沃利斯乘积

沃利斯乘積，又称沃利斯公式，由數學家約翰·沃利斯在1655年时发现。

\prod_{n = 1}^{\infty} \frac{2 n}{2 n - 1} \cdot \frac{2 n}{2 n + 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} \dots = \frac{π}{2} .

也可以写为: $\frac{(2 n)!!}{(2 n - 1)!!} \sim \sqrt{π n} (n \to \infty)$

當時證明

今日多數的微積分教科書透過比較 $\int_{0}^{π} \sin^{n} x d x$ 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。

\frac{\sin (x)}{x} = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) (1 - \frac{x^{2}}{9 π^{2}}) \dots = \prod_{n = 1}^{\infty} (1 - \frac{x^{2}}{n^{2} π^{2}}),

在x = π/2時

\frac{2}{π} = \prod_{n = 1}^{\infty} (1 - \frac{1}{4 n^{2}}) = (1 - \frac{1}{2^{2}}) (1 - \frac{1}{2^{2} \cdot 4}) (1 - \frac{1}{2^{2} \cdot 9}) \dots

\begin{matrix} \frac{π}{2} & = \prod_{n = 1}^{\infty} (\frac{4 n^{2}}{4 n^{2} - 1}) \\ = \prod_{n = 1}^{\infty} \frac{(2 n) (2 n)}{(2 n - 1) (2 n + 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} \dots \end{matrix}

嚴謹證明

先考慮不定積分 $\int \sin^{n} x d x$ 有

$\int \sin^{n} x d x$

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

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

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

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

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

故

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

$\int_{0}^{\frac{π}{2}} \sin^{n} x d x = \frac{n - 1}{n} \int_{0}^{\frac{π}{2}} \sin^{n - 2} x d x$

對整數m

$\int_{0}^{\frac{π}{2}} \sin^{2 m} x d x$

$= \frac{2 m - 1}{2 m} \int_{0}^{\frac{π}{2}} \sin^{2 m - 2} x d x$

$= \frac{2 m - 1}{2 m} \frac{2 m - 3}{2 m - 2} \int_{0}^{\frac{π}{2}} \sin^{2 m - 4} x d x$

$= . . .$

$= \frac{2 m - 1}{2 m} \frac{2 m - 3}{2 m - 2} . . . \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin^{0} x d x$

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

另一方面

$\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x$

$= \frac{2 m}{2 m + 1} \int_{0}^{\frac{π}{2}} \sin^{2 m - 1} x d x$

$= \frac{2 m}{2 m + 1} \frac{2 m - 2}{2 m - 1} \int_{0}^{\frac{π}{2}} \sin^{2 m - 3} x d x$

$= . . .$

$= \frac{2 m}{2 m + 1} \frac{2 m - 2}{2 m - 1} . . . \frac{2}{3} \int_{0}^{\frac{π}{2}} \sin x d x$

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

兩式相除得

$\frac{\int_{0}^{\frac{π}{2}} \sin^{2 m} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} = \frac{\frac{2 m - 1}{2 m} \frac{2 m - 3}{2 m - 2} . . . \frac{1}{2} \frac{π}{2}}{\frac{2 m}{2 m + 1} \frac{2 m - 2}{2 m - 1} . . . \frac{2}{3}}$

故

$\frac{π}{2} = \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5} . . . \frac{2 m}{2 m - 1} \frac{2 m}{2 m + 1} \frac{\int_{0}^{\frac{π}{2}} \sin^{2 m} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} = \frac{\int_{0}^{\frac{π}{2}} \sin^{2 m} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} \prod_{n = 1}^{m} \frac{2 n}{2 n - 1} \cdot \frac{2 n}{2 n + 1}$

又因為

$1 = \frac{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} < \frac{\int_{0}^{\frac{π}{2}} \sin^{2 m} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} < \frac{\int_{0}^{\frac{π}{2}} \sin^{2 m - 1} x d x}{\int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x d x} = \frac{2 m + 1}{2 m}$

由夾擠定理知

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

故

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

尋找 ζ(2)

我們可將上述的正弦乘積式化為泰勒级数：

x (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) (1 - \frac{x^{2}}{9 π^{2}}) \dots = x - \frac{1}{3!} x^{3} + \frac{1}{5!} x^{5} - \dots

沃利斯乘积

當時證明

嚴謹證明

尋找 ζ(2)

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