
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted:
Feb 18, 2013 12:40 PM


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(x1) = 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(x1) = 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=ba and the circle at f°k(x) indicates the k'th iterate. > > After that the problem can be attacked by the concept of Carlemanmatrixes > and their powers. Let F be th carlemanmatrix 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 Neumannseries, 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 sumofsqrt from > sqrt(a) to sqrt(b) in 1steps progression from its argument a. > > However, G has a zero in the topleft 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 LDUfactorize 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 rowcolumndotproducts (which is only making explicite > the helpless Pari/GPattempts 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 > columnfinite. > > 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 dotproduct > > s(a,b) = X(sqrt(a),sqrt(a)) * Q > > is divergent, but now it seems, that we can apply Eulersummation > for the evaluation of the divergent dotproduct. > The results are nice approximations for the first couple of > sums s(1,4), s(2,4) and some more. s(1,9) requires Eulersummation > 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 + ... + (N1)^a or 1 + 1/2^s + ... + 1/(N1)^s because in EulerMacLaurin summation for zeta(s), 1/1^s +1/2^s + ... + 1/(N1)^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^(1s)/(s1) [ Edwards].
But a "shortcut" to evaluate: 1/1^s +1/2^s + ... + 1/(N1)^s to highprecision would save time in the EulerMacLaurin summation for zeta(s), so this connects to algorithms to compute zeta(s) to highprecision, a muchstudied topic.
It seems pretty hard too ...
David
 dracut:/# lvm vgcfgrestore File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID 993: sh Please specify a *single* volume group to restore.

