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What is the growth rate of $$S_d(M)=\sum_{n=1}^M n^d e\left(\frac{n^2}{2M}\right)$$ where $M$ is an even integer. My numerical experiments show that $$\frac{S_d(M)}{M^{d+\frac{1}{2}}}\approx e^{\frac{\pi i}{4}}\cdot C(M)$$ where $C(M)\in \mathbf{R},$ $|C(M)|\le O(1).$

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  • $\begingroup$ What is $e(...)$ ? And what do you mean by $\approx$ when complex numbers are involved? $\endgroup$ Sep 21, 2021 at 14:42
  • $\begingroup$ look up (quadratic) Gauss sums and use summation by parts; en.wikipedia.org/wiki/Quadratic_Gauss_sum $\endgroup$
    – Conrad
    Sep 21, 2021 at 15:23
  • $\begingroup$ I think that there is some imaginary number $i$ missing in the definition of $S_d(M)$. $\endgroup$ Sep 21, 2021 at 15:57
  • $\begingroup$ @RolandBacher The OP is using the notation $e(z) = e^{2 \pi i z}$, so I don't think any complex numbers are missing. $\endgroup$
    – Random
    Sep 21, 2021 at 16:32
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    $\begingroup$ have you tried Poisson summation into the stationary phase method? I guess that might be what @Conrad is saying implicitly, I just didn’t see a way to treat the partial sums you’d face in partial summation using just knowledge of Gauss sums but admittedly I didn’t think too hard about it $\endgroup$
    – alpoge
    Sep 22, 2021 at 0:41

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