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

More Elementary Proof for Euler’s Sine Expansion

Our goal in this post is to provide an elementary proof for the real version of Euler’s famous Sine expansion.

I first prove that for any n that is a power of 2 and for any x\in\mathbb R,

\displaystyle{\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~(1)

by using the following more familiar identity

\displaystyle\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2

we can easily arrive to the following identity that is valid for n equal to any power of 2,

                    \displaystyle\sin(x)=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n

\displaystyle=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n

  \displaystyle=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)~~~~~~~~~~(2)

By considering what happens as x\to0 in the recent formula, we can obtain that for n a power of 2

\displaystyle n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n~~~~~~~~~~(3)

and then I replace (3) in (2) and prove (1).

Now I choose m<\frac n 2-1 and break up p_n(x) as follows

\displaystyle p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)~~~~~~~~~~(4)

Since for 0<\theta<\frac{\pi}2, \sin(\theta)>\frac2{\pi}\theta, we have

\displaystyle\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}

and if I choose m and n large enough such that \frac{x^2}{4m^2}<1, then

\displaystyle0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1

which implies that

\displaystyle0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1

Therefore from (4) we have

\displaystyle p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

Also we have

\displaystyle{\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m}

and once again by (4), we get

\displaystyle p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

To summarize, I have shown that

\displaystyle{(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~~~~~~(5)

Now taking n\to\infty in (5) I arrive to

\displaystyle(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)

After rearrangement and then taking absolute values, we get

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|

but since we have

\displaystyle{\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}}

I can write

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}

Finally since \lim_{m\to\infty}s_m=0, this inequality implies our goal, indeed

\boxed{\displaystyle\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x}

More Elementary Proof for Euler’s Sine Expansion