```Date: Feb 18, 2013 12:26 AM
Author: David Bernier
Subject: Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)

On 02/17/2013 02:50 AM, David Bernier wrote:> 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) .I've extended the series above:? sum(X=1,M,sqrt(X))-sum(X=1,16, (M^(5/2 - X))*V2[X])%733 = -0.2078862249773545660173067253970493022262685312876725376? M%734 = 1000000expression in M:-----------------2/3 *M^(3/2)1/2 *M^(1/2)1/24   *M^(-1/2)     0-1/1920 *M^(-5/2)    01/9216 *M^(-9/2)    0-11/163840 *M^(-13/2)    065/786432  *M^(-17/2)    0-223193/1321205760  *M^(-21/2)    052003/100663296   *M^(-21/2)Then to approximate  the sum of the square roots, oneadds the myterious constantC = -0.2078862249773545660173067253970493...to the above rational function in M^(1/2).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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