Timeline for Does $a_n=\prod^n_{k=1}(1-e^{k\alpha \pi i})$ converge to zero when $\alpha$ is irrational?

11 events

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Oct 21, 2020 at 4:47	history	edited	No One		edited tags
Jun 9, 2016 at 23:01	vote	accept	No One
May 1, 2016 at 15:32	answer	added	Lucia		timeline score: 24
Apr 19, 2016 at 22:32	history	edited	No One		edited tags
Apr 19, 2016 at 22:15	history	migrated			from math.stackexchange.com (revisions)
Mar 2, 2016 at 20:38	comment	added	No One		@A.S. Thank you! This idea using probability theory is nice! But I still can't see a formal rigorous proof.
Mar 2, 2016 at 6:08	answer	added	Sangchul Lee		timeline score: 15
Mar 2, 2016 at 5:42	comment	added	A.S.		To see that, note that $\int_0^{2\pi} \log(1-e^{ix})=0$, and the integrand has only one singularity. At the same time, empirical distribution of $b_k=k\alpha\pi\mod 2\pi$ converges to $U(0,2\pi)$. Hence, when the region around $0$ is suddenly overemphasized (a sudden very small $b_k$), $a_k$ gets very close to zero and starts to slowly return to its "average" of $1$ (which it should not "surpass" since convergence of the rectangle intergration method outside of $0$ is quadratic - weak point) which it reaches whenever density around $0$ gets de-emphasized (drops below "average").
Mar 2, 2016 at 5:21	comment	added	A.S.		There is no convergence - $\liminf a_n=0$ while $\limsup a_n\ge 1$.
Mar 2, 2016 at 1:27	answer	added	Robert Israel		timeline score: 10
Mar 2, 2016 at 0:44	history	asked	No One	CC BY-SA 3.0