Wikipedia:Reference desk/Mathematics

The Wikipedia Reference Desk covering the topic of mathematics.
WP:RD/MA Reference desk/Mathematics #eee #f5f5f5 #eee #aaa #aaa #aaa #00f #36b #000 #00f mathematics Wikipedia:Reference desk/Mathematics Mathematics WP:RD/MA

[edit] May 20

[edit] 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)

[edit] 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): u_{x x} = -9u , so Failed to parse (Missing texvc executable; please see math/README to configure.): u_{t t} = 9u

which has the general solution Failed to parse (Missing texvc executable;

please see math/README to configure.): u(x,t) = a(x)e^{3 t} + b(x)e^{-3 t}

(or equivalently Failed to parse (Missing texvc executable;

please see math/README to configure.): u(x,t) = c(x) \cosh(3 t) + d(x) \sinh(3 t) ). The initial conditions narrow that down to Failed to parse (Missing texvc executable; please see math/README to configure.): u(x,t) = \sin(3 x) \cosh(3 t)

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)

[edit] 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)

[edit] Fields

For the field of integers mod p, I cannot find an inverse element for all cases. Failed to parse (Missing texvc executable; please see math/README to configure.): 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)

[edit] 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): (x-r)^2+y^2=r^2

and the line has equation Failed to parse (Missing texvc executable;

please see math/README to configure.): y=(x+r)\tan\theta

(you can work this out: 2 points on the line are (-r, 0) and (0, Failed to parse (Missing texvc executable;

please see math/README to configure.): 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)

[edit] 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)

[edit] 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)

[edit] May 21

[edit] 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

Failed to parse (Missing texvc executable;

please see math/README to configure.): E = F \oplus G.

Let Failed to parse (Missing texvc executable; please see math/README to configure.): \{e_1,\ldots,e_n\}

be a basis for F and extend this to a basis B for E.  Let Failed to parse (Missing texvc executable;

please see math/README to configure.): \{\varepsilon_1,\ldots,\varepsilon_n\}

be the basis dual to Failed to parse (Missing texvc executable;

please see math/README to configure.): \{e_1,\ldots,e_n\} . By the Hahn-Banach theorem, each Failed to parse (Missing texvc executable; please see math/README to configure.): \varepsilon_i

extends to a continuous linear functional on E.  The intersection of the kernels of the Failed to parse (Missing texvc executable;

please see math/README to configure.): \varepsilon_i

is then a closed subspace G of E, and clearly Failed to parse (Missing texvc executable;

please see math/README to configure.): F \cap G = \{0\} . And if Failed to parse (Missing texvc executable; please see math/README to configure.): \beta \in B - \{e_1,\ldots,e_n\}

then

Failed to parse (Missing texvc executable;

please see math/README to configure.): \beta - \sum_i \varepsilon_i(\beta)e_i \in G,

so that Failed to parse (Missing texvc executable; please see math/README to configure.): \beta \in F + G . It follows that Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): v \in E

then Failed to parse (Missing texvc executable;

please see math/README to configure.): 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)

[edit] 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)

[edit] 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): p(A | B)\cdot p(B)=p(A\cap B) , the condition Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): 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 Failed to parse (Missing texvc executable; please see math/README to configure.): p(A | B)\cdot p(B)=p(A\cap B) , unless you observe that empirically or define Failed to parse (Missing texvc executable; please see math/README to configure.): 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).

[edit] 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)

[edit] May 22

[edit] 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.

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): dx=f'(t)\cdot dt

to get Failed to parse (Missing texvc executable;

please see math/README to configure.): \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)

[edit] 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): \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 Failed to parse (Missing texvc executable; please see math/README to configure.): \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)

[edit] 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): S

be your surface, which is symmetric under negation of Failed to parse (Missing texvc executable;

please see math/README to configure.): x

or Failed to parse (Missing texvc executable;

please see math/README to configure.): y . Let Failed to parse (Missing texvc executable; please see math/README to configure.): G(x, y, z) = (xz)i + 0j + (y)k

and Failed to parse (Missing texvc executable;

please see math/README to configure.): H(x, y, z) = 0i + (x)j + 0k . Flipping Failed to parse (Missing texvc executable; please see math/README to configure.): G

across the Failed to parse (Missing texvc executable;

please see math/README to configure.): x -Failed to parse (Missing texvc executable; please see math/README to configure.): y

plane, we obtain Failed to parse (Missing texvc executable;

please see math/README to configure.): (x(-z))i + 0j + (y)(-k) = -G

so Failed to parse (Missing texvc executable;

please see math/README to configure.): G

and Failed to parse (Missing texvc executable;

please see math/README to configure.): -G

have the same flux, which must be zero.  Flipping Failed to parse (Missing texvc executable;

please see math/README to configure.): H

across the Failed to parse (Missing texvc executable;

please see math/README to configure.): y -Failed to parse (Missing texvc executable; please see math/README to configure.): z

plane, we obtain Failed to parse (Missing texvc executable;

please see math/README to configure.): 0(-i) + (-x)j + 0k = -H , so similarly the flux of Failed to parse (Missing texvc executable; please see math/README to configure.): H

is 0.  But the flux of Failed to parse (Missing texvc executable;

please see math/README to configure.): F

is the sum of the fluxes of Failed to parse (Missing texvc executable;

please see math/README to configure.): G

and Failed to parse (Missing texvc executable;

please see math/README to configure.): 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 Failed to parse (Missing texvc executable;

please see math/README to configure.): \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)

[edit] 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.