Timeline for Does this method analytically continue gap series series?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2022 at 16:58 | comment | added | Sidharth Ghoshal | Additionally theres a typo in the derivative the $\frac{-1}{z^2}$ should be $\frac{1}{z^2}$ im still checking if this causes any problems. | |
| Dec 2, 2022 at 16:17 | comment | added | Sidharth Ghoshal | Yea I don't feel the second argument involving functional equations is valid either but not because of any typos. It's not clear how that same argument doesn't prove $\sum_{n=1}^{\infty} - \frac{1}{x^n}$ isn't the continuation. of $\sum_{n=0}^{\infty} x^n$. Since you have $$ \sum_{k=1, n=1}^{\infty, \infty} \mu(k)z^{kn} = \sum_{k=1}^{\infty} \frac{\mu(k)z^k}{1-z^k} = z$$ and similarly form $z^{-1}$ going the other way and differentiate the two. | |
| Dec 2, 2022 at 14:27 | comment | added | Sidharth Ghoshal | Ah theres also a typo it should be $g(z) = -f(z^{-1})+1$ | |
| Dec 2, 2022 at 14:10 | comment | added | Sidharth Ghoshal | No part of the exploration in “Conflicting Behavior on the Boundary” depended on $z^{n^3}$ and could’ve been for $z^n$ instead which would lead to the conclusion $-\sum 1/z^n$ isn’t a generalized analytic continuation of $\sum z^n $ | |
| Jul 18, 2021 at 6:42 | vote | accept | Caleb Briggs | ||
| Jul 17, 2021 at 21:49 | history | answered | Joseph Van Name | CC BY-SA 4.0 |