Timeline for Bounds for analytic circles
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 18 at 6:17 | vote | accept | Bo Jonsson | ||
| Mar 18 at 6:17 | |||||
| Mar 18 at 6:11 | vote | accept | Bo Jonsson | ||
| Mar 18 at 6:12 | |||||
| Mar 18 at 6:09 | vote | accept | Bo Jonsson | ||
| Mar 18 at 6:09 | |||||
| Mar 17 at 22:32 | comment | added | Felixson | Thank you. I will also stop here. Your point is cristal clear, but my impression is that @bojonsson might be assuming $s$ is on the critical line, or strip. I have posted an answer with some basic comments. | |
| Mar 17 at 0:57 | comment | added | GH from MO | @Felixson By Stirling's approximation, one can determine the size of $\xi(s)$ and $f(s)$ very precisely on the positive axis (as $s$ tends to infinity). There is no issue here, really. OK, I must stop now (this site is not designed and not meant for chatting). | |
| Mar 17 at 0:48 | comment | added | Felixson | Thank you @GHfromMO. Of course the bound is trivial for bounded p. But the differing speeds at which $\xi(s)$ and $f(s)$ grow to infinity faster than exponentially in the positive axis, is far from trivial... | |
| Mar 17 at 0:31 | comment | added | GH from MO | @Felixson If $p$ is bounded, then the $O(e^{\epsilon p})$ bound is true for trivial reasons. Indeed, if $|s|<c$, then $\xi(s)/s=O_c(1)$, hence $|\xi(s)|=O_c(|s|)=O_{c,\epsilon}(e^{\epsilon|s|})$. I will finish this topic here. | |
| Mar 17 at 0:19 | comment | added | Felixson | But it seems to me that the problem is still open for circles of sufficiently small radius, since the behaviors of $\xi(s)$ in short intervals is still a mystery. The question is trivial for large p, I agree... | |
| Mar 16 at 23:51 | comment | added | GH from MO | @Felixson If $s=p$ is a large positive number, then $\xi(p)$ exceeds $e^p$. More precisely, $\log\xi(p)$ is asymptotically $p\log p$. | |
| Mar 16 at 23:40 | comment | added | Felixson | @GHfromMO Ok thank you. I see your point then. Still I'm not convinced that the bound is not satisfied, but I must be missing something... | |
| Mar 16 at 23:26 | comment | added | GH from MO | @AlekseiKulikov Right. I fixed my post. | |
| Mar 16 at 23:17 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Mar 16 at 23:11 | comment | added | GH from MO | @Felixson My previous version was partly nonsense, sorry about that. I have updated my post. | |
| Mar 16 at 23:10 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Mar 16 at 23:05 | comment | added | Aleksei Kulikov | It is not of order $0$, but of order $1$ and minimal type. | |
| Mar 16 at 23:03 | comment | added | Felixson | It is well known that there exist certain entire functions of order 1, that satisfy the bound mentioned. That all entire functions of order 0 satisfy the bound, is also well known, as you mentioned. Are you sure that the xi function does not satisfy this bound? is this not an open problem still? | |
| Mar 16 at 22:52 | history | answered | GH from MO | CC BY-SA 4.0 |