Timeline for Conjecture about $\prod_{n=1}^{\infty}\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|=0$
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Jul 2, 2023 at 13:21 | comment | added | DesmosTutu | en.wikipedia.org/wiki/Littlewood_conjecture | |
| Oct 14, 2022 at 0:08 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
appended answer 432374 as supplemental
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| S Oct 11, 2022 at 12:06 | history | bounty ended | CommunityBot | ||
| S Oct 11, 2022 at 12:06 | history | notice removed | CommunityBot | ||
| Oct 6, 2022 at 19:47 | comment | added | Gerald Edgar | @SamHopkins There are limits like: number of questions asked per day, or per week, and so on. Perhaps depending on your reputation. Perhaps Erik has reached the limit in math.se, but not in MO. | |
| Oct 6, 2022 at 16:59 | comment | added | Sam Hopkins | Is it really true that there is a limit to the number of questions one can ask on MO? I have not heard of that... | |
| Oct 6, 2022 at 15:57 | history | edited | DesmosTutu | CC BY-SA 4.0 |
added 333 characters in body
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| Oct 5, 2022 at 15:11 | comment | added | Conrad | I think it's unclear that there is no subsequence of $n$'s for which $\left|\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right| $ goes to infinity; even if the irrationality measure of $\pi$ is $2$ there are stil infinitely many $n$ for which $|\cos n| <<1/n$ and then $|\tan (n+1)|$ is not small so the term in the bracket is $>> \delta >1$ infinitely many times | |
| Oct 5, 2022 at 14:50 | history | edited | DesmosTutu | CC BY-SA 4.0 |
added 260 characters in body
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| Oct 5, 2022 at 11:44 | comment | added | Fedor Petrov | again a different question, please stop doing this, ask new questions without removing old | |
| Oct 5, 2022 at 11:41 | history | edited | DesmosTutu | CC BY-SA 4.0 |
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| Oct 4, 2022 at 12:04 | history | edited | DesmosTutu | CC BY-SA 4.0 |
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| Oct 4, 2022 at 6:06 | comment | added | Fedor Petrov | You mean that the function must be logarithmically concave on each interval $(0,1/2)$ and $(1/2,1)$, right? | |
| Oct 3, 2022 at 19:54 | history | rollback | Stefan Kohl♦ |
Rollback to Revision 1
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| Oct 3, 2022 at 14:36 | history | edited | Richard Stanley | CC BY-SA 4.0 |
changed "does" to "do" (twice)
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| Oct 3, 2022 at 12:36 | history | edited | DesmosTutu | CC BY-SA 4.0 |
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| Oct 3, 2022 at 11:37 | comment | added | Pietro Majer | The angles $\alpha$ such that $\tan(\alpha)+1$ and $\tan(\alpha+1)+1$ have opposite signs are half of the interval $[0,\pi]$ (in measure). So the fraction should be negative half of the times (in density). Is there a reason why the effect of negative terms is stronger? | |
| Oct 3, 2022 at 10:34 | comment | added | Wojowu | Why did you completely overwrite the question instead of asking a new one?? | |
| S Oct 3, 2022 at 10:09 | history | bounty started | DesmosTutu | ||
| S Oct 3, 2022 at 10:09 | history | notice added | DesmosTutu | Authoritative reference needed | |
| Oct 3, 2022 at 10:09 | history | edited | DesmosTutu | CC BY-SA 4.0 |
deleted 1291 characters in body; edited tags; edited title
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| Feb 6, 2020 at 19:08 | comment | added | Iosif Pinelis | Have you proved that $h''\le q''$? | |
| Feb 6, 2020 at 18:38 | history | edited | GH from MO |
edited tags
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| Feb 6, 2020 at 18:37 | comment | added | GH from MO | Please ask a question explicitly. A claim is not a question. | |
| Feb 6, 2020 at 17:53 | history | asked | DesmosTutu | CC BY-SA 4.0 |