Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

May 20

Neg. Binom. GLM question

In doing a negative binomial glm in R, I found two oddities. Doing the analysis of deviance I get the message "Dispersion parameter for the negative binomial(0.553) family taken to be 1." I interpret this to mean no dispersion parameter, as in the case of a quasilikelihood, is estimated. Is this correct?

The second question is more detailed. In comparing two (nested, neg. binom.) models, the anova comparison, anova(model1,model2), shows a different log-likelihood than taking the deviance(model1)-deviance(model2). Not very different, mind you, about 0.1, which is more that I could attribute to rounding error. Any ideas? --TeaDrinker (talk) 01:57, 20 May 2008 (UTC)

Suggestions.

1. Are you using glm or something from the MASS package? (Or something else???) Whichever you are, look at the help page for the relevant function, especially where it talks about "family". But I would suspect your hunch is true regardless, that the dispersion is assumed fixed.
2. Two thoughts. You will want to make sure the fitting method for both models is the same, i.e., don't use REML for one and ML for the other; in fact, you should probably use ML for both and not REML (which is the default for some functions, BTW), as in many cases the comparison of REML-fitted models can be meaningless. Second, check the "type" argument of the relevant anova method. Make sure it will do the equivalent test to the comparison of the deviances, as some "types" do different tests.

Check also the R-Help mailing list if you need to. The readers there are very helpful and knowledgeable—just make sure you read the manuals that came with your R distribution, the relevant function help pages, and search the archives of that list before posting a question there. If you don't, they can be quite, er, uncivil.  ;-) Baccyak4H (Yak!) 03:20, 20 May 2008 (UTC)

Thanks! It is the glm.nb function from the MASS library; I was unaware of any ordinary glms being fit using reml; (although lme and glmm models I have seen it used). I may have to send it over to the r-help, thanks! -TeaDrinker (talk) 20:39, 20 May 2008 (UTC)

Partial differential equation

Heh...does anyone wanna help me out and solve the partial differential equation given in the last problem here? Since it IS a PDE, it is quite a lot of grunt work for not much benefit. But if you were to get (1/2)sin(3x)e^(3t)+(1/2)sin(3x)e^(-3t), I would be happy and grateful. My answer is a solution to the PDE, but I'd like to know if it's the general solution, as was asked for. --M1ss1ontomars2k4 (talk) 02:42, 20 May 2008 (UTC)

I think you got it. Obviously $u_{xx}=-9u$, so $u_{tt}=9u$ which has the general solution $u(x,t)=a(x)e^{3t}+b(x)e^{-3t}$ (or equivalently $u(x,t)=c(x)\cosh(3t)+d(x)\sinh(3t)$). The initial conditions narrow that down to $u(x,t)=\sin(3x)\cosh(3t)$ just like you have it. --Prestidigitator (talk) 05:59, 20 May 2008 (UTC)

Cool, thanks. I would never have thought of using a hyperbolic trig function to express it; that's neat. --M1ss1ontomars2k4 (talk) 02:21, 21 May 2008 (UTC)

Standard deviations and probabilities

Hi. I'm working on Donald Bradman, trying to get the article to FA. As long ago as Dec 2006, this edit was made, with no edit summary or discussion at the article talk page. It's never been challenged or referenced.

As the requirements of FAC mean that every claim needs to be referenced, and the user seems to have left the Project (with no email address enabled) I wondered whether the probability figures he's used derive from the (cited) standard deviation figures. As my maths is rubbish, I have no idea.

So, three questions:

1. Do the probability figures derive from the SD ones?
2. Are they accurate?
3. If we're so far at a "yes" and "yes" scenario, how can I possibly cite the probabilities?

Cheers --Dweller (talk) 10:41, 20 May 2008 (UTC)

Yes, the probabilities derive from the number of SDs. They are accurate (except that the first should be 185,000, I think) subject to the following caveats: firstly, it is not clear on what basis they have been rounded to their current values. Without access to the book in question, I don't know the precise SD figures, but assuming the current figures are accurate to 1 decimal place only, the probability figures are being given far too precisely: for example, the Don's probability should be '1 in somewhere between 147,000 and 233,000'. Secondly, they depend on the assumption that the underlying distribution is approximately normal. I do not know if this is a sensible assumption.

As for sourcing, I doubt you'll be able to find these specific figures anywhere if they're not present in Davis' work (which I don't have access to). Algebraist 11:30, 20 May 2008 (UTC)

Thanks. I don't think Davis includes the probabilities, but getting hold of a copy of the book has been a bit of a problem! Your comment of "'1 in somewhere between 147,000 and 233,000'"... does that derive from a log chart or something that I could reference? --Dweller (talk) 11:36, 20 May 2008 (UTC)

I couldn't find precise enough normal distribution tables online (though they doubtless exist somewhere), so I used this tool, linked from normal distribution. Algebraist 11:48, 20 May 2008 (UTC)

I had a reply from User:Macgruder. The maths is beyond (I mean way beyond) my poor brain:

You can work these out by number how many standard deviations you are from the mean

For example

Move than 2 [actually 1.96] standard deviations above mean is about 2.5% (1/40)

http://davidmlane.com/hyperstat/z_table.html Type 1.96 into the 'above' box

Or look at the x-numbers here:

http://photos1.blogger.com/blogger/4850/1908/1600/sd%20table1.jpg

So Pele is 3.7 ---> 0.000108 ----> 1/9259 (about 1/9300)

The Bradman one is so far out you need to find special website that doesn't have rounding errors, but I calculated it to be as 1/184,000

This is not 'original' research. This is a straight correspondence to the statistical meaning of Standard Deviation. —Preceding unsigned comment added by Macgruder (talk • contribs) 13:28, 20 May 2008 (UTC)

So I'm stuck. I don't know what figures to include or how to reference them. Please help! --Dweller (talk) 13:46, 20 May 2008 (UTC)

The above poster has a good point in that 4.4 is anywhere between 4.35 and 4.45, which would give a range.

You can use this

/*
4.35: 0.99999318792039  (1/147,000)
4.4:   0.99999458304695	(1/185,000)
Agreeing with 4.4 --> 0.000005413 from published tables ; see below

4.45: 0.99999570276356  (1/232,000)


*/

function normdist($X, $mean, $sigma) {
  $res = 0;

  $x = ($X - $mean) / $sigma;

  if ($x == 0) {
    $res = 0.5;
  } else {
    $oor2pi = 1 / (sqrt(2 * 3.14159265358979323846));
    $t = 1 / (1 + 0.2316419 * abs($x));
    $t *= $oor2pi
        * exp(-0.5 * $x * $x)
        * (0.31938153 + $t
            * (-0.356563782 + $t
                * (1.781477937 + $t
                    * (-1.821255978 + $t * 1.330274429))));

    if ($x >= 0) {
      $res = 1 - $t;
    } else {
      $res = $t;
    }
  }
  return $res;
}

refer to this table :

http://www.math.unb.ca/~knight/utility/NormTble.htm which agrees with my calculations. Look at the far 'right numbers'

If you link to that table it indicates that the calculations are by and large correct (if we consider 4.4 to be correct). There is no more need to cite a SD to probability than a conversion of Inches to cms. It's a factual conversion. Macgruder (talk) 14:09, 20 May 2008 (UTC)

I'd like to reiterate that this is not a straight factual conversion, it is a conversion predicated on the assumption that the data are normally distributed. This assumption is of course straightforwardly wrong in a strict sense (the data are all nonnegative) but it might be a sensible approximation. I do not have the means to discover whether or not it is. Algebraist 14:15, 20 May 2008 (UTC)

Agree with Algebraist. In practice a data point like this that is more than 4 SDs away from the mean is far too much of an outlier to provide any useful likelihood estimates. I doubt that the sample size was any near large enough to support the "1 in 184,000" conclusion. I would delete the whole "Probability" column and the paragraph immediately following the table, and just leave the SD figures, which presumably come straight from Davis. It is reasonable to state the SD figures, if they are given by Davis, but we should not embroider them with additional interpretations. Gandalf61 (talk) 14:25, 20 May 2008 (UTC)

That's exactly the conclusion User:The Rambling Man and I have just come to. The probabilities are just that bit too close to debatable (and are definitely very hard to source) for the strictures of WP:FA and I'm afraid they'll just have to go. We have sourced comment giving context for just how good people in other sports would have to be to reach such an exalted SD, which will do just fine. Thanks all, especially Macgruder who responded very quickly to my request for help. --Dweller (talk) 14:37, 20 May 2008 (UTC)

I mean 'factual' in the sense that (if the author is stating the statistics are a normal distribution) it is a straight interpretation of the number 4.4. The discussion of statistically what 4.4 means in terms of how it can be interpreted or how reliable it is, is not under discussion here. We are taking an authors conclusion and converting what it means. Any reliability issues with 4.4 must be sourced from the author. If the author states the 4.4 figure is not reliable then so can we. Of course, I can't find the original book so I don't know if he says it's normal or not. If in another article someone states X is 1.96 SDs above the normal we could state that this is 2.5% - it would not be appropriate to go into a discussion of sample bias, etc etc if the respected source doesn't do so. My feeling is that either the 1/x probabilities stay or the the whole table goes. If we feel that the authors original statistics are not a valid reference then fair enough. However, if we feel that the author's reference is valid and he himself does not discuss the 4.4 then there is no reason to remove it. [unless of course the author says that it's not a normal distribution]. These numbers make the S.D's clear to understand to a reader, and if we have problems with them anyway the whole table shouldn't be there.

Although thinking about it, major championships is obviously not Normal. It will heavily skewed towards zero, but some of the other stats like goals might be, and cricket scores might be too. I wish I could see the original source.

Macgruder (talk) 14:53, 20 May 2008 (UTC)

Algebraist is correct that it's not possible to score <0 in these statistics. Davis seems to have only cited the SDs. To show their impact in a practical sense, rather than looking at probabilities of finding others with such skill, he's applied the SDs to the leading stats in various sports, showing what a (say) baseball batting average would need to be to be Bradmanesque. --Dweller (talk) 15:06, 20 May 2008 (UTC)

Fields

For the field of integers mod p, I cannot find an inverse element for all cases. $x*x^{-1}=1$ where x = 0 does not seem to work. What have I missed? 81.187.252.174 (talk) 11:10, 20 May 2008 (UTC)

0 is the only element with no multiplicative inverse. You've missed nothing. –King Bee ^{(τ • γ)} 11:29, 20 May 2008 (UTC)

Hurf, didn't read the definition properly. Blargh burp 81.187.252.174 (talk) 12:14, 20 May 2008 (UTC)

Geometry question on angles

Assume a circle (call it A if you desire) on a line with radius r. Constructed at 2r from its center point (rightwards direction) is an identical circle B of radius r. Then assume an angle θ in the first circle, extending outwards as a line in the direction of the second circle. How does one find the point for the rightmost intersection of circle B's perimeter, expressed as a θ(b)? Ah, I will upload something to help illustrate my example! Right here! I'm thankful for any help. Scaller (talk) 16:09, 20 May 2008 (UTC)

The right circle has equation $(x-r)^{2}+y^{2}=r^{2}$ and the line has equation $y=(x+r)\tan \theta$ (you can work this out: 2 points on the line are (-r, 0) and (0, $r\tan \theta$), the latter by trigonometric identities.) Then it's just a simultaneous equation which may be able to be solved (haven't gone any further). x42bn6 Talk Mess 16:34, 20 May 2008 (UTC)

Alternatively, you could use the law of sines. You have a triangle with one known angle and two known sides. The law of sines, combined with the fact that angles in a triangle add up to 180 degrees should be enough to determine all the other angles and sides. That will then give you the point in polar coordinates with the centre of the second circle as origin, which you can then convert to Cartesian coordinates if you want. --Tango (talk) 16:43, 20 May 2008 (UTC)

Game show theorem

I came across the monthy hall problem some time ago, and I'm curious about how it applies to more complex situations. Anybody have a clue? Bastard Soap (talk) 21:42, 20 May 2008 (UTC)

What kind of more complex situations did you have in mind? There are all kinds of situations in probability were intuition turns out to be wildly incorrect. Take a look at Prosecutor's fallacy for one example. Also, if you haven't seen it already, we have an article on the Monty Hall problem. --Tango (talk) 21:55, 20 May 2008 (UTC)

I’ve seen this problem which is somewhat similar cause much debate (and I remember it because I was convinced I had the right answer for several hours until I figured out otherwise).

The situation is that there are 3 people, and they will enter the same room and receive a hat (either red or blue, each with a 50% chance). They cannot communicate whatsoever once in the room, but they can collaborate and determine a strategy before entering the room.

Then at the exact same time each of them is to guess what color their hat is (or they may choose not to guess at all). If at least one person guesses correctly and nobody guesses wrong, they win a prize.

The question is, given an optimal strategy, what is the probability that they will get the prize? GromXXVII (talk) 22:36, 20 May 2008 (UTC)

Give me a minute... --Tango (talk) 22:44, 20 May 2008 (UTC)

Nope, not happening. At least one person has to guess, and they have a 50% chance of getting it wrong and blowing everything, so there is no way to improve on 50%. However, if the answer was 50%, you wouldn't have asked the question... Damn you... Let me sleep on it. --Tango (talk) 22:50, 20 May 2008 (UTC)

75%. Turning my computer off and going to sleep: Take 2. --Tango (talk) 23:02, 20 May 2008 (UTC)

I had first confused this with another problem where after receiving the hats, the players are asked one by one if they know their own hat's color, and after at most 5 (IIRC) queries, someone will know. BTW, Tango's going to sleep quip is actually quite a cute coincidence, as a New York Times <SPOILER ALERT>article </SPOILER ALERT> covering the problem mentions someone going to bed after solving it and recognizing its relevance to coding theory as he fell asleep. And yes, the answer is 75% Baccyak4H (Yak!) 04:11, 21 May 2008 (UTC)

I'm having a hard time understanding the problem completely. Each one gets a hat (at the same time?), but can't see what color their own hat is? Can they see the color of each of the others' hats? --Prestidigitator (talk) 04:55, 21 May 2008 (UTC)

You have to assume that they can in order to give them a better than 50% chance of getting it right. And it took me a while to work out what the strategy is but I knew it had something to do with what to guess given what you see. (Here's a hint: no matter what the arrangement of hats, there will always be at least two people with the same colour. If you're a prisoner who can see two hats the same colour, or two hats of different colour, see if that affects your optimum strategy for guessing your own colour.) Confusing Manifestation(Say hi!) 06:42, 21 May 2008 (UTC)

Interesting how the game show contestants became prisoners there. --tcsetattr (talk / contribs) 07:10, 21 May 2008 (UTC)

Am I missing something? If all are given with 50% probability then whatever colour hats the other people have won't affect the probability of guessing your own colour. The obvious answer seems to be that they should decide that just one should guess having a 50% chance. -- Q Chris (talk) 07:36, 21 May 2008 (UTC)

Yeah, you're missing it. The external link given by Baccyak4H contains a pretty thorough spoiler which should have you slapping your forehead and saying "Of course!" --tcsetattr (talk / contribs) 07:58, 21 May 2008 (UTC)

You’re right that each person that guesses has a 50% chance of guessing right, or guessing wrong regardless of any other factors. But the problem has three people, and whether one person guesses right or wrong is not independent of whether somebody else does (in an optimal strategy, which too my knowledge results in 75%). GromXXVII (talk) 10:52, 21 May 2008 (UTC)

The key point to note is that the method which the hats were assigned was made explicit; there is a prior distribution on what the hats can be. While when reading the problem the description seems trivial, it turns out to be crucial. If no information about that is known then of course there is no improvement over 50%. So look at this as a conditional probability problem, and consider Thomas Bayes your friend... Baccyak4H (Yak!) 14:28, 21 May 2008 (UTC)

Or just right out all the possible combinations - worked for me! --Tango (talk) 15:17, 21 May 2008 (UTC)

Right, but that still presupposes the prior. <SPOILER ALERT>What if the prior was probability 1 for all hats being red? What is the chance that the strategy will work in that case? </SPOILER ALERT> Was your sleeping comment intentional, or just a coincidence, with respect to the article I linked to above? Baccyak4H (Yak!) 16:37, 21 May 2008 (UTC)

Well, yes, but we were given the prior. I wasn't suggesting you were wrong, just that there's a less technical approach. Pure coincidence - I read the problem just as I was about to turn the computer off and go to sleep, attempted it, gave up, turned the computer off, immeadiately realised the answer and turned the computer back on again! (I'm not entirely sure why... bragging rights, I guess.) --Tango (talk) 18:30, 21 May 2008 (UTC)

So it's not really correct that events which aren't linked don't "influence" each other? You can get the flow of the event by looking at other manifestations of it? Bastard Soap (talk) 08:40, 21 May 2008 (UTC)

Having read the spoiler I don't think that's right. The events don't influence eachother, the strategy just means that when a wrong guess is made it will be by all three people at the same time, whereas a correct guess will be made by just one. Since each geuess has a 50/50 chance three out of four times a correct guess will be made, one out of four three incorrect guesses will be made. -- Q Chris (talk) 10:54, 21 May 2008 (UTC)

Incidentally, is there a simple proof that I'm missing that the 75% strategy is optimal? I'm pretty sure it is, but I can't see how to prove it. --Tango (talk) 11:46, 21 May 2008 (UTC)

I saw more or less the same problem a while ago. Here's one form: you have (say) 1023 (this is a hint) people, each provided with a random hat as before. This time everyone must vote, and the group wins if the majority vote correctly. A variant allows weighted voting, so you can (say) cast 100 votes that you have a red hat. I was reminded of these because the forced voting clause makes it easy to prove the optimal strategies are indeed optimal, which I can't quite see how to do for the problem here. Algebraist 12:53, 21 May 2008 (UTC)

This might work...although I haven’t tried to actually prove the optimality before

Consider that regardless of the strategy everyone will guess wrong 50% of the time. If two people guess wrong together in a case, that provides correctness of at most 67%. If all three people guess wrong together in a case, then this provides correctness of at most 75%. This exhausts all cases. It also assumes that one can partition the sample space though, which isn’t always possible, but nonetheless should provide the sufficient upper bound. (for instance, I don’t think it’s possible to have a strategy which wins two thirds of the time because 8 isn’t divisible by 3). GromXXVII (talk) 16:42, 21 May 2008 (UTC)

It's not possible to have a deterministic strategy with 2/3 chance of winning, if you allow non-deterministic strategies, it should be doable. --Tango (talk) 18:25, 21 May 2008 (UTC)

Momentum and vectors - Mechanics

Hopefully this is okay here, I have an exam tomorrow and I'm going over past papers and am very confused :(

Two particles, A and B, are moving on a smooth horizontal surface. Particle A has mass m kg and velocity (5i - 3j) ms^-1. Particle B has mass 0.2kg and velocity (2i + 3j) ms^-1. The particles collide and form a single particle C with velocity (ki - 1j) ms^-1, where k is a constant.

Show that m = 0.1

I formed an expression for the total momentum of the two particles: m(5i - 3j) + 0.2(2i + 3j). Due to the conservation of momentum, the particle C has to have the same mass and momentum, right? So I set that expression equal to (m + 0.2)(ki + 1j). But now I appear to have two unknowns and am thoroughly stuck. Any pointers would be appreciated. naerii - talk 22:52, 20 May 2008 (UTC)

You have two unknowns, but if you look carefully, you also have two equations - one for i and one for j. --Tango (talk) 23:03, 20 May 2008 (UTC)

I love you tango :) naerii - talk 23:12, 20 May 2008 (UTC)

I love you, too! --Tango (talk) 11:47, 21 May 2008 (UTC)

Maybe this is a red herring and was just a typo when you restated the problem, but I think you introduced a sign error at some point in the velocity of C too. At one point you said -1j, and at another you used +1j. --Prestidigitator (talk) 05:02, 21 May 2008 (UTC)

May 21

Complements of finite-dimensional subspaces

Let E be a Banach space and F a finite-dimensional subspace. The claim is that there exists a closed subspace G such that

$E=F\oplus G.$

Let $\{e_{1},\ldots ,e_{n}\}$ be a basis for F and extend this to a basis B for E. Let $\{\varepsilon _{1},\ldots ,\varepsilon _{n}\}$ be the basis dual to $\{e_{1},\ldots ,e_{n}\}$. By the Hahn-Banach theorem, each $\varepsilon _{i}$ extends to a continuous linear functional on E. The intersection of the kernels of the $\varepsilon _{i}$ is then a closed subspace G of E, and clearly $F\cap G=\{0\}$. And if $\beta \in B-\{e_{1},\ldots ,e_{n}\}$ then

$\beta -\sum _{i}\varepsilon _{i}(\beta )e_{i}\in G,$

so that $\beta \in F+G$. It follows that $F+G=E\,$.

Is this the right approach?  — merge 08:54, 21 May 2008 (UTC)

Well, it works, though the use of the basis B is completely unnecessary. You could just end by saying 'if $v\in E$ then $v-\sum _{i}\varepsilon _{i}(v)e_{i}\in G$.' Algebraist 10:30, 21 May 2008 (UTC)

Good point. Thanks!  — merge 11:38, 21 May 2008 (UTC)

Oh, and the assumption that E is complete was unused. Dispose of it. Algebraist 12:22, 21 May 2008 (UTC)

Oh, that's interesting. The name of the theorem confuses me at times. I should know by now that Lang is cavalier with his hypotheses.  ;)  — merge 12:38, 21 May 2008 (UTC)

Unfortunately, doing functional analysis requires distinguishing a lot of things named after Stefan Banach. Algebraist 13:01, 21 May 2008 (UTC)

Ahaha. Well, that I can live with (although I do often rail against the mathematical tradition of naming things after people instead of descriptively). But Lang is always doing things like assuming "normed instead of metric" so that "we can write the distance in terms of the absolute value sign", or adding hypotheses because "this is the only case anyone cares about anyway." I suspect this is just another combination of his perverse sense of humour and love of dropping stealth exercises for the reader.  — merge 13:22, 21 May 2008 (UTC)

The problem with descriptive naming is that disambiguation can make things pretty unwieldy. My own preference is a combination: thus Hahn-Banach extension theorem, Tietze-Urysohn extension theorem, Carathéodory's extension theorem, etc. On the subject of weak hypotheses in exercises, this often serves the purpose of making the reader think more. In some cases (the five lemma springs to mind), the minimal hypotheses are the proof. Algebraist 13:50, 21 May 2008 (UTC)

Right. Stealth exercises.  ;)  — merge 14:16, 21 May 2008 (UTC)

To the best of my memory, it is also true for locally convex spaces. twma 11:07, 22 May 2008 (UTC)

Yeah. For LCTVSs you can prove the continuous-extension version of HB from the most basic version via messing around with Minkowski functionals. Algebraist 21:34, 22 May 2008 (UTC)

proof for a.b=lcm x gcd

could please give me the proof for the following equality

product of two natural numbers is equal to the product of their lcm and gcd

( where lcm stands for least common multiple and gcd for greatest common divisor )Kasiraoj (talk) 14:07, 21 May 2008 (UTC)

I would start by expressing a and b as products of primes (see Fundamental theorem of arithmetic), and then work out what the lcm and gcd are in terms of those primes, and it should follow from that. --Tango (talk) 14:32, 21 May 2008 (UTC)

This answer got me thinking about whether the Fundamental Theorem is actually necessary here. More precisely: firstly, does there exist an integral domain in which any pair of elements has a GCD and an LCM, but which is not a UFD? (edit:yes) Secondly, if there are such rings, does this result hold in them, i.e. is ab always an associate of [a,b](a,b)? Algebraist 15:13, 21 May 2008 (UTC)

In case you didn't already find it, see GCD domain.

Seems to me that once you have a GCD w of x and y (not necessarily unique) then z=xy/w is a multiple of x and of y so is a common multiple. And z must be a LCM because if we had a u that was a common multiple of x and y and also a divisor of z then v=w(z/u) is common divisor of x and y (because x/v = u/y and y/v = u/x) that is also multiple of w, which contradicts our assumption that w is a GCD. So for each GCD w there is a LCM xy/w (and vice versa, by reversing the above), even if it is not unique. Gandalf61 (talk) 16:04, 21 May 2008 (UTC)

I suspect you might be right, but your proof doesn't work. You've shown that z is a minimal common multiple, but not that it is a least common multiple. Algebraist 21:50, 21 May 2008 (UTC)

And of course we're assuming the existence of LCMs, so any minimal CM is an LCM. Thanks. Algebraist 21:52, 21 May 2008 (UTC)

That we're in a GCD domain is only assuming the existence of GCDs, not LCMs, as far as I can tell, so you need a slightly stronger assumption than just being in a GCD domain. (It may turn out to be equivalent, of course.) --Tango (talk) 12:35, 22 May 2008 (UTC)

Sorry, by we I mean me, in my initial question above. Our article doesn't state whether a GCD domain automatically has LCMs, and it should. Algebraist 12:39, 22 May 2008 (UTC)

Aren't we going in circles here ? If x and y are in a general commutative ring and w is a maximal common divisor of x and y (i.e. the only multiples of w that are also c.d.s are associates of w) then z=xy/w is a minimal common multiple of x and y (i.e. the only divisors of z that are also c.m.s are associates of z) and vice versa - as per my argument above (with a little tightening up to allow for associates). And if further x and y are in a GCD domain so that w is not just a maximal c.d. but is a GCD of x and y then z=xy/w is a LCM of x and y, and vice versa.

I am sure this must be in a standard text somewhere - I will add it to the GCD domain article when I have found a reference. Gandalf61 (talk) 13:06, 22 May 2008 (UTC)

My algebra's been slow lately, but finding a piece of paper has finally allowed me to prove that any GCD domain has (binary) LCMs. Unfortunately, GCD domain is completely unreferenced and my algebra textbook doesn't mention the things explicitly; time to go looking for a reference that does. Algebraist 13:26, 22 May 2008 (UTC)

I'm off to the Uni library in a bit anyway - I'll see if I can find anything. --Tango (talk) 13:29, 22 May 2008 (UTC)

Google books gave me a ref, which I've added. Curiously, in any ID, ({x,y} has an LCM) → ({x,y} has a GCD), but the converse fails. Algebraist 14:17, 22 May 2008 (UTC)

I was unable to find any books which mentioned GCD domains, I did however find one that briefly mentioned a "ring [or it might have been ID, I don't recall] with a greatest common divisor" - it seems the name is far from standard. The only relevant thing it mentioned about them was a theorem: An integral domain with a greatest common divisor is one with a least common multiple, and conversely. I think that's what you'd already worked out. It is, indeed, curious that every pair having a GCD implies every pair has an LCM, but a given pair having a GCD doesn't imply that that same pair has a LCM. (The book I found was very large, very old and very much falling apart, so I left it in the library, so don't ask any more questions!) --Tango (talk) 14:58, 22 May 2008 (UTC)

According to my source, 'GCD domain' was popularised by Irving Kaplansky's textbook in 1974, so it may now be standard. Other terms mentioned are 'pseudo-Bezout', 'HCF-ring', 'complete' and 'property BA'. Algebraist 15:06, 22 May 2008 (UTC)

The book I was reading could easily have pre-dated 1974. Odd that none of the other books I looked at mentioned that name, though - maybe Durham Uni library is very out-of-date! --Tango (talk) 15:26, 22 May 2008 (UTC)

set theories

hey ive got fundamental dobts...kindly some one give me a link or the required answers clearly... i wd b highly greatful ..

my doubt is that ...how can we represent two indepentent events on a venn diagram.if we do it by two intersecting circle .then what can we say about the condition of independence.p(A intersection B)=p(A)*p(B).how is this derived.

can we say mutually exclusive events as a special case of independent events. Reveal.mystery (talk) —Preceding comment was added at 15:52, 21 May 2008 (UTC)

Hi. First off, no you can't say mutually exclusive events are a special case of independent events, as by definition the two are not independent - if you have one, you cannot have the other. The proof of independence I admit I can't remember right now, but it is fairly logical when you think about it. If you have two events, neither of which influences the other, the probability of them both happening would be the probability of one happening multiplied by the probability of the other happening. -mattbuck (Talk) 17:35, 21 May 2008 (UTC)

The definition of independence is $p(A\cap B)=p(A)p(B)$, so there is nothing to prove. You may ask, why did we choose to call "independent" two events satisfying this. I think this boils down to the empirical observation that real-world events which seem independent to us tend to satisfy this rule. You can rephrase this in terms of Conditional probability, but that's just moving the problem elsewhere. -- Meni Rosenfeld (talk) 17:45, 21 May 2008 (UTC)

The definition of statistical independence can be written p(A|B) = p(A), meaning that the conditional probability of A given B is the same as the unconditional probability of A. So the probability of A is independent on whether B occured or not. This definition seems intuitively natural. As the conditional probability satisfies $p(A|B)\cdot p(B)=p(A\cap B)$, the condition $p(A\cap B)=p(A)\cdot p(B)$ is derived. In a unit square diagram, A may be drawn as a horizontal bar and B as a vertical bar crossing A, illustrating that the area of the intersection between A and B is the product of the areas of A and B. Bo Jacoby (talk) 19:44, 21 May 2008 (UTC).

Which, as Meni said, just move the problem to the definition of conditional probability. Sooner or later you just have to accept that those definitions seem to work - you can't prove everything, you have to start somewhere. --Tango (talk) 20:27, 21 May 2008 (UTC)

It is ok to use p(A|B) = p(A), rather than p(A intersection B)=p(A)*p(B) as the definition of independence. You don't just have to accept anything. Bo Jacoby (talk) 08:07, 22 May 2008 (UTC).

Of course you can define "A and B are independent if $P(A|B)=P(A)$" (I could quibble about what happens when the probabilities of A or B are zero, but never mind that). But how do you know that $p(A|B)\cdot p(B)=p(A\cap B)$, unless you observe that empirically or define $p(A|B)={\frac {P(A\cap B)}{P(B)}}$? And, why would you define the latter without empirically observing it? -- Meni Rosenfeld (talk) 11:07, 22 May 2008 (UTC)

No empirical observations are needed, (nor are they possible because the probability is a limit which is not accesible observationally), but a thought experiment: Consider an event A, (say, that a white dice show four or five), and an event B, (say, that a blue dice show six). The probability p(A) is the limit of the ratio between (the number of times you throw the white dice and it shows four or five) and (the number of times you throw the white dice altogether). The conditional probability p(A|B) is the limit of the ratio between (the number of times that you throw both dice and the white dice show four or five and the blue dice show six) and (the number of times that you throw both dice and the blue dice show six). Now, if the two events are independent in the non-technical sense of the word, that the result of the two dice do not depend on one another, then the two limits must be equal. So p(A)=p(A|B) if A and B are independent. Now define that A and B are statistically independent if p(A)=p(A|B). So independent events are also statistically independent. Consider the equation p(A|B)·p(B)=p(A and B). The left hand side is the limit of the ratio between (the number of times you throw both dice and the white one shows four or five and the blue one shows six) and (the number of times you throw both dice and the blue one shows six), multiplied by the limit of the ratio between (the number of times the blue dice shows six) and (the number of times you throw the blue dice all together). Using that the product of limits is the limit of the product, you get the limit of the ratio between (the number of times you throw both dice and the white one shows four or five and the blue one shows six) and (the number of times you throw the blue dice all together), which is equal to the right hand side p(A and B). Bo Jacoby (talk) 14:16, 22 May 2008 (UTC).

But again, you just move the problem a little further along. In order to justify that two dice will be independent of each other, you first assume that successive tosses of a single die will be independent, with that assumption and its definition swept under the carpet. Black Carrot (talk) 15:55, 22 May 2008 (UTC)

The very concept of the probability of an outcome of an experiment assumes that the experiment can be repeated indefinitely and that the outcome of the repetitions are mutually independent. Bo Jacoby (talk) 18:08, 22 May 2008 (UTC).

As I understand it, that's frequency probability, Bayesian probability is a little different. I don't think that's an issue for this discussion, where such frequencies are meaningful, but it's always good to be precise. --Tango (talk) 19:02, 22 May 2008 (UTC)

Yes, the 'probability of an outcome of an experiment', based on some hypothesis, is a frequentist probability. The Bayesian point of view is the opposite one, to estimate the credibility of a hypothesis, based on given outcomes of experiments. While the two interpretations of probability differ, the algebra is the same. Computing the probability of the outcome based on the hypothesis is deductive reasoning. Computing the credibility of the hypothesis based on the observations is called inductive reasoning. Bo Jacoby (talk) 06:18, 23 May 2008 (UTC).

Odd trig/geometry question

On a certain bike there are spokes that are 14 inches long. Each spoke forms an angle of 30^o with each of the two spokes beside it. What is the distance between the places where two spokes that are beside each other attach to the wheel? I think this has something to do with geometric mean, but I'm not sure and have no idea how to do it. *This is in fact not a homework. It was a bonus question on a test we had today* Zrs 12 (talk) 23:35, 21 May 2008 (UTC)

Assuming they come from the same point, you can treat them as radii. You then use the formula for the circumference (c = 2(pi)r), but you only want 30 degrees rather than 360, so multiply by 30/360. -mattbuck (Talk) 23:49, 21 May 2008 (UTC)

Hmm, I wonder why that tripped me up so badly. Is there any way to do this with triangles and trig? This was on a trig test so I have no idea why there would be something about circles on it. I like Mattbuck's method though. That's a lot simpler than what I was trying. Thanks, Zrs 12 (talk) 23:58, 21 May 2008 (UTC)

Mattbuck's answer gives the distance along the wheel, i.e. along an arc of a circle. If you want the straight-line distance, then trigonometry is the way to go, specifically the law of cosines. Algebraist 00:02, 22 May 2008 (UTC)

Hmm, how would you pull that off? I tried that on the test, but I couldn't figure out how to make it work. Please explain. Wait, crap. I feel stupid now. I misunderstood the problem and took it that the spokes didn't touch at any point. Well, that was a stupid mistake. Zrs 12 (talk) 01:42, 22 May 2008 (UTC)

Also check out chord (geometry). --Prestidigitator (talk) 03:54, 22 May 2008 (UTC)

Oh, I understand what's going on now. I just completely misunderstood the problem. Otherwise, I would've hadn't any trouble solving it. Thank you all for your help, Zrs 12 (talk) 13:24, 22 May 2008 (UTC)

May 22

Integrating parametric equations

Recently in Calculus II, we have started doing some integrals of functions defined by parametric equations (as well as in polar coordinates), my question is how does one prove that area under a parametric curve is given by the formula.

$A=\pm \int _{\alpha }^{\beta }g(t)f'(t)dt$

A math-wiki (talk) 06:23, 22 May 2008 (UTC)

f is the x-coordinate and g is the y-coordinate. Substitute $dx=f'(t)\cdot dt$ to get $\int _{\alpha }^{\beta }g(t)\cdot f'(t)\cdot dt=\int _{f(\alpha )}^{f(\beta )}y\cdot dx.$ Bo Jacoby (talk) 08:20, 22 May 2008 (UTC).

I suppose that works, but I really want to see a constructionist type proof (much like the Fundamental theorem of Calculus). 69.54.143.177 (talk) 23:22, 22 May 2008 (UTC)

Solution for Tan(x)=2x. {0<x<Pi/2}

Heyo, I am trying to solve Tan(x)=2x for x, but I have never encountered a function such as this before. The domain is {0<x<Pi/2}. My current efforts in solving it have given me an approximate value of x=1.1655612, but an exact value would be wonderful. Mathematica can't solve it, either, it would seem. I'm only a high school student, so it would be perfectly reasonable to assume that such a solution is beyond my capabilities (for now), so can anyone help me out, please? I assure you that this is not homework, but rather a curiousity I am following. Many thanks Vvitor (talk) 09:02, 22 May 2008 (UTC)

Well, there ic clearly only one solution in the range 0 < x < π/2, but I doubt that there is a closed-form expression for this solution - if there was, Mathematica would probably have found it. So a numerical approximation is the best that you will get. Newton's method works nicely if you take a starting value between 1 and 1.5. Starting further away, say between 0.7 and 0.9, Newton's method sometimes converges to other roots, sometimes cycles, and sometimes does not seem to converge at all. Perhaps someone might be interested in imaging the Newton fractal for this equation ? Gandalf61 (talk) 10:26, 22 May 2008 (UTC)

It's interesting you mention Newton's method. I was attempting find a root for Sin(x) between -Pi/2 and Pi/2, which is clearly zero, using Newton's method. I was investigating, however, the possibly of taking an initial value of x₁ in which the x₂ value would be -x₁. As such, the approximation would continually jump from negative to positive to negative again, etc.. The value of x that would seem to work is such that Tan(x)=2x. I know this sounds silly, and probably because it is, but I was awfully bored at the time :P. Thank you very much for your insight, I'll try and get a fairly accurate numerical approximation :) Vvitor (talk) 10:39, 22 May 2008 (UTC)

Silly, perhaps, but in a good way :). I agree that this problem is intriguing. Note that Mathematica will be very happy to provide a numerical approximation if you ask it nicely - The following command:

N[x /. FindRoot[Tan[x] - 2 x, {x, 1}, AccuracyGoal -> 55, WorkingPrecision -> 65], 50]

gives 1.1655611852072113068339179779585606691345388476931. -- Meni Rosenfeld (talk) 11:20, 22 May 2008 (UTC)

Perhaps it is worth mentioning that if the equation you wanted to solve was $\log x=2x$, there is a non-elementary function called Lambert W function which was invented for just this purpose. You could just as well invent a function that gives the solution to $\tan x=\alpha x$, but it seems no-one has deemed it important enough to give it a name. -- Meni Rosenfeld (talk) 11:33, 22 May 2008 (UTC)

Well, tan is actually composed of (complex) exponentials, so one might be able to use the Lambert W Function for that purpose. Paxinum (talk) 21:22, 22 May 2008 (UTC)

I think I've thought about that once and concluded that it is not actually possible. Too lazy to look at it again. -- Meni Rosenfeld (talk) 00:14, 23 May 2008 (UTC)

Your Mathematica-Fu is very impressive. Thank you very much! Vvitor (talk) 11:36, 22 May 2008 (UTC)

Someone has invented such a function: the tanc function. To follow the naming convention for inverse trig/hyperbolic function, the sought value would be atanc(2). I've only seen the name tanc used on MathWorld, and the name atanc was made up on the spot here, but both really should be standard since equations similar to tan(x)=x are so common in physics. Maybe if I implement both in mpmath and sympy, the world will follow... - Fredrik Johansson 10:56, 23 May 2008 (UTC)

Flux across surface.

Is the flux of the vector field F=(xz)i+(x)j+(y)k across the surface of the hemisphere of radius 5 oriented in the direction of the positive y-axis equal to Zero? I'm getting up to a string of double-integrals between 0 and pi, but each term has a sin function that results in everything ending up zero, but it seems an unlikely answer. —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 11:39, 22 May 2008 (UTC)

Yes, it is zero. No messy double integrals are necessary, by the way. Algebraist 12:09, 22 May 2008 (UTC)

Let $S$ be your surface, which is symmetric under negation of $x$ or $y$. Let $G(x,y,z)=(xz)i+0j+(y)k$ and $H(x,y,z)=0i+(x)j+0k$. Flipping $G$ across the $x$-$y$ plane, we obtain $(x(-z))i+0j+(y)(-k)=-G$; so $G$ and $-G$ have the same flux, which must be zero. Flipping $H$ across the $y$-$z$ plane, we obtain $0(-i)+(-x)j+0k=-H$, so similarly the flux of $H$ is 0. But the flux of $F$ is the sum of the fluxes of $G$ and $H$, so must be zero. Eric. 144.32.89.104 (talk) 12:23, 22 May 2008 (UTC)

I choose the lazy approach. The divergence theorem is very useful here. Note that $\nabla \cdot {\overrightarrow {\mathbf {F} }}$ is anti-symmetric about the plane z=0. --Prestidigitator (talk) 20:49, 22 May 2008 (UTC)

Yeah, that was my approach. Algebraist 21:19, 22 May 2008 (UTC)

Ah -- I had misread "surface of the hemisphere" as if it were "half of the surface of a sphere", which would be an unclosed surface. Eric. 217.42.199.10 (talk) 10:23, 23 May 2008 (UTC)

Actually, that was my reading too. Fortunately, the flux through the flat side of the hemisphere is obviously zero, so you get the same answer. Algebraist 11:03, 23 May 2008 (UTC)

Four-dimensional pyramid

I know that the four-dimensional equivalent of the tetrahedron is the pentachoron, which would resemble a tetrahedron dwindling to a point as it passed through our three-dimensional space, but what's the four-dimensional equivalent of a square-based Egyptian-type pyramid? Would its cross section(if it fell "face-first") be a cube, a triangular prism, or something else? 69.111.191.122 (talk) 15:00, 22 May 2008 (UTC)

I am not 100% sure about this, but I suspect there is some ambiguity about what the analog actually would be. The tetrahedron's point group T_d has triply degenerate symmetries, meaning that the three dimensions are in some sense equivalent or at least exchangeable. The pyramid (C_4v ) does not, only two of the dimensions (the ones defining the base) are exchangeable. So you could either make the next dimension a second "height" or a third "base" dimension. The resulting two figures would not be the same. If you made specific the type of analog you were considering, it would help. Baccyak4H (Yak!) 15:25, 22 May 2008 (UTC)

As Baccyak says, there are multiple things that could be called a 4-D pyramid. I would go with a cube going to a point (a square pyramid is a square going to a point, add one dimension to a square, you get a cube). I think that might be what Baccyak means by adding a third base dimension. I'm not quite sure what adding a 2nd height dimension would mean... --Tango (talk) 15:30, 22 May 2008 (UTC)

A square pyramid going to a point, maybe. Black Carrot (talk) 15:45, 22 May 2008 (UTC)

Re adding a second height dimension: in general, the join of two shapes S (in d dimensions) and T (in e dimensions) is formed by taking disjoint d-dimensional and e-dimensional affine subspaces of (d + e + 1)-dimensional space, placing S and T within them, and taking the convex hull of S and T as placed together in the larger dimensional space. For the geometry of the resulting shape it matters where S and T are placed but for the combinatorics of what faces the join has it doesn't matter. A three-dimensional pyramid over some base is the join of that base (2-dimensional) with a single point (0-dimensional). The 4-simplex and pyramid-over-cube you're talking about are joins of a regular tetrahedron or cube with a single point. But the 4-simplex is also the join of a triangle and a line segment, and you could consider the shape formed by the join of a square with a line segment as a generalization of a pyramid. This shape has as its facets two square pyramids (the joins of the square with the endpoints of the line segment) and four tetrahedra (the joins of the sides of the square with the whole line segment). Its Schlegel diagram can be formed as a square pyramid with a single vertex inside it connected by edges to all five pyramid corners. And, as Black Carrot suggests, it is also the join of a square pyramid and a single point. —David Eppstein (talk) 15:47, 22 May 2008 (UTC)

My 2c..

Starting with a shape S, there are many ways to increase its dimension. One way is to add a point in the next dimension. This gives a pyramid, whose base is the original shape.

• If S is a line segment, you get a triangle.
• If S is a triangle, you get a tetrahedron.
• If S is a square, you get a square pyramid.
• If S is a circle, you get a cone.

Let's call this operation, Adding a point, A. Another way is to move the shape in the direction of the new dimension, and see what it carves out. This gives you a prism, whose base is the original shape.

• If S is a line segment, you get a square.
• If S is a triangle, you get a triangular prism.
• If S is a square, you get a cube.
• If S is a circle, you get a cylinder.

Let's call this operation, Moving the shape, M. Now we can think of what you get if you apply A and M successively, starting with a line segment L. (Or, you could start with a single point if you prefer)

• AL is a triangle,
• ML is a square

Then, in three dimensions...

• AAL is a tetrahedron
• MAL is a triangular prism
• AML is a square pyramid
• MML is a cube

Your question is about the next step...

• AAAL is a pentachoron
• AMAL is a pyramid whose base is a triangular prism. Its "faces" has two tetrahedra, three square pyramids, and the base, a triangular prism.
• AAML is a pyramid whose base is a square pyramid. Its "faces" are four tetrahedra, and two square pyramids (including the base). In fact, once the shape is made, there's no way to identify which of the two square pyramids was the "original" base.
• AMML is a pyramid with a cube for its base. Its "faces" are 6 square pyramids and the cubic base.

Any of the last three could perhaps count as the "equivalent" of the square pyramid, but I'd vote for the last. Then, there are the prisms..

• MAAL is a prism with a tetrahedral base. Its faces are 4 triangular prisms, and the 2 tetrahedra.
• MMAL is a prism with a triangular prism for its base. Its faces include has 4 triangular prisms, in two opposite pairs, and three cubes. Either of the pairs of triangular prisms could count as the base.
• MAML is a prism with pyramids at the top and bottom. The sides consist of 1 cube, and 4 triangular prisms.
• MMML is the good ole' tesseract.

Hope that helps! mike40033 (talk) 02:10, 23 May 2008 (UTC)

Thanks, that's very informative! Are there any images, nets, or animations of those shapes? (Most of the 4D geometry pages I've seen just have the perfectly symmetrical ones and their stellations.) —Preceding unsigned comment added by 69.111.191.122 (talk) 00:44, 24 May 2008 (UTC)

That was a great answer - worth three cents, at least! --Tango (talk) 20:15, 23 May 2008 (UTC)

A simpler way to explain why there is an ambiguity is that we don't know the form of the extra dimension. In the case of a regular tetrahedron, it's reasonable to assume that the 4th dimensional analog is also a regular figure - so you get the tetrahedron that shrinks to a dot as the 4 dimensional shape translated through 3-space.

But a square-based pyramid is an irregular shape - there are any number of irregular 4 dimensional spaces which have pyramidal hyper-faces. It's kinda like if you were a 2D being - you could ask for 3D analogs of things like squares, circles and equilateral triangles - and get answers like "cube", "sphere" and "tetrahedron". But demanding the 3D analog of an irregular 2D figure would be an unanswerable question.

70.116.10.189 (talk) 15:43, 24 May 2008 (UTC)

Figuring "Final Average"

My professor said to figure my final average for the year, without the final exam, to:

1. Take my first quarter grade and double it
2. Take my second quarter grade and double it
3. Take my semester exam
4. Take my third quarter grade and double it
5. Take my fourth quarter grade and double it

Then I am to add these numbers together and divide by 10 to get my final average.

For example Q1=93%, Q2=90%, SE=93%, Q3=100%, and Q4=94%; my final average would be 85%.

What's the logic behind the doubling and dividing. Also how do I go from 90's (as in my example) to 85? Thanks...I'm really lost. §hep • ¡Talk to me! 22:34, 22 May 2008 (UTC)

My best guess would be that the semester exam is out of 200 points, not 100: but has equal weighting as the quarters. Thus to get equal weighting he gave them all a value of 200 points by doubling each quarter.

The key is that you’re dividing by 10 at the end, and so the original sum of (1) to (5) must be 1000. Assuming each quarter is out of 100, that’s how I deduced that the semester exam is probably out of 200.

(If everything were out of 100, then the maximum possible would be 900/10=90) GromXXVII (talk) 22:40, 22 May 2008 (UTC)

Sorry, that's percentage not points. I should have been more specific. §hep • ¡Talk to me! 22:46, 22 May 2008 (UTC)

Ok, the semester exam is worth half as much as the quarter grades, so you count each quarter grade twice. That means you've effectively got 9 marks to count, plus the final exam, makes the 10 you divide by. The reason your average is less than the individual marks is because it's assuming you get 0% on the final exam (which apparently has the same weight as the semester exam). If you want just your average for the bits you've taken so far, not including the final exam at all, just divide by 9 instead of 10. --Tango (talk) 22:56, 22 May 2008 (UTC)

Ahhh. Gotcha. Thanks! §hep • ¡Talk to me! 23:11, 22 May 2008 (UTC)

This type of average is called a weighted mean. Confusing Manifestation(Say hi!) 23:30, 22 May 2008 (UTC)

So we can deduce that the final exam has a weight of 10% in the final average. If you scoe x% in the final exame then your final average will be 85 + (x/10) %. So you can be certain of a final average of at least 85%, but even if you ace the final exam you can't get a final average better than 95%. Gandalf61 (talk) 08:49, 23 May 2008 (UTC)

May 23

wikipedia and power laws

I know that if you consider a network model of Wikipedia (in which nodes are pages, edges are links), then that network is (or is close to being) scale-free. That is to say, the network degree follows a power law. But I can't find any reference that says what the slope of the power law relationship is (the variable k in the power law article). Anyone have any thoughts? AndrewGNF (talk) 01:49, 23 May 2008 (UTC)

Well, let's gather some data. I clicked "random article" 20 times, and got articles with the following number of blue links (not counting boiler-plate text):

2, 3, 5, 5, 6, 6, 7, 7, 10, 11, 14, 17, 20, 27, 28, 30, 33, 41, 44, 55

These are outgoing links, not incoming links (since Wikipedia is a directed graph). Among articles with at least one link, the probability that an article has exactly one link is ${\frac {1}{\zeta (\gamma )}}$ (where $\gamma$ is the parameter in the scale-free network article). At $\gamma \geq 2$, we have ${\frac {1}{\zeta (\gamma )}}\geq {\frac {6}{\pi ^{2}}}=0.607927$, so at least a 61% chance of an article with exactly one link -- and since even in 20 random articles, there are no articles with exactly one link, we can be fairly sure that if Wikipedia is scale-free, then $\gamma$ is less than 2 -- probably very close to 1. Someone with more statistics knowledge (or more data) could probably conjure up a reasonable estimate for $\gamma$. Eric. 217.42.199.10 (talk) —Preceding comment was added at 10:50, 23 May 2008 (UTC)

Variation in statistics

I have a few questions regarding statistics and my textbook (Stirzaker) isn't helping.

Say you have a set of data that shows daily price movement in some commodity over ten years, with 3652 or so prices (one for each day).

Given that it is normally distributed, I understand that once you calculate the daily variance over N days (σ^2_N, that is sigma squared subscript N), you can find its 95% confidence interval: σ^2 ∈ $\left[{\frac {\sigma _{N}^{2}}{1+1.96{\sqrt {\frac {2}{N-1}}}}},{\frac {\sigma _{N}^{2}}{1-1.96{\sqrt {\frac {2}{N-1}}}}}\right]$

where N is the number of days (which in this data set would be 3652), and σ^2 is the daily variance.

(apologies for not using the html code but it 'fails to parse' the σ symbol)

I understand that you could find the yearly variance by multiplying the daily variance by 365 (and the annual standard deviation or 'volatility' by multiplying the daily standard deviation by sqrt(365).

But if you wanted to find the 95% confidence interval for the annual variance, how would you change the formula? Could you just replace σ^2 with 365*σ^2, and N=3652 with N=(3652/365)? In which case the N would be the number of years (10)? Because when I do that I end up with an awfully wide interval. Or would you let N=365, the number of days in one year?

I'm interested in seeing whether or not the annual volatility changes over time. I was thinking that once you had found the confidence interval of the annual variance, one could split the data set into smaller intervals, calculate the annual variance of each interval (by multiplying the variance of each interval by 365), and see if 95% of the resulting set of annual variances falls within the confidence interval. Would this be a valid method of tracking changes in annual volatility? —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 11:30, 23 May 2008 (UTC)

I TeXed your formula for readability. Algebraist 11:40, 23 May 2008 (UTC)

Counting problem

In how many ways can 15 students be seated in a row such that the 2 most talkative children never sit together? I've been given the answer 14! * 13, but I don't compute this. Seems like it should be 13! ways for every two positions that are not adjacent. 70.250.149.151 (talk) 19:27, 23 May 2008 (UTC)john

Well, if we considered it as a circle, then we place the most talkative student first (15 options). That leaves 12 places for the 2nd most talkative (can't use the seat taken, or the 2 on either side). After that you have 13 seats available which can be used in any order, so here the answer is 15.12.13! = 1.12E12.

In a row, things are more complicated - there are two cases to consider. First suppose that the most talkative student doesn't sit at the end (13 options). Then there are again 12 possible places for #2 to sit, and then we arrange the other 13 students as we like, so 13.12.13!. If we sit the most talkative student at the end (2 options), then there are 13 places for #2, and then 13! ways to arrange the rest, so a tital of 2.13.13! Putting this together, we have 13!(13.12 + 2.13) = 14.13.13! = 13.14! as required. Nice problem. -mattbuck (Talk) 19:47, 23 May 2008 (UTC)

Another way to see it: there's a one-to-one correspondence between pairs of nonadjacent seats chosen from a row of N and arbitrary pairs of seats chosen from a row of N−1 (just add/remove a seat between the chosen seats). So there are 14·13 choices for the two most talkative students and then 13! for the rest. -- BenRG (talk) 21:06, 23 May 2008 (UTC)

Instead you might want to consider the way to seat the two most talkative together. So making them a block of 2 which can be arrange 2 (or 2! ways) you have 14 items, 13 children and the two most talkitive together. So these are arranged 14!*2 ways. Then you can just minus this from the total 15! ways or arranging. I think this is a more simple way of getting your 13*14! answer. Hope this helps Rambo's Revenge (talk) 07:41, 24 May 2008 (UTC)

This is the lovely thing about combinatorics - no matter how clever you think you're being, there's always someone else who can find an easier way. That and it deals with proper numbers. But then, I like complex analysis so who am I to talk? -mattbuck (Talk) 11:03, 24 May 2008 (UTC)

What is the formula for exponential growth/decay?

What is the forumla for exponential growth/decay?  Marlith (Talk)  23:56, 23 May 2008 (UTC)

See exponential growth. Paragon12321 (talk) 01:32, 24 May 2008 (UTC)

May 24

Drawing Graphs (Graph Theory, not Functions)

I have an adjacency matrix (in Excel 07) from which I would like to generate a graph in order to be able to visualise the data nicely. Is there a program that will generate a graph from the adjacency matrix specified in any format Excel can output (e.g. xls, csv, etc.). I also need a couple of features, specifically my graph is a weighted (but undirected) graph so I would like to be able to visualise these weights either by colour or by thickness of the edge. In addition, it would be really great if the program would allow me to have a colour/size scale for the nodes based on how connected they were or some other number that I could assign to each node.

I have looked at graphviz and the GUI is basically non-existant and it looks like I would have to reenter all the data and I looked at an excel frontend for it but that wasn't very promising either. I also found a program called yEd but that doesn't seem to allow me to enter in an adjacency matrix. I don't want to do it manually because the graph is highly connected and has ~40 nodes. Any ideas?--AMorris (talk)●(contribs) 04:42, 24 May 2008 (UTC)

Hmmm. Even if you find appropriate graph generation software, will it be of much use to you ? A highly connected graph with ~40 nodes is very unlikely to be planar. Unless the graph has some special symmetries, I would think a diagram is going to be very messy, and visually almost indistinguishable from a complete graph. Gandalf61 (talk) 14:30, 24 May 2008 (UTC)

That's true, but I'm more looking to use the graph as just a broad visual representation of the data. The idea is not so much to visualise the existance of the connections, but more the weight of the connections either through colour or line thickness. Unfortunately all the programs seem to require manual entry or I would need to undergo a steep learning curve to use them. AMorris (talk)●(contribs) 10:03, 25 May 2008 (UTC)

Connected and Totally Bounded metric spaces

I've just finished a course in metric spaces, and connectedness was covered very briefly. My intuition tells me that every bounded connected space is totally bounded, but I'm having a hard time proving it or finding a counterexample (the only examples I know of a bounded but not totally bounded metric spaces are disconnected). Is this intuition true or false? And how might I go about proving it? --196.210.152.31 (talk) 07:51, 24 May 2008 (UTC)

Try looking at infinite-dimensional examples. Have you had any Banach space theory? --Trovatore (talk) 08:25, 24 May 2008 (UTC)

No, I have not, but I will take a look. Thank you. --196.210.152.31 (talk) 08:36, 24 May 2008 (UTC)

More directly, you can make any metric space bounded without making it totally bounded or changing its topology in the slightest. Just define a new distance to be the minimum of the old distance and 1. Algebraist 11:06, 24 May 2008 (UTC)

Surprise!

I have a logic problem that's been bothering me for years - it comes in the form of a true story:

I had a bet with my sister (which I lost) my penalty was that I had to take her to lunch. I ask her when we should meet up. Being a fun-loving person, she says "Well, today is Sunday - I'm going on vacation on Saturday morning so it has to be before then. Surprise me! Just show up at work around noon someday this week and we'll do lunch...but I want to be surprised - don't tell me you're coming, just show up."

So I could take her on Monday, Tuesday, Wednesday, Thursday or Friday. Let's stipulate that I always keep my promise, we're both intelligent, we think very much alike. Is it possible to surprise her?

On the face of it, it seems easy - I just need a random number. So maybe I should roll a dice, 1=Monday, 2=Tuesday...5=Friday, and on a 6, I re-roll. Easy - right?

But there's a problem: If I were to happen to roll a 5 (Friday - the last possible day) then on Friday morning my sister will think to herself: "He didn't take me to lunch yet - and he promised to do it this week - so today MUST be the day"...so it's not a surprise when I show up on Friday lunchtime - which means that Friday cannot be an acceptable result.

OK - so I have to choose between 1=Monday, 2=Tuesday, 3=Wednesday, 4=Thursday and re-roll the dice if I get a 5 or a 6. But now we have the same problem with Thursday. She knows (because we've stipulated that she thinks exactly like me) that Friday wouldn't be a surprise - so it's impossible. Hence: on Thursday morning, she knows that I can't leave it until Friday to take her to lunch because that wouldn't be a surprise - so it must be happening today...so again, it won't be a surprise. That means that Thursday is impossible too.

Sadly, now that Thursday is out of the question - so is Wednesday...and that means that Tuesday is impossible...and that only leaves Monday - and that won't be a surprise because it's the only possible day remaining.

Hence no day is truly (logically) a surprise.

This seems like a bogus argument - but I can't find a logical hole in it. Is it true that it's impossible to truly surprise someone under these circumstances?

70.116.10.189 (talk) 15:29, 24 May 2008 (UTC)

This problem is well known as the unexpected hanging paradox. That article explains some approaches that have been attempted to resolve the problem. Algebraist 16:13, 24 May 2008 (UTC)

Cool - it's good to know the mathematicians are earning their keep answering everyday problems! Many thanks - I'm off to read it carefully. SteveBaker (talk) 17:27, 24 May 2008 (UTC)

That's not fair. It is unreasonable to expect someone to assume the OP knew about the article after they'd typed out that scenario in such detail. Zain Ebrahim (talk) 17:41, 24 May 2008 (UTC)

Introduce a tiny bit of uncertainty. "Dear, dear sister. Please keep in mind that there is always the smallest probability that even the best made and most carefully thought-out promises cannot be kept (due to unexpected hospital stays for example), and that despite best efforts, it is always possible that I might have to take you to lunch after you get back from your trip." Then take her Tuesday. --Prestidigitator (talk) 19:11, 24 May 2008 (UTC)

The way I would look at it is like this: Since every day has been ruled out, there is the same probability (albeit zero) that it will occur on any day of that week. Since it must occur (because you always keep your promises), and it has an equal probability of occuring any day, then there is no certainty as to when it will occur or will not occur. Thus, any day (except Friday) it would be a suprise. If you take them any day but Friday, they don't know if you'll wait until the next day or not. (This is the first time I have heard of this paradox, so that may be flawed logic.) Ζρς ι'β' ¡hábleme! 19:19, 24 May 2008 (UTC)

But his/her sister is also intelligent (they think alike) so when she gets up on Thursday, she'll know that it has to be the day (because Friday is out) and therefore it's not a surprise. And the same reasoning can be applied to Wednesday, Tuesday and Monday. Zain Ebrahim (talk) 19:36, 24 May 2008 (UTC)

Ah, yep, I got it now. I thought about it one way and it was logical, then the other way, and it was illogical. Personally, I like Predtigitator's idea. Ζρς ι'β' ^¡hábleme! 20:19, 24 May 2008 (UTC)

Reduce the problem to two days (Thursday and Friday). Say your sister expects you to take her out on Thursday (since obviously you can't do it on Friday). But lunchtime comes and goes and you don't show up. Now what? If we allow her to expect to be taken out on Friday also, then your argument is valid, but all it shows is the trivial fact that if she expects to be taken out every day then she won't be surprised on the day that she is. If we don't allow her to expect to be taken out every day, then your argument breaks down immediately; you can surprise her on any day, including Friday, because there's a good chance she'll have blown her one chance at unsurprise before then.

That's one solution, anyway; it's not the only one because there's more than one way of interpreting English words like "expect". You can turn it into a problem about the slippery nature of belief (can your sister really make herself believe in Thursday after having reasoned as I just did?), but then it's a problem of psychology rather than mathematics. You can replace "expect" with "logically deduce" and turn it into a problem of formal logic. The article is misleading when it says that "no consensus on its correct resolution has yet been established", since one doesn't expect a consensus on the meaning of an utterance in the English language. If you ever find yourself in the condemned prisoner's position, keep in mind the parable of the dagger. Not that it'll save you from your fate. -- BenRG (talk) 22:08, 24 May 2008 (UTC)

I don't follow. When you say that, are you referring to:

• the fact that she can't expect to be taken out and be surprised, or
• if he/she doesn't take her out on Thursday, she'll expect to be taken out on Friday?

In either case, I think that we have to allow that. Zain Ebrahim (talk) 22:23, 24 May 2008 (UTC)

I reworded the paragraph to try to clarify it. It would probably have made more sense if I'd used "predict" instead of "expect", since that implies more certainty. It doesn't sound fair to predict that the lunch will happen every day and then claim to be vindicated when it finally does. It sounds somewhat fairer to expect it every day, and I'm happy to allow her to, but the problem becomes trivial if we do. It's all about the vagueness of the terms. Replace "expect" with "know". Define "know" such that one can't know something that doesn't come to pass, i.e. there are no possible worlds where your sister knows you're going to take her to lunch on day N and you don't do so. Then she can't know that you'll take her to lunch on any day, even Friday, since there's no law of physics compelling you to do so; thus you can fulfill the bargain by taking her out on any day, even Friday. Alternatively we could define "know" such that there are no worlds where all three of the following are true: she knows you'll take her to lunch on day N, you don't do so, and you ultimately fulfill the bargain. In that case she can know on Friday but not on any earlier day, since on Thursday she's forced to acknowledge the existence of a possible world where (for some reason) she doesn't know on Friday and you take her out then. Or we could define "know" even more broadly by allowing her to eliminate possible worlds based on plausible assumptions about her own future behavior. In that case she can (choose in advance to) "know" you'll take her to lunch every day, and so you can't fulfill the bargain. Once you define your terms precisely enough the problem can be solved, and the solution depends on the definition. If you don't define your terms and try to reason intuitively about your knowledge/beliefs/predictions then you're doomed, because they'll just keep flip-flopping. -- BenRG (talk) 00:10, 25 May 2008 (UTC)

Was the criterion that she can choose the day for the dinner also in the bet? If you promised in the bet that you'd take her to a dinner, but then later you just casually asked what day would be the best for her, then I think it might not be a break of your promise if you took her to a dinner but not at the day she said would be the best for her. It was nice of you that you asked her because obviously it would be bad if she had some other occupation at the time you scheduled the dinner, but that she can choose some almost impossible criterion for the day wasn't really part of the penalty. On the other hand, you could also try to show up on any day and surprise her in some way other than by the choice of the day. – b_jonas 11:35, 25 May 2008 (UTC)

Take her out on Friday. Monday morning, she will think that you have thought through the puzzle and will take her out on Monday because it is the "most" surprising. Tuesday morning, she will think that you are being clever and waiting until Tuesday. On Wednesday she will be yet more expectant, and on Thursday she will be 100% sure that that is the day. On Friday she will think you have forgotten. This way, you surprise her 5 times instead of just once.

BenRG's argument is very persuasive... also, that "parable of the dagger link" is great.

Where is your sister going? If it's very close, then you could contrive to take her to lunch on Saturday. (And if to you the week starts on Monday, Sunday is also an option.) Eric. 81.159.33.9 (talk) 17:53, 25 May 2008 (UTC)

Finding Nash equlibrium

Hi. I have functions $f_{i}:\mathbb {R} ^{p}\to \mathbb {R}$, for $1\leq i\leq p$. I am interested in finding the Nash equilibrium of the functions, that is, a point $x=(x_{1},\ldots ,x_{p})\in \mathbb {R} ^{p}$ such that for every $1\leq i\leq p$ and every $t\in \mathbb {R}$ we have $f_{i}(x_{1},\ldots ,x_{i-1},t,x_{i+1},\ldots ,x_{p})\leq f_{i}(x)$. Does anyone know a good algorithm for this? Ideally it should only evaluate the functions themselves and not their derivatives.

The naive method is of course to optimize each coordinate separately in every iteration. However, this can fail to converge, and at best converges linearly. I tried a basic google search, but could only find discussions of Nash equilibria for mixed strategies from a discrete space (then again, I have not yet fully mastered the most important skill of the modern world).

So, any suggestions? Thanks. -- Meni Rosenfeld (talk) 20:06, 24 May 2008 (UTC)

Roots of $f(t)$

I’d like to check my understanding here. I’m trying to write the r^th roots of a function of $t$ in a concise manner. Specifically say I want to write the set of all r^th roots of $f(t)$. Is it correct to write $\left\{\zeta _{r}^{n}{\sqrt[{r}]{f(t)}}:0\leq n\leq r-1\right\}$ where $\zeta _{r}$ is an r^th root of unity, and ${\sqrt[{r}]{f(t)}}$ is some r^th root of $f(t)?$ GromXXVII (talk) 21:51, 24 May 2008 (UTC)

Mostly yes. However:

1. $\zeta _{r}$ must be a primitive rth root of unity, not just any root.
2. If t was fixed you would just have $f(t)=a$, and then an adequate description of the set of all rth roots of a would be "$\left\{\zeta _{r}^{n}{\sqrt[{r}]{a}}:0\leq n\leq r-1\right\}$ where $\zeta _{r}$ is a primitive r^th root of unity, and ${\sqrt[{r}]{a}}$ is some r^th root of $a$", because there are only a finite number of choices and it doesn't matter which one you take. However, if you want this to apply to a variable t - that is, define a function $R:\mathbb {C} \to {\mathcal {P}}(\mathbb {C} ),\ t\mapsto \left\{\zeta _{r}^{n}{\sqrt[{r}]{f(t)}}:0\leq n\leq r-1\right\}$ - it gets a little trickier. Now "${\sqrt[{r}]{f(t)}}$ is some r^th root of $f(t)$" comprises an infinitude of choices, and with the way you have you have phrased this, it is not immediately obvious that this is at all possible. In this particular case it is of course a non-issue as providing an explicit choice function is trivial, but this goes to show you that the description is not perfect. A possible fix is to replace "${\sqrt[{r}]{f(t)}}$ is some r^th root of $f(t)$" with "${\sqrt[{r}]{}}$ is some branch of the rth root" (which we take to be known to exist).
3. If you want to be even more precise, you can specify explicitly that n must be an integer.

All that said, if all you want is a concise notation for the set of roots, why not just write $\{z\in \mathbb {C} :z^{r}=f(t)\}$? -- Meni Rosenfeld (talk) 22:16, 24 May 2008 (UTC)

In response to your last point: That's a set of complex numbers which depends on t, the OP seems to be looking for a set of functions. If you have a fixed t, then you're ok, but I'm assuming that's not the case the OP is interested in (if it was, why mention f at all, and not just give f(t) its own name?). --Tango (talk) 22:57, 24 May 2008 (UTC)

May 25

Game theory symbol meaning

To give a bit of context, here is a payoff matrix (well not a matrix because I don't know how to do them) for the Hawk-Dove game

Hawk meets Hawk - Both get -2 (fight)
Hawk meets Dove - Hawk gets 2, Dove gets 0 (steal)
Dove meets Dove - both get 1 (share)

My notes then show:

• Θ_D = λ + P_D * 1 + P_H * 0

• Θ_H = λ + P_D * 2 + P_H * -2

P_D is the probability of meeting a dove (the proportion of doves in the population). I'm wondering what exactly theta and lambda mean here. My instinct is that Θ_D would represent the expected pay-off for a dove, but the use of the lambda (which is apparently equal to 3) throws me off here. What is the point in lambda and what does it represent? --80.4.203.142 (talk) 12:22, 25 May 2008 (UTC)

variance of t-distribution

Why don't they standardise the t-distribution so that the variance is always 1? After all, it is already a general, standardised, distribution, because in getting a t-statistic, which comes from a t-distribution (typically if we add the assumption of the null hypothesis in practical applications), you divide by the standard error. Thanks in advance, 203.221.126.247 (talk) 12:59, 25 May 2008 (UTC)

Apart from anything else, there is the problem that a t₁ hasn't got a variance, and t₂ has variance infinity. Algebraist 16:37, 25 May 2008 (UTC)

It's because the distribution is interesting because it has a certain number of relationships with other distributions, (in particular, see the bottom of http://en.wikipedia.org/wiki/Student_t_distribution#Derivation) which would be broken if it's 'standardised'.--Fangz (talk) 22:32, 25 May 2008 (UTC)

May 26

Symbols question

What would you use ⋗ / ⋖ for? And what's the difference between ≆ and ≇? Thanks. 70.162.29.88 (talk) 03:49, 26 May 2008 (UTC)