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May 15, 2023 at 1:12 vote accept CommunityBot
May 15, 2023 at 1:12
May 14, 2023 at 22:47 vote accept CommunityBot
May 15, 2023 at 1:12
May 14, 2023 at 5:25 comment added Noam D. Elkies @SridharRamesh yes, as I expect to note in the next edit: $\eta(q) = q^{1/24} \prod_{n=1}^\infty (1 - q^n)$ is a modular form of weight $1/2$, so $\eta(e^{-2\pi/t}) = t^{1/2} \eta(e^{-2\pi t})$, and if $q=1/2$ then $2\pi t = \log 2$ makes $t$ small and $e^{-2\pi/t}$ tiny.
May 14, 2023 at 5:06 comment added Sridhar Ramesh Does the shockingly good approximation come from some nice reasoning?
May 14, 2023 at 4:16 history answered Noam D. Elkies CC BY-SA 4.0