Wallis' product exploration

[Gerson W. Barbosa]

(02-12-2020 10:01 PM)pinkman Wrote:  

(02-11-2020 10:48 PM)Gerson W. Barbosa Wrote:  Only linear convergence, but better than waiting forever for a just few digits.

Thank you for all, this is beautiful.

Thank you and your cousin for having started that conversation about the Wallis Product and posting here, otherwise I probably wouldn’t have tried. Unlike approximation polynomials, the equivalent continued fractions usually follow beautiful patterns. Once you have a continued fraction, those polynomials are just their successive approximants. For instance,

Simplify [1 + 1/(4 n + 3/(2 - 1/(4 n + 5/2)))]

-> (64 n^2 + 72 n + 23)/(64 n^2 + 56 n + 15)

= (n^2 + 9n/8 + 23/64)/(n^2 + 7n/8 + 15/64)

This technique had worked previously for other slowly convergent series, so I guessed it would work for Wallis as well.