Wikipedia:Reference desk/Archives/Mathematics/2006 July 20

Natural Logarithm
Does anyone know the inverse of xln(x)? I know, of course that itll be in terms of W(x), and that it is multiple-valued for negative numbers grater than -1/e, but beyond that, I've found nothing. -- He Who Is[ Talk ] 03:35, 20 July 2006 (UTC)
 * $$x \ln{x} = y \iff x = \frac{y}{W(y)}$$
 * Mathematica is your friend :). Or you can substitute x = et and use the definition of W to get x = eW(y), which appears to be equivalent. -- Meni Rosenfeld (talk) 08:11, 20 July 2006 (UTC)


 * Did anyone else notice this is the second time LambertW has arisen on this page in a few weeks? This really must be The World's Most Useful Special Function which Almost Nobody Knows.  (For anyone who falls into that group, $$w = z \exp(z)$$ iff $$ z = W(z)$$.  It is a multibranched function which has a real valued branch on the interval $$(-1/e, \infty)$$.  Some other places where I have seen it come up is in the Kruskal-Szekeres chart for the Schwarzschild vacuum solution in general relativity, a problem in mathematical biology, etc, etc....) ---CH 09:41, 23 July 2006 (UTC)

probability
$$Insert formula here$$ i have 42 numbers.i get to pick 7 numbers.what are the odds that the numbers i pick are 4,8,9,32,36,18,7.how can i intelligently increase my chances of picking in sets of seven nos.and all those numbers are there or atleast three numbers are there.


 * I would say that your chances of picking that set of numbers is 42!/7!35! = 1/26978328 (about 27 million to one against). Not sure what on earth the rest of your question actually means, but it looks like you are trying to improve your chances in a lottery, and the bad news is that you can't - sorry! Madmath789 09:40, 20 July 2006 (UTC)


 * Madmath's answer is assuming that order doesn't matter. If, however, this is a lottery (as suggested below), order does matter and the answer is 42!/35!, about 135*109 to 1. &mdash; Lomn | Talk 15:34, 20 July 2006 (UTC)


 * If this is a lottery, there is nothing you can do to make it more likely that you pick the winning numbers. However, there is a recognised technique for increasing your winnings if you do happen to get the right numbers. That is to pick numbers which fewer other people will pick, so you share the prize with less people. Overall though, I have to say I'd never have thought the government would find a way to tax insane optimism. Notinasnaid 09:42, 20 July 2006 (UTC)


 * I never considered it a tax on optimism so much as a tax on people who are bad at math. &mdash; Lomn | Talk 15:34, 20 July 2006 (UTC)


 * Since taxing any activity tends to reduce the level of that activity, let's just hope the lottery, AKA "stupidity tax", has the same effect. :-) StuRat 21:57, 20 July 2006 (UTC)


 * The way I figure it, the lottery reduces the tax burden on those of us who are mathematicians. 48v 05:49, 21 July 2006 (UTC)


 * Which 42 numbers do you have? The chance you pick a certain number depends on which are included. - Mgm|(talk) 11:54, 24 July 2006 (UTC)

Zero multiplied by infinity?
What is the product of zero and infinity? What if infinity is replaced by infinity squared? Is this value a real number? --Loool 10:53, 20 July 2006 (UTC)
 * The product of zero and infinity is undefined. See Infinity. — Mets 501  (talk) 12:47, 20 July 2006 (UTC)
 * But this section deals only with one interpretation of infinity - the extended real number line. The product is similarly undefined in some other contexts, such as the real projective line. However, in the contexts of set theory, measure theory and non-standard analysis, the product is generally taken to equal 0. So your question, as phrased, does not really have a "correct" answer - it depends on the context in which it is asked. In any case, however, the result will not change if you replace infinity with infinity squared. -- Meni Rosenfeld (talk) 13:58, 20 July 2006 (UTC)

I'm not the original poster, but can someone explain to me how having "some kind" of infinite 0 times would leave me with anything? I already have no box of infinite apples, no computer with infinite memory, and no piano with infinite keys. How can that sentence leave me with some apples, memory, or piano keys, whereas if I hadn't said that sentence you wouldn't know I have those???? Thank you. 82.131.187.228 18:44, 20 July 2006 (UTC).


 * I don't see where anybody has said that '"some kind" of infinity 0 times' would leave anything. It has been said that is either undefined or taken to be 0.  The best thing to remember with infinity is not to try to count with it.  --LarryMac 19:04, 20 July 2006 (UTC)


 * Yes, your argument basically shows that defining 0 * &infin; as anything but 0 would have weird results, and is therefore generally not recommended. However, in many cases, defining it as equal to 0 also leads to weird results, and therefore it is best left undefined. -- Meni Rosenfeld (talk) 19:48, 20 July 2006 (UTC)


 * What weird results? In measure theory, defining it to be zero works nicely.  (Cj67 02:51, 21 July 2006 (UTC))


 * see http://mathworld.wolfram.com/Indeterminate.html for a good explination of the baisic indefinite forms. 48v 05:47, 21 July 2006 (UTC)


 * I said, "in many cases"... Measure theory is not one of these cases. -- Meni Rosenfeld (talk) 06:28, 21 July 2006 (UTC)

New computing reference desk
Well the title says it all really. This is to announcement that there is a new section of the reference desk devoted to software, hardware and computer science at Reference desk/Computer so that those of you who want to can add it to their watchlists. if you want to comment on the wisdom/stupidity of the move please don't do it here do it here Theresa Knott | Taste the Korn 15:31, 20 July 2006 (UTC)


 * Judging by the distinct lack of enthusiasm for answering computing questions here, this is welcome news. Thanks. --KSmrqT 21:16, 20 July 2006 (UTC)

Does anyone know the name of this conjecture?
In pseudocode:

while n is not equal to 1: if n is even: n/2 else if n is odd: (n*3)+1

I've just been reading an online tutorial on Python, and its author claimed that it is yet unproved whether or not that will always eventually result with 1 or not. Unfortunately the author didn't name this conjecture.

I've tried searching but with no success -- I didn't know what to search exactly, it's difficult to describe something like that as a search term.

If anyone knows the name of this conjecture I'd appreciate it if you'd mention it, because if it's been proved I'd very much like to attempt to understand it, and if it hasn't I'd like to read about the history of it.

Many thanks.

That's the Collatz Conjecture. It's also often known as the Hailstone conjecture, because that operation forms a "Hailstone sequence". No counterexample has ever been found, (although some numbers climb thousands before descending to 1) but it is unproven at the moment. -- He Who Is[ Talk ] 20:05, 20 July 2006 (UTC)


 * Thanks for the quick reply!


 * I think you will also find that some mathematicians refer to it as a 'disease' rather than a conjecture, due to the amount of wasted time that can easily be spent working on it. Madmath789 20:27, 20 July 2006 (UTC)


 * You know, it's odd. They mention quite a few neat tricks and optimizations in there, but never do they mention that you can entirely ignore multiples of three, along with multiples of two - although a sequence might happen to start out at one (say 15), as a quick look at the rules can prove, it will almost immediately leave it and will never pass through a multiple of three again. Black Carrot 02:10, 23 July 2006 (UTC)


 * This famous conjecture was discussed in one of Martin Gardner's columns on Mathematical Games (or was it A. K. Dewdney?), and there has also been at least one article in the American Mathematical Monthly devoted to it. It is quite fascinating once you start thinking about it!---CH 09:35, 23 July 2006 (UTC)

Function approaching zero
It's been awhile since I've been in a math class, but this came up in discussion the other day: what is the name of a function that approaches, but never actually becomes, zero. If I'm not clear enough, ask a question. Thanks in advance! --Antonymous 20:00, 20 July 2006 (UTC)
 * You are probably looking for the word 'asymptotic', or maybe 'asymptote'? Madmath789 20:04, 20 July 2006 (UTC)

I believe he's looking for "Hyperbolic". While its not always the same thing, it would seem to me that most functions with that property are hyperbolas, and I don't know of any other better terminology. -- He Who Is[ Talk ] 20:07, 20 July 2006 (UTC)
 * Both are excellent answers, given the lack of info I included. I think asymptote is closer to what I was looking for. The topic came up when someone asked "What are the chances of you hooking up with your ex?" and I tried to convey that the odds approach zero with every passing day, but will probably never actually reach zero...I just didn't have the math lingo down and almost had to resort to drawing a graph! Any thoughts on what I should've said to be more clear ("The odds of that happening are clearly an asympototic function where x equals...")? --Antonymous 20:15, 20 July 2006 (UTC)
 * Yes, asymptotic would be the more correct answer, as it's the more general case. Hyperbolic functions usually (always?) have asymptotes, but not all functions with asymptotes are hyperbolic -- say, y=x-1 &mdash; Lomn | Talk 20:18, 20 July 2006 (UTC)
 * Also, if you like, I believe a function like my example would be an exponential inverse ("the odds are inversely exponential..."). Or maybe "negatively exponential". Or perhaps just "logarithmic", which also meets the condition you want, and that's a one-word answer (y = ln x). I'm not so good with the math terminology. &mdash; Lomn | Talk 20:19, 20 July 2006 (UTC)


 * Actually, the function y=x-1 is a hyperbola... But obviously, most functions aren't. Hyperbolic is definitely not the word I'd use for this. I think the best word you are looking for is limit, or the related terms "approaches" (you have used this word in your description; it is actually a good word to use by itself) and "converges". For example, "your chances of hooking up with your ex approaches 0 as time advances", or "the limit of the probability of hooking up with your ex as time approaches infinity is 0". This does not, however, rule out the posibility that the chances are 0; you have to add that info separately. -- Meni Rosenfeld (talk) 20:32, 20 July 2006 (UTC)


 * Most functions that asymptotically approach zero have nothing to do with a hyperbola. One of the best-known examples is the family of "bell curves", whose essential form is exp(−x2/σ2). It is trivial to construct other examples, such as (x2+1)/(x4+x3+1), or a sigmoid function. Nor need we use a pretty formula; for example, we have the function giving the density of prime numbers. No generic term comes to mind for functions that approach zero, nor is it clear that there is much need for one. --KSmrqT 21:13, 20 July 2006 (UTC)


 * How about "[asymptotically] vanishing"? Fredrik Johansson 21:18, 20 July 2006 (UTC)

Easy formula for calculation of correlation of 2 x 2 categories?
Some years ago I read in a statistics textbook a very simple formula that allows you to calculate the correlation between four categories. I've never been able to find mention of that formula again. (It might have had a name such as Somebody's Formula, but that could be just my imagination).

I don't remember the jargon to describe this, but you could use it in situations such as this (imaginary) one:

You want to find out the correlation between wearing hats and gender. So you observe people walk down a busy street for five minutes. You see 57 women wearing hats, 42 women not wearing hats, 23 men wearing hats, and 64 men not wearing hats. From this data you easily calculate a correlation.

Anyone know what the formula is please?

Thanks. --62.253.52.155 20:26, 20 July 2006 (UTC)


 * Just taking a guess here, but the percentage of women who wear hats is 100(57)/(57+42) or 57.57%, while the percentage of men wearing hats is 100(23)/(23+64) = 26.44%. I then get a correlation factor of 57.57-26.44 or 31.13%, meaning women are 31% more likely to wear hats than men. StuRat 21:46, 20 July 2006 (UTC)
 * Unfortunately, that's not correlation (and it would be [57-26]/26 or slightly over 100% more likely to wear hats). &mdash; Lomn | Talk 21:50, 20 July 2006 (UTC)

Sorry, what I meant was how to calculate the correlation coefficient. This is a number which is no more than plus 1 and no less than minus 1. So talking about percentages doesnt make any sense.

The article Correlation discusses the maths of it, including linking to another article about correlation. I am not a mathematician, so could anyone tell me if any of the formulas for, for example, an n x n case would simplify down for the 2 x 2 case please? Thanks again.

I've now had a search around on the internet. I did some web searching. It could be the Goodman – Kruskal Gamma

G = (C-D)/(C+D)

where C = concordant pairs and D = discordant pairs

except I don't know what concordant and discordant pairs mean - could anyone enlighten me please?

It could be also be the Phi coefficient.

Phi=ad-bc/(efgh)^0.5

where a, b, c and d correspond to the four cells of a 2 x 2 table, and e=a+b, f=c+d, g=a+c, and h=b+d

There is not any seperate Wikipedia article for Phi coefficient as far as I know, and the simple formula above is not quoted either. I do not think I am allowed to include URLs, but searching on Google for Phi coeficcient and Richard Lowry should bring you to a page that calculates the phi coefficient online. For the figures above, the result is +0.31.

it might also have been something connected with the Fisher Exact Probability Test:

(a/(a+b))-(c/(c+d))

but I am not sure what this actually measures or if it will always give a result ranging from plus 1 to minus 1.


 * I have calculated (a/(a+b))-(c/(c+d)) for the hat figures and it gives minus 0.316, the same as the phi coefficient. The fact that its minus rather than plus may have just been due to me forgetting what order I put the cells in previously. It could be that this is the formula I was trying to find. I seem to remember is being called Glass's formula, but this has a greater than 50% chance of merely being a trick of my memory after so many years. Just how reliable this method would be to calculate a correlation coefficient for a 2x2 table is, I don't know.

Math Riddle
Someone told me a riddle yesterday and maybe you could help me solve it.

Apperantly, there is a number less in quantity than 1 and higher in quantity than 0. Yet, this number is not 0, 1, or any number between 0 and 1.

I figured maybe the symbol pi, but this has a decimal equivalent between 0 and 1. I figured maybe some imaginary number or something like that...

Any ideas?

--69.138.61.168 20:26, 20 July 2006 (UTC)

Unless I read incorrectly, it can be mathematically shown that there is no such number. So I believe it may have nothing to do with math altogether. Since its being compared, it can't be complex. Less than one and greater than 0 implies:


 * $$0<x<1$$

Which is equivalent to saying that it is between 1 and 0, which contradicts the claim that it is not between the numbers. the only mathematical solution I can think of the infinity on the projective reals, which is both greater than and less than all numbers, nor is it between 1 and 0. -- He Who Is[ Talk ] 21:25, 20 July 2006 (UTC)


 * I'm thinking that by "greater in quantity than 0" they mean the absolute value is greater than zero. So, -1 would be an answer. StuRat 21:35, 20 July 2006 (UTC)


 * If that is the case, than, -1 is equal in quantity to 1. Since we're comparing absolute values, then it can be any nonzero complex number inside the unit circle. That makes sense. -- He Who Is[ Talk ] 22:12, 20 July 2006 (UTC)


 * That's what I first thought when I read it, that "in quantity" meant "in absolute value", in which case an imaginary number like .5i fits the bill. I suspect, however, that the riddle has lost something critical in translation, as can sometimes happen when a story's being repeated. The joke about the roof, for instance. Black Carrot 23:54, 20 July 2006 (UTC)


 * Oh, the question above gives me an idea. How about percentages? In quantity, 50% is less than proportion 1 and more than proportion 0, but 50% isn't (as far as most people know) actually between 0 and 1, since it has a 50 in it. Black Carrot 23:58, 20 July 2006 (UTC)


 * Hmmm... The only is that it could be argued that %50 is numerically equivalent to .5. Also, I don't think that it has a range of answers. Seeing as its a riddle, it probobaly has a single answer. Or at least a single intended answer, or countably many intended answers, along the the uncountably infinite other answers we've already found but aren't nessecarily the ones the person who told the riddle wants. -- He Who Is[ Talk ] 00:28, 21 July 2006 (UTC)


 * A guess: Could it be 25 cents, or a quarter? (Igny 01:22, 21 July 2006 (UTC))


 * Ooh! Ooh! I've got an answer! It's probably not the one the creator intended, but hey, there's nothing wrong with lateral thinking. Take a function that oscillates between two nonequal values between zero and one (say, sin(x)/4+1/2). Then both the actual value f(x), and the limit of f(x) as x->infinity, are without question between zero and one, but neither is expressible as a single number within that range. Black Carrot 04:20, 21 July 2006 (UTC)