An incomplete post about sphere volumes

The Endeavour 2013-03-19

This is an incomplete blog post. Maybe you can help finish it.

One of the formulas I’ve looked up the most is the volume of a ball in n dimensions. I needed it often enough to be aware of it, but not often enough to remember it. Here’s the formula:

$\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2} + 1\right)}r^n$

The factor of rⁿ is no surprise: of course the volume as a function of radius has to be proportional to rⁿ. So we can make the formula a little simpler by just remembering the formula for the volume of a unit ball.

Next, we can make the formula a simpler still by using factorials instead of the gamma function. If n is a non-negative integer, n! = Γ(n+1). We can use that to define factorial for non-integers. Then the volume of a unit ball is

$\frac{\pi^{\frac{n}{2}}}{\frac{n}{2}!}$

That’s easier to remember.

It’s also curious. The nth term in the series for e^x is xⁿ/n!, so the volumes of unit balls look like series for e^π except compressed, with each index n cut in half. The volumes are not the coefficients in the series for e^x, but could they be the coefficients in the series for another familiar function? To find out, let’s stick back in the factor of rⁿ and sum.

$\sum_{n=0}^\infty \frac{\pi^{\frac{n}{2}}}{\frac{n}{2}!} \,r^n$

This is the sum of the volumes of balls of radius r in all dimensions. That doesn’t make sense by itself, but you could also think of this as the generating function for the volumes of unit balls. So can we find a closed-form expression for the generating function? Yes:

$\sum_{n=0}^\infty \frac{\pi^{\frac{n}{2}}}{\frac{n}{2}!} \,r^n = \sqrt{\pi} r \exp(\pi r^2) (\mbox{erf}(\sqrt{\pi} r) + 1)$

If you work with probability, you probably find Φ more familiar than the error function (see notes relating these) and find exp(x²/2) more familiar than exp(x²). So you could rewrite the generating function as f(√(2π)r) where

$f(x) = \sqrt{2} x\exp(x^2/2) \Phi(x)$

That looks familiar, but I don’t know what to do with it.

I warned you this would an incomplete post. I feel like there’s an interesting connection to be made, but I’m not quite there. Any suggestions?