Timeline for Is regularization of infinite sums by analytic continuation unique?
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7 events
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| May 12, 2020 at 23:45 | comment | added | reuns | @Anixx At first the idea is that $\sum^L (-1)^n = \lim_{z\to 1} \sum_{n\ge 0} (-1)^n z^n$ is not the same as $\sum^A (-2)^n = \frac1{1+2z}|_{z=1}$. The former is a limit regularization and has good chances to be related from one method to the other while the latter is analytic continuation and usually is not related (except for zeta/power series with finitely many poles because one is the Mellin transform of the other). | |
| May 12, 2020 at 21:16 | comment | added | Anixx | @The thing is, the most of regularization methods AGREE on the same value. Alanytic continuation is less rigorous in this respect and I am sure you can find such function that it will give any value you want. | |
| May 12, 2020 at 21:11 | comment | added | MCH | Thanks. I understand that different transformations that assign values to infinite sums (Borel sums etc) may result in different values. But I'm curious to know if, for example for the sum S above, there are possible functions (other than Riemann zeta) that can be analytically continued to a value different from -1/12. | |
| May 12, 2020 at 18:43 | comment | added | Anixx | @LSpice analytic continuation does not belong to the set of the mutually-compatible methods that produce the same result when applicable (Abel, Cesaro, Borel, Dirichlet, etc). Analytic continuation without additional restrictions can yeld, in general, any result. | |
| May 12, 2020 at 18:39 | comment | added | LSpice | What is 'here' in "Analytic continuation in general does not belong here"? | |
| May 12, 2020 at 18:39 | history | edited | LSpice | CC BY-SA 4.0 |
Link to the other answer
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| May 12, 2020 at 18:23 | history | answered | Anixx | CC BY-SA 4.0 |