[math-fun] Re: Bernoulli numbers
Dear William Gosper,
I'm looking for a reference for the identity involving the Bernoulli numbers. Please look at (34) at http://mathworld.wolfram.com/BernoulliNumber.html It is said that it is yours formula, but a reference is not given. This identity is equivalent to an (alternating) zeta relation. It's interesting if you proved your formula using this re lation, or if you used only properties of the Bernoulli numbers. Thanks in advance, Sergey. Dear Sergey, thank you for your inquiry. I can't locate that identity in the Math-Fun archive, where I usually announce things. In fact, I often flame there against Bernoulli number identities, simply because they can almost always be generalized to Bernoulli polynomials in new variables. E.g., (34) is the special case x=y=1/2 of n ==== \ n > ( ) B (x) B (y) = / i n - i i ==== i = 0 n (y + x - 1) B (y + x - 1) - (n - 1) B (y + x - 1), n - 1 n which follows immediately from multiplying the generating functions, so I don't know why I'd've bothered with such a routine announcement. Similarly, the preceding (32) and (33) can be greatly generalized by multisecting the expansion (c150) (sum(binomial(n,k)*t^k*bernpoly(x,k),k,0,n),%% = hypersimp(%%)) n ==== \ k n 1 (d150) > binomial(n, k) t bernpoly(x, k) = t bernpoly(x + -, n) / t ==== k = 0 E.g., hexasecting gives ==== \ (d180) 6 > binomial(n, 6 k) bernpoly(x, 6 k) = / ==== k = 0 %i %pi n -------- 2 7 %pi n %i sqrt(3) 1 2 (realpart(2 %e cos(-------) bernpoly(x + ---------- + -, n)) 6 2 2 2 %i %pi n - ---------- 3 %i sqrt(3) 1 n - 1 + realpart(- n %e (x + ---------- - -) )) + bernpoly(x + 1, n) 2 2 n + (- 1) bernpoly(x - 1, n) plus the five others for 6k+1, 6k+2,... . --Bill Gosper
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R. William Gosper