andydude Wrote:\( \left[\begin{tabular}{c}
g_1 \\
g_2 \\
g_3
\end{tabular}\right] == \left[\begin{tabular}{ccc}
\frac{1}{d} & -\frac{1}{2} & \frac{d}{6} \\
0 & \frac{1}{2d} & -\frac{1}{2} \\
0 & 0 & \frac{1}{3d}
\end{tabular}\right] \)
Hmm, nice coincidence here. Did you notice, that the fractional entries are just the bernoulli-numbers, resp binomially weighted bernoulli-numbers? It would be more obvious, if your matrix were of bigger size (but I don't know, which iterative function that required). When I was fiddling with the pascal-matrix last year I found the inverse of the shifted pascal-matrix (where -before shifting- the diagonal is removed) is just a matrix containing the bernoulli-numbers as it was found by Faulhaber and J Bernoulli, giving coefficients (with a little modification) of bernoulli-polynomials. I found, that it is part of the eigensystem for the *column-signed* pascal-matrix. Perhaps you like to have a look at my according small treatise at p-matrix
Things converge...
Gottfried
Gottfried Helms, Kassel