三角恒等式

ZhuYue286

问问下面的等式如何证明:
\[\sum_{n=1}^{k-1}{\csc \frac{n\pi}{2k}}\cdot \csc \frac{\left( n+1 \right) \pi}{2k}=\csc ^2\frac{\pi}{2k}\cos \frac{\pi}{2k}
\]

kuing

由于我习惯将 `n` 视为定的，`k` 为变的，所以先将 1# 的 `n` 和 `k` 对换一下，即：
\[\sum_{k=1}^{n-1}\csc\frac{k\pi}{2n}\csc\frac{(k+1)\pi}{2n}=\csc^2\frac\pi{2n}\cos\frac\pi{2n}.\]

为方便书写，记
\[\theta_k=\frac{k\pi}{2n},\]
则由积化和差，可知
\begin{align*}
\LHS&=\sum_{k=1}^{n-1}\frac2{\cos\theta_1-\cos\theta_{2k+1}}\\
&=\frac2{\cos\theta_1-\cos\theta_3}+\frac2{\cos\theta_1-\cos\theta_5}+\frac2{\cos\theta_1-\cos\theta_7}+\cdots+\frac2{\cos\theta_1-\cos\theta_{2n-1}}.
\end{align*}

由于当 `k\in\{1,3,5,\ldots,2n-1\}` 时 `\cos(n\theta_k)=0`，而根据切比雪夫多项式，`\cos(n\theta)` 可展开为 `\cos\theta` 的 `n` 次多项式，且最高次项系数为 `2^{n-1}`，因此有恒等式
\[\cos(n\theta)=2^{n-1}(\cos\theta-\cos\theta_1)(\cos\theta-\cos\theta_3)\cdots(\cos\theta-\cos\theta_{2n-1}),\]
取对数有
\[\ln\frac{\cos(n\theta)}{2^{n-1}}=\ln(\cos\theta-\cos\theta_1)+\ln(\cos\theta-\cos\theta_3)+\cdots+\ln(\cos\theta-\cos\theta_{2n-1}),\]
对 `\theta` 求导得
\[-n\tan(n\theta)=\frac{-\sin\theta}{\cos\theta-\cos\theta_1}+\frac{-\sin\theta}{\cos\theta-\cos\theta_3}+\cdots+\frac{-\sin\theta}{\cos\theta-\cos\theta_{2n-1}},\]
右边第一项移到左边，整理得
\[\frac{n\tan(n\theta)}{\sin\theta}-\frac1{\cos\theta-\cos\theta_1}=\frac1{\cos\theta-\cos\theta_3}+\frac1{\cos\theta-\cos\theta_5}+\cdots+\frac1{\cos\theta-\cos\theta_{2n-1}},\]
只需计算出当 `\theta\to\theta_1` 时左边的极限值即可。

由泰勒展开式以及 `n\theta_1=\pi/2`，有
\begin{align*}
\LHS={}&\frac{n\bigl(1-\frac12n^2(\theta-\theta_1)^2+o(\theta-\theta_1)^2\bigr)}{\bigl(\sin\theta_1+\cos\theta_1\cdot(\theta-\theta_1)+o(\theta-\theta_1)\bigr)\bigl(-n(\theta-\theta_1)+o(\theta-\theta_1)^2\bigr)}\\
&-\frac1{-\sin\theta_1\cdot(\theta-\theta_1)-\frac12\cos\theta_1\cdot(\theta-\theta_1)^2+o(\theta-\theta_1)^2}\\
={}&\frac{-\frac n2\cos\theta_1\cdot(\theta-\theta_1)^2+o(\theta-\theta_1)^2}{-n\sin^2\theta_1\cdot(\theta-\theta_1)^2+o(\theta-\theta_1)^2},
\end{align*}
即得
\[\lim_{\theta\to\theta_1}\LHS=\frac{\cos\theta_1}{2\sin^2\theta_1}=\frac12\csc^2\frac\pi{2n}\cos\frac\pi{2n},\]
所以
\[\frac2{\cos\theta_1-\cos\theta_3}+\frac2{\cos\theta_1-\cos\theta_5}+\frac2{\cos\theta_1-\cos\theta_7}+\cdots+\frac2{\cos\theta_1-\cos\theta_{2n-1}}=\csc^2\frac\pi{2n}\cos\frac\pi{2n},\]
即得证。

kuing

按上述方法，可以推个一般的公式：

若 `f(x)=A(x-x_1)(x-x_2)\cdots(x-x_n)`，则
\[f'(x)=\left(\frac1{x-x_1}+\frac1{x-x_2}+\cdots+\frac1{x-x_n}\right)f(x),\]
则
\[\frac{f'(x)}{f(x)}-\frac1{x-x_1}=\frac1{x-x_2}+\frac1{x-x_3}+\cdots+\frac1{x-x_n},\]
由泰勒有
\begin{align*}
\LHS&=\frac{f'(x_1)+f''(x_1)(x-x_1)+o(x-x_1)}{f'(x_1)(x-x_1)+\frac12f''(x_1)(x-x_1)^2+o(x-x_1)^2}-\frac1{x-x_1}\\
&=\frac{\frac12f''(x_1)(x-x_1)^2+o(x-x_1)^2}{f'(x_1)(x-x_1)^2+o(x-x_1)^2},
\end{align*}
得到
\[\lim_{x\to x_1}\LHS=\frac{f''(x_1)}{2f'(x_1)},\]
所以
\[\bbox[#CFF,15px]{
\frac1{x_1-x_2}+\frac1{x_1-x_3}+\frac1{x_1-x_4}+\cdots+\frac1{x_1-x_n}=\frac{f''(x_1)}{2f'(x_1)}.
}\]

kuing

kuing 发表于 2024-4-25 15:26
按上述方法，可以推个一般的公式：

若 `f(x)=A(x-x_1)(x-x_2)\cdots(x-x_n)`，则

哦，其实完全不需要泰勒，我太笨了，设
\begin{align*}
f(x)&=A(x-x_1)(x-x_2)\cdots(x-x_n),\\
g(x)&=A(x-x_2)(x-x_3)\cdots(x-x_n),
\end{align*}
则
\begin{align*}
f'(x)&=\bigl((x-x_1)g(x)\bigr)'=g(x)+(x-x_1)g'(x),\\
f''(x)&=2g'(x)+(x-x_1)g''(x),
\end{align*}
得到
\begin{align*}
f'(x_1)&=g(x_1),\\
f''(x_1)&=2g'(x_1),
\end{align*}
所以
\[\frac1{x_1-x_2}+\frac1{x_1-x_3}+\frac1{x_1-x_4}+\cdots+\frac1{x_1-x_n}=\frac{g'(x_1)}{g(x_1)}=\frac{f''(x_1)}{2f'(x_1)}.\]

再进一步，多复合一个函数，也可以：若
\begin{align*}
f(x)&=A\bigl(\varphi(x)-x_1\bigr)\bigl(\varphi(x)-x_2\bigr)\cdots\bigl(\varphi(x)-x_n\bigr),\\
g(x)&=A\bigl(\varphi(x)-x_2\bigr)\bigl(\varphi(x)-x_3\bigr)\cdots\bigl(\varphi(x)-x_n\bigr),
\end{align*}
则
\begin{align*}
f'(x)&=\Bigl(\bigl(\varphi(x)-x_1\bigr)g(x)\Bigr)'=\varphi'(x)g(x)+\bigl(\varphi(x)-x_1\bigr)g'(x),\\
f''(x)&=\varphi''(x)g(x)+2\varphi'(x)g'(x)+\bigl(\varphi(x)-x_1\bigr)g''(x),
\end{align*}
设 `t` 满足 `\varphi(t)=x_1`，则
\begin{align*}
f'(t)&=\varphi'(t)g(t),\\
f''(t)&=\varphi''(t)g(t)+2\varphi'(t)g'(t),
\end{align*}
解得
\begin{align*}
g(t)&=\frac{f'(t)}{\varphi'(t)},\\
g'(t)&=\frac{f''(t)\varphi'(t)-\varphi''(t)f'(t)}{2\bigl(\varphi'(t)\bigr)^2}=\frac12\left.\left(\frac{f'(x)}{\varphi'(x)}\right)'\right|_{x=t},
\end{align*}
而
\[\frac{g'(x)}{g(x)}=\varphi'(x)\left(\frac1{\varphi(x)-x_2}+\frac1{\varphi(x)-x_3}+\cdots+\frac1{\varphi(x)-x_n}\right),\]
所以
\[\frac1{x_1-x_2}+\frac1{x_1-x_3}+\frac1{x_1-x_4}+\cdots+\frac1{x_1-x_n}=\frac{g'(t)}{g(t)\varphi'(t)},\]
代入上面的结果即得如下更一般的公式
\[\bbox[#CFF,15px]{
\frac1{x_1-x_2}+\frac1{x_1-x_3}+\frac1{x_1-x_4}+\cdots+\frac1{x_1-x_n}=\frac1{2f'(t)}\left.\left(\frac{f'(x)}{\varphi'(x)}\right)'\right|_{x=t}.
}\]

这样再回头看 2# 的恒等式
\[\cos(n\theta)=2^{n-1}(\cos\theta-\cos\theta_1)(\cos\theta-\cos\theta_3)\cdots(\cos\theta-\cos\theta_{2n-1}),\]
即 `f(x)=\cos(nx)`, `\varphi(x)=\cos x`, `t=\frac\pi{2n}`，有
\begin{align*}
\frac1{2f'(t)}\left.\left(\frac{f'(x)}{\varphi'(x)}\right)'\right|_{x=t}&=\frac1{-2n}\left.\left(\frac{n\sin(nx)}{\sin x}\right)'\right|_{x=t}\\
&=\frac1{-2}\cdot\frac{n\cos(nt)\sin t-\sin(nt)\cos t}{\sin^2t}\\
&=\frac{\cos t}{2\sin^2t},
\end{align*}
和 2# 的结果相同。