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