三角関数の無限乗積展開

三角関数の無限乗積展開

$$\sin z=z\displaystyle\prod_{n=1}^{\infty} \biggl(1-\displaystyle\frac{z^2}{n^2\pi^2 }\biggr)$$

$$\cos z=\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{(2n-1)^2 \pi^2}\biggr)$$

\(\sin z\)の展開に\(z=\displaystyle\frac{\pi}{2}\)を代入すると

\(1=\displaystyle\frac{\pi}{2}\displaystyle\prod_{n=1}^{\infty} \biggl(1-\displaystyle\frac{1}{4n^2}\biggr)=\displaystyle\frac{\pi}{2}\prod_{n=1}^{\infty} \displaystyle\frac{4n^2-1}{4n^2}\)　　より

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

これをウォリスの公式という。

導出

\(\sin z\)

\(f(z)\)が\(a_{1}, \cdots ,a_{n}\)で一位の零点を持つとき、\(\displaystyle\frac{f'(z)}{f(z)}\)は\(a_{1}, \cdots ,a_{n}\)で一位の極をもち、留数が\(1\)になる。このとき、\(\displaystyle\frac{f'(z)}{f(z)}\)の部分分数展開は

$$\displaystyle\frac{f'(z)}{f(z)}=\displaystyle\frac{f'(0)}{f(0)}+\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{z-a_{n}}+\displaystyle\frac{1}{a_{n}}\biggr)$$

積分すると

$$\log f(z)=\displaystyle\frac{f'(0)}{f(0)}z+\displaystyle\sum_{n=1}^{\infty} \biggl[\log\biggl(1-\displaystyle\frac{z}{a_{n}}\biggr)+\displaystyle\frac{z}{a_{n}}\biggr]+C$$

$$f(z)=A \mathrm{exp}\biggl(\displaystyle\frac{f'(0)}{f(0)} z\biggr) \displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{z}{a_{n}}\biggr)e^{\frac{z}{a_{n}}}$$

\(\displaystyle\frac{\sin z}{z}=\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{z}{n\pi}\biggr)e^{\frac{z}{n\pi}}\biggl(1+\displaystyle\frac{z}{n\pi}\biggr)e^{-\frac{z}{n\pi}}\)　※\(n\)は正負両方あるから２つある。

$$\sin z=z\displaystyle\prod_{n=1}^{\infty} \biggl(1-\displaystyle\frac{z^2}{n^2\pi^2 }\biggr)$$

\(\cos z\)

\(\cos z=\displaystyle\frac{\sin 2z}{2\sin z}=\displaystyle\frac{2z\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{n^2\pi^2}\biggr)}{2z\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{z^2}{n^2\pi^2}\biggr)}=\displaystyle\frac{2z\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{n^2\pi^2}\biggr)}{2z\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{(2n^2)\pi^2}\biggr)}\)

\(=\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{(2n-1)^2 \pi^2}\biggr)\)　　\(n\)が奇数の部分だけ残るから

$$\cos z=\displaystyle\prod_{n=1}^{\infty}\biggl(1-\displaystyle\frac{4z^2}{(2n-1)^2 \pi^2}\biggr)$$