4
$\begingroup$

Let $\theta, \phi \in [0,1)$, and consider the sums $$ S_n(\theta,\phi)=\sum_{k=0}^n \log|e^{2\pi i (k\theta+\phi)}-1|. $$ The possible boundedness from above of such sums plays a key role in analyzing the expansiveness of certain algebraic actions of the discrete Heisenberg group, which is why I'm interested. However, some numerical experiments indicate that for typical $\theta$ and $\phi$ these sums occasionally get as large as a constant times $\log n$.

Question: Is it true that for almost every pair $\theta,\phi$ there is a $c=c(\theta,\phi)$ with $0<c<\infty$ such that $$ \limsup_{n\to\infty} \frac{1}{\log n} S_n(\theta,\phi) = c\ ? $$

$\endgroup$
  • $\begingroup$ This can only hold if $\int_{-\pi}^{\pi} \ln |e^{i\varphi}-1|\, d\varphi = 0$ (by the ergodic theorem), so I guess this is clear?! $\endgroup$ – Christian Remling Sep 6 '14 at 18:58
  • $\begingroup$ On second thoughts, this is indeed clear since the integrand is harmonic. $\endgroup$ – Christian Remling Sep 6 '14 at 19:01
  • $\begingroup$ Did you ever resolve this question, Doug? $\endgroup$ – Anthony Quas Oct 29 '16 at 2:01
  • $\begingroup$ No, although the numerical evidence was pretty convincing, at least for some "random" values of $\theta$ and $\phi$. Are you asking for any particular reason? $\endgroup$ – Douglas Lind Nov 3 '16 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.