(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<all<=1) you get a set of eigenvalues in AS (0<all<1), which makes the associated powerseries for AS(x) nice-behaved.
Gottfried Helms, Kassel

