Wallis' product exploration

02132020, 12:17 AM
Post: #20




RE: Wallis' product exploration
(02122020 10:01 PM)pinkman Wrote:(02112020 10:48 PM)Gerson W. Barbosa Wrote: Only linear convergence, but better than waiting forever for a just few digits. Thank you and your cousin for having started that conversation about the Wallis Product and posting here, otherwise I probably wouldnâ€™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. 

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