12Â¥: 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}}

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13Â¥: Originally posted by hank612 at 2016-05-30 13:11:20
ʵÔÚ²»ÏëÐÁÐÁ¿à¿à´òÁË°ëÌìµÄ×Ö£¬È´±ä³É ¡±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} ...

13Â¥: Originally posted by hank612 at 2016-05-30 13:11:20
ʵÔÚ²»ÏëÐÁÐÁ¿à¿à´òÁË°ëÌìµÄ×Ö£¬È´±ä³É ¡±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} ...

2Â¥: Originally posted by 001gqs at 2016-05-06 20:15:40
³ýÁ˸´±äÀïÃæµÄWeierstrass factorization Thm, ÎÒ±ðÎÞËû·¨¡£

3Â¥: Originally posted by maxman at 2016-05-07 10:51:37
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7Â¥: Originally posted by hank612 at 2016-05-30 08:22:21
ÓÉ Wallis ÎÞÏ޳˻ý \prod_{k=1}^{\infty}\frac{4k^2}{4k^2-1}=\frac{\pi}{2}  £¨Ò²¿ÉÒÔÓÉ OutÖÐ x=\frac{1}{2} Ò»ÑÛ¿´³ö£©

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

7Â¥: Originally posted by hank612 at 2016-05-30 08:22:21
ÓÉ Wallis ÎÞÏ޳˻ý \prod_{k=1}^{\infty}\frac{4k^2}{4k^2-1}=\frac{\pi}{2}  £¨Ò²¿ÉÒÔÓÉ OutÖÐ x=\frac{1}{2} Ò»ÑÛ¿´³ö£©

µÃ³ö
\prod_{k=1}^{\infty}(1-\frac{x}{4k^2-1})\prod_{k=1}^{\infty}\frac{4k^2-1}{ ...

8Â¥: Originally posted by gold2007 at 2016-05-30 08:32:52
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