沃利斯乘積,又稱沃利斯公式,由數學家約翰·沃利斯在1655年時發現。
![{\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)
當時證明[編輯]
今日多數的微積分教科書透過比較
在n是奇數或是偶數,甚至是接近無窮大的情況下,發現即使將n增加一就會發生不一樣的情形。在那時,微積分尚未存在,而且有關數學收斂的分析工具也還未俱全,所以完成這證明較現今有相當的難度。從現在來看,從歐拉公式中的正弦展開式得到此乘積是必然的結果。
![{\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時
![{\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)
嚴謹證明[編輯]
先考慮不定積分
有
故
對整數m
另一方面
兩式相除得
故
又因為
由夾擠定理知
故
尋找 ζ(2)[編輯]
我們可將上述的正弦乘積式化為泰勒級數:
![{\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)