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

= January 30 =

Generalization of integral 2
It is known $$\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$.

Consider the following generalization

$$\int_0^1 x^a(1-x)^b f(x) dx$$

where $$f$$ satisfies $$\int_0^1 f(x)dx = 1$$ 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: $$\lim_{n\rightarrow\infty}\frac{n^{-1/2}(1/2)^n}{\int_0^1 x^{n/2}(1-x)^{n/2} f(x) dx}$$ --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: 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 $$\int_0^1 x^{n/2}(1-x)^{n/2} f(x) dx$$ 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:
 * $$\lim_{n->\infty} \frac {\int_0^1 x^{n/2}(1-x)^{n/2} f(x) dx}{\int_0^1 x^{n/2}(1-x)^{n/2} dx} = f\left(\tfrac 1 2\right)$$
 * — Arthur Rubin (talk) 17:48, 3 February 2013 (UTC)