$\sin{x}$ 的无穷乘积式

观察以下两个式子:

\begin{matrix} \begin{aligned} \sin 3 x = 3 \sin x - 4 \sin^{3} x \\ \sin 5 x = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x \\ \dots \end{aligned} \end{matrix}

可以发现

\sin (2 n + 1) x = \sin x \cdot P (\sin^{2} x)

$\mathrm{P}(x)$ $x$ $n$ 次多项式.

$\lim_{x\to0}\dfrac{\sin{(2n+1)x}}{\sin{x}}=2n+1$ $\mathrm{P}(x)$ $2n+1$ .

$\sin{(2n+1)x}$ $\dfrac{k\pi}{2n+1},k\in\mathbb{Z}$ $\sin^2{\dfrac{k\pi}{2n+1}},k=1,2,\cdots,n$ $\mathrm{P}(x)$ $n$ 个根.

\begin{aligned} P (x) = (2 n + 1) (1 - \frac{x}{\sin^{2} \frac{π}{2 n + 1}}) (1 - \frac{x}{\sin^{2} \frac{2 π}{2 n + 1}}) \dots (1 - \frac{x}{\sin^{2} \frac{n π}{2 n + 1}}) \\ P (x) = (2 n + 1) \prod_{k = 1}^{n} (1 - \frac{x}{\sin^{2} \frac{k π}{2 n + 1}}) \\ \frac{\sin (2 n + 1) x}{\sin x} = (2 n + 1) \prod_{k = 1}^{n} (1 - \frac{\sin^{2} x}{\sin^{2} \frac{k π}{2 n + 1}}) \\ \frac{\sin x}{(2 n + 1) \sin \frac{1}{2 n + 1} x} = \prod_{k = 1}^{n} (1 - \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}}) \\ \frac{\sin x}{(2 n + 1) \sin \frac{1}{2 n + 1} x \prod_{k = 1}^{m} (1 - \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}})} = \prod_{k = m + 1}^{n} (1 - \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}}) \end{aligned}

$n$ $n\to\infty$ .

\frac{\sin x}{x \prod_{k = 1}^{m} (1 - \frac{x^{2}}{k^{2} π^{2}})}

对于右边，有这一不等式

\begin{matrix} \begin{aligned} \frac{2}{π} x < \sin x < x, x \in (0, \frac{π}{2}) \\ \sin^{2} \frac{1}{2 n + 1} x < \frac{x^{2}}{(2 n + 1)^{2}} \\ \sin^{2} \frac{k π}{2 n + 1} > \frac{4 k^{2}}{(2 n + 1)^{2}} \\ \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}} < \frac{x^{2}}{4 k^{2}} \end{aligned} \end{matrix}

于是就有

\begin{aligned} 1 > \prod_{k = m + 1}^{n} (1 - \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}}) > \prod_{k = m + 1}^{n} (1 - \frac{x^{2}}{4 k^{2}}) > \prod_{k = m + 1}^{\infty} (1 - \frac{x^{2}}{4 k^{2}}) \\ 1 > \prod_{k = m + 1}^{n} (1 - \frac{\sin^{2} \frac{1}{2 n + 1} x}{\sin^{2} \frac{k π}{2 n + 1}}) > \prod_{k = m + 1}^{\infty} (1 - \frac{x^{2}}{4 k^{2}}) \end{aligned}

$n$ $n\to\infty$ 的极限可得

1 > \frac{\sin x}{x \prod_{k = 1}^{m} (1 - \frac{x^{2}}{k^{2} π^{2}})} > \prod_{k = m + 1}^{\infty} (1 - \frac{x^{2}}{4 k^{2}})

$m$ $m\to\infty$ $1$ ，所以根据极限夹逼准则，可得

\frac{\sin x}{x} = \prod_{k = 1}^{\infty} (1 - \frac{x^{2}}{k^{2} π^{2}})

\sin x = x \prod_{k = 1}^{\infty} (1 - \frac{x^{2}}{k^{2} π^{2}})

$\cfrac{\sin{x}}{x }$ $\mathrm{sinc}\,x$ .

sin⁡x\sin{x}的无穷乘积式