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Consider $$S_N:=\sum_{n=1}^N \exp\left(2\pi i\left(\frac12n^2x+\alpha n\right)\right)$$

where $\alpha$ is irrational. For certain $x$ (say integer) we can get that this is bounded for all $N$. I am curious, for what $x\in \mathbb R$ is this sum "small" - that is $S_N=o(\sqrt N)$?

Is the Lebesgue measure of such $x$ positive?

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  • $\begingroup$ Call the sum $S_{N;\alpha}(x)$. Using Chebyshev's inequality, it's straightforward to show that $m(\{x \in [0,2] :|S_{N;\alpha}(x)| \geq C\}) \leq \frac{2N}{C^2}$. Maybe one can do better by considering higher moments? $\endgroup$
    – Peter Humphries
    Commented Aug 2, 2023 at 21:26
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    $\begingroup$ A useful keyword is Weyl sum, see for example here: en.wikipedia.org/wiki/Weyl%27s_inequality_(number_theory) $\endgroup$
    – Christian Remling
    Commented Aug 2, 2023 at 22:30
  • $\begingroup$ This question is partially not well-defined: what is $\alpha$? Do you want the answer as a function of $\alpha$? $\endgroup$
    – mathworker21
    Commented Aug 3, 2023 at 17:20
  • $\begingroup$ @mathworker21 I would be interested in the answer for any irrational $\alpha$. $\endgroup$
    – user479223
    Commented Aug 3, 2023 at 17:23
  • $\begingroup$ How about almost every? $\endgroup$
    – mathworker21
    Commented Aug 3, 2023 at 17:27

1 Answer 1

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Ok, apologies if this is overkill, but this paper shows that for almost every (irrational) $\alpha \in \mathbb{R}$, only a measure $0$ set of $x \in \mathbb{R}$ satisfy $S_{N,\alpha}(x) = o(\sqrt{N})$ as $N \to \infty$.

Indeed, that paper,

Metric theory of Weyl sums
Changhao Chen, Bryce Kerr, James Maynard, Igor Shparlinski

shows that there is an absolute constant $c > 0$ such that almost every pair $(x,\alpha) \in \mathbb{R}^2$ has $|S_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$. It follows from basic measure theory that for almost every $\alpha \in \mathbb{R}$, it holds for almost every $x \in \mathbb{R}$ that $|S_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$.

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  • $\begingroup$ This is very helpful, thank you. I just want to make sure I have this right. Does this imply that for infinitely many $N$, we have that the Lebesgue measure of $x\in [0,1]$ so that $|S_N|/\sqrt N > c$ is $1$? $\endgroup$
    – user479223
    Commented Aug 3, 2023 at 18:01
  • $\begingroup$ @user479223 No, I don't see why it would imply that (apriori). $\endgroup$
    – mathworker21
    Commented Aug 3, 2023 at 18:06

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