Timeline for Does $a_n=\prod^n_{k=1}(1-e^{k\alpha \pi i})$ converge to zero when $\alpha$ is irrational?
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| Oct 15, 2021 at 17:16 | history | edited | Lucia | CC BY-SA 4.0 |
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| Oct 28, 2020 at 3:13 | comment | added | No One | Now at the end of your proof, you may have to replace the in $q\exp(-2\pi q^2 |\beta| \log q)$ by $q\exp(-c\pi q^2 |\beta| \log q)$. Since this time we only have $|\beta| \le \frac{1}{\sqrt{5} q^2}$, your $q^{0.9}$ will have to become a negative power of $q$ going to zero.I understand that you may be very busy and have no time to look at this again after more than four years. But this problem has drawn my attention again recently and I do have a feeling that $\limsup$ depends on $\alpha$ by some computational experiments. | |
| Oct 28, 2020 at 3:12 | comment | added | No One | I checked you argument again for the general case when $|\beta| \le \frac{1}{\sqrt{5} q^2}$. Between your equations (3) and (4), $\pi q^2 |\beta|/2 < \pi/2\sqrt{5}$ and instead of $1-\eta\ge \exp(-2\eta)$, you want to have something like $1-\eta\ge \exp(-c\eta)$, where the best $c$ you can take is $-\log(1-\pi/2\sqrt{5})/(\pi/2\sqrt{5})$ which is about $12/7$ (but anyway this $c$ has to be greater than $1$!). | |
| Jun 9, 2016 at 23:01 | vote | accept | No One | ||
| Jun 7, 2016 at 22:41 | comment | added | Lucia | No -- it's just some careful bookkeeping in the above argument. It's been a little while, but when I wrote this up I checked that everything goes through with $|\alpha- a/q| \le 1/(\sqrt{5}q^2)$ (as mentioned in the answer). | |
| Jun 7, 2016 at 22:36 | comment | added | No One | Thank you for the answer! Is the proof of the general result about (1) much harder than this? Where can I find the proof of it? | |
| May 1, 2016 at 21:50 | history | edited | Lucia | CC BY-SA 3.0 |
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| May 1, 2016 at 15:32 | history | answered | Lucia | CC BY-SA 3.0 |