Date: Feb 17, 2013 2:50 AM
Author: David Bernier
Subject: Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
On 02/17/2013 02:38 AM, David Bernier wrote:

> 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.

oops. There's also an N^(-1/2) term.

> 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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