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(12C+) Bernoulli Number
07-27-2019, 12:41 PM (This post was last modified: 07-27-2019 10:41 PM by Albert Chan.)
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RE: (12C+) Bernoulli Number
(07-27-2019 06:41 AM)Gamo Wrote:  Formula use to calculate Bernoulli Number

B(n) = [2(2n)! ÷ ((2^2n) - 1)(Pi^2n)] [1 + (1/3^2n) + (1/5^2n) + ...]

Hi, Gamo

I think there is a typo: B(n) should be abs(B(2n))

Can you provide source ?

From https://wstein.org/edu/fall05/168/projec...rnproj.pdf

zeta(x) = sum(k^-x, where k = 1 to ∞) = 1 / product(1 - p^-x, where p is prime)

B(2n) = (-1)^(n+1) * 2*(2n)!/(2 pi)^(2n) * zeta(2n)
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Messages In This Thread
(12C+) Bernoulli Number - Gamo - 07-27-2019, 06:41 AM
RE: (12C+) Bernoulli Number - Albert Chan - 07-27-2019 12:41 PM
RE: (12C+) Bernoulli Number - Gamo - 07-27-2019, 01:40 PM
RE: (12C+) Bernoulli Number - John Keith - 07-27-2019, 07:49 PM
RE: (12C+) Bernoulli Number - Albert Chan - 07-28-2019, 12:02 AM
RE: (12C+) Bernoulli Number - John Keith - 07-28-2019, 11:21 AM
RE: (12C+) Bernoulli Number - Albert Chan - 07-28-2019, 01:08 AM
RE: (12C+) Bernoulli Number - Gamo - 07-28-2019, 02:29 AM
RE: (12C+) Bernoulli Number - Albert Chan - 07-31-2019, 05:14 PM



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