Timeline for If the generating function summation and zeta regularized sum of a divergent series exist, do they always coincide?
Current License: CC BY-SA 4.0
25 events
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| Oct 14 at 13:05 | comment | added | Max Lonysa Muller | The following paper by Navas et al. (2021) may be relevant: sciencedirect.com/science/article/abs/pii/S0022247X20307034 | |
| Oct 11, 2020 at 10:24 | comment | added | reuns | @MaxMuller You need to clarify the analyticity-meromorphicity conditions you are assuming for your functions. | |
| Oct 10, 2020 at 16:18 | answer | added | darkl | timeline score: 5 | |
| Oct 10, 2020 at 16:03 | comment | added | darkl | @Max Muller yeah, exactly. You should be able to express your Fibonacci Dirichlet series as a linear combination of geometric Dirichlet series, perhaps these are more studied in the literature? | |
| Oct 10, 2020 at 15:52 | comment | added | Max Lonysa Muller | Ah, I think I understand the example of the Riemann zeta function. One then considers the generating function of $f(y) = \sum_{n \geq 1} y^{n} $, which amounts to $\frac{y}{1-y} $, after which you plug in $y=e^{-x}$ to obtain $f(e^{-x}) = \frac{1}{e^{x}-1}$, right? Hence the choice of $f$ in the example. | |
| Oct 10, 2020 at 15:41 | comment | added | Max Lonysa Muller | @darkl Hmm alright. I don't completely understand how this works yet, but if you can show it does indeed amount to the appropriate series and that the limit $x \to 0$ exists and is finite, then I suppose it could be a fitting method. | |
| Oct 10, 2020 at 12:57 | comment | added | darkl | The idea is that the Mellin transform of $e^{-kt}$ is $\Gamma\left( x \right)\frac{1}{k^x}$ (by the definition of the Gamma function and by changing variable). So if you apply the Mellin transform to $g(x)$ you get $\sum_{k=0}^{\infty} \Gamma\left( x \right) \frac{f_k}{k^x}$. See also this Wikipedia page for a similar argument for the Zeta function. | |
| Oct 10, 2020 at 10:35 | comment | added | Max Lonysa Muller | @darkl Maybe, could you elaborate? | |
| S Oct 10, 2020 at 3:17 | history | suggested | gmvh |
Added top-level tag
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| Oct 9, 2020 at 23:06 | comment | added | darkl | Regarding your question about the Fibonacci Dirichlet series, can't you just take $\frac{1}{\Gamma\left( x \right)} \int_0^{\infty} g(e^{-t}) t^{x} \frac{dt}{t}$ where $g$ is the function you defined? | |
| Oct 9, 2020 at 20:57 | review | Suggested edits | |||
| S Oct 10, 2020 at 3:17 | |||||
| Oct 9, 2020 at 20:39 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
corrected the title
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| S Sep 6, 2014 at 17:04 | history | bounty ended | CommunityBot | ||
| S Sep 6, 2014 at 17:04 | history | notice removed | CommunityBot | ||
| S Aug 29, 2014 at 15:42 | history | bounty started | Max Lonysa Muller | ||
| S Aug 29, 2014 at 15:42 | history | notice added | Max Lonysa Muller | Draw attention | |
| Aug 29, 2014 at 15:42 | comment | added | Max Lonysa Muller | @DavidSpeyer Although perhaps similar arguments given in answers to your question can be used to answer mine? | |
| Aug 24, 2014 at 21:13 | comment | added | David E Speyer | Ah, you are right. Related, but not the same. | |
| Aug 24, 2014 at 17:26 | comment | added | Max Lonysa Muller | @DavidSpeyer No, I don't think it's a duplicate. I think we have a different idea of what zeta summation is. You are actually doing summation by means of the analytic continuation of the corresponding dirichlet series. Therefore, you take the limit $ \lim_{x \to 0 } \sum_{n=1}^{\infty} \frac{a_{n}}{n^{x}} $. However, I take the limit $\lim_{x \to -1 } \sum_{n=1}^{\infty} { {a_{n}}^{-x}} $ . The methods are confused quite often with one another, though. | |
| Aug 24, 2014 at 16:21 | comment | added | David E Speyer | Duplicate of mathoverflow.net/questions/19410/… | |
| Aug 24, 2014 at 16:13 | comment | added | paul garrett | A quibble: The way the first example is worded may give a misleading impression, blurring "taking a limit" with "evaluation of meromorphic continuation at a point". If the meromorphic continuation has the desired point at the edge of the circle of convergence, then a non-tangential limit should give the same outcome, yes, but the idea is not restricted to that case... | |
| Aug 24, 2014 at 16:08 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed two typos in the formula for $z(x)$
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| Aug 24, 2014 at 15:33 | history | edited | Max Lonysa Muller | CC BY-SA 3.0 |
improved spelling
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| Aug 24, 2014 at 15:14 | comment | added | Max Lonysa Muller | Cross-posted from math.stackexchange.com/questions/906980/… | |
| Aug 24, 2014 at 15:13 | history | asked | Max Lonysa Muller | CC BY-SA 3.0 |