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It is well known that for each $m\in\mathbb{N}$ $$\lim_{N\to\infty}\frac1N\sum_{n=1}^Ne^{2\pi i\sqrt{nm}}=0$$ My question is whether there is some uniformity in the variable $m$. More precisely, is it true that $$\lim_{N\to\infty}\frac1N\sum_{n=1}^Ne^{2\pi i\sqrt{nN}}=0?$$ I would be particularly interested in a power saving of the form $$\lim_{N\to\infty}\frac1{N^{1-\epsilon}}\sum_{n=1}^Ne^{2\pi i\sqrt{nN}}=0$$ for some $\epsilon>0$.

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2 Answers 2

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I write $e(x)=e^{2\pi i x}$. Here is a naive bound via Kuzmin's estimate, which says that if you have a sequence $c_n$ with monotonic differences $\delta_n$ all in some interval $[k+\epsilon, k+1-\epsilon]$, then the entire exponential sum is small:

$$|\sum_n e(c_n)| \ll \epsilon^{-1}$$

If we write $c_n = \sqrt{nN}$ then in Elkies' notation

$$\delta_n = c_n-c_{n-1} = \frac{\sqrt{N}}{\sqrt{n} + \sqrt{n-1}}$$

Now pick an $\epsilon>0$ and break up the interval $[1,N]$ into two pieces - those numbers where the fractional part of $\delta_n$ is within $\epsilon$ of an integer, and everything else. The possible integers go up to $O(\sqrt{N})$, and for each $m \ll \sqrt{N}$ it is easy to check that the number of $|\delta_n - m| < \epsilon$ is $\ll \epsilon N/m^2$. Cutting off the interval $[1, \epsilon N]$ to be safe, we can say that the total area of these exceptional intervals is $O(\epsilon N)$.

On the other hand the rest of the $c_n$ are broken up into $O(\sqrt{N})$ intervals with sum at most $\epsilon^{-1}$ each. Thus,

$$|\sum_n e(c_n)| \ll \epsilon N + \epsilon^{-1} \sqrt{N}.$$

Picking $\epsilon = N^{-1/4}$, this should give a power-saving, for any $\delta>0$:

$$\lim_{N\rightarrow\infty} \frac{1}{N^{\frac{3}{4}+\delta}}\sum_{n\le N} e(\sqrt{nN}) \rightarrow 0.$$

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The answer given by Xiaoyu He uses the same arguments as usual van der Corput's theorem (see Theorem 2.2 from Graham, S. W. & Kolesnik, G. "van der Corput's method of exponential sums"):

Suppose that $f$ is a real valued function with two continuous derivatives on $I$. Suppose also that there is some $\lambda > 0 $ and some $a > 1$ such that $\lambda < |f''(x)| < a\lambda$ on $I$. Then $$\sum_{n\in I} e(f(n))\ll a|I|\lambda^{1/2} + \lambda^{-1/2}.$$

So we can try to apply this theorem directly. For $f(x)=\sqrt{ N x}$ on the interval $I(X)=[X,2X)$ we have $f''(x)\asymp\lambda =N^{1/2}X^{-3/2}$. Van der Corput's theorem gives the estimate $$\sum_{n\in I(X)} e(f(n))\ll N^{1/4}X^{1/4} + N^{-1/4}X^{3/4}.$$ Summation over $X=1,2,4,8,\ldots,2^k\asymp N$ leads to the upper bound $$\sum_{n=1}^N e(f(n))\ll N^{1/2}.$$

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