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:
>> 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
>> 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 . 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
> S_r(a,b) = sum zeta(-r+c) * binomial(r,c) *((a+1)^r - b^r)
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
>  http://go.helms-net.de/math/binomial_new/
>  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 that
f(x) - f(x-1) = sqrt(x) for any x in [a, oo) ?
Ramanujan wrote a paper on sum of consecutive square roots:
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
Please specify a *single* volume group to restore.