Timeline for References on Taylor series expansion of Riemann xi function
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2020 at 23:42 | history | edited | GH from MO |
edited tags; edited tags
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| Jan 8, 2020 at 18:11 | answer | added | J. Gélinas | timeline score: 3 | |
| Sep 14, 2016 at 19:45 | vote | accept | mike | ||
| Sep 14, 2016 at 16:28 | answer | added | juan | timeline score: 4 | |
| Mar 10, 2016 at 19:51 | comment | added | user88789 | I have a pdf file of the proof that a_2n > a_2n+1. I can send you a copy of this proof if you are still interested. Send request to [email protected] Unfortunately, MathOverflow does not input pdf files. I do not have time to covert the file to MathJax format. | |
| Jan 8, 2016 at 7:50 | comment | added | reuns | knowing that $\xi(s)$ is an entire function, there is no doubt that $a_{2n}$ goes to $0$ quite fast. for the sign, it should be more complicated, at least than $\frac{\xi(s)}{s(s-1)} = \int_0^\infty x^{s-1}\Phi(x) dx$ which is much simpler, also even around $\Re(s) = 1/2$, but has a radius of convergence of $1/2$ at $1/2$. | |
| Jan 7, 2016 at 22:44 | comment | added | Matthias Ludewig | Of course he is^^ | |
| S Jan 7, 2016 at 20:54 | history | suggested | Tadashi |
Added relevant tag
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| Jan 7, 2016 at 20:13 | review | Suggested edits | |||
| S Jan 7, 2016 at 20:54 | |||||
| Jan 7, 2016 at 18:56 | comment | added | user85015 | I have a short proof that a_2n greater than a_2n+2 for all n. Are you interested? | |
| Oct 27, 2013 at 22:57 | history | edited | mike | CC BY-SA 3.0 |
deleted 9 characters in body
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| May 27, 2013 at 20:14 | history | asked | mike | CC BY-SA 3.0 |