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<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.

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