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I am interested in the exponential sum

$$\sum_{n=1}^X \frac{e(c_1n^2+c_2 n)}{1-e(c_1n)}$$

where $c_2$ is irrational and $e(x)=e^{2\pi i x}$. I know if the denominator is not there, this is a Weyl sum with a lot of bounds.

Is there anything that can be done to relate this sum to a Weyl sum or get some other bounds?

Edit, one thing that might be nice is to get a bound like

$$\left|\sum_{n=1}^X \frac{e(c_1n^2+c_2 n)}{1-e(c_1n)}\right|\ll \left|\sum_{n=1}^X e(c_1n^2+c_2 n) \right|\max_{1\leq n\leq N} \frac{1}{|1-e(c_1n)|}.$$

Is a bound like this possible?

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    $\begingroup$ You write that $c_2$ is irrational, but say nothing about $c_1$. If $c_1$ is a rational number $a/b$ with $0 < b \leq X$ then at least one summand is infinite. In general the sum may be dominated by terms for which $c_1 n$ is close to an integer, which means that the size of the sum may hinge on small rational approximations to $c_1$. $\endgroup$ Commented May 30, 2023 at 19:01
  • $\begingroup$ @NoamD.Elkies Thank you, you are correct. Here's a question. If we know that $c_1$ is so that $|1-e(c_1 n)|>\epsilon$ for all $n\in \{1,...,X\}$, then can we get a bound in terms of $\epsilon$? $\endgroup$ Commented May 30, 2023 at 19:10
  • $\begingroup$ You can approximate $1/ (1- e (c_1n ))$ by a polynomial in $e(c_1 n)$ with an error term which is usually small but large for $c_1$ close to an integer. Each monomial in the polynomial gives a Weyl sum and you can then bound the error term separately (depending on rational approximations to $c_1$). $\endgroup$ Commented May 30, 2023 at 19:10
  • $\begingroup$ @WillSawin Are you suggesting to write something like $\frac{1}{1-e(c_1 n)}=e(c_1 n)+O(e(2 c_1 n))$ for example? How would you handle the error term? I am fine with the first finitely many, but how do we control the error in terms of rational approximation to $c_1$? $\endgroup$ Commented May 30, 2023 at 19:19
  • $\begingroup$ A much better estimate is $\frac{1}{ 1- e(c_1n )} = \sum_{k=0}^d e( c_1 n k ) \frac{d-k}{d} + O\left( \frac{ 1 - e ( d c_1 n )}{ (1- e(c_1n))^2}\right)$ but, thinking about it more, this seems to only be good enough to get an improvement of the "trivial" bound (which is not so trivial in this case) by a constant factor. $\endgroup$ Commented May 30, 2023 at 21:03

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Under your assumption that no $n$ from $1$ to $X$ has $|1 -e(c_1n)|<\epsilon$, we have an upper bound for your sum of the form $2 \epsilon^{-1} \log X + O(X)$.

This is the "trivial bound" in that it just comes from estimating $$\sum_{n=1}^X \frac{1}{ |1 - e(c_1 n )|}.$$

By the pidgeonhole principle, the number of $n$ with $e( c_1 n)$ at an angle less than $\theta$ from $1$ is at most $ 2 \theta /\epsilon$ since otherwise there would be $2$ such $n$ with angles within an interval of length less than $\epsilon$ and then the absolute value of their difference would have an angle less than $\epsilon$ and thus have $|1 -e(c_1n) |< \epsilon$.

If $\theta_1,\dots, \theta_X$ are any set of angles such that the number with distance less than $\theta$ from $0$ is at most $2\theta/\epsilon$ then

$$ \sum_{n=1}^{X} \frac{1}{ |1 -e(\theta_n)|} \leq \sum_{n=1}^{X} \frac{1}{ |1 -e ( n \epsilon/2))|}$$

because $|1- e(c_1 n )|$ is an increasing function of the angular difference, and this sum is well-approximated by the integral

$$ 2 \epsilon^{-1} \int_{\epsilon/2}^{ X \epsilon/2} \frac{ d \theta}{ | 1- e(\theta)|}$$

and since $\frac{1}{ 1- e(\theta)} = \frac{1}{ \theta} + O(1)$ , the integral is bounded by $\log X + O(X \epsilon)$, giving the claim.

Improving this bound by a significant factor seems to require estimates at many different scales, since the integral producing the $\log X$ term has equal contributions from many different scales. I suspect this can be done in such a way that the estimate at each scale is essentially a Weyl sum over a Bohr set, which probably means that cancellation can indeed be found under hypotheses on $c_1,c_2$, but one couldn't hope to save more than the $\log X$ factor as one is unlikely to get a bound for the overall sum greater than the bound for a single term.

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  • $\begingroup$ So is the point to save on the even more trivial bound $X/\epsilon$? You put the $\epsilon$ in the numerator but get an additive log term? $\endgroup$ Commented May 31, 2023 at 0:34
  • $\begingroup$ @user479223 No, we multiply this by $\epsilon$. The bound we get is stated at the beginning, and represents a saving of $X/\log X$ over the even more trivial bound. $\endgroup$ Commented May 31, 2023 at 0:50
  • $\begingroup$ Thank you. This is helpful. $\endgroup$ Commented May 31, 2023 at 0:56
  • $\begingroup$ @user479223 So this indeed gets a bound better than that requested in your question - sort of. One never gets a better bound for the original sum that $\sqrt{X}$ so your requested bound is $\sqrt{X} \epsilon^{-1}$, which this bound easily beats. However, the original sum is probably quite small some of the time (and sometimes even zero) if $c_1$ and $c_2$ are chosen exactly right so the bound you request isn't literally true. $\endgroup$ Commented May 31, 2023 at 1:09
  • $\begingroup$ Thank you so much for your help. $\endgroup$ Commented May 31, 2023 at 1:16

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