From: Robin John Chapman Subject: Re: Summation Research Question Date: Wed, 03 Feb 1999 14:25:00 +1100 Newsgroups: sci.math Keywords: Bernoulli numbers Steve Schooler wrote: > > Request HELP on private math research. Am using induction to generate > formula for SUM: 1**k + 2**k + ... + n**k. Letting (r,s) denote > r!/[(s!)(r-s)!] and letting Sub(r) denote the rth coefficient > (subscripting), have determined that SUM = > > n**(k+1) * Sub(k+1) * [1/(k+1)] + > n**(k) * Sub(k) * (k,0) + > n**(k-1) * Sub(k-1) * (k,1) + > n**(k-2) * Sub(k-2) * (k,2) + > n**(k-3) * Sub(k-3) * (k,3) + > n**(k-4) * Sub(k-4) * (k,4) + > .. > n**(1) * Sub(1) * (k,k-1) > > where Sub(k+1) = 1, Sub(k) = 1/2, Sub(k-1) = 1/12, Sub(k-2) = 0, > Sub(k-3) = -1/120, Sub(k-4) = 0, ... > > For example, SUM 1**5 + 2**5 + ... + n**5 = > > n**6 * 1 * 1/6 + > n**5 * 1/2 * 1 + > n**4 * 1/12 * 5 + > n**2 * -1/120 * 10 > > SO FAR SO GOOD, but WHY do the coefficients (bernoulli #'s: 1, 1/2, > 1/12, 0, -1/120, 0, ...) follow the pattern: + 0 - 0 ...? > Can this pattern be established via Matrix theory or DIFFERENCES? There are lots of ways of determining the signs of the Bernoulli numbers. They are usually defined by t/(e^t - 1) = sum_{n=0}^infinity B_n t^n/n! (so what you have are the B_n/n! s). As t/(e^t - 1) + t/2 = (t/2)(e^t + 1)/(e^t -1) is an even function then B_1 = -1/2 and all other B_n for n odd vanish. After a bit of rearrangement we get (e^t + 1)/(e^t - 1) = 2 sum_{m=0}^infinfty B_{2m} t^{2m-1}/(2m)!. Replacing t by 2t gives (e^2t + 1)/(e^2t - 1) = 2 sum_{m=0}^infinfty B_{2m} 2^{2m-1} t^{2m-1}/(2m)!. Some cunning manipulation gives (1/2) tanh(t/2) = (e^t - 1)/(e^t + 1) = sum_{m=1}^infinity B_{2m} (4^m - 1) t^{2m-1}. Replacing t by it (i = sqrt(-1)) gives (1/2) tan(t/2) = sum_{m=1}^infinity (-1)^{m-1} B_{2m} (4^m-1) t^{2m-1}. We want to show that all the coefficients of this series are positive. Equivalently we must show that all the coefficients of odd powers in the power series for tan(t) are positive. We find the k-th coefficient by differentiating tan(t) k times, evaluating at zero and dividing by k!. But the serivative of tan is sec^2 and that of sec is sec tan. It's easily see that the k-th derivative of tan is a homogeneous polynomial of degree k+1 in tan and sec with non-negative coefficients, and that for k odd the coefficient of sec^{k+1} is positive. This shows that when evaluated at zero the k-th derivative of tan is positive for odd k and this completes the argument. -- Robin Chapman + "Going to the chemist in Department of Mathematics, DICS - Australia can be more Macquarie University + exciting than going to NSW 2109, Australia - a nightclub in Wales." rchapman@mpce.mq.edu.au + Howard Jacobson, http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Summation Research Question Date: 3 Feb 1999 01:31:12 -0500 Newsgroups: sci.math In article , Steve Schooler wrote: >Request HELP on private math research. Am using induction to generate >formula for SUM: 1**k + 2**k + ... + n**k. [formulas deleted] >WHY do the coefficients (bernoulli #'s: 1, 1/2, >1/12, 0, -1/120, 0, ...) follow the pattern: + 0 - 0 ...? >Can this pattern be established via Matrix theory or DIFFERENCES? > >Haven't found any number theory literature on point. REQUEST book >references, &/or math analysis. ... Thanks An excellent reference: Ronald L. Graham, Donald E. Knuth, Oren Patashnik: Concrete Mathematics, A Foundation for Computer Science Addison-Wesley Publishing Company 1994 ISBN 0-201-55802-5 Section 6.5 discusses Bernoulli Numbers, Section 7.4 has Special Generating Functions. An answer to your question (pattern of B_n) is obtained from the generating function and from its properties for purely imaginary parameter. We have G(z) = sum(n>=0) B_n * z^n/n! summed up: G(z) = z / (exp(z) - 1) , abs(z) < pi and after some algebra, G(i*t) = i*t/2 + (t/2) * cot(t/2) Now, if you expand (t/2) * cot(t/2) into partial fractions, and the fractions into power series, you will find that G(i*t) - i*t/2 is a power series in t^2 with positive coefficients. That will explain the alternating signs (since i^(2*n) = (-1)^n ). Hope it helps, ZVK(Slavek).