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Jul 15, 2021 at 15:00 comment added Dispersion By Poisson summation, you can show that $S(s) -\sqrt{2\pi}s=\sqrt{2\pi}s\sum_{n\neq 0} e^{-2\pi^2 n^2 s^2} \neq 0$, which also explains why equality seems to hold numerically, because the right hand side is exponentially small (up to an overall factor of $s$ in front).
Jul 15, 2021 at 14:49 comment added mathworker21 I don't get it. How did you answer the question? You just said the identity is false, but you didn't explain why...
Jul 15, 2021 at 11:14 vote accept Iosif Pinelis
Jul 15, 2021 at 6:54 history answered Dispersion CC BY-SA 4.0