(12C+) Bernoulli Number
07-31-2019, 05:14 PM (This post was last modified: 07-31-2019 11:04 PM by Albert Chan.)
Post: #9
 Albert Chan Senior Member Posts: 1,992 Joined: Jul 2018
RE: (12C+) Bernoulli Number
Knowing the pattern of sk(n) = (Σi^k, k=0 to n-1) = n^(k+1)/(k+1) - n^k/2 + k/12 * n^(k-1) + 0 * x^(k-2) + ...,
we can get Bernoulli constants another way.

Example: to get upto B(6), s6(n) = n^7/7 - n^6/2 + n^5/2 + a*n^3 + b*n, where a, b are unknown

→ a*n^2 + b = s6(n)/n - n^4*(n^2/7 - (n-1)/2)

For n=1, eqn1 = a + b = 0/1 - 1*(1/7 - 0/2) = -1/7
For n=2, eqn2 = 4a + b = 1/2 - 16*(4/7 - 1/2) = -9/14

eqn2 - eqn1: 3a = -9/14 + 2/14 = -1/2 → a = -1/6

B(6) = b = -1/7 - a = 1/42
B(4) = a * 3 / $$\binom{6}{4}$$ = -1/30
B(2) = (1/2) * 5 / $$\binom{6}{2}$$ = 1/6
B(1) = (-1/2) * 6 / $$\binom{6}{1}$$ = -1/2
B(0) = (1/7) * 7 / $$\binom{6}{0}$$ = 1
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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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