Timeline for Is there a proof that all the following expressions for Ramanujan's summation are equal?
Current License: CC BY-SA 3.0
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| Jan 28, 2017 at 14:45 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jan 28, 2017 at 7:11 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 27, 2016 at 7:17 | comment | added | reuns | Everything is explained there :) The definition is $g(z)\in O^{2\pi}$ if $g(z)$ is analytic and $|g(z)| < C e^{(2\pi-\epsilon) |z|}$ for $Re(z) > 1-\epsilon$.The idea is that $O^{2\pi}$ makes the contour integral $\int_\gamma \frac{g(z)}{e^{2\pi z}-1}dz$ well-defined. Then for $f(z) \in O^{2\pi}$ : $\displaystyle\overset{\mathfrak{R}}{\sum_{n\ge 1}} f(n) = R_f(1) $ where $R_f(z) \in O^{2\pi}$ and is the unique solution of $R_f(z)-R_f(x+1) = f(z)$ such that $\int_1^2 R_f(z)dz = 0$... | |
| Nov 26, 2016 at 17:29 | comment | added | Sylvain JULIEN | @Michael Hardy : yes it is, but the wikipedia article is a bit unclear as it mentions notations it doesn't explain. | |
| Nov 26, 2016 at 17:28 | comment | added | Michael Hardy | @Anixx : Is that a "summation method" for "divergent series"? Maybe you should "remind" us of its definition. $\qquad$ | |
| Nov 26, 2016 at 17:25 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Nov 26, 2016 at 15:09 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 26, 2016 at 15:08 | comment | added | Anixx | @T. Amdeberhan R symbolizes Ramanujan summation. | |
| Nov 26, 2016 at 5:55 | comment | added | T. Amdeberhan | Where is the dependence of the right-side on $R$? | |
| Nov 26, 2016 at 1:28 | history | asked | Anixx | CC BY-SA 3.0 |