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November 12

Strange LaTeX feature

When rendered in PNG, the code \sqrt 3 results in ${\sqrt {3}}$ while \sqrt{3} results in ${\sqrt {3}}$. The latter image has additional white space before the root symbol. Why?

When you replace 3 with 1 or 2 the resulting pictures do not have any leading space:
${\sqrt {1}}$       ${\sqrt {1}}$       ${\sqrt {2}}$       ${\sqrt {2}}$
CiaPan (talk) 13:06, 12 November 2008 (UTC)

Your two allegedly different square roots of three look exactly identical on the two browsers on which I've viewed them. What browser are you using? Michael Hardy (talk) 23:59, 13 November 2008 (UTC)

Probably they are fixed now. They look identical here too now, which they didn't when the question was asked. -- Jao (talk) 00:08, 14 November 2008 (UTC)

The LaTeX images are cached based on a hash (SHA1 MD5, I think) of the markup. Apparently there's something funny going on with the cached image for \sqrt{3}. Changing it to, say, \sqrt{{3}} again produces the expected rendering: ${\sqrt {3}}$. We've had other problems like that recently; perhaps someone should simply purge the whole LaTeX image cache. —Ilmari Karonen (talk) 16:31, 12 November 2008 (UTC)

I've asked on IRC for something to purge it, hopefully it will be fixed soon. --Tango (talk) 16:46, 12 November 2008 (UTC)

Root corrected to radical. CiaPan (talk) 19:47, 12 November 2008 (UTC)

I just changed it back. First off, you shouldn't be editing what other people say, unless it's something like adding a wikilink. Secondly, root is perfectly correct. They are called square roots, cube roots, whatever. The symbol is the root symbol. -mattbuck (Talk) 20:32, 12 November 2008 (UTC)

I agree, don't edit other people's signed comments. Doing so means that their signature follows something that they didn't write. I also agree that "root" is perfectly correct. --Tango (talk) 22:42, 12 November 2008 (UTC)

Are you two referring to this edit? The only comment I see edited there is CiaPan's own; am I just confused? -- Jao (talk) 00:08, 14 November 2008 (UTC)

I did not change what other people say. YOU did. --CiaPan (talk) 07:06, 14 November 2008 (UTC)

It seems Mattbuck missed the fact that you were the OP (and I didn't think to look, sorry!). --Tango (talk) 19:56, 14 November 2008 (UTC)

BTW, now I see both images displayed in my question looking the same. Possibly the cache has been cleaned and now both expressions (with digit in braces {3} and with 3 alone) render identical. --CiaPan (talk) 07:11, 14 November 2008 (UTC)

Excellent, someone managed to fix it! The sysadmins seemed a little unsure of whether they would be able to (I didn't understand half of what they said, unfortunately). --Tango (talk) 19:56, 14 November 2008 (UTC)

Very improbable election results

Suppose we have a population of N people who must vote for President A or President B (no other options). If N is a large number it is very improbable that the difference d between the votes of the two presidents will be 0 or 1 (unless we assume very particular - and improbable - hypotheses about the structure of the population). It is also improbable that the difference d between the votes of the two presidents is too close to 0 (say less then ε). Can we try to give an estimate of this probability given N and ε in any typical real situation?--Pokipsy76 (talk) 14:17, 12 November 2008 (UTC)

A very simple model would be assuming that everyone is as likely to vote for A as for B. In that case, we would have a binomial distribution with $p=0.5$, which for large $N$ will be close enough to a normal distribution (the number of votes for A is distributed around $0.5N$ with a standard deviation of $0.5{\sqrt {N}}$), which would make it easy to answer the questions. Obviously, this assumption is drawn from thin air, but I can't really think of a better one, unless we actually know something about the population. -- Jao (talk) 14:51, 12 November 2008 (UTC)

Ok, it seems reasonable at least for an estimate, however a 0.5 distribution would in any case give a very unrealistically small (and improbable) difference between the votes of the two presidents. This model doesn't say anything about how improbable is to have such a small difference.--Pokipsy76 (talk) 15:12, 12 November 2008 (UTC)

Illustrating this with a particular case of N=10000 and $p=0.5$, the number of votes for A will have, to a sufficient approximation, a normal distribution with mean 5000 and SD 50. The probability that each candidate gets the same number of votes will come from considering the area below the normal curve between the values 4999.5 and 5000.5, with respective z values -0.01 and 0.01, and likewise for any other distribution of the total vote.→81.151.247.46 (talk) 16:07, 12 November 2008 (UTC)

In case the OP or other readers aren't familiar with basic stats, this gives a probability around 1%. For 10000 votes differing by 0 or 1 gives around a 2% chance. While you can adjust the bernoulli probability p=0.5 a bit, note that if it differs significantly from 0.5, the probability of one of the candidates winning is too low for the election to be more than a formality. At least in the voting theory chapters I've read, in a two party election p=0.5 is always approximately the right choice since the two parties will adapt their stance until they both reach this equilibrium. The wikipedia article landslide victory seems to think already p=0.6 is a huge margin.

Using the normal distribution, however, it looks like the probability of winning drops quite dramatically as p moves from 0.5. For instance with 10000 votes, p=0.52, I get the non-favored choice has less than a millionth of a percent chance of winning. Can someone confirm that? It seems reasonable given that 2% of the vote is then 4 standard deviations. Of course the problem just gets worse with more votes. JackSchmidt (talk) 17:09, 12 November 2008 (UTC)

I don't see why you'd assume that people vote randomly. Most people vote a straight party ticket who vote at all, and even among the swing voters I doubt the decision is anything like random. The magic statistic fairies only give the right answer to people who check their assumptions. Black Carrot (talk) 18:01, 12 November 2008 (UTC)

Actually, as near as I can tell, the model where they vote randomly, and the model where people decide randomly whether to vote or not (or are included in the random poll) are nearly identical. Assuming a certain percentage decide to vote, but unexpected influences alter that percentage according to a normal distribution, also does not appear to significantly change the analysis. It seems to me that if the number of voters who will vote is large, then the election almost certainly reflects the true population's preferences, and so if the preference is anything distant from p=0.5, then the election outcome is certain already from large enough polls. JackSchmidt (talk) 19:19, 12 November 2008 (UTC)

Let's put it in this way

Suppose we can prove that with the binomial distribution with p=0.5 the probability of a less than d% difference in a population of N individual is less than 0.000001.

1. Can we say that the same extimate holds for any typical real polupation of N individuals?
2. If the answer is "yes" we can reasonably think that the estimate is however not optimal, can we do better?--Pokipsy76 (talk) 18:04, 12 November 2008 (UTC)

The binomial distribution is appropriate if not everyone votes. As you have mandated that everyone votes, then there is no use for probability here. The difference between the votes is simply the difference between the size of the two camps. I don't see any reason to think that any particular number between 0 and N inclusive cannot be the number of people who favor choice A. Assuming the two choices can influence the vote and are equally adept at doing so, the end result after years of influence is likely to be that N/2 favor each choice.

If you are asking for historical information, then probably the math ref desk is the wrong place. As I mentioned above, landslide victory gives some historical data whether the difference is large. You could probably find an article on the 2004 US presidential election that points to historical data on other close elections. JackSchmidt (talk) 19:19, 12 November 2008 (UTC)

The point is that even if "any particular number between 0 and N inclusive can be the number of people who favor choice A" it is still true that small intervals near 0 - say for example (-0.001%,0.001%) - will have a small probability, therefore some sets of outcomes with small differences are almost impossible.--Pokipsy76 (talk) 19:37, 12 November 2008 (UTC)

I assume you mean "small intervals near N/2". And yes, they will be almost impossible, but no more so than any other difference. Assuming that N/2 is truly the mean of the distribution (which, in the long run and all other things equal, could be accomplished by the effect described by JackSchmidt), a difference of 0 (assuming an even number of voters) is more likely than a difference of 2, which is in turn more likely than a difference of 4, etc. It's just that among a million votes or more, there's so many possible differences that even the most likely ones become very implausible when singled out. -- Jao (talk) 19:54, 12 November 2008 (UTC)

What you say is correct, in fact the focus should be on the set of events (in this case a small interval near 0) and not on the single events.--Pokipsy76 (talk) 20:08, 12 November 2008 (UTC)

• Probability theory does not say something is impossible. It tells us what we do not know, and what we can expect. The chances of me being hit by a car on Friday 13 are very low. However, large numbers of people get hit by cars, and Friday 13 is no exception. So I could make a list of people who get hit by a car on Friday 13th, and worry about that. It's the same with elections: you would not expect a small difference between A and B always, but it can easily happen sometimes. Have you got some historical data you would want to analyse? Kaaskop6666 (talk) 20:00, 12 November 2008 (UTC)

Yes it happen sometimes because the probabity is small but not too small. There are events that do not happen "sometimes" but just in almost impossible cases (for example a quantum fluctuation which creates an exact copy of yourself ).--Pokipsy76 (talk) 20:08, 12 November 2008 (UTC)

I would expect marginal wins quite often, because the party that is expected to loose will work like hell to win. Kaaskop6666 (talk) 20:03, 12 November 2008 (UTC)

You obviously can expect marginal wins quite often, but not "too" marginal wins quite often.--Pokipsy76 (talk) 20:08, 12 November 2008 (UTC)

Of course, when defining "marginal" you would probably do so by looking at the historical distribution of margins, so it's all circular. --Tango (talk) 00:01, 13 November 2008 (UTC)

November 13

What is the purpose of a quadratic equation?

My Mathematics education ended after three years of (honors level) classes in high school. One of the "fun" things to do was to solve quadratic equations. For the rest of my life I have never seen or used one, though, and I never knew what they are for, when or why they are used. I recently read a book by Carl Sagan who described the series of learning steps that would be required before one could even understand the vocabulary of quantum physics. Is there a progression in math that was never taught to me, that would explain to me the point of leaning Quad equations, or the Calculus, other than for their own sake in school?Skinjohn (talk) 07:53, 13 November 2008 (UTC)

There are many real world situations where quadratic equations are used. probably the most common involves the relationship between distance speed and acceleration. If you are an engineer designing a production line in a factory, you will need to solve such equations to determine how to safely and quickly get product from A to B. There are millions of related examples. -- SGBailey (talk) 09:00, 13 November 2008 (UTC)

An ancient example from geometry: The area of a rectangle is 12 and the perimeter is 14, what are the sides? This problem leads to the two equations x·y = 12 and x+y+x+y = 14. From the second equation you get y = 7−x, which inserted into the first equation gives a quadratic equation x·(7−x) = 12. It is interesting, though, that your simple question is hard to answer. Solving quadratic equations is the mathematicians tool. A craftsmans tool is not appreciated by those who do not know the craft. You cannot explain a screwdriver to a person who do not know a screw. The progression of math, which you have never been taught, is the history of mathematics. The quadratic equation is from antiquity while calculus is from the renaissance. Bo Jacoby (talk) 10:31, 13 November 2008 (UTC).

The comparison with tools seems pretty apt to me. Certainly, one can get through life just fine without ever solving a quadratic equation (outside school, anyway), just as one can get by just fine without ever using a chisel or a hacksaw. Even so, on the off chance that you ever find yourself needing to solve a quadratic equation or carve a hole in a piece of wood, it's good to know that such tools exist and maybe even have some idea how to use them. Mind you, I'm a mathematics student myself, and I've never really got around to fully memorizing the quadratic formula. I probably should, but generally it's enough for me to know that there is a formula that I can look up (online or in a book) if and when I need it. I don't actually own a chisel, either — but if I need one, I know where to get one. —Ilmari Karonen (talk) 12:28, 13 November 2008 (UTC)

Look it up? It's quicker to just re-derive it. Algebraist 13:12, 13 November 2008 (UTC)

Or just complete the square - the quadratic formula is just a way to skip ahead to the final answer, there's nothing wrong with going through it step by step. --Tango (talk) 15:24, 13 November 2008 (UTC)

Quadratic equations have no real purpose in higher mathematics but can be useful for finding values relating to projectile motion or, in general, basic physics. It can also be useful in calculus. Topology Expert (talk) 07:24, 14 November 2008 (UTC)

I'm not sure I'd agree with that, I find quadratic equations come up all the time in various branches of mathematics. Just today I was doing some affine geometry which involved the characteristic polynomial of a 2x2 matrix - a quadratic equation. --Tango (talk) 19:54, 14 November 2008 (UTC)

I'd certainly disagree with that. The entire study of elliptic PDEs, especially generalized solutions to them, often involves comparisons with particular quadratic functions. RayAYang (talk) 04:54, 18 November 2008 (UTC)

Oldest known unsolved mathematical conjecture? (Possibly with legitimate references?)Taffycaker (talk) 13:12, 13 November 2008 (UTC)

Hello wiki volunteers, my question is: What is the oldest known unsolved mathematical conjecture?(Possibly with legitimate references?) I have Google searched this, and read through wikipedia and the internet which has many claims (ie goldbach, twin prime, etc) but rarely with the appropriate references so can verify their claims. One exception is for twin primes conjecture which seems to be located in Euclid, though searching for Euclid's references of it doesn't verify the claim, so could any of you please provide the proper answer since there's always some guy claiming to prove the oldest conjecture (ie Kepler's conjecture/Hales) but isn't really certain that it is the oldest, which is annoying. Thanks for your time. Taffycaker (talk) 13:12, 13 November 2008 (UTC)

Does an unproved assertion count as a conjecture? According to this, Nicomachus claims in his Introduction to Arithmetic that all perfect numbers are even, which would beat Goldbach and Kepler. Algebraist 13:44, 13 November 2008 (UTC)

The Guinness Book of Records (2001 edition) lists Goldbach's conjecture as the "longest standing unresolved maths problem". Hut 8.5 18:33, 13 November 2008 (UTC)

I don't know whether Euclid wrote down conjectures but I guess he (and other ancient Greeks) privately speculated about things closely related to his writings and tried to prove them, for example whether there are infinitely many Mersenne primes. He proved the infinitude of primes and proved that Mersenne primes generate perfect numbers, so it would appear unnatural to not at least think about infinitude of Mersenne primes and whether they generate all perfect numbers (it was later proven they generate all even perfect numbers and the existence of odd perfect numbers is unknown). PrimeHunter (talk) 18:56, 13 November 2008 (UTC)

pi

what is pi to the nearest 20 digits?Fern 24 15:58, 13 November 2008 (UTC)

Oh yeah, and in roman numerals to the nearest 5 digits?Fern 24 16:00, 13 November 2008 (UTC)

According to pi, π=3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510... Roman numerals don't seem to have involved a systematic way of writing fractions. Algebraist 16:04, 13 November 2008 (UTC)

Although this may seem a good question, it is not really much use to know what pi is. You do not need to know what pi is (correct to 20 d.p) to do mathematics and nor do you need to know it correct to 2 d.p. Perhaps something a little more non-trivial would be to show that there is no polynomial equation with rational co-efficients such that pi is a solution to that equation (try it!).

Topology Expert (talk) 07:19, 14 November 2008 (UTC)

But do not expect to succeed. Algebraist 11:53, 14 November 2008 (UTC)

As for the Roman numeral approximation, ³⁵⁵/₁₁₃, should do it, as that's good to 7 significant figures. In Roman numerals, that would be ^CCCLV/_CXIII. StuRat (talk) 19:17, 16 November 2008 (UTC)

Did the Romans write fractions like that? --Tango (talk) 20:22, 16 November 2008 (UTC)

In fact, we have a section on Roman numeral fractions, Roman numeral#Fractions. According to that, pi would be approximately III•ЄƧ»»»»», that is 1+1+1+1/12+1/24+1/72+1/1728+1/1728+1/1728+1/1728+1/1728≈3.1418. --Tango (talk) 20:32, 16 November 2008 (UTC)

Dominoes game strategy

There's a game I have on my iPhone called Dominosa and I can't think of a strategy to use to solve it besides random guessing. You start off with an grid of numbers (0-3, 0-6, 0-9 depending on difficulty) and you need to create a grid of unique dominoes. They'll turn red if you've got a duplicate piece already, and it will look like this. Any tips? -- MacAddct1984 ^{(talk • contribs)} 16:22, 13 November 2008 (UTC)

I don't know the game, but are you putting the numbers on the grid or are the numbers there and you are pairing them off? Assuming the latter, which is a pencil and paper game I have come across, I always start by (a) listing all the dominoes (see triangular numbers of 4, 7 and 10; (b) look for the unique doubles; (c) look for each other domino being unique; (d) Look for places that already placed dominos force the shape of another domino. (When you scan the grid for (say) domino 1-2, there may be three places it could go. Once another domino is placed, this scan may be affected and you have to redo it. In some puzzles, you have to say "1-2" can be here or there. If here then this and this follow if there then that and that follow. One of the two will probably lead to a dead end, so ideally you'd do this in your head to prevent redness. -- SGBailey (talk) 23:18, 13 November 2008 (UTC)

Go through the 28 patterns from (0,0) to (6,6) one at a time and see if each appears more than once on the grid - if so, pass on. Mark those that are unique on the grid and repeat this cycle until all have been found. With random positioning of the 28 dominoes it's unlikely that any other strategy is worthwhile, given that it's a fairly small problem anyway. That's for the 0-6 case, if I've understood you correctly, the others are similar. It's not random guessing, it's systematic search. In the empty grid you gave, (0,0) for example appears only once. Enclosing that pair immediately removes one (0,1), one (0,4), two (0,7)s and two (0,9)s from consideration…81.132.235.188 (talk) 23:41, 13 November 2008 (UTC)

And there's only one (8,8) pair (in the lower right corner), which immediately implies (2,5) to the right of it.
And there is only one place, where one end of (3,3) can be (it's the third digit from top and from left edge of the board), so — although we can't say in advance which 3 will be the other end — we know for sure the (3,0) pair is somwhere else. :) --CiaPan (talk) 11:17, 14 November 2008 (UTC)

Just for your convenience: in the version I've played, from Simon Tatham's Portable Puzzle Collection, you can add lines between numbers by right-clicking, indicating that you know these two can't form a domino together. This helps a lot with solving. Perhaps the iPhone's version has this option, too? Is it actually the same implementation, ported to the iPhone? Oliphaunt (talk) 17:36, 14 November 2008 (UTC)

Oh wow, that's awesome, I'm going to have to grab that for my computer. Yup, it's pretty much a direct port of that; Simon Tatham is [listed as one of the people] behind the project. And yup, you can draw lines splitting up possibilities. -- MacAddct1984 ^{(talk • contribs)} 19:04, 14 November 2008 (UTC)

I guess it really is simply a matter of looking for unique combinations and keep eliminating possibilities. It just seems like such a tedious way to go about it, especially for the largest grid size. -- MacAddct1984 ^{(talk • contribs)} 19:04, 14 November 2008 (UTC)

Curve Sketching Program

Does anyone know a good computer program for curve sketching (graphing). I use the description curve sketching because I am interested in a program that is smart enough to know the important features of a graph (e.g. intercepts, stationary points, critical points, etc.) and mark them on the graph rather than just generating a curve and putting unimportant numbers on the axes.AMorris (talk)●(contribs) 23:30, 13 November 2008 (UTC)

I suggest you either use a graphics calculator (hate those animals) or use NuCalc (maybe useful for calculus).

Topology Expert (talk) 07:21, 14 November 2008 (UTC)

November 14

Fractal circle

Sorry if this is a dumb question, my math is old and crackin'. Ever since encountering Mandelbrot in the library, I've been wondering if one could describe circles with a fractal equation instead of Pi. The obvious advantage of such an application would be that it would be scalable and should thus be superior to Pi. Just to clarify I'm not looking for a fractal circle as in add smaller and smaller circles, but something that would have a circle as result (sort of like:for radius r element size z add another element x with displacement y, repeat till you get a circle}. If this could be done, what would it look like and if yes, why should we not replace existing "pi equations" with it? 76.97.245.5 (talk) 00:43, 14 November 2008 (UTC)

Are you acquainted with complex numbers and the exponential definition of trig functions? That might give you some ideas. —Tamfang (talk) 07:25, 14 November 2008 (UTC)

Thanks that will give me a starting point. I vaguely recall thinking "Why didn't they teach us this to begin with" when we got to "higher math" way back when. Unfortunately they didn't go into great detail and I've forgotten most of that since then. I have a suspicion that I will come back and ask "Why do we bother with pi." after I've read up on the subject (again). Thks. for your help.76.97.245.5 (talk) 10:03, 14 November 2008 (UTC)

Although a circle is not, strictly speaking, a fractal, it can be viewed as the limiting curve of a series of regular polygons. Archimedes obtained an approximate value for π in Measurement of a Circle by calculating the perimeter of a 96-sided regular polygon, and Chinese mathematician Liu Hui obtained a more accurate value using a 192-sided polygon. Our article on Liu Hui's π algorithm describes how the method can be extended to approximate π as closely as you like by calculting the perimier of a polygon with 3(2ⁿ) sides, which only involves taking repeated square roots. Gandalf61 (talk) 10:54, 14 November 2008 (UTC)

I was not looking for a way to calculate π or calculate it more precisely. The error using a certain value of pi is not related to the size of the circle in question. I was rather trying to replace π with a way of describing circles for which any errors introduced are related to the size of the object. I haven't had time to look into the complex number avenue suggested by tamnfang in detail, but it looks promising. 76.97.245.5 (talk) 05:44, 16 November 2008 (UTC)

I'm not sure I understand. You can describe a circle using x²+y²=R² (where R is the radius). No 'pi' anywhere in sight. Pi only crops up when you want the circumpherence or area of the circle. SteveBaker (talk) 06:28, 16 November 2008 (UTC)

Or "all points p such that d(p,0)=R" (where d is the distance function, or metric), you don't even need coordinates. --Tango (talk) 13:57, 16 November 2008 (UTC)

If your question is, "are there any fractally things that can output a circle", take a look at Indra's Pearls. Black Carrot (talk) 21:18, 20 November 2008 (UTC)

what kinds of math can and can't you do orally (no symbols)

what kinds of math can and can't a person of exceptional intelligence understand all orally (no symbols)?

relatedly, are there any blind mathematicians? —Preceding unsigned comment added by 83.199.126.76 (talk) 03:40, 14 November 2008 (UTC)

Wikipedia has a very incomplete list of blind mathematicians and scientists. Leonhard Euler (not on the list) was not blind from birth, but was nearly blind in his right eye by the time he was in his early thirties and later lost sight in his other eye due to a cataract. Bernard Morin (on the list) is a topologist who lost his sight as a child. He helped exhibit an eversion of the sphere (Smale had earlier proved existence). If you would like more information, the article The World of Blind Mathematicians in the Notices of the AMS is worth reading. Michael Slone (talk) 04:26, 14 November 2008 (UTC)

Let's separate a few concepts and misconceptions here:

• Blind mathematicians are not restricted to verbal communication. There are mathematical "dialects" of Braille, such as Nemeth Braille and GS8 Braille, which, I believe, use systems of markup similar to LaTex.
• Verbal communication involves symbols just as much as written communication - the spoken phrase "one hundred and twenty three" and the written string of digits "123" are both symbols for the number 123.
• If the question is really "what kinds of math can a blind person do", then I believe the answer must be any kind, at any level - a mathematician's most productive activity is thinking, not reading or writing. The AMS article mentioned above makes a good case for its claim that mathematics is, if anything, more accessible for blind people than other professions. Gandalf61 (talk) 09:57, 14 November 2008 (UTC)

You might be interested in Learning styles#Other models Different people think in different ways. Despite what lots of people write one can still think without words, I can see it might be a primary means of thought for most people who write books though :) Dmcq (talk) 13:50, 14 November 2008 (UTC)

anything ever proved true that was almost certainly false (or vice versa?)

Have there ever been a proofs of the truth or falseness of a conjecture that through empirical testing had been almost certain to have the opposite truth value? —Preceding unsigned comment added by 83.199.126.76 (talk) 14:28, 14 November 2008 (UTC)

Borsuk's conjecture...? --CiaPan (talk) 15:14, 14 November 2008 (UTC)

Gödel's Incompleteness Theorem was a surprise. Bo Jacoby (talk) 15:17, 14 November 2008 (UTC).

Skewes' numbers are linked to such kind of conjectures.--Pokipsy76 (talk) 16:44, 14 November 2008 (UTC)

first mathematician. left-handed mathematician.

who is the first mathematician known by name?

of the truly famous mathematicians, are any left-handed? —Preceding unsigned comment added by 83.199.126.76 (talk) 18:06, 14 November 2008 (UTC)

The first name I see mentioned in History of mathematics is the Greek mathematician (and philosopher), Thales. --Tango (talk) 19:38, 14 November 2008 (UTC)

Wikipedia is quite keen on purging lists of left handed people so its hard to find this info here. Anyway Charles Dodgson aka (Lewis Carroll) Alan Turing, and Nicole Oresme[1] were lefthanded mathematicians. There is speculation that Pythagoras was a lefty[2]. There is some very speculative research disproportionate number of people with high mathematical ability are left handed.[3]--Salix (talk): 23:47, 14 November 2008 (UTC) (a lefty)

What?! Why would it purge this, it's way more relevant, notable and incontrovertible, not to mention well-documented with copious references, than ethnicity or nationality or.... How do they possibly justify this? —Preceding unsigned comment added by 83.199.126.76 (talk) 01:08, 15 November 2008 (UTC)

I suspect that they have some sinister motive. :-) StuRat (talk) 19:00, 16 November 2008 (UTC)

C'mon Stu, lets call a spade a spade ;-) -hydnjo talk 03:09, 17 November 2008 (UTC)

it's just that when they made sure all the information was right, nothing was left.Gzuckier (talk) 19:18, 19 November 2008 (UTC)

I heard it was all due to the actions of one evil deletionist named Dexter. StuRat (talk) 00:58, 20 November 2008 (UTC)

O(n log n) Polynomial Derivative Evaluation?

I was wondering if there's an algorithm for calculating the value of a degree-n polynomial and its n derivatives at a given point in O(n log n) time. The extended Horner's rule does it in O(n^2) time, but there are a large number of algorithms that are nominally O(n^2) which can be evaluated recursively to get O(n log n) time. I've tried making algorithms that subdivide high and low terms and that subdivide even and odd terms, and I can't get it to work. --Zemyla^t 23:48, 14 November 2008 (UTC)

Yesterday, reading your post, I doubted it, but not anymore. If the point you're considering is zero, we get the answer immediately, looking at the polynomial as a Taylor series,

$p(x)=\sum _{j=0}^{n}p_{j}x^{j}\Rightarrow p^{(j)}(0)=p_{j}j!{\text{.}}$

Now let's look at the problem not just at zero, but at a constant $a$ instead. That is, we're interested in $p^{(j)}(a)\ \forall j\in \{0,1,\dots ,n\}{\text{.}}$ $p(a)$ can be expressed as a polynomial at zero,

$p(a)=\left.\sum _{j=0}^{n}p_{j}(x+a)^{j}\right|_{x=0}{\text{.}}$

If we knew the coefficients before $x^{j}$ in that expression, we would immediately know the $p^{(j)}(a)$ we're looking for. Let's expand it using the Binomial theorem and the Heaviside step function ($H(x\geq 0)\equiv 1,\ H(x<0)\equiv 0$):

$\sum _{j=0}^{n}p_{j}(x+a)^{j}=\sum _{j=0}^{n}p_{j}\sum _{k=0}^{j}x^{k}a^{j-k}{j \choose k}=\sum _{j=0}^{n}p_{j}\sum _{k=0}^{n}x^{k}a^{j-k}{j \choose k}H(j-k)=$

$=\sum _{j=0}^{n}p_{j}\sum _{k=0}^{n}x^{k}a^{j-k}{\frac {j!}{k!(j-k)!}}H(j-k)=\sum _{k=0}^{n}{\frac {1}{k!}}x^{k}\sum _{j=0}^{n}(j!p_{j}){\frac {a^{j-k}H(j-k)}{(j-k)!}}{\text{.}}$

We've implicitly defined

${\frac {H(x)}{x!}}\equiv 0\ \forall x<0{\text{.}}$

Defining $f_{j}=j!p_{j}$ and $g_{j}={\frac {a^{-j}H(-j)}{(-j)!}}$ it's clear that $p^{(j)}(a)$ is calculable as the discrete convolution $(f*g)_{j}{\text{.}}$ Both $f$'s and $g$'s relevant parts are sequences of length $n+1$ and computable in $O(n)$ time. The convolution, on the other hand, can be calculated by some fast convolution algorithm in $O(n\log n)$ time, the goal sought! In particular, both sequences can be fast-Fourier-transformed in $O(n\log n)$, the transformed ones multiplied together in $O(n)$ and finally inverse-transformed in $O(n\log n)$ again. — Pt _(T) 13:16, 18 November 2008 (UTC)

November 15

Countable compactness => Lindelof (under what conditions)

This may be a silly question but please give me some sleep (and an answer too if possible):

Is there a countably compact, separable, first countable space that is not Lindelof?

Seperable, first countable spaces that are not Lindelof are easy to find (for instance the Sorgenfrey plane). But does adding countable compactness ensure the Lindelof condition? It seems (if the example I have in mind, is correct) that one can find a counterexample assuming the continuum hypotheses. Otherwise, the problem is a 'little' less trivial. I find it amazing that an 'unrelated axiom' (countable compactness) can make finding a counterexample a lot harder.

If this problem is so obvious and I have not been thinking properly a reference would be much appreciated (or a proof if it is a 'textbook theorem').

Thanks in advance.

(Some immediate 'narrowing down' can be done by eliminating topological groups and more importantly metric spaces but this 'narrowing down' is not that useful if my guess that the answer to the question is 'no', is correct).

Topology Expert (talk) 12:40, 15 November 2008 (UTC)

Have you checked Counterexamples in Topology? I'm in a different city from my copy at the moment. Algebraist 12:51, 15 November 2008 (UTC)

Yep, I checked there (first) but it is not there. Topology Expert (talk) 08:09, 16 November 2008 (UTC)

Number of terms

Could you please explain in detail how to deduce the formula to find the number of terms in the expansion of ( 1+x+x^2+x^3+... )^n (in simplified form)Kasiraoj (talk) 15:12, 15 November 2008 (UTC)

To what power of x is the part in brackets meant to go up to? If it goes on forever (which is what having "..." at the end usually means), then there will be infinitely many terms in the expansion. If you mean it to stop at a certain point, say x^m, then the expansion will have mn+1 terms. The lowest order term is going to be 1 (=1ⁿ) and the highest order term is going to be x^mn (=(x^m)^n), and there aren't going to be any terms missing inbetween since any integer between 0 and mn can be made by adding together n integers from 0 to m (just add together as many m's as you can without going over, then add on the amount necessary to get to the desired number, and then make the rest of them 0). --Tango (talk) 16:36, 15 November 2008 (UTC)

If $|x|<1$ then the sum of the geometric series inside the parenthesis is ${\frac {1}{1-x}}$ so the value of the expression in question can be found in closed form. Topology Expert (talk) 08:14, 16 November 2008 (UTC)

I guess that depends on whether you consider a Laurent polynomial to be a simpler form than a power series, which probably depends on context. --Tango (talk) 13:48, 16 November 2008 (UTC)

November 16

definition and usage of "identity"

Mathworld says: An identity is a mathematical relationship equating one quantity to another (which may initially appear to be different).

Wikipedia says: An identity is an equality that remains true regardless of the values of any variables that appear within it

I guess this is distinguished from mere "equations" in that $\sin ^{2}x+\cos ^{2}x=1\qquad \qquad \!$ is an identity whereas ƒ(x) = sin(x) is an equation.

So my question is, both these definitions seem to apply to both Euler's formula ($e^{ix}=\cos x+i\sin x\!$) and Euler's identity ($e^{i\pi }=-1\,\!$), making them both identities. So is there a particular reason why Euler's identity is called such?

Thanks, --VectorField (talk) 07:03, 16 November 2008 (UTC)

I converted the mathematical symbols into LaTex for you (hopefully this makes it easier for other editors to read your message). Topology Expert (talk) 08:08, 16 November 2008 (UTC)

The "formula" is clearly an identity, it's a equation that holds for all values of x. The "identity" on the other hand just equates two constants, so I guess it is trivially an identity (it holds for all values of all the variables that appear, since there are no variables), but it's not what we usually mean when we talk about identities. I think the problem is that Euler just has far too many things named after him (interestingly, I think he was actually involved in the discovery of some of them, which may be unique in mathematics!) so you have to come up with different names to give them which results in stretching the definitions a bit to find something that hasn't already been used. --Tango (talk) 13:55, 16 November 2008 (UTC)

Yes, precisely... List of topics named after Leonhard Euler is a rather long article! Eric. 131.215.158.179 (talk) 00:40, 17 November 2008 (UTC)

1x1 matrix determinant

what is the determinant of a 1x1 matrix? ----The Successor of Physics 14:46, 16 November 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

The number in the determinant. Now if you'd asked about a 0x0 determinant you might have got an argument - I'll stick my neck out and say 1. Dmcq (talk) 14:50, 16 November 2008 (UTC)

Just to be contrary, I'd say 0, a determinant is a sum over permutations, there are no permutations of no elements and an empty sum is 0. --Tango (talk) 15:25, 16 November 2008 (UTC)

There is exactly one permutation of a zero-element set (the empty function), and so the sum-over-permutations is the sum of one term, that term being the empty product, which is 1. More signally, det is by definition a homomorphism from the general linear group of the space to the multiplicative group of the underlying field, and in particular must map the identity map to 1. Algebraist 15:32, 16 November 2008 (UTC)

I'll confess, I'd never before heard of "the empty function"... --Tango (talk) 15:55, 16 November 2008 (UTC)

I've never seen a definition of function that doesn't include it. It isn't always emphasized, but it's always there. Algebraist 15:58, 16 November 2008 (UTC)

Sure, when you think about it, the empty set is a set so there must be a function from it to other sets, I'd just never thought about it. --Tango (talk) 16:20, 16 November 2008 (UTC)

The empty function is somewhat important in the category of sets, since it makes the empty set an initial object of that category. Aenar (talk) 22:19, 16 November 2008 (UTC)

It's more basic than that: if for some reason we disallowed the empty function, then the category of sets wouldn't even be a category, since the empty set would lack an identity morphism. Algebraist 23:26, 16 November 2008 (UTC)

On a different note, if the determinant of a 0x0 matrix was not 1, then we wouldn't have $\det(A\oplus B)=\det(A)\det(B)$. 76.126.116.54 (talk) 04:29, 17 November 2008 (UTC)

I guess you mean matrix multiply, and 0=0×0. By the way I just had a look at the determinant article and it actually does mention the 1x1 and 0x0 determinants. It even has a specific reference for saying the 0x0 determinant is 1! Now that's a thorough job. Dmcq (talk) 09:25, 17 November 2008 (UTC)

No, he used the symbol for direct sum, and I'm sure he meant direct sum. Algebraist 11:34, 17 November 2008 (UTC)

Thanks, yes it would follow then. Never thought of it actually meaning that. Dmcq (talk) 11:57, 17 November 2008 (UTC)

Yes, it should be very unconfortable to deny the status of mathematical object to the empty function. Notice that all "trivial" objects like the empty map, the number "0", the identical transformation,... had their big trubles to be accepted as the others, and in fact they came as the last, historically. Even "1" was looked on as a suspicious number initially (because, after all, it does not require the plural form!). As to det of the 0x0 matrix, all definitions and all formulas agree on 1 as its value. For instance, it's a a sum of exactly 0!=1 term(s) (indicized by the empty permutation), the term beeing an empty product (thus 1). --PMajer (talk) 12:01, 17 November 2008 (UTC)

The Derivative

A couple of us can't seem to settle this argument so we turn to wikipedia experts to resolve this issue. What is the derivative of f(x) with respect to x if
$f(x)=\int _{a(x)}^{b(x)}g(x,t)dt$.
Thanks!--130.166.165.22 (talk) 22:15, 16 November 2008 (UTC)

You should look at Leibniz integral rule#Limits that are variable. Eric. 131.215.158.179 (talk) 22:33, 16 November 2008 (UTC)

So, $f'(x)=g(x,b(x))b'(x)-g(x,a(x))a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial g(x,t)}{\partial x}}dt$? Wohoooooo! Thanks!--130.166.165.22 (talk) 22:43, 16 November 2008 (UTC)

November 17

Polynomials

Are things like the mod function or the floor function allowed in polynomials? --Melab^±1 01:37, 17 November 2008 (UTC)

Well, a polynomial is just a function of the form $\sum \limits _{n}a_{n}x^{n}$ - you get a polynomial mod p quite often in various branches of mathematics, but such functions aren't really polynomials themselves, so not really. -mattbuck (Talk) 02:19, 17 November 2008 (UTC)

The answer is definitely not (for polynomials as in algebra). See the page polynomial for the formal definition.

Topology Expert (talk) 05:48, 17 November 2008 (UTC)

Polynomials over the integers mod p are polynomials, you can have polynomials over any ring. I don't think that's what the OP is talking about, though. --Tango (talk) 11:32, 17 November 2008 (UTC)

I have seen this term being used so often. What does 'OP' mean?

(My answer assumed that the person who asked the question was referring to calculus).

Topology Expert (talk) 11:43, 17 November 2008 (UTC)

OP here stands for original poster. Algebraist 11:51, 17 November 2008 (UTC)

It's hard to determine what the question really means, but if the OP means whether it's possible to define a mod function on a polynomial ring analogous to one of the possible mod functions in the ring of integers, then sure, why not? Polynomial division is even nicer than integer division as the remainders are made unique more naturally. Similarly, if we want to define floor as a function from rational expressions to polynomials, analogous to the floor mapping Q to Z, then why not? -- Jao (talk) 12:03, 17 November 2008 (UTC)

How would you define floor on rational expressions? Algebraist 13:27, 17 November 2008 (UTC)

Expand as a Laurent polynomial and truncate the negative order terms? That seems analogous to taking a decimal expansion and truncating it at the decimal point. It's going to depend on what point you expand around, of course. --Tango (talk) 15:56, 17 November 2008 (UTC)

Hang on, that could give a power series rather than a polynomial, couldn't it? I guess it's more analogous to taking the fractional part of a rational number. There are lots of analogies between (Laurent) polynomials and integers (rational numbers), but you have to reverse the order in some sense (positive powers of X in a (Laurent) polynomial correspond to negative powers of 10 in the decimal expansion of the rational number). --Tango (talk) 16:01, 17 November 2008 (UTC)

I had simply thought to define $\lfloor {\frac {f(x)}{g(x)}}\rfloor$ as the polynomial $q(x)$ such that $f(x)=q(x)g(x)+r(x)$ with $\deg r(x)<\deg g(x)$. I don't know how much sense it would make to call it floor, or how useful it would be, but there's at least a clear analogy with the Q → Z floor function. -- Jao (talk) 17:21, 17 November 2008 (UTC)

No not in Calculus or rings an 8th grade definition of a polynomial. —Preceding unsigned comment added by Melab-1 (talk • contribs) 21:56, 17 November 2008 (UTC)

Could you give an example of what you mean, then? I suspect the answer is "no". A polynomial is just something of the form a+bx+cx²+...+zxⁿ, where all the a,b,c,...,z are numbers (usually rational numbers if it isn't specified), anything else, and it isn't a polynomial. --Tango (talk) 00:50, 18 November 2008 (UTC)

Matrix

What are matrices really for? The article on wikipedia doesn't really explain that. ----The Successor of Physics 13:22, 17 November 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

Well, they have many uses. Possibly the simplest use is in geometry. Any n-dimensional transformation can be written as an n by n matrix T, so if you want to find out what happens when you apply the transformation to some vector, you take T(v) and multiply out the matrices to get your new vector. -mattbuck (Talk) 13:53, 17 November 2008 (UTC)

Linear transformation. --Tango (talk) 15:52, 17 November 2008 (UTC)

Matrices can be used to model real-world data as well. For example, if you have three types of shoes at a store, and want to model this for men and women, you could have a 3x2 matrix showing how many men and women bought each type of shoe. Then, to this 3x2 matrix, matrix operations can be applied to model different things. For example, you could have a matrix showing the cost for each type of shoe (perhaps a 1x3) and then multiply this matrix by your original matrix to give the total money men and women spent. Hope this helps, Ζρς ι'β' ^¡hábleme! 00:07, 18 November 2008 (UTC)

Markov chains are pretty sweet (see the subsection about Markov chains with finite state space). 76.126.116.54 (talk) 05:17, 18 November 2008 (UTC)

Mathmatic

The perimeter of a rhombus is 100 cm if one of its diagonals is 14 cm then the area of the rhombus is? —Preceding unsigned comment added by Nebul (talk • contribs) 14:59, 17 November 2008 (UTC)

This looks like homework to me, so I'll suggest looking at Rhombus#Area, and leave it at that. --Tango (talk) 15:46, 17 November 2008 (UTC)

Basic mechanics

It's a long time since I studied this, so wonder if I've got it right. Does statical equilibrium correspond to the position of lowest potential energy? Obviously true for a weight on a string or a ball at the bottom of a bowl, but it seems to work for more complicated systems as well. For example, A and B are fixed points at the same height, A has a light rod freely pivoted at it, B has a light string attached to it. The string passes through a smooth eye at the free end of the rod and has a weight at the end. The system will be in equilibrium when the two portions of string make equal angles with the rod and, using a numerical case to simplify things, some trig and calculus gave exactly the same position of the rod for the weight to be at its lowest. So if I'm right, equilibrium can be determined either by resolving forces/taking moments, or by minimising potential energy. In a particular case, could one way be simpler than the other?…81.154.108.92 (talk) 15:12, 17 November 2008 (UTC)

Stable equilibria correspond to *local* minima of potential energy (unstable equilibria correspond to local maxima, or inflection points), it doesn't necessarily have to be the overall minimum. Finding the potential energy and differentiating (twice, to check stability) can often be much easier than resolving forces, etc., especially for more complicated systems (you can use conservation of energy, so you know if kinetic energy is at a maximum, potential energy must be at a minimum, for example). --Tango (talk) 15:51, 17 November 2008 (UTC)

Agreed. Now, for a real-word example: A bowl of cereal on the table is in equilibrium, even though it would have less potential energy when knocked on the floor. The reason, as stated above, is that you would need to momentarily increase the potential energy to lift the cereal out of the bowl. StuRat (talk) 18:04, 17 November 2008 (UTC)

Real Analysis: Inverse of continuous function of interior set

Hi, can anyone please help with solution for this problem from:

Considering a function f:R->R, how can we show that f is continuous iff for every subset B $f^{-1}$(int(B) ⊆ int($f^{-1}$(B). Thank You. —Preceding unsigned comment added by 130.166.160.76 (talk) 21:06, 17 November 2008 (UTC)

This characterisation of continuity (which works for arbitrary topological spaces, not just R), is fairly easily shown to be equivalent to the standard definition of continuity for functions on top spaces. It's even mentioned in our article continuous function (topology). Algebraist 21:12, 17 November 2008 (UTC)

Emacs + LaTex on Windows XP

Any LaTeX head fancy helping this person out:

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Computing#Setting_up_LaTex —Preceding unsigned comment added by 78.86.164.115 (talk) 21:16, 17 November 2008 (UTC)