For which $\alpha,\beta$ is $\sum_n \text{sign} \sin(\alpha n^2 + \beta)$ bounded?

Question

I was thinking about Proving that the function $f(N)=\sum_{n=1}^N \cos(n^2)$ is unbounded and it is linked to the distribution of $n^2 \,\%\, 2 \pi$ where $\%$ is thought of as the "real modulo" i.e. $a \% b = a - n b$ where $n$ is chosen such that $a \% b$ is positive and smaller than $b$.

Now the question is for which $\alpha$ the distribution $\alpha n^2 \% 2 \pi$ is balanced in the sense that the difference between points in the interval $[0;\pi)$ and the points in the interval $[\pi; 2 \pi)$ stays finite.

This should be equivalent to the question in the title

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

A proof of the unbounded-ness of the partial sums of $\sin(n^2)$ can be found here.
The key steps are

Van Der Corput's trick
Dirichlet's hyperbola method to estimate the average order of $d_1$ and $d_2$
Weyl's inequality.

$\sin(n^2)$ and $\text{sign}\sin(n^2)$ are pretty much the same thing under a distributional point of view (we may say that $\sin(n^2)$ is just a mollified version of $\text{sign}\sin(n^2)$) and I would expect that if $\alpha\neq 0$ the contribute of $\beta$ in the actual problem is completely irrelevant. By adapting the previous method to the present question, the most reasonable conjecture is

$$ \sum_{n=1}^{N}\text{sign}\sin(\alpha n^2+\beta)\quad\text{bounded}\quad\Longrightarrow \quad\color{red}{\alpha\in \pi\mathbb{Q}} $$

since in the opposite case ($\alpha\not\in\mathbb{Q}$) the LHS is expected to be $\gg\sqrt{N}$ for an infinite number of $N$s.

Of course $\beta$ ultimately does play a role: If $\alpha=0\in\pi\Bbb Q$, then the partial sums are bounded iff $\beta\in\pi\Bbb Z$. — Hagen von Eitzen, Feb 01 '17 at 20:09
@HagenvonEitzen: all right, fixed. Anyway, I do not think the OP is really interested in the $\alpha=0$ case. — Jack D'Aurizio, Feb 01 '17 at 20:11

For which $\alpha,\beta$ is $\sum_n \text{sign} \sin(\alpha n^2 + \beta)$ bounded?

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