不用留数证明周期函数可展开为cot

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计算函数$\displaystyle R(z):=\frac{e^{i z} \cot z \sin (a+2 z)}{\sin (b+z) \sin (c+z) \sin (d+z) \sin (e+z)}$的Partial fractions

令$\displaystyle R(z)=\frac{A}{\sin z}+\frac{B}{\sin (b+z)}+\frac{C}{\sin (c+z)}+\frac{D}{\sin (d+z)}+\frac{E}{\sin (e+z)}$
计算留数$A,B,C,D,E$：Partial Fraction Decompositions and Trigonometric Sum Identities
$\begin{aligned} & A=\lim _{z \rightarrow 0} z R(z)=\frac{\sin a}{\sin b \sin c \sin d \sin e}, \\ & B=\lim _{z \rightarrow-b}(b+z) R(z)=\frac{-e^{-b i} \cot b \sin (a-2 b)}{\sin (c-b) \sin (d-b) \sin (e-b)}, \\ & C=\lim _{z \rightarrow-c}(c+z) R(z)=\frac{-e^{-c i} \cot c \sin (a-2 c)}{\sin (b-c) \sin (d-c) \sin (e-c)}, \\ & D=\lim _{z \rightarrow-d}(d+z) R(z)=\frac{-e^{-d i} \cot d \sin (a-2 d)}{\sin (b-d) \sin (c-d) \sin (e-d)}, \\ & E=\lim _{z \rightarrow-e}(e+z) R(z)=\frac{-e^{-e i} \cot e \sin (a-2 e)}{\sin (b-e) \sin (c-e) \sin (d-e)} .\end{aligned}$
代入$\displaystyle R(z)=\frac{A}{\sin z}+\frac{B}{\sin (b+z)}+\frac{C}{\sin (c+z)}+\frac{D}{\sin (d+z)}+\frac{E}{\sin (e+z)}$
乘以 $\sin z$，然后令 $z = M i$ 且 $M → +∞$，我们得到三角和恒等式：\begin{aligned}
\frac{\sin a}{\sin b \sin c \sin d \sin e} & =\frac{\cot b \sin (a-2 b)}{\sin (c-b) \sin (d-b) \sin (e-b)} \\
& +\frac{\cot c \sin (a-2 c)}{\sin (b-c) \sin (d-c) \sin (e-c)} \\
& +\frac{\cot d \sin (a-2 d)}{\sin (b-d) \sin (c-d) \sin (e-d)} \\
& +\frac{\cot e \sin (a-2 e)}{\sin (b-e) \sin (c-e) \sin (d-e)} .
\end{aligned}

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1. a,b,c,d,e=symbols('a b c d e')
2. simplify(sin(a)/(sin(b)*sin(c)*sin(d)*sin(e))-cot(b)*sin(a-2*b)/(sin(c-b)*sin(d-b)*sin(e-b))-cot(c)*sin(a-2*c)/(sin(b-c)*sin(d-c)*sin(e-c))-cot(d)*sin(a-2*d)/(sin(b-d)*sin(c-d)*sin(e-d))-cot(e)*sin(a-2*e)/(sin(b-e)*sin(c-e)*sin(d-e)))

复制代码

ZCos666

似乎有
\[ \dfrac{\sin{\theta_0}}{\displaystyle\prod_{k=1}^{n}\sin{\theta_k}}=\sum_{i=1}^{n}\dfrac{\cot{\theta_i}\sin(\theta_0-(n-2)\theta_i)}{\displaystyle\prod_{j\ge1,j\ne i}^{n}\sin(\theta_j-\theta_i)} \]
转化成指数形式，即
\[ \dfrac{x_0-\dfrac{1}{x_0}}{\displaystyle\prod_{k=1}^{n}\left(x_k-\dfrac{1}{x_k}\right)}=\sum_{i=1}^{n}\dfrac{\dfrac{x_i^2+1}{x_i^2-1}\left(\dfrac{x_0}{x_i^{n-2}}-\dfrac{x_i^{n-2}}{x_0}\right)}{\displaystyle\prod_{j\ge1,j\ne i}^{n}\left(\dfrac{x_j}{x_i}-\dfrac{x_i}{x_j}\right)} \]

ZCos666

ZCos666 发表于 2025-2-20 23:45
似乎有
\[ \dfrac{\sin{\theta_0}}{\displaystyle\prod_{k=1}^{n}\sin{\theta_k}}=\sum_{i=1}^{n}\dfrac{\c ...

只须证
\[ \dfrac{2}{\displaystyle\prod_{k=1}^{n}\left(x_k-\dfrac{1}{x_k}\right)}=\sum_{i=1}^{n}\dfrac{\dfrac{x_i^2+1}{x_i^2-1}\cdot\dfrac{1}{x_i^{t}}}{\displaystyle\prod_{j\ge1,j\ne i}^{n}\left(\dfrac{x_j}{x_i}-\dfrac{x_i}{x_j}\right)} \]
提出分母的分母，可以约掉，即证
\[ \dfrac{2}{\displaystyle\prod_{k=1}^{n}(x_k^2-1)}=\sum_{i=1}^{n}\dfrac{\dfrac{x_i^2+1}{x_i^2-1}\cdot\dfrac{1}{x_i^t}}{\displaystyle\prod_{j\ge1,j\ne i}^{n}(x_j^2-x_i^2)} \]
整理一下式子，即证
\[ 2=\sum_{i=1}^{n}(x_i^2+1)\cdot x_i^{n-2-t}\prod_{j\ge1,j\ne i}^{n}\dfrac{1-x_j^2}{x_i^2-x_j^2} \]
设多项式$P(x)$满足$P(x^2)=(x^2+1)x^{n-2-t}$，则$P(1)=2$，右式即拉格朗日插值
那么恒等式在$t=n-2,n-4,\cdots,-(n-2)$时都成立

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一般的定理：
周期函数的余切展开式

Let $f(z)$ be a periodic function of $z$, analytic except at a certain number of simple poles; for convenience, let $\pi$ be the period of $f(z)$ so that $f(z)=f(z+\pi)$.

Let $z=x+i y$ and let $f(z) \rightarrow \ell$ uniformly with respect to $x$ as $y \rightarrow+\infty$, when $0 \leq x \leq \pi$; similarly let $f(z) \rightarrow \ell^{\prime}$ uniformly as $y \rightarrow-\infty$. Let the poles of $f(z)$ in the strip $0<x \leq \pi$ be at $a_1, a_2, \ldots, a_n$; and let the residues at them be $c_1, c_2, \ldots, c_n$. Further, let $A B C D$ be a rectangle whose corners are⁷ $-i \rho, \pi-i \rho, \pi+i \rho^{\prime}$ and $i \rho^{\prime}$ in order.

Consider $\frac{1}{2 \pi i} \int f(t) \cot (t-z) d t$ taken round this rectangle; the residue of the integrand at $a_r$ is $c_r \cot \left(a_r-z\right)$, and the residue at $z$ is $f(z)$. Also the integrals along $D A$ and $C B$ cancel on account of the periodicity of the integrand; and as $\rho \rightarrow \infty$, the integrand on $A B$ tends uniformly to $i \ell^{\prime}$, while as $\rho^{\prime} \rightarrow \infty$ the integrand on $C D$ tends uniformly to $-i \ell$; therefore
\[
\frac{1}{2}\left(\ell+\ell^{\prime}\right)=f(z)+\sum_{r=1}^n c_r \cot \left(a_r-z\right) .
\]
That is to say, we have the expansion
\[
f(z)=\frac{1}{2}\left(\ell+\ell^{\prime}\right)+\sum_{r=1}^n c_r \cot \left(z-a_r\right) .
\]
___________
⁷ If any of the poles are on $x=\pi$, shift the rectangle slightly to the right; $\rho, \rho^{\prime}$ are to be taken so large that $a_1, a_2, \ldots, a_n$ are inside the rectangle.

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E. T. Whittaker and G. N. Watson. A Course of Modern Analysis. Cambridge University Press (2021) page 141
原书的标题 7.7 The expansion of a class of periodic functions in a series of cotangents

Example 7.7.1见8#
Example 7.7.2 当$n=2$时是
\begin{multline*}\frac{\sin (x-b_1) \sin (x-b_2)}{\sin (x-a_1) \sin (x-a_2)}=\frac{\sin (a_1-b_1) \sin (a_1-b_2)}{\sin (a_1-a_2)}\cot (x-a_1)\\+\frac{\sin (a_2-b_1) \sin (a_2-b_2)}{\sin (a_2-a_1)}\cot (x-a_2)+\cos (a_1+a_2-b_1-b_2)\end{multline*}

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由这帖知$\lim_{y→\infty}\cot \left(x+i y\right)=-i$和$\lim_{y→-\infty}\cot \left(x+i y\right)=i$
所以$\ell=(-i)^n,\ell'=i^n,$
当$n$为偶数时,$\frac{(-i)^n+i^n}2=(-1)^{n/2}$
当$n$为奇数时,$\frac{(-i)^n+i^n}2=0$
这就解释了Example 7.7.1后面的常数.
$\cot(x-a_1)$在$a_1$是1阶奇点,留数为1,所以$\cot \left(x-a_{1}\right) \cot \left(x-a_{2}\right) \cdots \cot \left(x-a_{n}\right)$在$a_1$的留数为$\cot \left(a_1-a_2\right) \cot \left(a_1-a_3\right) \cdots \cot \left(a_1-a_{n}\right)$
这就解释了那些系数$c_i$

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$\lim_{y\to -\infty }\frac{\sin (x+iy-b_1)}{\sin (x+iy-a_1)}=e^{i(a_1-b_1)}$, 所以$\ell'=e^{-i(a_1+\cdots+a_n-b_1-\cdots-b_n)}$.
$\lim_{y\to \infty }\frac{\sin (x+iy-b_1)}{\sin (x+iy-a_1)}=e^{-i(a_1-b_1)}$, 所以$\ell=e^{i(a_1+\cdots+a_n-b_1-\cdots-b_n)}$.
这就解释了Example 7.7.2后面的常数.
$\frac1{\sin(x-a_1)}$在$a_1$是1阶奇点, 留数为1, 所以$\frac{\sin \left(x-b_{1}\right) \sin \left(x-b_{2}\right) \cdots \sin \left(x-b_{n}\right)}{\sin \left(x-a_{1}\right) \sin \left(x-a_{2}\right) \cdots \sin \left(x-a_{n}\right)}$在$a_1$的留数为$\frac{\sin \left(a_{1}-b_{1}\right) \cdots \sin \left(a_{1}-b_{n}\right)}{\sin \left(a_{1}-a_{2}\right) \cdots \sin \left(a_{1}-a_{n}\right)}$
这就解释了那些系数$c_i$
Ok

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Hermite's cotangent identity
Warren P. Johnson, "Trigonometric Identities à la Hermite", American Mathematical Monthly, volume 117, number 4, April 2010, pages 311–327
Lemma 1. 复数 $a_1$ 与 $a_2$ 之差不为 $\pi$ 的整数倍, 则
\[
\cot \left(z-a_1\right) \cot \left(z-a_2\right)=\cot \left(a_1-a_2\right) \cot \left(z-a_1\right)+\cot \left(a_2-a_1\right) \cot \left(z-a_2\right)-1 .
\]

Proof

If $u$ and $v$ do not differ by an integer multiple of $\pi$, then
\[
\cot (u-v)=\frac{\cos u \cos v+\sin u \sin v}{\sin u \cos v-\cos u \sin v} \times \frac{\csc u \csc v}{\csc u \csc v}=\frac{\cot u \cot v+1}{\cot v-\cot u} .
\]
Set $u=z-a_1$ and $v=z-a_2$ and rearrange, and Lemma 1 follows.

Theorem 1 (Hermite). 复数 $a_1, a_2, \ldots, a_n$, 每两数之差不为 $\pi$ 的整数倍. 令
\[
A_{n, k}=\prod_{\substack{1 \leq j \leq n \\ j \neq k}} \cot \left(a_k-a_j\right),
\]
其中, 空乘积 $A_{1,1}$ 等于1. 则
\[
\cot \left(z-a_1\right) \cot \left(z-a_2\right) \cdots \cot \left(z-a_n\right)=\cos \frac{n \pi}{2}+\sum_{k=1}^n A_{n, k} \cot \left(z-a_k\right)
\]

Proof

For $n=1$ Hermite's theorem is trivial, and the case $n=2$ is Lemma 1. Assume the theorem holds up to $n$. Using Lemma 1 in the second step, we have
\begin{aligned}
&\cot \left(z-a_1\right) \cot \left(z-a_2\right) \cdots \cot \left(z-a_n\right) \cot \left(z-a_{n+1}\right) \\
=& \cos \frac{n \pi}{2} \cot \left(z-a_{n+1}\right)+\sum_{k=1}^n A_{n, k} \cot \left(z-a_k\right) \cot \left(z-a_{n+1}\right) \\
=& \cos \frac{n \pi}{2} \cot \left(z-a_{n+1}\right) \\
&+\sum_{k=1}^n A_{n, k}\left\{\cot \left(a_{n+1}-a_k\right) \cot \left(z-a_{n+1}\right)+\cot \left(a_k-a_{n+1}\right) \cot \left(z-a_k\right)-1\right\} \\
=& \cot \left(z-a_{n+1}\right)\left\{\cos \frac{n \pi}{2}+\sum_{k=1}^n A_{n, k} \cot \left(a_{n+1}-a_k\right)\right\} \\
&+\sum_{k=1}^n A_{n, k} \cot \left(a_k-a_{n+1}\right) \cot \left(z-a_k\right)\\&-\sum_{k=1}^n A_{n, k} .
\end{aligned}By the induction hypothesis, the first of these three terms equals
\[
\cot \left(z-a_{n+1}\right) \cot \left(a_{n+1}-a_1\right) \cot \left(a_{n+1}-a_2\right) \cdots \cot \left(a_{n+1}-a_n\right),
\]
which is $A_{n+1, n+1} \cot \left(z-a_{n+1}\right)$. This combines nicely with the second term, since $A_{n, k} \cot \left(a_k-a_{n+1}\right)=A_{n+1, k}$ for $1 \leq k \leq n$, so now we have
\begin{align}
&\cot \left(z-a_1\right) \cot \left(z-a_2\right) \cdots \cot \left(z-a_n\right) \cot \left(z-a_{n+1}\right) \nonumber\\
&=\sum_{k=1}^{n+1} A_{n+1, k} \cot \left(z-a_k\right)-\sum_{k=1}^n A_{n, k}\label2
\end{align}
But the induction hypothesis also implies
\begin{align}
\sum_{k=1}^n A_{n, k} &=\sum_{k=1}^{n-1} A_{n, k}+\cot \left(a_n-a_1\right) \cdots \cot \left(a_n-a_{n-1}\right)\nonumber \\
&=\sum_{k=1}^{n-1} A_{n, k}+\cos \frac{(n-1) \pi}{2}+\sum_{k=1}^{n-1} A_{n-1, k} \cot \left(a_n-a_k\right) .\label3
\end{align}
If $k<n$ then $A_{n, k}=\cot \left(a_k-a_n\right) A_{n-1, k}$, so the sums on the right side of \eqref{3} cancel, leaving us with
\begin{equation}\label4
\sum_{k=1}^n A_{n, k}=\cos \frac{(n-1) \pi}{2}=\sin \frac{n \pi}{2}=-\cos \frac{(n+1) \pi}{2} .
\end{equation}
Substituting \eqref{4} in \eqref{2} we have Hermite's theorem with $n+1$ in place of $n$.

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Example 7.7.2也能不用留数证明吗?