Timeline for Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2013 at 5:47 | history | undeleted | Kim Morrison | ||
| Jun 21, 2013 at 6:13 | history | deleted | user631 | ||
| Apr 18, 2012 at 16:33 | history | edited | user631 | CC BY-SA 3.0 |
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| Mar 6, 2012 at 18:30 | comment | added | Marc Palm | @Agno: the logarithmic derivative is the tool to count zeros and it is always available. | |
| Feb 25, 2012 at 3:19 | comment | added | user631 | @GH: Rouché? Touché! | |
| Feb 25, 2012 at 0:39 | comment | added | GH from MO | @Agno: Rouché's theorem is contained in basic textbooks, and this is all you need (actually a slight generalization of it). Using this you can shorten the above proof to a few lines (e.g. no integrals), see my comment above. | |
| Feb 25, 2012 at 0:33 | comment | added | Agno | Very impressive, although I honestly have to say that fully understanding the proof is beyond my math skills. Still got the goosebumps from reading it though :-) The proof does induce two follow up questions: 1) could the function $\Gamma(s)^2 - \Gamma(1-s)^2$ be uniquely represented by an infinite product involving its 'complex' zeros (via Weierstrass factorization)? 2) is there a function for locating the zeros (similar to $Z(t)$ for the Riemann non trivial zeros)? Thanks. | |
| Feb 25, 2012 at 0:15 | comment | added | GH from MO | Wonderful. You can simplify and strengthen the proof by using a generalized Rouché's theorem. This tells us that $\Gamma(z)+\theta\cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$ when $|\Gamma(1-z)|<|\Gamma(z)|$ holds on the boundary. This is equivalent to $\pi/|\sin(\pi z)|<|\Gamma(z)|^2$, hence it suffices to have $\pi<|\Gamma(z)|^2$ on $\partial C_n$. It seems that the last inequality holds for $n\geq 5$. | |
| Feb 24, 2012 at 23:57 | vote | accept | Agno | ||
| Feb 25, 2012 at 0:37 | |||||
| Feb 24, 2012 at 23:57 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Feb 24, 2012 at 23:46 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Feb 24, 2012 at 18:44 | history | answered | user631 | CC BY-SA 3.0 |