Testwiki:Reference desk/Archives/Mathematics/2016 June 6

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June 6

Expected distance between point inside a hypercube and a vertex

Let $I_n=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}...\underset{0}{\overset{1}{\int }}\sqrt{x_1^2+x_2^2+...x_n^2}d{x}_1...d{x}_n$. $I_1=1/2,I_2=(\sqrt{2}+{\sinh }^{-1}1)/3,I_3\approx 0.960591956455053$. What would be the best way to numerically evaluate this integral in higher dimensions? Or, if there's a simple closed form what is it? 24.255.17.182 (talk) 22:01, 6 June 2016 (UTC)

Use generalized spherical coordinates? See n-sphere#Spherical coordinates. Not that this is gonna make things much easier though - I'm sure that a hypercube's boundaries can be expressed as trigonometric functions in spherical coordinates, and integrating trigonometric functions can always be done symbolically. For example, using the notation from that article, we could write $x_1=1$ as $r=\sec ({\varphi }_1)$.--Jasper Deng (talk) 00:18, 7 June 2016 (UTC)