---------------- Date: Sat, 24 Jan 1998 03:11-0500 Reply-To: rwg@NEWTON.macsyma.com Subject: Bernoulli recurrences To: math-fun@optima.CS.Arizona.EDU Message-Id: <"19980124081150.3.rwg@SWEATHOUSE"@SWEATHOUSE.macsyma.com> Bernoulli numbers and polynomials are often defined by a "full history recurrence" whose calculation involves all the predecessors down to B_0. At the other extreme is Ramanujan's *first* order recurrence, but it involves the infinite product over primes for a ratio of zetas. A "one sixth history recurrence" for the polynomials is [slightly simplified from original email] sum(binomial(n+2,n-6*k)*bernpoly(y,n-6*k)/(2*k+1),k,0,floor(n/6)) = (y^(n+2)+(y-1)^(n+2))/2-'realpart((y+(-1)^(2/3))^(n+2)) n floor(-) 6 ==== binom(n + 2, n - 6 k) B (y) \ n - 6 k > --------------------------------- / 2 k + 1 ==== k = 0 n + 2 n + 2 y + (y - 1) 2/3 n + 2 = --------------------- - Re((y + (- 1) ) ) 2 (for real y. Otherwise expand the Re first.) For y=-1/2, this give a neat recurrence for the Bernoulli numbers: j floor(-) 6 6 k - j + 1 ==== binomial(j + 2, 6 k + 2) B (2 - 1) \ j - 6 k
---------------------------------------------------- =
/ 2 k + 1 ==== k = 0 j/2 j/2 + 1 %pi j ((- 1) + 3 ) cos(-----) 2 -------------------------------- . j + 2 2 Can we do better than every 6th? ------------------------ The mystery was: Why aren't there more of these? Where are the rest? Answer: This was a fluke! What's usually on the rhs is a linear combination of Bernpolys(y + various roots of unity), but in this case, the Bernpolys (nearly) cancel in contiguous pairs due to the integer separations between 1^(k/6). The underlying identity is really 6*x^2*'sum(binomial(n,6*k+2)*x^(6*k)*bernpoly(y,n-6*k-2),k,0,floor(n/6)) = -(bernpoly(y+(-1)^(2/3)*x,n)+bernpoly(y-(-1)^(2/3)*x,n))/(-1)^(1/3)-(-1)^(1/3)*(bernpoly(y+(-1)^(1/3)*x,n)+bernpoly(y-(-1)^(1/3)*x,n))+bernpoly(y+x,n)+bernpoly(y-x,n) n floor(-) 6 ==== 2 \ 6 k 6 x > binomial(n, 6 k + 2) x B (y) / n - 6 k - 2 ==== k = 0 2/3 2/3 B (y + (- 1) x) + B (y - (- 1) x) n n = - --------------------------------------- + B (y + x) + B (y - x) 1/3 n n (- 1) 1/3 1/3 1/3 - (- 1) (B (y + (- 1) x) + B (y - (- 1) x)). n n The 2k+1 in the denominator was a red herring from a gratuitous integration. Similar identities with any desired sparseness can be had from multisecting the standard identity x^n*sum(binom(n,k)*bernpoly(y,k)/x^k,k,0,n) = bernpoly(y+x,n) n ==== \ n - k > binomial(n, k) x B (y) = B (y + x) / k n ==== k = 0 and linearly combining eadem(w*x) for w = various roots of unity. --rwg Actually, the Continuum Hypothesis is false. Between C and 2^aleph_0 lies the number of ways to spell "Chebyshev".