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Let $f(x)=\log|2\sin(x/2)|$ (the normalizing factor $2$ is chosen to have the average over the period equal to $0$). Does there exist $a>0$ such that all sums $\sum_{k=1}^n f(ak)\ge 0$? The computations (run up to the values of $n$ where I could not rely on the floating point precision any more) show that it may be the case even for $a=2\pi\sqrt 2$ but I'll be happy with any $a$ (or with a proof that no such $a$ exists).

This question came up in our joint study of Leja interpolation points with Volodymyr Andriyevskyy but the connection is a bit too convoluted to be explained here :-)

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    $\begingroup$ have u seen B6 and solutions? $\endgroup$
    – mathworker21
    Commented Oct 15, 2021 at 0:50
  • $\begingroup$ @mathworker21 No. I agree that it has the same spirit but I do not immediately see whether the techniques there yield anything in my case :-) $\endgroup$
    – fedja
    Commented Oct 15, 2021 at 4:08
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    $\begingroup$ Yuval Peres and David Ralston have some papers on a closely related topic. They deal with continuous functions (unlike your function with a singularity), and prove the existence of what they call heavy points: these are exactly the points you are looking for where all sums are non-negative. $\endgroup$
    – Anthony Quas
    Commented Oct 15, 2021 at 5:00
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    $\begingroup$ I don't know about non-negativity, but this beautiful paper shows that there is a uniform lower bound for this sum in the case $a = \pi(\sqrt{5}-1)$. There's also a lot of references, in particular about the continued fraction expansion of the potential $a$. $\endgroup$
    – Aleksei Kulikov
    Commented Oct 15, 2021 at 6:40
  • $\begingroup$ Just a remark: judging by the outcome of numerical experiments, the choice $a = 2\pi\sqrt2$ seems to maximize the infimum of the sequence of sums (as you certainly know, but it might be interesting for the others). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 15, 2021 at 11:57

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The answer to this question is yes, which is proved in the following paper that appeared on arXiv today. Specifically, part $(ii)$ of Theorem 2 of this paper, in the case $b = 1$, says that for $\beta = \frac{\sqrt{5}-1}{2}$ the smallest term of the sequence $$P_N(\beta) = \prod\limits_{r = 1}^N 2|\sin(\pi r\beta)|$$ is the first term $P_1(\beta)$ (this is the same as the sequence in OP up to denoting $a = 2\pi \beta$ and taking the exponent). Since $$P_1(\beta) = 2|\sin(\pi \beta)| = 1,86\ldots > 1,$$ the sequence in the OP for $a = 2\pi \beta$ is strictly positive and even has a uniform poisitve lower bound $\log(1,86\ldots) = 0,62\ldots$ .

Note also that the case $b = 2$ of the same Theorem gives us $\beta = \sqrt{2}-1$. By periodicity of sine, it's the same as $\beta' = \sqrt{2}$, which translates to $a = 2\pi \sqrt{2}$ from the OP. So, for this $a$ infimum is also the first term, which is also strictly positive.

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