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