Date: Feb 17, 2013 2:38 AM
Author: David Bernier
Subject: Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)

On 02/16/2013 01:42 AM, David Bernier wrote:
> 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?
>
> My first thought would be that the Euler-MacLaurin
> summation method might be applicable.
>
> Above, if k^a is the k'th term, a = 1/2 .

[...]

Numerical experiments suggest a pattern of
excellent approximations.

There's a series involving N^(3/2), N^(1/2),
N^(-5/2), N^(-9/2) and a constant term C.

To get rid of the unkwown C, I take the difference
of the sum of square roots of integers up to N and
a smaller number N' .

For example, N = 2000, N' = 1000:

Below, A is in fact sum_{k=1001 ... 2000} sqrt(k) :

A = (sum(X=1,2000,sqrt(X)) - sum(X=1,1000,sqrt(X)));

Below, B is the approximation broken over 5 lines:

B= (2/3)*(2000^( 1.5)-1000^( 1.5))\
+(1/2)*(2000^( 0.5)-1000^( 0.5))\
+(1/24)*(2000^(-0.5)-1000^(-0.5))\
+(-1/1920)*(2000^(-2.5)-1000^(-2.5))\
+(1/9216)*(2000^(-4.5)-1000^(-4.5));


? A - B
%280 = 2.09965132898428157559493347219264943224 E-24

So, | A - B | < 1/(10^23) .

David Bernier

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