# 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.

(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. Lianxin says:

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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