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May 4[edit]

Detecting infinite moments in data[edit]

Is there some reasonably convincing way to test data to see if it came from a distribution with no first (or second, etc.) moment? Some information might suggest it, such as a large estimated second or higher moment, but is there something more clever? Thanks, --TeaDrinker 00:16, 4 May 2007 (UTC)[reply]

I don't think so. Take the random variable obtained by throwing a coin until the value obtained changes, giving an initial run of length L. If this was a run of heads, the outcome is 2^L. If it was a run of tails, the outcome is −2^L. The distribution of this r.v. has no first moment. But the probability that the average of even a large sample (say N = 10⁶) is alarmingly large in magnitude is actually fairly small. If you modify the r.v. by taking a cut-off maximum value for L of 32, then in all likelihood you'd have to keep throwing coins for a couple of centuries before the data would reveal a difference between the modified (momentous) distribution and the original one. --Lambiam ^Talk 07:26, 4 May 2007 (UTC)[reply]

If there is reason to suspect that a particular distribution with infinite moments is at work (e.g. Cauchy with pdf b/pi((x-m)²+b²)), a goodness of fit test (e.g. Kolmogorov-Smirnov) would be appropriate. —The preceding unsigned comment was added by 81.154.108.219 (talk) 18:24, 4 May 2007 (UTC).[reply]

Thanks for the replies! --TeaDrinker 20:48, 4 May 2007 (UTC)[reply]

28, the largest number[edit]

I'd seen a website a long time ago that humourously 'proved' that there was no number larger than 28. I've looked for it in about every different way I can think of to no result =/ anyone know where it is? -- Phoeba Wright^OBJECTION! 12:27, 4 May 2007 (UTC)[reply]

I haven't heard of that one, but can prove that 1 is the largest integer. For if n is the largest integer, then n>=n^2. But n^2>n for n not 1 or 0, and 1 is greater than 0. Thus n=1. Algebraist 12:39, 4 May 2007 (UTC)[reply]

I know that Lou Costello "proved" that 13 times 7 is 28 (three different ways, in fact; in the movie In The Navy), but I'd never seen a proof that 28 is the largest number. Here is a page with that "proof" though - [1]. --LarryMac 14:19, 4 May 2007 (UTC)[reply]

Hmm... I don't like it. The 28 is arbitrary. Can anyone think of an argument for 28 being largest that only applies to 28? Algebraist 15:31, 4 May 2007 (UTC)[reply]

That's a cute trick. The trick is the inequality. The problem is this step:

-x <= 2(2*7) square both sides.

(-x)^2 <= 2^2(2*7)^2

You can't square both sides of an inequality and expect it to hold. For example:

-2<1\,

(-2)^{2}=(-1)^{2}\cdot 4\not <1^{2}=1

(-2)^{2}\not <1^{2}

.

So yea,

x<y\not \Rightarrow x^{2}<y^{2}

. This is essentially the same as noting that

a<b\not \Rightarrow -a<-b

or

a<b\not \Rightarrow |a|<|b|

. —Ben FrantzDale 15:56, 4 May 2007 (UTC)[reply]

user: lost elim tree im working this out but im pretty sure none of those numbers are the greatest numbers 29 is an example lol

Beta function limits[edit]

When carrying out double and triple integration using, say, a change of variable to spherical coordinates, I'm often left with an integral with a product of high powers of sines and cosines. I understand this can be dealt with easily using the trigonometric form of the Beta function, and its relationship with the Gamma function, but what about when the integral is not over [0,π/2]? The work I'm doing seems to imply its possible to do it for other upper limits, such as π and 2π, but I'm unsure how the Beta function can be extended to deal with this. Does it involve considering the symmetry of the trig functions? Thanks, Icthyos 14:53, 4 May 2007 (UTC)[reply]

Using Euler's formula any product of sines and cosines can be rewritten as a weighted sum of exponentials, which can again be rewritten as a sum of sines and cosines. Would that help? There's also the incomplete Beta function, which doesn't look immediately helpful. --Lambiam ^Talk 17:00, 4 May 2007 (UTC)[reply]

Maybe I'm being really naive, but why not use power reduction? You know,

Sine	$\sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}$	$\sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}$
Cosine	$\cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}$	$\cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}$
Other	$\sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}$

Hmm?--Kirby♥time 23:22, 4 May 2007 (UTC)[reply]

This will work as well to reduce the powers, but you will have to apply these formulas recursively to get rid of high powers, whereas the method of reducing the powers through Euler's formula does everything at once, and in the same effort even gets rid of products with different arguments, as in

\sin \alpha \,\sin \beta \,\sin \gamma =-{\frac {1}{4}}\left(\sin(\alpha {+}\beta {+}\gamma )-\sin(\alpha {+}\beta {-}\gamma )-\sin(\alpha {-}\beta {+}\gamma )+\sin(\alpha {-}\beta {-}\gamma )\right).

--Lambiam ^Talk 06:16, 5 May 2007 (UTC)[reply]