My Math Forum Beta function

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November 27th, 2013, 03:19 AM   #1
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Beta function

Hi guys. I have to solve this problem. I even have a hint, but i can't solve it for days. Plesae help.
[attachment=0:2n1gmzh6]PrtScr capture.jpg[/attachment:2n1gmzh6]
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 July 7th, 2014, 05:11 AM #2 Newbie   Joined: Jul 2014 From: jordan Posts: 7 Thanks: 0 Since we have $B\left( {m,n} \right) = \int_0^1 {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} = \int_0^{1/2} {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} + \int_{1/2}^1 {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt}$ then we may obtain the following estimates (1) $\int_0^{1/2} {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} \ge \int_0^{1/2} {t^{m - 1} t^{n - 1} dt} = \int_0^{1/2} {t^{m + n - 2} dt} = \left. {\frac{{t^{m + n - 1} }}{{m + n - 1}}} \right|_0^{1/2} = \frac{1}{{\left( {m + n - 1} \right)2^{m + n - 1} }}$ (2) $\int_{1/2}^1 {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} \ge \int_{1/2}^1 {\left( {1 - t} \right)^{m - 1} \left( {1 - t} \right)^{n - 1} dt} = \int_{1/2}^1 {\left( {1 - t} \right)^{m + n - 2} dt} = \left. { - \frac{{\left( {1 - t} \right)^{m + n - 1} }}{{m + n - 1}}} \right|_{1/2}^1 =\frac{1}{{\left( {m + n - 1} \right)2^{m + n - 1} }}$ (3) $\int_{1/2}^1 {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} \le \int_{1/2}^1 {t^{m - 1} t^{n - 1} dt} = \int_{1/2}^1 {t^{m + n - 2} dt} = \left. {\frac{{t^{m + n - 1} }}{{m + n - 1}}} \right|_{1/2}^1 = \frac{{1 - 2^{1 - m - n} }}{{m + n - 1}}$ and (4)$\int_0^{1/2} {t^{m - 1} \left( {1 - t} \right)^{n - 1} dt} \le \int_0^{1/2} {\left( {1 - t} \right)^{m - 1} \left( {1 - t} \right)^{n - 1} dt} = \int_0^{1/2} {\left( {1 - t} \right)^{m + n - 2} dt} \\ = - \left. {\frac{{\left( {1 - t} \right)^{m + n - 1} }}{{m + n - 1}}} \right|_0^{1/2} = \frac{{1 - 2^{1 - m - n} }}{{m + n - 1}}$ Adding the inequalities (1) to (2) and (3) to (4) we get $$\frac{2}{{\left( {m + n - 1} \right)2^{m + n - 1} }} \le B\left( {m,n} \right) \le 2\frac{{1 - 2^{1 - m - n} }}{{m + n - 1}}$$ Finally, since $\frac{1}{{\left( {m + n - 1} \right)2^{m + n - 1} }} \le \frac{2}{{\left( {m + n - 1} \right)2^{m + n - 1} }}$, then $$\frac{1}{{\left( {m + n - 1} \right)2^{m + n - 1} }}\le \frac{2}{{\left( {m + n - 1} \right)2^{m + n - 1} }} \le B\left( {m,n} \right) \le 2\frac{{1 - 2^{1 - m - n} }}{{m + n - 1}}$$ In all above cases, of course we assume that $m+n\ne1$ for all $m,n$

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