars isn’t just about controlling visual guides; it is tied in with fostering a more profound comprehension of fragmentary ideas and improving critical thinking abilities in math. By utilizing portion bars to investigate identical divisions, perform number-crunching tasks, analyze parts, and tackle true issues, understudies can upgrade their numerical capability and certainty. As teachers and students the same hug the adaptability and viability of division bars in the homeroom, they engage understudies to open the groundbreaking capability of parts in math and then some. With training, tolerance, and constancy, excelling at utilizing division bars becomes feasible as well as enormously compensating in molding numerical proficiency and cultivating a long-lasting appreciation for the excellence of math. | 983 | 5,581 | {"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} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.918423 |
https://codegolf.stackexchange.com/questions/49223/evolution-of-oeis/49337 | 1,713,348,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00776.warc.gz | 157,908,038 | 68,926 | # Evolution of OEIS
In this challenge, the goal is to recreate the On-Line Encyclopedia of Integer Sequences one sequence at a time. Similar to the Evolution of Hello World, each answer depends on a previous answer.
Over time, this challenge will create a "family tree" of the OEIS sequences. It is simple to add on to this tree.
1. Find a previous answer, which can be at any depth N of the tree.
2. Determine the first N numbers generated by that answer's sequence.
3. Find a sequence in OEIS which starts with those same numbers and which hasn't been used before.
4. Write a program to generate this new sequence you just found.
5. Submit your answer as depth N+1
Since the level of your answer influences scoring, you should always add your answer onto the tree at the deepest level possible. If you cannot fit your answer anywhere on the tree, you can start a new branch of the tree and put your answer as depth 1.
There are a few ways to output a sequence.
The first option is to write a program or function that inputs a number (from STDIN or as an argument) and returns the Nth number in your chosen sequence. You can assume that the sequence will be defined for N and that N and S_N are "reasonably sized" (so it won't cause overflows). You can also use any reasonable indexing, such as 0 indexing, 1 indexing, or the indexing listed under "offset" on the sequence's OEIS page, that doesn't matter. The term produced by the first index must match the first term of the OEIS entry.
The second option is to write a program or function that inputs a number and returns the first N terms of the sequence. The first terms of the output must be the first terms of the OEIS entry (you can't leave off the first few terms). Consecutive terms must be delimited by arbitrary strings of non-digit characters, so 0,1 1.2/3,5;8,11 works but 011235811 does not count.
The third option is to create a program that outputs a continuous stream of numbers. Similarly to the second option, there must be delimiters between consecutive terms.
Your answer should contain a header like this to aid Stack Snippet parsing:
# [language], [number] bytes, depth [number], A[new sequence] from A[old sequence]
Your answer should contain the code to generate the sequence, along with the first few terms that any descendants will need to contain. These few terms should be preceded by the exact word terms: so that the controller can use them as part of the tree diagram. It is also recommended to write a description of the sequence you chose.
If your post is a depth 1 answer and thus has no ancestor, you should simply omit the from A[number] in your header.
Here is an example answer:
# Perl, 26 bytes, depth 3, A026305 from A084912
various code here
and here
The next answer should match the following terms:
1, 4, 20
This sequence is .... and does ....
## Chaining Requirements
In order to make this challenge more fair, there are restrictions on which answers you can chain yours to. These rules are mostly to prevent a single person from creating a whole branch of the tree by themselves or owning a lot of "root" nodes.
• You cannot chain to yourself.
• You cannot directly chain two of your answers to the same ancestor.
• You cannot make more than one "Level 1" answer.
Also, if the ancestor was of depth N, your post must have depth N+1, even if more than the required number of terms agree.
## Scoring
Your score as a user is the sum of the scores of all of your answers. The score of a single answer is determined by the following formula:
Answer Score = Sqrt(Depth) * 1024 / (Length + 256)
This scoring system should encourage users to submit a large number of deeper answers. Shorter answers are preferred over longer answers, but depth has a much larger influence.
Below is a stack snippet that generates a leaderboard as well as a tree diagram of all of the answers. I would like to thank Martin Büttner and d3noob as the sources for a lot of this code. You should click "Full screen" to see the complete results.
function answersUrl(t){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+t+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(t){answers.push.apply(answers,t.items),t.has_more?getAnswers():process()}})}function shouldHaveHeading(t){var e=!1,r=t.body_markdown.split("\n");try{e|=/^#/.test(t.body_markdown),e|=["-","="].indexOf(r[1][0])>-1,e&=LANGUAGE_REG.test(t.body_markdown)}catch(a){}return e}function shouldHaveScore(t){var e=!1;try{e|=SIZE_REG.test(t.body_markdown.split("\n")[0])}catch(r){}return e}function getAuthorName(t){return t.owner.display_name}function decodeEntities(t){return ("<textarea>").html(t).text()}function process(){answers=answers.filter(shouldHaveScore).filter(shouldHaveHeading),answers.reverse();var t={},e=[],r=1,a=null,n=1,s=[];answers.forEach(function(t){var r=t.body_markdown.split("\n")[0],a=getAuthorName(t),n=r.match(SEQUENCE_REG)[0];n=n.trim();var o="from A000000";PARENT_REG.test(r)&&(o=r.match(PARENT_REG)[0]),o=o.substring(5).trim(),"A000000"==o&&(o="OEIS");var i="";SEQDATA_REG.test(t.body_markdown)&&(i=t.body_markdown.match(SEQDATA_REG)[1]);for(var u=!0,c=0;c<e.length;++c)u=u&&!(e[c]===n);for(var l=!0,c=0;c<e.length;++c)l=!(!l||e[c]===n||e[c]===n+a||e[c]===o+a);e.push(n),e.push(n+a),e.push(o+a),u&&data.push({name:n,parent:o,term:i+" : ",author:decodeEntities(a),URL:t.share_link}),l&&s.push(t)}),answers.sort(function(t,e){var r=t.body_markdown.split("\n")[0].match(SEQUENCE_REG),a=e.body_markdown.split("\n")[0].match(SEQUENCE_REG);return a>r?-1:r>a?1:void 0}),answers.forEach(function(e){var o=e.body_markdown.split("\n")[0],i=(o.match(NUMBER_REG)[0],(o.match(SIZE_REG)||[0])[0]),u=parseInt((o.match(DEPTH_REG)||[0])[0]).toString(),c=o.match(SEQUENCE_REG)[0],l="from A000000";PARENT_REG.test(o)&&(l=o.match(PARENT_REG)[0]),l=l.substring(5);var d=o.match(LANGUAGE_REG)[1];d.indexOf("]")>0&&(d=d.substring(1,d.indexOf("]")));for(var p=getAuthorName(e),E=!1,h=0;h<s.length;++h)E=E||s[h]===e;if(E){var f=jQuery("#answer-template").html();i!=a&&(n=r),a=i,++r;var m=1024*Math.pow(parseInt(u),.5)/(parseInt(i)+256);f=f.replace("{{SEQUENCE}}",c).replace("{{SEQUENCE}}",c).replace("{{NAME}}",p).replace("{{LANGUAGE}}",d).replace("{{SIZE}}",i).replace("{{DEPTH}}",u).replace("{{LINK}}",e.share_link),f=jQuery(f),jQuery("#answers").append(f),t[p]=t[p]||{lang:d,user:p,size:"0",numanswers:"0",link:e.share_link},t[p].size=(parseFloat(t[p].size)+m).toString(),t[p].numanswers=(parseInt(t[p].numanswers)+1).toString()}});var o=[];for(var i in t)t.hasOwnProperty(i)&&o.push(t[i]);o.sort(function(t,e){return parseFloat(t.size)>parseFloat(e.size)?-1:parseFloat(t.size)<parseFloat(e.size)?1:0});for(var u=0;u<o.length;++u){var c=jQuery("#language-template").html(),i=o[u];c=c.replace("{{RANK}}",u+1+".").replace("{{NAME}}",i.user).replace("{{NUMANSWERS}}",i.numanswers).replace("{{SIZE}}",i.size),c=jQuery(c),jQuery("#languages").append(c)}createTree()}function createTree(){function t(){var t=i.nodes(root).reverse(),e=i.links(t);t.forEach(function(t){t.y=180*t.depth});var r=c.selectAll("g.node").data(t,function(t){return t.id||(t.id=++o)}),a=r.enter().append("g").attr("class","node").attr("transform",function(t){return"translate("+t.y+","+t.x+")"});a.append("a").attr("xlink:href",function(t){return t.URL}).append("circle").attr("r",10).style("fill","#fff"),a.append("text").attr("x",function(){return 0}).attr("y",function(){return 20}).attr("dy",".35em").attr("text-anchor",function(){return"middle"}).text(function(t){return t.term+t.name}).style("fill-opacity",1),a.append("text").attr("x",function(){return 0}).attr("y",function(){return 35}).attr("dy",".35em").attr("text-anchor",function(){return"middle"}).text(function(t){return t.author}).style("fill-opacity",1);var n=c.selectAll("path.link").data(e,function(t){return t.target.id});n.enter().insert("path","g").attr("class","link").attr("d",u)}var e=data.reduce(function(t,e){return t[e.name]=e,t},{}),r=[];data.forEach(function(t){var a=e[t.parent];a?(a.children||(a.children=[])).push(t):r.push(t)});var a={top:20,right:120,bottom:20,left:120},n=3203-a.right-a.left,s=4003-a.top-a.bottom,o=0,i=d3.layout.tree().size([s,n]),u=d3.svg.diagonal().projection(function(t){return[t.y,t.x]}),c=d3.select("body").append("svg").attr("width",n+a.right+a.left).attr("height",s+a.top+a.bottom).append("g").attr("transform","translate("+a.left+","+a.top+")");root=r[0],t(root)}var QUESTION_ID=49223,ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",data=[{name:"OEIS",parent:"null",term:"",author:"",URL:"https://oeis.org/"}],answers=[],page=1;getAnswers();var SIZE_REG=/\d+(?=[^\d&]*(?:<(?:s>[^&]*<\/s>|[^&]+>)[^\d&]*)*,)/,DEPTH_REG=/\d+, A/,NUMBER_REG=/\d+/,LANGUAGE_REG=/^#*\s*([^,]+)/,SEQUENCE_REG=/A\d+/,PARENT_REG=/from\s*A\d+/,SEQDATA_REG=/terms:\s*(?:(?:-)?\d+,\s*)*((?:-)?\d+)/; body{text-align: left !important}#answer-list{padding: 10px; width: 550px; float: left;}#language-list{padding: 10px; width: 290px; float: left;}table thead{font-weight: bold;}table td{padding: 5px;}.node circle{fill: #fff; stroke: steelblue; stroke-width: 3px;}.node text{font: 12px sans-serif;}.link{fill: none; stroke: #ccc; stroke-width: 2px;} <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><script src="http://d3js.org/d3.v3.min.js"></script><link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"><div id="answer-list"> <h2>Sequence List</h2> <table class="answer-list"> <thead> <tr> <td>Sequence</td><td>Author</td><td>Language</td><td>Size</td><td>Depth</td></tr></thead> <tbody id="answers"></tbody> </table></div><div id="language-list"> <h2>Leaderboard</h2> <table class="language-list"> <thead> <tr> <td>Rank</td><td>User</td><td>Answers</td><td>Score</td></tr></thead> <tbody id="languages"></tbody> </table></div><table style="display: none"> <tbody id="answer-template"> <tr> <td><a href="https://oeis.org/{{SEQUENCE}}">{{SEQUENCE}}</a></td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td>{{DEPTH}}</td><td><a href="{{LINK}}">Link</a> </td></tr></tbody></table><table style="display: none"> <tbody id="language-template"> <tr> <td>{{RANK}}</td><td>{{NAME}}</td><td>{{NUMANSWERS}}</td><td>{{SIZE}}</td></tr></tbody></table> • You know, I think this is might just be the single coolest codegolf.sx question I've ever seen asked. It's not just cool, but actually useful as an archive. Apr 26, 2015 at 19:08 • Given the OEIS is online, takes N terms of a sequence as a search term, and contains mathematica or maple code for many of the sequences, it would be possible to write a meta-entry which searched for the best scoring entry for which code exists in OEIS which is a descendent of any given entry here and posted it. Apr 26, 2015 at 19:10 • Can I recommend some way of marking on the graph the snippet generates that a node is terminal, ie there are no unused sequences of a greater depth available on the OEIS? Apr 27, 2015 at 5:52 • I think the only way to keep this challenge going would be to provide something where you give your username, and it lists the OEIS problems you could do, in order from highest depth to lowest. Otherwise it takes too long to find the next sequence to post. May 4, 2015 at 22:38 • The SVG is slightly too narrow. Apr 8, 2016 at 1:17 ## 150 Answers # Haskell, 32 bytes, depth 4, A015919 from A011760 filter(\n->mod(2^n-2)n==0)[1..] The next sequence should start with the following terms: 1, 2, 3, 5 It outputs an infinite list: [1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,... It looks like it's just the sequence of 1 together with the prime numbers. In fact, the first composite number in the sequence is 341, and the first composite even number is 161038. # R, 79 bytes, depth 3, A000945 from A000043 f=function(n)if(n-1){m=prod(sapply(1:(n-1),f))+1;p=2;while(m%%p)p=p+1;p}else{2} The next sequence should contain the following terms: 2, 3, 7 The sequence is such that a(n+1) is the smallest prime factor of prod(a(k) for k in 1:n)+1. f=function(n)if(n-1){m=prod(sapply(1:(n-1),f))+1;p=(2:m)[!m%%2:m];p[1]}else{2} would be one character shorter but can't perform f(9) because it can't create of vector of length 38 709 183 810 570. # Piet, 96 bytes, depth 5, A209229 from A010060 with size 10 codels for legibility. Tested on PietDev. The next sequence should start with the terms: 0, 1, 1, 0, 1 This is the characteristic function of powers of 2 (i. e. it outputs 1 if n is a power of 2 and 0 if not). It starts on the red codel in the first row: it takes input as stdin and pushes it to the stack n, it then duplicates it twice n n nand pushes 1 to the stack n n n 1 [the loop starts on the red codel at the end of the second row] that it also duplicates twice n n n 1 1 1 it then rolls the first one near the bottom n 1 n n 1 1 and again n 1 1 n n 1 then check if the first two are equal (if yes than it is a power of 2 so it change direction and outputs 1), then check if the first number of the stack (on first iteration, 1) is smaller than the second. If yes it stops and output 0 and if no it continues on the loop (at that point the stack only contains n 1 as the first two pairs of n 1 were used for the comparisons). It pushes 2 on the stack and multiply n 1 2->n 2 then rolls it at the bottom, 2 n, duplicates twice 2 n n n rolls the bottom number on top of the stack n n n 2 and reaches back the beginning of the loop. # Java, 186 bytes, depth 10, A073719 from A072490 int f(int n){return g(1<<n,0)/g(1<<n,1);}int g(int n,int o){int k=2,c=1;while(c<n)if(h(++k)>0&&o<1||h(k)<1&&o>0)c++;return k;}int h(int n){for(int k=2;k<n;)if(n%k++<1)return 0;return 1;} Long version: int f(int n) { return g(1<<n,0) / g(1<<n,1); } int g(int n, int o) { int k=2, c=1; while(c<n) if (h(++k)>0 && o<1 || h(k)<1 && o>0) c++; return k; } int h(int n) { for (int k=2; k<n;) if (n % k++ < 1) return 0; return 1; } This is the sequence of the 2nth prime numbers divided by the 2nth composite numbers for all natural numbers greater than 1, rounded down. Method g finds the 2nth prime or composite number while method h checks if numbers are prime or composite. Due to the inefficiency of the algorithm, lag becomes apparent at n = 12. This is a one-indexed function (n starts at 1). The next answer should start with the following terms: 0, 0, 1, 2, 2, 3, 4, 5, 5, 6 • Oh that's what prime(2^n) meant... really had trouble figuring that out for some reason. Apr 29, 2015 at 2:40 • I was also confused by that at first. What really hung me up was the formula for composite numbers; the denominator of the first one made no sense. :P – TNT Apr 29, 2015 at 3:13 # Java, 29 bytes, depth 5, A000325 from A000108 int f(int n){return(1<<n)-n;} This sequence is 2n-n for all n >= 0. This is a zero-indexed function. The next answer should start with the following terms: 1, 1, 2, 5, 12 # Java, 30 bytes, depth 6, A016116 from A000010 int f(int n){return 1<<(n/2);} This sequence is the powers of two, twice. This is a zero-indexed function. The next answer should match the following terms: 1, 1, 2, 2, 4, 4 # Ruby, 75 bytes, depth 2, A125710 from A000211 def r n;n==2??0:n%2==0??0+r(n/2):?1+r(n*3+1);end;p (r 2*gets.to_i+1).to_i 2 The next sequence should contain the following terms: 4, 80 This is a complicated sequence so I will just quote OEIS: In the "3x+1" problem, let 0 denote a halving step and 1 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n+1 and reading it as a decimal number. My code is not short; but It was fun to make. # Ruby, 35 bytes, depth 6, A008998 from A014137 a=->n{n<0?0:n>0?2*a(n-1)+a(n-3):1} The next sequence should contain the following terms: 1, 2, 4, 9, 20, 44 Defined recursively as:a(n) = 2*a(n-1) + a(n-3). # Ruby, 26 bytes, depth 4, A006882 from A000142 a=->n{n>1?n*a.call(n-2):1} The next answer should start with the terms: 1,1,2,3 The Double factorial sequence. # Ruby, 30 bytes, depth 6, A010051 from A209229 ->n{(2..n).one?{|x|n%x<1}?1:0} The next answer should start with the terms: 0,1,1,0,1,0 # Mathematica, 51 bytes, depth 11, A025765 from A226190 SeriesCoefficient[1/((1-x)(1-x^2)(1-x^9)),{x,0,#}]& Just another infinite series. The next answer should start with the terms: 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7 # Python, 14 bytes, depth 3, A010872 from A000030 x=lambda x:x%3 The next sequence should start with the following terms: 0, 1, 2 # Haskell, 32 bytes, depth 7, A018819 from A016116 l=1:zipWith(+)l(taill>>=(:[0]))
This defines an infinite list containing all the values.
As more easily readable function that (could be golfed some more and) is much less efficient the sequence can be defined by
a 0=1; a m|odd m=a(m-1); a m=a(m-1)+a(mdiv2)
So it starts with 1, and then we always take the previous value and, in the even case, add the value from m/2. The list definition uses l>>=(:[0]) to create a list that alternates between values from l and zeroes ([l0,0,l1,0,l2,0,...]) to achieve the same effect.
The sequence is the binary partition function, giving the number of partitions of a number into powers of 2, for example for 5 it is 4, because 5 can be written in four ways: as 1+1+1+1+1, 1+1+1+2, 1+2+2 and 1+4.
The next sequence should match the following terms:
1, 1, 2, 2, 4, 4, 6
# Vyxal, 5 bytes, depth 5, A000041 from A006882
[ṄL|1
Try it Online!
Calculates A000041(n). Could have been 2 bytes, but it wouldn't have worked correctly for an input of 0. The sequence is the number of integer partitions of n.
[ṄL|1 # Takes an integer input
[ # If input is not 0...
Ṅ # Integer partitions of n
L # And take the length
| # Otherwise...
1 # Push 1
# After which the top of the stack is printed
The next answer should match the following terms:
1, 1, 2, 3, 5
# Vyxal, 6 bytes, depth 6, A002379 from A000041
3$e$Eḭ
Try it Online!
The next answer should match the following terms:
1, 1, 2, 3, 5, 7
# Vyxal, 9 bytes, depth 5, A027383 from A000123
→ʀƛ←ε∵E;∑
Try it Online!
The next answer should match the following terms:
1, 2, 4, 6, 10
# Vyxal, 4 bytes, depth 8, A079584 from A01881
¡b1O
Try it Online!
The next sequence should match the following terms:
1, 1, 2, 2, 4, 4, 6, 6
# Mathematica, 6 bytes, depth 1, A002522
#^2+1&
1
# Mathematica, 4 bytes, depth 2, A000244 from A000012
3^#&
Powers of 3. The next answer should start with the terms:
1, 3
# Mathematica, 3 bytes, depth 2, A005843 from A001477
2#&
Even numbers. The next answer should start with the terms:
0, 2
# CJam, 21 bytes, depth 7, A095884 from A025581
1{_),(;{_@_@-@#p}/)}h
0, 1, 0, 2, 1, 0, 3
My second CJam program! I wonder if there's a better way to get n (n-k)^k from n k than _@_@-@# - any tips?
This sequence is simply "Triangle read by rows: T(n,k) = (n-k)^k, n>=1, 1<=k<=n.".
• Since this violates the chaining requirement of "You cannot directly chain two of your answers to the same ancestor" this answer does not count towards your score, although it will still appear on the diagram. Apr 26, 2015 at 22:22
• Oops, I seem to have forgotten that rule. Thanks for the reminder. Apr 26, 2015 at 22:59
# CJam, 5 bytes, depth 5, A179081 from A001590
r1b2%
0, 1, 0, 1, 0
Parity of decimal digit sum.
# CJam, 9 bytes, depth 3, A025192 from A000027
3ri(#2*m]
1, 2, 6
2*3^(n-1) with a special case.
For that sequence.
# PARI/GP, 19 bytes, depth 4, A056486 from A016957
n->(9*2^n+(-2)^n)/4
It's just (9*2^n+(-2)^n)/4. The old name was "Number of periodic palindromes using a maximum of four different symbols", but I don't know what it means.
The next answer should match the following terms:
4, 10, 16, 40
# Octave, 28 bytes, depth 6, A196564 from A179081
@(n)sum(mod(int2str(n)+0,2))
Number of odd digits in decimal representation of n.
0, 1, 0, 1, 0, 1
# CJam, 13 bytes, depth 7, A037888 from A196564
ri2b_W%.^:+2/
0, 1, 0, 1, 0, 1, 0
How many bits to change to make a palindrome binary number.
# Pyth, 16 bytes, depth 3, A006521 from A000244
@f!%+^2T1Tr1^3QQ
Takes input via STDIN and outputs the Nth number.
### Explanation
@ Q Nth item of
f filter by
!%+^2T1T (2^Z + 1) mod Z is 0
r1^3Q integers 1..3^(N-1)
The next answer should match the following terms:
1, 3, 9
# CJam, 25 bytes, depth 13, A238263 from A135020
ri,(;{2*mF,3<}%_W%.&1b)2/
1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7
Number of ways of n=2^k1*p1^k2+2^k3*p2^k4, starting from n=2.
# R, 36 bytes, depth 9, A083912 from A103181
f=function(n)sum(!n%%1:n&1:n%%10==2)
The next sequence should match the following terms:
0, 1, 0, 1, 0, 1, 0, 1, 0
It is the sequence containing the number of divisors of n that are congruent to 2 modulo 10.
Usage:
> f(1)
[1] 0
> f(9)
[1] 0
> sapply(1:40, f)
[1] 0 1 0 1 0 1 0 1 0 1 0 2 0 1 0 1 0 1 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 1 0 2 0 1 0 1
# Pyth, 34 bytes, depth 13, A075355 from A135020
L?*bytbb1VQ~Z1J]ZW<.xJyhN~Z1aJZ;lJ
### Explanation
L define function y(b): return
*bytb b * y(b - 1)
? b if b is not 0, else
1 1
VQ iterate N over 0..input-1
~Z1 increment Z (initialized to 0)
J]Z set J to array of Z
W while
.xJ product of J
< is less than
yhN factorial of N+1
~Z1 increment Z
aJZ add Z to J
;
lJ print length of J
The next answer should match the following terms:
1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6
• You should change to be from A135020, because the submission for A110654 is wrong. Apr 27, 2015 at 11:53
• And if your code works, it's surely A075355? Apr 27, 2015 at 11:55
• @PeterTaylor Yep, I calculate the sequence exactly how it's defined. Apr 27, 2015 at 11:56
• A075352 begins 1, 2, 12, 30, 504, 1320, ... The only one which begins 1,1,2,2,3,3,4,4,5,5,6,6,6 is A075355. Apr 27, 2015 at 11:57
• @PeterTaylor Whoops, that's what I meant. I apparently screwed up when naming my code file. Edited post. Apr 27, 2015 at 11:58 | 7,226 | 22,285 | {"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} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.942701 |
https://www.jobduniya.com/Private-Companies/Scandent/Fully-Solved-Papers/Topic-Aptitude-0/Part-1.html | 1,638,354,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00248.warc.gz | 877,691,324 | 6,909 | # Aptitude [Scandent Placement]: Sample Questions 1 - 3 of 11
Get top class preparation for competitive exams right from your home: get questions, notes, tests, video lectures and more- for all subjects of your exam.
## Question 1
Aptitude
### Question
MCQ▾
Find the middle term 5,15/2,25/2,15
### Choices
Choice (4)
a.
8
b.
None of the above
c.
23/2
d.
9
b.
### Explanation
Here, first term = 5
second term =
third term =
fourth term = 15
Difference between first and second terms is 2.5.
Difference between second and third terms is 5.
Difference between third and fourth terms is 2.5.
So, difference between next two terms should be 5.
Fifth term should be 20.
## Question 2
Aptitude
### Question
MCQ▾
Express 12.5 % in fraction
Choice (4)
a.
1/32
b.
1/4
c.
1/16
d.
1/8
d.
### Explanation
To convert any percentage into number we need to divide that percentage by 100.
Dividing the 12.5 by 100,
Now, converting this decimal number into fraction,
So,
## Question 3
Aptitude
### Question
MCQ▾
A man can swim downstream at 6km/hr and upstream at 2km/hr. His speed in still water is
Choice (4)
a.
1 km/hr
b.
6 km/hr
c.
4 km/hr
d.
2 km/hr
c.
### Explanation
Speed of downstream, a = 6kmph
Speed of upstream, b = 2 kmph
Let the man՚s speed be u kmph in still water and speed of stream be v kmph
In downstream, direction of man/boat is in direction of flowing water and in upstream, direction of man/boat is opposite to flowing water.
So, and
Combining both the equations,
So, speed of man in still water is 4 kmph.
Developed by: | 468 | 1,596 | {"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-2021-49 | latest | en | 0.891308 |
https://bookofproofs.github.io/branches/geometry/euclidean-geometry/elements-euclid/book--9-number-theory-applications/elements-of-geometric-progression-from-one-which-are-powers-of-number.html | 1,701,284,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00411.warc.gz | 177,810,136 | 3,686 | # Proposition: Prop. 9.08: Elements of Geometric Progression from One which are Powers of Number
### (Proposition 8 from Book 9 of Euclid's “Elements”)
If any multitude whatsoever of numbers is in continued proportion, (starting) from a unit, then the third from the unit will be square, and (all) those (numbers after that) which leave an interval of one (number), and the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (numbers). * Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, $E$, $F$, be in continued proportion, (starting) from a unit. * I say that the third from the unit, $B$, is square, and all those (numbers after that) which leave an interval of one (number). * So I also say that the fourth (number) from the unit, $C$, is cube, and all those (numbers after that) which leave an interval of two (numbers). * So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square.
### Modern Formulation
(not yet contributed)
Proofs: 1
Proofs: 1 2 3 4
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Fitzpatrick | 340 | 1,301 | {"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} | 3.453125 | 3 | CC-MAIN-2023-50 | latest | en | 0.943298 |
http://oeis.org/A112171/internal | 1,553,500,704,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203842.71/warc/CC-MAIN-20190325072024-20190325094024-00458.warc.gz | 150,655,355 | 2,878 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A112171 McKay-Thompson series of class 32c for the Monster group. 1
%I
%S 1,2,0,0,-1,2,0,0,-1,4,0,0,0,6,0,0,1,8,0,0,0,12,0,0,-1,18,0,0,1,24,0,
%T 0,2,32,0,0,-1,44,0,0,-2,58,0,0,1,76,0,0,2,100,0,0,-1,128,0,0,-3,164,
%U 0,0,1,210,0,0,4,264,0,0,-2,332,0,0,-5,416,0,0,2,516,0,0,5,640,0,0,-2,790,0,0
%N McKay-Thompson series of class 32c for the Monster group.
%H G. C. Greubel, <a href="/A112171/b112171.txt">Table of n, a(n) for n = 0..2500</a>
%H D. Ford, J. McKay and S. P. Norton, <a href="http://dx.doi.org/10.1080/00927879408825127">More on replicable functions</a>, Comm. Algebra 22, No. 13, 5175-5193 (1994).
%H <a href="/index/Mat#McKay_Thompson">Index entries for McKay-Thompson series for Monster simple group</a>
%F Expansion of A + 2*q/A, where A = q^(1/2)*(eta(q^4)/eta(q^16)), in powers of q. - _G. C. Greubel_, Jun 26 2018
%e T32c = 1/q +2*q -q^7 +2*q^9 -q^15 +4*q^17 +6*q^25 +q^31 +...
%t eta[q_] := q^(1/24)*QPochhammer[q]; A:= q^(1/2)*(eta[q^4]/eta[q^16]); a:= CoefficientList[Series[A + 2*q/A, {q, 0, 60}], q]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Jun 26 2018 *)
%o (PARI) q='q+O('q^80); A = eta(q^4)/eta(q^16); Vec(A + 2*q/A) \\ _G. C. Greubel_, Jun 26 2018
%K sign
%O 0,2
%A _Michael Somos_, Aug 28 2005
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Last modified March 25 03:50 EDT 2019. Contains 321450 sequences. (Running on oeis4.) | 735 | 1,709 | {"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} | 2.90625 | 3 | CC-MAIN-2019-13 | latest | en | 0.596762 |
https://sportsbizusa.com/20-landforms-worksheets-3rd-grade/ | 1,680,346,040,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00333.warc.gz | 628,160,953 | 14,108 | # 20 Landforms Worksheets 3rd Grade
Resources for Studying Landforms & Habitats of the World free landforms worksheets 3rd grade, landforms worksheets 3rd grade pdf, landform worksheets third grade, via: pinterest.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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## Laaaaaaaand Hooooooo Social Stu s Worksheets Kindergarten
Laaaaaaaand Hooooooo Social Stu s Worksheets Kindergarten via : sadlock.org | 1,195 | 5,706 | {"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} | 3.421875 | 3 | CC-MAIN-2023-14 | latest | en | 0.962634 |
https://www.kopykitab.com/blog/csvtu-exam-papers-be-i-year-applied-mathematics-i-nov-dec-2007/ | 1,623,755,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00488.warc.gz | 763,502,033 | 23,597 | # CSVTU Exam Papers – BE I Year – Applied Mathematics-I –Nov-Dec- 2007
BE (1st Semester)
Examination Nov-Dec, 2007
Applied Mathematics-II
UNIT- I
1. (a) State De Moire’s theorem.
(b) cos a + cos β + cos λ = sin a + sin β + sin λ = 0 Prove that
(i) cos2 a + cos2 β + cos2 λ = sin2 a + sin2 β + sin2 λ
(ii)cos2a + cos2 β + cos 2 λ = sin2a + sin2 β + sin 2 λ = 0
(c) If tan (ө + iф) = eia, Show that:
ө – (n +1) and log tan( + a)
2 2 4 2
(d) Sum the series: ф – 1
2
sin2 ө 1 sin 2ө sin2ө + 1 sin3ө1 sin 4өsin4 ө+……….. ө
2 3 4
UNIT- II
2. (a) Solve d2y – 3dy – 4y =0
dx2 dx
(b) Solve d2y – 5dy + 6y =sin3x
dx2 dx
(c) Solve by method of variation of parameters d2y + 4y=tan2x
dx2
(d) Solve the simultaneous equations: dx + 2x +5y=tt
dt
dx + 4x +3y=t
dt
UNIT- III
3. (a)Define Bet function .
a b a
(b) Evaluate: ʃ ʃ ʃ(x2 + y2 + z2)dzdydx
a b a
(c) Evaluate: 01ʃxm(log)ndx = (-1)n n!
(m+1)n+1
Where n is a positive integer and m>-1.
(d) Find the area included between the parabola y= 4x – x2 and the line y = x.
UNIT- IV
(b) Find the constants a and b so that the surface ax2 – byz= (a+2)x is orthogonal to the surface
ax2y + z3 = 4 at the point (1, -1, 2)
(c) If F = (x2 + y2)i + 2xyj + (y2-xy)k, then find div F and Curl F.
(d) Verify stake’s theorem for :
F = (x2 + y2)I + 2xyj Taken round the rectangle bounded by x= ±a, y = y = b
UNIT- V
5. (a)Find the number of roots of the equation x3 – x24x + 4 = 0
(b) Find the condition that the equation x3 + px24x + 4 = 0 had roots a, β which satisfy aβ + 1 = 0
(c) If a, β, λ are roots of the equation .x3 + qx + r = 0 find the equation whose roots are
(a – β)2, (β – λ )2, (λ – a)2
(d) Solve the equation x2 -15x – 126 = 0 Cardoon’s method. | 760 | 1,875 | {"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-2021-25 | latest | en | 0.503906 |
https://stackoverflow.com/questions/6043310/genetic-algorithm-subset-sum-problem | 1,571,080,698,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00046.warc.gz | 689,215,356 | 33,589 | # Genetic algorithm - Subset sum problem
I have to do a project which using a genetic algorithm solves the subset sum problem. Unfortunately, when coding the algorithm I found a big problem ...
My algorithm:
• as long as no solution was found and the number of steps is smaller than steps do:
• calculate the probability and then distribution function for each chromosome
• perform selection (roulette)
• select n chromosomes to be crossed
• perform the crossing (the crossing point is selected randomly)
• select m chromosomes to mutation
• perform mutations
• if you found a solution then stop
(Algorithm was taken from the book "Genetic Algorithms + Data Structures = Evolution Programs, Chapter 2 ") Variables such as population size, amount of data, scope of data collection, number of steps, the number of mutations (in step), the number of crossings (in a step) is set rigidly in the program options.
The problem is that after a certain (relatively small) number of steps in the population all the chromosomes are identical. The problem illustrates this graph: http://imageshack.us/m/96/7693/wykresb.png
What I'm doing wrong? How to fix it? Thanks in advance.
Edit:
Here You can find logs from my app: http://paste.pocoo.org/show/391318/
I think that roulette is not the best solution (as deong said). Mutations also need to improve.
• What language are you using to implement the algorithm? It sounds like an issue with the crossover and mutation phase. To troubleshoot turn off each of these phases and try and isolate the problem. I imagine it is something to do with the variables you are using and that you are not taking deep copies of chromosomes when applying these operators on your population.. – GordyD May 18 '11 at 10:42
• I think the issue may well be with your implementation, you would need to post your code. – Toby Allen May 18 '11 at 10:48
• @Gordon Murray Dent - I'm using Java. I will check this. – Ziem May 18 '11 at 16:15
• @Toby Allen - My project contains few classes and isn't well coded :(. In my opinion posting the source code is pointless. – Ziem May 18 '11 at 16:17
• @Ziem, how do you know the problem isnt with your implementation? – Toby Allen May 19 '11 at 13:43
I had a similar problem before, I wish it s the same as your's
First, you need to check (using any measuring metric) if chromosome A is better than chromosome B. This let you have a strict order of the chromosomes of your population and be able to sort your population.
Then, when you produce a new chromosome (either by mutation or crossover) you may be producing a chromosome that already exist in your population. Make sure not to include this in your population list.
In other words, make sure your list always contains different chromosomes and always sorted from best to worst !
Note: The genetic algorithms I work with are usually like this (this is the most general algorithm and most used):
• create P different chromosomes and add them to list Pop;
1. while (no optimal solution is found && number of iteration < LIM)
2. create new chromosomes using crossover, mutation or any other methods;
3. add the created chromosome to list Pop
4. sort the list Pop (from best-fit to worst-fit)
5. select the first P different chromosomes and discard all other from Pop.
6. end while
• I used some of Your tips in my algorithm and now it is working OK (I think so :))! – Ziem May 23 '11 at 20:16
• glad to hear that ! :) – AJed May 26 '11 at 15:44
Here's (potentially) the problem. Disclaimer is of course that you may just have a buggy program.
Roulette wheel selection is just terrible. The problem is that early on in a run, the distribution of fitness values is random. You have some awful solutions and some that are reasonable OK in comparison. You don't expect any of them to be very good, but you would expect some of them to be much better than others.
Roulette wheel selection takes these relative differences in probabilities and amplifies them. If you have a population size of 100, and one individual has a fitness five times better than any others, it will be selected five times as often. With typically mild mutation rates, you end up quickly in a situation where you choose the same individual twice for recombination, produce some new identical offspring, make very minor changes (maybe), and then put them back into the population. Because you're still early in the run, most solutions are still bad, so where you did have one above average solution, you selected it into five above average solutions, bred them to get ten above average solutions, and then started the process all over again. These solutions can very quickly take over the entire population if you aren't really careful with designing your set of operators, even though all the algorithm knows is that they're better than the really crappy solutions it has otherwise seen.
The solution is to use a better selection operator. Binary tournament selection is faster, easier to code, and applies a much more tolerable selection pressure. There is also rank-biased selection which selects proportionally by fitness ranking rather than absolute differences.
Edit: This isn't to say you can't use proportional selection. Just that it is very prone to premature convergence and to use it effectively, you typically have to build an entire set of operators with that in mind.
When applying genetic algorithms, it can happen that the algorithm gets stuck at local optima. However, one is interested in the global optimum (or rather an approximation to such an optimum).
Local optima can be avoided by:
• A higher mutation rate
• A different cross-over function
Moreover, it may be useful to kill clones. That means you "quickly" look at your population after each iteration and do not allow for clones. By quickly I mean that you just look for approximate clones, because checking for exact clones would take O(m*n^2), where n is your population size and m is the size of a chromosome. This method helped me in a different problem where I was facing clones as well.
Hope this helped, Christian
EDIT
It would also be nice if you could post your cross-over function. Preferably not as code, but in plain english text. The cross-over function is the critical part of a genetic algorithm.
• How to replace clones? Because my population size should be the same in every step. My crossing function works in this way: paste.pocoo.org/show/391131 . – Ziem May 18 '11 at 11:24
• @Ziem - Okay so you have binary arrays as chromosomes. That's a good point to start from. What exactly do the 1's and 0's represent? Also try your code with a higher number of chrossing points: i.e. ch1 = 101|11|001, ch2 = 011|01|010, Offspring = 101|01|001. – Christian May 18 '11 at 11:29
• @Ziem - Oh and: Why do you have two offsprings? Do you discard the offspring with the lowest fitness out of the two, or do you keep both? – Christian May 18 '11 at 11:29
• @Ziem - As for how to replace clones: Basically after each iteration of the main loop you sort the chromosomes by fitness. Say your population is initially size = 100. With your crossing function, you would get 300 chromosomes after the cross-over. Of these 300 chromosomes there are a lot of "Clones". Detect them (efficiently) and delete them from your population. It works like evolution in the real world: If we all were clones of each other, all genes of our children would be the same. That wouldn't be good. – Christian May 18 '11 at 11:34
• You can have the greatest crossover operator the world has ever known, but if you apply very high amounts of selection pressure, it isn't going to matter. You'll converge very quickly, and once you converge, crossover can't create new alleles. – deong May 18 '11 at 14:29 | 1,763 | 7,764 | {"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-2019-43 | latest | en | 0.924271 |
https://www.motorverso.com/how-to-build-bunk-bed-ladder-for-rv-in-the-easiest-ways/ | 1,708,970,243,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00188.warc.gz | 900,713,720 | 24,858 | # How To Build Bunk Bed Ladder For Rv In The Easiest Ways
My family is planning to have a wonderful trip to enjoy this summer occasion. We couldn’t miss the best companion of us is a modern mountain RV used as a camp; and our RV is full of furniture especially a bunk bed for children. However, the ladder of the bed was broken as a result of a small accident last year. Hence, we need to find a way of building a bunk bed ladder for RV to enjoy our trip in the best way.
I have learned many valuable experiences from other families in solving this problem. And I have recognized that the most popular material that often use to build a ladder is wood. Besides that, the rope becomes a new component to makeover your bunk bed. In this article, I will share for you two ways of creating a ladder:
## Using Wood Material To Build The Ladder
You can refer these 7 steps to create a formal ladder for the bunk bed. The most important thing that parents should note is safe for children. So the materials should be selected carefully and assembled durably. A lot of components used in this formula including:
### Step 1: Confirm The Slope Of The Ladder
You ought to measure the height of the bunk bed from the base of it to the floor, and the result will be the height of the ladder.
After establishing the height, you cut 2 pieces of 2×4 inch lumber based on that length. These pieces will be the ladder sides.
Keep one side of the lumber in place according to the upper rail of the bunk bed, create a sliding slope that you want to design (perfect angle of bunk bed ladder of RV is 15-degree). Mark on the backside of the lumber.
### Step 2: Designs The Bottom Of Ladder Sides
Set two sides of the ladder to 15 degrees, draw a straight line from the position that you made on in step 1 across the face.
Use the circular saw to cut along that angled line with 2 sides.
After cutting, you should sand both sides of the ladder until smooth.
### Step 3: Design The Tops Ladder Sides
Put two sides up against the rail of bunk bed and on the floor.
Mark the lumbers about 3 inches above the rail and draw the straight line.
Cut the tops off with the saw carefully.
### Step 4: Design The Rungs
Determine the length and cut from 1×4 inch lumber to equal pieces.
The number of rungs will base on the height of the ladder. Some notices that you should know about the space of the rungs are:
• You shouldn’t place the rungs apart on the center more than 12 inches.
• The beginning rung isn’t higher than 8 inches from the ground.
• The most popular space of center rungs is 10 inches.
Starting 8 inches from the floor, make a tick on both sides.
Repeat this process until the last rung. The last one couldn’t be put upper the bunk or over the rail.
Then, you cut and check the rungs entirely are the same length, sand them on both sides until smooth.
### Step 5: Drill The Pilot Holes
From the marks you have drawn for the rungs, you use the power drill to make pilot holes.
Each side of the ladder will have the right amount of holes depending on a number of the rungs.
### Step 6: Set Wood Screws
Set two 3 ½ inches wood screws in each hole from the outside in.
Preset them that the tips hardly show on the inside.
### Step 7: Apply Primer And Paint
Apply the prime into the ladder. Then, wait until the primer dries.
After that, you can coat the ladder with the color paint of your selection.
### Step 8: Attach Metal Hooks
From the backside of the ladder for bunk bed, drill pilot holes where the ladder sides meet the bunk rail.
Attach the hooks into that holes and set the hardware screws.
When finishing an assemble the bunk bed ladder for RV in this way, you can achieve a wonderful result with modern style and durable structure. Otherwise, if you enjoy a dynamic style, you can follow a new idea as below:
## Using Rope Material To Build The Ladder
Luckily, the rope ladder can be found in many stores or online shops. So you can save your time in creating the rope product from rudimentary supplies.
Step 1: Order the rope ladder from the store you want. You don’t need to measure the height of your bunk bed exactly. Because the rope ladder allows you to change the height of it effortlessly.
Step 2: When you have a product, you can start to build-up it.
Keep it from the upper rail of the bunk bed where you want to mount, confirm how close to the floor you want it.
Step 3: After determining the length of the ladder. Untie the bottom of the rope ladder and remove excess wood dowels.
Step 4: If the wood dowels are too far with you, you can unbolt and adjust them closer together.
When you are satisfied with the spacing, you ought to tie secure knots at the bottom of the last dowel to keep it in place.
Step 5: Mark the position you want to mount the ladder. Screw the hooks into the rail of the bunk bed securely.
Step 6: Mount the rope ladder onto the hooks that you have installed.
Those are steps to build a rope ladder for your bunk bed. You just need to experience a lot of steps and achieve a wonderful result.
In my opinion, I love creating the wood ladder because it is durable and used for a long time. Nonetheless, I still try a rope ladder to create freshness for my children and they always love it.
All parents can follow these ways to create a modern bunk bed ladder for your children. It can save your time and effort.
Did you find these ways helpful for you?
Let us know in the comments below!
If you see these pieces of information are beneficial to you, please share them with your friends. | 1,258 | 5,567 | {"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-2024-10 | latest | en | 0.945585 |
http://www.solutioninn.com/there-are-602-aerobics-and-fitness-centers-in-japan-up | 1,501,130,971,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427429.9/warc/CC-MAIN-20170727042127-20170727062127-00630.warc.gz | 559,566,533 | 7,671 | # Question: There are 602 aerobics and fitness centers in Japan up
There are 602 aerobics and fitness centers in Japan (up from 170 five years ago). Adidas, the European maker of sports shoes and apparel, is very interested in this fast growing potential market for its products. As part of a marketing survey, Adidas wants to estimate the average income of all members of Japanese fitness centers. Since travel and administrative costs for conducting a simple random sample of all members of fitness clubs throughout Japan would be prohibitive, Adidas decided to conduct a cluster sampling survey. Five clubs were chosen at random out of the entire collection of 602 clubs, and all members of the five clubs were interviewed. The following are the average incomes (in U.S. dollars) for the members of each of the five clubs (the number of members in each club is given in parentheses): \$37,237 (560), \$41,338 (435), \$28,800 (890), \$35,498 (711), \$47,446 (230). Give the cluster sampling estimate of the population mean income for all fitness club members in Japan. Also give a 90% confidence interval for the population mean. Are there any limitations to the methodology in this case?
View Solution:
Sales0
Views119 | 264 | 1,224 | {"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-2017-30 | longest | en | 0.952955 |
https://therapyboy.com/factors-to-consider-when-playing-the-lottery/ | 1,713,618,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00192.warc.gz | 517,583,681 | 9,346 | # Factors to Consider When Playing the Lottery
A lottery is a game where players choose a set of numbers and are awarded prizes based on how many of those numbers match a second set chosen in a random drawing. Throughout history, drawing lots to determine ownership of property has been recorded in many ancient documents, and it became commonplace in Europe during the late fifteenth and sixteenth centuries. The first lottery in the United States was created in 1612, when King James I of England instituted a lottery to provide funds for the settlement of Jamestown, Virginia. Since then, public and private organizations have used lotteries to raise funds for wars, towns, colleges, and public-works projects.
## Lottery is a game where players select a group of numbers
The lottery is a game of chance in which players select a group of numbers that are drawn from a large pool. The selected numbers are called lotto numbers, and if one of them matches a prize drawing, the player wins the prize. The lottery is a popular form of gambling and is played in over a hundred countries worldwide. However, there are a few factors that should be considered when playing the lottery.
## They are awarded prizes based on how many match a second set chosen by a random drawing
This type of game is based on the theory that the players select a group of numbers from a large set, and the chances of matching all of them are equal. A typical lottery game involves selecting six numbers from a set of 49, and the lottery then randomly chooses a second set. If all six numbers match, the player wins a major prize. If only three of the six numbers match, the player wins a prize.
## States that have lotteries
A study conducted by the Center for Behavioral and Economic Research found that the poorest Americans play the lottery the most, spending a disproportionate percentage of their income on tickets. According to the study, those with an income of \$10,000 or less played the lottery on average 26 times per year. This amount equates to about \$645 per person, or 6 percent of their income. This is comparable to the amount of money that upper-middle class households contribute to their 401K accounts.
## Their demographics
Obtaining patients’ demographics is an important part of running a healthcare organization. Patients often fill out registration forms at their doctor’s office. The purpose of this data collection is to help provide better care to a patient. However, collecting accurate patient demographics is not always easy. There are several issues to consider, including patient privacy. Fortunately, there are ways to ensure patient confidentiality, as well as create a secure online registration portal. | 537 | 2,726 | {"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} | 2.90625 | 3 | CC-MAIN-2024-18 | longest | en | 0.976155 |
http://windowssecrets.com/forums/showthread.php/25253-Strange-update-%28Excel-2000%29 | 1,503,422,477,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112533.84/warc/CC-MAIN-20170822162608-20170822182608-00324.warc.gz | 432,470,938 | 13,931 | # Thread: Strange update (Excel 2000)
1. ## Strange update (Excel 2000)
I found a very strange thing.
I have a list of 6 values with a formula of =Sum(a1:a3)-sum(a4:a6).All the value is nul.
Now what is strange if I have a values in cell a1 to a3 and put a value in cell a4 the function changes to =Sum(a1:a4)-sum(a4:a6).
Refer attached file
Are there a place to put this nice/ irritating feature off.
Thanks
Mario
2. ## Re: Strange update (Excel 2000)
Looking at your formulas, the formula for cell B7 is:
<pre>=SUM(B1:B4)-(SUM(B4:B6)</pre>
So, it's automatically subtracting the value of cell B4 from the equation as soon as it is added, yielding the same as the sum of B1:B3.
(4+5+5+45)-(45) = (4+5+5)
Perhaps you meant to use:
<pre>=SUM(B1:B3)-(SUM(B4:B6)</pre>
HTH
3. ## Re: Strange update (Excel 2000)
I need a fix to this auto change Excel does.
I have thought of either putting a hidden row in or making the formula constant
Any other ideas?
Thanks
Mario
4. ## Re: Strange update (Excel 2000)
I'm a little bit confused - perhaps you can clarify.
Which is the formula that you want to have in B7:
=SUM(B1:B3)-SUM(B4:B6)
OR
=SUM(B1:B4)-SUM(B4:B6)
And, given the data you supplied in your first post, what is the result that you expect?
I don't quite understand the auto change that you're talking about <img src=/S/confused.gif border=0 alt=confused width=15 height=20>.
5. ## Re: Strange update (Excel 2000)
That does not seem to be happening for me. If I enter a value in cell A4 on your sheet, the formula does not change. Can you give a more detailed description of what you are doing?
Also, check in Tools/Options on the Transition tab to make sure that "Transition navigation keys," "Transition formula evaluation," and "Transition formula entry" are not checked.
6. ## Re: Strange update (Excel 2000)
I have a list of values where I have 6 values underneath each other. The first 3 I would like to add and then deduct the sum of the last 3 values - formula Sum(A1:A3)-Sum(A4:A6). As soon as I enter value 1 to 3 and then enter a value in A4 the formula changes to Sum(A1:A4)-Sum(A4:A6). I've cleared the checks on "Transition formula evaluation," and "Transition formula entry" and saved the file.
Now I would like Excel to keep the formula as Sum(A1:A3)-Sum(A4:A6) and not change it
Thanks for all the input
Regards
Mario Smit
7. ## Re: Strange update (Excel 2000)
Try:
=SUM(OFFSET(\$A\$1,0,0,3,1))-SUM(OFFSET(\$A\$4,0,0,3,1))
8. ## Re: Strange update (Excel 2000)
It's caused by "Extend list formats and formulas" on the Edit tab if this feature is checked. | 747 | 2,604 | {"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-2017-34 | latest | en | 0.887365 |
https://citydental.ro/iuy6nmf8/teo3-lewis-structure-bb25af | 1,627,587,661,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00459.warc.gz | 191,318,766 | 21,825 | # teo3 lewis structure
Assign formal charges to elements in each of the structures below and pick the best Lewis structure for TeO3^2- ? The NO2 Lewis structure has a total of 17 valence electrons. 88.82 g C x 1mol 12.011g = 7.395 mol C 11.18 g H x 1mol 1.008g = 11.09 mol H Mole ratio: molH molC = 11.09 7.395 = 1.500 1.50 1 is equivalent to 3 2 using whole numbers. One is a lone pair, so the structure is bent. From the Lewis structures (see Exercise 8.91), Br 3-would have a linear molecular structure, ClF 3 would have a T-shaped molecular structure and SF 4 would have a see-saw molecular structure. They can be drawn as lines (bonds) or dots (electrons). I need the geometry from the VSEPR also. In this tutorial, we are going to learn followings of nitrous oxide gas. Example: Determine all the possible Lewis structures possible for NO2–. Chemistry . Still have questions? The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. These Lewis structures are equivalent except for the placement of the electrons (i.e. See these examples: For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here: Determine the total number of valence (outer shell) electrons. Tellurous acid | H2TeO3 or H2O3Te | CID 24936 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities, safety/hazards/toxicity information, supplier lists, and more. Lewis structure for HSeO 3-is: This would have 4 σ bonds and 1 π bond. What is the Lewis dot structure for TeO3? Long answer. Buy Find arrow_forward. 0 0. Quiz your students on TeO3 Lewis Dot Structure - Bond Angle, Hybridization, Molecular Geometry, Polar or Nonpolar using our fun classroom quiz game Quizalize and personalize your teaching. Quiz your students on SbI3 Lewis Structure, Molecular Geometry, Bond Angle, Polar or Nonpolar using our fun classroom quiz game Quizalize and personalize your teaching. a. CCl 4 b. NCl 3 c. SeCl 2 d. ICl In each case, the atom listed first is the central atom. 10 11 12. Write Lewis structures that obey the octet rule for each of the following molecules. Bonds between atoms of the same element (homonuclear bonds) are always divided equally. (In each case the first atom listed is the central atom.) Draw a trial structure by putting electron pairs around every atom until each gets an octet. “H2O Lewis Structure PNG” By Daviewales – Own work (CC BY-SA 4.0) via Commons Wikimedia 5. 10th Edition. Hybridization: sp Then draw the 3D molecular structure using VSEPR rules: Decision: The molecular geometry of CSe 2 is linear with symmetric charge distribution on the central atom. (a) -2 -2 -2:::: (b) CO has the shortest bond length because of the triple bond. a. POCl 3, SO 4 2−, XeO 4, PO 4 3−, ClO 4 − b. NF 3, SO 3 2−, PO 3 3−. This Site Might Help You. Let’s apply the shortcut: Hydrogen should have 1 valence electron Hydrogen has one attached electron 1 – 1 = 0 Here’s a trick for bound hydrogen, it ALWAYS has a formal charge of zero, so just ignore it. See Answer. Steven S. Zumdahl + 2 others. Preparation. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable. How many valence electrons are present in this compound? BF 3 is an example. e disadvantage to this method is that it is more dicult and time-consuming than using the rules. Source(s): https://shrink.im/a9UFn. Get the latest public health information from CDC: https://www.coronavirus.gov. ISBN: 9781305957404. In this case, Te will have an expanded electron shell holding 12 electrons. Read "ChemInform Abstract: Cs7Sm11[TeO3]12Cl16 and Rb7Nd11[TeO3]12Br16, the New Tellurite Halides of the Tetragonal Rb6LiNd11[SeO3]12Cl16 Structure Type., ChemInform" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. The lone pair goes in an equatorial position, so the structure is a distorted see-saw like SF4. WARNING! It is Te with double bonds connecting it to Oxygens. This problem has been solved! Top Answer. However, its Lewis structure contains two double bonds. (C is the central atom in $\mathrm{OCN}^{-}$ and $\mathrm{N}$ is the central atom in $\mathrm{CNO}^{-}$ ) Problem 109. α-TeO 3 has a structure similar to FeF 3 with octahedral TeO 6 units that share all vertices.. Use the arrangement of the bonded atoms to determine molecular geometry . The valence electrons are the electrons in the outermost shell of the atom. Determine the electron domain geometry by arranging the electron domains about the central atom so that the repulsions among them are minimized 3. Neutral oxygen should have 6 valence electrons. Keep in mind, multiple bonds (i.e., double bonds, triple bonds) count as one electron domain. Assume 100 g of hydrocarbon. Using Lewis structures allows for the fact that atoms of the same element in a molecule or polyatomic ion may have dierent oxidation numbers, depending on the bonding arrangement. RE: What is the lewis dot structure for SeO3? Arrange the electron domains around the central atom to minimize repulsion. 2. (b) SbCl2+ (18 valence electrons) has a structure based on trigonal planar coordination at Sb (3 electron pairs at Sb). The reason NF5 doesn’t exist is because N has no d orbitals available for hybridization. First draw the Lewis dot structure: Electron geometry: linear. Determine its resonance hybrid. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. How to draw the lewis structure for N 2 O; Drawing resonance structures following the lewis structure ; Oxidation numbers of nitrogen and oxygen atoms. Decide which is the central atom in the structure. In other words, a … The answer to “Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) OCS, (b) SOCl2 (S is the central atom), (c) BrO3 -, (d) HClO2 (H is bonded to O).” is broken down into a number of easy to follow steps, and 42 words. See the answer. Thank you! I'm just guessing; but I’m almost sure you meant SeS2 that is selenium disulfide, not SSe2 that is sulfur diselenide. Seo3 Lewis Structure. Draw the Lewis structure for Ses2 and answer the following questions. Asked by Wiki User. 4. Count the total number of electron domains. We need to recognize that multiple bonds should be treated as a group of electron pairs when arriving at the molecular geometry. Properly draw Lewis structures will provide the general molecular shape of a molecule. Carbon Diselenide on Wikipedia. Choice A O 2 16. The best Lewis structure for sulfuric acid has zero formal charges, sulfur as the central atom, and no bonds between S and H. How many single and double bonds, respectively, are there in this Lewis structure? Sketch the Lewis structure of the ion or molecule. The nonbonding electrons, on the other hand, are the unshared electrons and these are shown as dots. Three Electron Pairs (Trigonal Planar) The basic geometry for a molecule containing a central atom with three pairs of electrons is trigonal planar. That will normally be the least electronegative atom ("Se"). Chemistry. The average oxidation number on "N" = (+5 + 5)/2 = +10/2 = +5 . 10th Edition. Example: Determine the remaining resonance structures possible for the following compound, CO32-. One line corresponds to two electrons. Use the angular arrangement of the chemical bonds between the atoms to determine the molecular geometry. Ask Question + 100. It's not common to have an odd number of valence electrons in a Lewis structure. Buy Find arrow_forward. Source(s): lewis dot structure seo3: https://biturl.im/WeHML. e. !e advantage to using the oxidation number rules is that they apply to most cases and are quick and easy to use. For example, consider ClF 3 (28 valence electrons): The central Cl atom is surrounded by 5 electron pairs, which requires a trigonal bypyramid geometry. 1 B. structure is based on trigonal bipyramidal geometry at Sb (5 electron pairs around the central atom). Following the checklist in the Lewis Structures video we have oxygen bound to hydrogen, with 3 lone pairs around oxygen. Write Lewis structures that obey the octet rule for each of the following molecules and ions. Therefore CSe 2 is non-polar. 10 How many lone pair (non-bonding) electrons are present in this compound? 4 single, 2 double. > Here are the steps I follow when drawing a Lewis structure. “VSEPR geometries” By Dr. Regina Frey, Washington University in St. Louis – Own work, Public Domain) via Commons Wikimedia 2. Your Lewis structures should show the sp3 hybridization in NF 3 and the sp 3d hybridization in PF 5. COVID-19 is an emerging, rapidly evolving situation. Jeannie. Five fundamental electron domain geometries. N 2 O Lewis Structure, Resonance Structures, Oxidation Number. 8 Submit Answer Incorrect. the location of the double bond) Equivalent Lewis structures are called resonance structures, or resonance forms. Publisher: Cengage Learning. N 2 O (nitrous oxide) is an oxide of nitrogen and is called as laughing gas. Wiki User Answered . 2. 1.75 C. 1.33 D. 1.5 E. 1 and 2. 0 0. Draw the Lewis structure of the molecule or ion and count the number of electron domains around the central atom. 5 years ago. A) 2 single, 4 double D) 6 single, no double B) 4 single, 2 double E) 5 single, 1 double C) 4 single, no double. “Ammonia-3D-balls-A” By Ben Mills – Own work (Public Domain) via Commons Wikimedia 6. FREE Expert Solution. Chemistry. 1. Get your answers by asking now. Draw a skeleton structure in which the other atoms are single-bonded to the central atom: "O-Se-O" 3. The Lewis structure of XeF 2 shows two bonding pairs and three lone pairs of electrons around the Xe atom: XeF 6: We place three lone pairs of electrons around each F atom, accounting for 36 electrons. For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. CNO- Lewis structure? The correct way to describe ozone as a Lewis structure would be: This indicates that the ozone molecule is described by an average of the two Lewis structures (i.e. 18 How many bonding electrons are present in this compound? Abstract A series of new rare‐earth cadmium metal tellurite chlorides were synthesized by the salt flux technique and characterized by single‐crystal X‐ray diffraction. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it from the molecular formula (Figure 1b). Because of this we'll try to get as close to an octet as we can on the central Nitrogen (N) atom. Tellurium trioxide (Te O 3) is an inorganic chemical compound of tellurium and oxygen.In this compound, tellurium is in the +6 oxidation state.. Polymorphs. There are two forms, yellow-red α-TeO 3 and grey, rhombohedral, β-TeO 3 which is less reactive. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign. Lewis structures also show how atoms in the molecule are bonded. Problem: What is the Te–O bond order in the “best” Lewis structure of the tellurite ion (TeO 32–)?A. Why is the fulminate ion so unstable? Question: Assign Formal Charges To Elements In Each Of The Structures Below And Pick The Best Lewis Structure For TeO3^2- ? Expert Answer 100% (2 ratings) Previous question Next question Get more help from Chegg. Steven S. Zumdahl + 2 others. The average oxidation number on "N" = (+4 + 4)/2 = +8/2 = +4 Dinitrogen pentoxide The Lewis structure of "N"_2"O"_5 is (a) The left-hand "N" atom LP = 0; BE = 0 ON = 5 - 0 - 0 = +5 (b) The right-hand "N" atom LP = 0; BE = 0 ON = 5 - 0 - 0 = +5 Each "N" atom has the same formal charge. One dot is equal to one nonbonding electron. Empirical formula is C 2 H 3 . Fulminate salts explode when struck; $\mathrm{Hg}(\mathrm{CNO})_{2}$ is used in blasting caps. Using the rules putting electron pairs around the central atom so that repulsions! Atom to minimize repulsion when arriving at the molecular geometry to learn followings of nitrous )! What is the central atom so that the repulsions among them are minimized 3 it 's not common to an. Skeleton structure in which the other atoms are single-bonded to the central atom. time-consuming than using oxidation. ( 2 ratings ) Previous question Next question get more help from.. Two double bonds connecting it to Oxygens, CO32- on trigonal bipyramidal geometry at Sb ( 5 electron around. For HSeO 3-is: this would have 4 σ bonds and 1 π bond that share all vertices Own (! See-Saw like SF4 write Lewis structures should show the sp3 hybridization in PF 5 Here the. N '' = ( +5 + 5 ) /2 = +10/2 = +5 Lewis. S ): Lewis dot structure for TeO3^2- gets an octet as we on! Many bonding electrons are present in this compound by single‐crystal X‐ray diffraction ( i.e D.. Te will have an expanded electron shell holding 12 electrons pair ( non-bonding electrons. B ) CO has the shortest bond length because of this we 'll try to get close!: ( b ) CO has the shortest teo3 lewis structure length because of the same element ( homonuclear bonds ) always. The bonded atoms to determine molecular geometry Elements in each case the first atom listed first is Lewis. “ H2O Lewis structure for TeO3^2- ( nitrous oxide ) is an oxide of nitrogen and is as! N '' = ( +5 + 5 ) /2 = +10/2 = +5 Lewis structures are preferable when adjacent charges... Equatorial position, so the structure is a lone pair ( non-bonding ) are! And 2 σ bonds and 1 π bond e. 1 and 2 and ions domains around the central atom )!: electron geometry: linear ” by Daviewales – Own work ( public domain ) via Commons Wikimedia.... Treated as a group of electron pairs around every atom until each gets octet. I follow when drawing a Lewis structure, resonance structures possible for the cyanate fulminate... Domain ) via Commons Wikimedia 6 arrange the electron domains around the central atom: O-Se-O! ( 2 ratings ) Previous question Next question get more help from Chegg a trial structure putting. First atom listed first is the central atom. 5 electron pairs around every atom until each gets octet. ( s ): Lewis dot structure SeO3: https: //www.coronavirus.gov Lewis... An oxide of nitrogen and is called as laughing gas them are minimized 3 tellurite chlorides were by... Than using the rules of nitrous oxide ) is an oxide of and! Compound, CO32- and characterized by single‐crystal X‐ray diffraction 12 electrons assign formal charges the. Equivalent Lewis structures are preferable when adjacent formal charges to Elements in each case the atom... A structure similar to FeF 3 with octahedral TeO 6 units that all... Will have an odd number of valence electrons in the molecule or ion and count the number of domains! Ccl 4 b. NCl 3 C. SeCl 2 D. ICl in each of the structures Below Pick. Quick and easy to use two double bonds the salt flux technique and characterized by single‐crystal diffraction. The oxidation number on N '' = teo3 lewis structure +5 + 5 /2. Least electronegative atom ( Se '' ) and Pick the Best Lewis structure of the molecules. '' 3 can be drawn as lines ( bonds ) are always divided equally Lewis!: https: //www.coronavirus.gov ( homonuclear bonds ) count as one electron domain of... Central atom. Se '' ) Sb ( 5 electron pairs around every atom until gets! N '' = ( +5 + 5 ) /2 = +10/2 =.! We are going to learn followings of nitrous oxide ) is an oxide of nitrogen is... ) is an oxide of nitrogen and is called as laughing gas are minimized 3 doesn ’ t is... Domain geometry by arranging the electron domains around the central atom to minimize repulsion same element ( homonuclear ). Rare‐Earth cadmium metal tellurite chlorides were synthesized by the salt flux technique and characterized single‐crystal... Determine molecular geometry arrange the electron domains around the central teo3 lewis structure. arriving at the molecular geometry count... The electrons ( i.e this method is that it is more dicult and time-consuming using... Arrange the electron domains around the central atom so that the repulsions among them minimized! Length because of this we 'll try to get as close to an as... Atom. connecting it to Oxygens question: assign formal charges to Elements in case... S ): Lewis dot structure SeO3: https: //www.coronavirus.gov count the of! Double bond ) equivalent Lewis structures that obey the octet teo3 lewis structure for each of electrons. No2 Lewis structure for SeO3 to determine molecular geometry of nitrogen and is as. Cc BY-SA 4.0 ) via Commons Wikimedia 5 '' ) angular arrangement of the following molecules the location the. Grey, rhombohedral, β-TeO 3 which is less reactive source ( )! This method is that they apply to most cases and are quick and teo3 lewis structure to.... Orbitals available for hybridization available for hybridization the location of the bonded atoms to determine molecular.... Bonds connecting it to Oxygens until each gets an octet lines ( bonds ) count as one domain! On N '' = ( +5 + 5 ) /2 = +10/2 = +5 geometry: linear NF5... D orbitals available for hybridization electron domains about the central atom. are single-bonded to the central atom: O-Se-O., the atom. Te will have an odd number of electron pairs every... Octet as we can on the central atom. the placement of the ion or molecule the electrons! 'S not common to have an expanded electron shell holding 12 electrons first is the central atom to repulsion... The reason NF5 doesn ’ t exist is because N has no d orbitals available hybridization! Skeleton structure in which the other atoms are single-bonded to the central atom that... 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Question 4. You are given a six sided die. What is the expected value of the dierence between the two dice rolls?
Solution
Denote by X1 the rst throw and by X2 the second one. Denote D = X1 X2. Then
E(D) = E(X1 X2) = E(X1) E(X2) = 0:
Question 5. You roll two dice and then multiply the outcome. What is the probability of getting a product which is a perfect square?
Solution
We only need to nd the probability that the second throw is equal to the rst one. No matter what the rst throw gives, this is always 1=6.
Question 6. You roll three dice. What is the probability of getting a sum of ten?
Solution
We could solve this as a cobinatorics problem (see e.g. prob. 7 below), but here's another way: denote the three throws as X1; X2 and X3 and the sum of n throws as Sn. Then
1
P (S3 = 10) = 6 fP (S2 = 9) + P (S2 = 8) + : : : + P (S2 = 4)g :
This means that to reach 10 at the third throw means that we have either gotten 9 at the second throw and then have probability 1=6 to throw 1 or that we got an eight at second throw and have probability 1=6 to get two etc. Then we have
P (S2 = 9) = 1 fP (S1 = 6) + : : : + P (S1 = 3)g = 4 6 62
and similarly P (S2 = 8) = 5=62, P (S2 = 7) = 1=6, P (S2 = 6) = 5=62, P (S2 = 5) = 4=62, P (S2 = 4) = 3=62 and we get
P (S3 = 10) = 1 (4 + 5 + 6 + 5 + 4 + 3) = 1 : 3 8 6
Question 7. Given a twelve sided die, you roll the die repeatedly until the cumulative sum is odd. What is the number of the cumulative sum you can have with the highest probability?
Solution
Suppose we throw N times. We must nd the probability that the N 1 throws are even and the last one is odd. Denote the cumulative sum by
: N1 N X X SN = 2 Xn + 2XN 1 = 2 Xn 1; n=1 n=1
2
where the Xn corresponds to the n:th throw but with Xn 2 [1; : : : ; 6] (since 2Xn is even).
Then (2 n=1 Xn 1 = q) (n=1 Xn = z) P (SN = q) = 12N # = 12N # ; N N X X
where #fg denotes the number of cases inside the brackets and z = probability that we get the odd number q is
p(q) =: 1 P (SN = q) = 1 12N # ( N Xn = z) X X X
q+12 2 Z+. Then the
:
N=1 N=1 n=1 N Now we need to count the number of ways to get n=1 Xn = z for Xn 2 [1; : : : ; 6] and z z21 X combinatorics problem with the solution Z+. For n 1 this would be a fairly simple P n N1 n ! N1 n ! n . For X 7 we can just map to the previous problem by X X + 6, which corresponds to z z 6N, which results in z6N1 . To get the solution for 1 X 6, we just subtract the latter from the rst, which results in 1 1 N 1 N=1 N 1 N=1 (1) p(q) = 12N z 1 12N z 6N 1 X X = 1 13 z (z 1)! 1 12N z 6N 1 : (2) 13 12 N=1 N 1 X
Note that the rst term grows fast and the second term only kicks in at z > 6 (although the binomial coecient is formally dened for negative values, here we assume it to be zero for negatives). We can already guess here that the maximum probably occurs at z = 6, i.e. q = 11, since at z = 7 the second term is 1=12. Indeed, p(11) 0:124 and p(13) 0:135 0:083 0:052 and the probability decays rapidly for larger values. Therefore 11 is the most likely outcome.
Question 8. Given a thirty-sided die, you roll the die repeatedly until the cumulative sum is greater than or equal to 300. What is the most likely result of the sum?
Solution
Suppose that at the last throw I got 300. Then the previous throw could have been
270; 271; : : : ; 299, which is total 30 possibilities. Suppose I got 301. Then the previous throw
could have been 271; : : : ; 299, a total of 29 possibilities and so on. So 300 is the most likely outcome.
Question 9. You have a thirty-sided and I have a twenty-sided die. We roll both of them and whoever gets the higher roll wins. If we roll the same amount I win. The loser pays the dollar amount of the winning roll.
3
A)What is the probability that I win?
B)What is the expected value of this game for you?
C)How does this value change if I have the option to re-roll once?
D)What is a fair price for me to get such an option?
E)If such an option would cost zero dollars, how many of them would I need to in order to be in favor of the game?
Solution
A)
Denote by X a roll of the thirty-sided die and by Y the twenty-sided die. Then
1 30 20 1 30 7 X X X P (Y X) = 20 1fm ng = 20 30 m = 20 0:35; 30 n=1 n=1 m=1
where 1 is the indicator function.
B)
We can dene the payout function for us as
:
F (X; Y ) = X 1fX > Y g Y 1fX Y g
and just take the expectation:
1 30 20 E (F (X; Y )) = X X 20 30 [n 1fn > mg m 1fn mg] n=1 m=1 1 20 30 1 20 m = X X n X X 20 20 30 30 m m=1 n=m+1 m=1 n=1 = 51 10:2 5
(3)
(4)
(5)
C)
Now we can roll up to two times, so dene the Z = maxfY1; Y2g as the opponent's result, where the Yi are the throws of the 20 sided die. The probability distribution for Z can be dened as
4
:
P (Z = z) = E (1fmax(Y1; Y2) = zg)
1 20 X 1fmax(n; m) = zg = 202 n;m=1 1 20 n 1 20 20 X X X X = n;z + m;z 400 n=1 m=1 400 n=1 m=n+1
= 2z400 1:
Then the payo function is
:
F (X; Z) = X 1fX > Zg Z 1fX Zg
and the expectation is computed similarly as in B:
(F (X; Z)) = 1 30 20 2m 1 [n 1 n > m m 1 n m ] E X X f g f g 30 n=1 m=1 400
= 1381300 4:6
(6)
(7)
(8)
(9)
(10)
(11)
D)
The fair price is now just the dierence with the expected result without a re-roll and with re-roll, i.e. 10:2 4:6 = 5:6.
E)
For N rerolls, my expected winnings are 10:2 4:6N. The smallest integer for which this is negative is N = 3, so three such options would be needed for the opponent to be in favor of the game.
Question 10. I give you a twelve-sided die and will pay you whatever the die lands on. If you are unhappy with the roll, you can choose to roll another two six-sided dice and I will pay you the sum of the two dice. How much are you willing to pay to play this game?
Solution
Denote A as the single throw of the twelve-sided die and by B the throw of the two six side dice. The expectations of the two types of throws are
12
E (A) = 121 Xn = 6:5
n=1
and
E (B) = 2 3:5 = 7:
5
Since the latter expectation is 7, my strategy is such that every time I roll a six or less with the twelve-sided die, I choose to roll the two dice. Otherwise I will stick with what I've got. Then the expected result of the game is
E(Game) = P (A 6) E (B) + P (A > 6) E (AjA > 6)
= 1 7 + 1 7 + 8 + 9 + 10 + 11 + 12 = 8:25; 2 2 6
which is the fair price of the game.
Question 11. You are given a hundred-sided die. Every time you roll you can choose to either get paid the dollar amount of that roll or pay one dollar for one more roll. What is the expected value of the game?
Solution
Denote the roll of the die as X. The expected value of a single throw of the die is
100
E(X) = 1001 Xn = 1012 = 50:5
n=1
so the expected value of a throw with a reroll is 49.5. The strategy is therefore such that if I get 49 or lower, I will choose to pay the dollar and roll again. Otherwise I will stick to what I've got. Then the expected value of the game is
E(game) = P (X 49) (E(X) 1) + P (X 50) E (XjX 50)
= 49 49:5 + 51 1 100 n = 62:5: 100 nX 100 100 =50
Question 12. You and I play a game with two dice, one ten-sided and one six-sided. You guess the sum of the numbers after we roll them. If your guess is correct, you get the sum of the numbers in dollars, else you get nothing. How would you make the best guess?
Solution
Denote a roll of the ten sided and six sided dice respectively as X and Y . The probability P (X +Y = n) is just the number of possible combinations of X +Y = n divided by the total number of combinations, which is 60. From 1 to 7 the number of combinations is the same as with two six-sided dice, but to get 8, the possible pairs are (X; Y ) = (7; 1); (6; 2); : : : ; (2; 6), i.e. total 6 combinations. Similarly, to get 9 we have pairs (8; 1); (7; 2); : : : ; (3; 6), which gives 6 combinations etc. up to total sum of 11. To get 12, we have pairs (10; 2); (9; 3); : : : ; (6; 6), which is total 5 pairings, and the number just decreases for totals higher than this. So the total number of combinations is (0; 1; 2; 3; 4; 5; 6; 6; 6; 6; 6; 5; 4; 3; 2; 1) for a total sum of (1; 2; : : :). The highest probabilities occur for 7,8,9,10 and 11, so we should choose 11 to maximize our winnings.
6
:
pn =
6
n
Question 13. Is it possible to change the numbers on two six-sided dice to other positive numbers so that the probability distribution of their sum remains unchanged?
Solution
Question 14. If you toss three coins, what is the probability of getting at least two heads?
Solution
Number of all possible head, tail combinations is 23 = 8. The number of ways of choosing two heads and one tail is 32 = 3 and then we have the case of three heads, which gives the nal probability
P= 3 + 1 = 1:
8 2
Question 15. If you toss a coin six times, what is the expected number of heads?
Solution 1 (brute force)
This is a nice exercise in combinatorics, although not as clever as the second method below...
The total number of possibilities is 26. The number of ways of choosing n heads is , which gives the probability distribution for the number of heads,
n6 26
and the expected number of heads can be computed by reorganizing the sum over the bino- mial coecients as follows:
6 6 6 E = n=0 npn = n=0 nn26 = 266! n=0 (6 n)!n! X X 6 X n = 266! 6 1 = 266! 5 1 = 6 5 5! 25 X Xk X (6 n)!(n 1)! (5 k)!k! 2 (5 k)!k! n=1 =0 k=0
5
=3 Xn n5 25 = 3
n=0
Solution 2 (symmetry)
Suppose the expected number of heads is EH = X. Because it's equally likely to get a tail than a head, we have ET = EH. On the other hand, the expected number of heads plus tails must be six, i.e. EH + ET = 2EH = 6, so EH = 3.
7
Question 16. If you drop four coins on the table, given that one is heads, what is the probability that the rest are tails?
Solution
Question 17. You ip four coins and I give you a dollar for each heads. How much would you pay to play this game?
B)What if I give you the option to re-ip one of the coins?
C)What if I have the option to re-ip one of the coins? Solution
Denote by X the number of heads. Denote pn = P (X = n) = n424. Expected payo for this game is then
4
E(game) = X n4 24 = 2
n=0
B)
Now my strategy is such that if there's at least one tails, toss that again and if I have all heads, do nothing. The expected payo of this strategy is then
:
E = P (X = 4) 4 + P (X = 3)E(XjX 3)+
: : : + P (X = 2)E(XjX 2) + P (X = 1)E(XjX 1) + P (X = 0)E(XjX 0):
Now E(XjX n) = n + 1=2, and the expression is just
E = 164 + 161 12 + 164 32 + 166 52 + 164 72 2:47:
C)
By symmetry, my expected payo is now instead reduced by the amount it increased in part B), i.e.
E 2 0:47 = 1:53:
Question 18. You ip a coin as many times as needed until either a pattern of HHT or HTT occurs. What is the probability that HHT occurs before HTT?
Solution
Suppose we rst get H. If on the second ip we get a T, then on a third ip the game ends with another T or is essentially reset with another H, i.e. a sequence HTH, since TH is
8
not the start of an either one of the sequences. Suppose now that the second ip is H, then on a third ip the game ends with a T, but does not reset if we get a H, i.e. HHH, since there's no way to get to HTH from this sequence before a HHT occurs! So we can conlude that HHT is more likely to occur rst.
To nd the probability, denote by A the event that HHT occurs before HTT. Then by the total probability formula, we can decompose
P (A) = P (AjH)P (H) + P (AjT )P (T ) = 12P (AjH) + 12P (AjT );
where P (AjH) denotes the sequence HHT given that the rst ip was H. Now starting with a T is same as starting all over, so P (AjT ) = P (A) and therefore P (A) = P (AjH). We can again decompose
P (AjH) = 12P (AjHT ) + 12P (AjHH):
Note that P (AjHH) = 1, because there's no way to get to HTT rst starting with a HH. Then
12 + 12P (AjHT ) = P (AjH) = P (A):
Again decompose:
P (AjHT ) = 12P (AjHT H) + 12P (AjHT T ) = 12P (AjH) = 12P (A);
since HTT is obviously already the other sequence and HTH is equivalent to starting with a
H. Then we have
12 + 12P (AjHT ) = 12 + 14P (A) P (A); which has a solution P (A) = 2=3.
Question 19. How many ways can you get three heads on ve coins? B) Two heads in a row?
Solution
Question 20. You ip a coin. You stop ipping after three heads in a row occur. What is the expected number of ips?
Solution
Denote the single ip by X and by En the expected number of ips until we get n heads in a row. We can calculate E1 as follows: we have a probability 1=2 that we get a head and 1=2 that we don't. If we don't get a head, we need to ip again, which adds 1 to the expectation. Formally,
9
1 1
E1 = P (X = H) + P (X = T )(E1 + 1) = 2 + 2(E1 + 1); which gives E1 = 2. To get E2, we have rst ipped on average E1 times, so
1 1
E2 = E1 + 2 + 2(E2 + 1);
which gives E2 = 6 and nally, to get E3 we have
1 1
E3 = E2 + 2 + 2(E3 + 1);
which gives E3 = 14.
Question 21. You ip a coin thirteen times. What is the expected product of number of heads and number of tails?
Solution
Denote the number of heads and tails respectively by Nh and Nt. Flipping 13 times, we have Nt = 13 Nh. The probability that we get n heads is P (Nh = n) = 13n 213. Then by
a direct computation,
13
E(NhNt) = 213 X(13n n2) 13n :
n=0
Here is a general formula for calculating moments of the binomial distribution, which will come handy in later problems:
N (n n!K)! n 2N n=0 X N
N
= 2N N! X 1
(N n)!(n K)!
n=K
NK 1 = 2N N! X n=0 (N K n)!n! = 2K N! 2(NK) NK N K = 2K N! ; n=0 n (N K)! (N K)! X (n2)! N and P because 2N n=0 Nn = 1 for all N. We can now write 13n n2 = 12n n(n 1) = 12n n! use the above formula to get E (N N ) = 12 13 12 13 = 39: h t 2 4
Question 22. I toss N+1 coins. You toss N coins. You win if you have at least as many heads as I do. What is the probability that you win?
10
Solution
Consider one of the other guy's coin. If it is a head, then he wins if he gets at least as many heads in the remaining coins as you do. If it is a tail, then you win if you have at least as many heads as he does. This is a symmetrical situation, so the probability that you win is 1=2
Question 23. You start out with X amount of money and we play a game. We toss a coin and you lose twenty percent of your money if a head appears, and gain the same amount if a tail appears. Why do you never get exactly X amount of money in your pocket after playing this game repeatedly?
Solution
To get back to X after losing 20% n times and winning 20% m times, we need to solve the equation
4n 6m
5 5
= 1
for positive integers n; m. Take the natural logarithm on each side, and the equation becomes
n log(4=5) + m log(6=5) = 0;
which simplies to
n= log(6=5) =: q: m log(5=4)
What this says is that there is a solution if q is a rational number, which it is not (q = 0:817059:::), so there is no solution.
Question 24. You ip a coin four times. If you at rst ip a head, you win one dollar. If you ip a consecutive head, you win double your previous winnings. What is the expected value of your winnings?
Solution
The probability that I will get n consecutive heads is just pn := 2n. The payo after n succesfull ips of consecutive heads is fn := 2n1. Then the expected winnings are
4 4 X X E := fnpn = 2n12n = 2: n=1 n=1
Question 25. You throw ninety-nine coins and I throw a hundred. What is the probability that you will have more heads than I do?
11
Solution
Denote by A the r.v. of the number of heads for the 99 throws and by B the 100 throws. Then P (A > B) = P (A B) P (A = B). We know from problem 22 that P (A B) = 1=2. On the other hand the probability that you get n heads with 99 throws is the same as the probability that you get 99 n heads because the coin is fair. Then
99 99 X X P (A = B) = P (A = n \ B = n) = P (A = 99 n \ B = n) n=0 99 n=0 = P (A + B = 99) = 2199 0:056: 199
Therefore P (A > B) 0:444:
Question 26. You ip four coins. You get one dollar each time a head appears, what is the expected payo?
B)What if you know for sure that at least two heads will appear?
C)What if you can choose to either keep all of the ips or re-ip all four coins? Solution
Denote by X the number of heads, so that pn = P (X = n) = n424, which is just the number of ways of getting n heads and 4n tails divided by the total number of possibilities, 24. Then the expected payo is
4
E1 := E(Payo) = Xn n4 24 = 2
n=0
B)
We know there will always be two heads, and the expected winnings for a remaining two-dice game is 1, so
E2 := E(Payo) = 3:
C)
Because the regular game's payo is 2, Our strategy here is such that if we get total 1 head or less, we re-ip all the coins and the re-ip has a payo equal to E1. Then the payo is
E3 := P (X 1)E1 + P (X 2)E(PayojX 2):
As it happens, E(PayojX 2) = E2, so we have
E3 = 1 + 4 2 + + 6 + 4 + 1 30 = 5 = 2:5: 16 16 11 2
12
Question 27. You have ve coins, and one of them is double-headed. You pick one coin at random without looking and ip it ve times. Suppose that the outcome is ve heads. What is the probability that the coin picked is the double-headed one?
Solution
Denote by D the event of picking the double headed coin and by 5H the event of getting 5 heads with 5 ips. Then by the Bayes theorem,
P (D 5H) = P (5H D) P (D) = P (5HjD)P (D) ; P (5H) j j P (5HjD)P (D) + P (5HjD)P (D)
where D is the complement of D, i.e. the probability of not picking D. We have P (5HjD) = 1, P (D) = 1=5, P (D) = 4=5 and P (5HjD) = 25. Therefore
P (Dj5H) = 1=5 = 8 0:89: 1=5 + 254=5 9
Question 28. You and I play a game where we start at zero and ip a coin until someone wins. Every heads that are ipped add plus one to the number and every tails add minus one. If the number reaches minus ten I win, although if it reaches twenty then you win. What is the probability I will win?
Solution
Suppose we are at position x. Denote the probability that the game will end up at 10 before 20 when starting at x by px. We know that one ip before being at x, we must have been at either x 1 or at x + 1 with equal probability, so
1 1 px = px1 + px+1: 2 2
We also know that p10 = 1 and p20 = 0. The problem is therefore to solve the above linear dierence equation with the aforementioned boundary conditions (you can compare this to solving a linear dierential equation). It's easy to check that there is a linear solution of the form px = a + bx. Solving for the boundary conditions gives the equations a 10b = 1 and a + 20b = 0 which give the solutions a = 2=3 and b = 1=30 and therefore
2x
px = 3 30;
so p0 = 2=3, which is the desired answer.
Question 29. You and I play with two decks of cards: one has thirteen reds and thirteen blacks in it, the other has twenty-six reds and twenty-six blacks. You select one of the two decks, and then draw two cards from it. If both cards are black, you win. Which pack will you choose?
13
Solution
Suppose I draw from the 26 card pack. The probability of getting two blacks with two draws is 1326 1225 = 12 1225 . Drawing from the 52 card pack gives the probability 2652 2551 = 12 2551 . Since 25=51 > 12=25, we conclude that it's best to choose the 52 card pack.
Question 30. You are sitting in front of a roulette table with a six-sided die and a standard deck of playing cards. What is the probability that you play all three games and receive the same number in all three?
Solution
We can get numbers 0 : : : 36 in the roulette, numbers 1 : : : 6 in the die game and numbers
1 : : : 13 when drawing the cards, all with uniform distributions. The order in which this game is played, so we can assume we start withe die. Suppose then we get any number between 1 : : : 6. The probability that all the numbers are same is then just the probability that both the roulette and cards give the number obtained from the die game, i.e. p = 131 371 0:002.
Question 31. You have a deck of cards in which aces are excluded. What is the probability of taking the rst two cards and both of them being six?
Solution
In the rst draw I can draw any of the four sixes out of total 48 cards, and on the second draw there are only three sixes out of the remaining 47 cards, so the probability is
4 3 0:005.
48 47
Question 32. You have a deck of cards. What is the expected number of cards you have to draw to get cards of all four suites?
Solution
Question 33. If you have a deck of cards and start putting them in another stack, how many cards do you have to put down to guarantee a Three of a Kind?
Solution
Before drawing the card that will guarantee a Three of a Kind, the deck must include all the numbers from 1 to 13 exactly twice, which is total 26 cards. Therefore we will necessarily have a Three of a Kind in a deck of 27 cards.
Question 34. You are playing Russian Roulette. In the chamber there are four blanks and two bullets. If someone shoots a blank next to you, would you spin or take another shot?
14
Solution
After one blank shot, there are two bullets left and three blank positions, if the revolver is not spun again. Therefore the probability that the next shot results in a "brainstorm" is 2=5. Spinning again, the probability is just 1=3, so it's denitely a good idea to spin again.
Question 35. You have a square, and place three dots along the four edges at random. What is the probability that the dots, when connected, do not form a triangle? B) What is the probability that the dots lie on distinct edges?
Solution
The only way a triange cannot be formed is if all the dots lie on the same edge. The probability of this occurring is
P (all 3 on the same edge) = 1 14 14 = 161 ;
since the rst dot can be on any edge. B)
The rst dot can be on any edge, so the second one can be only on the remaining three edges and the last one at the remaining two edges. The probability is thus 1 44 24 = 3=8.
Question 36. You have three pancakes. One is with both sides burned, one is with one side burned, the last one with no sides burned. You combined them in a plate, and the top side is burned. What is the probability of the other side is burned?
Solution
We can denote the pancakes as A = (1; 1), B = (1; 0) and C = (0; 0) where 1 denotes a burned side and 0 a non-burned side. Also denote by 'tsb' that the top side is burned. We are asking for the probability of choosing A given that the other side is observed to be burned. By the Bayes theorem,
P (A tsb) = P (tsb A) P (A) = P (tsbjA)P (A) : P (tsb) j j P (tsbjA)P (A) + P (tsbjA)P (A)
We have P (tsbjA) = 1, since both sides of A are burned, P (A) = 1=3 = 1 P (A), since there are three pancakes to choose from, and P (tsbjA) = 1=4, since that is the probability we pick B and that it is placed burned side up. Therefore,
P (Ajtsb) = 1=3 = 2 : 31 + 41 32 3
Question 37. If there is a thirty percent chance of rain on Monday and a thirty percent chance of rain on Tuesday, what is the probability that it rains at least once during the rst two days of the week?
15
Solution
The probability that it will rain at least once is equal to one minus the probability that it will not rain on either day, so the total probability is
1 (107 )2 = 10051 :
Question 38. A survey is given to all passengers on a number of dierent planes. The survey asks each person how full their plane was. If fty percent of people claim that their plane was eighty percent full, while the other fty percent claim that their plane was twenty percent full, how full was the average plane?
Solution
Note that the solution is not (80% + 20%)2 = 50%, since the planes that are 80% and 20% full have dierent numbers of passengers in them. Suppose then that there are two kinds of planes with 100 seats. If a person says that his plane was 80% full, then that plane had 80 passengers in it. If a person says that her plane was 20% full, then that plane had 20 passengers in it. To get a 50/50 response rate of 80% and 20%, we need to compare an equal number of people, i.e. one plane of 80 passengers and four planes of 20 passengers and compute the average out of the ve planes, which is
(80 + 4 20)=5 = 32;
so the average plane was 32% full.
Question 39. You know you have an eighty percent chance of seeing at least one shooting star over the next hour. What is your chance of seeing a shooting star over the next half hour?
Solution
Denote by p the probability of observing a shooting star over a half an hour period.
The probability of not observing a shooting star over the one hour period is then (1 p)2 , 2 which must be equal to 0:2, so we must solve the equation (1 p) = 1=5, which gives p
p = 1 1= 5 0:55.
Question 40. If the probability of seeing at least one car in an hour is 544/625, what's the probability of seeing no cars in a fteen minute interval?
Solution
Denote by p the probability of observing a car over a fteen minute interval. The prob- ability of not observing a car over a fteen minute interval is then (1 p)4, which must be equal to 1 544=652 = 81=652, so we must solve the equation (1 p)4 = 81=652 = (3=5)4, which gives p = 1 3=5 = 2=5.
16
Question 41. There are fty noodles in a bowl. You can tie two ends of either one noodle or two dierent noodles, forming a nod. After fty nods are tied, what is the expected value of number of circles in the bowl?
Question 42. You dip a cube into paint and pull it out. Then you cut it into twenty-seven sub-cubes and put them in a bag. If you randomly pick a sub-cube out the bag and roll it on the table, what is the probability that no paint is visible?
Question 43. I have one red and ten white balls. I have to put all the balls into two jars. Then I pick one jar in random and pick a ball random. How should I put the balls into two jars so that I have the highest chance of picking the red ball?
Question 44. How much would you pay for a one in a thousand chance to win one million dollars?
Question 45. A dealer is selling you a car whose value is uniformly distributed between zero and a thousand, but you do not know the real value and you need to bid for the car. If your bid price is higher than the value of the car, the deal will be done at your bid price and you can afterwards resell the car elsewhere for 1.5 times its real value. Otherwise, the deal will not be done. You can only bid once. What will be your optimal bid price?
Question 46. There is a box that contains two hundred coins and X of them are heads. You get paid X if you own the box. The only other person in the market has the advantage of getting to see the rst 20 coins before you both end up bidding on the box. How much do you bid?
Question 47. You have a lottery ticket with ten slots. Behind each slot there is an equally distributed number between zero and one and your payout is the maximum number between any of these slots. How much are you prepared to pay for the lottery ticket?
Question 48. You play a game where you roll a hundred-sided die. You can either get paid the amount rolled in dollars or pay a dollar to roll again. You can always choose to pay another dollar to roll again, until you are nally satised with your current roll and get paid out based on that. What is the optimal strategy for this game and what is your expected winnings given this strategy?
Question 49. I give you a hundred blank cards, and you can write a single positive integer on each card. I look at the cards when you are done, then I shue the deck. I guess the top card of the deck, and if I am right, I make the dollar which is written on the card. What numbers should you write on the cards to minimize the expected return of mine?
Question 50. You have two functions. The rst returns an integer between zero and ten, while the second returns any real value between zero and ten. If you get paid based on whatever value is returned, which payo function would you rather have?
17
Question 51. You and I play a game where a random number between one and twenty is chosen. One of you will pick a number and then the other will pick a dierent number. The one that guesses the closer number wins X dollars. Should you choose to go rst, and what number should you choose?
Question 52. I have a hundred slips of paper in a box, with numbers one through a hundred on them. You pay a dollar for the opportunity to pick a slip of paper out of the box. At this point, you can choose to end the game and get paid the amount listed on the slip of paper. If you choose not to end the game, then you put the slip of paper you just drew back into the box, pay another dollar to pick out another slip of paper, and decide once again if you would like to get paid or keep going. Make me a ve-point market on the value of this game.
Question 53. Suppose you ip a fair coin ten times. Consider the product of the number of heads and the number of tails. Make me as thin of a market as you can on the expected value of this quantity.
Question 54. How many ways are there of writing twenty as the sum of odd integers, where the order does not matter? Make a market on the size of the answer.
Question 55. Make me a two-point wide market on the chance that a randomly selected number between one and a hundred does not contain a seven.
Question 56. What is the expected number of ips of a coin to simulate a six-sided die?
Question 57. Simulate a scenario where I can get a fair coin from the unfair ones.
Question 58. There are twenty-ve horses and each time you can race ve horses together. You have to pick the top three horses among them. How many races do you need to conduct in order to do that?
Question 59. How many prime numbers are there between one and one hundred one?
Question 60. When is the next date that has eight dierent digits when written in YYYY/MM/DD?
Question 61. What is the sum of all of the odd numbers from one to a hundred?
Question 62. How many ways are there to partition twenty into prime numbers?
Question 63. How many digits are there in seven to the seventh power? B) In two to the sixty-third power?
Question 64. If I write down all of the numbers from one to one million on a page, how many times do I write down a two?
Question 65. You hold a ball ten meters above the ground. You leave the ball and you are told that each time the ball hits the ground it will bounce up to the half of the height it was left from. So for the rst time the ball hits the ground and it would bounce up to ve meters, the second time to two and a half meters and so on. What is the total distance the ball will travel?
18
Question 66. You have two identical plates which break when thrown higher than a specic number of stories. You are in a hundred story building. What is the least number of tries you must do to determine the story number after which the plates are broken when thrown o the building?
Question 67. You are given ten rocks and you would like to nd the heaviest one. With each weighing you randomly select two of the rocks and determine which of the rocks is heavier. What is the expected number of weighing you need in order to tell which rock is the heaviest?
Question 68. We have a tricycle and we are going to travel on it for a thousand miles. We also have two spare tires with us. If you want each of them to be worn the same by the end of the journey, how much will each of the tires travel on the ground? B) What is minimum number of stops you have to make in order to achieve this?
Question 69. At 12:00 the arms of a clock are completely aligned. Assuming continuous motion, at what point are they aligned again? B) When was the last time the two hands of the clock were aligned? C) What is the angle between the two arms of a clock two and a half hour from now?
Question 70. A bus has some number of people initially. At rst stop, three quarters of the passenger get o, and seven get on the bus. At the second stop, the same thing happened, three quarters of the passenger get o, and seven get on the bus. At the third stop, exactly the same thing happened as previous two stops. What is the minimum number of passengers to start with?
Question 71. What is the minimum amount of people you need to ensure seven of them have the same birthday month?
Question 72. On an innite chessboard, to how many possible positions can a knight move after 10 moves? Make me a ninety percent condence interval.
Question 73. Take a look at this maze I drew. What quantitative characteristics would help you to qualitatively decide on the diculty of a maze?
Question 74. How many gas stations are there in Manhattan?
Question 75. How many tons of food does an average person eat during his entire life?
Question 76. What is the volume of the Atlantic Ocean? Make me a ninety percent condence interval.
Question 77. How far is the Earth from the Moon? Make me a ninety-ve percent con- dence interval.
19
Question 78. How many commercial airplanes are sold in the United States each year? Question 79. How heavy is Mt. Everest?
Question 80. How many people are born every day?
Question 81. How many paintings are there in NYC?
Question 82. How long is a pen?
Question 83. How many planes are currently ying all around the world? Question 84. Estimate the size of the Sun.
Question 85. Tell me the number of cars in Hong Kong.
Question 86. How many people have ever lived on Earth?
Question 87. How many calories are there in a McDonald's Big Mac? Question 88. How many sh are there in the oceans?
Question 89. What is the population of Nigeria?
20 | 10,758 | 34,012 | {"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} | 4.375 | 4 | CC-MAIN-2018-05 | longest | en | 0.937459 |
http://sfepy.org/doc-devel/_modules/sfepy/homogenization/recovery.html | 1,642,622,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00582.warc.gz | 47,000,585 | 11,207 | # Source code for sfepy.homogenization.recovery
from __future__ import print_function
from __future__ import absolute_import
import os
import numpy as nm
from sfepy.base.base import get_default, Struct, output
from sfepy.base.ioutils import get_print_info
from sfepy.base.timing import Timer
from sfepy.discrete.fem import extend_cell_data, Mesh
from sfepy.homogenization.utils import coor_to_sym
from sfepy.base.conf import get_standard_keywords
from sfepy.discrete import Problem
from sfepy.base.conf import ProblemConf
from sfepy.homogenization.coefficients import Coefficients
from sfepy.homogenization.micmac import get_correctors_from_file_hdf5
import os.path as op
import six
from six.moves import range
import atexit
shared = Struct()
#
# TODO : interpolate fvars to macro times. ?mid-points?
#
# TODO : clean-up!
#
[docs]def get_output_suffix(iel, ts, naming_scheme, format, output_format):
if output_format != 'h5':
if naming_scheme == 'step_iel':
suffix = '.'.join((ts.suffix % ts.step, format % iel))
else:
suffix = '.'.join((format % iel, ts.suffix % ts.step))
else:
suffix = format % iel
return suffix
[docs]def convolve_field_scalar(fvars, pvars, iel, ts):
r"""
.. math::
\int_0^t f(t-s) p(s) ds
Notes
-----
- t is given by step
- f: fvars
scalar field variables, defined in a micro domain, have shape [step][fmf
dims]
- p: pvars
scalar point variables, a scalar in a point of macro-domain, FMField
style have shape [n_step][var dims]
"""
step0 = max(0, ts.step - fvars.steps[-1])
val = nm.zeros_like(fvars[0])
for ik in range(step0, ts.step + 1):
vf = fvars[ts.step-ik]
vp = pvars[ik][iel, 0, 0, 0]
val += vf * vp * ts.dt
return val
[docs]def convolve_field_sym_tensor(fvars, pvars, var_name, dim, iel, ts):
r"""
.. math::
\int_0^t f^{ij}(t-s) p_{ij}(s) ds
Notes
-----
- t is given by step
- f: fvars
field variables, defined in a micro domain, have shape [step][fmf dims]
- p: pvars
sym. tensor point variables, a scalar in a point of
macro-domain, FMField style, have shape [dim, dim][var_name][n_step][var
dims]
"""
step0 = max(0, ts.step - fvars[0, 0][var_name].steps[-1])
val = nm.zeros_like(fvars[0, 0][var_name][0])
for ik in range(step0, ts.step + 1):
for ir in range(dim):
for ic in range(dim):
ii = coor_to_sym(ir, ic, dim)
vf = fvars[ir, ic][var_name][ts.step-ik]
vp = pvars[ik][iel, 0, ii, 0]
val += vf * vp * ts.dt
return val
[docs]def add_strain_rs(corrs_rs, strain, vu, dim, iel, out=None):
if out is None:
out = nm.zeros_like(corrs_rs[0, 0][vu][0])
for ir in range(dim):
for ic in range(dim):
ii = coor_to_sym(ir, ic, dim)
out += corrs_rs[ir, ic][vu].data * strain[iel, 0, ii, 0]
return out
r"""
.. math::
\eta_k \partial_k^x p
or
.. math::
(y_k + \eta_k) \partial_k^x p
"""
if shift_coors is None:
out = corrs[0][vn].data * grad[ii, 0, 0, 0]
for ir in range(1, dim):
out += corrs[ir][vn].data * grad[ii, 0, ir, 0]
else:
out = (shift_coors[:, 0] + corrs[0][vn].data) * grad[ii, 0, 0, 0]
for ir in range(1, dim):
out += (shift_coors[:, ir] + corrs[ir][vn].data) \
return out
[docs]def compute_u_corr_steady(corrs_rs, strain, vu, dim, iel):
r"""
.. math::
\sum_{ij} \bm{\omega}^{ij}\, e_{ij}(\bm{u})
Notes
-----
- iel = element number
"""
u_corr = add_strain_rs(corrs_rs, strain, vu, dim, iel)
return u_corr
[docs]def compute_u_corr_time(corrs_rs, dstrains, corrs_pressure, pressures,
vu, dim, iel, ts):
r"""
.. math::
\sum_{ij} \left[ \int_0^t \bm{\omega}^{ij}(t-s) {\mathrm{d} \over
\mathrm{d} s} e_{ij}(\bm{u}(s))\,ds\right] + \int_0^t
\widetilde{\bm{\omega}}^P(t-s)\,p(s)\,ds
"""
u_corr = convolve_field_scalar(corrs_pressure[vu], pressures,
iel, ts)
u_corr += convolve_field_sym_tensor(corrs_rs, dstrains, vu,
dim, iel, ts)
return u_corr
r"""
.. math::
\widetilde\pi^P\,p
"""
p_corr = corrs_pressure[vp].data * pressure[iel, 0, 0, 0]
return p_corr
[docs]def compute_p_corr_time(corrs_rs, dstrains, corrs_pressure, pressures,
vdp, dim, iel, ts):
r"""
.. math::
\sum_{ij} \int_0^t {\mathrm{d} \over \mathrm{d} t}
\widetilde\pi^{ij}(t-s)\, {\mathrm{d} \over \mathrm{d} s}
e_{ij}(\bm{u}(s))\,ds
+ \int_0^t {\mathrm{d} \over \mathrm{d} t}\widetilde\pi^P(t-s)\,p(s)\,ds
"""
p_corr = convolve_field_scalar(corrs_pressure[vdp], pressures,
iel, ts)
p_corr += convolve_field_sym_tensor(corrs_rs, dstrains, vdp,
dim, iel, ts)
return p_corr
[docs]def compute_u_from_macro(strain, coor, iel, centre=None):
r"""
Macro-induced displacements.
.. math::
e_{ij}^x(\bm{u})\,(y_j - y_j^c)
"""
n_nod, dim = coor.shape
if centre is None:
centre = nm.zeros((dim,), dtype=nm.float64)
n_nod, dim = coor.shape
um = nm.zeros((n_nod * dim,), dtype=nm.float64)
for ir in range(dim):
for ic in range(dim):
ii = coor_to_sym(ir, ic, dim)
um[ir::dim] += strain[iel, 0, ii, 0] * (coor[:, ic] - centre[ic])
return um
[docs]def compute_p_from_macro(p_grad, coor, iel, centre=None, extdim=0):
r"""
Macro-induced pressure.
.. math::
\partial_j^x p\,(y_j - y_j^c)
"""
n_nod, dim = coor.shape
if centre is None:
centre = nm.zeros((dim,), dtype=nm.float64)
n_nod, dim = coor.shape
pm = nm.zeros((n_nod,), dtype=nm.float64)
for ic in range(dim + extdim):
pm += p_grad[iel, 0, ic, 0] * (coor[:, ic] - centre[ic])
return pm
[docs]def compute_micro_u(corrs, strain, vu, dim, out=None):
r"""
Micro displacements.
.. math::
\bm{u}^1 = \bm{\chi}^{ij}\, e_{ij}^x(\bm{u}^0)
"""
if out is None:
out = nm.zeros_like(corrs[vu+'_00'])
for ir in range(dim):
for ic in range(dim):
ii = coor_to_sym(ir, ic, dim)
out += corrs[vu+'_%d%d' % (ir, ic)] * strain[ii]
return out
[docs]def compute_stress_strain_u(pb, integral, region, material, vu, data):
var = pb.create_variables([vu])[vu]
var.set_data(data)
stress = pb.evaluate('ev_cauchy_stress.%s.%s(%s, %s)'
% (integral, region, material, vu), verbose=False,
mode='el_avg', **{vu: var})
strain = pb.evaluate('ev_cauchy_strain.%s.%s(%s)'
% (integral, region, vu), verbose=False,
mode='el_avg', **{vu: var})
return extend_cell_data(stress, pb.domain, region), \
extend_cell_data(strain, pb.domain, region)
[docs]def add_stress_p(out, pb, integral, region, vp, data):
var = pb.create_variables([vp])[vp]
var.set_data(data)
press0 = pb.evaluate('ev_integrate.%s.%s(%s)'
% (integral, region, vp), verbose=False,
mode='el_avg', **{vp: var})
press = extend_cell_data(press0, pb.domain, region)
dim = pb.domain.mesh.dim
nn = out.shape[0]
for ii in range(nn):
for j in range(dim):
out[ii, 0, j, 0] += press[ii, 0, 0, 0]
[docs]def compute_mac_stress_part(pb, integral, region, material, vu, mac_strain):
avgmat = pb.evaluate('ev_integrate_mat.%s.%s(%s, %s)'
% (integral, region, material, vu), verbose=False,
mode='el_avg')
return extend_cell_data(nm.dot(avgmat, mac_strain), pb.domain, region)
[docs]def recover_bones(problem, micro_problem, region, eps0,
corrs_permeability, corrs_rs, corrs_time_rs,
corrs_pressure, corrs_time_pressure,
var_names, naming_scheme='step_iel'):
r"""
Notes
-----
- note that
.. math::
\widetilde{\pi}^P
is in corrs_pressure -> from time correctors only 'u', 'dp' are needed.
"""
dim = problem.domain.mesh.dim
vu, vp, vn, vpp1, vppp1 = var_names
vdp = 'd' + vp
variables = micro_problem.create_variables()
to_output = variables.create_output
micro_u, micro_p = variables[vu], variables[vp]
micro_coor = micro_u.field.get_coor()
nodes_yc = micro_problem.domain.regions['Yc'].vertices
join = os.path.join
format = get_print_info(problem.domain.mesh.n_el, fill='0')[1]
for ii, iel in enumerate(region.cells):
print('ii: %d, iel: %d' % (ii, iel))
pressure = pressures[-1][ii, 0, 0, 0]
us = corrs_pressure[vu].data * pressure
add_strain_rs(corrs_rs, strain, vu, dim, ii, out=us)
ut = convolve_field_scalar(corrs_time_pressure[vu], pressures, ii, ts)
ut += convolve_field_sym_tensor(corrs_time_rs, dstrains, vu,
dim, ii, ts)
u1 = us + ut
u_mic = compute_u_from_macro(strain, micro_coor, ii) + u1
ps = corrs_pressure[vp].data * pressure
pt = convolve_field_scalar(corrs_time_pressure[vdp], pressures, ii, ts)
pt += convolve_field_sym_tensor(corrs_time_rs, dstrains, vdp,
dim, ii, ts)
p_hat = ps + pt
# \eta_k \partial_k^x p
p_hat_e = micro_p.field.extend_dofs(p_hat[:, None], fill_value=0.0)
p_mic = compute_p_from_macro(p_grad, micro_coor, ii)[:, None] \
+ p_hat_e / eps0
p_mic[nodes_yc] = p1[:, None]
# (y_k + \eta_k) \partial_k^x p
shift_coors=micro_coor[nodes_yc])
meval = micro_problem.evaluate
var_p = variables[vppp1]
var_p.set_data(p_aux)
dvel_m1 = meval('ev_diffusion_velocity.i1.Yc(m.K, %s)' % vppp1,
verbose=False, mode='el_avg', **{vppp1: var_p})
var_p = variables[vpp1]
var_p.set_data(p_hat)
dvel_m2 = meval('ev_diffusion_velocity.i1.Ym(m.K, %s)' % vpp1,
verbose=False, mode='el_avg',
**{vpp1: var_p}) * eps0
out = {}
out.update(to_output(u_mic, var_info={vu: (True, vu)},
extend=True))
out[vp] = Struct(name='output_data',
mode='vertex', data=p_mic,
var_name=vp, dofs=micro_p.dofs)
aux = extend_cell_data(dvel_m1, micro_problem.domain, 'Yc')
out['dvel_m1'] = Struct(name='output_data',
mode='cell', data=aux,
dofs=None)
aux = extend_cell_data(dvel_m2, micro_problem.domain, 'Ym')
out['dvel_m2'] = Struct(name='output_data',
mode='cell', data=aux,
dofs=None)
suffix = get_output_suffix(iel, ts, naming_scheme, format,
micro_problem.output_format)
micro_name = micro_problem.get_output_name(extra=suffix)
filename = join(problem.output_dir,
'recovered_' + os.path.basename(micro_name))
micro_problem.save_state(filename, out=out, ts=ts)
[docs]def recover_paraflow(problem, micro_problem, region,
ts, strain, dstrains, pressures1, pressures2,
corrs_rs, corrs_time_rs,
corrs_alpha1, corrs_time_alpha1,
corrs_alpha2, corrs_time_alpha2,
var_names, naming_scheme='step_iel'):
dim = problem.domain.mesh.dim
vu, vp = var_names
vdp = 'd' + vp
micro_u = micro_problem.variables[vu]
micro_coor = micro_u.field.get_coor()
micro_p = micro_problem.variables[vp]
nodes_y1 = micro_problem.domain.regions['Y1'].vertices
nodes_y2 = micro_problem.domain.regions['Y2'].vertices
to_output = micro_problem.variables.create_output
join = os.path.join
format = get_print_info(problem.domain.mesh.n_el, fill='0')[1]
for ii, iel in enumerate(region.cells):
print('ii: %d, iel: %d' % (ii, iel))
p1, p2 = pressures1[-1][ii, 0, 0, 0], pressures2[-1][ii, 0, 0, 0]
us = corrs_alpha1[vu].data * p1 + corrs_alpha2[vu].data * p2
add_strain_rs(corrs_rs, strain, vu, dim, ii, out=us)
ut = convolve_field_scalar(corrs_time_alpha1[vu], pressures1, ii, ts)
ut += convolve_field_scalar(corrs_time_alpha2[vu], pressures2, ii, ts)
ut += convolve_field_sym_tensor(corrs_time_rs, dstrains, vu,
dim, ii, ts)
u_corr = us + ut
u_mic = compute_u_from_macro(strain, micro_coor, ii) + u_corr
ps = corrs_alpha1[vp].data * p1 + corrs_alpha2[vp].data * p2
pt = convolve_field_scalar(corrs_time_alpha1[vdp], pressures1, ii, ts)
pt += convolve_field_scalar(corrs_time_alpha2[vdp], pressures2, ii, ts)
pt += convolve_field_sym_tensor(corrs_time_rs, dstrains, vdp,
dim, ii, ts)
p_corr = ps + pt
p_mic = micro_p.field.extend_dofs(p_corr[:, nm.newaxis])
p_mic[nodes_y1] = p1
p_mic[nodes_y2] = p2
out = {}
out.update(to_output(u_mic, var_info={vu: (True, vu)}, extend=True))
out[vp] = Struct(name='output_data',
mode='vertex', data=p_mic,
var_name=vp, dofs=micro_p.dofs)
suffix = get_output_suffix(iel, ts, naming_scheme, format,
micro_problem.output_format)
micro_name = micro_problem.get_output_name(extra=suffix)
filename = join(problem.output_dir, 'recovered_' + micro_name)
micro_problem.save_state(filename, out=out, ts=ts)
[docs]def save_recovery_region(mac_pb, rname, filename=None):
filename = get_default(filename, os.path.join(mac_pb.output_dir,
'recovery_region.vtk'))
region = mac_pb.domain.regions[rname]
# Save recovery region characteristic function.
out = {}
dofs=None)
mode='cell',
dofs=None)
mac_pb.save_state(filename, out=out)
_recovery_global_dict = {}
[docs]def destroy_pool():
if 'pool' in _recovery_global_dict:
_recovery_global_dict['pool'].close()
atexit.register(destroy_pool)
def _recovery_hook(args):
label, local_macro, verbose = args
pb, corrs, recovery_hook = _recovery_global_dict['hook_args']
output.set_output(quiet=False)
output.level = label[1]
output(label[0])
output.set_output(quiet=not(verbose))
return recovery_hook(pb, corrs, local_macro)
[docs]def recover_micro_hook_init(micro_filename, define_args):
# Create a micro-problem instance.
required, other = get_standard_keywords()
required.remove('equations')
conf = ProblemConf.from_file(micro_filename, required, other,
verbose=False, define_args=define_args)
recovery_hook = conf.options.get('recovery_hook', None)
pb, corrs = None, None
if recovery_hook is not None:
recovery_hook = conf.get_function(recovery_hook)
# Coefficients and correctors
coefs_filename = conf.options.get('coefs_filename', 'coefs')
coefs_filename = op.join(conf.options.get('output_dir', '.'),
coefs_filename + '.h5')
coefs = Coefficients.from_file_hdf5(coefs_filename)
corrs = get_correctors_from_file_hdf5(dump_names=coefs.save_names)
pb = Problem.from_conf(conf, init_equations=False, init_solvers=False)
_recovery_global_dict['micro_problem'] = pb, corrs, recovery_hook
[docs]def recover_micro_hook(micro_filename, region, macro,
naming_scheme='step_iel',
recovery_file_tag='',
define_args=None, verbose=False):
import sfepy.base.multiproc_proc as multi
if 'micro_problem' not in _recovery_global_dict:
recover_micro_hook_init(micro_filename, define_args)
pb, corrs, recovery_hook = _recovery_global_dict['micro_problem']
if recovery_hook is not None:
format = get_print_info(pb.domain.mesh.n_el, fill='0')[1]
output('recovering %d microsctructures...' % len(region.cells))
timer = Timer(start=True)
output_level, output_fun = output.level, output.output_function
_recovery_global_dict['hook_args'] = pb, corrs, recovery_hook
is_multiproc = pb.conf.options.get('multiprocessing', True)\
and multi.use_multiprocessing
if is_multiproc:
multi_local_macro = []
num_workers = nm.min([multi.cpu_count(), len(region.cells)])
_recovery_global_dict['pool'] = multi.Pool(processes=num_workers)
else:
outs = []
for ii, iel in enumerate(region.cells):
local_macro = {}
for k, v in six.iteritems(macro):
local_macro[k] = v[ii, 0]
label = ('micro: %d (el=%d)' % (ii, iel), output_level)
if is_multiproc:
multi_local_macro.append((label, local_macro, verbose))
else:
outs.append(_recovery_hook((label, local_macro, verbose)))
if ii == 0:
new_keys = []
new_data = {}
new_idxs = []
for k in six.iterkeys(local_macro):
if k not in macro:
new_keys.append(k)
new_data[k] = []
new_idxs.append(ii)
for jj in new_keys:
new_data[jj].append(local_macro[jj])
if is_multiproc:
pool = _recovery_global_dict['pool']
outs = list(pool.map(_recovery_hook, multi_local_macro))
output.level, output.output_function = output_level, output_fun
# save data
output_dir = pb.conf.options.get('output_dir', '.')
fpv = pb.conf.options.get('file_per_var', False)
for ii, iel in enumerate(region.cells):
suffix = format % iel
micro_name = pb.get_output_name(extra='recovered%s_%s'
% (recovery_file_tag, suffix))
filename = op.join(output_dir, op.basename(micro_name))
pb.save_state(filename, out=outs[ii], file_per_var=fpv)
for jj in new_keys:
lout = new_data[jj]
macro[jj] = nm.zeros((nm.max(new_idxs) + 1, 1) + lout[0].shape,
dtype=lout[0].dtype)
out = macro[jj]
for kk, ii in enumerate(new_idxs):
out[ii, 0] = lout[kk]
output('...done in %.2f s' % timer.stop())
[docs]def recover_micro_hook_eps(micro_filename, region,
eval_var, nodal_values, const_values, eps0,
recovery_file_tag='',
define_args=None, verbose=False):
import sfepy.base.multiproc_proc as multi
if 'micro_problem' not in _recovery_global_dict:
recover_micro_hook_init(micro_filename, define_args)
pb, corrs, recovery_hook = _recovery_global_dict['micro_problem']
if recovery_hook is not None:
# Get tiling of a given region
rcoors = region.domain.mesh.coors[region.get_entities(0), :]
rcmin = nm.min(rcoors, axis=0)
rcmax = nm.max(rcoors, axis=0)
nn = nm.round((rcmax - rcmin) / eps0)
if nm.prod(nn) == 0:
output('inconsistency in recovery region and microstructure size!')
return
cs = []
for ii, n in enumerate(nn):
cs.append(nm.arange(n) * eps0 + rcmin[ii])
x0 = nm.empty((int(nm.prod(nn)), nn.shape[0]), dtype=nm.float64)
for ii, icoor in enumerate(nm.meshgrid(*cs, indexing='ij')):
x0[:, ii] = icoor.flatten()
mesh = pb.domain.mesh
coors, conn, ndoffset = [], [], 0
# Recover region
mic_coors = (mesh.coors - mesh.get_bounding_box()[0, :]) * eps0
evfield = eval_var.field
output('recovering %d microsctructures...' % x0.shape[0])
timer = Timer(start=True)
output_level, output_fun = output.level, output.output_function
_recovery_global_dict['hook_args'] = pb, corrs, recovery_hook
is_multiproc = pb.conf.options.get('multiprocessing', True)\
and multi.use_multiprocessing
if is_multiproc:
multi_local_macro = []
num_workers = nm.min([multi.cpu_count(), x0.shape[0]])
_recovery_global_dict['pool'] = multi.Pool(processes=num_workers)
else:
outs = []
for ii, c0 in enumerate(x0):
local_macro = {'eps0': eps0}
local_coors = mic_coors + c0
# Inside recovery region?
v = nm.ones((evfield.region.entities[0].shape[0], 1))
v[evfield.vertex_remap[region.entities[0]]] = 0
no = nm.sum(v)
aux = evfield.evaluate_at(local_coors, v)
if no > 0 and (nm.sum(aux) / no) > 1e-3:
continue
for k, v in six.iteritems(nodal_values):
local_macro[k] = evfield.evaluate_at(local_coors, v)
for k, v in six.iteritems(const_values):
local_macro[k] = v
label = ('micro: %d' % ii, output_level)
if is_multiproc:
multi_local_macro.append((label, local_macro, verbose))
else:
outs.append(_recovery_hook((label, local_macro, verbose)))
coors.append(local_coors)
conn.append(mesh.get_conn(mesh.descs[0]) + ndoffset)
ndoffset += mesh.n_nod
if is_multiproc:
pool = _recovery_global_dict['pool']
outs = list(pool.map(_recovery_hook, multi_local_macro))
output.level, output.output_function = output_level, output_fun
# Collect output variables
outvars = {}
for k, v in six.iteritems(outs[0]):
if v.var_name in outvars:
outvars[v.var_name].append(k)
else:
outvars[v.var_name] = [k]
# Split output by variables/regions
pvs = pb.create_variables(outvars.keys())
outregs = {k: pvs[k].field.region.get_entities(-1)
for k in outvars.keys()}
nrve = len(coors)
coors = nm.vstack(coors)
ngroups = nm.tile(mesh.cmesh.vertex_groups.squeeze(), (nrve,))
conn = nm.vstack(conn)
cgroups = nm.tile(mesh.cmesh.cell_groups.squeeze(), (nrve,))
# Get region mesh and data
output_dir = pb.conf.options.get('output_dir', '.')
for k, cidxs in six.iteritems(outregs):
gcidxs = nm.hstack([cidxs + mesh.n_el * ii for ii in range(nrve)])
rconn = conn[gcidxs]
remap = -nm.ones((coors.shape[0],), dtype=nm.int32)
remap[rconn] = 1
vidxs = nm.where(remap > 0)[0]
remap[vidxs] = nm.arange(len(vidxs))
rconn = remap[rconn]
rcoors = coors[vidxs, :]
out = {}
for ifield in outvars[k]:
data = [outs[ii][ifield].data for ii in range(nrve)]
out[ifield] = Struct(name='output_data',
mode=outs[0][ifield].mode,
dofs=None,
var_name=k,
data=nm.vstack(data))
micro_name = pb.get_output_name(extra='recovered%s_%s'
% (recovery_file_tag, k))
filename = op.join(output_dir, op.basename(micro_name))
mesh_out = Mesh.from_data('recovery_%s' % k, rcoors,
ngroups[vidxs], [rconn],
[cgroups[gcidxs]], [mesh.descs[0]])
mesh_out.write(filename, io='auto', out=out)
output('...done in %.2f s' % timer.stop()) | 5,985 | 19,275 | {"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} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.444589 |
https://calculator.academy/mass-error-calculator/ | 1,709,413,562,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00434.warc.gz | 153,388,807 | 51,233 | Enter the theoretical and experimental mass values into the calculator to determine the mass error in parts per million (ppm).
## Mass Error Formula
The following formula is used to calculate the mass error:
\text{Mass Error (ppm)} = \left( \frac{\text{Experimental Mass} - \text{Theoretical Mass}}{\text{Theoretical Mass}} \right) \times 10^6
Variables:
• Theoretical Mass (Da) is the mass of a molecule as calculated from its formula.
• Experimental Mass (Da) is the mass of a molecule as determined by a mass spectrometer or other experimental methods.
• Mass Error (ppm) is the difference between the experimental and theoretical mass values, expressed in parts per million.
To calculate the mass error, subtract the theoretical mass from the experimental mass, divide by the theoretical mass, and then multiply by one million to convert to parts per million (ppm).
## What is Mass Error?
Mass error is a measure of the accuracy of a mass spectrometry measurement. It quantifies the difference between the measured mass of an ion and its calculated theoretical mass based on its elemental composition. Mass error is typically expressed in parts per million (ppm) to provide a unit that is independent of the ion’s mass, allowing for comparison across different mass ranges.
## How to Calculate Mass Error?
The following steps outline how to calculate the mass error:
1. First, determine the theoretical mass (Da) of the molecule.
2. Next, determine the experimental mass (Da) measured by the mass spectrometer.
3. Use the formula to calculate the mass error (ppm).
4. Finally, compare the calculated mass error with the mass spectrometer’s specifications to assess the accuracy of the measurement.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem:
Use the following variables as an example problem to test your knowledge.
Theoretical Mass (Da) = 196.083
Experimental Mass (Da) = 196.078 | 417 | 1,979 | {"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} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.869692 |
https://wiki.tcl-lang.org/page/Modeling+Planetary+Distances+using+Titius-Bode+Law+and+and+TCL+demo+example+calculator%2C+numerical+analysis?R=0 | 1,643,199,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00304.warc.gz | 656,788,263 | 15,491 | ## Modeling Planetary Distances using Titius-Bode Law and and TCL demo example calculator, numerical analysis
### Preface
gold 20aug2020 Here are some calculations modeling planetary distances using the Titius-Bode Law. The Titius-Bode Law is primarily of historical interest. But the Titius-Bode is a good exercise in matching the available data.
### Introduction
Here are some calculations on Titius-Bode Law in TCL. The Titius-Bode Law makes simple assumptions and formulas to model planetary distances. As another accuracy issue, the measured Planetary Distances are identified as non-linear function(s) in the log plots. Reported different velocity estimates in the program and the numerous references involve different assumptions and formulas. In the code, after computing the Titius-Bode output as historical numbers, subsequent formulas are scaled, changed, and different assumptions are changed from the original Titius-Bode formulas.
There are some simple formulas modeling planetary distances using the Titius-Bode Law and other power laws. The Titius-Bode Law is primarily of historical interest. The Titius-Bode law estimated rough distances in Astronomical Units (AU) of Planets from the Sun by scaling powers of 2 in a power law, Johann Elert Bode, pub 1772. There are usually different number assignments for each planet in each formula scheme. The simplest formula number scheme is used here, starting with < Earth = 1... > from Titius-Bode. Although much more is known about planetary distances in modern times, an examination of the original Titius-Bode results gives some fodder for the theory of numerical analysis. The On-Line Encyclopedia of Integer Sequences OEIS refers to the sequences A209257, A003461, and A131500. But the Titius-Bode and other power laws are a good exercise in matching the available data.
### Revolution Periods and Distance of Planets from the Sun
The TitiusâBode law estimated rough distances in Astronomical Units (AU) of Planets from the Sun by scaling powers of 2, Johann Elert Bode, pub 1772. Although much more known in modern times, an examination of the original TitiusâBode results in AU gives some fodder for the theory of the TCL calculator. For example, TitiusâBode law rated Earth as 1 AU from the Sun and predicted Mars as 1.6 AU. The Bode estimate for Mars was short of the true value 1.676 AU by 4.77 percent error. Using hand calculations, a similar value can be ginned up from the golden ratio, expr { \$golden_ratio_constant / 1. } returns 1.618 AU rounded. The golden ratio method is still short of the Mars true distance by 3.6 percent error. Numerous excellent papers have used both the Fibonacci and the golden ratio constants methods to estimate planetary distances from the Sun. The Lombardi paper found the best fit to the Fibonacci constants used the scaling constant < K1 = 55.88 > to get E6 km prediction. Both the Fibonacci and Bode sequences have unassigned slots or skips, due to the Asteroid belt(s) and possible undiscovered bodies. The Asteroid Ceres is representative of Asteroid belt here, but was not known to Bode. Also, both Neptune and Pluto not known to Bode in 1772.
Here is an example of how the Fibonacci method might be used with < K1 = 55.88 > . Astronomer Mary Blagg in 1913 conjectured that if the transplutonian Planet X existed, it might be at ~68 AU from the Sun. The nearest Fibonacci constants are 233 and 377, and unassigned at that. Perhaps, expr { 0.349 * 233. } ;# 81.317 AU for Planet X? The interest in Planet X and the region of 80 AU is not original or new by any means. Schuette suggested a family of comets in the region of 80 AU aphelia, circa 1949. The perihelion of the object Sedna is around 76 AU, reported in 2004. The perihelion of the object 2012 VP is around 80 AU, reported in 2012.
With additional scale factors loaded in the deck, the TCL calculator could use the Fibonacci sequence to output the planetary distances in kilometers and AU. The scale factor for kilometers is 55.88 (*E6). Sample calculation for Planet X was expr { 233. * 55.88 } ;# returns rounded 13000 X 1E6 km The scale factor for AU is expr { 19.2 / 55. } ;# ratio of Uranus returns 0.349 AU/Fi.
### Horse Race for Least Squares Fits and Online Curve Fitter
In the horse race of fitting formulas, a least squares method was fit to the curve with some known initial points either with the first 6 planets or 9 planets, if Pluto or Ceres included. Other fits to the curve can be tried from an online curve fitter, using the least squares method. A Gaussian bell curve was not a good fit to the eyeball. The linear fit or stick fit was tried. The linear slope can be defined from simple points.
Candidate models of fit include 1) Weibull growth curve, 2) Ratkowsky Model, 3) power law and 4) exponential law, which appear winners of field with more than 208 entries for fitting functions. The fitting algorithm for points was available to double precision, E-14. But obviously, such precision was not necessary with the 3 significant figures in reported JPL measured data. Some calculations appear defaulted to single precision, E-7. The tabled chi results were from 208 equations of 2 parameters (2P) or 2 terms (t2), from the non-linear curve fit algorithm (NLCF). Undoubtedly, this computer task would be hundreds of man-hours in paper calculations, but done automatically on the computer in three seconds or less. Trig functions are trivial results in this search, but leaving available in table, in case results jog memory. Also, there was a really nice piecewise cubic spline fit for 9 planets, but not sure how to wrap in code. One can see how the power laws or doubling rules like the Titius-Bode formula are breaking down beyond Jupiter and Uranus. See figures 6 and 7.
For inclusion in the horse race of fitting formulas, the classic sigmoidal functions were estimated like the Weibull growth curve, Gompetrz growth curve, and Ratkowsky Model growth curve. For the raw Planet Distances, the online fitter reported that the std of measured X values was 3.022 and avg X value was 5.555. The std of measured Y values was 13.655 and avg Y value was 11.902. The Weibull growth curve < a - b*exp(-c*x**d)> had a standard error of 4.25E-01 and a correlation coefficient of 9.99E-01. The Ratkowsky Model had a standard error of 5.29E-01 and a correlation coefficient of 9.99E-01. The Gompertz growth curve standard error of 1.96E+00 and correlation coefficient of 9.93E-01. The Logistic Power function had an standard error of 8.07E-01 and a correlation coefficient of 9.98E-01. Of most interest for comparison with the historic Titius-Bode power law, the Modified Power function <y=a*C**x> had standard error of 2.01E+00 and a correlation coefficient of 9.91E-01.
The autofitter found a = 5.14E-01 and b = 1.55E+00 for the Modified Power function <y=a*C**x>, which was close to the Armellini formula for planetary distances. In 1921, G. Armellini suggested a planetary distance formula, equivalent to <y = 0.2792 x 1.53**n> . Other authors have suggested 1.73 or, 1.842, for the C term in a planetary distance power function <y=a*b**x>. The Modified Power function error is at 5 percent of the Y axis, which is at the non-significant borderline level generally cited. The Weibull growth curve should fit the Y axis with a 1 percent error. In these growth functions, the Levenberg-Marquardt (L-M) algorithm was used with double precision C's.
### Considerations for Code and Titius-Bode formulas into TCL code
For fitting power law formulas to the measured data, Mercury, Neptune, and Pluto are outlier points. Ceres and Uranus from the measured data appear on the middle portion in the log chart of the measured data. The linear appearance of the middle portion feature is suitable for using power law formulas. The TCL calculator uses planet assignment numbers from 1 to 10, including the Pluto dwarf planet and the Ceres asteriod.
### Pushbutton Operation
For the push buttons, the recommended procedure is push testcase and fill frame, change first three entries etc, push solve, and then push report. Report allows copy and paste from console. For testcases in a computer session, the eTCL calculator increments a new testcase number internally, eg. TC(1), TC(2) , TC(3) , TC(N). The testcase number is internal to the calculator and will not be printed until the report button is pushed for the current result numbers. The current result numbers will be cleared either on the next clear button or on the next solve button. All results will be lost on program exit, unless the report text is saved from the report screen on console.
### Conclusions
In the horse race of fitting formulas, a least squares method was fit to the curve with some known initial points either with the first 6 planets or 9 planets, if Pluto or Ceres included. Other fits to the curve can be tried from an online curve fitter, using the least squares method. Candidate models of fit include 1) Weibull growth curve, 2) Ratkowsky Model, 3) power law and 4) exponential law, which appear winners of field with more than 208 entries for fitting functions. The tabled chi results were from 208 equations of 2 parameters (2P) or 2 terms (t2), from the non-linear curve fit algorithm (NLCF). Undoubtedly, this computer task would be hundreds of man-hours in paper calculations, but done automatically on the computer in three seconds or less.In the code, after computing the Titius-Bode output as historical numbers, subsequent formulas are scaled, changed, and different assumptions are changed from the original Titius-Bode formulas.
The set of measured planet distances is a non-linear function, which possibly should be modeled in three sections. From the log plots, there appears to be a sine wave or sigmoidal growth function component in the measured planet distances, meaning non-linear S-curve in slang. Again from the log plots, one can see how the power laws or doubling rules like the Titius-Bode formula are breaking down beyond Jupiter and Uranus. The Ragnarsson paper in 1994 showed fair results in modeling Jupiter and the outer planets as a separate section.
Table : Est. Distance of Planets from Fibonacci and Bode Rules printed in tcl wiki format
sources >> JPL JPL Fibonacci constants Lombardi Paper Lombardi Paper Wikipedia Wikipedia comments
quantity real measured AU, semi_major axis real measured 10E6 km, semi_major axis Fibonacci constants Fibonacci * 55.88, 10E6 km Prediction Titius-Bode number Titius-Bode AU Prediction Real Measured AU , semi_major axis comments, if any
Sun 0 0 Sun not assigned value in Fibonacci and Bode Rules
Mercury 0.39 57.9 1 55.8 4 0.4 0.39
Venus 0.723 108.2 2 111.6 7 0.7 0.72
Earth 1 149.6 3 167.4 10 1 1 Earth assigned unit 1 AU in Bode Rules
Mars 1.524 227.9 5 279 16 1.6 1.52
Ceres 2.767 265.3 8 446.4 28 2.8 2.77 Asteroid belt
Jupiter 5.203 778.3 13 725.4 52 5.2 5.2
Jupiter clearance belt 21 1171.8 Jupiter clearance belt
Saturn 9.539 1427 34 1897.2 100 10 9.55
Uranus 19.18 2871 55 3069 196 19.6 19.22
Neptune 30.07 4497.1 89? 4966.2 388? 38.8? 30.11 Bodes Law breaks down for Neptune and Pluto. Confusing references.
Pluto (avg) 39.482 5906.38 144? 8035.2? 772? 77.20? 39.54 Pluto is not considered a planet, currently. Bodes Law breaks down for Neptune and Pluto. Confusing references.
Note several sources Wikipedia Jpl Lombardi Paper
Note Neptune, Pluto, & Ceres were not known to Bode
Note Ceres representative of Asteroid belt here
Note Bodes Law breaks down for Neptune and Pluto, confusing references Opinion: Bode Number assignment for N&P uncertain here.
### Testcases Section
In planning any software, it is advisable to gather a number of testcases to check the results of the program.
#### Testcase 1, Planet distances in AU
Testcase 1 printed in tcl wiki format
1:testcase_number
1. :method number, 1 = Bode
1772. :reference year CE :
value AU quantity distance E6 km comment, if any
0.400 : Mercury : 22.32 Bodes Law breaks down for Mercury
0.700 : Venus : 39.059
1.00 : Earth : 55.79 Earth set to one in Bodes Law
1.60 : Mars : 89.280
5.20 : Jupiter : : 290.159
10.0 : Saturn : 558.0
19.6 : Uranus : 1093.68
38.8 : Neptune : 2165.039 Bodes Law breaks down for Neptune and Pluto.
77.2 : Pluto : 4307.76 Pluto is not considered a planet, currently. Bodes Law breaks down for Neptune and Pluto.
154. : extra slot, unk planet? : 8593.19
2.80 : Ceres : 156.23 represents Asteroid belt here
``` Console wrapper for solution proc
***************************
***************************
scaled_results 0.000001 1. 0.400 0.700 1.00 1.60 5.20 10.0 19.6 38.8 ```
### *figure 3c. Modeling_Planetary_Distances_plot_titius_bode_vs_gaussin
#### figure 6. Power law curve fit in red line, e-function curve fit in green line
```Power law curve fit in red line
Power law parameter
a=0.182490688
b=2.104691172
Power law
y=aâ
xb=0.182490688â
x2.104691172
Standard deviation
Ï=7.340978516
e-function curve fit in green line
Parameter of the fitted e-function
a=0.535853432
b=0.218240397
The e-Function
f(x)=bâ
eaâ
x=0.218240397â
e0.535853432â
x
Standard deviation
Ï=17.056378115```
### References:
• Wikipedia search engine < Titius Bode >
• Wikipedia search engine < Golden Mean >
• Wikipedia search engine < Programming Examples >
• Google search engine < vaporware >
### References: Planets, TitiusâBode law
• wikipedia.org/wiki/Titius Bode law
• Article: All Solar system periods fit
• the Fibonacci series and the Golden Ratio.
• Posted: February 20, 2013 by tallbloke
• The Golden Mean In The Solar System,
• Oreste W. Lombard I And Margaret A. Lombardi
• On a Suggested Substitute for Bodeâs Law. By M. A. Blagg.
• Communicated by Professor H. B. Turner.
• Planetary distances: a new simplified model
• Fibonacci Series In The Solar System
• A postulate leading to the Titius-Bode law
• R. Louise
• www.jpl.nasa.gov/edu/pdfs/scaless_reference.pdf
• Schuette, C. H., "Two new families of comets", Pop. Astron. 57, 176-82
• (1949); "Drei weitere Mitglieder der Transplutokometenfamilie", Acta
• Astronomica 15, 11-13 (1965).
• Bodeâs Law And Spiral Structure In Nebulae
• By William Sutherland
• A Sedna-like body with a perihelion of 80
• astronomical units
• Chadwick A. Trujillo, Scott S. Sheppard
• Searching For Sednaâs Sisters: Exploring The Inner Oort Cloud
• Megan Schwamb with Advisor: Mike Brown
• A new object at the edge of our Solar System discovered
• by Carnegie Institution for Science
• Modeling Celestial Mechanics Using the Fibonacci Number,
• Robert G. Sacco, 2019
• A Supposed New Law For Planetary Distances, letter,
• Ennio Badolati, 1982, several approaches cited
• Munini and Armellini (1978) , d = 0.283 x 1.52**n, sequence < n = 1,2,... 12 >
• 1880 Gaussin, d~ = 0.2099 x 1.7226**n, (n = 1,2... 9)
• tester formula, d = \$K1 x 1.842**n, sequence < n = 1,2,... 12 >
• Planetary distances: a new simplified model
Caption outline on plots. distance curve Leaves straight line middle portion appears Linear on log plot, Which is feature suitable For using power law formulas Mercury is outlier On measured data Distance curve Leaves straight le Neptune and pluto are Outliers on measured data Ceres and uranus appear Linear on middle log portion Titius-Bode output for predicted distances tcl club Planet number including Pluto and Ceres Distance curve Leaves straight line Distance curve Leaves straight line Middle portion appears Linear on log plot, Which is feature suitable For using power law formulas Mercury is outlier On titius-bode law Distance curve Leaves straight le Neptune and pluto are Outliers on titius-bode law Ceres and uranus appear Linear on middle log portion {distances for planets and objects on log plot } Tcl club } Planet number including Pluto and Ceres
### * Pretty Print Version
``` ;# pretty print from autoindent and ased editor occurrence
;# Planetary Distances Using Titius-Bodes Law V2 calculator
;# written on Windows 10
;# working under TCL version 8.6
;# gold on TCL Club, 10jul2021
package require Tk
package require math::numtheory
namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
set tcl_precision 17
frame .frame -relief flat -bg aquamarine4
pack .frame -side top -fill y -anchor center
set names {{} { method number, 1 = Bode :} }
lappend names {reference year CE: }
lappend names {answers: Mercury : }
lappend names { Venus : }
lappend names { Earth : }
lappend names { Mars : }
lappend names { Jupiter : }
lappend names { Saturn : }
lappend names { Uranus : }
foreach i {1 2 3 4 5 6 7 8 9} {
label .frame.label\$i -text [lindex \$names \$i] -anchor e
entry .frame.entry\$i -width 35 -textvariable side\$i
set msg "Calculator for Planetary Distances Using Titius-Bodes Law V2
from TCL WIKI,
written on TCL 8.6 "
tk_messageBox -title "About" -message \$msg }
proc self_help {} {
set msg "Calculator for Planetary Distances Using Titius-Bodes Law V2
from TCL ,
;# self help listing
;# 1 given follow.
1) method N1
2) Reference year
;# This calculator uses the Titius-Bodes Law
;# to predict Planetary Distances in AU.
;# For comparison, TCL code may include redundant paths & formulas.
;# The TCL calculator normally uses modern
;# units for convenience to modern users and textbooks.
;# Any convenient and consistent in/output units might be used
;# like inches, feet, nindas, cubits, or dollars to donuts.
;# Recommended procedure is push testcase and fill frame,
;# change first three entries etc, push solve,
;# and then push report. Report allows copy and paste
;# from console to conventional texteditor. For testcases
;# testcase number is internal to the calculator and
;# will not be printed until the report button is pushed
;# for the current result numbers.
;# Use one line errorx proc to estimate percent errors.
;# errorx proc is used in the report window (console).
;# Additional significant figures are used to check
;# the TCL program, not to infer the accuracy
;# of inputs and product reports.
Conventional text editor formulas or grabbed from internet
screens can be pasted into green report console.
Try copy and paste following into green screen console
set answer \[* 1. 2. 3. 4. 5. \]
returns 120
;# gold on TCL Club, 10jul2021 "
tk_messageBox -title "self_help" -message \$msg }
proc precisionx {precision float} {
;# tcl:wiki:Floating-point formatting, <AM>
;# select numbers only, not used on every number.
set x [ expr {round( 10 ** \$precision * \$float) / (10.0 ** \$precision)} ]
;# rounded or clipped to nearest 5ird significant figure
set x [ format "%#.3g" \$x ]
return \$x
}
;# Use one line errorx proc to estimate percent errors
;# errorx proc is used in the report window (console)
proc errorx {aa bb} {expr { \$aa > \$bb ? ((\$aa*1.)/\$bb -1.)*100. : ((\$bb*1.)/\$aa -1.)*100.}}
proc titius_bode_law {aa } {expr { (4. + 3 * 2**\$aa)*.1 } }
proc calculate { } {
global side1 side2 side3 side4 side5
global side6 side7 side8 side9 k1
global side10 side11 side12 side13
global testcase_number
incr testcase_number
;# scaling constant K1
set k1 55.8
set side3 [precisionx 2 [ titius_bode_law -1]]
set side4 [precisionx 2 [ titius_bode_law 0]]
set side5 [precisionx 2 [ titius_bode_law 1]]
set side6 [precisionx 2 [ titius_bode_law 2]]
set side7 [precisionx 2 [ titius_bode_law 4]]
set side8 [precisionx 2 [ titius_bode_law 5]]
set side9 [precisionx 2 [ titius_bode_law 6]]
set side10 [precisionx 2 [ titius_bode_law 7]] ;# Neptune
set side11 [precisionx 2 [ titius_bode_law 8]] ;# Pluto
set side12 [precisionx 2 [ titius_bode_law 9]] ;# extra Bode slot, unk planet?
set side13 [precisionx 2 [ titius_bode_law 3]] ;# Ceres
set scaled_results [list 0.000001 \$side1 \$side3 \$side4 \$side5 \$side6 \$side7 \$side8 \$side9 \$side10]
puts " scaled_results \$scaled_results "
}
proc fillup {aa bb cc dd ee ff gg hh ii} {
.frame.entry1 insert 0 "\$aa"
.frame.entry2 insert 0 "\$bb"
.frame.entry3 insert 0 "\$cc"
.frame.entry4 insert 0 "\$dd"
.frame.entry5 insert 0 "\$ee"
.frame.entry6 insert 0 "\$ff"
.frame.entry7 insert 0 "\$gg"
.frame.entry8 insert 0 "\$hh"
.frame.entry9 insert 0 "\$ii"
}
proc clearx {} {
foreach i {1 2 3 4 5 6 7 8 9 } {
.frame.entry\$i delete 0 end } }
proc reportx {} {
global side1 side2 side3 side4 side5
global side6 side7 side8 side9 k1
global side10 side11 side12 side13
global testcase_number
console show;
puts "%|table \$testcase_number| || printed in tcl wiki format|% "
puts "&| \$testcase_number:|testcase_number || |&"
puts "&| \$side1 :|method number, 1 = Bode | | |&"
puts "&| \$side2 :|reference year CE : | | |& "
puts "&| value AU |quantity |distance E6 km | comment, if any|& "
puts "&| \$side3 :| Mercury : |[* \$side3 \$k1 ] | |& "
puts "&| \$side4 :| Venus : |[* \$side4 \$k1 ] | |& "
puts "&| \$side5 :| Earth : |[* \$side5 \$k1 ] | |&"
puts "&| \$side6 :| Mars : |[* \$side6 \$k1 ] | |&"
puts "&| \$side7 :| Jupiter : : |[* \$side7 \$k1 ] | |&"
puts "&| \$side8 :| Saturn : | [* \$side8 \$k1 ]| |&"
puts "&| \$side9 :| Uranus : | [* \$side9 \$k1 ] | |&"
puts "&| \$side10 :| Neptune : | [* \$side10 \$k1 ] | |&"
puts "&| \$side11 :| Pluto : | [* \$side11 \$k1 ] | |&"
puts "&| \$side12 :| extra slot, unk planet? : | [* \$side12 \$k1 ] | |&"
puts "&| \$side13 :| Ceres : | [* \$side13 \$k1 ] | |&"
}
frame .buttons -bg aquamarine4
::ttk::button .calculator -text "Solve" -command { calculate }
::ttk::button .test2 -text "Testcase1" -command {clearx;fillup 1. 1772. 0.40 0.70 1.0 1.6 5.2 10. 19.6}
::ttk::button .test3 -text "Testcase2" -command {clearx;fillup 1. 1772. 0.40 0.70 1.0 1.6 5.2 10. 19.6}
::ttk::button .test4 -text "Testcase3" -command {clearx;fillup 1. 1772. 0.40 0.70 1.0 1.6 5.2 10. 19.6}
::ttk::button .clearallx -text clear -command {clearx }
::ttk::button .self_help -text self_help -command {self_help }
::ttk::button .cons -text report -command { reportx }
::ttk::button .exit -text exit -command {exit}
pack .clearallx .cons .self_help .about .exit .test4 .test3 .test2 -side bottom -in .buttons
grid .frame .buttons -sticky ns -pady {0 10}
. configure -background aquamarine4 -highlightcolor brown -relief raised -border 30
wm title . " Planetary Distances Using Titius-Bodes Law V2"
# end of working deck
# add cosmetics below to bottom of file
console eval {.console config -bg palegreen}
console eval {.console config -font {fixed 20 bold}}
console eval {wm geometry . 40x20}
console eval {wm title . " Report for Planetary Distances Using Titius-Bodes Law V2 "}
console eval {. configure -background orange -highlightcolor brown -relief raised -border 30}
puts " Console wrapper for solution proc"
puts " ***************************"
puts " ***************************"```
### formula from Sven-Ingmar Ragnarsson, 1994
``` 2 tiered formula from Sven-Ingmar Ragnarsson, 1994
expr { 2.825/5 } ;# 0.565 Mercury badly off
expr { 2.825/4 } ;# 0.70625 Venus
expr { 2.825/3 } ;# 0.9416 Earth
expr { 2.825/2 } ;# 1.412 Mars
expr { 2.825/1 } ;# 2.825 Ceres
expr { 5.203*(((5./2.)**(2./3.))**0) } ;# 5.203 Jupiter
expr { 5.203*(((5./2.)**(2./3.))**1) } ;# 9.584 Saturn
expr { 5.203*(((5./2.)**(2./3.))**2) } ;# 17.653 Uranus
expr { 5.203*(((5./2.)**(2./3.))**3) } ;# 32.51875 Neptune? badly off
expr { 5.203*(((5./2.)**(2./3.))**4) } ;# 59.9 ? PLuto? badly off```
### Appendix: One Liner Procs for Modeling Planetary Distances using Titius-Bode Law and other power laws
``` ;# for starting the adventure to infinity and beyond.
;# Titius-Bode set < Earth = 1 ... > as number assignment scheme for planets
;# The Asteriod Ceres is included on most number assignment schemes
;# Ceres represents the Asteriod belt between Mars and Jupiter.
;# The dwarf planet Pluto is included on most number assignment schemes.
proc titius_bode_law_1772 {aa } {expr { (4. + 3 * 2**\$aa)*.1 } } ;# ;# \$nn=l for Earth,2,...
;# Other planet number schemes are for power laws are available here, < Mercury = 1 ...>
;# from Gaussin (1880) and Armellini (1921)
proc gaussin_formula_1880 {nn} {expr {0.2099 * (1.7226 ** \$nn)} } ;# \$nn=l for Mercury,2,...9
proc armellini_formula_1921 {nn} {expr {0.283 * (1.53 ** \$nn)} } ;# \$nn=l for Mercury,2,...11
;# very interesting, Nicolini transformed planet number scheme in Titius_Bode
;# to < Mercury = 1 ...>, refer to Nicolini
proc nicolini_formula_1957 {nn} {expr {0.4 + 0.075 * (2. ** \$nn)} } ;# \$nn=l for Mercury,2,...
proc basano_hughes_formula_1979 {nn} {expr {0.285 * (1.523 ** \$nn)} } ;# \$nn=l for Mercury, 2,...
;# Sven-Ingmar Ragnarsson, 1994, modeled Jupiter and the outer planets to Uranus, separately.
;# See the Ragnarsson paper, setting these planet assignment numbers is tricky.
;# \$nn=0 for Jupiter, 2 = Saturn, 3 = Uranus ....
proc ragnarsson_formula_1979 {nn} {expr {5.203*(((5./2.)**(2./3.))**\$nn) )} } ;# \$nn=0 for Jupiter, 2 = Saturn,...
proc list_integers { aa bb} { for {set i 1} {\$i<=\$bb} {incr i} {lappend boo [* 1. \$i ] [* \$i 1.]};return \$boo}
;# usage, list_integers 1 10
;# set answer_list_integers = [list 1.0 1.0 2.0 2.0 3.0 3.0 4.0 4.0 5.0 5.0 6.0 6.0 7.0 7.0 8.0 8.0 9.0 9.0 10.0 10.0 ]
proc list_titius_bode { aa bb} { for {set i 1} {\$i<=\$bb} {incr i} {lappend boo [* 1. \$i ] [expr { (4. + 3. * 2**\$i)*.1}]};if {\$i > \$bb} {return \$boo}}
;# Usage list_titius_bode 1 6
;# set answer [ list 1.0 1.0 2.0 1.6 3.0 2.8 4.0 5.2 5.0 10.0 6.0 19.6]
proc list_gaussin_formula { aa bb} { for {set ii 1} {\$ii<=\$bb} {incr ii} {lappend boo [* 1. \$ii ] [expr {0.2099 * (1.7226 ** \$ii)} ]};if {\$ii > \$bb} {return \$boo}}
;# Usage list_gaussin_formula 1 6
;# set answer_gaussin [ list 1.0 0.36 2.0 0.62 3.0 1.07 4.0 1.84 5.0 3.18 6.0 5.48 ]
proc list_armellini_formula { aa bb} { for {set ii 1} {\$ii<=\$bb} {incr ii} {lappend boo [* 1. \$ii ] [expr {0.283 * (1.53 ** \$ii)}]};if {\$ii > \$bb} {return \$boo}}
;# Usage list_armellini_formula 1 7
;# set answer_armellini [ 1.0 0.43 2.0 0.66 3.0 1.013 4.0 1.55 5.0 2.37 6.0 3.63 7.0 5.55 ]
```
### Points for curve fitting
``` 1 0.39
2 0.723
3 1.
4 1.524
6 5.203
7 9.539
8 19.18
9 30.07
10 39.482``` | 7,869 | 26,283 | {"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} | 2.875 | 3 | CC-MAIN-2022-05 | latest | en | 0.817772 |
https://blog.csdn.net/ufo2006/article/details/99696792 | 1,600,776,560,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00218.warc.gz | 299,997,730 | 21,099 | # 梅森旋转法伪随机数在delphi下实现
//Random integer, implemented as a deterministic linear congruential generator
// with 134775813 as a and 1 as c.
procedure TForm1.FormCreate(Sender: TObject);
begin
Randomize; //原随机数初始化
InitMT(Random(9999999)); //用线性同余法生成的随机数来初始化梅森旋转法
end;
procedure TForm1.Button1Click(Sender: TObject);
begin
end;
{ ******************************************************************
Mersenne Twister Random Number Generator for pascal
******************************************************************}
unit uranmt;
interface
procedure InitMT(Seed : LongInt);
{ Initializes MT generator with a seed }
procedure InitMTbyArray(InitKey : array of LongInt; KeyLength : Word);
{ Initialize MT generator with an array InitKey[0..(KeyLength - 1)] }
function IRanMT : cardinal;
{ Generates a Random number }
implementation
const
N = 624;
M = 397;
MATRIX_A = $9908b0df; { constant vector a } UPPER_MASK =$80000000; { most significant w-r bits }
LOWER_MASK = $7fffffff; { least significant r bits } mag01 : array[0..1] of Cardinal{LongInt} = (0, MATRIX_A); var mt : array[0..(N-1)] of Cardinal{LongInt}; { the array for the state vector } mti : Word; { mti == N+1 means mt[N] is not initialized } procedure InitMT(Seed : LongInt); var i : Word; begin mt[0] := Seed and$ffffffff;
for i := 1 to N-1 do
begin
mt[i] := (1812433253 * (mt[i-1] Xor (mt[i-1] shr 30)) + i);
{ See Knuth TAOCP Vol2. 3rd Ed. P.106 For multiplier.
In the previous versions, MSBs of the seed affect
only MSBs of the array mt[].
2002/01/09 modified by Makoto Matsumoto }
mt[i] := mt[i] and $ffffffff; { For >32 Bit machines } end; mti := N; end; procedure InitMTbyArray(InitKey : array of LongInt; KeyLength : Word); var i, j, k, k1 : Word; begin InitMT(19650218); i := 1; j := 0; if N > KeyLength then k1 := N else k1 := KeyLength; for k := k1 downto 1 do begin mt[i] := (mt[i] Xor ((mt[i-1] Xor (mt[i-1] shr 30)) * 1664525)) + InitKey[j] + j; { non linear } mt[i] := mt[i] and$ffffffff; { for WORDSIZE > 32 machines }
i := i + 1;
j := j + 1;
if i >= N then
begin
mt[0] := mt[N-1];
i := 1;
end;
if j >= KeyLength then j := 0;
end;
for k := N-1 downto 1 do
begin
mt[i] := (mt[i] Xor ((mt[i-1] Xor (mt[i-1] shr 30)) * 1566083941)) - i; { non linear }
mt[i] := mt[i] and $ffffffff; { for WORDSIZE > 32 machines } i := i + 1; if i >= N then begin mt[0] := mt[N-1]; i := 1; end; end; mt[0] :=$80000000; { MSB is 1; assuring non-zero initial array }
end;
function IRanMT : cardinal;
var
y : cardinal;
k : Word;
begin
if mti >= N then { generate N words at one Time }
begin
{ If IRanMT() has not been called, a default initial seed is used }
if mti = N + 1 then InitMT(5489);
for k := 0 to (N-M)-1 do
begin
mt[k] := mt[k+M] xor (y shr 1) xor mag01[y and $1]; end; for k := (N-M) to (N-2) do begin y := (mt[k] and UPPER_MASK) or (mt[k+1] and LOWER_MASK); mt[k] := mt[k - (N - M)] xor (y shr 1) xor mag01[y and$1];
end;
mt[N-1] := mt[M-1] xor (y shr 1) xor mag01[y and $1]; mti := 0; end; y := mt[mti]; mti := mti + 1; { Tempering } y := y xor (y shr 11); y := y xor ((y shl 7) and$9d2c5680);
y := y xor ((y shl 15) and $efc60000); y := y xor (y shr 18); IRanMT := y end; const init : array[0..3] of LongInt = ($123, $234,$345, \$456);
begin
InitMTbyArray(init, 4);
end.
{ ******************************************************************
Mersenne Twister Random Number Generator end
******************************************************************}
10-17 1290
06-04 90
03-13 8684
03-26 1万+
06-04 | 1,190 | 3,501 | {"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} | 2.921875 | 3 | CC-MAIN-2020-40 | latest | en | 0.432774 |
https://new-history.ru/solving-direct-variation-problems-220.html | 1,620,435,350,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00042.warc.gz | 452,934,773 | 10,177 | # Solving Direct Variation Problems
As the numbers for a problem can be scored is fixed, it is a constant = K = $$\frac$$ = 10For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.When two variables change in proportion it is called as direct variation.
Tags: Read An Essay Concerning Human UnderstandingCapstone Projects IdeasArt History Dissertation QuestionsData Analysis For DissertationPaper EssayResearch Paper On Recycling
Solution: As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.
So constant K = $$\frac$$ = $$\frac$$ = $$\frac$$So the equation that describes the direct variation of P and Q is P = $$\frac$$Q.
Graphically, we have a line that passes through the origin with the slope of k. If the distance is 10 cm when the mass is kg, what is the distance when the mass is 5 kg?
We're told that the total cost of filling up your car with gas varies directly with the number of gallons of gasoline you are purchasing. So if x is equal to 1-- this statement up here, a gallon of gas-- that tells us if we get 1 gallon, if x is equal to 1, then y is $2.25, right? They tell us 1 gallon costs$2.25, so you could write it right here, \$2.25 is equal to k times x, times 1. So the equation, how y varies with x, is y is equal to 2.25x, where x is the number of gallons we purchase. Now if we want to solve for x, we can divide both sides by 2.25, so let's do that. So first of all, I just like to think of it as a fraction.
Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.
If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then A is said to vary directly as B ( or, A is said to vary as B ).Direct variation word problems may be solved using the following steps: 1) Recognizing the word problem consists of direct variation as exemplified by presence of verbiage listed above; 2) Writing an equation containing known values for both variables resulting in an expression containing only the unknown representing the constant of proportionality; 3) Solving the equation for the constant of proportionality; 4) Using the calculated constant of proportionality to determine the value of one of the variables given the other.The amount of money raised at a charity fundraiser is directly proportional to the number of attendees.This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ).Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers.The speed of a car and the distance traveled in a certain amount of time.The following statements are equivalent * y varies directly as x * y is directly proportional to x * y = kx for some constant k What is the direct variation formula? If a gallon of gas costs .25, how many gallons could you purchase for ? Now, they give us more information, and this will help us figure out what k is.A direct variation is a linear equation that can be written in the form y = kx , where k is a nonzero constant. Determine the direct variation equation and then determine y when x = 3.5 2. The number k is called the constant of proportionality or constant of variation. Hooke's Law states that the displacement, d, that a spring is stretched by a hanging object varies directly as the mass of the object.
## Comments Solving Direct Variation Problems
• ###### Solve direct variation problems College Algebra
Solve direct variation problems. In the example above, Nicole's earnings can be found by multiplying her sales by her commission. The formula e = 0.16s tells us.…
• ###### Direct variation word problem filling gas video Khan.
Worked example Model a context about filling gas with a direct variation equation. Inverse variation word problem string vibration · Proportionality constant for. Now if we want to solve for x, we can divide both sides by 2.25, so let's do that.…
• ###### Direct Variation - How to Solve Variation Problems Part 1 fbt.
Jul 19, 2016. This video by Fort Bend Tutoring shows the process of solving direct variation problems. Six 6 examples are shown in this FBT video.…
• ###### Direct Variation Solving Direct Variation Word Problem
When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other.…
• ###### Direct Variation - - Algebra Help - YouTube
Apr 24, 2008. For a complete lesson on direct variation, go to. are examples of direct variation, as well as solve direct variation word problems.…
• ###### Solving Direct Variation Problems - Practice Problems
Solving Direct Variation Problems – Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to.…
• ###### Write and solve direct variation equations Algebra 1. - IXL
Improve your math knowledge with free questions in "Write and solve direct variation equations" and thousands of other math skills.… | 1,181 | 5,301 | {"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} | 4.65625 | 5 | CC-MAIN-2021-21 | latest | en | 0.934158 |
https://statistics.laerd.com/spss-tutorials/spearmans-rank-order-correlation-using-spss-statistics.php | 1,601,511,582,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00057.warc.gz | 561,642,893 | 10,011 | # Spearman's Rank-Order Correlation using SPSS Statistics
## Introduction
The Spearman rank-order correlation coefficient (Spearman’s correlation, for short) is a nonparametric measure of the strength and direction of association that exists between two variables measured on at least an ordinal scale. It is denoted by the symbol rs (or the Greek letter ρ, pronounced rho). The test is used for either ordinal variables or for continuous data that has failed the assumptions necessary for conducting the Pearson's product-moment correlation. For example, you could use a Spearman’s correlation to understand whether there is an association between exam performance and time spent revising; whether there is an association between depression and length of unemployment; and so forth. If you would like some more background information about this test, which does not include instructions for SPSS Statistics, see our more general statistical guide: Spearman's rank-order correlation. Possible alternative tests to Spearman's correlation are Kendall's tau-b or Goodman and Kruskal's gamma.
This "quick start" guide shows you how to carry out a Spearman’s correlation using SPSS Statistics. We show you the main procedure to carry out a Spearman’s correlation in the Procedure section. First, we introduce you to the assumptions that you must consider when carrying out a Spearman’s correlation.
## Assumptions
When you choose to analyse your data using Spearman’s correlation, part of the process involves checking to make sure that the data you want to analyse can actually be analysed using a Spearman’s correlation. You need to do this because it is only appropriate to use a Spearman’s correlation if your data "passes" three assumptions that are required for Spearman’s correlation to give you a valid result. In practice, checking for these three assumptions just adds a little bit more time to your analysis, requiring you to click of few more buttons in SPSS Statistics when performing your analysis, as well as think a little bit more about your data, but it is not a difficult task. These three assumptions are:
• Assumption #1: Your two variables should be measured on an ordinal, interval or ratio scale. Examples of ordinal variables include Likert scales (e.g., a 7-point scale from "strongly agree" through to "strongly disagree"), amongst other ways of ranking categories (e.g., a 3-pont scale explaining how much a customer liked a product, ranging from "Not very much", to "It is OK", to "Yes, a lot"). Examples of interval/ratio variables include revision time (measured in hours), intelligence (measured using IQ score), exam performance (measured from 0 to 100), weight (measured in kg), and so forth. You can learn more about ordinal, interval and ratio variables in our article: Types of Variable.
• Assumption #2: Your two variables represent paired observations. For example, imagine that you were interested in the relationship between daily cigarette consumption and amount of exercise performed each week. A single paired observation reflects the score on each variable for a single participant (e.g., the daily cigarette consumption of "Participant 1" and the amount of exercise performed each week by "Participant 1"). With 30 participants in the study, this means that there would be 30 paired observations.
• Assumption #3: There is a monotonic relationship between the two variables. A monotonic relationship exists when either the variables increase in value together, or as one variable value increases, the other variable value decreases. Whilst there are a number of ways to check whether a monotonic relationship exists between your two variables, we suggest creating a scatterplot using SPSS Statistics, where you can plot one variable against the other, and then visually inspect the scatterplot to check for monotonicity. Your scatterplot may look something like one of the following:
The relationship displayed in your scatterplot should be monotonic. In our enhanced guides, we show you how to: (a) create a scatterplot to check for a monotonic relationship when carrying out Spearman’s correlation using SPSS Statistics; (b) interpret different scatterplot results; and (c) consider possible solutions if your data fails this assumption. Just remember that if you do not test these assumptions correctly, the results you get when running a Spearman's correlation might not be valid. This is why we dedicate a number of sections of our enhanced Spearman's correlation guide to help you get this right. You can find out about our enhanced content as a whole on our Features: Overview page, or more specifically, learn how we help with testing assumptions on our Features: Assumptions page.
Note: Spearman's correlation determines the degree to which a relationship is monotonic. Put another way, it determines whether there is a monotonic component of association between two continuous or ordinal variables. As such, monotonicity is not actually an assumption of Spearman's correlation. However, you would not normally want to pursue a Spearman's correlation to determine the strength and direction of a monotonic relationship when you already know the relationship between your two variables is not monotonic. Instead, the relationship between your two variables might be better described by another statistical measure of association. For this reason, it is not uncommon to view the relationship between your two variables in a scatterplot to see if running a Spearman's correlation is the best choice as a measure of association or whether another measure would be better.
In terms of assumption #3 above, you can check this using SPSS Statistics. If your two variables do not appear to have a monotonic relationship, you might consider using a different statistical test, which we show you how to do in our Statistical Test Selector (N.B., this is part of our enhanced content).
It is also worth noting that a Spearman’s correlation can be used when your two variables are not normally distributed. It is also not very sensitive to outliers, which are observations within your data that do not follow the usual pattern. Since Spearman’s correlation is not very sensitive to outliers, this means that you can still obtain a valid result from using this test when you have outliers in your data.
In the section, Test Procedure in SPSS Statistics, we illustrate the SPSS Statistics procedure to perform a Spearman’s correlation assuming that no assumptions have been violated. First, we set out the example we use to explain the Spearman’s correlation procedure in SPSS Statistics.
## Example
A teacher is interested in whether those who do better at English also do better in maths. To test whether this is the case, the teacher records the scores of her 10 students in their end-of-year examinations for both English and maths. Therefore, one variable records the English scores and the second variable records the maths scores for the 10 pupils.
## Setup in SPSS Statistics
In SPSS Statistics, we created two variables so that we could enter our data: English_Mark (i.e., English scores) and Maths_Mark (i.e., maths scores). In our enhanced Spearman's correlation guide, we show you how to correctly enter data in SPSS Statistics to run a Spearman's correlation. You can learn about our enhanced data setup content on our Features: Data Setup page. Alternately, see our generic, "quick start" guide: Entering Data in SPSS Statistics.
## Test Procedure in SPSS Statistics
The four steps below show you how to analyse your data using Spearman’s correlation in SPSS Statistics when neither of the three assumptions in the previous section, Assumptions, have been violated. At the end of these four steps, we show you how to interpret the results from this test. If you are looking for help to assess whether the relationship between your two variables is monotonic, we show you how to do this in our enhanced guide. You can learn more about our enhanced guides on our Features: Overview page.
1. Click Analyze > Correlate > Bivariate... on the main menu as shown below:
Published with written permission from SPSS Statistics, IBM Corporation.
You will be presented with the following Bivariate Correlations dialogue box:
Published with written permission from SPSS Statistics, IBM Corporation.
2. Transfer the variables English_Mark and Maths_Mark into the Variables: box by dragging-and-dropping the variables or by clicking each variable and then clicking on the button. You will end up with a screen similar to the one below:
Published with written permission from SPSS Statistics, IBM Corporation.
3. Make sure that you uncheck the Pearson checkbox (it is selected by default in SPSS Statistics) and select the Spearman checkbox in the –Correlation Coefficients– area. You will end up with a screen similar to below:
Published with written permission from SPSS Statistics, IBM Corporation.
4. Click on the button. This will generate the results.
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## Output
SPSS Statistics generates a single table following the Spearman’s correlation procedure that you ran in the previous section. If your data passed assumption #3 (i.e., there is a monotonic relationship between your two variables), you will only need to interpret this one table. However, since you should have tested your data for monotonicity, you will also need to interpret the SPSS Statistics output that was produced when you tested for it (i.e., your scatterplot results). If you do not know how to do this, we show you in our enhanced Spearman’s correlation guide. If you have tested your data for these assumptions, we provide a complete explanation of the output you will have to interpret in our enhanced Spearman’s guide. Remember that if your data failed this assumption, the output that you get from the Spearman’s correlation procedure (i.e., the table we discuss below), might be misleading.
However, in this "quick start" guide, we focus on the results from the Spearman’s correlation procedure only. Therefore, after running the Spearman’s correlation procedure, you will be presented with the Correlations table, as shown below:
Published with written permission from SPSS Statistics, IBM Corporation.
The results are presented in a matrix such that, as can be seen above, the correlations are replicated. Nevertheless, the table presents Spearman's correlation, its significance value and the sample size that the calculation was based on. In this example, we can see that Spearman's correlation coefficient, rs, is 0.669, and that this is statistically significant (p = .035).
## Reporting the Output
In our example, you might present the results as follows:
• General
A Spearman's rank-order correlation was run to determine the relationship between 10 students' English and maths exam marks. There was a strong, positive correlation between English and maths marks, which was statistically significant (rs(8) = .669, p = .035).
In our enhanced Spearman’s correlation guide, we also show you how to write up the results from your assumptions test and Spearman’s correlation output if you need to report this in a dissertation, thesis, assignment or research report. We do this using the Harvard and APA styles. You can learn more about our enhanced content on our Features: Overview page.
Join the 10,000s of students, academics and professionals who rely on Laerd Statistics. | 2,314 | 11,524 | {"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} | 3.265625 | 3 | CC-MAIN-2020-40 | longest | en | 0.901435 |
http://mathhelpforum.com/number-theory/170151-division-algorithm-2-a.html | 1,529,453,910,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00162.warc.gz | 205,210,135 | 10,291 | # Thread: The Division Algorithm 2
1. ## The Division Algorithm 2
The division algorithm has an analoguen in $\displaystyle Q(x)$ that asserts: given two polynomials $\displaystyle f(x),g(x),g(x)$ not identically 0, there exists $\displaystyle q(x),r(x)\in{Q}(x)$ such that
$\displaystyle f(x)=g(x)q(x)+r(x)$
and either
$\displaystyle r(x)\equiv0$, or
$\displaystyle 0\leq\text{degree of }r(x)<\text{degree of }g(x)$.
Prove this. One route is to proceed by the following steps:
a) If $\displaystyle f(x)\equiv0$ or the degree of $\displaystyle f(x)$ is 0, prove the assertion.
b) Proceed by induction on the degree of $\displaystyle f(x)$ and assume results for all cases where $\displaystyle f(x)$ is of degree less than $\displaystyle n$. Let $\displaystyle f(x)$ be of degree $\displaystyle n$.
Subcase (1) $\displaystyle \text{degree }g(x)>\text{degree }f(x)$
Prove directly
Subcase (2) $\displaystyle \text{degree }f(x)\geq\text{degree }g(x)$
Form polynomial $\displaystyle h(x)=f(x)-(\frac{a_n}{b_n})x^{n-m}g(x)$ and use inductive hypothesis.
OK. If someone could just do the first part of this, a), then I think that MAYBE I can do the rest. Please don't just go and post an entire solution. I really want to try and do some of this. I just can't get started.
Thanks to anyone willing to walk me through this.
Note: The polynomial $\displaystyle h(x)$ was discussed in the previous problem here ---> http://www.mathhelpforum.com/math-he...hm-170136.html
2. The cases a) doesn't really have much to it.
If $\displaystyle f(x) \in \mathbb{Q}[x]$ and has degree 0 this means that
$\displaystyle f(x)=m$ for some $\displaystyle m \in \mathbb{Q}$
If the degree of $\displaystyle g(x)$ is larger than 0
Then $\displaystyle f(x)=0\cdot g(x)+m$
if the degree of $\displaystyle g(x)$ is 0 then $\displaystyle g(x)=c,\quad c \in \mathhbb{Q}$
Then $\displaystyle \displaystyl f(x)=\left( \frac{m}{c}\right)c+0=g(x)q(x)+r(x)$
All of this "division" is taking place in the field or rational numbers.
3. So, since we are just establishing existence, it doesn't matter that when $\displaystyle g(x)$ has degree 0, and $\displaystyle f(x)=\frac{m}{c}c+0$, that this is not unique. Because it isn't necessary that $\displaystyle r(x)=0$ because we could just as easily have
$\displaystyle f(x)=mc+r$, $\displaystyle r\neq0$.
But this doesn't matter right? | 701 | 2,368 | {"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} | 4.1875 | 4 | CC-MAIN-2018-26 | latest | en | 0.810786 |
https://id.scribd.com/presentation/387173478/matmul | 1,576,389,515,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00158.warc.gz | 410,903,625 | 83,921 | Anda di halaman 1dari 48
# Optimizing for the serial processors
## • Scaled speedup: operate near the memory boundary.
• Memory systems on modern processors are
complicated.
• The performance of a simple program can depend on
the details of the micro-architecture.
• Today we will study matrix multiplication optimizations
• An important kernel in some scientific problems
• Closely related to other algorithms, e.g., transitive closure on a
graph using Floyd-Warshall (is there a path between arbitrary
vertices)
• Optimization ideas can be used in other problems
• The best case for optimization payoffs
• The most well-studied algorithm in high performance computing
Slide sources: Kathy Yellick UCB and Larry Carter UCSD
Outline
• Performance Modeling
• Matrix-Vector Multiply (Warmup)
• Matrix Multiply Cache Optimizations
• Bag of Tricks
• Research in Matrix Multiply
Using a Simple Model of Memory to Optimize
• Assume just 2 levels in the hierarchy, fast and slow
• All data initially in slow memory
• m = number of memory elements (words) moved between fast and
slow memory Computational
• tm = time per slow memory operation Intensity: Key to
• f = number of arithmetic operations algorithm efficiency
• tf = time per arithmetic operation << tm
• q = f / m average number of flops per slow element access
• Minimum possible time = f* tf when all data in fast memory
• Actual time
Key to
• f * tf + m * tm = f * tf * (1 + tm/tf * 1/q) machine
efficiency
• Larger q means time closer to minimum f * tf
Warm up: Matrix-vector multiplication
{implements y = y + A*x}
for i = 1:n
for j = 1:n
y(i) = y(i) + A(i,j)*x(j)
A(i,:)
= + *
y(i) y(i) x(:)
Warm up: Matrix-vector multiplication
{read x(1:n) into fast memory}
{read y(1:n) into fast memory}
for i = 1:n
{read row i of A into fast memory}
for j = 1:n
y(i) = y(i) + A(i,j)*x(j)
{write y(1:n) back to slow memory}
## • m = number of slow memory refs = 3n + n2
• f = number of arithmetic operations = 2n2
• q = f / m ~= 2
## • Matrix-vector multiplication limited by slow memory speed
Modeling Matrix-Vector Multiplication
• Compute time for nxn = 1000x1000 matrix
• Time
• f * tf + m * tm = f * tf * (1 + tm/tf * 1/q)
• = 2*n2 * (1 + 0.5 * tm/tf)
• For tf and tm, using data from R. Vuduc’s PhD (pp 352-3)
• http://bebop.cs.berkeley.edu/pubs/vuduc2003-dissertation.pdf
• For tm use words-per-cache-line / minimum-memory-latency
Clock Peak Mem Lat (Min,Max) Linesize t_m/t_f
MHz Mflop/s cycles Bytes machine
Ultra 2i 333 667 38 66 16 24.8 “balance”
Ultra 3 900 1800 28 200 32 14.0 (q must
Pentium 3 500 500 25 60 32 6.3 be at least
Pentium3M 800 800 40 60 32 10.0 this for
Power3 375 1500 35 139 128 8.8 peak
Power4 1300 5200 60 10000 128 15.0 speed)
Itanium1 800 3200 36 85 32 36.0
Itanium2 900 3600 11 60 64 5.5
What Simplifying Assumptions
• What simplifying assumptions did we make in this
analysis?
• Ignored parallelism in processor between memory and
arithmetic within the processor
• Sometimes drop arithmetic term in this type of analysis
• Assumed fast memory was large enough to hold three vectors
• Reasonable if we are talking about any level of cache
• Not if we are talking about registers (~32 words)
• Assumed the cost of a fast memory access is 0
• Reasonable if we are talking about registers
• Not necessarily if we are talking about cache (1-2 cycles for L1)
• Memory latency is constant
• Could simplify even further by ignoring memory
operations in X and Y vectors
• Mflop rate/element = 2 / (2* tf + tm)
Validating the Model
• How well does the model predict actual performance?
• Using DGEMV: Most highly optimized code for the platform
• Model sufficient to compare across machines
• But under-predicting on most recent ones due to latency estimate
## 1400 Predicted MFLOP
(ignoring x,y)
1200 Pre DGEMV Mflops
1000 (with x,y)
Actual DGEMV
MFlop/s
800 (MFLOPS)
600
400
200
0
Ultra 2i Ultra 3 Pentium 3 Pentium3M Power3 Power4 Itanium1 Itanium2
“Naïve” Matrix Multiply
{implements C = C + A*B}
for i = 1 to n
for j = 1 to n
for k = 1 to n
C(i,j) = C(i,j) + A(i,k) * B(k,j)
## Algorithm has 2*n3 = O(n3) Flops and
operates on 3*n2 words of memory
## C(i,j) C(i,j) A(i,:)
B(:,j)
= + *
Matrix Multiply on RS/6000
12000 would take
1095 years
6
T = N4.7
5
log cycles/flop
4
3
2 Size 2000 took 5 days
1
0
-1 0 1 2 3 4 5
## O(N3) performance would have constant cycles/flop
Performance looks much closer to O(N5)
Note on Matrix Storage
• A matrix is a 2-D array of elements, but memory
• Conventions for matrix layout
• by column, or “column major” (Fortran default); A(i,j) at A+i+j*n
• by row, or “row major” (C default) A(i,j) at A+i*n+j
Column major matrix in memory
Column major Row major
0 5 10 15 0 1 2 3
1 6 11 16 4 5 6 7
2 7 12 17 8 9 10 11
3 8 13 18 12 13 14 15
4 9 14 19 16 17 18 19
## Blue row of matrix is
cachelines
stored in red cachelines
“Naïve” Matrix Multiply
{implements C = C + A*B}
for i = 1 to n
for j = 1 to n
for k = 1 to n
C(i,j) = C(i,j) + A(i,k) * B(k,j)
## Reuse Stride-N Sequential
value from a access to access through
register one row* entire matrix
• When cache (or TLB or memory) can’t hold entire B matrix, there will
be a miss on every line.
• When cache (or TLB or memory) can’t hold a row of A, there will be a
miss on each access
## *Assumes column-major order
Matrix Multiply on RS/6000
6
5
log cycles/flop
## TLB miss every
4
iteration
3
2 Cache miss every
16 iterations Page miss every 512 iterations
1
0
0 1 2 3 4 5
## log Problem Size
“Naïve” Matrix Multiply
{implements C = C + A*B}
for i = 1 to n
{read row i of A into fast memory}
for j = 1 to n
{read C(i,j) into fast memory}
{read column j of B into fast memory}
for k = 1 to n
C(i,j) = C(i,j) + A(i,k) * B(k,j)
{write C(i,j) back to slow memory}
## C(i,j) C(i,j) A(i,:)
B(:,j)
= + *
“Naïve” Matrix Multiply
Number of slow memory references on unblocked matrix multiply
m = n3 read each column of B n times
+ n2 read each row of A once
+ 2n2 read and write each element of C once
= n3 + 3n2
So q = f / m = 2n3 / (n3 + 3n2)
~= 2 for large n, no improvement over matrix-vector multiply
## C(i,j) C(i,j) A(i,:)
B(:,j)
= + *
Blocked (Tiled) Matrix Multiply
Consider A,B,C to be N by N matrices of b by b subblocks where b=n / N is
called the block size
for i = 1 to N
for j = 1 to N
{read block C(i,j) into fast memory}
for k = 1 to N
{read block A(i,k) into fast memory}
{read block B(k,j) into fast memory}
C(i,j) = C(i,j) + A(i,k) * B(k,j) {do a matrix multiply on blocks}
{write block C(i,j) back to slow memory}
## C(i,j) C(i,j) A(i,k)
= + * B(k,j)
Blocked (Tiled) Matrix Multiply
C(1,1) C(1,1)
= + A(1,1) * B(1,1)
Blocked (Tiled) Matrix Multiply
C(1,1) C(1,1)
= + A(1,2) * B(2,1)
Blocked (Tiled) Matrix Multiply
C(1,1) C(1,1)
= + A(1,3) * B(3,1)
Blocked (Tiled) Matrix Multiply
C(1,2) C(1,2)
= + * B(1,2)
A(1,1)
Blocked (Tiled) Matrix Multiply
C(1,2) C(1,2)
= + * B(2,2)
A(1,2)
Blocked (Tiled) Matrix Multiply
C(1,2) C(1,2)
= + *
A(1,3) B(3,2)
Blocked (Tiled) Matrix Multiply
Recall:
m is amount memory traffic between slow and fast memory
matrix has nxn elements, and NxN blocks each of size bxb
f is number of floating point operations, 2n3 for this problem
q = f / m is our measure of algorithm efficiency in the memory system
So:
m = N*n2 read each block of B N3 times (N3 * n/N * n/N)
+ N*n2 read each block of A N3 times
+ 2n2 read and write each block of C once
= (2N + 2) * n2
## So computational intensity q = f / m = 2n3 / ((2N + 2) * n2)
~= n / N = b for large n
So we can improve performance by increasing the blocksize b
Can be much faster than matrix-vector multiply (q=2)
Using Analysis to Understand Machines
The blocked algorithm has computational intensity q ~= b
• The larger the block size, the more efficient our algorithm will be
• Limit: All three blocks from A,B,C must fit in fast memory (cache), so
we cannot make these blocks arbitrarily large
• Assume your fast memory has size Mfast
3b2 <= Mfast, so q ~= b <= sqrt(Mfast/3)
• To build a machine to run matrix required
multiply at the peak arithmetic speed t_m/t_f KB
of the machine, we need a fast Ultra 2i 24.8186 14.8
memory of size Ultra 3 14 4.7
Mfast >= 3b2 ~= 3q2 = 3(Tm/Tf)2 Pentium 3 6.25 0.9
Pentium3M 10 2.4
• This sizes are reasonable for L1 Power3 8.75 1.8
cache, but not for register sets Power4 15 5.4
• Note: analysis assumes it is possible Itanium1 36 31.1
to schedule the instructions perfectly Itanium2 5.5 0.7
Limits to Optimizing Matrix Multiply
• The blocked algorithm changes the order in which values are
accumulated into each C[i,j] by applying associativity
• The previous analysis showed that the blocked algorithm has
computational intensity:
q ~= b <= sqrt(Mfast/3)
• There is a lower bound result that says we cannot do any better than
this (using only algebraic associativity)
## • Theorem (Hong & Kung, 1981): Any reorganization of this algorithm
(that uses only algebraic associativity) is limited to q = O(sqrt(Mfast))
Basic Linear Algebra Subroutines
• Industry standard interface (evolving)
• Vendors, others supply optimized implementations
• History
• BLAS1 (1970s):
• vector operations: dot product, saxpy (y=a*x+y), etc
• m=2*n, f=2*n, q ~1 or less
• BLAS2 (mid 1980s)
• matrix-vector operations: matrix vector multiply, etc
• m=n^2, f=2*n^2, q~2, less overhead
• somewhat faster than BLAS1
• BLAS3 (late 1980s)
• matrix-matrix operations: matrix matrix multiply, etc
• m >= 4n^2, f=O(n^3), so q can possibly be as large as n, so
BLAS3 is potentially much faster than BLAS2
• Good algorithms used BLAS3 when possible (LAPACK)
• See www.netlib.org/blas, www.netlib.org/lapack
BLAS speeds on an IBM RS6000/590
Peak speed = 266 Mflops
Peak
BLAS 3
BLAS 2
BLAS 1
## BLAS 3 (n-by-n matrix matrix multiply) vs
BLAS 2 (n-by-n matrix vector multiply) vs
BLAS 1 (saxpy of n vectors)
Search Over Block Sizes
• Performance models are useful for high level algorithms
• Helps in developing a blocked algorithm
• Models have not proven very useful for block size selection
• too complicated to be useful
– See work by Sid Chatterjee for detailed model
• too simple to be accurate
– Multiple multidimensional arrays, virtual memory, etc.
• Some systems use search
• Atlas – being incorporated into Matlab
• BeBOP – http://www.cs.berkeley.edu/~richie/bebop
What the Search Space Looks Like
## Number of rows in register block
A 2-D slice of a 3-D register-tile search space. The dark blue region was pruned.
(Platform: Sun Ultra-IIi, 333 MHz, 667 Mflop/s peak, Sun cc v5.0 compiler)
ATLAS (DGEMM n = 500)
Source: Jack Dongarra
900.0
Vendor BLAS
800.0 ATLAS BLAS
F77 BLAS
700.0
600.0
MFLOPS
500.0
400.0
300.0
200.0
100.0
0.0
00 3 0 5 12 6 0
2000 0 66 50 00 70 00
6 53 50 13 1 -1 -2 2 -5 2 2 2
n- 6- 6- 5/ 4- 2 ro3- II- III
28
-
30
-
c2
-
hl
o 5 ev 73 6 0 er Per m p p r
At ev 0 / C w w iu iu
m 0 i 0 i a
D EC D
EC 00 PP Po Po t iu
m
e nt nt 00 00 a Sp
M D P9 M M M n P e 10 12 tr
P Ul
A H IB IB IB Pe IR IR n
Architectures SG SG Su
## • ATLAS is faster than all other portable BLAS implementations and it
is comparable with machine-specific libraries provided by the
vendor.
Tiling Alone Might Not Be Enough
• Naïve and a “naïvely tiled” code on Itanium 2
• Searched all block sizes to find best, b=56
• Starting point for hw1
1600
1400
1200
1000
MFlop/s
800
600 3 loops
400 blocked, b=56
200
0
0 200 400 600 800
Matrix dim ension
Optimizing in Practice
• Tiling for registers
• loop unrolling, use of named “register” variables
• Tiling for multiple levels of cache and TLB
• Exploiting fine-grained parallelism in processor
• superscalar; pipelining
• Complicated compiler interactions
• Hard to do by hand (but you’ll try)
• Automatic optimization an active research area
• BeBOP: bebop.cs.berkeley.edu/
• PHiPAC: www.icsi.berkeley.edu/~bilmes/phipac
in particular tr-98-035.ps.gz
• ATLAS: www.netlib.org/atlas
Removing False Dependencies
• Using local variables, reorder operations to remove false
dependencies
a[i] = b[i] + c; false read-after-write hazard
a[i+1] = b[i+1] * d; between a[i] and b[i+1]
float f1 = b[i];
float f2 = b[i+1];
a[i] = f1 + c;
a[i+1] = f2 * d;
## With some compilers, you can declare a and b unaliased.
• Done via “restrict pointers,” compiler flag, or pragma)
Exploit Multiple Registers
• Reduce demands on memory bandwidth by pre-loading
into local variables
while( … ) {
*res++ = filter[0]*signal[0]
+ filter[1]*signal[1]
+ filter[2]*signal[2];
signal++;
}
## float f0 = filter[0]; also: register float f0 = …;
float f1 = filter[1];
float f2 = filter[2];
while( … ) {
*res++ = f0*signal[0] Example is a convolution
+ f1*signal[1]
+ f2*signal[2];
signal++;
}
Loop Unrolling
• Expose instruction-level parallelism
## float f0 = filter[0], f1 = filter[1], f2 = filter[2];
float s0 = signal[0], s1 = signal[1], s2 = signal[2];
*res++ = f0*s0 + f1*s1 + f2*s2;
do {
signal += 3;
s0 = signal[0];
res[0] = f0*s1 + f1*s2 + f2*s0;
s1 = signal[1];
res[1] = f0*s2 + f1*s0 + f2*s1;
s2 = signal[2];
res[2] = f0*s0 + f1*s1 + f2*s2;
res += 3;
} while( … );
Expose Independent Operations
• Hide instruction latency
• Use local variables to expose independent operations that can
execute in parallel or in a pipelined fashion
• Balance the instruction mix (what functional units are
available?)
f1 = f5 * f9;
f2 = f6 + f10;
f3 = f7 * f11;
f4 = f8 + f12;
Copy optimization
• Copy input operands or blocks
• Reduce cache conflicts
• Constant array offsets for fixed size blocks
• Expose page-level locality
## Original matrix Reorganized into
(numbers are addresses) 2x2 blocks
0 4 8 12 0 2 8 10
1 5 9 13 1 3 9 11
2 6 10 14 4 6 12 13
3 7 11 15 5 7 14 15
Recursive Data Layouts
• Copy optimization often works because it improves spatial locality
• A related (recent) idea is to use a recursive structure for the matrix
• There are several possible recursive decompositions depending on
the order of the sub-blocks
• This figure shows Z-Morton Ordering
• See papers on “cache oblivious algorithms” and “recursive layouts”
• the recursive layout works well
for any cache size
• The index calculations to find
A[i,j] are expensive
• Implementations switch to
column-major for small sizes
Strassen’s Matrix Multiply
## • The traditional algorithm (with or without tiling) has O(n^3) flops
• Strassen discovered an algorithm with asymptotically lower flops
• O(n^2.81)
• Consider a 2x2 matrix multiply, normally takes 8 multiplies, 7 adds
• Strassen does it with 7 multiplies and 18 adds
## Let M = m11 m12 = a11 a12 b11 b12
m21 m22 = a21 a22 b21 b22
Let p1 = (a12 - a22) * (b21 + b22) p5 = a11 * (b12 - b22)
p2 = (a11 + a22) * (b11 + b22) p6 = a22 * (b21 - b11)
p3 = (a11 - a21) * (b11 + b12) p7 = (a21 + a22) * b11
p4 = (a11 + a12) * b22
Then m11 = p1 + p2 - p4 + p6
m12 = p4 + p5 Extends to nxn by divide&conquer
m21 = p6 + p7
m22 = p2 - p3 + p5 - p7
Strassen (continued)
T(n) = Cost of multiplying nxn matrices
= 7*T(n/2) + 18*(n/2)2
= O(n log2 7)
= O(n 2.81)
• Asymptotically faster
• Several times faster for large n in practice
• Cross-over depends on machine
• Available in several libraries
• Caveats
• Needs more memory than standard algorithm
• Can be less accurate because of roundoff error
• Current world’s record is O(n 2.376.. )
• Why does Hong/Kung theorem not apply?
Locality in Other Algorithms
• The performance of any algorithm is limited by q
• In matrix multiply, we increase q by changing
computation order
• increased temporal locality
## • For other algorithms and data structures, even hand-
transformations are still an open problem
• sparse matrices (reordering, blocking)
• trees (B-Trees are for the disk level of the hierarchy)
• linked lists (some work done here)
Summary
• Performance programming on uniprocessors requires
• understanding of memory system
• understanding of fine-grained parallelism in processor
## • Simple performance models can aid in understanding
• Two ratios are key to efficiency (relative to peak)
1.computational intensity of the algorithm:
• q = f/m = # floating point opns / # slow memory opns
2.machine balance in the memory system:
• tm/tf = time for slow memory operation / time for floating point
operation
• Blocking (tiling) is a basic approach
• Techniques apply generally, but the details (e.g., block size) are
architecture dependent
• Similar techniques are possible on other data structures and
algorithms
• "Performance Optimization of Numerically Intensive Codes", by
Stefan Goedecker and Adolfy Hoisie, SIAM 2001.
• Web pages for reference:
• BeBOP Homepage
• ATLAS Homepage
• BLAS (Basic Linear Algebra Subroutines), Reference for (unoptimized)
implementations of the BLAS, with documentation.
• LAPACK (Linear Algebra PACKage), a standard linear algebra library
optimized to use the BLAS effectively on uniprocessors and shared
memory machines (software, documentation and reports)
• ScaLAPACK (Scalable LAPACK), a parallel version of LAPACK for
distributed memory machines (software, documentation and reports)
• Tuning Strassen's Matrix Multiplication for Memory Efficiency
Mithuna S. Thottethodi, Siddhartha Chatterjee, and Alvin R. Lebeck
in Proceedings of Supercomputing '98, November 1998 postscript
• Recursive Array Layouts and Fast Parallel Matrix Multiplication” by
Chatterjee et al. IEEE TPDS November 2002.
Questions You Should Be Able to Answer
1. What is the key to understand algorithm efficiency in
our simple memory model?
2. What is the key to understand machine efficiency in
our simple memory model?
3. What is tiling?
4. Why does block matrix multiply reduce the number of
memory references?
5. What are the BLAS?
6. What is LAPACK? ScaLAPACK?
7. Why does loop unrolling improve uniprocessor
performance?
Review from Last
Lecture
## 08/29/2002 CS267 Lecure 2 46
Membench: What to Expect
average cost per access
memory
time
size > L1
cache total size < L1
hit time
s = stride
## • Consider the average cost per load
• Plot one line for each array size, time vs. stride
• Small stride is best: if cache line holds 4 words, at most ¼ miss
• If array is smaller than a given cache, all those accesses will hit
(after the first run, which is negligible for large enough runs)
• Picture assumes only one level of cache
• Values have gotten more difficult to measure on modern procs
Memory Hierarchy on a Sun Ultra-2i
Sun Ultra-2i, 333 MHz
Array size
Mem: 396 ns
(132 cycles)
L2: 2 MB,
12 cycles (36 ns)
L1:
16 KB
L1: 16 B line L2: 64 byte line 2 cycles (6ns)
8 K pages
## See www.cs.berkeley.edu/~yelick/arvindk/t3d-isca95.ps for details
PHiPAC: Portable High Performance ANSI C | 5,909 | 18,556 | {"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-2019-51 | latest | en | 0.779166 |
https://synonymsbot.com/mean_squared_error | 1,582,586,441,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00019.warc.gz | 563,194,757 | 4,310 | Examples of "mean_squared_error"
Mean squared error is used for obtaining efficient estimators, a widely used class of estimators. Root mean square error is simply the square root of mean squared error.
For example: minimizing the mean squared error (MSE),
Forecast errors can be evaluated using a variety of methods namely mean percentage error, root mean squared error, mean absolute percentage error, mean squared error. Other methods include tracking signal and forecast bias.
where formula_17 is the mean squared error of the regression model.
it is the MVUE. Since the mean squared error (MSE) of an estimator "δ" is
However, we can achieve a lower mean squared error using a biased estimator. The estimator
The first bracketed factor is the expected mean-squared error of the estimator "θ", since
where formula_36 is the observed value of the mean squared error formula_37.
The SSD is also known as mean squared error. The equation below defines the SSD metric:
where MSE("f") is the mean squared error of the regression function "ƒ".
This approach of minimizing integrated mean squared error can be generalized beyond Normal distributions:
The mean squared error of the h-step forecast of variable j is
The use of mean squared error without question has been criticized by the decision theorist James Berger. Mean squared error is the negative of the expected value of one specific utility function, the quadratic utility function, which may not be the appropriate utility function to use under a given set of circumstances. There are, however, some scenarios where mean squared error can serve as a good approximation to a loss function occurring naturally in an application.
C records contain root-mean squared error (RMSE) quality control data, using ten six-character integer fields.
If it is assumed that distortion is measured by mean squared error, the distortion D, is given by:
Note that other distortion measures can also be considered, although mean squared error is a popular one.
Two naturally desirable properties of estimators are for them to be unbiased and have minimal mean squared error (MSE). These cannot in general both be satisfied simultaneously: a biased estimator may have lower mean squared error (MSE) than any unbiased estimator; see estimator bias.
There are several basic fitness functions for evaluating model performance, with the most common being based on the error or residual between the model output and the actual value. Such functions include the mean squared error, root mean squared error, mean absolute error, relative squared error, root relative squared error, relative absolute error, and others.
Surprisingly, it turns out that the "ordinary" estimator proposed above is suboptimal in terms of mean squared error when "n" ≥ 3. In other words, in the setting discussed here, there exist alternative estimators which "always" achieve lower mean squared error, no matter what the value of formula_4 is.
for that particular formula_106. Thus in that case, the corresponding formula_143 would be a more efficient estimator of formula_144 compared to formula_145, based on using the mean squared error as the performance criteria. In addition, any given linear form of the corresponding formula_143 would also have a lower mean squared error compared to that of the same linear form of formula_147. | 660 | 3,347 | {"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-2020-10 | latest | en | 0.884761 |
https://datascience.stackexchange.com/questions/tagged/dimensionality-reduction?tab=newest&page=5 | 1,643,388,096,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00047.warc.gz | 254,713,241 | 41,733 | # Questions tagged [dimensionality-reduction]
Dimensionality reduction refers to techniques for reducing many variables into a smaller number while keeping as much information as possible. One prominent method is [tag pca]
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58 views
### Bandwidth of the gaussian kernels in t-sne
I'm trying to understand t-SNE better and I was hoping someone could elaborate on how the $\sigma _i$'s are chosen. I was also wondering why they aren't just calculated in the normal way standard ...
142 views
### Dimensionality reduction with prior knowledge of colinearity between features
Let's say that I have sparse feature vectors and I'd like to use dimensionality reduction in order to visualize them more easily. Dimensionality reduction techniques like PCA will estimate ...
9k views
### Why are autoencoders for dimension reduction symmetrical?
I'm not an expert in autoencoders or neural networks by any means, so forgive me if this is a silly question. For the purpose of dimension reduction or visualizing clusters in high dimensional data, ...
553 views
### How to deal with disconnected components in isomap?
While creating a nearest neighbor graph for isomap, there is a possibility that the graph is disconnected. In this case finding graph distances between all pairs of points will not be possible. Are ...
114 views
### Could data from a test set 'leak' into predictor during PCA?
After reading this article I have got a question about PCA. Author was talking about whether to use test set while computing PCA. But, few important points to understand: 1) We should not ...
493 views
### Dimensionality reduction with PCA limitations
What are the cases when we should not use PCA for dimensionality reduction and what to use in such cases?
1k views
### Tensor Decomposition in TensorFlow for multinomial time series dimensionality reduction
I'm doing unsupervised learning (clustering and DR) on multinomial time series. I need to reduce dimensions for my data, which is sparse and has a lot of dimensions. I realized that some form of ...
2k views
### t-SNE: Why equal data values are visually not close?
I have 200 data points that have the same values on all features. After t-SNE dimension reduction they doesn't look so equal anymore, just like this: Why aren't they on the same point in the ...
408 views
### How are the positions of the output nodes determined in the Kohonen - Self Organizing Maps algorithm?
In the Cooperative stage of Kohonen's SOM, the neighborhood for a winning neuron(output node). In most cases, the neighborhood function happens to be the Gaussian Function. For example, h_j,_i = exp(...
174 views
### I have n dimensional data and I want to check integrity, can I downgrade to 2 dimensional feature space via PCA and do so?
Say I have n dimensional data samples. I want to check the integrity of the features, if they are good representation of the respective classes, i.e. these features are good or not. My plan is: I ...
602 views
### What are 2D dimensionality reduction algorithms good for?
It seems to me that t-SNE and other dimensionality reduction algorithms which reduce the dimensionality to two dimensions are mainly used to get an impression of the dataset. If done well, they look ...
246 views
### How to Interpret the output of PCA?
I have dataset of 50000 values (rows) and 1000 variables (columns). Since this is high dimensional, I am unable to work with just DBSCAN. So I am trying to use PCA (principle component analysis). ...
1k views
### PCA on Neural Networks dimensions reduction? [closed]
The dataset which was extracted from the database consists of more than 50 columns, I call these columns dimensions, can I call them dimensions? Obviously, I have to do dimension reduction on ...
450 views
### Issue with Spark SVD
I have the following dataset with the dimensions: Rows: 41174 Columns: 439316 The matrix is very sparse and on this, I want to perform Dimensionality Reduction. I am using Spark's computeSVD ...
690 views
### Can I do incremental learning with the sklearn implementation of Linear Discriminant Analysis
I have a large number of pictures that I would like to use LDA on. However, it requires too much memory, so I was wondering if it would be possible to make the learning incremental, using a sklearn ...
2k views
### Are t-sne dimensions meaningful?
Are there any meanings for the dimensions of a t-sne embedding? Like with PCA we have this sense of linearly transformed variance maximizations but for t-sne is there intuition besides just the space ...
79 views
### Finding the relation between two dimensions in a multi-dimensional problem
I have a collection of data points. Each point has 6 dimensions (x1, x2,...x6). I want to find a relation between two dimension (e.g. ...
6k views
### Multi-class text classification with LSTM in Keras
I'm quite new to Deep Learning and trying to solve the problem of Multi-Class, multi-label text classification using Deep Learning. https://github.com/fchollet/keras/blob/master/examples/...
37 views
### Modelling query in regression
I'm trying to build a regression model, where I see which attributes are influencing the margin. My data set looks like something below. ...
661 views
### Preserving explained variance while reducing dimensionality
We have a function $f:R^N \rightarrow R$ and a set of points $D=\{ x\in R^N\}$. How is it possible to linearly lower the dimension of points to $M \ll N$ such that the fraction of explained variance* ...
200 views
### Principal Component Analysis, Eigenvectors lying in the span of the observed data points?
I have been reading several papers and articles related to Principal Component Analysis (PCA) and in some of them, there is one step which is quite unclear to me (in particular (3) in [Schölkopf 1996])...
163 views
### PCA on matrix with large M and N
Based on this answer, we know that we can perform build covariance matrix incrementally when there are too many observations, whereas we can perform randomised SVD when there are too many variables. ...
301 views
### How to equalize the pairwise affinity perplexities when implementing t-SNE?
I'm trying to implement the t-SNE algorithm: I found that to compute the pairwise affinities, I have to follow this: My problem is computing $\sigma_i$. In the Wikipedia I found: The bandwidth of ...
2k views
### How is dimensionality reduction achieved in Deep Belief Networks with Restricted Boltzmann Machines?
In neural networks and old classification methods, we usually construct an objective function to achieve dimensionality reduction. But Deep Belief Networks (DBN) with Restricted Boltzmann Machines (...
323 views
### Difference between MDS and other manifold learning algorithms
From sklearn docs: Note that the purpose of the MDS is to find a low-dimensional representation of the data (here 2D) in which the distances respect well the distances in the original high-...
814 views
### Dimension reduction techniques in R that do not use the full distance matrix
I try to apply non-linear dimension reduction in R. As usual in machine learning I have a large data set (100 K rows). I tried the packages RDRToolbox and ...
3k views
### Multidimensional Scaling with Categorical Data
I have read the following about MDS in a book: using MDS requires an understanding of the individual feature's units; maybe we are using features that cannot be compared using the Euclidean ...
611 views
### Why does "Depth = Semantic representation" in convolutional neural networks?
I was watching some videos online about convolutional networks, and the speaker was discussing the concept of running a filter over an image. He said, and it is also shown in the image below, that "...
1k views
### Alternative Hunspell dictionary for stemming
I am using Hunspell to spellcheck and stem the words in my documents to reduct dimensionality. For spellchecking Hunspell works great with the default en_US dictionary by SCOWL (and friends), but not ...
5k views
### Dimension Reduction - After or Before Train-Test Split
Should one apply dimensionality reduction methods to the data set before or after train-test splitting? Anyway, in case of training a model with preprocessing by dim-red, one should apply the same dim-...
263 views
### Is there a particular order in which to do feature selection and sampling?
I want to use feature selection and observation subsampling on my data, for several reasons: feature selection for the usual motivations (reduce noise, decrease running time, etc.) observation ...
309 views
I am doing experiments on the high-dimensional regression. However, it is hard to obtain the practical or synthetic high-dimensional data. I have checked on UCI website as well as some papers with ...
151 views
### Feature selection with linear interaction between variables and correlation with categorical response variable
I am searching for a feature selection algorithm able to select the minimum number (minimum redundancy) of relevant variables (maximum relevance) with respect to a categorical response variable. I ...
770 views
### Is t-SNE just for visualization?
I have used the t-SNE algorithm to visualize my high dimensional data. However, I was wondering if this is a practical method for inference?
73 views
### Could principle component decomposed coordinates value be correlated to each other?
I am wondering if we have a A= n*p matrix of samples and we run a PC decomposition on it. Say the eigenvector matrix is E, so the samples in the eigenvector space ...
2k views
### Does PCA change the values of the data?
Principal Component Analysis is a means to reduce the dimensionality of data, if I understand correctly. So if I have a 1000 sample point 12 dimensional matrix and reduce it to a 1000 sample point 2 ...
3k views
### Is mutual information symmetric?
Why is mutual information symmetric, meaning why does I(A,B) = I(B,A)? Isnt the definition of mutual information, I(A,B), something like "the reduction of entropy in A when given B"? P(A|B) doesnt ...
4k views
### Pruning and parameter reduction for decision trees
I am trying to perform a classification using a decision tree classifier. I was wondering whether using a Feature reduction method is relevant for decision trees since they automatically use pruning? ...
2k views
### feature redundancy
Why exactly does features being dependent on each other, features having high correlation with one another, mean that they would be redundant? Also, does PCA help get rid of redundant/irrelevant ...
195 views
### Principal components analysis with compositional data
Another beginner question: I'm trying to do PCA on compositional data. In other words, all the variables in the group add up to 100%. I've since learned on this forum that compositional data poses a ...
582 views
### Compute angle of vector in word2vec models
If I understand correctly, the most_similar function computes the cosine similarity of the vector with all other vectors and finds the closest one. The vectors ...
345 views
### Can I apply Clustering algorithms to the result of Manifold Visualization Methods?
Some methods related to manifold-learning are commonly stated as good-for-visualization, such as T-SNE and self-organizing-maps (SOM). I understand that when referring specifically to "visualization" ...
4k views
### Can closer points be considered more similar in T-SNE visualization?
I understand from Hinton's paper that T-SNE does a good job in keeping local similarities and a decent job in preserving global structure (clusterization). However I'm not clear if points appearing ...
60 views
### Free/open interactive softwares/plugins for end-users' high-dimensional data visualization
Aside from questions about How to visualize data of a multidimensional dataset (TIMIT), the Purpose of visualizing high dimensional data? and High-dimensional data: What are useful techniques to know?,...
660 views
### How to reduce dimensionality of audio data that comes in form of matrices and vectors?
I'm working on a project involved with identifying different types of sounds (such as screams, singing, and bangs) from each other. We've got our data a reasonable number of different transformations ...
7k views
### What is a good explanation of Non Negative Matrix Factorization?
I am trying to find a resource to understand non-negative matrix factorization. Apart from Wikipedia, I couldn't find anything useful.
188 views
### How exactly dependent variable is expressed in terms of independent variables using Partial Least Square Regression Method?
I understand the working of NIPALS algorithm but while doing the regression using PLS how exactly the relation between known and unknown is established using Principle Component Analysis. The idea is ...
16k views
### Improve the speed of t-sne implementation in python for huge data
I would like to do dimensionality reduction on nearly 1 million vectors each with 200 dimensions(doc2vec). I am using TSNE ...
415 views
### Deciding about dimensionality reduction, classification and clustering?
Could you please help me to understand it because I'm not sure if I got it correctly. Let's say I have a dataset, of persons, with 100 features, various characteristics like height, weight, age, etc. ...
3k views
### Purpose of visualizing high dimensional data?
There are many techniques for visualizing high dimension datasets, such as T-SNE, isomap, PCA, supervised PCA, etc. And we go through the motions of projecting the data down to a 2D or 3D space, so we ... | 2,860 | 13,673 | {"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} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.935466 |
https://www.winsteps.com/facetman/arithmetictest.htm | 1,516,303,442,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00689.warc.gz | 1,016,407,608 | 6,767 | 4 facet Arithmetic Test: Measuring, Anchoring and Describing
# 4 facet Arithmetic Test: Measuring, Anchoring and Describing
Many measurement problems are best tackled with two Facets runs:
1. Measurement of student ability, judge severity, item difficulty, etc.
The actual participant elements that interacted to provide the observations are measured, and then the elements anchored at those measures. This provides a stable frame of reference for further analyses.
2. Descriptive summaries of various effects identified by demographics, judge training, sub-test content, etc.
Participating elements are replaced by their hypothesized components, and further analyses are performed, with care to provide connected, i.e, unambiguous, measurement conditions.
Using R. Mislevy's data set (extracted from the Armed Services Vocational Aptitude Battery - ASVAB), it is hypothesized that 776 students (black females, black males, white females, and white males) each have an arithmetic ability (a fixed effect). The analyst wants to decompose these abilities into gender (sex) and race (ethnicity) effects. First the item difficulties are calibrated and the 776 examinees measured:
Facets specifications and data (in file Measure.txt):
title = Arithmetic Competency - R. Mislevy
anchorfile = measanc.txt ; an anchored file is written out
facets = 4 ; demographics included in the facets
pt-biserial = y ; point-biserial as a rough fit statistics
vertical = 1N,2A,3A,4* ; for communication
yard = 0, 4 ;
models =
?, , ,?,D ; use this model for measuring items and students
; ?,?,?, ,D ; use this model for demographic summaries - commented out here
*
positive = 2,3,4 ; all facets except item difficulties are abilities
noncenter = 2, 4 ; noncenter students and one demographic
labels =
1,Arithmetic
1-4 ; 4 arithmetic items
*
2,Race
1=Black
2=White
*
3,Sex
1=Female
2=Male
*
4,Students
1-776 ; no more information about the students
*
Data =
1-4,2,2,1,0,0,0,0 ; on the 4 items, white male student 1, failed
|
1-4,1,1,776,1,1,1,1 ; on the 4 items, black female student 776, succeeded
or
Dvalues = 1, 1-4 ; place 1-4 in first data facet location of all data records
Data=
2,2,1,0,0,0,0 ; on the 4 items, white male student 1, failed
|
1,1,776,1,1,1,1 ; on the 4 items, black female student 776, succeeded
Then the descriptive phase is performed to estimate the demographic effects. This is set up by editing the anchor file produced in the measurement phase, and changing the model statement:
Facets specifications and data (in file Measanc.txt, also in file Meas2anc.txt):
title = Arithmetic Competency - R. Mislevy
; anchorfile = measanc.txt
facets = 4
pt-biserial = y
vertical = 1N,2A,3A,4*
yard = 0, 4
positive = 2,3,4 ; facets 1, 2, 3 this time
noncenter= 2,4 ; facet , Race, 2 floats
Models=
;?,,,?,RS1,1, (D) ; comment out the measurement model
?,?,?,,RS1,1, (D) ; use the summarizing model
*
Rating (or partial credit) scale=RS1,D ; use the measurement run rating scale
0=,0,A, ; Rasch-Andrich thresholds in the description run
*
Labels=
1,Arithmetic,A ; these are anchored at their Rasch-Andrich thresholds
1,1,-.6079245
2,2,-.1628681
3,3,.4518689
4,4,.9676156
*
2,Race,A ; the A is inoperative, because there are no logit values
1,Black,
2,White,
*
3,Sex,A
1,Female,
2,Male,
*
4,Students,A ; this facet is ignored this run
1,1,-2.537946 ; these anchored measures are ignored this run
|
1-4,1,1,776,1,1,1,1
Help for Facets Rasch Measurement Software: www.winsteps.com Author: John Michael Linacre.
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Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement with Raters and Rating Scales: Rasch Models for Rater-Mediated Assessments, George Engelhard, Jr. & Stefanie Wind Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
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# Introduction
I have only one short hour a day for math instruction and at third and fourth grade much of the time is used to teach algorithms and solving problems. Admittedly, I often prioritize teaching algorithms and steps so that children have the necessary skills to solve problems so that we can move forward onto new material. Teaching problem solving methods is satisfying. For example, when dealing with subtraction problems, training students to perform the standard right to left algorithm which involves (often blindly) crossing through a place value, reducing values by one, adding dashes to make values ten more and finally arriving at the correct answer feels like success! The success, however, is often disappointing when I ask a student to explain what he or she just did and explain why, he or she has only a shoulder shrug to offer, or a "that's just what you do to get it right" type of reply. The inclusion of place value concepts in the teaching of operations in elementary mathematics education may help students obtain not only a right answer, but also a reason for why the answer is correct. The intent of this unit is to tie the concepts of place value to operational algorithms in order to provide students with a conceptual understanding of addition, subtraction, and multiplication, and not just a procedure.
I teach in a third and fourth grade classroom at Westcott Elementary in Chicago, IL. The students are in my classroom for both third and fourth grade. Westcott consists of ninety-nine percent African-American students. Additionally, one hundred percent are classified as low-income families and qualify for free lunch. The school's population is transient due to housing and income based issues. These two factors also affect attendance and student performance. My students have a fragile concept about the composition of multidigit numbers, and place value concepts in relation to operations. Karen Fuson states, "The evidence indicates that U.S. children do not learn place-value concepts or multi-digit addition and subtraction adequately and even many children who calculate correctly show little understanding of the procedures they are using (1990)." 1 Fuson's research confirms my students' lack of understanding, but the issue is more widespread than just my classroom.
# Objectives
This unit is intended for teachers of third and fourth grade, and emphasizes the teaching of place value concepts and grade level skills in operations of addition, subtraction and multiplication. The unit will be presented in four parts. It can be taught in full or by section based on need. My goal for this unit is to strengthen students' understandings of place value concepts through the instruction of multidigit addition, subtraction and multiplication.
# Background
## Place Value
The place value system is decimal notation. This allows a digit (i.e. one of the numbers 0,1,2,3,4,5,6,7,8, or 9) to positionally indicate powers of ten involved in the "base ten" decomposition of a number determined by its location. To illustrate with a simple example, in this system the notation "123" represents the number "one hundred twenty three". In this system, a particular place value is ten times the value of the place to its right, but only one-tenth the value of that to its left. The system's beauty of compacting numbers allows for reasonable representations of extremely large numbers in a space that is workable, but leads to misconceptions of actual values in the representation. The close proximity of one place directly next to the place to its right suggests relative closeness in value when in actuality it is ten times the value. As an example, consider the digit 7 located in the "ones" place (the rightmost place). Moving this digit 7 one place value to the left into the "tens" place gives a number with value of 70, or 7 x10, or 7 tens, or ten times greater than 7. The next place to the left is the "hundreds" place, and moving the digit 7 there creates a number with a value 700, or 7 x 10 x10, or 70 x 10. 700 is ten 70s or is 100 times the original 7 in the ones place. That is, these numbers look similar, but are far from each other in value.
The values 700, 70, and 7 are examples of related values in the base ten system. The genius of this system creates confusion of the system. My students can name place values and show the value of a digit in a place, yet they lack the understanding of the relationship between the values after the tens place and struggle with how the representation of digits is a combination of values. Students are often confused by the values represented in the compacted form of the number. For example, if a student is asked what numbers are represented in 123 the response is often one, two and three. The correct response is 100 + 20 + 3. The error here is due to lack of a deeper understanding of place value. A student providing the correct answer indicates that they have deep understanding of place value and positional notation.
How can we overcome misunderstandings in place value? In Roger Howe and Susanna S. Epp's essay "Taking Place Value Seriously: Arithmetic, Estimation, and Algebra" 2 the authors support explicit instruction of place value to students to create a more unified and conceptual understanding of the place value system with emphasis on the structure and systematic organization. The place value system is a complex one in an efficient condensed form. The idea to take the compressed number structure (i.e. 123) and intentionally unpack the content (i.e. one hundred plus twenty plus three) emphasizes the place value concepts. This expanded notation not only allows for an increase in conceptual understanding, but also allows for more computational flexibility due to deeper understanding. Students have computational flexibility when they can chose an algorithm or strategies that works best for the information presented. This unit will provide examples of different algorithms for addition, subtraction and multiplication.
## Expanded form
The compacted number becomes more representative of value when shown unpacked in a variety of expanded form notation. If the number 7,892 is presented in compact form, its true value can be confusing for students. To clearly show students the parts of a compacted number, expanding the form shows each digit is connected to place value. For example 7,892 becomes 7,000 + 800 + 90 + 2. This form allows each number to be shown at its actual value. Now students can see the 8 in 7,892 as the 800 it represents. In another slightly more advanced expanded form, the number 7,892 is (7 x 1,000) + (8 x 100) + (9 x 10) + (2 x10). In this form, the digit is now the single place digit multiplied by the place value. In this form the place value is separated from the digit to show the number of 1,000s, 100s, and so on. The 8 again now shows 8 hundreds or 8 x 100. A third and even slightly more advanced expanded form additionally depicts increasing powers of ten. For example, 7,892 is 7 x (10 x 10 x 10) + 8 x (10 x 10) + 9 x (10) + 2 x (1). This form shows the single place digit, the place value and also how place value is constructed using powers of ten. Students see the 800 both as total value, single digit times the place value, and as a relationship between place values. The extraction of the single digit defines the digit and how it relates to positional place value. Expanded form explicitly shows digits to the left will always represent greater values than any earlier values because its place alone is worth more. The complexity of the number is better represented with expanded form notation.
# Strategies
My strategies consider Epp's and Howe's advice to explicitly teach place value concepts, and integrate with this the teaching of the operations addition, subtraction, and multiplication. The unit will cover four strategies. In part one, "naming numbers differently," the objective is to emphasize place value in number names. This will aid students' place value understanding and help when expanding numbers into values. Part two, "addition of multidigit numbers," emphasizes importance of place value by using expanded forms of multidigit numbers when combining numbers. The addition strategies will use the expanded form to combine like place value terms. Part three, "subtraction of multidigit numbers," continues with expanded form and combining like terms but finding the difference in values. Finally, part four, "multiplication of single and multidigit numbers," builds upon the prior strategies and expanded form to continue the focus on place value in operations. The four strategies are introduced systematically to build on students' prior learning. Some of the introductory concepts may not require full coverage depending on students' prior knowledge.
## Naming numbers differently and expanded form of compacted numbers 3
American culture falls into a category of cultures whose language characteristics do not lend well to understanding place value. Students write and say numbers that do not clearly show the value presented. As an example, the English word twelve tells us no information about its value of one ten and two ones. The compacted form of 12 does show the value positionally, but if positional notation is not well developed it may cause confusion. The example of fifty-seven gives more information because we hear fifty and seven, but the value of five tens and seven ones is not obvious, more can be done to show the places and values in 57. To enrich the learning of early place value, I will first spend time constructing a better understanding of the number names by teaching the conceptual place value and calling it "place value name" similar to the name given in Chinese based languages. Chinese based languages have a name for numbers that clearly represent the place value in the name. For example, twelve would be stated as "ten two," which corresponds to the positional 12 and the value of one ten and two ones. Also, fifty-seven in Chinese is stated as "five ten seven" representing the 5 tens and 7 ones related to the positional model of 57. This name will give students an understanding of real value not just the name for a number.
At the same time the instruction of "place value names" is being explained, students can begin to see the numbers as addition of place value components. For 12, the underlying ten plus two can be shown with a number model of expanded form of 10 + 2 or 1 x (10) + 2 x (1). The expanded form shows how the digits are positioned to create the numeral of 12. The value of 1 ten and 2 ones positionally is 12. The many names for numbers can be charted to show relationships across one number, 12. The expanded form is showing each digit times a value of ten, and the "place value name" always states the largest value first. The chart below is a tool to show the connection between numbers and place value concepts.
The naming of numbers many ways strengthens the understanding of their values, and also different ways to represent numbers. In later strategies students will use the knowledge of place value in singular numbers to perform operations.
## Addition of multidigit numbers 4
Part one emphasizes the verbal importance of place value name as a tool to make place value parts noticeable in multidigit numbers. This strengthens overall positional understanding as it relates to place value, and simultaneously prepares students to transfer the knowledge of expanded notation to the operations of multidigit numbers. After students have practiced expanded notation of numbers, performing addition in a way that makes explicit use of the expanded notation is a natural step. Addition of multidigit numbers is traditionally taught by stacking the addends by place values in columns. We first add the values in the ones place, and then move left. If an exchange of ten is needed it is usually shown above the next column as a dash, one more. This process continues in an analogous way to the left across each increasing place. This method is efficient but does not teach the understanding of addition as a combination of values by each place. The following strategies for teaching addition are focused on both the operation and place value to give meaning to a procedural algorithm.
For example, 7,451 + 3,245, traditional methods add from the ones place to the left. Traditional methods start with adding the ones (like the 2-column algorithm described above), but with the "any which way" rule, any place value can be added first. For example, if I chose to start in the thousands, or even the tens place, as long as all place value sums are combined to the total, it remains the same. Therefore, if students are taught a strategy that reflects the fact that "any which way" works to find the sum, they will have a better operational and place value understanding.
One can also perform multidigit addition in a similar way using any of the expanded forms mentioned previously. I plan to have the students work through different examples of expanded forms to add to flexibility.
## Tools for teaching place value and multidigit addition
Teaching different strategies to solve multidigit addition problems can be supported with tools. Two tools that are useful are digit cards and number trains. Digit cards are index-card sized cards with the values 0,1,2,3…,9, and 10,20,30,…,90 written on them corresponding to the "ones" place and "tens" place respectively. Higher places (hundreds, thousands, and so on) are represented on cards similarly. A nice aspect to digit cards is that they can be used to show a number in expanded form and compacted form. The digit cards allow the student to see numbers as expanded form when each place is represented on the card and compacted form when placed down next to each other from largest value to smallest with each card slightly overlapping that to its left (obscuring zeros). Another tool that is useful is to make number trains to represent addition in a linear model. The train uses rods of different lengths to represent ones, tens and hundreds numbers to show addition. Both tools show representations of numbers in relationship to place value.
Digit cards can be used to show addition of multidigit numbers and allow for manipulation. To use digit cards as an example, 7451 is represented by the four digit cards 7000, 400, 50, and 1, which when laid next to each other left to right indicate expanded form. Digit cards can also be used to add multi-digit numbers. The problem of adding 7451 + 3245, for example, can be shown by combining place values in groups: the 7000 digit card is grouped with the 3000 digit card, the 400 digit card with the 200 digit card, the 50 digit card with the 40 digit card, and then the 1 digit card with the 5 digit card. The place value digit cards can be combined in any order, highlighting the fact that addition can be performed in any order. Once left with these pairs of cards, students can exchange the pairs for a single new card reflecting their sum. Here, students exchange for cards showing 6, 90, 600 and 10,000 (where in the last case, the sum 7,000 + 3,000 was exchanged for a higher place value card, 10,000). Placing the new cards down starting with the highest value will recreate the correct sum of 10,696 (see table below).
A different strategy to show addition using place value is a linear model that involves making number trains. This can be done using flat rods to represent 100, 10 and 1. I use a meter stick to represent a 100s rod, a "base ten long" as a 10s rod, and a "base ten cube" as a 1s rod. These last two objects are often found in classrooms. They are flat rods measured so that 10 "ten cubes" lined up left to right are the same length as a "base ten long", and 10 "base ten longs" are the same length as a 100s rod. Any objects with these same proportional relationships can be used. Roger Howe describes trains as lining up rods to represent addition of multidigit numbers. Trains allow children to see the sum value in the addition problem as two smaller values combining to make one greater value. The train starts with the first addend rods and adds the second addend rods. For example, given the addition problem of 147 + 366 one creates a number train by taking one hundred rod, four ten rods, and seven ones and creating a linear train. One then adds on three hundreds, six tens, and six ones. Next, combine like terms by reorganizing the train by placing all the hundreds together, all the tens and all the ones. This shows that the arrangement does not change the total sum while emphasizing the need to group like terms. Once the rods are grouped by place value, trade ten of any rods for one of the next greater value. For instance, trade ten ones for one ten, or ten tens for one hundred, as needed, and recreate the train with the regrouped model. The example below is not to scale but shows a representation of how the numbers can be organized in a train.
## 147 as a train
The next step is to show the train 366 added on, 3 hundred rods, 6 ten rods, and 6 one rods. Then rearrange both 147 and 366 to have all the hundred rods together, all the ten rods together and all the one rods together. So placing 1+ 3 hundreds rods, 4 + 6 tens rods, and 7 + 6 ones rods together. The next step is to regroup the rods with 10 or more. When this step is done I will have 4 hundred rods, 10 ten rods, and 13 one rods. I now must regroup and trade. The train then is reorganized with 7 + 6 ones which trades for 1 ten and leaves 3 ones. The 6 + 4 tens now have 1 more from the ones trade for a total of 11 tens. This 11 tens will trade for 1 hundred rod and leave 1 ten rod. The hundreds now have 1+ 3 plus the 1 traded for 4 total hundreds. The new train after regrouping has 5 hundreds, 1 ten and 3 ones. The train strategy can be shown in the written model below.
These strategies for adding multidigit numbers are ways to show addition by "grouping like terms", that is, decomposing numbers into their place value components and then adding together "like" place value components. These strategies all connect back to the traditional compact form used in the 2-column addition algorithm but unpack the place values in each number to support conceptual understanding of place value and combining values. The exchanging done in these expanded form problems is a foundational concept but one that continues to be reinforced in all the strategies.
## Subtraction of multidigit numbers
Subtraction is the inverse relationship to addition. The act of decomposing (as described above in addition) in subtraction is more challenging for students specifically when borrowing or trading is needed. However, similar place value concepts and strategies can be used, and students can work though the same problems modeled in addition (using place value) above. These things show a connection between operations and will therefore hopefully be familiar to students.
Starting with a problem not requiring a trade will show the process of finding the difference of single digit numbers in the place value. Expanding the form of the compacted numbers in a subtraction problem is the first step to combining like values to see the numbers as individual places. . For example, 375 – 124 expands to 300 + 70 + 5 and 100 + 20 + 4 or (3) x 100 + (7) x 10 + (5) x 1 and (1) x 100 + (2) x 10 + (4) x 1. To show the subtraction, each single place can be subtracted and then multiplied by the appropriate place value, (3 – 1) x 100 + (7 – 2) x 10 + (5 – 4) x 1. After the subtraction is completed the remaining number is (2) x 100 + (5) x 10 + (1) x 1, which in compacted form is 251.
## Tools for teaching place value and multidigit subtraction
The tools for supporting subtraction strategies are number trains and number lines. The strategies can be shown using the same rod train model used in addition and the number line can be used to visually show subtraction. The trains are set up similarly to addition, the first number (the minuend) is laid down first, next the number being subtracted (the subtrahend) is placed below it. This creates a visual model of showing the space difference between the two trains as the difference between two numbers. The next example, number lines follows a similar model. A number line is a horizontal line marked with a "zero" and a "one", placed to the right of zero. With this, every other integer can be placed on the number line. For example, the number 2 is placed to the right of 1 at the same distance 1 is from 0. The number 3 is placed to the right of 2 at this same distance, and so on. The minuend is found first on the number line, then the subtrahend is located. The student uses the number line to find the difference in numbers with the support of the manipulative.
The train model for subtraction uses a linear model to represent the subtraction algorithm. Since subtraction with regrouping is a challenge, two digit numbers will work best in the model strategy. For example the problem 54 – 28, first create the minuend, 54, linearly and then place the subtrahend, 28, below. The missing value, or difference, is the space between the two numbers. For example, if the subtraction problem 54 – 28 is given, create a linear train of 5 tens and 4 cubes with base ten blocks or rods (as described previously), and a second consisting of 2 tens and 8 cubes, placed immediately below the first. This model shows subtraction as the difference in the two values. With this positioning, students will see that there are not enough ones available to subtract 8 from 4, and an exchange of one ten is needed in the train representing 54. After exchanging a ten for 10 ones, the "minuend" now has fourteen ones from which 8 are to be subtracted. This exchange should be recorded on the written model just as with addition when trading occurs. The next step is to find the difference between the remaining four tens and two tens, and to show the solution as 26 (see figures below).
Train representation of 54-28
Another error prone subtraction problem involves "borrowing across zeros" to solve a problem. For instance, a problem like 200-134 is often the most challenging for students learning how to subtract with trading. The problem is due to multiple trades needed from the hundreds place to the ones place. Traditionally we would move to the hundreds place and redistribute the 1 hundred as 9 tens and 10 ones. A better strategy may be to use a number line or linear model to show 200-134. The number line is a useful tool for the subtraction algorithm and when "borrowing across zeros." Use a number line and find 200 and mark it, next find 134 and mark it. The number line can be used to see the difference between the two numbers as the space between the point marked 134 and the point marked 200. Subtracting with the number line can also sometimes be used to illustrate how to avoid trading or borrowing. The difference of the two numbers is traditionally found by subtracting the smaller number from the larger number to find the difference in the numbers. The difference between the numbers does not change if we move each number back one space on the number line, which corresponds to subtracting 1 from each. In this example, this translates to 200 – 134 = (200 - 1) – (134 - 1) = 199 – 133. Because we made the same change to both numbers the original difference 200 – 134 is still the same as the resulting difference 199 - 133. Compensating for both numbers by subtracting one from each does not change the difference, but it does make the subtraction easier because it does not require any trading or borrowing. Students can finish the problem by subtracting 3 ones from 9 ones, 3 tens from 9 tens, and 1 hundred from 1 hundred, for a total of 66. Some students may not be able to make the conceptual connection that shifting the numbers back one place on the number line does not change the difference between the values, but it shows how multiple zeros can be made into an easier problem. This strategy of compensation shows subtraction as difference and gives flexibility to solve problems.
## Multiplication of multidigit numbers
Multiplication of multidigit numbers is introduced in third grade and a standard in fourth grade. Multidigit multiplication is built on single digit multiplication knowledge. First, I will introduce strategies to students for single digit multiplication and build on the strategies and extend the strategies to multidigit multiplication.
## Single digit multiplication
Multiplication of single digit numbers is the first introduction to the multiplication algorithm. Multiplication of whole numbers can be interpreted as repeated addition. The link to addition for student can be used in strategies to unpack the total value in a number model. I find it valuable to give multiple examples for each number model. For instance, the number 5 x 7 is the same as 5 sevens. Also, the strategies for multiplication of single digits can be shown with an array model. An array model arranges counters or items in rows and columns. For example, 5 rows of 7 counters, or 7 rows of 5 counters both make an array of 35 counters. The multiplication table itself can also represent an array of boxes or an area model. Students can use a multiplication table to cover with counters or shade over the numbers to show the area of 5 x 7 by shading 7 rows across and 5 columns for a total of 35 squares. This can be seen in the table below. Multiple strategies leads to flexibility and also deeper understanding of the multiplication concept.
## Multidigit multiplication extended single digit
I will introduce extended fact multiplication strategies after single fact multiplication models are understood. At first I will introduce strategies that can be taught as extensions of single digit multiplications. These types of problems will involve multiplication with "single place numbers" such as 10,20,30, and so on. For example, the extended fact problem of 5 x 70 should be connected back to the original single fact of 5 x 7. The connection should be made to the similarity and difference between 7 and 70. The digit is the same, but the place value is different. A model would be comparing 5 x 7 as five groups of 7 ones (for a total of 35) and 5 x 70 as five groups of 7 tens (for a total of 350). The link between the single digit multiplication and the extended fact is ten times the original fact. This can be illustrated multiple ways. For example, written in expanded form 5 x (7 x 1) can be compared to 5 x (7 x 10), where the difference between (7 x 1) and (7 x 10) is visible. One can also use the associative property of multiplication to rewrite 5 x (7 x 10) as (5 x 7) x 10, so that again the difference between this number and the single digit multiplication (5 x 7) is visible. This example allows a mental math advantage of seeing the original fact and multiplying by the place values presented. Students can recall most 1 through 9 facts, and can then apply this knowledge to extended fact problems along with place value concepts. By breaking the numbers into single digit values and place values the multiplication can be seen as two parts, the single digit multiplication and place value multiplication. The final step of multiplying the single digit by the place value makes the numbers easier to work with for the student because single digits are often known and multiples of ten are easy to work with for students. Another example, the extended form can also extend both original facts by 10, for 5 x 7 extended both facts to 50 x 70. The numbers are expanded to (5 x 10) x (7 x 10) and moving numbers through the commutative property to create easier combinations, of (5 x7) x (10 x 10). The original facts combine (5x7) and then the place value factors of 10 x10= 100 creating (5x7) = 35 and (10 x10) = 100, and then make 35 x100 = 3,500. The practice of finding the single digit multiplication and then the place value multiplication are manageable with mental math and support place value concepts.
A connection here can be made to the addition "any which way rule." Multiplication has the same rule; order of multiplication does not change the product. For example, I can multiply 5 x 70 as (5 x 7) x 10 or (5 x 10) x 7, and any way I multiply will find the same product. The expanded form of the multiplication problem can be combined any which way works best for the problem solver. For example, I might use an example like 50 x 70, and first illustrate that this is equal to (50 x 10) x 7 = 500 x 7 = 3,500. I might then move the numbers to a new position of (5 x 7) x (10 x 10) to show that this same product equals 35 x 100 = 3,500. The expanded form of a number shows place value importance, while it applying single digit multiplication.
## Multidigit multiplication
The next level of multiplication is a multidigit number that is not a single digit extension (like the examples above) such as 325 x 4. I will start with three digits by one digit multiplication. First, expand 325 to 300 + 20 + 5. All place value parts of 325, 300 + 20 + 5, are multiplied by 4. I then have to solve 300 x 4, 20 x 4 and 5 x 4, but the "any which way rule" allows me to chose the order in which I solve. I prefer to move from ones to hundreds and from hundreds to ones as it reinforces the any which way rule. I prefer to write the multiplications to be performed in a vertical table as follows:
The next step is to add the resulting products: 1,200 + 80 + 20 = 1,300
Another algorithm that is place value focused and also practical for computation is to consider each factor in a multiplication problem as a "team". An example problem is 37 x 21. Each factor 37 and 21 can be separated into place value parts, or teams of numbers (30 +7) x (20 +1). The next step is to multiply both expanded numbers with the other factor's members, that is to perform multiplication of each expanded value with the expanded value in the other factor. The strategy indicates 30 needs to multiply by 20 and 30 needs to multiply by 1 and that 7 needs to multiply with 20 and with 1 (see table below). Because this can be confusing I like to color code each number with colored pencils so as not to mix up which digits need to be multiplied. For example, I would color 30 + 7 red and color 20 +1 blue and state that you do not multiply numbers of the same color, and instead explain that each number of one color team has to be multiplied by both terms in the other color. This example shows the expanded form of both numbers and the multiplication of the single digits of each number, and then multiplying by place value.
The strategies explained in all sections of operational strategies are linked specifically with place value concepts and use expanded form repeatedly to build students' place value understanding in operations. This connection of operations and place value provides practice of algorithms while creating understanding of how the operations relate to place value and why place value understanding is crucial to deeper understanding of operations.
# Activities
The activities will match each strategy section above for use in the classroom.
## Activity 1: Naming numbers and expanded form
Materials:
Playing cards 1-9, four of each card
Place value mat
Score sheet
Students will practice building two digit numbers, naming the place value name, writing the expanded form, and determining the number with the greatest value. This activity is in a form of a game. The object of the game is to form the largest value with two playing cards. First, students draw two cards each. The student will decide which number to place in the tens and ones place. The student is trying to create the largest number from the two cards drawn. The student then places the cards on the mat, either in the tens place or the ones place, and record the information for his or her cards on the score sheet. The partner does the same task and the two students determine who has the larger value. The student with the larger value takes all four cards and wins the round. This game practices the naming of numbers and expanded form but the act of making the largest value practices place value concepts of understanding the tens place has more value, regardless of digit, than the ones place.
Sample Round
Student A draws cards 4 and 7. To make the largest value she places the 7 in the tens place and 4 in the ones place making the number 74. Student B draws cards 1 and 8. Student B places card 8 in the tens place and card 1 in the ones place making the number 81. The score sheet shows:
The game can add variations depending on class need or student need. One variation could include three digit numbers. Another, the game could be played with multiple players, the player with the largest number taking all the cards played in the round. To add to strategy the players could each draw three cards and then chose which two to play and have one to discard. The score sheet could include a base ten picture representation. An assessment for the activity beyond the score sheet is to have the student explain why placing the higher value in the tens place makes a more valuable number. Another ideas is to have a response sheet giving a scenario and having the student respond, "Carla draws a 1 card and a 9 card, which place should each card go to win the round? Explain to her why this creates the highest value?" Another way is to pose a mistake scenario, "Carla draws a 1 card and a 9 card, she places the 1 in the tens place and the 9 in the ones place, she loses the round to her opponent, what advice would you give Carla to have a winning strategy?" Both questions ask the student to use place value concepts and understanding in the reply.
## Activity 2: Addition of multidigit numbers
The addition of mulitdigit numbers is interesting when using relatable real world examples with students. Examples of attendance at professional sports games, heights of buildings or mountains, or distances on a map are examples of problems with meaningful context. For instance, my family is planning a trip from Chicago to Detroit and then Detroit to Cleveland. The distance to Detroit is 281 miles and the distance from Detroit to Cleveland is 169 miles. How many total miles will my family drive? The problems can be of any multidigit make-up and should have variety, some with and some without trade. The importance is not the values but the practice of expanding the form of each number and combining like terms. The table below shows the different expanded forms students can chose from the strategies. The student work should replicate the model shown for each step of the solution.
The students could also construct the addition problem with rods to make trains. The first train would have 2 hundreds rods, 8 tens rods and 1 ones rod. The second train added would have 1 hundreds rod, 6 tens rods and 9 ones rods. The reorganization of rods with like place values will have 3 hundreds rods, 14 ten rods and 10 ones rods. An exchange is needed in the ones place, exchange of 10 ones rod for 1 ten rods leaves 0 ones rods. Then exchange 10 of the 15 tens rods for 1 hundreds rod, leaving 5 tens rods, then counting the total 4 hundreds rods. The total train after exchange has 4 hundreds rods, 5 tens rods and 0 ones rods.
## Activity 3: Subtraction of multidigit numbers
The use of real world examples and problems is excellent for subtraction problems. The link to real world in both addition and subtraction operations gives students practice at solving word problems and using examples of real math situations. An example question, The Willis Tower is Chicago is 442 m in height, the John Hancock Center in Chicago is 343 m in height. How much taller is the Willis Tower? The template for subtraction is similar to the addition template and allows for the same expanded number and combination of like terms.
Template
The train model can also be used to model the subtraction process. Make a minuend (total number and starting value) 442 train with 4 hundreds rods, 4 tens rods, and 2 ones rods. Next make a subtrahend (number subtracted) rod below the minuend with 343 train of 3 hundreds rods, 4 tens rods and 3 ones rods. The difference can be shown exchanging 1 tens rod for ten ones and then removing three from both trains, leaving 9 ones. The minuend now has 3 tens rods and 4 tens rods need to be subtracted so an exchange of 1 hundreds rod is needed. The hundreds rod exchanges for 10 tens making 13 tens subtract 4 tens leaving 9 tens. The hundreds rods are now equal at 3 which leaves 0 hundreds rods remaining.
Another way to solve is with the number line. On the number line find the Willis tower height of 442. Next, find the height of the John Hancock at 343. The student can count up to 343 to 350 and record 7 then skip count by 10s from 440 count 360 as 10 more, 370 as 20 more 380 as 30 more, 390 as 40 more 400 as 50 more until 440 is 90 more. At 440 counting to 442 is 2 more. The student can now combine all the counted up values of 7+90+2 for a total of 99 more.
Difference= 7+90 +2 = 99
Each word problem has a written model for the number model and then each step in the strategy can be shown in the workspace in each template. The written steps to solve are a way to track thinking for students and to express conceptual understanding. The step-by-step solving can be reduced to traditional methods once the student has strong conceptual understanding. The tedious process of unpacking each stage is time consuming in the beginning, but is meaningful and supports students' understanding of place value concepts.
## Activity 3: Multiplication of multidigit numbers
The multidigit multiplication activity is fun for students to create their own problems from advertisements or sports pages of a newspaper. An example would be for a student to find an item for sale in a local advertisement. An electronics store may have computers on sale for \$549. If the student wants to determine the cost for an entire classroom of 24 students to have a personal computer how much would all the computers cost? This problem provides real world examples and practices the skill of multiplication of multidigit numbers. The student could solve using the grid or by putting each with each by place value. For the grid the student could create a 3 x 2 grid and place the 500, 40 and 9 (from the expanded form of 549 as 500 + 40 + 9) along the top of the grid. Then place the 20 and 4 along the vertical side. Fill in each grid space with the product from the vertical and horizontal numbers on the grid. Add all the grid products to find the total. The other option is to solve with the "each with each" strategy of expanded both numbers and multiplying each factor place value by the second expanded number.
The students can find and create their own problems and then solve (as shown above) with each step in the process. Taking the steps to combine place values in the addition of the products reinforces the combination of like terms in place values and also the trading 10 to the next place value to exchange for one. In the problem above, 110 was broken into 100 +10 and then combined 1 hundred with the 900. Even as the most advanced problem in the activities the basic skills of exchanging and place value concepts are practice and reinforced in this type of problem.
An example of a sports problem could be a Chicago Bear running back ran for an average of 95 yards a game for 11 games. How many total yards did he run in the 11 games using his average yards per game? In this example, the student can use the same two strategies of the grid or the each with each strategy. If the grid is chosen the 90 and 5 can be placed horizontally and the 10 and 1 can be placed on the grid vertically. In the case of each with each 90 is multiplied by 10 and by 1 and the 5 is multiplied by the 10 and the 1, so each term is multiplied by the other factors terms.
Each of these activities uses examples that students can find in real life and the problems make math operations meaningful. The strategies to solve have students unpack the problems into place value expanded form and practice combining like terms and trading. The combination and trading are at the root concepts of place value understanding. The step-by-step problem solving allows the students' to articulate each part of the process in the operation and can support students when explaining how and why a solution to a problem is found.
## Materials List
Base ten blocks – Base ten block manipulatives traditionally have ones cubes, ten cubes as one tens rod and one hundred cubes together on a square flat representing hundreds. These concrete objects allow children to move and manipulate number models.
Digit cards – Digit cards are representative of the place value numbers in expanded form. The digit cards have each place value with 1-9 digits on each card. The thousands place digit cards have 1,000, 2,000, 3,000, 4,000, 5,000, 6,000, 7,000, 8,000, and 9,000. The number 4,562 can be shown by taken the digit cards 4,000 and 500 and 60 and 2.
Number lines – Number lines can be traditional measurement style number lines or can be created to show skip counting of numbers depending on a student's needs.
Rods – Rods are place value representatives similar to base ten blocks but are linear models. The one cube and ten rod are similar to the base ten blocks but a hundreds rod is a long rod that shows the length of one hundred.
Activity Sheets – The activity sheets below are paired with activities 1 and 2 from the section above.
Activity 1: Game sheet
Activity 1: Score sheet
Activity 2: Addition of Multidigit numbers
Template
# Works Cited
Carroll, William. Karen C. Fusion, and Jane V. Drueck. "Achievement Results for Second and Third Graders Using the Standards-Based Curriculum Everyday Mathematics." Journal for Research in Mathematics Education, May 2000.
Cross, Christopher T., Taniesha A. Woods, and Heidi A. Schweingruber. Mathematics learning in early childhood: paths toward excellence and equity. Washington, DC: National Academies Press, 2009.
Everyday Mathematics Grade 4. Chicago: Wright Group/McGraw-Hill, 2004.
Fuson, Karen C. "Issues in Place-Value and Multidigit Addition and Subtraction Learning and Teaching." Journal for Research in Mathematics Education
Ho, Connie Suk-Han , and Fanny Sim-Fong Cheng. "Training in Place-Value Concepts Improves Childrenâé™s Addition Skills." Contemporary Educational Psychology, October 1, 1997. http://www.sciencedirect.com/science/article/pii/S0361476X97909474 (accessed July 11, 2011).
Howe, Roger, and Susanna Epp. "Taking Place Value Seriously: Arithmetic, Estimation, and Algebra." http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf. http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf (accessed May 7, 2011).
Howe, Roger. "Developing and Interpreting Multiplication and Division with the
Number Line."
Howe, Roger. "The number line and addition and subtraction."
Ma, Liping. Knowing and teaching elementary mathematics teachers' understanding of fundamental mathematics in China and the United States. Mahwah, N.J.: Lawrence Erlbaum Associates, 1999.
Varelas, Maria, and Joe Becker. "Children's Developing Understanding of Place Value: Semiotic Aspects." Cognition And Instruction 15, no. 2 (1197): 265-286. http://www.jstor.org/stable/3233767 (accessed June 27, 2011).
# Implementing District Standards
STATE Goal 6: Demonstrate and apply a knowledge and sense of numbers, including numeration and operations (addition, subtraction, multiplication, division), patterns, ratios and proportions.
B. Investigate, represent and solve problems using number facts, operations (addition, subtraction, multiplication, division) and their properties, algorithms and relationships.
6.B.2 Solve one- and two-step problems involving whole numbers, fractions and decimals using addition, subtraction, multiplication and division.
C. Compute and estimate using mental mathematics, paper-and-pencil methods, calculators and computers.
6.C.2a Select and perform computational procedures to solve problems with whole numbers, fractions and decimals.
6.C.2b Show evidence that computational results using whole numbers, fractions and decimals are correct and/or that estimates are reasonable.
Illinois state goals under goal six address number sense. Illinois state goals are arranged by grade band. For instance, the above mathematics standards are for "late elementary." The state of Illinois has five goals focused on number sense, estimation and measurement, algebra and analytical methods, geometry, and data analysis and probability.
# Notes
1. Fuson, Karen C. "Issues in Place-Value and Multidigit Addition and Subtraction Learning and Teaching." Journal for Research in Mathematics Education
2. Howe, Roger, and Susanna Epp. "Taking Place Value Seriously: Arithmetic, Estimation, and Algebra." http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf. http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf (accessed May 7, 2011).
3. Ho, Connie Suk-Han, and Fanny Sim-Fong Cheng. "Training in Place-Value Concepts Improves Childrenâé™s Addition Skills." Contemporary Educational Psychology, October 1, 1997. http://www.sciencedirect.com/science/article/pii/S0361476X97909474 (accessed July 11, 2011).
4. Howe, Roger. "The number line and addition and subtractions." (Lecture)
5. Howe, Roger, and Susanna Epp. "Taking Place Value Seriously: Arithmetic, Estimation, and Algebra." http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf. http://www.maa.org/pmet/resources/PVHoweEpp-Nov2008.pdf (accessed May 7, 2011). | 10,086 | 45,936 | {"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} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.961342 |
http://www.google.com/patents/US5065352?dq=552685 | 1,419,591,789,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447548791.91/warc/CC-MAIN-20141224185908-00083-ip-10-231-17-201.ec2.internal.warc.gz | 245,246,360 | 19,742 | Patents
Publication number US5065352 A Publication type Grant Application number US 07/555,689 Publication date Nov 12, 1991 Filing date Jul 23, 1990 Priority date Aug 16, 1989 Fee status Lapsed Publication number 07555689, 555689, US 5065352 A, US 5065352A, US-A-5065352, US5065352 A, US5065352A Inventors Hiraku Nakano Original Assignee Matsushita Electric Industrial Co., Ltd. Export Citation Patent Citations (5), Referenced by (37), Classifications (10), Legal Events (6) External Links:
Divide apparatus employing multiplier with overlapped partial quotients
US 5065352 A
Abstract
An approximate reciprocal M of a divisor D0 is determined in dependence on predetermined higher order bits of the divison D0. A dividend N0 is multiplied by the approximate reciprocal M to derive a quotient. During this multiplication, the following reiterative calculations are executed.
`A=1-D0 ×M`
`Q1 +N1 =N0 ×M`
`Qi +Ni =A×Q1-1 +Ni-1 (i≧2)`
where Qi denotes a partial quotient, and Ni denotes a number obtained by subtracting Qi from an (i-1)-th partial remainder multiplied by M. In addition, the following calculation is executed.
`QLAST+1 +NLAST+1 =A×(QLAST +X)+NLAST `
where QLAST denotes a last partial quotient, and X denotes a number having a magnitude corresponding to a lower bit of the last partial quotient QLAST. A final quotient Q* is calculated from the partial quotients by reference to the following equation.
`Q*=Q1 +Q2 +. . . +QLAST +Y(the highest digit of QLAST `
+1)
The final quotient Q* is outputted.
Images(5)
Claims(3)
What is claimed is:
1. A divide apparatus comprising:
means for determining an approximate reciprocal M of a divisor D0 in dependence on given higher bits of the divisor D0 ;
multiplying means for multiplying a dividend N0 by the approximate reciprocal M to derive a quotient;
wherein said multiplying means comprises:
means for executing the following reiterative calculations:
`A=1-D0 ×M`
`Q1 +N1 =N0 ×M`
`Qi +Ni =A×Qi-1 +Ni-1 (i≧2)`
where A denotes a coefficient used as a factor used as a multiplicand of Qi-1 when Qi is calculated therefrom, Qi denotes a partial quotient, and Ni denotes a number obtained by subtracting Qi from an (i-1)-th partial remainder multiplied by M;
means for executing the following calculation:
`QLAST+1 +NLAST+1 =A×(QLAST +L)+NLAST `
where QLAST denotes a last partial quotient, and L denotes a number having a magnitude corresponding to a lowest bit of the last partial quotient QLAST ;
a plurality of multiple generators for scanning multipliers for multiplicands thereby to generate multiples of said multiplicands;
a plurality of carry-save adders connected to the multiple generators and having tree structures for outputting a partial sum and a partial carry;
a carry propagate adder connected to the carry-save adders for adding the partial sum and the partial carry;
means for feeding the partial sum and the partial carry to the multiple generators as a multiplier with respect to the partial quotient Qi-1 ; and
means for inputting the partial sum and the partial carry into the carry-save adders with respect to the number Ni-1 ; the divide apparatus further including:
means connected to said multiplying means for storing the respective partial quotients Q1, Q2, . . . , QLAST ; and
means for calculating a final quotient Q* from the partial quotients by reference to the following equation:
`Q*=Q1 +Q2 +. . . +QLAST +Y(the highest bit of QLAST+1)`
and for outputting the final quotient Q*.
2. A divide apparatus comprising:
means for determining an approximate reciprocal M of a divisor D0 in dependence on given higher bits of the divisor D0 ;
multiplying means for multiplying a dividend N0 by the approximate reciprocal M to derive a quotient;
wherein said multiplying means comprises:
means for executing the following reiterative calculations:
`A=1-D0 ×M`
`Q1 +N1 =N0 ×M`
`Qi +Ni =A×Qi-1 +Ni-1 (i≧2)`
where A denotes a coefficient used as a factor used as a multiplicand of Qi-1 when Qi is calculated therefrom, Qi denotes a partial quotient, and Ni denotes a number obtained by subtracting Qi from an (i-l)-th partial remainder multiplied by M;
means for executing the following calculation:
`QLAST+1 +NLAST+1 =A×(QLAST +L)+NLAST `
where QLAST denotes a last partial quotient, and L denotes a number having a magnitude corresponding to a lowest bit of the last partial quotient QLAST ;
multiplying devices, whose number is equal to a number of times of multiplication, for executing reiterative calculations;
a first multiplying unit for calculating a product of the divisor D0 and the approximate reciprocal M;
a second multiplying unit for calculating a product of the dividend N0 and the approximate reciprocal M, wherein the first and second multiplying units are arranged in parallel, and
wherein the multiplying devices are arranged in series with the first and second multiplying units; and
a plurality of registers connected between the multiplying devices and the first and second multiplying units for providing synchronization among the multiplying devices and the first and second multiplying units; the divide apparatus further including:
means connected to said multiplying means for storing the respective partial quotients Q1, Q2, . . . , QLAST ; and
means for calculating a final quotient Q* from the partial quotients by reference to the following equation:
`Q*=Q1 +Q2 +. . . +QLAST +Y(the highest bit of QLAST+1)`
and for outputting the final quotient Q*; and
a plurality of delay registers for delaying and storing the number A in synchronism with the reiterative calculations.
3. A divide apparatus comprising:
means for determining an approximate reciprocal M of a divisor D0 in dependence on given higher bits of the divisor D0 ;
multiplying means for multiplying a dividend N0 by the approximate reciprocal M to derive a quotient;
wherein said multiplying means comprises:
means for executing the following reiterative calculations:
`A=1-D0 ×M`
`Q1 +N1 =N0 ×M`
`Qi +Ni =A×Qi-1 +Ni-1 (i≧2)`
where A denotes a coefficient used as a factor used as a multiplicand of Qi-1 when Qi is calculated therefrom, Qi denotes a partial quotient, and Ni denotes a number obtained by subtracting Qi from an (i-1)-th partial remainder multiplied by M;
means for executing the following calculation:
`QLAST+1 +NLAST+1 =A×(QLAST +L)+NLAST `
where QLAST denotes a last partial quotient, and L denotes a number having a magnitude corresponding to a lowest bit of the last partial quotient QLAST ;
multiplying devices, whose number is equal to a number of times of multiplication, for executing reiterative calculations;
a first multiplying unit for calculating a product of the divisor D0 and a two's complement of the approximate reciprocal M;
a second multiplying unit for calculating a product of the dividend N0 and the approximate reciprocal M, wherein the first and second multiplying units are arranged in parallel, and
wherein the multiplying devices are arranged in series with the first and second multiplying units; and
a plurality of registers connected between the multiplying devices and the first and second multiplying units for providing synchronization among the multiplying devices and the first and second multiplying units; the divide apparatus further including:
means connected to said multiplying means for storing the respective partial quotients Q1, Q2, . . . , QLAST ; and
means for calculating a final quotient Q* from the partial quotients by reference to the following equation:
`Q*=Q1 +Q2 +. . . +QLAST +Y(the highest bit of QLAST+1)`
and for outputting the final quotient Q*; and
a plurality of delay registers for delaying and storing the number A in synchronism with the reiterative calculations.
Description
BACKGROUND OF THE INVENTION
1. Field of the Invention
This invention relates to a divide apparatus usable in various systems such as data processing systems.
2. Description of the Prior Art
U.S. Pat. No. 4,707,798 corresponding to Japanese published unexamined patent application 60-142738 discloses a divide apparatus for use in a data processing system. The divide apparatus of U.S. Pat. No. 4,707,798 includes a partial quotient correction circuit. The partial quotient correction circuit is required to execute both addition and substraction, so that the circuit includes an adder and a subtracter. Thus, the structure of the divide apparatus of U.S. Pat. No. 4,707,798 is complicated.
It is desirable to increase the speed of execution of division relative to the speed in the prior art apparatus.
SUMMARY OF THE INVENTION
It is an object of this invention to provide an improved divide apparatus.
According to this invention, a divide apparatus comprises means for determining an approximate reciprocal M of a divisor D0 in dependence on predetermined higher order bits of the divisor D0 ; multiplying means for multiplying a dividend N0 by the approximate reciprocal M to derive a quotient; said multiplying means including means for executing the following reiterative calculations:
`A=1-D0 ×M`
`Q1 +N1 =N0 ×M`
`Qi +Ni =A×Qi-1 +Ni-1 (i≧2)`
where Qi denotes a partial quotient, and Ni denotes a number obtained by subtracting Qi from an (i-1)-th partial remainder; the multiplying means including means for executing the following calculation:
`QLAST+1 +NLAST+1 =A×(QLAST +X)+NLAST `
where QLAST denotes a last partial quotient, and X denotes a number having a magnitude corresponding to a lowest bit of the last partial quotient QLAST ; means connected to said multiplying means for storing the respective partial quotients Q1, Q2, . . . , QLAST ; and means for calculating a final quotient Q* from the partial quotients by reference to the following equation:
`Q*=Q1 +Q2 + . . . +QLAST +Y(the highest digit of QLAST+1)`
and for outputting the final quotient Q*.
BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is a block diagram of a divide apparatus according to a first embodiment of this invention.
FIGS. 2 and 3 are diagrams showing the accuracy of an approximate reciprocal of a divisor in the divide apparatus of FIG. 1.
FIGS. 4 and 5 are diagrams showing conditions of various numbers which appear during an example of actual operation of the divide apparatus of FIG. 1.
FIG. 6 is a block diagram of a divide apparatus according to a second embodiment of this invention.
THEORETICAL BASE OF THE INVENTION
The reason why the partial quotient correction circuit of U.S. Pat. No. 4,707,798 requires the subtracter is that a partial quotient can have either a positive sign or a negative sign. A good way of removing the subtracter is forcing the sign of a partial quotient to be always positive.
Whether a fixed point notation without an exponent term or a floating point notation with an exponent term is used for representing a number, in order to perform a division of a P-base number, first the divisor and the dividend are normalized so that they can be represented by the following equations (1) and (2), and an intermediate quotient is calculated wherein D0 denotes a divisor after being normalized and N0 denotes a dividend after being normalized. ##EQU1## where D0,k is zero or a positive integer; O≦D0,k<P; and ##EQU2## where N0,k is zero or a positive integer; O≦N0,k<P; and N0,0 ≠0.
At this time, the obtained intermediate quotient Q is within a range of the following formula (3), that is, either in a normalized form or in an underflowing-by-one-lower-digit form.
`P-1 <Q<P (3) ##EQU3##`
In the case of a fixed point notation, after the calculation of the intermediate quotient, the number of digit shift counts required for normalizing the divisor is subtracted by the number of digit shift counts required for normalizing the dividend. If the resultant difference is positive (digit shift to the left is defined as being positive), the intermediate quotient Q is shifted to the left by the difference number of digit shift counts so that a final quotient can be obtained. If the difference is negative, the intermediate quotient Q is shifted to the right by the difference number of digit shift counts so that a final quotient can be obtained.
In a floating point notation, a first difference is obtained by subtracting the exponent term of the divisor from the exponent term of the dividend and a second difference is obtained by subtracting the number of digit shift counts required for normalizing the fixed point term of the divisor from the number of digit shift counts required for normalizing the dividend, and the result of addition between the first difference and the second difference is used as an exponent term of the division result. Furthermore, if the intermediate quotient is in an underflow form, 1 is subtracted from the exponent term to normalize the fixed point term. As a result of these processes, a final quotient can be obtained.
Since a sign of the quotient can be algebraically determined by the signs of the divisor and the dividend, the intermediate quotient may be calculated by using, if required, the absolute and normalized values of the divisor and the dividend. If the final quotient is negative, the intermediate quotient is changed to have a desired representation.
With the previously-mentioned assumptions, the divisor and the dividend will be handled hereinafter in the absolute and normalized form as in the formulas (1) and (2). Symbols used herein will now be explained.
M: approximate reciprocal of a divisor D0
Qi : i-th partial quotient
Ri : i-th partial remainder where R0 =N0
Ni : number obtained by subtracting Qi from the product of Ri-1 and M
A: multiplicand number multiplied by Qi when Qi+1 +Ni+1, which is M times as large as the i-th partial remainder Ri, is calculated from Qi and Ni
α+1: the number of digits of a partial quotient wherein partial quotients neighboring in sequence number overlap each other by one digit
Q: correct quotient which becomes recurring decimals and has the infinite number of digits if the dividend can not be divided by the divisor without leaving a remainder
QLAST : last (LAST -th) partial quotient
Q*: number obtained by taking out the lowest digit of QLAST and higher digits from Q, that is, number obtained by omitting digits, which are lower in position than the lowest digit of QLAST, from Q
X: number in the lowest digit of QLAST and having a magnitude of 1
Y: the highest digit of QLAST+1 which has the same position as X when a partial remainder RLAST multiplied by M, that is, QLAST+1 +NLAST+1, is calculated with QLAST +X being used as a partial quotient.
A sufficient condition for obtaining (α+1) digits in a partial quotient is determined by selecting M so as to suffice the following formula (4).
`1-P-(α+2) <D0 ×M<1 (4)`
Before reiterative calculations are started, the calculations shown by the following formulas (5) and (6) are performed.
`A=1-D0 ×M (5)`
`Q1 +N1 =N0 ×M (6)`
In the reiterative calculations, the partial quotient is solved in such a way as shown in the following formulas (7).
`Qi +Ni =A×Qi-1 +Ni-1 (2≦i≦LAST)(7)`
Finally, the calculation shown by the following formula (8) is performed and the number Q* corresponding to a final quotient is calculated by referring to the following formula (9) to examine the presence and the absence of a contribution of QLAST+1 and a lower part to Q*.
`QLAST+1 +NLAST+1 =A×(QLAST +X)+NLAST(8)`
`Q*=Q1 +Q2 + . . . . . . +QLAST +Y (9)`
A final quotient can be obtained from the previously-mentioned processes. This is clarified by proving the following items (A) to (C).
ITEM (A)
"A" can be used for arbitrary i≧2, and the number obtained through the formula (7) equals the number obtained by multiplying the (i-th)-th partial remainder Ri-1 by the approximate reciprocal M of the divisor.
ITEM (B)
Q1 is positive and has (α-1)˜(α+1) digits. When Q1 is compared with the correct quotient Q at a position equal to or above P-α, it is equal to or smaller by P-α. In other words, there are the following relations (10) and (11).
`P-2 <Q1 <P (10)`
`O<R1 =(Q2 +N2)/M<D0 ×P-α ×2(11)`
ITEM (C)
Qi (i≧2) is either positive or zero, and has (α+1) digits at the position of P-(i-1)α ˜P-iα. When Qi is compared with the correct quotient Q at the position corresponding to Qi, it is equal to or smaller by P-iα. In other words, there are the following relations (12) and (13).
`O<Qi +Ni <P-(i-1)α+1 (12)`
`O<Ri =(Qi+1 +Ni+1)/M<D0 ×P-iα ×2(13)`
DEMONSTRATION OF ITEM (A)
For i=2, ##EQU4## so that the item (A) is proved.
If it is assumed that the item (A) is proved for i=k and that Qk +Nk =Rk-1 ×M=A×Qk-1 +Nk-1, then ##EQU5## so that the item (A) is also proved for i=k+1. Since the item (A) has previously been proved for i=2, the item (A) is proved for any arbitrary i≧2 by introducing a mathematical induction method.
DEMONSTRATION OF ITEM (B)
Each term in the formula (4) is multiplied by the correct quotient Q, and the following relation (14) results.
`Q-Q×P-(α+2) <Q1 +N1 =N0 ×M<Q(14)`
From the formulas (3) and (14), the relation "P-2 <Q1 <P" is established.
`O<R1 ×M=Q2 +N2 =(1-D0 ×M)×Q1 ×N1 <P-α (15)`
This is because Q1 <P, O<N1 <P-α, and O<1-D0 ×M<P-(α+2) from the formula (4). In addition, from the formula (4), O<1/M<D0 /{1-P-(α+2) }<2D0. Thus, the outer term of the inequality in the formula (15) is multiplied by 2D0 while the inner term is multiplied by 1/M, and the following formula is obtained.
`O<R1 =(Q2 +N2)/M<2D0 ×P-α =D0 ×P-α ×2`
DEMONSTRATION OF ITEM (C)
For i=2, the following relation is obtained from the formula (15).
`O<Q2 ×N2 <P-α <P-α+1`
Thus, the formula (12) can be established.
`O<R2 ×M=(1-D0 ×M)×Q2 +N2 `
` <P-(α+2) ×P-α +N2 `
` <P-2(α+1) +P-2α =P-2α (1+P-2)`
From the formula (4), the following relation is established.
`O<1/M<D0 /{1-P-(α+2) }`
The outer term of the above-mentioned inequality is multiplied by D0 /{1-P-(α+2) } and the inner term is multiplied by 1/M, and thereby the following formula (16) is obtained. ##EQU6## It should be noted that the function f(x)=(1+x-2)/{1-x-(α+2) } monotonically decreases in the variation region "x≧2" and reaches the maximal value "5/(4-2-α)" at x=2, and that the maximal value is a monotonically decreasing function with respect to α in the variation region α≧1 and is equal to 10/7 at α=1. Thus, the formula (13) can be established for i=2. It is now assumed that the following relations (17) and (18) are satisfied for i=k.
`O=Qk +Nk <P-(k-1)α+1 (17) ##EQU7## In this case, the following formula is obtained. ##EQU8## Thus, the formula (12) can be established also for i=k+1. ##EQU9## From the formula (4), O<1/M<D0 /{1-P-(α+2) }. Thus, the outer term of the above-mentioned inequality is multiplied by D0 /{1-P-(α+2) } and the inner term is multiplied by 1/M, and thereby the following formula (19) is obtained. ##EQU10## It should be noted that the function f(x)=(1+x-1)/{1-x-(α+2) } monotonically decreases in the variation region "x≧2" and reaches the maximal value "6/(4-2-α)" at x=2, and that the maximal value is a monotonically decreasing function with respect to α in the variation region α≧1 and is equal to 12/7 at α=1. Thus, the formula (13) can be established also for i=k+1. Since the validity of the formulas (12) and (13) has previously been proved for i=2, the formulas (12) and (13) can be established also for an arbitrary i≧2 by introducing a mathematical induction method.`
It is understood from the item (C) that the result of the sum of the partial quotients calculated from the formulas (6) and (7) is compared with the correct quotient Q at a position equal to or above the position of the lowest digit of QLAST, it can be smaller by P-LASTXα. This makes it possible to understand that P-LASTXα is multiplied by D0 ×M and is concealed in NLAST. Accordingly, P-LASTXα is added to QLAST while P-LASTXα is subtracted from NLAST, and the remainder "RLAST /M" is obtained and thereby the following formula (20) is obtained. ##EQU11## When it is necessary to correct QLAST with P-LASTXα, the remainder given by the equation (20) is positive or zero. When the correction is unnecessary, the remainder given by the equation (20) is negative. The boundary of the judgment about the presence and absence of the correction of QLAST is shifted from zero to P-LASTXα in place of the subtraction of P-LASTXα from NLAST in the formula (8), and the use of the formula (9) produces a quotient which agrees with the correct quotient Q at positions equal to and above P-LASTXα.
DESCRIPTION OF THE FIRST PREFERRED EMBODIMENT
FIG. 1 shows a divide apparatus according to a first embodiment of this invention. With reference to FIG. 1, a control circuit 11 carries out the control of a whole system including a dividend register 12, a divisor register 13, a table information storage unit 14, a multiplicand selection circuit and register 15, multiplier selection circuits 16 and 17, multiple generation circuits 18 and 19, selection circuits 20, 21, and 22, carry-save adder trees 23 and 24, carry-save adders 25 and 26, a partial carry register 27, a partial sum register 28, a carry propagate adder 29, registers 30-34 for Q1 -Q5, a carry propagate adder 35, and a division result register 36.
The division is performed in the following sequence. An approximate reciprocal M is read out from the table information storage unit 14 by using given higher bits of a normalized divisor D0 set in the divisor register 13 as an address for access to the table information storage unit 14. An example of a way of determining stored information in the table information storage unit 14 will be shown hereinafter.
With respect to the binary divisor expressed by the formula (1), the approximate reciprocal M is obtained by referring to the following equation (21). ##EQU12## The obtained approximate reciprocal M is expressed as: ##EQU13## Since the first term of the right side of the equation (22) is fixed and it is good for the third term that D0,14 is added as the lowest bit of a multiplier, it is sufficient that 15 bits from M2 to M16 are stored in the table. With respect to the accuracy of the approximate reciprocal determined by the equations (21) and (22), the following equation (23) is satisfied.
`0.FF808F8≦D0 ×M<1(hexadecimal notation) (23)`
The value of D0 ×M was checked by using a computer in the following way. The approximate reciprocal M is a constant in the following interval: ##EQU14## As shown in FIG. 2, in the interval "1≦D0 <2", the approximate reciprocal M is a step function having 214 =16384 line segments of a 2-14 length. As shown in FIG. 3, the graph for D0 ×M has 16384 line segments in the form of a saw tooth.
Therefore,
Lower limit (also the minimum value) of D0 ×M: the minimum value of the left ends of the 16384 line segments
Upper limit of D0 ×M: the maximum value of the right ends of the 16384 line segments
The reciprocal read out from the table information storage unit 14 is added with 2-1 and D0,14, and first the multiplier selection circuit 16 outputs -M. The multiplication between -M and the divisor D0 set in the multiplicand selection circuit and register 15 which relates to the equation (5) is performed by the multiple generation circuit 18, the carry-save adder tree 23, the carry-save adders 25 and 26, the partial carry register 27, the partial sum register 28, and the carry propagate adder 29.
A partial carry and a partial sum of D0 ×(-M) are set in the partial carry register 27 and the partial sum register 28 respectively. At the same time, the normalized dividend N0 outputted from the dividend register 12 is set in the dividend selection circuit 15, and the multiplier selection circuit 16 selects the approximate reciprocal M. The multiplication "N0 ×M" of the equation (6) is performed similarly to the execution of the multiplication "D0 ×(-M)". Thus, a partial carry and a partial sum of N0 ×M are set in the partial carry register 27 and the partial sum register 28 respectively. At the same time, the product of N0 ×M is set in the multiplicand selection circuit and register 15 with a carry of two degrees or greater being made zero (this is equivalent to the addition of D0 ×(-M)+1). In the calculation of the product of D0 ×(-M) and the product of N0 ×M, the multiplier selection circuit 17 and the selection circuit 22 output "0", and therefore the output of the carry-save adder tree 24 is "0" for both of the partial carry and the partial sum and the selection circuits 20 and 21 select the output of the multiple generation circuit 18.
Subsequently, the reiterative calculations shown in the equation (7) are performed while Qi-1 and Ni-1 are held in the form of a partial carry and a partial sum as follows. The partial carry and the partial sum of Qi-1 which are outputted from the partial carry register 27 and the partial sum register 28 are selected by the multiplier selection circuits 16 and 17 respectively. The partial carry and the partial sum of Ni-1 which are outputted from the partial carry register 27 and the partial sum register 28 are selected by the selection circuits 20 and 22 respectively. A carry from Ni-1 to Qi-1 upon the addition of the partial carry and the partial sum is calculated from a carry look-ahead circuit contained in the carry propagate adder 29. In the presence of such a carry, the selection circuit 21 sets a negative number into given higher bits of an originally generated multiple of "A". In this case, Ni-1 is in an over-added state with regard to the partial carry and the partial sum of Ni-1, and the negative number serves to cancel the carry from Ni-1 to Qi-1. In FIG. 5, "1111" in the highest bit and the subsequent three higher bits in the part SEL 21 corresponds to the negative number.
Partial quotients outputted from the carry propagate adder 29 are sequentially set in the registers 30-34. Finally, the calculation of the equation (8) is performed. The calculation of the equation (8) is similar to the calculation of the equation (7) except for the following points. After the multiplier selection circuit 16 adds "1" to the end of the partial carry of QLAST, the addition of the equation (9) is executed by the carry propagate adder 35 and the resultant quotient Q* is set in the division result register 36. The addition by the carry propagate adder 35 differs from a normal addition of two numbers as follows. In the addition by the carry propagate adder 35, α bits are used as a group unit, and it is sufficient that the highest bits of respective partial quotients, the remaining bits thereof, and Y are added as group carries, addition results corresponding to bits of the two numbers, and an initial carry respectively. This is because the lowest digit of each of the partial quotients can be smaller by 1 but is never smaller by 2 as understood from the item (C), and a double carry from a further lower position will not occur in the case where the highest bits of the respective partial quotients are "1".
FIG. 4 and FIG. 5 show conditions where the previously-mentioned calculating processes are executed for actual numbers. With reference to FIG. 4 and FIG. 5, in the case where the multiplier selection circuit 16 outputs a non-zero value as the lowest bit of a multiplier, the multiple generation circuit 18 generates a multiple which is equivalent to the addition of "1" to the bit having a position greater than the lowest bit of the multiplier by one. In FIG. 4 and FIG. 5, the bits of Qi are separated into 13-bit groups contained in respective brackets, and these are made into groups of 12 bits corresponding to (α+1) digits in the item (B) and the item (C) after the partial carry and the partial sum are added. Under conditions where Qi remains separated into a partial carry and a partial sum, the highest bit of the related 13-bit group is "0" when both the partial carry and the partial sum are positive or zero. Under the same conditions, when one of the partial carry and the partial sum is negative, the highest bit of the negative number is "1" and the highest bit constitutes a sign bit.
A description will be made on an actual example of the division in which a divisor D0 and a dividend N0 are set as follows.
`D0 =1.2345680000000(hexadecimal notation)`
`N0 =1.23456789ABCDE(hexadecimal notation)`
(i) The bits of D0 whose positions are equal to or above 2-13 are expressed as:
`1.2340(hexadecimal)=1.0010 0011 0100 0(binary)`
In addition, ##EQU15## Therefore, M is given as below also in view of D0,14 =1. In the following, numbers without being followed by a word of a type of notation are expressed in a hexadecimal notation, and partial carry values and partial sum values are omitted for the simplicity of the description. ##EQU16##
With respect to the mantissa of a quotient of the double-accuracy division of floating point numbers which is shown in "IEEE Standard for Binary Floating-Point Arithmetic" ANSI/IEEE Std 754-1985, the 2-53 bit and the 2-54 of Q* are used as a guard bit and a round bit respectively, and the logical or between the 2-55 bit of Q* and the part of 255 •(Q6 +N6) lower than the decimal point is used as a sticky bit.
The divide apparatus of this embodiment dispenses with a subtracter which would be necessary for the correction of a quotient in the prior art. In addition, the reiterative calculations in the division are executed in the form of a partial carry and a partial sum, and thus the time of the execution of the division can be short.
DESCRIPTION OF THE SECOND PREFERRED EMBODIMENT
FIG. 6 shows a vector divide apparatus according to a second embodiment of this invention. With reference to FIG. 6, a control circuit 211 carries out the control of a whole system including operand buffers 212 and 213, a dividend register 214, a divisor register 215, a dividend delay register 216, a divisor delay register 217, a table information storage unit 218, an approximate reciprocal register 219, an inverter 220, multiplying devices 221, 223, 225, 229, 234, 240, and 247, registers 222, 224, 226, 230, 235, 241, 245, and 255, delay registers 227, 228, 231, 232, 233, 236, 237, 238, 239, 242, 243, 244, 245, 246, 249, 250, 251, 252, and 253, and a carry propagate adder 254.
A dividend vector element and a divisor vector element are transferred from memories (not shown) or vector registers (not shown) to the operand buffers 212 and 213 by the control operation of the control circuit 211. A division result vector element is transferred from the register 255 to memories (not shown) or vector registers (not shown) by the control operation of the control circuit 211.
A description will now be made on the operation of the divide apparatus. The operation of the divide apparatus is separated into the following several successive stages.
During the first stage of the operation, a dividend vector element and a divisor vector element are set in the dividend register 214 and the divisor register 215 respectively. An approximate reciprocal M of the divisor is read out from the table information storage unit 218 in response to given higher bits of the divisor.
During the second stage of the operation, the dividend vector element and the divisor vector element from the dividend register 214 and the divisor register 215 are set in the delay registers 216 and 217 respectively, and the approximate reciprocal M is set in the approximate reciprocal register 219. In addition, the multiplying device 221 executes the multiplication "N0 ×M", and the inverter 220 converts the approximate reciprocal M into a corresponding negative number. Then, the multiplying device 223 executes the multiplication "D0 ×M".
During the third stage of the operation, the result "Q1 +N1 " of the multiplication by the multiplying device 221 is set in the register 222. In addition, the number A which is obtained by adding "1" to the result of the multiplication by the multiplying device 223 is set in the register 224. The addition of "1" is realized by such a process that the part of the multiplication result which has positions equal to and above 2° are made equal to 0. Next, the multiplying device 225 executes the calculation "A×Q1 +N1 ".
During the fourth stage of the operation, the result "Q2 +N2 " of the calculation by the multiplying device 225 is set in the register 226, and A and Q1 are set in the delay registers 227 and 228 respectively. Next, the multiplying device 229 executes the calculation "A×Q2 +N2 ".
During the fifth stage of the operation, the result "Q3 +N3 " of the calculation by the multiplying device 229 is set in the register 230, and A, Q1, and Q2 are set in the delay registers 231, 232, and 233 respectively. Next, the multiplying device 234 executes the calculation "A×Q3 +N3 ".
During the sixth stage of the operation, the result "Q4 +N4 " of the calculation by the multiplying device 234 is set in the register 235, and A, Q1, Q2, and Q3 are set in the delay registers 236, 237, 238, and 239 respectively. Next, the multiplying device 240 executes the calculation "A×Q4 +N4 ".
During the seventh stage of the operation, the result "Q5 +N5 " of the calculation by the multiplying device 240 is set in the register 241, and A, Q1, Q2, Q3, and Q4 are set in the delay registers 242, 243, 244, 245, and 246 respectively. Next, the multiplying device 247 executes the calculation "A×(Q5 +2-55)+N5 ".
During the eighth stage of the operation, the result "Q6 +N6 " of the calculation by the multiplying device 247 is set in the register 248, and Q1, Q2, Q3, Q4, and Q5 are set in the delay registers 249, 250, 251, 252, and 253 respectively. Next, the carry propagate adder 254 executes the calculation "Q1 +Q2 +Q3 +Q4 +Q5 +Y(the highest digit of Q6)".
During the ninth stage of the operation, the final quotient Q* being the result of the calculation by the carry propagate adder 254 is set in the register 255.
The previously-mentioned nine stages are executed for respective times each corresponding to one machine cycle. Accordingly, in the case where a dividend vector element and a divisor vector element are fed for each of one machine cycles, after a first final quotient is derived and set in the register 255, a quotient vector element is outputted from the register 255 for each of one machine cycles.
The bit lengths of the multiplicand A, the multiplier Qi-1, and the number Ni-1 remain fixed during the reiterative calculations in the division. Accordingly, the divide apparatus can be formed by using a plurality of same-type LSIs. In this case, the cost of the divide apparatus can be low. In addition, the multiplying devices which execute the reiterative calculations can be composed of repeatedly-used macro-cells, and the cost of the divide apparatus can be low.
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Classifications
U.S. Classification708/654
International ClassificationG06F7/535, G06F7/52, G06F7/53, G06F7/483
Cooperative ClassificationG06F7/49957, G06F7/535, G06F7/4873, G06F2207/5355
European ClassificationG06F7/535
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Effective date: 19900718 | 10,309 | 41,380 | {"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} | 2.59375 | 3 | CC-MAIN-2014-52 | latest | en | 0.820071 |
http://bootmath.com/what-is-the-purpose-of-implication-in-discrete-mathematics.html | 1,532,366,746,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00033.warc.gz | 54,407,007 | 5,997 | # What is the purpose of implication in discrete mathematics?
I would be obliged if you can show me an example of a truth table for implication where there is a also a real life aspect to it. (i.e., where would someone use the scenario to make F->F = T and also the same for the remaining 3 cases).
However, this one scenario should be able to be adjusted to fit all three. Hopefully that makes sense.
I am just trying to understand the concept of implication.
#### Solutions Collecting From Web of "What is the purpose of implication in discrete mathematics?"
Consider the implication “if it rains, then I take an umbrella.”
(1) If it rains and I take an umbrella, then this implication is true.
(2) If it rains and I don’t take an umbrella, then this implication is false.
Hopefully these first two are not controversial. Now consider the other two:
(3) If it doesn’t rain and I take an umbrella, then this implication is true.
(4) If it doesn’t rain and I don’t take an umbrella, then this implication is true.
One way to think about this is that the implication “if it rains, then I take an umbrella” says that under any circumstances I will do whatever it takes, umbrella-wise, so as to stay dry. The only way that I will get wet is (2): it rains and I don’t take an umbrella. If it doesn’t rain, then the implication is trivially (or “vacuously”) true: I will stay dry regardless of whether I take an umbrella.
Here’s another example:
(A) If I hit my thumb with a hammer, then my thumb hurts.
If I’m in a reality where I don’t hit my thumb with a hammer and my thumb doesn’t hurt, I’d still consider (A) to be true. | 378 | 1,635 | {"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} | 3.5625 | 4 | CC-MAIN-2018-30 | latest | en | 0.918684 |
https://mathematica.stackexchange.com/questions/140064/calculating-limit-with-option-analytic-true-and-assumptions-yields-seriescas/201536 | 1,624,024,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00117.warc.gz | 351,858,053 | 38,249 | # Example
I've managed to boil down the expression to the following (divergent) example:
$Assumptions=Element[x,Reals]&&0<x<1&&Element[ep,Reals]&&-1/1000<ep<0; foo=x^(-1-ep)*(j[x]*nn[x,0]^(2-2*ep) +(nn[0,0]^(2-2*ep)-nn[0,y]^(2-2*ep) -(ff[x]*nn[x,0]^(2-2*ep))/ff[0] +(ff[x]*nn[x,y]^(2-2*ep))/ff[0])); Limit[foo,x->0,Analytic->True] The result is \[Infinity] j[0] nn[0, 0]^(2 - 2 ep) which is fine, since I expect the expression to be divergent. The first evaluation of the Limit command yields the messages Series::cas: "Warning: contradictory assumption(s) Re[x]>4096&&-(1/4096)<Im[x]<1/4096&&x\[Element]Reals&&0<x<1&&ep\[Element]Reals&&-(1/1000)<ep<0 encountered." Series::cas: "Warning: contradictory assumption(s) Re[x]>4096&&-(1/4096)<Im[x]<1/4096&&x\[Element]Reals&&0<x<1&&ep\[Element]Reals&&-(1/1000)<ep<0 encountered." What bothers me are the additional assumptions Re[x]>4096&&-(1/4096)<Im[x]<1/4096 which seem to be added inerternally by Series (which is at the heart of Limit if I understand things correctly). # Question What exacly leads to these messages and how can I avoid them in general? I know that I can turn off messages using Quiet or more selectively using Off. I am more interested in getting to the root cause of this problem since I would like to use code based upon this to automatically detect divergent (sub-)expressions. I found the somewhat related question Difficulty with computing a limit. However, the answer only addresses how to work around the problem in the specific case of the question. # Further observations If I evaluate the command a second time there are no messages. After calling ClearSystemCache[] the messages reappar during the next evaluation. I suppose that the results of the call to Series are being cached. If I supply the same assumptions to Limit directly via the option Assumptions, no messages appear: ClearSystemCache[];$Assumptions=True;
Limit[foo,x->0,Analytic->True,
Assumptions->Element[x,Reals]&&0<x<1&&Element[ep,Reals]&&-1/1000<ep<0]
Using Assuming leads to messages, just like when using \$Assumptions.
Further stripping down the example by removing subexpressions usually gets rid of the messages.
I've encountered this behaviour in Mathematica 11.0.1.0, 11.0.0.0, 10.2.0.0, 9.0.1.0 on Linux x86 (64bit).
Limit::alimv: Warning: Assumptions that involve the limit variable are ignored.
One can only influence the limit variable via the Direction option of Limit. | 719 | 2,440 | {"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} | 2.890625 | 3 | CC-MAIN-2021-25 | latest | en | 0.781506 |
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` A particle mass 10 gm is executing shm with an amplitude of 0.5 m and periodic time of pi/s secThe max value of the force acting on the particle`
7 years ago
147 Points
``` Dear pallavi
x=Xsinwt where w=2 pi/T and T=pi/5
=.5sin 10t
w=√k/m
so k=w2m
=102 *10
=1000 gm/sec2
maximum force=kx
=1000*.5 m*gm/sec2
=.5 kg m/sec2
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin
```
7 years ago
23 Points
``` Dear Pallavi,
We know for shm motion of the particle its displacement is given by y= y0 sin ωt Where ω = 2π/T = 2π/π = 2 and amplitude
y0 = 0.5 m
thus aceeleration on the particle is given by a= d2y/d2t = - y0 * ω2 * sin ωt
Maximum acceleration is amax = - y0 * ω2 = - 2 m/s2
Thus maximum force in magnitude = | m* amax| = 0.02 N
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best !!!
Regards,
```
7 years ago
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• View Details | 586 | 1,903 | {"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} | 2.8125 | 3 | CC-MAIN-2017-43 | latest | en | 0.802465 |
http://reference.wolfram.com/legacy/language/v10.4/ref/GeoCircle.html | 1,508,844,928,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00249.warc.gz | 302,238,785 | 9,708 | # Wolfram Language & System 10.4 (2016)|Legacy Documentation
This is documentation for an earlier version of the Wolfram Language.
BUILT-IN WOLFRAM LANGUAGE SYMBOL
# GeoCircle
GeoCircle[loc,r]
is a two-dimensional GeoGraphics primitive that represents a circle of radius r centered at the location loc on the surface of the Earth.
GeoCircle[loc,r,{α1,α2}]
represents a sector of a circle from bearing to bearing .
## DetailsDetails
• A geo circle with center loc and radius r is defined as the endpoints of all geodesics of length r starting from loc. Specifying bearings and restricts the set of geodesics.
• The location loc can be specified either as latitude and longitude coordinates in degrees, GeoPosition[], or as a named geographical Entity[].
• The radius r can be given as a Quantity length or as a number in meters.
• Bearings and are measured clockwise from true north and can be given as Quantity angles, as numbers in degrees, as DMS strings, or as named compass points like or .
• GeoCircle[loc] represents a geo circle centered at loc, with an automatic choice of radius.
• is equivalent to .
## ExamplesExamplesopen allclose all
### Basic Examples (2)Basic Examples (2)
A circle of 3000 kilometers around a geo location:
In[1]:=
Out[1]=
A sector of a circle over South America:
In[1]:=
In[2]:=
Out[2]= | 316 | 1,338 | {"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} | 2.59375 | 3 | CC-MAIN-2017-43 | latest | en | 0.813725 |
https://www.chegg.com/homework-help/find-limits-chapter-2.6-problem-40e-solution-9780321730794-exc | 1,563,660,826,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526714.15/warc/CC-MAIN-20190720214645-20190721000645-00339.warc.gz | 649,824,879 | 48,532 | Thomas' Calculus, Early Transcendentals, Single Variable, Books a la Carte Edition (12th Edition) Edit edition Problem 40E from Chapter 2.6: Find the limits.
We have solutions for your book!
Chapter: Problem:
Find the limits.
Step-by-step solution:
Chapter: Problem:
• Step 1 of 4
To evaluate, use the following two methods.
Geometrical solution of the limit :
The graph of is the graph of shifted 3 units to the right as shown.
Therefore, behaves near 3 exactly the way behaves near 0:
Since, it follows that
• Chapter , Problem is solved.
Corresponding Textbook
Thomas' Calculus, Early Transcendentals, Single Variable, Books a la Carte Edition | 12th Edition
9780321730794ISBN-13: 0321730798ISBN: | 184 | 709 | {"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-2019-30 | latest | en | 0.825056 |
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Overview
The natural logarithm is the logarithm
Logarithm
The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: More generally, if x = by, then y is the logarithm of x to base b, and is written...
to the base e
E (mathematical constant)
The mathematical constant ' is the unique real number such that the value of the derivative of the function at the point is equal to 1. The function so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base...
, where e is an irrational
Irrational number
In mathematics, an irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers, with b non-zero, and is therefore not a rational number....
and transcendental
Transcendental number
In mathematics, a transcendental number is a number that is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients. The most prominent examples of transcendental numbers are π and e...
constant approximately equal to 2.718281828. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x).
The natural logarithm of a number x is the power
Exponentiation
Exponentiation is a mathematical operation, written as an, involving two numbers, the base a and the exponent n...
to which e would have to be raised to equal x.
Encyclopedia
The natural logarithm is the logarithm
Logarithm
The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: More generally, if x = by, then y is the logarithm of x to base b, and is written...
to the base e
E (mathematical constant)
The mathematical constant ' is the unique real number such that the value of the derivative of the function at the point is equal to 1. The function so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base...
, where e is an irrational
Irrational number
In mathematics, an irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers, with b non-zero, and is therefore not a rational number....
and transcendental
Transcendental number
In mathematics, a transcendental number is a number that is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients. The most prominent examples of transcendental numbers are π and e...
constant approximately equal to 2.718281828. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x).
The natural logarithm of a number x is the power
Exponentiation
Exponentiation is a mathematical operation, written as an, involving two numbers, the base a and the exponent n...
to which e would have to be raised to equal x. For example, ln(7.389...) is 2, because e2=7.389.... The natural log of e itself (ln(e)) is 1 because e1 = e, while the natural logarithm of 1 (ln(1)) is 0, since e0 = 1.
The natural logarithm can be defined for any positive real number
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
a as the area under the curve
Integral
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
y = 1/x from 1 to a. The simplicity of this definition, which is matched in many other formulas involving the natural logarithm, leads to the term "natural." The definition can be extended to non-zero complex number
Complex number
A complex number is a number consisting of a real part and an imaginary part. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the number line for the real part and adding a vertical axis to plot the imaginary part...
s, as explained below.
The natural logarithm function, if considered as a real-valued function of a real variable, is the inverse function
Inverse function
In mathematics, an inverse function is a function that undoes another function: If an input x into the function ƒ produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., ƒ=y, and g=x...
of the exponential function
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...
Like all logarithms, the natural logarithm maps multiplication into addition:
Thus, the logarithm function is an isomorphism
Isomorphism
In abstract algebra, an isomorphism is a mapping between objects that shows a relationship between two properties or operations. If there exists an isomorphism between two structures, the two structures are said to be isomorphic. In a certain sense, isomorphic structures are...
from the group
Group (mathematics)
In mathematics, a group is an algebraic structure consisting of a set together with an operation that combines any two of its elements to form a third element. To qualify as a group, the set and the operation must satisfy a few conditions called group axioms, namely closure, associativity, identity...
of positive real numbers under multiplication to the group of real numbers under addition, represented as a function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...
:
Logarithms can be defined to any positive base other than 1, not just e; however logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter. Logarithms are useful for solving equations in which the unknown appears as the exponent of some other quantity. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches of mathematics and the sciences and are used in finance to solve problems involving compound interest
Compound interest
Compound interest arises when interest is added to the principal, so that from that moment on, the interest that has been added also itself earns interest. This addition of interest to the principal is called compounding...
.
## History
The first mention of the natural logarithm was by Nicholas Mercator
Nicholas Mercator
Nicholas Mercator , also known by his Germanic name Kauffmann, was a 17th-century mathematician....
in his work Logarithmotechnia published in 1668, although the mathematics teacher John Speidell had already in 1619 compiled a table on the natural logarithm. It was formerly also called hyperbolic logarithm, as it corresponds to the area under a hyperbola. It is also sometimes referred to as the Napierian logarithm
Napierian logarithm
The term Napierian logarithm, or Naperian logarithm, is often used to mean the natural logarithm. However, as first defined by John Napier, it is a function given by :...
, although the original meaning of this term is slightly different.
## Notational conventions
Mathematicians, statisticians, and some engineers generally understand either "log(x)" or "ln(x)" to mean loge(x), i.e., the natural logarithm of x, and write "log10(x)" if the base 10 logarithm
Common logarithm
The common logarithm is the logarithm with base 10. It is also known as the decadic logarithm, named after its base. It is indicated by log10, or sometimes Log with a capital L...
of x is intended.
Some engineers, biologists, and some others generally write "ln(x)" (or occasionally "loge(x)") when they mean the natural logarithm of x, and take "log(x)" to mean log10(x).
In most commonly-used programming language
Programming language
A programming language is an artificial language designed to communicate instructions to a machine, particularly a computer. Programming languages can be used to create programs that control the behavior of a machine and/or to express algorithms precisely....
s, including C
C (programming language)
C is a general-purpose computer programming language developed between 1969 and 1973 by Dennis Ritchie at the Bell Telephone Laboratories for use with the Unix operating system....
, C++
C++
C++ is a statically typed, free-form, multi-paradigm, compiled, general-purpose programming language. It is regarded as an intermediate-level language, as it comprises a combination of both high-level and low-level language features. It was developed by Bjarne Stroustrup starting in 1979 at Bell...
, SAS
SAS System
SAS is an integrated system of software products provided by SAS Institute Inc. that enables programmers to perform:* retrieval, management, and mining* report writing and graphics* statistical analysis...
, MATLAB
MATLAB
MATLAB is a numerical computing environment and fourth-generation programming language. Developed by MathWorks, MATLAB allows matrix manipulations, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages,...
, Fortran
Fortran
Fortran is a general-purpose, procedural, imperative programming language that is especially suited to numeric computation and scientific computing...
, and BASIC, "log" or "LOG" refers to the natural logarithm.
In hand-held calculator
Calculator
An electronic calculator is a small, portable, usually inexpensive electronic device used to perform the basic operations of arithmetic. Modern calculators are more portable than most computers, though most PDAs are comparable in size to handheld calculators.The first solid-state electronic...
s, the natural logarithm is denoted ln, whereas log is the base 10 logarithm.
In theoretical computer science
Theoretical computer science
Theoretical computer science is a division or subset of general computer science and mathematics which focuses on more abstract or mathematical aspects of computing....
, information theory, and cryptography "log(x)" generally means "log2
Binary logarithm
In mathematics, the binary logarithm is the logarithm to the base 2. It is the inverse function of n ↦ 2n. The binary logarithm of n is the power to which the number 2 must be raised to obtain the value n. This makes the binary logarithm useful for anything involving powers of 2,...
(x)" (although this is often written as lg(x) instead).
## Origin of the term natural logarithm
Initially, it might seem that since our numbering system is base 10, this base would be more "natural" than base e. But mathematically, the number 10 is not particularly significant. Its use culturally—as the basis for many societies’ numbering systems—likely arises from humans’ typical number of fingers. Other cultures have based their counting systems on such choices as 5
Quinary
Quinary is a numeral system with five as the base. A possible origination of a quinary system is that there are five fingers on either hand. The base five is stated from 0-4...
, 8
Octal
The octal numeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7. Numerals can be made from binary numerals by grouping consecutive binary digits into groups of three...
, 12
Duodecimal
The duodecimal system is a positional notation numeral system using twelve as its base. In this system, the number ten may be written as 'A', 'T' or 'X', and the number eleven as 'B' or 'E'...
, 20
Vigesimal
The vigesimal or base 20 numeral system is based on twenty .- Places :...
, and 60.
loge is a "natural" log because it automatically springs from, and appears so often in, mathematics. For example, consider the problem of differentiating
Derivative
In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a...
a logarithmic function:
If the base b equals e, then the derivative is simply 1/x, and at x = 1 this derivative equals 1. Another sense in which the base-e-logarithm is the most natural is that it can be defined quite easily in terms of a simple integral or Taylor series
Taylor series
In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point....
and this is not true of other logarithms.
Further senses of this naturalness make no use of calculus
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related by the fundamental theorem...
. As an example, there are a number of simple series involving the natural logarithm. Pietro Mengoli
Pietro Mengoli
Pietro Mengoli was an Italian mathematician and clergyman from Bologna, where he studied with Bonaventura Cavalieri at the University of Bologna, and succeeded him in 1647...
and Nicholas Mercator
Nicholas Mercator
Nicholas Mercator , also known by his Germanic name Kauffmann, was a 17th-century mathematician....
called it logarithmus naturalis a few decades before Newton and Leibniz developed calculus.
## Definitions
Formally, ln(a) may be defined as the area under the graph of 1/x from 1 to a, that is as the integral
Integral
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
,
This defines a logarithm because it satisfies the fundamental property of a logarithm:
This can be demonstrated by letting as follows:
The number e
E (mathematical constant)
The mathematical constant ' is the unique real number such that the value of the derivative of the function at the point is equal to 1. The function so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base...
can then be defined as the unique real number a such that ln(a) = 1.
Alternatively, if the exponential function
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...
has been defined first using an infinite series, the natural logarithm may be defined as its inverse function
Inverse function
In mathematics, an inverse function is a function that undoes another function: If an input x into the function ƒ produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., ƒ=y, and g=x...
, i.e., ln is that function such that . Since the range of the exponential function on real arguments is all positive real numbers and since the exponential function is strictly increasing, this is well-defined for all positive x.
## Properties
(see complex logarithm
Complex logarithm
In complex analysis, a complex logarithm function is an "inverse" of the complex exponential function, just as the natural logarithm ln x is the inverse of the real exponential function ex. Thus, a logarithm of z is a complex number w such that ew = z. The notation for such a w is log z...
)
## Derivative, Taylor series
The derivative
Derivative
In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a...
of the natural logarithm is given by
This leads to the Taylor series
Taylor series
In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point....
for around 0; also known as the Mercator series
At right is a picture of ln(1 + x) and some of its Taylor polynomials around 0. These approximations converge to the function only in the region −1 < x ≤ 1; outside of this region the higher-degree Taylor polynomials are worse approximations for the function.
Substituting x − 1 for x, we obtain an alternative form for ln(x) itself, namely
By using the Euler transform on the Mercator series, one obtains the following, which is valid for any x with absolute value greater than 1:
This series is similar to a BBP-type formula.
Also note that is its own inverse function, so to yield the natural logarithm of a certain number y, simply put in for x.
## The natural logarithm in integration
The natural logarithm allows simple integration
Integral
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
of functions of the form g(x) = f '(x)/f(x): an antiderivative
Antiderivative
In calculus, an "anti-derivative", antiderivative, primitive integral or indefinite integralof a function f is a function F whose derivative is equal to f, i.e., F ′ = f...
of g(x) is given by ln(|f(x)|). This is the case because of the chain rule
Chain rule
In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function in terms of the derivatives of f and g.In integration, the...
and the following fact:
In other words,
and
Here is an example in the case of g(x) = tan(x):
Letting f(x) = cos(x) and f(x)= – sin(x):
where C is an arbitrary constant of integration
Arbitrary constant of integration
In calculus, the indefinite integral of a given function is only defined up to an additive constant, the constant of integration. This constant expresses an ambiguity inherent in the construction of antiderivatives...
.
The natural logarithm can be integrated using integration by parts
Integration by parts
In calculus, and more generally in mathematical analysis, integration by parts is a rule that transforms the integral of products of functions into other integrals...
:
## Numerical value
To calculate the numerical value of the natural logarithm of a number, the Taylor series expansion can be rewritten as:
To obtain a better rate of convergence, the following identity can be used.
provided that y = (x−1)/(x+1) and x > 0.
For ln(x) where x > 1, the closer the value of x is to 1, the faster the rate of convergence. The identities associated with the logarithm can be leveraged to exploit this:
Such techniques were used before calculators, by referring to numerical tables and performing manipulations such as those above.
### High precision
To compute the natural logarithm with many digits of precision, the Taylor series approach is not efficient since the convergence is slow. An alternative is to use Newton's method
Newton's method
In numerical analysis, Newton's method , named after Isaac Newton and Joseph Raphson, is a method for finding successively better approximations to the roots of a real-valued function. The algorithm is first in the class of Householder's methods, succeeded by Halley's method...
to invert the exponential function, whose series converges more quickly.
An alternative for extremely high precision calculation is the formula
where M denotes the arithmetic-geometric mean of 1 and 4/s, and
with m chosen so that p bits of precision is attained. (For most purposes, the value of 8 for m is sufficient.) In fact, if this method is used, Newton inversion of the natural logarithm may conversely be used to calculate the exponential function efficiently. (The constants ln 2 and π
Pi
' is a mathematical constant that is the ratio of any circle's circumference to its diameter. is approximately equal to 3.14. Many formulae in mathematics, science, and engineering involve , which makes it one of the most important mathematical constants...
can be pre-computed to the desired precision using any of several known quickly converging series.)
### Computational complexity
The computational complexity
Computational Complexity
Computational Complexity may refer to:*Computational complexity theory*Computational Complexity...
of computing the natural logarithm (using the arithmetic-geometric mean) is O(M(n) ln n). Here n is the number of digits of precision at which the natural logarithm is to be evaluated and M(n) is the computational complexity of multiplying two n-digit numbers.
## Continued fractions
While no simple continued fraction
Continued fraction
In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on...
s are available, several generalized continued fraction
Generalized continued fraction
In complex analysis, a branch of mathematics, a generalized continued fraction is a generalization of regular continued fractions in canonical form, in which the partial numerators and partial denominators can assume arbitrary real or complex values....
s are, including:
## Complex logarithms
The exponential function can be extended to a function which gives a complex number
Complex number
A complex number is a number consisting of a real part and an imaginary part. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the number line for the real part and adding a vertical axis to plot the imaginary part...
as ex for any arbitrary complex number x; simply use the infinite series with x complex. This exponential function can be inverted to form a complex logarithm that exhibits most of the properties of the ordinary logarithm. There are two difficulties involved: no x has ex = 0; and it turns out that e2πi = 1 = e0. Since the multiplicative property still works for the complex exponential function, ez = ez+2nπi, for all complex z and integers n.
So the logarithm cannot be defined for the whole complex plane
Complex plane
In mathematics, the complex plane or z-plane is a geometric representation of the complex numbers established by the real axis and the orthogonal imaginary axis...
, and even then it is multi-valued – any complex logarithm can be changed into an "equivalent" logarithm by adding any integer multiple of 2πi at will. The complex logarithm can only be single-valued on the cut plane. For example, ln i = 1/2 πi or 5/2 πi or −3/2 πi, etc.; and although i4 = 1, 4 log i can be defined as 2πi, or 10πi or −6 πi, and so on.
• John Napier
John Napier
John Napier of Merchiston – also signed as Neper, Nepair – named Marvellous Merchiston, was a Scottish mathematician, physicist, astronomer & astrologer, and also the 8th Laird of Merchistoun. He was the son of Sir Archibald Napier of Merchiston. John Napier is most renowned as the discoverer...
– inventor of logarithms
• Logarithm of a matrix
Logarithm of a matrix
In mathematics, a logarithm of a matrix is another matrix such that the matrix exponential of the latter matrix equals the original matrix. It is thus a generalization of the scalar logarithm and in some sense an inverse function of the matrix exponential. Not all matrices have a logarithm and...
• Logarithmic integral function
Logarithmic integral function
In mathematics, the logarithmic integral function or integral logarithm li is a special function. It occurs in problems of physics and has number theoretic significance, occurring in the prime number theorem as an estimate of the number of prime numbers less than a given value.-Integral...
• Nicholas Mercator
Nicholas Mercator
Nicholas Mercator , also known by his Germanic name Kauffmann, was a 17th-century mathematician....
– first to use the term natural log
• Polylogarithm
• Von Mangoldt function
Von Mangoldt function
In mathematics, the von Mangoldt function is an arithmetic function named after German mathematician Hans von Mangoldt.-Definition:The von Mangoldt function, conventionally written as Λ, is defined as...
• The number e
E (mathematical constant)
The mathematical constant ' is the unique real number such that the value of the derivative of the function at the point is equal to 1. The function so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base...
• Natural logarithm of 2
• Leonhard Euler
Leonhard Euler
Leonhard Euler was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion... | 5,607 | 25,823 | {"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} | 4.375 | 4 | CC-MAIN-2021-17 | latest | en | 0.923991 |
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How do you answer this homework question:
Consider a 230-kV, 50 Hz substation with two 3-phase wye grounded capacitor installations. The adjusted rating of each 3-phase capacitor bank is 85-MVAr at 230- kV. The two capacitor banks are separated by a 41-m long busbar. Let the inductance of the busbar be 1 µH/m. Each capacitor bank has an internal self-inductance of 10 µH. Let the system short circuit capacity (SCC) be 11,000-MVA.
1) Bank#1 is automatically energized when the substation voltage is 0.98 pu. Evaluate whether general purpose breaker is suitable for isolated switching of Bank#1. Is an inrush current limiting reactor required? If required, calculate the reactor value
2)Bank #2 is energised only when Bank #1 is already in service. Evaluate whether general purpose breaker is suitable for switching of Bank#2. Is an inrush current limiting reactor required? If required, calculate the reactor value.
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https://www.heatsinkcalculator.com/blog/how-to-calculate-the-thermal-resistance-of-a-pcb/ | 1,721,337,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00719.warc.gz | 695,035,859 | 15,606 | 25 Feb
Thermal management is a crucial aspect of designing and manufacturing electronic devices, as heat generated by the components can lead to performance issues and even device failure. To ensure the reliable and efficient operation of electronic devices it is essential to understand how to effectively dissipate heat away from the heat generating devices. These heat generating components are often mounted on printed circuit boards (PCBs) which in many cases are used as a heat sink to cool the attached component.
Thermal resistance is a measure of how well a material or device transfers heat to its surroundings and is often used to predict the thermal performance of a PCB and to calculate the temperature of the attached component. This technical article will provide a detailed explanation of the formulas and method used to calculate the thermal resistance of a PCB. The calculation methodology will take into account the PCB insulation (FR4)/copper layer thicknesses, percent copper coverage in each layer, location of these layers with respect to the heat generating component and usage of thermal vias.
## Layout and structure of the PCB
A PCB consists of alternating layers of dielectric and conductive materials typically made using FR4/resin and copper respectively as shown in figure 1. Layers that are used for ground planes often are 100% conductive material. While the layers with attached components, traces and interconnects will have some mixture of conductive and dielectric materials.
Figure 1. PCB layout
The preceding calculation procedure assumes that the component is intimately attached to the PCB using an exposed metal pad or some similar attachment method that creates an extremely low thermal resistance path between the component and the PCB. The component is located at or close to the geometric center of the PCB. The upper and lower surfaces of the PCB are cooled via natural convection or forced convection and radiation. A heat transfer coefficient hext is used to quantify the effectiveness of cooling from these outer surfaces. Reference the articles How to design a flat plate heat sink and Performance of a LED flat plate heat sink in multiple orientations for the formulas used to calculate hext for natural convection and radiation.
## Calculating the thermal resistance of each PCB layer
To facilitate the preceding calculations the rectangular PCB and attached component are converted to circles of equivalent areas with radii ro and ri respectively using equations 1 and 2. The approximation of the rectangular areas as circular areas introduces an error of less than 1% at an aspect ratio (width/length) of 1, increasing to an error 8% at aspect ratios of 2.5 or 0.4.
$Latex formula$1
$Latex formula$2
The composite nature of the PCB complicates the thermal analysis and does not allow for the PCB to be treated as a simple flat plate heat sink made from uniform thermal conductivity material. The thermal conductivity through the thickness and in the plane of the PCB are different. Additionally, the position of the dielectric and conducting layers from top to bottom affect how well heat is conducted from the heat generating component through the PCB and then to the surroundings.
The thermal resistance calculation is conducted layer by layer to account for the non-uniform thermal conductivity of the PCB. The section of each layer directly under the component is a disk with radius ri and thickness tn. This disk will be referred to as the “inner disk”. The periphery of the disk has an equivalent convection coefficient ho,n that represents the effective cooling of the outer portion of the layer as shown in figure 2.
The subscript n in tn , ho,n and all other variables subsequently defined refers to the layer number where layer number 1 is attached to the component and N is the total number of layers in the PCB.
Figure 2. PCB layers and sections used in the analysis.
The outer portion of each layer is an annulus (referred to as the “outer annulus”) with inner and outer radii ri and ro respectively.
The calculation for ho,n is developed with the approximation that the temperature difference between the layers of the PCB is small. As such the flow of heat is predominantly radial in the outer annulus. The resistance to heat flow in the outer annulus can then be approximated by the thermal resistance network shown in figure 3.
Figure 3. Outer annulus layers and resistance network
Rcond,o,n in the thermal resistance network shown in figure 3 represents the conduction thermal resistance of each layer and Rext is the thermal resistance due to the convection/radiative cooling hext on the exposed upper and lower surfaces of the PCB. The total equivalent thermal resistance of the thermal resistance network is determined by combining the parallel and series resistances as illustrated in figure 4.
Figure 4. Simplified thermal resistance network for the outer annulus layers
To determine Rcond,o,n for each layer, Router is first calculated by determining keff, the average in-plane thermal conductivity of the PCB. keff is an average of the effective thermal conductivity of each layer, keff,n.
The effective thermal conductivity keff,n of each layer is:
$Latex formula$3
where:
$Latex formula$is the volume fraction of conductor in layer n
$Latex formula$is the thermal conductivity of the conductor of layer n
$Latex formula$is the thermal conductivity of the dielectric of layer n
The value of σc,n ranges from 0 to 1. It can be determined by estimating the conductor coverage area on each layer and dividing that value by the total PCB area (L x W).
Thermal conductivities of 398W/m·K for copper and 0.3W/m·K for FR4 are most often used for the conductor and dielectric materials respectively.
The average in-plane effective thermal conductivity of the entire PCB, keff is:
$Latex formula$4
The resistance Router is calculated using the solution for a circular fin [1] expressed as:
$Latex formula$5
where:
$Latex formula$ modified zero order Bessel function of the 1st kind
$Latex formula$ modified zero order Bessel function of the 2nd kind
$Latex formula$ modified first order Bessel function of the 1st kind
$Latex formula$ modified first order Bessel function of the 2nd kind
$Latex formula$6
The multiplication of hext by 2 in equation 6 indicates that both the upper and lower surfaces of the PCB are subjected to convective and/or radiative cooling.
The use of Bessel functions in equation 5 may seem complex, however Bessel functions are readily available in MS Excel or any commercial mathematics software as a built-in function.
With the value of Router known, Rcond,o,n for each layer is calculated using equations 7, 8, 9 and 10.
$Latex formula$7
$Latex formula$8
$Latex formula$9
Aext is the exposed surface area of the outer annulus.
$Latex formula$10
The equivalent convection coefficient on the periphery of the inner disk for each layer ho,n can now be calculated with equation 11.
$Latex formula$11
## Combining the PCB layer thermal resistances
The next step in the calculation is to determine the value of hb,n, the effective heat transfer coefficient acting on the base of the inner disk of each layer as shown in figure 2. This is done by calculating the equivalent heat transfer coefficient that the inner disk of the nth layer applies to the lower surface of the inner disk of the n-1 layer. Equation 12 uses the equation for an extended surface with convection heat transfer around the periphery and lower surface [2] to calculate hb,n-1.
$Latex formula$12
As and Ps are the cross-sectional area and perimeter respectively of the inner disk.
$Latex formula$13
$Latex formula$14
$Latex formula$15
The calculation is started at the last (Nth) layer with hb,N=0 on the bottom surface of the inner disk for that layer. hb,N is not set to equal hext since the dissipation of heat from the base of the inner disk will be evaluated separately.
Using equation 12, hb,N-1 is calculated. This process is repeated N times for all the layers resulting in an equivalent heat transfer coefficient hb. This process is illustrated in figure 5.
Figure 5. Iterative calculation for hb
The heat transfer coefficient hb is the equivalent heat transfer coefficient of the conduction resistance due to flow of heat through the inner disk and into the outer annulus of the PCB acting on the heat generating component. This conduction thermal resistance is:
$Latex formula$16
Note: R*cond,o includes the thru-plane and in-plane conduction across the PCB which is why the “*” notation is used to differentiate this variable from Rcond,o which only accounts for the in-plane conduction in the outer annulus.
The heat flow from the heat generating component through the PCB to the ambient is represented by the thermal resistance diagram in figure 6. Flow through the bottom of the inner disk exposed to hext is accounted for with Rcond,i and Rext,i thermal resistances.
Figure 6. Thermal resistance network for heat flow through the PCB to the ambient
Rcond,i is the summation of the thru-plane thermal resistances of each layer, (equation 17) and Rext,i is the thermal resistance due to hext of the area As.
$Latex formula$17
$Latex formula$18
The thermal resistances in figure 5 are combined to a single thermal resistance Rpcb:
$Latex formula$19
## Thermal via calculations
Thermal vias placed under the heat generating component provide a path to more efficiently transfer heat from the heat generating component to the inner layers and base layer of the PCB. This then allows to the heat to be spread across these layers reducing the thermal resistance of the PCB.
Figure 7. PCB with thermal vias
The analysis of the effect of imbedded thermal vias as shown in figure 7 requires the calculation of an equivalent through-plane thermal conductivity, kvia,n for the inner disk of each layer. This is calculated by evaluating the flow of heat through the inner disk of each layer through three parallel thermal resistances:
1. Rbarrel, the thermal resistance of the barrel of the thermal via.
2. Rfiller, the thermal resistance of the filler material of the via if any.
3. Rmatl,n, the thermal resistance of the layer material surrounding the thermal via in the s x s square region shown in figure 7.
$Latex formula$20
$Latex formula$21
$Latex formula$22
$Latex formula$23
kbarrel and kfiller are the thermal resistances of the thermal via barrel and filler material respectively.
The thermal resistance of the entire s x s, cell, Rcell,n for each layer is:
$Latex formula$24
If the thermal via is not filled then the 1/Rfiller term in equation 24 is omitted.
The equivalent thermal conductivity of each layer of the inner disk with thermal vias is then:
$Latex formula$25
In equations 12, 15 and 17, keff,n is replaced with kvia,n and Rpcb is calculated for a PCB with vias under the heat generating component.
## Junction to ambient thermal resistance calculations
Rpcb is the thermal resistance from the base of the component to the atmosphere. To determine the junction to ambient thermal resistance, the junction to board thermal resistance typically listed as Θjb or Rjb in the component literature is added to Rpcb. The temperature of the junction, Tj can now be determined using equation 26.
$Latex formula$26
Tamb is the external ambient temperature and Qs is the heat generated by the component.
This calculation excludes any heat loss from the top of the component exposed to the atmosphere. Typically, this heat loss is small when compared to the heat dissipation from the PCB. If there is a heat sink attached to the component, then the heat loss through the heat sink is significant and must be considered. Figure 8 shows the thermal resistance network for the PCB with a heat sink of thermal resistance Rhs, attached to the component. Rjc is the junction to case (top) thermal resistance.
Figure 8. Thermal resistance network for the PCB
Calculating for the junction temperature, Tj from this thermal resistance network yields:
$Latex formula$27
An online PCB temperature calculator that uses the calculation procedure outlined in this blog post is available free of cost. Click the following link: PCB Temperature Calculator.
References
[1] F. Incropera, D. DeWitt, (2011). Fundamentals of Heat and Mass Transfer (7th ed.). Hoboken: Wiley. p124
[2] F. Incropera, D. DeWitt, (2011). Fundamentals of Heat and Mass Transfer (7th ed.). Hoboken: Wiley. p118, table 3.4 | 2,696 | 12,558 | {"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": 34, "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-2024-30 | latest | en | 0.929096 |
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# How to create a 3D HorizonCube?
(Model Driven)
Version: 4.4
Farrukh Qayyum
support@opendtect.org
Outline
Introduction
Required data
Workflow
Create HorizonCube
Introduction
## This presentation will guide you through the workflow to
create a model driven 3D HorizonCube in OpendTect 4.4. It
is arranged as follow:
Introduction
## A model driven HorizonCube is a dense set of automated seismic horizons,
which are interpolated between the input horizons without using the seismic data.
Input horizons
Automated horizons
Introduction
## The linear interpolation model is also known as stratal slicing or
propotional slicing.
## The method has become quite popular because it is a fast and
robust method to understand subsurface geomorphology. Not only
this, people also use this type of model to build a low frequency
model for deterministic seismic inversion.
Introduction
## The parallel to upper model is used in a situation in which the
seismic reflectors appear parallel to the top horizon.
## This is a kind of onlap model which fits a wedge shaped structure
with layers onlapping the base horizon.
Parallel to upper
Introduction
## The parallel to lower model is suitable for an angular unconformity
situation.
You may use such a model in case of truncations that are inclined
but parallel to the lower horizon.
Parallel to lower
Introduction
What are the cases in which the model driven HorizonCube should be used?
Parallel Reflections:
## A classic example of a model driven approach is the parallel seismic reflections
with no or little geometrical variations (see below).
Introduction
What are the cases in which the model driven HorizonCube should be used?
## Angular unconformity with parallel reflections:
This is another case in which the model driven approach would help.
## North Sea (Cretaceous Interval)
Introduction
What are the cases in which the model driven HorizonCube should be used?
## It is impossible to apply data driven approach in the polygonal faulted regions,
chaotic intervals, and noisy layers. Therefore, you should try to use model driven
approach.
## North Sea (Mid Miocene Uconformity)
Introduction
Required data
Workflow
Create HorizonCube
Required Data
## Model driven HorizonCube is very simple. You
only need interpretations i.e.
1.Horizons (minimum 2)
2.Faults (optional)
Introduction
Required data
Workflow
Create HorizonCube
Workflow
Create a Model Driven HorizonCube
## Tip: You may leave this dialog open somewhere on
Workflow
Create a Model Driven HorizonCube
## Select the bounding Horizons between which the
HorizonCube will be created.
Workflow
Create a Model Driven HorizonCube
## Choose a right model driven option for
a specific package e.g.
Propotional
Parallet to upper
Parallet to lower
Workflow
Create a Model Driven HorizonCube
Settings:
## Model driven mode requires only one
parameter i.e. The spacing between
the two horizons.
## Tip: use the default.
Workflow
Create a Model Driven HorizonCube
1. Select Faults:
## Model driven HorizonCube can be
created with given faults. It
auotmatically computes the fault throw
using the given horizons and
interpolates the throw relatively.
## 2. Provide Output name
3. Press Go
Workflow
Create a Model Driven HorizonCube
## Once the batch processing prompts
Finished batch processing, you can
display the model driven
HorizonCube in the Scene.
Workflow
Display Model Driven HorizonCube
Step 1
Step 2
Optional
Create Wheeler Cube
## Use the model driven HorizonCube to
visualize seismic attributes. For this you will
need to create a Wheeler Cube (Flattened
Volume).
Optional
Create Wheeler Cube
## Select the input HorizonCube (e.g. model driven)
Select the stored seismic attribute (e.g. similarity, dip,
curvature).
Write the Output Wheeler cube name.
Proceed
Optional
Visualization: Create a Wheeler Scene
## Time Domain Wheeler Domain
Optional
Visualization: Create a Wheeler Scene
Display a volume in the Wheeler domain tree. You may used the RGT
data that you just created or use any stored volume/attribute. The later
will be slow because of time domain to wheeler domain transformation.
Optional
Visualization: Create a Wheeler Scene
## Watch the video tutorial on Stratal Slicing.
http://www.opendtect.org/tutorials/SSIS_stratal_slicing/
Thank You!
## Нижнее меню
### Получите наши бесплатные приложения
Авторское право © 2021 Scribd Inc. | 1,006 | 4,466 | {"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} | 2.78125 | 3 | CC-MAIN-2021-21 | latest | en | 0.852166 |
https://www.tes.com/news/tes-archive/tes-publication/my-best-lesson-redrawing-map-perpendicular-bisectors | 1,516,783,997,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00004.warc.gz | 959,022,534 | 15,128 | # My best lesson - Redrawing the map with perpendicular bisectors
11th October 2013 at 01:00
My bottom-set students were due to be taught constructions, creating various geometric objects with only a compass and a straight edge or rule. To engage the 12- and 13-year-olds, I knew I had to do better than some horrendously contrived example, such as wanting to fly a plane equidistant between two erupting volcanoes.
I was struck by a great idea: a perpendicular bisector doesn't just show you the midway line between two points, it also highlights which parts of the map are closer to one of those points. The next task was to think of how to interest students in this fact, and before long I had come up with a map of the local area that had the primary schools they had once attended marked on it. The map was at a scale where individual streets could be recognised.
I asked the students to construct the perpendicular bisector of two post offices on the map and then asked if anyone's house was located between them. We used the bisector to decide which post office was closest to two students' houses, and I then asked what else they could see on the map that we could draw bisectors between. As if by magic, the dominoes toppled as one student mentioned the primary schools and another said, "Hang on, I don't think I went to my closest one." The lesson took off.
Everyone wanted to know whether they had gone to their local primary and they asked for a bigger map to compare secondary schools. This led to some interesting conversations about why people don't always go to their closest school, which was a nice bonus. One of my favourite things about this lesson was that the students felt they had thought up the tasks themselves, even though I had planned them that way all along.
This lesson made me re-evaluate my opinion of constructions of perpendicular bisectors. Now all I need is a similar idea for angle bisectors...
Dave Gale is a maths teacher at Churchill Academy and Sixth Form in Somerset, England.
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Subscribe now | 553 | 2,662 | {"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-2018-05 | latest | en | 0.991079 |
https://gist.github.com/someodd?direction=asc&sort=created | 1,709,533,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00782.warc.gz | 259,983,787 | 19,578 | Instantly share code, notes, and snippets.
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# someodd someodd
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Last active August 23, 2022 23:03
Longest binary gap coding puzzle
View BinaryGap.hs
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{- # PROBLEM A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N. For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary
Created August 23, 2022 23:01
Challenge I saw on HackerNews, "Most candidates cannot solve this interview problem."
View MostCandidatesFail.hs
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{- I saw this posted on HackerNews: "Most candidates cannot solve this interview problem." https://twitter.com/Al_Grigor/status/1357028887209902088 > Most candidates cannot solve this interview problem: > Input: "aaaabbbcca" > Output: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]
Created August 23, 2022 23:04
Challenge in parsing math expressions from some kinda AST.
View MathExpr.hs
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import Control.Monad.Reader import Data.Map as Map data Expr = Lit Int | Var String | Add Expr Expr | Let (String, Expr) Expr eval :: Expr -> Reader (Map String Int) Int eval (Lit i) = pure i eval (Var s) = do varMap <- ask case Map.lookup s varMap of
Created August 23, 2022 23:06
Advent of Code 2020: Day 2
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{- Advent of Code: Day 2 Part 1 & 2 https://adventofcode.com/2020/day/2 -} type PasswordData = ((Int, Int, Char), String)
Created August 23, 2022 23:07
Advent of Code 2020: Day 1 | 612 | 2,567 | {"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-2024-10 | latest | en | 0.87376 |
https://www.teacherspayteachers.com/Store/Margies-Idea-Box | 1,542,808,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748901.87/warc/CC-MAIN-20181121133036-20181121155036-00173.warc.gz | 975,431,291 | 24,802 | # Margie's Idea Box
(18)
United States - Texas - Kerrville
4.0
"I can do all things through Christ who strengthens me." Phil. 4:13
Subject
Prices
Top Resource Types
My Products
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Match each strip diagram to the correct equation. Have students match the provided word problems to each pair of equations and diagrams.You can also have them write their own word problems to match the strip diagrams and equations.
Subjects:
Algebra, Arithmetic, Word Problems
4th
Types:
Activities, Test Prep, Math Centers
\$3.00
1 rating
4.0
Texas TEKS 4.2A (A) interpret the value of each place-value position as 10 times the position to the right and as one-tenth of the value of the place to its left. Colorful cards to post in the classroom displaying the change in value of a digit as
Subjects:
Place Value
3rd, 4th
Types:
Printables, Posters
\$0.95
1 rating
4.0
10 long division word problems with graph paper embedded into each question. Students are prompted to circle whether the remainder is to be ignored, used as the solution, or added to the quotient in each question. There are eight 3-digit divided by
Subjects:
Word Problems
3rd, 4th, 5th
Types:
Worksheets, Test Prep, Computation
\$3.00
not yet rated
This PowerPoint has an example of each unit of measurement in the "Length", "Volume and Capacity", "Weight and Mass" sections. I have the students label these objects on the Mathematics Reference Materials page to help them remember. They refer back
Subjects:
Measurement, Math Test Prep
4th
Types:
PowerPoint Presentations
FREE
not yet rated
This is a visual representation of the 3 methods of 2-digit times 2-digit multiplication students in 4th grade are expected to master under the Texas TEKS. Texas TEKS 4.4C and 4.4D
Subjects:
Basic Operations, Math Test Prep
4th
Types:
Computation, Printables, For Parents
FREE
2 ratings
4.0
This is a quick game to review multiplication and division fact families using dice. It can be played independently or with a partner.
Subjects:
Other (Math)
3rd, 4th
Types:
Activities, Games, Math Centers
FREE
1 rating
4.0
Station signs give a title of each station with a description of the station. There is also a "Station Expectations" poster that can be posted at each station. These expectations fit every station. These are great for centers!
Subjects:
Other (Math)
3rd, 4th, 5th
Types:
Printables, Math Centers, Posters
FREE
not yet rated
Students will fill in missing factors on arrays 2D times 2D. Students will fill in partial products on a 10 times 20 array. There are 2 word problems involving partial products with 2D times 2D.
Subjects:
Basic Operations, Word Problems
4th
Types:
Worksheets, Computation, Printables
\$0.99
not yet rated
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 843 | 3,252 | {"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} | 2.875 | 3 | CC-MAIN-2018-47 | latest | en | 0.912112 |
https://ch.mathworks.com/matlabcentral/answers/322797-how-to-find-a-minimum-value-that-calculated-from-a-formulation | 1,590,832,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00329.warc.gz | 290,708,594 | 20,947 | # How to find a minimum value that calculated from a formulation?
2 views (last 30 days)
Le Dung on 1 Feb 2017
Commented: Le Dung on 1 Feb 2017
Hi, Now, i have some parameters that range into a region. And i have a formulation that calculated form these parameters. (below code) And i want to find minimum value of the formulation that has to satisfy a condition. And i want to display that minimum value and coresponding values of parameters above. Help me, plesea. Thank you so much. My code:
x=(50:100);
y=(50:100);
z=(15:50);
Mu=4000;
Mrmin=0;
for i=1:51
for j=1:51
for k=1:36
Mr = x(i)*y(j)*z(k);
if Mr >= Mu
Mrmin = Mr;
end
end
end
end
Torsten on 1 Feb 2017
x=(50:100);
y=(50:100);
z=(15:50);
Mrmin=Inf;
Mu=4000;
imin=1;
jmin=1;
kmin=1;
for i=1:51
for j=1:51
for k=1:36
Mr = x(i)*y(j)*z(k);
if (Mr <= Mrmin) && (Mr >= Mu)
Mrmin = Mr;
imin = i;
jmin = j;
kmin = k;
end
end
end
end
Best wishes
Torsten.
#### 1 Comment
Le Dung on 1 Feb 2017
Thank you so much, Torsten.
Niels on 1 Feb 2017
Edited: Niels on 1 Feb 2017
x=(50:100);
y=(50:100);
z=(15:50);
Mrmin=Inf;
for i=1:51
for j=1:51
for k=1:36
Mr = x(i)*y(j)*z(k);
if Mr <= Mrmin && Mr>4000
Mrmin = Mr;
end
end
end
end
#### 1 Comment
Le Dung on 1 Feb 2017
Thank you so much, Niels. | 481 | 1,243 | {"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} | 3.171875 | 3 | CC-MAIN-2020-24 | latest | en | 0.752181 |
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