What is the asymptotic order of $$ \int_0^1 \left| \sum_{n=1}^N e^{2 \pi i n^2 x} \right| ~dx $$ as $N \to \infty$. This should be known, but I cannot find it in the literature.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ See mathoverflow.net/questions/208415/… for a similar question. $\endgroup$– Vaclav KotesovecJun 4, 2015 at 10:02
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Jurkat and van Horne showed that the $L^1$ norm is asymptotic to a constant times $\sqrt{N}$ (see Theorems 4 and 5 in their paper which compute all moments). For other related work see Jurkat and van Horne and Marklof. Finally Vaughan and Wooley considered Weyl sums for powers larger than $2$, and formulated some conjectures -- the case for squares behaves differently from higher powers.
-
$\begingroup$ The answer is conjectured to be $\frac{1}{2}\sqrt{\pi} \sqrt{N}$, and the numerical evidence seems to support this conjecture (see the cited paper of Vaughan and Wooley). The situation is not different for squares versus higher powers for the first moment. The differences appear for fourth moments, since the major arc contribution takes over for squares at the fourth moment, and this has a logarithmic factor. $\endgroup$– tdwJun 4, 2015 at 10:37
-
$\begingroup$ @tdw: Dear Trevor, From the works ofJurkat and van Horne and Marklof, the quadratic Weyl sums have a distribution that is not Gaussian. So the constant in the moments, I don't think needs to match your conjecture. The constant they get is by averaging moments of a theta function over a fundamental domain. It is possible that for the first moment this could evaluate to your conjectured value, but I don't see why. In any case, the distribution is not Gaussian, which seems quite different from other powers. Am I missing something? $\endgroup$– LuciaJun 4, 2015 at 13:29
-
$\begingroup$ I agree! I had forgotten what Jurkat and van Horne had proved. I guess the numerics are misleading for small N. $\endgroup$– tdwJun 4, 2015 at 15:09