查看“沃利斯乘积”的源代码

數學家[[約翰·沃利斯]]在1655年寫下了今日有名的'''沃利斯乘積'''。
:<math> 
\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}.
</math>

== 當時證明 ==
今日多數的微積分教科書透過比較<math> \int_0^\pi \sin^nxdx</math>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[[欧拉公式]]中的正弦展開式得到此乘積是必然的結果。
:<math>\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),</math>
在''x'' = π/2時
:<math>
\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 
</math>
:<math>\begin{align}
\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{align}
</math>

== 嚴謹證明 ==

先考慮不定積分<math> \int \sin^nxdx </math>有

<math> \int \sin^nxdx</math> 

<math>=- \int \sin^{n-1}xd\ cosx </math>

<math>=- \ cosx \sin^{n-1}x + \int \ cosxd\sin^{n-1}x </math>

<math>=- \ cosx \sin^{n-1}x + \int (n-1)\sin^{n-2}x\ cos^2 xdx</math>

<math>=- \ cosx \sin^{n-1}x + (n-1)\int \sin^{n-2}x(1 - \ sin^2 x)dx</math>

<math>=- \ cosx \sin^{n-1}x + (n-1)\int \sin^{n-2}xdx - (n-1)\int \sin^n x dx</math>

故

<math> \int \sin^nxdx =-\frac{1}{n} \ cosx \sin^{n-1}x + \frac{n-1}{n}\int \sin^{n-2}xdx</math>

<math> \int_0^\frac{\pi}{2} \sin^nxdx = \frac{n-1}{n}\int_0^\frac{\pi}{2} \sin^{n-2}xdx</math>

對整數m

<math> \int_0^\frac{\pi}{2} \sin^{2m}xdx </math>

<math>= \frac{2m-1}{2m}\int_0^\frac{\pi}{2} \sin^{2m-2}xdx</math>

<math>= \frac{2m-1}{2m}\frac{2m-3}{2m-2}\int_0^\frac{\pi}{2} \sin^{2m-4}xdx</math>

<math>= ...</math>

<math>= \frac{2m-1}{2m}\frac{2m-3}{2m-2}...\frac{1}{2}\int_0^\frac{\pi}{2} \sin^{0}xdx</math>

<math>= \frac{2m-1}{2m}\frac{2m-3}{2m-2}...\frac{1}{2}\frac{\pi}{2}</math>

另一方面

<math> \int_0^\frac{\pi}{2} \sin^{2m+1}xdx </math>

<math>= \frac{2m}{2m+1}\int_0^\frac{\pi}{2} \sin^{2m-1}xdx</math>

<math>= \frac{2m}{2m+1}\frac{2m-2}{2m-1}\int_0^\frac{\pi}{2} \sin^{2m-3}xdx</math>

<math>= ...</math>

<math>= \frac{2m}{2m+1}\frac{2m-2}{2m-1}...\frac{2}{3}\int_0^\frac{\pi}{2} \sin xdx</math>

<math>= \frac{2m}{2m+1}\frac{2m-2}{2m-1}...\frac{2}{3}</math>

兩式相除得

<math> \frac{\int_0^\frac{\pi}{2} \sin^{2m}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx} = \frac{\frac{2m-1}{2m}\frac{2m-3}{2m-2}...\frac{1}{2}\frac{\pi}{2}}{ \frac{2m}{2m+1}\frac{2m-2}{2m-1}...\frac{2}{3}}</math>

故

<math>\frac{\pi}{2} = \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5}... \frac{2m}{2m-1}\frac{2m}{2m+1} \frac{\int_0^\frac{\pi}{2} \sin^{2m}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx} =\frac{\int_0^\frac{\pi}{2} \sin^{2m}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx} \prod_{n=1}^{m} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1 }</math>

又因為

<math>1 = \frac{\int_0^\frac{\pi}{2} \sin^{2m+1}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx} >\frac{\int_0^\frac{\pi}{2} \sin^{2m}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx}> \frac{\int_0^\frac{\pi}{2} \sin^{2m-1}xdx}{ \int_0^\frac{\pi}{2} \sin^{2m+1}xdx} = \frac{2m+1}{2m}</math>


由[[夾擠定理]]知

<math>\lim_{m \to \infty}1 = \lim_{m \to \infty} \frac{2m+1}{2m} = 1</math>

故

<math> 
\frac{\pi}{2} = \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 </math>

==尋找 &zeta;(2)==
我們可將上述的正弦乘積式化為[[泰勒级数]]：
:<math>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</math>

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