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)}


{\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<c_n} 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:

Logarithm of approximation 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.

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.