(HP15C)(HP67)(HP41C) Bernoulli Polynomials

09052023, 07:09 PM
(This post was last modified: 09112023 08:13 PM by Albert Chan.)
Post: #23




RE: (HP15C)(HP67)(HP41C) Bernoulli Polynomials
(09052023 05:57 PM)John Keith Wrote: This is quite interesting, I haven't seen this method before. The algorithm is simply rewriting B(2n) as function of zeta(2n), and B(2n) denominator = product(p, such that (p1)2n) It does not involve complex numbers. Complex number form is just more compact. B(2n) = Re(2 (2n)! / (2*pi*i)^(2n)) * zeta(2n) (2*pi*i)^(2n) = (2*pi)^(2n) * i^(2n) = (2*pi)^(2n) * (1)^n (09032023 01:20 AM)Albert Chan Wrote: Update 1: halved d, and use floor instead of round Redo the same examples, with update #1, #2 ideas. Here, we add sign afterwards, but it could be added to b beforehand. lua> m = 12 lua> b, d = 2*fac(m)/pi^m, 2^m1 lua> b, d 1036.4980453216303 4095 lua> _ + b*3^m 1036.4999956755648 lua>  1036.5 / 4095  = B(12) 0.2531135531135531 lua> m = 16 lua> b, d = 2*fac(m)/pi^m, 2^m1 lua> b, d 464784.48919973 65535 lua> _ + b*3^m 464784.49999694 lua>  464784.5 / 65535  = B(16) 7.092156862745098 Note: next corrections (if needed) are b*5^m, b*7^m, ... 

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