# Prove that the sum of division of two squared sin function is 1

When I numerically compute the below summation it is always $$1$$. How can I prove this? $$N$$ is an integer number and $$k$$ is an integer number between $$0$$ to $$N-1$$ and $$x$$ is real number between $$0$$ and $$0.5$$ $${\frac{1}{N^2}\sum_{l=0}^{N-1}\frac{\sin^2(\pi(l-k+x))}{\sin^2(\frac{\pi}{N}(l-k+x))}}$$

Many Thanks

• I didn't try tro prove it, but you may show that its derivative vanishes and then compute the expression for a given value of $x$. Mar 18 at 18:40

It is sufficient to handle the case $$k=0$$, i.e. to show that $$\sum_{n=0}^{N-1}\frac1{\sin^2\frac\pi N(n+x)}=\frac{N^2}{\sin^2\pi x},$$ since the general case follows if you put $$l=(n+k)\bmod N$$.

Recall that $$\cos Nt$$ is a polynomial in $$\cos t$$ (of degree $$N$$). Hence $$\sin^2 Nt-\sin^2\pi x=P_N(\sin^2 t)$$ with a polynomial $$P_N$$ of degree $$N$$ (whose coefficients may depend on $$x$$).

Let $$t_n=(n+x)\pi/N$$ for $$0\leqslant n. Then $$\sin^2 Nt_n=\sin^2\pi x$$, hence $$y_n=\sin^2 t_n$$ are roots of $$P_N(y)$$, and in fact different ones (because of $$0, and more generally $$2x\notin\mathbb{Z}$$). Thus $$\sin^2 Nt-\sin^2\pi x=A(x)\prod_{n=0}^{N-1}\left(\sin^2 t-\sin^2\frac\pi N(n+x)\right),$$ where $$A(x)$$ depends on $$x$$ only. Taking $$\partial/\partial t$$, we get $$\frac{N\sin 2Nt}{\sin^2 Nt-\sin^2\pi x}=\sum_{n=0}^{N-1}\frac{\sin 2t}{\sin^2 t-\sin^2\frac\pi N(n+x)}.$$

It remains to divide this by $$\sin 2t$$ and take $$t\to 0$$.

• brilliant solution! Mar 24 at 8:52