Timeline for Alternating power series $\sum_{k\geq 0}(-1)^k z^{(2k+1)^2}$
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2017 at 14:42 | comment | added | js21 | Only asymptotic. Check the growth of the coefficients. | |
| Jun 17, 2017 at 9:18 | comment | added | Menglin | @js21: A question about the convergence of your series: is the expansion only asymptotic or convergent in some sense near 0? | |
| Jun 10, 2017 at 1:28 | answer | added | Christian Remling | timeline score: 5 | |
| Jun 9, 2017 at 12:22 | answer | added | მამუკა ჯიბლაძე | timeline score: 2 | |
| Jun 9, 2017 at 12:16 | comment | added | მამუკა ჯიბლაძე | Tried to make some plots, looks like it cannot be continued beyond $\operatorname{Re}(x)>0$ | |
| Jun 9, 2017 at 9:02 | comment | added | Menglin | @Christian Remling: Thanks for the update. Perhaps the possibility of analytic continuation should also be included as part of the question. I guess that's also what misled Bemte before. | |
| Jun 8, 2017 at 23:28 | comment | added | Christian Remling | @მამუკაჯიბლაძე: Indeed. I also made (while editing) the very similar observation that "supposing" that $f(x)$ has a holomorphic continuation to a neighborhood of zero won't help if that is actually false. | |
| Jun 8, 2017 at 19:20 | comment | added | მამუკა ჯიბლაძე | @ChristianRemling your edit directed my attention to one circumstance - it might well be that $f$ is not actually defined in a neighborhood of $0$ at all and the sought value should be understood as the limiting value as $x$ tends to $0$ along the real axis. It is not even clear to me whether one would obtain the same value if $x$ goes to $0$ tangentially - say, $x=t^2+it$ and $t\to0$ (from the positive direction). The latter limit might not even exist. | |
| Jun 8, 2017 at 18:47 | comment | added | Christian Remling | @Menglin: I've made some stylistic changes, in an attempt to state more clearly what I think you wanted to ask (maybe it's just me, but I found your earlier version mildly confusing). Feel free to go back to your version of course if you don't like this. | |
| Jun 8, 2017 at 18:45 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jun 8, 2017 at 12:10 | comment | added | მამუკა ჯიბლაძე | @js21 ...so I believe you should make an answer :) | |
| Jun 8, 2017 at 12:01 | comment | added | მამუკა ჯიბლაძე | Btw they also give an alternative expression for the same: $$ \sum_{n\geqslant0}\left(\zeta(-2n,\frac14)-\zeta(-2n,\frac34)\right)\frac{(-16x)^n}{n!} $$ | |
| Jun 8, 2017 at 12:00 | comment | added | მამუკა ჯიბლაძე | Calculations in the paper "A $q$-series identity and the arithmetic of Hurwitz zeta functions" by Coogan and Ono (PAMS 131 (2003), 719-724) confirm (I believe) calculations by @js21 in the comments. They have (see their (1.3)) $$ \sum_{k\geqslant0}(-1)^ke^{-(2k+1)^2x}=\frac {e^{-x}}{1+e^{-2x}}\sum_{n\geqslant0}\frac{(e^{-2x};e^{-4x})_ne^{-2nx}}{(-e^{-6x};e^{-4x})_n}=\frac12+\frac12x+\frac54x^2+...=\sum_{n\geqslant0}L(\chi_{-1},-2n)\frac{(-x)^n}{n!}, $$ exactly as in the above comment. | |
| Jun 8, 2017 at 11:53 | answer | added | Fedor Petrov | timeline score: 1 | |
| Jun 8, 2017 at 10:52 | history | edited | YCor |
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| Jun 8, 2017 at 10:11 | comment | added | js21 | (continued) I should have written $a_n = (-1)^n n!^{-1} L(-2n, \chi)$ instead. Note that $L(-2n,\chi)$ is equal to half of the $(2n)-$th Euler number $E_{2n}$ (which is $= (2n)! \times$ coefficient of $x^{2n}$ in the hyperbolic sechant $\mathrm{sech}(x)$). in particular $a_1 = \frac{1}{2}$. | |
| Jun 8, 2017 at 10:03 | history | edited | Menglin | CC BY-SA 3.0 |
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| Jun 8, 2017 at 9:59 | comment | added | js21 | @Menglin: a formal expansion of the exponentials suggests that $a_n = (-1)^n n!^{-1} L(-n,\chi)$ where $\chi $ is the nontrivial Dirichlet characer of odrer $4$. One can rigorously prove an asymptotic expansion $f(x) = \sum_{n=0}^N a_n x^n + o(x^N)$, with the $a_n$ as above, by writing the exponentials as an inverse Mellin transform of $\Gamma(s)$, and by moving the line of integration to the left using the residue theorem. | |
| Jun 8, 2017 at 9:58 | comment | added | Dirk | I would compute $f'(0) = \sum_{k \geq 0} (-1)^{k+1}(2k+1)^2$ and therefore conclude that yes, $f'(0)$ does not exist. Ok, I thought it existed first, so I might have been a little to fast in judging, yes. But still, you should explain how you come to assume that $f$ even has an expansion in the neighborhood of $0$, as this would give $a_1$ to be a non-existing/non-converging (?) number? | |
| Jun 8, 2017 at 9:54 | comment | added | Menglin | I don't see how as the result is divergent in doing so. | |
| Jun 8, 2017 at 9:53 | comment | added | js21 | @Bemte: I am afraid you may have judged this question too quickly. | |
| Jun 8, 2017 at 9:51 | history | edited | Menglin | CC BY-SA 3.0 |
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| Jun 8, 2017 at 9:48 | review | Close votes | |||
| Jun 8, 2017 at 11:02 | |||||
| Jun 8, 2017 at 9:38 | comment | added | Dirk | Just differentiate $f(x)$ element-wise, then input $x = 0$ (then add all the formal "what I did was ok, because..."). As mathoverflow deals with research level questions, you might want to try at math.SE. However, note that also there you will need your own approaches. | |
| Jun 8, 2017 at 9:33 | review | First posts | |||
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| Jun 8, 2017 at 9:29 | history | asked | Menglin | CC BY-SA 3.0 |