9

I am trying to prove that the function $$f(N)=\sum_{n=1}^N \cos(n^2)$$ is unbounded. Admittedly, I don't know for certain that it is, but someone told me, and I would like to see it for myself. First, I thought it might be because the square numbers mod $2\pi$ was not uniformly distributed, but numerics seem to imply that they are. My next thought is then that it has to do with the fact that we might have positive or negative subsequences in the sum of arbitrary length. However, what is a good strategy for proving this kind of statement?

Jonnaraev
  • 137
  • well $f(N) \leq N$. However I can't prove that $\lim_{N \to \infty}f(N)$ is unbounded... –  Jan 04 '17 at 13:10
  • Yes, you are right. It is the second thing I am trying to show :) – Jonnaraev Jan 04 '17 at 13:13
  • It is perhaps worth noting that, informally, if the values are "randomly distributed in $[-1,1]$", we would expect arbitrarily large and small numbers to appear in the sequence of partial series with probability 1. Note that this is not true if you remove the square. – Mees de Vries Jan 04 '17 at 13:24
  • Mees de Vries: Yes, I had the same thought. It could be seen as a sort of generalized coin-flipping, where we know the fluctuations of the sum can be arbitrarily large. However, I am unsure as to how far the "uniform distribution" approach can be taken, since we are looking at a specific realization of the sequence.. Can you elaborate on what changes if we remove the square? – Jonnaraev Jan 04 '17 at 13:56
  • It seems to me after plotting a few things, that whenever $\cos(n^2)$ is small ($\approx \epsilon)$ then $|\cos((n+1)^2)|\approx 1$... – Gabriel Romon Jan 04 '17 at 13:58
  • Euler Mac Laurin doesn't work here, right? – tired Jan 04 '17 at 14:56
  • @Jonnaraev, we have $$\sum_{k = 1}^n \cos(k) = \frac{\sin(n + \tfrac12) - \sin(\tfrac12)}{2\sin(\tfrac12)}.$$ – Mees de Vries Jan 04 '17 at 15:00
  • If $N$ is reasonably small, $f(N)$ exhibits an almost $355$-periodic behaviour (so it is bounded) since $355$ is the numerator of a convergent of $\pi$. You are essentially asking for a converse of Van Der Corput inequality. By setting $$ C_N=\sum_{n=-N}^{N}e^{in^2}=1+2\sum_{n=1}^{N}\cos(n^2): $$ – Jack D'Aurizio Jan 04 '17 at 21:28
  • we have $$ \left|C_N\right|^2 = C_N\overline{C_N} =\sum_{n,m\in[-N,N]}e^{i(n^2-m^2)}=\color{red}{(2N+1)}+!!!!!\color{blue}{\sum_{\substack{n,m\in[-N,N]\n\neq m}}e^{i(n^2-m^2)}}.$$ where the magnitude of the blue term depends on $\left|\frac{N}{\pi}\right|$. For larger $N$s, I would expect a $103993$ almost-periodicity and a slightly larger $\max|f(N)|$. We should deeply study the structure of the continued fraction of $\pi$ and deduce that $f(N)$ is unbounded along these lines. – Jack D'Aurizio Jan 04 '17 at 21:29
  • 1
    By the Erdos-Turan inequality (https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tur%C3%A1n_inequality) it follows that we just need a decent lower bound for the discrepancy of the ${\cos(n^2)}_{n\geq 1}$ sequence to disprove that $f$ is bounded. – Jack D'Aurizio Jan 04 '17 at 23:46

0 Answers0