标签:point math form ali style ber aop overflow call
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也就是求\[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{\prod\limits_{k = 0}^n {\left( {a + kd} \right)} }}} .\]
解.首先有$$\prod_{k=0}^n{\frac{1}{a+kd}}=\frac{\Gamma \left( \frac{a}{d} \right)}{d^{n+1}\Gamma \left( \frac{a}{d}+n+1 \right)},$$
又因为$$\gamma \left( s,x \right) =\sum_{k=0}^{\infty}{\frac{x^se^{-x}x^k}{s\left( s+1 \right) ...\left( s+k \right)}}=x^s\,\Gamma \left( s \right) \,e^{-x}\sum_{k=0}^{\infty}{\frac{x^k}{\Gamma \left( s+k+1 \right)}},$$我们有
$g(x) = x^a f(x^d)$ satifies $g‘(x) = x^{a-1} + x^{d-1} g(x)$. Solve the associated differential equation and conclude.
令$a\in (0,\pi)$,设$n$为正整数.证明$$\int_0^{\pi}{\frac{\cos \left( nx \right) -\cos \left( na \right)}{\cos x-\cos a}dx}=\pi \frac{\sin \left( na \right)}{\sin a}.$$
Since
Thus $$ f_{2n}\left(\frac {\pi}{2n + 1}\right) < \frac {\pi}{2n + 1}\cdot\frac {2\pi}{2n + 1} - \frac {\pi}{2(2n + 1)} < 0.$$
标签:point math form ali style ber aop overflow call
原文地址:https://www.cnblogs.com/Eufisky/p/11229174.html