COMPUTING SCIENCE

# An Adventure in the *N*th Dimension

On the mystery of a ball that fills a box, but vanishes in the vastness of higher dimensions

# What’s So Special About the 5-Ball?

I was taken by surprise when I learned that the volume of a unit *n*-ball goes to zero as *n* goes to infinity; I had expected the opposite. But something else surprised me even more—the fact that the volume function is not monotonic. Either a steady increase or a steady decrease seemed more plausible than having the volume grow for a while, then reach a peak at some finite value of *n*, and thereafter decline. This behavior singles out a particular dimension for special attention. What is it about five-dimensional space that allows a unit 5-ball to spread out more expansively than any other *n*-ball?

I can offer an answer, although it doesn’t really explain much. The answer is that everything depends on the value of π. Because π is a little more than 3, the volume peak comes in five dimensions; if π were equal to 17, say, the unit ball with maximum volume would be found in a space with 33 dimensions.

To see how π comes to have this role, we’ll have to return to the formula for *n*-ball volume. We can get a rough sense of the function’s behavior from a simplified version of the formula. In the first place, if we are interested only in the unit ball, then *r* is always equal to 1, and the *r*^{n} term can be ignored. That leaves a power of π in the numerator and a gamma function in the denominator. If we consider only even values of *n*, so that *n*/2 is always an integer, we can replace the gamma function with a factorial. For brevity, let *m*=*n*/2; then all that remains of the formula is this ratio: π^{m}/*m*!.

The simplified formula says that the *n*-ball volume is determined by a race between π^{m} and *m*!. Initially, for the smallest values of *m*, π^{m} sprints ahead; for example, at *m*=2 we have π^{2}≈10, which is greater than 2!=2. In the long run, however, *m*! will surely win this race. Both π^{m} and *m*! are products of *m* factors, but in π^{m} the factors are all equal to π, whereas in *m*! they range from 1 up to *m*. Numerically, *m*! first exceeds π^{m} when *m*=7, and thereafter the factorial grows very much larger.

This simplified analysis accounts for the major features of the volume curve, at least in a qualitative way. The volume of a unit ball has to go to zero in infinite-dimensional space because zero is the limit of the ratio π^{m}/*m*!. In low dimensions, on the other hand, the ratio is increasing with *m*. And if it’s going uphill for small *m* and downhill for large *m*, there must be some intermediate value where the function reaches a maximum.

To get a quantitative fix on the location of maximum, we must return to the formula in its original form and consider odd as well as even numbers of dimensions. Indeed, we can take a step beyond mere integer dimensions. Because the gamma function is defined for all real numbers, we can treat dimension as a continuous variable and ask with finer resolution where the maximum volume occurs. A numerical solution to this calculus problem—found with further help from Mathematica—shows a peak in the volume curve at *n*≈5.2569464; at this point the unit ball has a volume of 5.2777680.

With a closely related formula, we can also calculate the surface area of an *n*-ball. Like the volume, this quantity reaches a peak and then falls away to zero. The maximum is at *n*≈7.2569464, or in other words two dimensions larger than the volume peak.

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**Of Possible Interest**

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