
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 17, 2013 2:50 AM


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 17001713 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 EulerMacLaurin >> 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 E24 > > So,  A  B  < 1/(10^23) . > > David Bernier >
 dracut:/# lvm vgcfgrestore File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh Please specify a *single* volume group to restore.

