Sum of all Spheres

I ran across a video this afternoon that explains that the sum of volumes of all even-dimensional unit spheres equals eπ.

Why is that? Define vol(n) to be the volume of the unit sphere in dimension n. Then

\mathrm{vol}(n) = \frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2} + 1\right)}

and so the sum of the volumes of all even dimensional spheres is

\sum_{k=0}^\infty \mathrm{vol}(2k) = \sum_{k=0}^\infty \frac{\pi^k}{k!} = \exp(\pi)

But what if you wanted to sum the volumes of all odd dimensional unit spheres? Or all dimensions, even and odd?

The answers to all these questions are easy in terms of the Mittag-Leffler function that I blogged about a while back. This function is defined as

E_{\alpha, \beta}(x) = \sum_{k=0}^\infty \frac{x^k}{\Gamma(\alpha k+\beta)}

and reduces to the exponential function when α = β = 1.

The sum of the volumes of all unit spheres of all dimensions is E1/2, 1(√π). And from the post mentioned above,

E_{1/2, 1}(x) = \exp(x^2) \, \mbox{erfc}(-x)

where erfc is the complementary error function. So the sum of all volumes of spheres is exp(π) erfc(-√π).

Now erfc(-√π) ≈ 1.9878 and so this says the sum of the volumes of spheres of all dimensions is about twice the sum of the even dimensional spheres alone. And so the sum of the odd dimensional unit sphere volumes is almost the same as the sum of the even dimensional ones.

By the way, you could easily answer other questions about sums of sphere volumes in terms of the Mittag-Leffler function. For example, if you want to add up the volumes in all dimensions that are a multiple of 3, you get E3/2, 13/2).

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11 thoughts on “Sum of all Spheres

  1. But all of those volumes are measured in different units. This is like saying that one meter plus one kilogram is 2… at the same time that one meter plus two pounds is 3.

  2. If you think of each term as the proportion of space a sphere takes up inside a box of the same dimension, then each term is dimensionless. So if you replace “volume” with “proportion” everything is quantitatively the same, and there’s no problem with dimensional analysis.

  3. Tony Zbaraschuk

    So the two sums are about the same — is there a way of proving that they are, or aren’t, the same?

  4. @Tony: Since the sum of all spheres is exp(pi) erfc(-sqrt(pi)) and the sum of the even dimensional spheres is exp(pi), the volume of the odd dimensional spheres is exactly (erfc(-sqrt(pi)) – 1) exp(pi).

    The volume of the odd dimensional spheres is less than the even dimensional spheres because (erfc(-sqrt(pi)) – 1) < 1.

  5. I agree with Andrei: Sometimes a good title justifies the post, no matter the content. Especially for Tom Clancy fans.

  6. Oh man, your title has been itching away at the back of my brain ever since I read this post on the day it was posted. And the Clancy connection finally clicked today. Well done! I love a pun that sits innocently and ticks away, only to “go off” in the reader’s mind later.

  7. I told a colleague about this result, and he had the same reaction as Michael Watts — namely, that you can’t add these volumes because each is in different units.

    The ‘proportion’ argument would work if the volume of the containing box was always 1, but it isn’t — it’s 2^n, because 1 is the radius of the ball, not the diameter. The sum of the proportions is exp(pi/2), no?

  8. Federico Provvedi

    You can also generalize the problem for a p-normed n-sphere, using:
    E( 1 / p, 2 * r * (1 / p)! ) as sum all volumes for all dimensions of a p-normed n-ball with radius r. :)

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