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}<z_{-m+1}<...<z_{-1}<z_0<z_1<...<z_{m-1}<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 Product

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}}

One Finite Sum