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While studying characters of a finite field and the Polya-Vinogradov inequality, I've found some nice identities (verified by simulations) that I'm not sure how to prove. They seem to be related to Chebyshev's polynomials of the second kind.

The identities are:

$$ \sum_{a=1}^{q-1} \left( \frac{\sin (\frac{\pi N a}{q})}{\sin (\frac{\pi a}{q})}\right)^2 = N(q-N) \tag{1}$$

$$ \sum_{a=1}^{q-1} (-1)^a \frac{\sin (\frac{\pi N a}{q})}{\sin (\frac{\pi a}{q})} = -N \cdot 1_{N+q \equiv 1 \mod 2}\tag{2}$$

Another function that interests me is the following:

$$ f(N,q,c) = \sum_{a=1}^{q-1} (-1)^{a+ac} \frac{\sin (\frac{\pi N a}{q})}{\sin (\frac{\pi a}{q})} \frac{\sin (\frac{\pi N ac}{q})}{\sin (\frac{\pi ac}{q})}\tag{3}$$ For $c=\pm 1$ it coincides with $1$. How does it behave in general?

$N,q$ and $C$ are positive integers satisfying $N<q$.

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You might be interested to see how thees relate to the equations of diffraction from multiple slits. – nbubis Apr 21 '12 at 14:25
I guess the natural idea is to express the sines in terms of roots of unity. – franz lemmermeyer Apr 21 '12 at 16:06
These vaguely remind me of some identities that Barry McCoy and I found when studying the chiral Potts model. (See equation 3.3 in These were proved by Gervois and Mehta in "Some trigonometric identities encountered by McCoy and Orrick", J. Math. Phys. 36 (1995). I suspect their techniques would be useful to you, but unfortunately their paper doesn't seem to be freely available online. Feel free to contact me if you have trouble locating it, or wish to discuss further. – Will Orrick Apr 21 '12 at 17:44 I have an alternative derivation for the first identity and proved identity two. A closed formula for the third question is obtained for the special case where, $q$ is odd and $c=2$. These results will appear soon as parts of a paper I am writing. Noureddine Chair – user33862 Jun 17 '12 at 12:06
up vote 2 down vote accepted

The second identity may be proved using some known trigonometrical power sums, combined with the representation of the binomial coefficients by the residue operator. To carry out this computation, the sum may be written as $$S_{2}(N)=\sum_{s\geq0}(-1)^{s}\binom {N-s-1}{s}2^{N-2s-1}\sum_{a=1}^{q-1}(-1)^a\cos^{N-2s-1}(a\pi/q),$$ so, one has to evaluate the sum over $a$, this sum turns out to depend on both $N$, $q$. From the classic textbook by I. J. Schwatt, An Introduction to the Operations with Series. Philadelphia, (1924), using the formulas Eq (113), and Eq (114), page 222, then, we may show that the sum is different from $0$ for $N$ odd, $ q$ even, and vanishes for $N$ odd, $ q$ odd. If $N$ is even, then, the above sum is non-vanishing only for $q$ odd. Using the residue representation of the binomial coefficients $$\binom {n}{k}=\hbox{res}_w (1+w){^n}{w^{-k-1}},$$ together with the definition of the normalized Chebysheve polynomial of the second kind, then, we prove the following identities,

$$ S_{2}(l)=\sum_{a=1}^{q-1}(-1)^a\frac{\sin(aN\pi/q)}{\sin(a\pi/q)}=-2\hbox{res}_{w=0} \frac{1}{w^{N-1}}\frac{1}{(1-w)^2}-\hbox{res}_{w=0} \frac{1}{w^N}\frac{1}{(1-w)}=-(2N-1),$$ for $N$ odd, $ q$ even,

$$ S_{2}(N)=\sum_{a=1}^{q-1}(-1)^a\frac{\sin(aN\pi/q)}{\sin(a\pi/q)}=-2\hbox{res}_{w=0} \frac{1}{w^{N}}\frac{1}{(1-w)^2}=-2N,$$ for $N$ even, $ q$ odd. Now, consider the following alternating sum $$ S_{3}(q,N,c=2)=\sum_{a=1}^{q-1}(-1)^{a}\frac{\sin(aN\pi/q)}{\sin(a\pi/q)}\frac{\sin(2aN\pi/q)}{\sin(2a\pi/q)}. $$ Here, $q$ is assumed to be odd, then, it can be shown that the sum is non-vanishing only for $N$ even. By using similar steps as in the previous case, one can show that the sum $ S_{3}(q,2N-1,c=2)$, has the following closed form; $$-\frac{1}{2}3N(3N-1)+\frac{1}{2}N(N-1)-N+q\Big(3N-\frac{(q+1)}{2}+\frac{1}{2}(1-(-1)^{N-\frac{(q+1)}{2}}\Big), $$ where the last term contributes only for $ 3N > \frac{(q+1)}{2}$.

For the explicit computations, see the author 's recent work; Trigonometrical sums connected with one-dimensional lattice, the chiral Potts model, and number theory using the residue operator, arXiv:1206.6673v1 [math-ph]

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A partial answer:

To prove (1), follow Franz Lemmermeyer's suggestion. Let $\omega=e^{i\pi/q}$ and write the left-hand side as $$ S_1=\sum_{a=1}^{q-1}\left[\frac{\omega^{aN}-\omega^{-aN}}{\omega^a-\omega^{-a}}\right]=\sum_{a=1}^{q-1}\left[\sum_{j=1}^N \omega^{a(N+1-2j)}\right]^2=\sum_{a=1}^{q-1}\sum_{j=1}^N \omega^{a(N+1-2j)}\sum_{k=1}^N\omega^{a(N+1-2k)}. $$ Reordering the sums gives $$ S_1=\sum_{j=1}^N\sum_{k=1}^N\sum_{a=1}^{q-1}\omega^{2a(N+1-j-k)}=-N^2+\sum_{j=1}^N\sum_{k=1}^N\sum_{a=0}^{q-1}\omega^{2a(N+1-j-k)}. $$ The innermost sum equals 0 unless $j+k=N+1$, in which case it equals $q$. Since the double sum over $j$ and $k$ contains $N$ terms in which the condition $j+k=N+1$ holds, we have $$ S_1=-N^2+Nq. $$

The second one should work out similarly. I'll have to think more about the third.

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