12楼: Originally posted by hank612 at 2016-05-30 13:03:53
\prod_{n=0}^{\infty}(1+\frac{x}{4n+1})(1-\frac{x}{4n+3})=\prod_{n=0}^{\infty}\frac{1-(\frac{(x-1)/4}{(n+1/2)})^2}{1-(\frac{1/4}{n+1/2})^2}=\frac{\cos{\pi\frac{x-1}{4}}}{\cos{\pi\frac{1}{4}}

这里用 ...

13楼: Originally posted by hank612 at 2016-05-30 13:11:20
实在不想辛辛苦苦打了半天的字，却变成 ”Invalid Equation“

\prod_{n=0}^{\infty}(1+\frac{x}{4n+1})\cdot(1-\frac{x}{4n+3})

=\prod_{n=0}^{\infty}\frac{1-(\frac{(x-1)/4}{(n+1/2)})^2}{1-(\frac{1/4} ...

13楼: Originally posted by hank612 at 2016-05-30 13:11:20
实在不想辛辛苦苦打了半天的字，却变成 ”Invalid Equation“

\prod_{n=0}^{\infty}(1+\frac{x}{4n+1})\cdot(1-\frac{x}{4n+3})

=\prod_{n=0}^{\infty}\frac{1-(\frac{(x-1)/4}{(n+1/2)})^2}{1-(\frac{1/4} ...

2楼: Originally posted by 001gqs at 2016-05-06 20:15:40
除了复变里面的Weierstrass factorization Thm, 我别无他法。

3楼: Originally posted by maxman at 2016-05-07 10:51:37
可以用留数定理推导出

7楼: Originally posted by hank612 at 2016-05-30 08:22:21
由 Wallis 无限乘积 \prod_{k=1}^{\infty}\frac{4k^2}{4k^2-1}=\frac{\pi}{2}  （也可以由 Out中 x=\frac{1}{2} 一眼看出）

得出
\prod_{k=1}^{\infty}(1-\frac{x}{4k^2-1})\prod_{k=1}^{\infty}\frac{4k^2-1}{ ...

7楼: Originally posted by hank612 at 2016-05-30 08:22:21
由 Wallis 无限乘积 \prod_{k=1}^{\infty}\frac{4k^2}{4k^2-1}=\frac{\pi}{2}  （也可以由 Out中 x=\frac{1}{2} 一眼看出）

得出
\prod_{k=1}^{\infty}(1-\frac{x}{4k^2-1})\prod_{k=1}^{\infty}\frac{4k^2-1}{ ...

8楼: Originally posted by gold2007 at 2016-05-30 08:32:52
您和楼主对于无穷级数的理解令人仰望。
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