Skip to main content
14 events
when toggle format what by license comment
Apr 20, 2020 at 6:35 comment added Anixx I already asked this question here: math.stackexchange.com/questions/1176034/… There are some answers.
Apr 19, 2020 at 18:04 answer added Anixx timeline score: 0
Apr 18, 2020 at 3:32 answer added Anixx timeline score: 5
Apr 6, 2020 at 7:35 review Close votes
Apr 6, 2020 at 22:23
Apr 6, 2020 at 0:02 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Mar 7, 2020 at 10:01 comment added Oscar Cunningham According to wikipedia, $\Gamma(\frac{1}{2}-m) = \frac{(-4)^mm!}{(2m)!}\sqrt{\pi}$. Substituting this in to the above formula gives $|B^{-(2m+1)}| = \frac{(2m)!}{(-\frac{\pi}{4})^mm!}$.
Mar 7, 2020 at 2:43 comment added Sam Hopkins Maybe en.wikipedia.org/wiki/Reflection_formula is relevant here?
Mar 6, 2020 at 22:54 history edited YCor CC BY-SA 4.0
edited tags
Mar 6, 2020 at 20:46 comment added Bazin Here is a funny formula: with $\nu !=\Gamma (\nu+1)$, we have $$ \sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\frac{π^\nu}{\nu !} R^{2\nu}=e^{π R^2} +\sum_{k\in \mathbb N}\frac{π^{k+\frac12}}{(k+\frac12)!} R^{2k+1} =\sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\vert \mathbb B^{2\nu}\vert R^{2\nu} =\sum_{n\in \mathbb N} \vert \mathbb B^{n}\vert R^{n}. $$
Mar 6, 2020 at 10:12 comment added gmvh It can occur in the dimensional regularization of Feynman diagram integrals in quantum field theory, but I don't think it has too much significance there.
Mar 4, 2020 at 15:28 comment added James Propp Pure curiosity.
Mar 4, 2020 at 14:22 comment added Piotr Hajlasz You should explain the context why you are asking this question to make it more relevant.
Mar 4, 2020 at 2:50 review Close votes
Mar 4, 2020 at 16:43
Mar 4, 2020 at 1:12 history asked James Propp CC BY-SA 4.0