Timeline for Convergence of zeta Euler product with additional term
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2022 at 13:17 | comment | added | Conrad | You can add additional exponential factors to increase the convergence range but not sure if you gain much - the usual approach is to use more controllable factors (molifiers) that restrict the zeroes of the result rather than cancel them out together and get estimates for things if interest them | |
| Apr 27, 2022 at 13:11 | comment | added | Bertrand | Ok, I see, clear. Thanks. | |
| Apr 27, 2022 at 12:59 | comment | added | Conrad | No, $P$ has no zeroes, while the zeroes of $\zeta$ are cancelled by the poles of $B$: a simple example of this is $\xi$ who is entire and a product of $\zeta$ and $\Gamma$ and some non zero powers and for which the trivial zeroes of $\zeta$ cancel the poles of $\Gamma$ | |
| Apr 27, 2022 at 12:57 | comment | added | Bertrand | Yes, you are right. And can we write that $P(s)=\zeta(s) B(s)$ in this strip ? It would mean the zeros of $P(s)$ in this strip are same as Zeta ? Strange for me. | |
| Apr 27, 2022 at 12:54 | history | edited | Bertrand | CC BY-SA 4.0 |
added 20 characters in body
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| Apr 27, 2022 at 12:52 | comment | added | Conrad | Better put, $B(s)$ would be analytic except at the zeroes of $\zeta$ where it has poles | |
| Apr 27, 2022 at 12:50 | comment | added | Bertrand | Yes, I would say you are right. | |
| Apr 26, 2022 at 14:30 | comment | added | Bma | If $P(s)$ is indeed analytic for $\frac{1}{2} < \sigma \le 1$ (note the non-strict inequality on the right), then is the definition $B(s) = P(s) / \zeta(s)$ not an analytic continuation of $B(s)$ to that region? | |
| Apr 26, 2022 at 14:26 | history | edited | YCor | CC BY-SA 4.0 |
formatting, changed tags
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| Apr 26, 2022 at 13:08 | history | asked | Bertrand | CC BY-SA 4.0 |