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.

1 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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