(10/31/2009, 02:40 PM)Gottfried Wrote: Check, for base x=1/2 I get AS(x) = 0.938253002500
Yes, with the average of the two series, I get AS(0.5) = 0.938253.
The way that I got the coefficients is slightly different than your method. I did this:
Let \( \mathbf{B} \) be a matrix defined by \( B_{jk} = \frac{1}{j!} \text{spow}_k^{(j)}(1) \), and let
\(
\begin{tabular}{rl}
f(x)
& = \sum_{k=1}^\infty f_k (x - 1)^k \\
& = \sum_{k=1}^\infty g_k ({}^{k}x) \\
F &= (f_0, f_1, f_2, ...)^T \\
G &= (g_0, g_1, g_2, ...)^T \\
\end{tabular}
\)
then
\( \mathbf{B}.F = G \)
so I thought, if we know G (1, -1, 1, -1, ...), then
\( F = \mathbf{B}^{-1}G \)
and when the matrix size is even I get the first series, and when the matrix size is odd, I get the second series.

