沃利斯乘积

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數學家約翰·沃利斯在1655年寫下了今日有名的沃利斯乘積

 
\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}.

當時證明[编辑]

今日多數的微積分教科書透過比較 \int_0^\pi \sin^nxdxn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。

\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),

x = π/2時


\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
\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}

嚴謹證明[编辑]

先考慮不定積分 \int \sin^nxdx

 \int \sin^nxdx

=- \int \sin^{n-1}xd\ cosx

=- \ cosx \sin^{n-1}x + \int \ cosxd\sin^{n-1}x

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

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

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

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

 \int_0^\frac{\pi}{2} \sin^nxdx = \frac{n-1}{n}\int_0^\frac{\pi}{2} \sin^{n-2}xdx

對整數m

 \int_0^\frac{\pi}{2} \sin^{2m}xdx

= \frac{2m-1}{2m}\int_0^\frac{\pi}{2} \sin^{2m-2}xdx

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

= ...

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

= \frac{2m-1}{2m}\frac{2m-3}{2m-2}...\frac{1}{2}\frac{\pi}{2}

另一方面

 \int_0^\frac{\pi}{2} \sin^{2m+1}xdx

= \frac{2m}{2m+1}\int_0^\frac{\pi}{2} \sin^{2m-1}xdx

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

= ...

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

= \frac{2m}{2m+1}\frac{2m-2}{2m-1}...\frac{2}{3}

兩式相除得

 \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}}

\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 }

又因為

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}


夾擠定理

\lim_{m \to \infty}1 = \lim_{m \to \infty} \frac{2m+1}{2m} = 1

 
\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

尋找 ζ(2)[编辑]

我們可將上述的正弦乘積式化為泰勒级数

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