Timeline for Are all zeros of $\dfrac{\Gamma'}{\Gamma^2}(s) \pm \dfrac{\Gamma'}{\Gamma^2}(1-s)$ either real or on the line with $\Re(s)=\frac12$?
Current License: CC BY-SA 3.0
6 events
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| S Dec 9, 2017 at 9:56 | comment | added | Antonio DJC | then you can use it to confirm your conjecture. In the case of the plus (top) sign, it follows easily that $s=1/2$ is a zero. | |
| S Dec 9, 2017 at 9:56 | comment | added | Antonio DJC | Using the gamma function's reflection formula, you can rewrite your expression as \begin{aligned}D\Big[-\frac{1}{\Gamma (s)}\pm\frac1{\Gamma(1-s)}\Big]&=D\Big[\frac{-\Gamma(1-s)\pm\Gamma(s)}{\pi\csc\pi s}\Big]\\ &=-\cos\pi s\;[\Gamma(1-s)\mp\Gamma(s)]+\pi^{-1}\sin\pi s\;[\Gamma'(1-s)\pm\Gamma'(s)]\end{aligned}so if the answer you cited can be extended to $\sin\pi s\,(\Gamma'(1-s)\mp\Gamma'(s))$, (cont.) | |
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 24, 2016 at 9:07 | comment | added | joro | X-Ray for the plus sign: s31.postimg.org/x81i3jp7v/agno2.png Doesn't show complex zeros off the line. | |
| Jul 24, 2016 at 9:06 | comment | added | joro | The bug is in my code. "derivative=1" doesn't compute the derivative as I wrongly thought. Will delete very soon. | |
| Jul 23, 2016 at 16:06 | history | asked | Agno | CC BY-SA 3.0 |
