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 = 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.