Archimedean volume ratio

And although he made many excellent discoveries, he is said to have asked his kinsmen and friends to place over the grave where he should be buried a cylinder enclosing a sphere, with an inscription giving the proportion by which the containing solid exceeds the contained.

Sphere and cylinder

(Quote from Plutarch; see references on the Archimedes site by Chris Rorres. Image from Wikipedia)

The proportion is 3:2, as Archimedes discovered. More generally, let A_n be the ratio of the volumes of n-dimensional unit ball B^n and of the circumscribed cylinder B^{n-1}\times [-1,1]. This sequence begins with A_1=1, A_2=\pi/4, A_3=2/3, A_4=3\pi/16, A_5=8/15, A_6=5\pi/12\approx 0.49\dots. Note that beginning with 6 dimensions, the ball occupies less than half of the cylinder. It is not hard to see that the sequence monotonically decreases to zero. Indeed, the ratio of (n-1)-dimensional cross-sections by the hyperplane x_{n}=t, for -1\le t\le 1, is (1-t^2)^{(n-1)/2}. Hence A_n = \frac{1}{2} I_{(n-1)/2} where I_p:=\int_{-1}^1 (1-t^2)^p\,dt. The integrand monotonically converges to 0 a.e. as p\to\infty, and therefore I_p does the same.

The Archimedean ratio A_n is rational when n is odd, and is a rational multiple of \pi when n is even. Indeed, integrating I_p by parts, we get
\displaystyle I_p = \int_{-1}^1 (1-t^2)^p\,dt = 2p \int_{-1}^1 t^2(1-t^2)^{p-1}\,dt = 2p(I_{p-1}-I_{p})
hence I_p = \frac{2p}{2p+1}I_{p-1}. In terms of A_n = \frac{1}{2} I_{(n-1)/2}, we have A_{n}= \frac{n-1}{n}A_{n-2}. Since A_1=1 and A_2=\pi/4 are known (the former being trivial and the latter being one of definitions of \pi), the claim follows.

Combining the monotonicity with the recurrence relation, we find that \displaystyle \frac{n-1}{n} =\frac{A_{n}}{A_{n-2}} \le \frac{A_{n}}{A_{n-1}}\le 1. Hence \displaystyle \frac{A_{n}}{A_{n-1}} \to 1, which yields the Wallis product formula for \pi. Indeed, the recurrence shows that for n even, \displaystyle  \frac{A_{n}}{A_{n-1}} is a particular rational multiple of \displaystyle \frac{A_{2}}{A_{1}} = \frac{\pi}{4}.

The above is essentially the inductive proof of Wallis’s formula I remember from my calculus (/real analysis) class. Unfortunately, between integration by parts and the squeeze lemma we lose geometry. In geometric terms, Wallis’s formula says that the volumes of B^{n+1}\times B^{n-1} are B^n\times B^n are asymptotically equivalent: their ratio tends to 1. Is there a way to see that these solids have a large overlap? Or, to begin with, that B^{n+1}\times B^{n-1} has the smaller volume? I don’t see this even with B^3\times B^1 vs B^2\times B^2. By the way, the latter is the bidisc from complex analysis.

Aside: Wallis’s formula for statistics students by Byron Schmuland presents a nice 2-page proof (which uses the relation between \pi and the Gaussian).

Added: Another way to generalize the Archimedean volume ratio to higher dimensions is to divide the volume of B^{2n+1} by the volume of B^{n}\times B^{n+1}. The quotient is always rational, namely \displaystyle \frac{(n+1)!}{(2n+1)!!}. The sequence begins with \displaystyle 1, \frac23, \frac25, \frac{8}{35}, \frac{8}{63}, \frac{16}{231}, \frac{16}{429} and obviously tends to zero. It does not appear to be related to \pi.

One thought on “Archimedean volume ratio”

  1. Note sure if this is rigorous, but we may think B^3 as an imaginary number w_1 (its projection to the imaginary plane) with another real number (the real part of imaginary number w_2). Then
    B^3 \times B^1 = \{(w_1,w_2) | Re(w_1)^2+Im(w_1)^2+Re(w_2)^2<1, Im(w_2)^2<1\}
    bidisc = \{(w_1,w_2) | Re(w_1)^2+Im(w_1)^2<1, Re(w_2)^2+Im(w_2)^2<1\}
    That is, for a fixed w_1 on the unit disc, w_2 is confined to lie on a rectangle with width 2(\sqrt{1-|w_1|^2}) and height 2 for B^3 \times B^1, and on the unit disc for bidisc. Intuitively the rectangle shrinks as w_1 moves away from the origin, so B^3 \times B^1 should span a smaller volume than bidisc.
    V(B^3 \times B^1)=\int_0^{2 \pi} \int_0^1 4 \sqrt{1-r^2} r dr d\theta = 8/3 \pi
    V(B^2 \times B^2)=\pi^2
    And V(B^3 \times B^1)/V(B^2 \times B^2)=8/(3\pi)=A_3/A_2

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