Timeline for Central limit theorem for irrational rotations
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2023 at 12:39 | comment | added | Christophe Leuridan | What does the symbol $\asymp$ Unelss $\alpha = 1$, the sums are bounded above, so a lower bound $cn$ where $c>0$ is a positive constant is impossible. | |
| Dec 1, 2023 at 21:12 | comment | added | Nikita Sidorov | @ChristopheLeuridan see my answer | |
| Nov 30, 2023 at 13:34 | vote | accept | Nikita Sidorov | ||
| Nov 29, 2023 at 12:07 | comment | added | Christophe Leuridan | @NikitaSidorov The question has been corrected a bit. I modified my answer accordingly. | |
| Nov 29, 2023 at 12:07 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
added 525 characters in body
|
| Nov 28, 2023 at 15:53 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
deleted 111 characters in body
|
| Nov 28, 2023 at 14:11 | comment | added | Ronnie Pavlov | I still don't understand. The sums you work with are bounded (in $n$). So their real parts are also bounded, so your numerator is the logarithm of a bounded quantity. Therefore, it is bounded from above, and so dividing by $\log n$ cannot approach $1/2$. Can you either describe what you think is wrong (maybe we're both making a mistake!) or accept his answer? The whole removal of your previous comments and changing the question and adding a large bounty and asking for an authoritative reference is strange given that it's a one-line proof which has already been given. | |
| Nov 28, 2023 at 11:08 | comment | added | Nikita Sidorov | $\Re z$ stands for the real part of $z$. Log is well defined. | |
| Nov 25, 2023 at 13:54 | comment | added | Ronnie Pavlov | I don't understand the weird responses you're giving. Christophe pointed out that your sums are bounded in $\mathbb{C}$, so there's no version of log where they can grow like $1/2 \log n$. The whole talk about log of negative number seems immaterial. What am I missing? | |
| Nov 25, 2023 at 12:40 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
added 1 character in body
|
| Nov 25, 2023 at 12:39 | comment | added | Christophe Leuridan | The question should at least be reformulated, since the logarithm of a possibly negative number is not well-defined. If the true question is « Does the CLT [work for all $z$] ?», please give a correct statement of what it means. | |
| Nov 25, 2023 at 0:30 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
added 34 characters in body
|
| Nov 24, 2023 at 21:30 | history | answered | Christophe Leuridan | CC BY-SA 4.0 |