Volume of Sphere in Higher Dimension

Question

Considering Gaussian shape we can find the volume of n dimensional volume of a sphere. My intention is to find the volume using Monte Carlo method.

Using the gaussian integral I have found the formula

$$V_n = \frac{\pi^{(n/2)}}{\Gamma(\frac{n}{2}+1)}$$

I know I have to define the random function to evaluate the integral but I'm not sure how to do it. Any explicit explanation would be a huge help.

I would be grateful if anyone can give a python code for finding volume, so i can check more intuitively.

Answer 1

[This started out as a comment but it was too long, so I made it an answer, although it's more a roadmap than anything else.]

You should use your system library's random function: defining your own is fraught with peril (bear in mind that the system routine may have problems as well - consult the "Numerical Recipes" book for some details, but be aware that the subject is deep).

You need it to return uniformly distributed random numbers between $-1.0$ and $1.0$. Each n-tuple of random numbers corresponds to a point inside the n-dimensional hypercube of volume $2^n$. The ratio of those points inside the n-dimensional ball to the total number of points will then be roughly the same as the ratio of the volume of the ball to that of the cube. You figure out whether the point is inside the ball by calculating its Euclidean distance from the origin and seeing whether it's less than or equal to the radius (which is 1). Do that for as many points as you think you'll need to get a reasonable estimate of the volume. There is no good way to guess that: in most instances, you'll have to do it for some number of points and then repeat with twice as many (or ten times as many or $2^n$ times as many) points and seeing whether the estimate seems to converge. This is a naive approach: there are probably better approaches but for an initial acquaintance with Monte Carlo, it's probably good enough.

Lather, rinse, repeat for each $n$ of interest.

Also bear in mind that Monte Carlo is not fast: in fact, it's the method of last resort when you don't know of a better way. And the higher the dimension, the more darts you'll have to throw to get a reasonable estimate, so it gets slower as $n$ gets bigger.

Answer 2

This is a javascript example ( you can translate to Phyton ). I enclose the sphere in a cube of side $2$. $\vec{0}$ is the common origin of the cube and the sphere.

It runs in a terminal with node as

$\texttt{node nSphere0.js}$

"use strict"; // nSphere0.js. Felix Marin. 02-april-17.
/*
https://math.stackexchange.com/questions/2261623/volume-of-sphere-in-higher-dimension/2261779#2261779
*/

function nSphereVolume(dim,iter)
{
 var      i = 0;
 var      j = null;
 var      r = null;
 var total0 = null;
 var total1 = 0;

 while (i < iter) {
       total0 = 0;
       for ( j = 0 ; j < dim ; ++j ) {
           r = 2.0*Math.random() - 1.0;
           total0 += r*r;
       }

       if (total0 < 1.0) ++total1;
       ++i;
 }

 return Math.pow(2.0,dim)*(total1/iter);
}

console.log("Volume = " + nSphereVolume(3,10000)/(4.0*Math.PI/3.0));

A 'typical run' yields

$\texttt{1.0061138882497258}$

The example is for a $3$D-sphere where I already divided by $\displaystyle{4\pi \over 3}$. The exact result must be $\displaystyle{\color{#f00}{1}}$.