|
|
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 18, 2013 2:34 PM
|
|
Am 18.02.2013 18:40 schrieb David Bernier:
> On 02/18/2013 07:17 AM, Gottfried Helms wrote:
>> I've to correct some obvious typing errors:
>>
>> Am 18.02.2013 09:38 schrieb David Bernier:
>>> On 02/18/2013 02:26 AM, Gottfried Helms wrote:
>>>>
>>>>
>>>
>>> The summatory polynomial for the k'th powers of 1, 2, ... n,
>>> P(x), has the property that P(x) - P(x-1) = x^k,
>>> at least for positive integers x.
>>> I assume k is a positive integer.
>>>
>>> So, does there exist a continuous f: [a, oo) such that
>>> f(x) - f(x-1) = sqrt(x) for any x in [a, oo) ?
>>>
>>> Ramanujan wrote a paper on sum of consecutive square roots:
>>>
>>> http://oeis.org/A025224/internal
>>>
>>> david bernier
>>>
>> Hmm, my usual tools give only much diverging series for which my
>> procedures of divergent summation do not work well if I use -say-
>> only 64 or 128 terms for the power series.
>> But I've now converted the problem into one which employs the rationale:
>>
>> f(x) = sqrt(1+x^2)
>> such that f(1) -> sqrt(2) f(f(1)) = sqrt(3) and so on such that we
>> can write
>> S(a,b) = sqrt(a) + f(sqrt(a)) + f(f(sqrt(a))) + ... + f...f(sqrt(a))
>> = f°0(sqrt(a)) + f°1(sqrt(a)) + f°2(sqrt(a)) + ... + f°d(sqrt(a))
>>
>> where d=b-a and the circle at f°k(x) indicates the k'th iterate.
>>
>> After that the problem can be attacked by the concept of Carleman-matrixes
>> and their powers. Let F be th carleman-matrix for the function f(x),
>> then let G = I - F then, if we could invert G in the sense that
>>
>> M = G^-1 = I + F + F^2 + F^3 + F^4 + ... (see Neumann-series, wikipedia)
>>
>> then we had a solution in terms of a formal power series for
>>
>> m(x) = x + f(x) + f°2(x) + f°3(x) + ....
>>
>> and m(sqrt(a)) - m(sqrt(b)) would give the required sum-of-sqrt from
>> sqrt(a) to sqrt(b) in 1-steps progression from its argument a.
>>
>> However, G has a zero in the top-left entry(and the whole first column)
>> and cannot be inverted.
>> Now there is a proposal which I've seen a couple of times that we invert
>> simply the upper square submatrix after removing the first (empty) column
>> in G, let's call this H, and this gives often an -at least- usable
>> approximate solution, if not arbitrarily exact.
>>
>> But again - trying this using Pari/GP leads to nonconclusive results; the
>> inversion process seems to run in divergent series again.
>>
>> Here we have now the possibility to LDU-factorize H into triangular
>> factors, which each can be inverted, so we have formally
>>
>> H = L * D * U
>> M = H^-1 = U^-1 * (D^-1 * L^-1)
>>
>> We cannot perform that multiplication due to still strong divergences
>> when evaluating the row-column-dotproducts (which is only making explicite
>> the helpless Pari/GP-attempts for inverting H)
>>
>> But we can use the upper triangular U^-1 in the sense of a "change of base"-
>> operation. Our goal is to have M such that we can write
>>
>> (V(sqrt(a)) - V(sqrt(b))) * M[,2] = s(a,b+1) = sqrt(a)+sqrt(a+1)+...+sqrt(b)
>>
>> where V(x) means a vandermondevector V(x) = [1,x,x^2,x^3,x^4,....]
>>
>> But now we can proceed from the formal formula
>>
>> (V(sqrt(a)) - V(sqrt(b))) * M[,2]
>> = (V(sqrt(a)) - V(sqrt(a))) * U^-1 * ( D^-1 * L^-1)[,2]
>>
>> and can compute
>>
>> X(sqrt(a),sqrt(a)) = (V(sqrt(a)) - V(sqrt(a))) * U^-1
>>
>> exactly to any truncation size because U^-1 is upper triangular and thus
>> column-finite.
>>
>> Then we can as well do
>>
>> Q = ( D^-1 * L^-1)[,2]
>>
>> which is -besides the truncation to finite size- an exact expression.
>>
>> Still we have, that the dot-product
>>
>> s(a,b) = X(sqrt(a),sqrt(a)) * Q
>>
>> is divergent, but now it seems, that we can apply Euler-summation
>> for the evaluation of the divergent dot-product.
>> The results are nice approximations for the first couple of
>> sums s(1,4), s(2,4) and some more. s(1,9) requires Euler-summation
>> of some higher order such that I get
>>
>> s(1,9) ~ 16.3060
>> (where 16.3060005260 is exact to the first 11 digits)
>>
>> or
>> s(5,10)= sqrt(5)+sqrt(6)+ ... + sqrt(10)
>> ~ 16.32199
>> where 16.3220138163 is exact to the first 11 digits)
>>
>>
>> Don't know how to do better at the moment; surely if the composition
>> of the entries in the matrices were better known to me, one could
>> make better, possibly even analytical or at least less diverget, expressions.
>>
>> (But I've not enough time to step into this deeper, I'm afraid)
>
> Hi Gottfried,
>
> The exponent for the natural numbers that I'm most interested
> in is a = -s , with s = 1/2 + i*t , t real in [0, oo) as:
>
> 1^a + 2^a + ... + (N-1)^a
> or
> 1 + 1/2^s + ... + 1/(N-1)^s because in Euler-MacLaurin
> summation for zeta(s),
> 1/1^s +1/2^s + ... + 1/(N-1)^s is computed, for some N > |t| .
>
> The other term is one a sum involving the Bernoulli numbers, and
> and also (1/2) N^(-s) + N^(1-s)/(s-1) [ Edwards].
>
> But a "shortcut" to evaluate:
> 1/1^s +1/2^s + ... + 1/(N-1)^s to high-precision would
> save time in the Euler-MacLaurin summation for zeta(s),
> so this connects to algorithms to compute zeta(s) to
> high-precision, a much-studied topic.
>
> It seems pretty hard too ...
>
> David
>
>
>
Hmm, after reading that I remember I've tried to accelerate
the computation of zeta/eta(s) with arguments in that area
by Euler-summation of adapted (namely complex) order. It seemed
to me, that this a veritable "poor-mans" acceleration for the
actual computations.
Actually it comes out to be
n 1
eta(s) ~ sum c(s,o,k) * --- *(-1)^k
k=1 k^s
for some sufficient large (finite) n and the c(s,o,k) coefficients
gotten by the formulae of the Euler-summation of an optimal
order o, best adapted for the argument s.
And then zeta(s) = eta(s)/(1-2*2^-s)
I need not high n, say n~100, for very usable approximations for
my needs.
I have made the influence of the orders o for the summation process
visible in a excel-sheet, where the o's can continuously be changed
and then the trajectory of the resulting partial sums becomes "longer"
or "shorter" (needing greater or smaller n to come farther away or nearer
to the final result) depending on o. Very impressive in my view!
If this is interesting I could upload that excel-sheet to my webspace.
(It's of "garage"-type, just for the experimenter, no nice brushed appearance)
Gottfried Helms
|
|
