## Improving the Wallis product

The Wallis product for ${\pi}$, as seen on Wikipedia, is

${\displaystyle 2\prod_{k=1}^\infty \frac{4k^2}{4k^2-1} = \pi \qquad \qquad (1)}$

Historical significance of this formula nonwithstanding, one has to admit that this is not a good way to approximate ${\pi}$. For example, the product up to ${k=10}$ is

${\displaystyle 2\,\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8 \cdot 10 \cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16\cdot 16\cdot 18 \cdot 18\cdot 20\cdot 20}{1\cdot 3\cdot 3\cdot 5 \cdot 5\cdot 7\cdot 7\cdot 9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15\cdot 17\cdot 17\cdot 19\cdot 19\cdot 21} =\frac{137438953472}{44801898141} }$

And all we get for this effort is the lousy approximation ${\pi\approx \mathbf{3.0677}}$.

But it turns out that (1) can be dramatically improved with a little tweak. First, let us rewrite partial products in (1) in terms of double factorials. This can be done in two ways: either

${\displaystyle 2\prod_{k=1}^n \frac{4k^2}{4k^2-1} = (4n+2) \left(\frac{(2n)!!}{(2n+1)!!}\right)^2 \qquad \qquad (2)}$

or

${\displaystyle 2\prod_{k=1}^n \frac{4k^2}{4k^2-1} = \frac{2}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \qquad \qquad (3)}$

Seeing how badly (2) underestimates ${\pi}$, it is natural to bump it up: replace ${4n+2}$ with ${4n+3}$:

${\displaystyle \pi \approx b_n= (4n+3) \left(\frac{(2n)!!}{(2n+1)!!}\right)^2 \qquad \qquad (4)}$

Now with ${n=10}$ we get ${\mathbf{3.1407}}$ instead of ${\mathbf{3.0677}}$. The error is down by two orders of magnitude, and all we had to do was to replace the factor of ${4n+2=42}$ with ${4n+3=43}$. In particular, the size of numerator and denominator hardly changed:

${\displaystyle b_{10}=43\, \frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8 \cdot 10 \cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16\cdot 16\cdot 18 \cdot 18\cdot 20\cdot 20}{3\cdot 3\cdot 5 \cdot 5\cdot 7\cdot 7\cdot 9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15\cdot 17\cdot 17\cdot 19\cdot 19\cdot 21\cdot 21} }$

Approximation (4) differs from (2) by additional term ${\left(\frac{(2n)!!}{(2n+1)!!}\right)^2}$, which decreases to zero. Therefore, it is not obvious whether the sequence ${b_n}$ is increasing. To prove that it is, observe that the ratio ${b_{n+1}/b_n}$ is

${\displaystyle \frac{4n+7}{4n+3}\left(\frac{2n+2}{2n+3}\right)^2}$

which is greater than 1 because

${\displaystyle (4n+7)(2n+2)^2 - (4n+3)(2n+3)^2 = 1 >0 }$

Sweet cancellation here. Incidentally, it shows that if we used ${4n+3+\epsilon}$ instead of ${4n+3}$, the sequence would overshoot ${\pi}$ and no longer be increasing.

The formula (3) can be similarly improved. The fraction ${2/(2n+1)}$ is secretly ${4/(4n+2)}$, which should be replaced with ${4/(4n+1)}$. The resulting approximation for ${\pi}$

${\displaystyle c_n = \frac{4}{4n+1} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \qquad \qquad (5)}$

is about as good as ${b_n}$, but it approaches ${\pi}$ from above. For example, ${c_{10}\approx \mathbf{3.1425}}$.

The proof that ${c_n}$ is decreasing is familiar: the ratio ${c_{n+1}/c_n}$ is

${\displaystyle \frac{4n+1}{4n+5}\left(\frac{2n+2}{2n+1}\right)^2}$

which is less than 1 because

${\displaystyle (4n+1)(2n+2)^2 - (4n+5)(2n+1)^2 = -1 <0 }$

Sweet cancellation once again.

Thus, ${b_n<\pi for all ${n}$. The midpoint of this containing interval provides an even better approximation: for example, ${(b_{10}+c_{10})/2 \approx \mathbf{3.1416}}$. The plot below displays the quality of approximation as logarithm of the absolute error:

• yellow dots show the error of Wallis partial products (2)-(3)
• blue is the error of ${b_n}$
• red is for ${c_n}$
• black is for ${(b_n+c_n)/2}$

And all we had to do was to replace ${4n+2}$ with ${4n+3}$ or ${4n+1}$ in the right places.

## 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 $\pi$.