Testwiki:Reference desk/Archives/Mathematics/2013 January 30

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < January 29 ! width="25%" align="center"|<< Dec | January | Feb >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 30

Generalization of integral 2

It is known $\underset{0}{\overset{1}{\int }}{x}^a(1-x{)}^{b}dx=\frac{\mathrm{\Gamma }(a+1)\mathrm{\Gamma }(b+1)}{\mathrm{\Gamma }(a+b+2)}$.

Consider the following generalization

$\underset{0}{\overset{1}{\int }}{x}^a(1-x{)}^{b}f(x)dx$

where $f$ satisfies $\underset{0}{\overset{1}{\int }}f(x)dx=1$ and $f(x)\ge 0$ for all x.

What is the result then? --AnalysisAlgebra (talk) 11:04, 30 January 2013 (UTC)

You might find Cauchy–Schwarz inequality useful, otherwise I don't think there is much one can say. Dmcq (talk) 23:26, 30 January 2013 (UTC)

OK. I want to find this limit: $\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{-1/2}(1/2{)}^n}{\underset{0}{\overset{1}{\int }}{x}^{n/2}(1-x{)}^{n/2}f(x)dx}$ --AnalysisAlgebra (talk) 02:54, 31 January 2013 (UTC)

Hint: what is the maximum value of x(1-x) in that interval, and what value of x does it occur at? Looie496 (talk) 03:38, 31 January 2013 (UTC)

1/4 and 1/2, respectively. Are you trying to bound the expression or something? But that tells you nothing about the maximum value of $x^{n/2}(1-x{)}^{n/2}f(x)$ with the f in there.--AnalysisAlgebra (talk) 05:28, 31 January 2013 (UTC)

As n tends to infinity with f remaining constant, $x(1-x)$ becomes a more significant factor in determining $x^{n/2}(1-x{)}^{n/2}f(x)$. -- Meni Rosenfeld (talk) 05:46, 31 January 2013 (UTC)

Is that also true for any $\alpha$ in $x^{\alpha n}(1-x{)}^{(1-\alpha )n}f(x)$--AnalysisAlgebra (talk) 05:57, 31 January 2013 (UTC)

Then the dominant factor is $x^{\alpha }(1-x{)}^{1-\alpha }$. -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)

Are you saying the limit is independent of $f$? The limit is $\sqrt{2/\pi }$ if $f(x)=1$ but that's not the case if for example $f(x)=3x^2$?--AnalysisAlgebra (talk) 09:33, 31 January 2013 (UTC)

It depends on f, but only on its behavior in a specific area. -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)

Or the integral can be done exactly using the above identity by just substituting f for its taylor series evaluated at either x=0 or x=1. As a note, naively treating the distribution as a nascent delta could lead you to erroneously conclude that if f(1/2)=0 the integral is zero. — Preceding unsigned comment added by 123.136.64.14 (talk) 06:13, 31 January 2013 (UTC)

Only if you know $f$, and also only if $f$ is analytic. --AnalysisAlgebra (talk) 09:33, 31 January 2013 (UTC)

@User:AnalysisAlgebra: Please do not remove other's comments at Help Desk/Math as you did here: [1] This is considered bad form. See here for the guideline. El duderino ^(abides) 11:36, 31 January 2013 (UTC)

If the function is of the form $x^{\alpha }(1-x{)}^{\beta }$ then the integral is a beta function. So write your function as a linear combination of such functions, $f(x)=\sum _i{k}_i{x}^{{\alpha }_i}(1-x{)}^{{\beta }_i}$ . Bo Jacoby (talk) 12:41, 31 January 2013 (UTC).

What if $f$ can't be written as a linear combo of those functions? Even if you allow an infinite sum, I'm pretty sure there are functions satisfying the conditions above which are not linear combinations of such functions.--AnalysisAlgebra (talk) 12:47, 31 January 2013 (UTC)

I doubt you'll find any reasonable closed form for the general case. The limit however should be quite easy. -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)

If the function f is continuous, then the Stone–Weierstrass theorem tells you that it can be approximated by polynomials. So don't be pretty sure that it can't be done. Bo Jacoby (talk) 08:27, 1 February 2013 (UTC).

So, I was right, then. --AnalysisAlgebra (talk) 09:27, 1 February 2013 (UTC)

No, you were wrong. Bo Jacoby (talk) 13:01, 1 February 2013 (UTC).

If f is sufficiently differentiable at 1/2, the asymptotics of $\underset{0}{\overset{1}{\int }}{x}^{n/2}(1-x{)}^{n/2}f(x)dx$ should only depend on the even derivatives of f at 1/2, and probably only the first nonzero (even) derivative. — Arthur Rubin (talk) 17:32, 3 February 2013 (UTC)

When put that way, the result is obvious. If f is differentiable at 1/2, and integrable over (0,1), then:

$\underset{n->\mathrm{\infty }}{\lim }\frac{\underset{0}{\overset{1}{\int }}{x}^{n/2}(1-x{)}^{n/2}f(x)dx}{\underset{0}{\overset{1}{\int }}{x}^{n/2}(1-x{)}^{n/2}dx}=f\left(\frac{1}{2}\right)$

— Arthur Rubin (talk) 17:48, 3 February 2013 (UTC)