4 ms·
Clever approach! You can extend it to all even dimensions by following the same approach to get: P[point inside 2n-ball] = (π/4)^n P[x1 + x2 + ... + xn < 1 | x
by clipsy 4y ago
Clever approach! You can extend it to all even dimensions by following the same approach to get:
P[point inside 2n-ball] = (π/4)^n P[x1 + x2 + ... + xn < 1 | x ~ U([0,1]^n)]
Instead of the area below a triangle we have the volume inside a unit simplex, or 1/n!, and of course the volume of the 2n-cube is 2^(2n) or 4^n, giving the 2n-ball’s volume as:
(π/4)^n 4^n / n! = π^n / n!
- bumbledraven 4y agoUnfortunately, I haven’t been able to figure out how to find the volume of a unit n-simplex without calculus or linear algebra. Any suggestions? Simple single-variable calculus would be acceptable, like finding the area under the curve y = 1 – √x when x ranges from 0 to 1, but I’d prefer to avoid nested integrals and trigonometric integrals and derivatives.