Timeline for Is the left derivative of this theta function zero at $1$?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 18, 2023 at 16:52 | vote | accept | Iosif Pinelis | ||
| Oct 17, 2023 at 1:39 | comment | added | Iosif Pinelis | @GerryMyerson : I have now explained this. | |
| Oct 17, 2023 at 1:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 220 characters in body
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| Oct 17, 2023 at 0:51 | comment | added | Gerry Myerson | I'm confused. The title asks about the derivative, but the body just asks about the function. | |
| Oct 17, 2023 at 0:12 | history | edited | Michael Hardy | CC BY-SA 4.0 |
deleted 5 characters in body
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| Oct 16, 2023 at 21:54 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 156 characters in body
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| Oct 16, 2023 at 18:57 | answer | added | Iosif Pinelis | timeline score: 5 | |
| Oct 16, 2023 at 17:57 | comment | added | Sidharth Ghoshal | @IosifPinelis yes, the reason I mentioned it is that the boundary of lacunary series still often respect these questionable “identities”. See my comments at the answer of this question: mathoverflow.net/questions/403882/… the regular Jacobi Theta function $\sum_{n=0}^{\infty} x^{n^2}$, a lacunary function as well takes on the value $\frac{1}{2}$ at $x=-1$ and $1-1+1-1… = \frac{1}{2}$ which would be the unrigorous evaluation of the theta function at that point. | |
| Oct 16, 2023 at 17:39 | comment | added | Aleksei Kulikov | Well, Jacobi identities allows you to transform $x\to 1$ into $x\to 0$ I believe, and when $x\to 0$ the asymptotics of the derivative is easy to calculate, so maybe by doing the change of variables you get the asymptotics as $x\to 1$? | |
| S Oct 16, 2023 at 17:36 | history | suggested | mathworker21 | CC BY-SA 4.0 |
Changed "it it" to "is it". I also changed the first "this" in the title to "the", though, since I don't understand the title, this edit might be bad.
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| Oct 16, 2023 at 17:36 | comment | added | Iosif Pinelis | @GeraldEdgar : Thank you for your comment. My numerics did suggest this question. | |
| Oct 16, 2023 at 17:35 | comment | added | Iosif Pinelis | @AlekseiKulikov : Thank you for your comment. Do you have a more specific suggestion, with a particular identity and a possible way to use it? | |
| Oct 16, 2023 at 17:34 | comment | added | Iosif Pinelis | @SidharthGhoshal : Thank you for your comment. In the Wikipedia article, they seem to have $0$ for the Abelian sum. Here it is a different, lacunary power series. | |
| Oct 16, 2023 at 17:24 | comment | added | Aleksei Kulikov | What does the Jacobi identity give here? | |
| Oct 16, 2023 at 17:22 | comment | added | Sidharth Ghoshal | Yes, notice that the value is $-1 + 4 - 9 + 16 ... = -1 ( 1 - 4 + 9 - 16 ... ) = -1 *(0)$ per en.wikipedia.org/wiki/… | |
| Oct 16, 2023 at 17:17 | review | Suggested edits | |||
| S Oct 16, 2023 at 17:36 | |||||
| Oct 16, 2023 at 16:48 | comment | added | Gerald Edgar | It seems correct, with very rapid convergence. For $x=0.99$ the value is around $-10^{-101}$. | |
| Oct 16, 2023 at 16:33 | history | asked | Iosif Pinelis | CC BY-SA 4.0 |