I encountered an interesting identity when doing physics homework, that is, $$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
I encountered an interesting identity when doing physics homework, that is, $$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
From my answer here: Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$
Convert this to the problem of finding
$$\sum_{n=1}^N \frac{1}{1 - \cos \frac{2\pi n}{N}}$$
Convert this to the problem of getting a Chebyshev polynomial of which the roots are $\cos \frac{2\pi n}{N}$ (perhaps after using $\cos \frac{2 \pi (N-n)}{N} = \cos \frac{2 \pi n}{N}$ and getting a polynomial for a subset)
And using the fact the for any polynomial $P(x)$ with roots $r_i$ we have that
$$ \sum_{j=1}^{n} \frac{1}{x - r_j} = \frac{P'(x)}{P(x)}$$
Another (very similar approach) would be to start with
$$\frac{2}{\sin^2 x} = \frac{1}{1+\cos x} + \frac{1}{1-\cos x}$$ Note: I haven't worked out the details, but I am pretty sure this would work.
Let's denote $ S_N=\sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } $
Consider an equality:
$$\frac{\pi^2}{\sin^2(\pi s)}=\int_0^{\infty}\frac{x^{s-1}}{1-x}\ln\frac{1}{x}dx;0
Because of $0<\frac{n}{N}<1$, the integral form applies. Thus:
$$S_N=\frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\sum_{n=1}^{N-1}x^{\frac{n}{N}-1}dx=$$
$$= \frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{x^{\frac{1}{N}-1}-1}{1-x^{\frac{1}{N}}}dx=$$
$$=\frac{N^2}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx=$$
$$=2\frac{N^2}{{\pi}^2} \int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{3} $$
because of $$\int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{N^2}\frac{{\pi}^2}{6}$$