
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 18, 2013 12:26 AM


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 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) .
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 = 1000000
expression in M:  2/3 *M^(3/2) 1/2 *M^(1/2) 1/24 *M^(1/2) 0 1/1920 *M^(5/2) 0 1/9216 *M^(9/2) 0 11/163840 *M^(13/2) 0 65/786432 *M^(17/2) 0 223193/1321205760 *M^(21/2) 0 52003/100663296 *M^(21/2)
Then to approximate the sum of the square roots, one adds the myterious constant
C = 0.2078862249773545660173067253970493...
to the above rational function in M^(1/2).
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.

