The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Replies: 9   Last Post: Feb 18, 2013 2:34 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Gottfried Helms

Posts: 1,926
Registered: 12/6/04
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted: Feb 18, 2013 2:26 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

S_r(a,b) = sum zeta(-r+c) * binomial(r,c) *((a+1)^r - b^r)

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.
If you are happy with approximations like in your examples,
this all will not of much help/inspiration though, I'm afraid..

Gottfried Helms


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.