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:

The factor of *r*^{n} is no surprise: of course the volume as a function of radius has to be proportional to *r*^{n}. 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

That’s easier to remember.

It’s also curious. The *n*th term in the series for *e*^{x} is *x*^{n}/*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 *r*^{n} and sum.

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:

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

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?

## {{ parent.title || parent.header.title}}

## {{ parent.tldr }}

## {{ parent.linkDescription }}

{{ parent.urlSource.name }}