Timeline for show that $ \frac{\Gamma(\frac{1}{24})\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})} = \sqrt{3}\cdot \sqrt{2 + \sqrt{3}} $
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2017 at 4:47 | answer | added | Raimundas Vidunas | timeline score: 1 | |
| Sep 12, 2016 at 0:29 | vote | accept | john mangual | ||
| Sep 12, 2016 at 0:17 | answer | added | Noam D. Elkies | timeline score: 27 | |
| Sep 9, 2016 at 13:30 | vote | accept | john mangual | ||
| Sep 12, 2016 at 0:29 | |||||
| Sep 9, 2016 at 4:52 | answer | added | Y. Zhao | timeline score: 13 | |
| Sep 6, 2016 at 15:59 | comment | added | Gerald Edgar | See mathoverflow.net/q/7616/454 and its references | |
| Sep 6, 2016 at 15:25 | answer | added | Igor Rivin | timeline score: 6 | |
| Sep 6, 2016 at 14:09 | history | edited | john mangual | CC BY-SA 3.0 |
added 554 characters in body
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| Sep 6, 2016 at 13:52 | comment | added | john mangual | @FedorPetrov or the Chowla-Selberg formula -- which I have no idea about -- I would like to at least main the spirit of these reflection and multiplication formulas. | |
| Sep 6, 2016 at 13:36 | comment | added | Fedor Petrov | RHS equals $\sin \pi/3 \sin 5\pi/12:\sin^2\pi/6=\Gamma(1/12)^2\Gamma^2(11/12):\Gamma(1/3)\Gamma(2/3)\Gamma(5/12)\Gamma(7/12)$. So, maybe this identity follows from many reflection formulae and the formulae for $\Gamma(nz)$ via $\Gamma(z+k/n),k=0..(n-1)$? | |
| Sep 6, 2016 at 13:16 | history | edited | john mangual | CC BY-SA 3.0 |
added 309 characters in body
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| Sep 6, 2016 at 13:00 | history | asked | john mangual | CC BY-SA 3.0 |