Chebyshev多项式的分解

hbghlyj

Table of integrals, series and products I S Gradshteĭn; I M Ryzhik; Alan Jeffrey; Daniel Zwillinger
1.39 The representation of cosines and sines of multiples of the angle as finite products
1.391

• $\displaystyle \sin n x=n \sin x \cos x \prod_{k=1}^{\frac{n-2}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{k \pi}{n}}\right)$    [$n$ is even]
• $\displaystyle\cos n x=\prod_{k=1}^{\frac{n}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{(2 k-1) \pi}{2 n}}\right)$    [$n$ is even]
• $\displaystyle\sin n x=n \sin x \prod_{k=1}^{\frac{n-1}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{k \pi}{n}}\right)$    [$n$ is odd]
• $\displaystyle\cos n x=\cos x \prod_{k=1}^{\frac{n-1}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{(2 k-1) \pi}{2 n}}\right)$    [$n$ is odd]
  

由Chebyshev method知

• ${\sin nx\over\cos x}=U_{n-1}(\sin x)$, 其中$U_{n-1}$为$n-1$次多项式
• $\cos nx=T_n(\cos x)$, 其中$T_n$是$n$次多项式

1的证明: 设$n$为偶数,
若$\sin(nx)=0$则$x=\frac{k \pi}{n},\;k=0,\pm1,\cdots,\pm\frac{n-2}2,\frac n2$
则$\sin x=0,\pm1,\pm\sin\frac{k \pi}{n},\;k=1,\cdots,\frac{n-2}2$
所以$U_{n-1}(x)$的根是$0,\pm\sin\frac{k \pi}{n},\;k=1,\cdots,\frac{n-2}2$
所以$\displaystyle U_{n-1}(x)=C \sin x \prod_{k=1}^{\frac{n-2}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{k \pi}{n}}\right)$, 其中$C$是常数
因为$\lim_{x\to0}\frac{\sin nx}{\sin x}=n$, 所以$C=n$. 这就证明了1.
同理可证2,3,4.

hbghlyj

最后一步证明$C=n$也可以这样做:
由递推公式$\begin{aligned}U_{0}(x)&=1\\U_{1}(x)&=2x\\U_{n+1}(x)&=2x\,U_{n}(x)-U_{n-1}(x)\end{aligned}$知 $U_{n-1}(x)$ 的首项系数是 $2^{n-1}$.
再使用$\displaystyle\prod_{k=1}^{n-2\over2}\sin^2\left(\frac{\pi k}{n}\right)=\frac{n}{2^{n-1}}$
得到$C=n$.

hbghlyj

见这个回答
\begin{align*}
\sin(nx)&=\frac{e^{inx}-e^{-inx}}{2i}&\\\\
&=\frac{e^{-inx}}{2i}\left(e^{i2nx}-1\right)\\\\
&=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x}-e^{-i2\pi k/n})\\\\
&=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}\left(\color{blue}{e^{ix}}\color{red}{e^{-i\pi k/n}}\color{green}{(2i)}\sin\left(x+\frac{k \pi}{n}\right)\right)\\\\
&=\frac{e^{-inx}}{2i} \color{blue}{e^{inx}} \color{red}{e^{-i(\pi/n)n(n-1)/2}}\color{green}{(2i)^n}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)\\\\
&=2^{n-1}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)
\end{align*}
对$k=\frac{n+2}2,\cdots,n-1$使用$\sin\left(x+\frac{k \pi}{n}\right)=\sin\left(\frac{(n-k) \pi}{n}-x\right)$得\[\sin n x=2^{n-1}\sin x \cos x \prod_{k=1}^{\frac{n-2}{2}}\sin\left(\frac{k \pi}n-x\right)\sin\left(x+\frac{k \pi}n\right)\]
使用 $\sin^2\alpha-\sin^2\beta=\sin (\alpha-\beta) \sin (\alpha+\beta)$ 得 $\displaystyle \sin n x=2^{n-1}\sin x \cos x \prod_{k=1}^{\frac{n-2}{2}}\left(\sin ^2 \frac{k \pi}n-\sin ^2 x\right)$
使用 $\displaystyle\prod_{k=1}^{n-2\over2}\sin^2\left(\frac{k \pi}{n}\right)=\frac n{2^{n-1}}$ 得 $\displaystyle \sin n x=n \sin x \cos x \prod_{k=1}^{\frac{n-2}{2}}\left(1-\frac{\sin ^2 x}{\sin ^2 \frac{k \pi}{n}}\right)$

hbghlyj

观察$\displaystyle \sin(nx)=2^{n-1}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)$
$x+\frac{k \pi}{n},k=0,\cdots,n-1$形成一个半圆.
如果是1/4圆乘积有没有闭形式呢? 见这个回答
$$\prod_{k=1}^{n-1}\sin\left(\frac{k \pi}{2n}\right)=2^{1-n}\sqrt{n}$$
好像论坛上见过这个公式 @kuing