沃利斯乘积

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。

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

也可以写为:$\frac{(2n)

## 当时证明

今日多数的微积分教科书透过比较$\int_0^\pi \sin^nxdx$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。

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

在x = π/2时

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

$$\begin{align} \frac{\pi}{2} &{}= \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) \\ &{}= \prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots \end{align}$$

## 严谨证明

先考虑不定积分$\int \sin^nxdx$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

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

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

对整数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

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

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

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

$$= ...$$

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

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

两式相除得

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

故

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

又因为

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

由夹挤定理知

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

故

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

## 寻找 ζ(2)

我们可将上述的正弦乘积式化为泰勒级数：

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