沃利斯乘积,又称沃利斯公式,由数学家约翰·沃利斯在1655年时发现。
![{\displaystyle \prod _{n=1}^{\infty }{\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fb6f531dc241b623a91633957d60671f0e9adbf)
当时证明[编辑]
今日多数的微积分教科书透过比较
在n是奇数或是偶数,甚至是接近无穷大的情况下,发现即使将n增加一就会发生不一样的情形。在那时,微积分尚未存在,而且有关数学收敛的分析工具也还未俱全,所以完成这证明较现今有相当的难度。从现在来看,从欧拉公式中的正弦展开式得到此乘积是必然的结果。
![{\displaystyle {\frac {\sin(x)}{x}}=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ce517726457c46aa0bdc7b7390c655780fdc950)
在x = π/2时
![{\displaystyle {\frac {2}{\pi }}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)=\left(1-{\frac {1}{2^{2}}}\right)\left(1-{\frac {1}{2^{2}\cdot 4}}\right)\left(1-{\frac {1}{2^{2}\cdot 9}}\right)\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ac1e2ce6d5224338f480cc94f694783f9e3913e)
![{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&{}=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\&{}=\prod _{n=1}^{\infty }{\frac {(2n)(2n)}{(2n-1)(2n+1)}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f57febd9e773988f808ef637c81a94484ec5a9c0)
严谨证明[编辑]
先考虑不定积分
有
故
对整数m
另一方面
两式相除得
故
又因为
由夹挤定理知
故
寻找 ζ(2)[编辑]
我们可将上述的正弦乘积式化为泰勒级数:
![{\displaystyle x\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots =x-{\frac {1}{3!}}x^{3}+{\frac {1}{5!}}x^{5}-\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f323e69d8a67c18f069c064cfd4f491462e2c18)