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Consider the following sum (with $a$ being a real number and $N$ an even integer) $$S(a, N) = \sum_{m=1}^{N/2} \frac{4}{N+1} \sin^2\left( \frac{2\pi m}{N+1} \right)\sin \left( 2 a \cos \left( \frac{2\pi m}{N+1} \right) \right) $$

How to get approximate evaluation of this sum when $N \gg 1$?

I tried to turn $S(a, N)$ into the integral $I(a, N)$, but the integral is zero due to symmetry about $x=\pi/2$

$$I(a, N) = \int_{0}^{\pi} \frac{2}{\pi} \sin^2 \left(x \right) \sin \left( 2 a \cos \left( x \right) \right) \mathrm{d}x = 0 $$

So, how to find an approximate expression for $S(a, N)$ when $N$ is very large?

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    $\begingroup$ Combine symmetric terms (for $m$ and $N/2+1-m$) $\endgroup$ Apr 3 at 10:28
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    $\begingroup$ @FedorPetrov Sorry, I should have said that $N$ is even so that $N/2$ is an integer. How to proceed is this case? $\endgroup$
    – Nigel1
    Apr 3 at 10:51
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    $\begingroup$ That's what I guessed. Still pair symmetric terms. They do not cancel completely. $\endgroup$ Apr 3 at 11:18

1 Answer 1

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First, we rewrite the sum as a sum over the full period $$ S(a,N)=\frac{2}{N+1}\sum_{j=1}^{N+1} \sin^2\left( \frac{2\pi j}{N+1} \right)\sin \left( 2 a \cos \left( \frac{2\pi j}{N+1} \right) \right). $$ Denote $$ g(k,m)=\sum _{j=1}^m \cos ^{k}\left(\frac{2 \pi j}{m}\right). $$ Then the coefficients $a^{2k+1}$ of the Taylor series expansion of $S(a,N)$ is $$ \frac{2(-1)^k}{N+1}\frac{2^{2k+1}}{(2k+1)!}\left(g(2k+1,N+1)-g(2k+3,N+1)\right). $$ It is known that (see e.g. this article ) $$ g(k,m)=\frac{m}{2^k}\sum_{r=-\lfloor k/m\rfloor,\,rm+k\,\mathrm{even}}^{\lfloor k/m\rfloor}\binom{k}{\tfrac{rm+k}{2}}. $$ In particular $$ g(2k+1,2n+1)=0,\qquad k=0,1,2,\ldots,n-1.\quad (n=N/2) $$ Observe that this is also a consequence of Chebyshev-Gauss quadrature.

Now, summation with the above formula shows that coefficient of $a^{2k+1}$ in the Taylor series expansion of $S(a,N)$ with $k=0,1,2,\ldots,n-2$ vanish. The first non-vanishing term corresponds to $k=n-1$: \begin{align} S(a,N)&\approx (-1)^{N/2}\frac{2(2 a)^{N-1}}{N!}g(N+1,N+1)\\ &=(-1)^{N/2}\frac{2(2 a)^{N-1}}{N!}\frac{2(N+1)}{2^{N+1}}\\ &\approx \frac{(-1)^{N/2} a^{N-1}}{(N-1)!}. \end{align} This approximate formula has been confirmed numerically: With increasing $N$, the ratio of the sum and the approximate expression seems to approach $1$.

The higher order coefficients probably can be estimated too with the formulas above, and they will be much smaller than the leading coefficient. This is left as an exercise.

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