I was wondering whether anyone knows how to approach the following two generalizations of the quadratic Gauss sum:

Given integers r,s with gcd(r,s)=1 and integers a,b,N

$F(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs})$

$G(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{r})(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs})$

Note that removing the sine terms and the sign, setting a = 2, N = 4, r = 1 and s = prime gives the classical quadratic Gauss sum.

Some experimentation suggests that

$F(r,s,N,a,b) = 0$ for all integers b,N, r,s if a is even and (r,s) =1 and

$G(r,s,N,a,b) = 0$ for all a,b,N and r,s with (r,s) =1

Is there a good reason for these sums to vanish? Or a clean proof/reference?

Is it possible to evaluate F in the case a = 1? It seems to be non-zero then.

I tried reducing to the original Gauss sum by completing the square but this seems to get quite ugly.

More generally, do such Gauss-like sums have a more natural generalization that turns up somewhere?

Thanks

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Vinogradov did a lot with generalizations of gauss sums. Nothing like you have there. I read some physics papers where they dealt with sums like yours using finite fouler transform methods. –  Charlie Frohman May 13 '10 at 3:17

In $G$, it looks like the term with $w=k$ cancels the term with $w=rsa-k$, at least under certain parity assumptions on the variables. Maybe it's worth having a look at that.

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I think this should just be computations. Define as usual $e(x) = \exp(2 \pi i x)$. Define $$f(r,s,N,a,b) = \sum_{w=1}^{rsa} e\left(\frac{N}{4 rs} w^2 + \left(\frac{1}{2s}+\frac{b}{2} \right) w\right)$$ Then $F(a,r,s,N,a,b) = \frac{1}{2i } (f(r,s,N,a,b) - f(r,s,N,a, -b))$, at least if I didn't make any computational mistakes.

Now, start with $a = 1$. Then you can use, the method described in http://en.wikipedia.org/wiki/Quadratic_Gauss_sum#Generalized_quadratic_Gauss_sums .

Next, one has to understand what happens if one passes from $a$ to $a + 1$. For this compute $$f(r,s,N,a+1,b) - f(r,s,N,a,b)$$ I guess, the result should be of the form $z f(\hat r,\hat s, \hat N, 1, \hat b)$ with $|z| = 1$.

Note: I began summing at $1$ on purpose, so one sums over the group $\mathbb{Z}_{rsa}$, which seems like the correct choice ...

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