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 Tetra-series andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/31/2009, 11:01 AM (This post was last modified: 10/31/2009, 11:07 AM by andydude.) Quote: $AS(x+1) = 1 + x^2 + \frac{1}{2}x^3 + \frac{4}{3}x^4 + \frac{19}{12}x^5 + \cdots$ Wait... the series above only works for even approximants, I also got: $AS(x+1) = -x - x^3 - x^4 - \frac{29}{12}x^5 - \frac{13}{4}x^6 + \cdots$ for odd approximants, and they can't both be right... what am I doing wrong? Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 10/31/2009, 12:58 PM (10/31/2009, 10:38 AM)andydude Wrote: This is probably an old result. For the function $AS(x) = \sum_{k=0}^{\infty} (-1)^k ({}^{k}x)$ I found using Carleman matrices that $AS(x+1) = 1 + x^2 + \frac{1}{2}x^3 + \frac{4}{3}x^4 + \frac{19}{12}x^5 + \cdots$ is this related to the series above? Hmmm, don't recognize this coefficients. Would you mind to show more of your computations? Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 10/31/2009, 01:25 PM (This post was last modified: 10/31/2009, 01:32 PM by Gottfried.) (10/31/2009, 11:01 AM)andydude Wrote: Quote: $AS(x+1) = 1 + x^2 + \frac{1}{2}x^3 + \frac{4}{3}x^4 + \frac{19}{12}x^5 + \cdots$ Wait... the series above only works for even approximants, I also got: $AS(x+1) = -x - x^3 - x^4 - \frac{29}{12}x^5 - \frac{13}{4}x^6 + \cdots$ for odd approximants, and they can't both be right... what am I doing wrong? Alternating summing of coefficients requires averaging if the sequence to be summed is not converging fast enough. For example, for the coefficients at the linear term the alternating sum of iterates give 1x-1x+1x-1x... for the coefficient in AS and to get 0.5 x here in the limit one must employ cesaro or Euler-summation. Note, that the same problem strengthens if the sequence of coefficients at some x have also a growthrate (is divergent) Then for each coefficient you need an appropriate order for Cesaro-/Eulersum. If you use powers of a *triangular* Carleman-matrix X for the iterates, then you can try the geometric series for matrices AS = I - X + X^2 - X^3 + ... - ... = (I + X)^-1 in many cases and use the coefficients of the second row/column. If X is not triangular you have a weak chance, that you get a usable approximate, since the eigenvalues of AS may behave nicer than that of X, because for an eigenvalue x>1 the eigenvalue in AS is 1/(1+x) which is smaller than 1, and from a set of increasing eigenvalues (all >=1) as well from one of decreasing eigenvalues (0 infinity, then by definition, AS(x) = S(x,0) - S(x,1) + S(x,2) - ... + // Euler-sum Since the coefficients of any height converge to that of the S(x)-series I compute the difference D(x,h) = S(x,h) - S(x) and rewrite AS(x) = D(x,0) - D(x,1) + D(x,2) ... + aeta(0)*S(x) where aeta(0) is the alternating zeta-series zeta(0) meaning aeta(0) = 1-1+1-1+1-... = 1/2 Because the coefficients with index k

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