I would like to estimate the following sum

$\sum_{N <n \leq 2N}e(vn^{l})$, $l \geq 1$ constant(not integer) and $v$ is a parameter(integer) that doesn't grow too fast(a small power of N).

The first idea may be to apply Weyl-Van der Corput inequality several times($[l]$ times) then invoke the theory of exponential pairs. But then, I get terms like $N^{1-1/2^{l}}$, which is not sufficient for my purposes. The other idea may be to apply Vinagradov's method. But this time a limitation sourced by h appears. Is there any other way to estimate the above sum?

  • $\begingroup$ Yes, but whether it works for you depends on the bound you need. Since to tell exactly what estimate is sufficient is much easier than to run and verify the argument and to make a long post heavy with LaTeX, I suggest you make the first move :). $\endgroup$
    – fedja
    Commented Aug 31, 2013 at 15:58
  • $\begingroup$ The sequence $\{n^{\theta}\}$ is equidistributed, hence the full character sum tends to zero (normalized properly). As you are looking at half of the sum, it is a bit harder to analyse. If you assuming that $\theta$ is rational, if think there's a good chance to get proper asymptotics (probably by clever ergodic theory),see for example the Elkies-McMullen result about gaps in $\sqrt{n}$. $\endgroup$
    – Asaf
    Commented Aug 31, 2013 at 17:05
  • $\begingroup$ If non-integrality of $\theta l$ is important, it should be edited into the body of the question, rather than hidden in the comments. But why isn't $\theta l$ just written as a single constant, $q$, say, with the condition that $q$ is not an integer? And what is this $\delta$ that turns up, unexplained? $\endgroup$ Commented Sep 1, 2013 at 1:24
  • $\begingroup$ Yes you are right, $\theta$ is a bit important(but not that much) so to make sure that $l \theta$ is not an integer. $\endgroup$ Commented Sep 1, 2013 at 5:48

1 Answer 1


There is the article Primes in Special Intervals and Additive Problems with Such Numbers of Maris Changa. He considered exponential sum with $n^l$ $(l>1)$ and estimated them using double sums $$n^l\to n^l\left(1+\frac{xy}n\right)^l$$ and Vinogradov's mean value theorem.

Will it help?

  • $\begingroup$ Yes it did. After making the above change of variable the original sums turns out to be a Weyl type sum which is easily handled. $\endgroup$ Commented Jun 17, 2014 at 10:29

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