
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 18, 2013 2:26 AM


Am 16.02.2013 07:42 schrieb David Bernier: > The Bernoulli numbers can be used to compute for example > 1^10 + 2^10 + ... + 1000^10 . > > Jakob Bernoulli wrote around 17001713 that he had computed > the sum of the 10th powers of the integers 1 through 1000, > with the result: > 91409924241424243424241924242500 > > in less than "one half of a quarter hour" ... > > Suppose we change the exponent from 10 to 1/2, so the sum > is then: > sqrt(1) + sqrt(2) + ... sqrt(1000). > > Or, more generally, > sqrt(1) + sqrt(2) + ... sqrt(N) , N some largish positive > integer. > > Can Bernoulli numbers or some generalization be used > to compute that efficiently and accurately? > I've done an exploration of the integrals of the Bernoulli polynomials which I called for convenience Zetapolynomials and which I studied as a matrix of coefficients, which I call "ZETA"matrix [2]. Each row r gives the coefficients for the sums of like powers with exponent r, so we get the polynomials for r=0,1,2,3,... in one aggregate of numbers. It is then natural to generalize the creationrule for that matrix to fractional rowindexes. However, this method gives then no more polynomials but series (which is not what you want, sorry...). That series have the form
infty S_r(a,b) = sum zeta(r+c) * binomial(r,c) *((a+1)^r  b^r) k=0
where the definition of the binomials is also generalized to fractional r (the cases, when r+c=1 must be handled by replacing zeta(1)/gamma(0) by +1 or 1, don't recall the required sign at the moment) It gives then the sum for the r'th powers from the bases a to b in steps by 1 and for the natural numbers r give the Bernoullipolynomials in the original form of Faulhaber. If you are happy with approximations like in your examples, this all will not of much help/inspiration though, I'm afraid..
Gottfried Helms
[1] http://go.helmsnet.de/math/binomial_new/ [2] http://go.helmsnet.de/math/binomial_new/04_3_SummingOfLikePowers.pdf

