```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 ofexcellent 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 differenceof the sum of square roots of integers up to N anda 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-24So, | A - B | < 1/(10^23) .David Bernier-- dracut:/# lvm vgcfgrestoreFile descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh   Please specify a *single* volume group to restore.
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