I have the following exponential sum:

$\sum _{M<n\leq N}e\left (x/n^2\right )=\sum f(n),$

say, where $M$ and $N$ are something like $x^{1/4}$ and $x^{1/2}$.

My question is basically, how do I bound this?

I know of two methods used to bound exponential sums, the Weil way and the Van der Corput way. I think the first way is not really applicable here, but I would have guessed, based just on how it looks, that the Van der Corput method would have worked. However, that way would give, I believe, a bound of the type

$\frac {f'(N)+f'(M)}{(f''_{inf})^{1/2}}$

where $f''_{inf}\leq f''(n)$ for the whole range of summation.

This bound however is worse than trivial, so there's something happening there which I don't understand and which means the sum really is a different type of sum. Therefore, how should I bound it? There clearly is some oscillation, so it should be possible to say something. And also, as a curiosity question, why do we get absolutely no information on this sum using the Van der Corput method?

Thanks very much.


1 Answer 1


Actually, $|\sum_{M < n \leq N} e(x/n^2)| \sim c \sqrt{x}$, where $c \approx 0.016151690 + 0.0738060263i$.

To see this, write $$\sum_{M < n \leq N} e(x/n^2) = \sum_{M < n \leq \epsilon \sqrt{x}} e(x/n^2)+\sum_{\epsilon \sqrt{x} < n \leq N} e(x/n^2),$$ where $\epsilon = \frac{1}{x^{1/6}}$ (say, actually $\epsilon = 1/\log{x}$ is enough). Use the trivial bound on the first sum and use Euler-Maclaurin summation on the second sum. The point is that the second sum, is well approximated by $\int_{\epsilon \sqrt{x}}^N e(x/t^2)dt = \frac{1}{2} \int_1^{1/ \epsilon^2} e(u)/u^{3/2}du$.

The problem with Van der Corput is when $n \asymp\sqrt{x}$, as the exponential sum is not very oscillatory. The reason why Van der Corput does not work is that it gives the best results when $|f''(n)| \asymp 1/(N-M)$, but when $M,N \asymp \sqrt{x}$, in your problem we have $|f''(n)| \asymp 1/(N-M)^2$, which is too small. Actually if you are looking for a bound where $N \leq x^{1/2 - \delta}$ for any fixed $\delta > 0$, then Van der Corput, along with a dyadic decomposition, does give you nontrivial bounds.

  • $\begingroup$ Hi. Thanks very much for your answer, it helped me a lot. I had tried to use Euler-Maclaurin but I thought you couldn't do anything with the integral term :/ Actually, my $N$ is something closer to $x^{3/8}$ (I didn't think this was too important initially) but I still didn't think Van der Corput brought anything (see my bound). My problem was, I think, that I wasn't doing the dyadic split (I need to learn when this works) - so thanks very very much for pointing this out. I believe I then get a bound, for my case of $N=x^{3/8}$, of order $x^{1/4}$, which is what I want :) $\endgroup$
    – tomos
    Commented Jan 12, 2016 at 0:05

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