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 that

f(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/internal

david bernier

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