3楼: Originally posted by hank612 at 2014-10-28 01:52:50
众所周知 Weierstarss-Hadamard 亚纯函数无穷连乘展开式
\frac{\sin{z}}{z}=\prod_{n=1}^{\infty} (1-\frac{z^2}{n^2\pi^2})
那么, 让 z=i\pi , 因为双曲正弦定义 \sinh{z}=-i\sin(iz)=\frac{e^z-e^{-z}}{2}, 立 ...

2楼: Originally posted by 陈义chenyi at 2014-10-27 11:19:53
打错了
n→∞  (1+1/1^2)(1+/2^2)(1+/3^2)(1+1/4^2)……=(e^π-e^(-π))/(2π)