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Nov 23, 2020 at 6:14 comment added user64494 See that paper.
Nov 21, 2020 at 18:12 comment added flinty Based on Roman's closed form in the previous question: Limit[2^(n - 1)*Gamma[n/2]^2/(Sqrt[\[Pi]]*Gamma[n - 1/2]), n -> \[Infinity]] is Sqrt[2].
Nov 21, 2020 at 17:53 vote accept Penelope Benenati
Nov 21, 2020 at 17:50 comment added flinty In the limit for large $n$ the mean distance and distribution converge to the same results as the $n$-sphere distances. In other words, in high dimensions a single coordinate doesn't make much difference to the distances. Also I suspect for both spherical and hemispherical cases, the mean converges to $\sqrt{2}$ for high $n$ but that's just a guess.
Nov 21, 2020 at 17:45 history answered flinty CC BY-SA 4.0