Timeline for What are the analytic properties of Dirichlet Euler products restricted to arithmetic progressions?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2010 at 7:24 | comment | added | aghitza | Thanks Peter, that's giving me a very good starting point. | |
| Jun 15, 2010 at 7:23 | vote | accept | aghitza | ||
| Jun 13, 2010 at 9:07 | comment | added | Wadim Zudilin | It seems that the author asks for meromorphic continuation... Terence Tao's paper is arxiv.org/pdf/0908.4323 . | |
| Jun 13, 2010 at 9:06 | history | edited | Peter Humphries | CC BY-SA 2.5 |
added 1577 characters in body
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| Jun 13, 2010 at 8:44 | comment | added | Peter Humphries | Indeed, it's pretty obvious that $\zeta_{\mathcal{P}}(s)$ is going to have a singularity at $s=1$, so any continuation is going to at best be meromorphic. | |
| Jun 13, 2010 at 8:34 | comment | added | Wadim Zudilin | If one writes $$\log\biggl(1-\frac1{p^s}\biggr)=-\frac1{p^s}+O(p^{-2s})$$ for $p$ sufficiently large, the result you mention implies that the sum $$\sum_{p\equiv a\pmod{q}}\frac1{p^s}$$ has $\Re(s)=1$ as natural boundary for analytic continuation. But then the same is valid for $\sum_pp^{-s}$. Is this correct? | |
| Jun 13, 2010 at 8:03 | history | answered | Peter Humphries | CC BY-SA 2.5 |