Here $p$ is a known integer constant ($p > 0$).
I know that this is true for a fact (checked numerically in Matlab and it holds), but I'm just not able to prove it. I noticed a similar problem related to the sum of reciprocal of sine-squared (Sum of the reciprocal of sine squared), but I'm getting nowhere with this problem.
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$\begingroup$ My bad - the RHS of the equation should be $\frac{N}{2}$, not $\frac{N+1}{2}$ - changed it in the title $\endgroup$– R.K.Commented Feb 12, 2014 at 17:07
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3 Answers
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I can prove that the sum is equal to $N/2$. Here, just recognize that
$$\sin^2{y} = -\frac14 (e^{i y}-e^{-i y})^2 = \frac12 - \frac14 (e^{i 2 y} + e^{-i 2 y}) = \frac12 - \frac12 \Re{[e^{i 2 y}]}$$
so the sum is
$$\begin{align}&= \frac12(N-1) - \frac12 \Re{\sum_{n=1}^{N-1} e^{i 2 p \pi n/N}}\\ &= \frac12(N-1) - \frac12 \Re{\frac{e^{i 2 \pi p/N}-e^{i 2 \pi p}}{1-e^{i 2 \pi p/N}}}\end{align}$$
That last expression is $-1$ because $p \in \mathbb{Z}$. The result follows.
answered Feb 12, 2014 at 17:10
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$\begingroup$ this might be a dumb question - Isn't $\sum_{n=1}^{N-1}{r^n} = \frac{1-r^N}{1-r}$? So the exponential sum would be $\frac{1 - e^{i2p\pi}}{1-e^{i2p\pi/N}}$. How are you getting $e^{i2p\pi/N}$ in the numerator instead of 1? $\endgroup$– R.K.Commented Feb 12, 2014 at 17:44
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$\begingroup$ @R.K.: Your expression is valid when the sum starts at $n=0$. $\endgroup$ Commented Feb 12, 2014 at 17:45
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HINT:
Use $\displaystyle\cos2A=1-2\sin^2A\iff\sin^2A=\frac{1-\cos2A}2$
Then utilize
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
or
answered Feb 12, 2014 at 17:33
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Use the fact $\sin(x) = \frac{1}{2i} [e^{xi}-e^{-xi}]$ (by Euler's formula), expand, and use geometric series.
