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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