Re: Surfaces of hyperspheres
- From: "Michael Hennebry" <hennebry@xxxxxxxxxxxxxxxxxxxxx>
- Date: 3 May 2006 09:24:31 -0700
Hans Georg Schaathun wrote:
The answer I need to answer is the following:
Draw uniformly at random a point of norm 1 in Euclidean n-space.
What is the probability that the norm of its projection on the
x-axis is greater than alpha. (I.e. that the x-coordinate
is greater than alpha in abs. value.)
Since otherwise the answer is obvious,
let's assume that alpha is in the open interval (0, 1).
We want twice the probability that the x-coordinate > alpha.
This is proportional to the n-volume of the
appropriate section of the unit n-sphere.
It is the convex hull of the union of the origin
and the lens-shaped portion
{ p in R^n: |p|<=1 & p[1]> alpha} .
The n-volume is
alpha*S(n-1)*(1-alpha^2)^((n-1)/2)/n +
1
INT S(n-1)*(1-x^2)^((n-1)/2) dx ,
alpha
where S(n-1) is the n-1-volume
of a unit n-1-sphere.
The first term is n-volume of a cone
defined by the origin and an n-1-sphere.
The other is the n-volume of the lens-sphaped portion.
.
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