Let $q$ be a large prime and $\delta\in(0,1)$. Let $k$ be an integer which is not a multiple of $q$. Define $e(x)=e^{-2\pi ix}$. Can we get any non-trivial upper bounds for $$\sum_{a=1}^{q^{1-\delta}}e(\frac{k}{q}a) $$? A bound like $q^{1-\delta-\epsilon}$ would be good enough. I have searched literature in Weyl's sum and did not find any good bounds.
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5$\begingroup$ As written, this is just a finite geometric series, which has an elementary closed form that can be used to estimate the size of the sum. Is that really the sum you mean? $\endgroup$– Noam D. ElkiesCommented Mar 7, 2014 at 23:32
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$\begingroup$ Yes. I have tried the geometric series formula and was unable to use that form to get a non-trivial upper bound in terms of a power of q. Maybe I need to take the magnitude of k into consideration. $\endgroup$– Tony BCommented Mar 8, 2014 at 21:43
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In general no non-trivial bound exists. Suppose e.g. that $a=1$. Then for all $k$ in the range of summation we have $\frac{ka}{q}\in[0, q^{-\delta}]$, thus $e(\frac{ka}{q})=1+\mathcal{O}(q^{-\delta})$.
If you have more information on $a$ you might use $$ \sum_{a=1}^{N}e(\frac{ak}{q}) = \frac{1-e(\frac{(N+1)a}{q})}{1-e(\frac{k}{q})}\ll\frac{q}{a}, $$ which is non-trivial for $a>q^\delta$.
answered Mar 14, 2014 at 21:34
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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$\begingroup$ Thank you for your answer. It seems that you messed up with a and k somewhere. $\endgroup$– Tony BCommented Mar 14, 2014 at 23:04
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$\begingroup$ @Dong He means suppose $k=1$. $\endgroup$ Commented Apr 13, 2014 at 22:34