# One Finite Product

In this post, I wish to find the value of some finite product. I begin by defining a function $F_n(z)$ as follows and trying to find it’s roots

$\displaystyle F_n(z)=\frac1{2i}\left((1+\frac{iz}n)^n-(1-\frac{iz}n)^n\right)=$ $\displaystyle\frac{(-1)^m}{n^n}$ $\displaystyle z^n+\cdots+$ $\displaystyle1$ $\displaystyle z$

Let $z=n\tan(\theta)$, then we have

$\displaystyle{1+\frac{iz}n=1+i\tan\theta=1+i\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}(\cos\theta+i\sin\theta)=\sec\theta e^{i\theta}}$

and similarly $1-\frac{iz}n=\sec\theta e^{-i\theta}$. So we have

$\displaystyle F_n(n\tan\theta)=\frac1{2i}\sec^n\theta(e^{in\theta}-e^{-in\theta})=\sec^n\theta\sin n\theta$

The sine function vanishes at integer multiples of $\pi$, so it follows that $F_n(n\tan\theta)=0$ where $n\theta=k\pi$ for all integers $k$, that is, for $\theta=k\pi/n$ for all $k\in\mathbb Z$. Thus, $F_n(z_k)=0$ for

$\displaystyle z_k=n\tan\frac{k\pi}n=(2m+1)\tan\frac{k\pi}{2m+1}$

where we recall that $n=2m+1$. Since $\tan\theta$ is strictly increasing on the interval $(-\pi/2,\pi/2)$, it follows that

$\displaystyle{z_{-m}

moreover, since tangent is an odd function, we have $z_{-k}=-z_k$ for each $k$. In particular we have found $2m+1=n$ distinct roots of $F_n(z)$, so as a consequence of the fundamental theorem of algebra, we can write

$\displaystyle F_n(z)=$ $\displaystyle\frac{(-1)^m}{n^n}$ $\displaystyle{z\prod_{k=1}^m(z^2-z_k^2)=\frac{(-1)^m}{n^n}z\prod_{k=1}^m(z^2-n^2\tan^2\frac{k\pi}n)}$

$\displaystyle=\frac{(-1)^m}{n^n}\prod_{k=1}^m(-n^2\tan^2\frac{k\pi}n)z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)$

$\displaystyle=$ $\displaystyle\frac1 n\prod_{k=1}^m\tan^2\frac{k\pi}{n}$ $\displaystyle z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)$

So we have $\frac1{2m+1}\prod_{k=1}^m\tan^2\frac{k\pi}{2m+1}=1$, and

$\boxed{\displaystyle\prod_{k=1}^m\tan\frac{k\pi}{2m+1}=\sqrt{2m+1}}$

# One Finite Sum

In this post I want to find the value of the following finite sum,

$\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}$

I use the following amazing identity that you can see it’s proof in my previous posts

$\displaystyle\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}$

By this identity, I can write

$\displaystyle\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}$

$\displaystyle=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}$

$\displaystyle=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}$

$\displaystyle=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}$

and this follows that

$\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}$

Hence,

$\boxed{\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}}$