```Date: Feb 18, 2013 3:38 AM
Author: David Bernier
Subject: Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)

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 1700-1713 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 Zeta-polynomials> 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 creation-rule for that> matrix to fractional row-indexes. 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 Bernoulli-polynomials 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.helms-net.de/math/binomial_new/> [2] http://go.helms-net.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(x-1) = 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 thatf(x) - f(x-1) = sqrt(x) for any x in [a, oo) ?Ramanujan wrote a paper on sum of consecutive square roots:http://oeis.org/A025224/internaldavid bernier-- dracut:/# lvm vgcfgrestoreFile descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh   Please specify a *single* volume group to restore.
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