沃利斯乘積

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。

$$\prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}.$$

也可以寫為:$\frac{(2n)

## 當時證明

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

$$\frac{\sin(x)}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots = \prod_{n = 1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right),$$

在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

先考慮不定積分$\int \sin^nxdx$有

$$\int \sin^nxdx$$

$$=- \int \sin^{n-1}xd\cos x$$

$$=- \cos x \sin^{n-1}x + \int \cos xd\sin^{n-1}x$$

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

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

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

故

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

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

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

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

故

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

又因為

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

由夾擠定理知

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

故

$$\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots$$

## 尋找 ζ(2)

我們可將上述的正弦乘積式化為泰勒級數：

$$x\left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots$$