
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 18, 2013 3:38 AM


On 02/18/2013 02:26 AM, Gottfried Helms wrote: > > > 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 >
Are c and k related?
> 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.
Maybe I should look up Faulhaber's formula.
> 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 [...]
The summatory polynomial for the k'th powers of 1, 2, ... n, P(x), has the property that P(x)  P(x1) = x^k, at least for positive integers x. I assume k is a positive integer.
So, does there exist a continuous f: [a, oo) such that f(x)  f(x1) = sqrt(x) for any x in [a, oo) ?
Ramanujan wrote a paper on sum of consecutive square roots:
http://oeis.org/A025224/internal
david bernier
 dracut:/# lvm vgcfgrestore File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh Please specify a *single* volume group to restore.

