Timeline for Continuation up to zero of a Dirichlet series with bounded coefficients
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2010 at 22:04 | vote | accept | Anweshi | ||
| Jan 9, 2010 at 23:15 | answer | added | Pete L. Clark | timeline score: 3 | |
| Jan 9, 2010 at 22:48 | answer | added | engelbrekt | timeline score: 4 | |
| Jan 9, 2010 at 22:45 | comment | added | Anweshi | @Pete. Then how do you establish continuation of the Dedekind zeta function up to zero? | |
| Jan 9, 2010 at 22:06 | comment | added | Pete L. Clark | Surely (meaning of course, that I am not completely sure) this is false. A strategy for constructing a counterexample would be to take an infinite sum of Riemann zeta-like functions, each one having a single pole in $0 < \Re(s) < 1$, in such a way so that the set of poles has an accumulation point in, say, $\Re(s) \geq \frac{1}{2}$. | |
| Jan 9, 2010 at 21:34 | history | edited | Anweshi | CC BY-SA 2.5 |
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| Jan 9, 2010 at 21:33 | history | rollback | Anweshi |
Rollback to Revision 3
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| Jan 9, 2010 at 21:23 | history | edited | Anweshi | CC BY-SA 2.5 |
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| Jan 9, 2010 at 20:36 | history | edited | Anweshi |
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| Jan 9, 2010 at 20:30 | history | edited | Anweshi | CC BY-SA 2.5 |
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| Jan 9, 2010 at 20:23 | history | asked | Anweshi | CC BY-SA 2.5 |