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While trying to prove some identities for generating functions, I ended up needing to show that

$$\sum_{p=1}^n \exp\left(\frac{i\pi p l}{2m}\right)\prod_{\substack{k=1\\k\neq p}}^n\frac{1}{\sin\left(\frac{\pi (k-p)}{2m}\right)} \stackrel{?}{=} 0$$

for integers $m \geq 1$, $2\leq n\leq 2m$, and $l = -n+2,-n+4,\ldots n-2$. Is this identity known? I have checked it to be valid for small values, but so far I have been unable to prove the general case.

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Let us consider the case when $n$ is odd. Let $$P(x):=\exp\left(\frac{ixl}{2}\right)\left(\exp\left(\frac{ix}{2}\right)-\exp\left(\frac{-ix}{2}\right)\right),$$ and notice that the degree of $P$ as an exponential polynomial equals $$\max\left(\left|\frac{l+1}{2}\right|,\left|\frac{l-1}{2}\right|\right)\leq\frac{n-1}{2}.$$ Therefore this is the unique trigonometric polynomial of degree at most $\frac{n-1}{2}$ passing through any given $n$ distinct points. Let these given points be $x_p=\frac{\pi p}{m}$ for $p\in \{1,\dots,n\}$. Then, the trigonometric version of Lagrange's interpolation formula tells us that $P(x)$ can be written as $$P(x)=\sum _{p=1}^n P(x_p)\prod_{\substack{k=1\\k\neq p}}^n\frac{\sin\frac{1}{2}(x_k-x)}{\sin\frac{1}{2}(x_k-x_p)}.$$ By plugging $x=0$ we obtain $$0=\sum _{p=1}^n \exp\left(\frac{i\pi p l}{2m}\right)2i\prod_{k=1}^n\sin\left(\frac{\pi k}{2m}\right)\prod_{\substack{k=1\\k\neq p}}^n\frac{1}{\sin\left(\frac{\pi (k-p)}{2m}\right)},$$ and by dividing by the middle factors the OP's identity follows.

The case of even $n$ can be done similarly.

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  • $\begingroup$ Can you provide more detail? The OP's expression also depends on $n$, so I don't see immediately how such a uniform choice of $p(x)$ can solve the problem. $\endgroup$ – GH from MO Oct 14 '15 at 21:46
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    $\begingroup$ I added some more details to clarify the notations and the polynomial which implies the identity. $\endgroup$ – Gjergji Zaimi Oct 15 '15 at 0:05
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    $\begingroup$ Thank you, I finally got it, and I like it very much. (I added more detail and synchronized with the OP's notation, I hope you don't mind.) $\endgroup$ – GH from MO Oct 15 '15 at 1:39
  • $\begingroup$ That is a really nice argument, thanks! I hadn't come across the trigonometric version of Lagrange's interpolation formula yet. $\endgroup$ – Timothy Budd Oct 15 '15 at 7:12
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    $\begingroup$ For the record, after a considerable delay this answer has been used as part of the proof of Proposition 17 in arXiv:1809.02012. Thanks again! $\endgroup$ – Timothy Budd Sep 7 '18 at 13:18

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