Applying the iteration methods to simpler functions

[andydude]

You are confusing Bell matrices and Carleman matrices again. The (generalized) Bell matrix of \( f(x) = cx+d \) is
\( \left[\begin{tabular}{ccc}
1 & d & d^2 \\
0 & c & 2cd \\
0 & 0 & c^2
\end{tabular}\right] \)
whereas the Carleman matrix (is
\( \left[\begin{tabular}{ccc}
1 & 0 & 0 \\
d & c & 0 \\
d^2 & 2cd & c^2
\end{tabular}\right] \)
all the other stuff is right, though, and interesting. I like your notation, but I'm having trouble following your series notation.

I think the example of \( f(x) = cx+d \) is an interesting one, and I intend on looking into it, but first for the purposes of illustration, your simplest example is the nicest, for example, addition \( f(x) = x+d \). Taking the Bell matrix:
\( \mathbf{B}_x[x + d] = \left[\begin{tabular}{ccc}
1 & d & d^2 & d^3 \\
0 & 1 & 2d & 3d^2 \\
0 & 0 & 1 & 3d \\
0 & 0 & 0 & 1
\end{tabular}\right] \)
and subtracting the identity matrix:
\( \mathbf{B}_x[x + d] - \mathbf{I}= \left[\begin{tabular}{ccc}
0 & d & d^2 & d^3 \\
0 & 0 & 2d & 3d^2 \\
0 & 0 & 0 & 3d \\
0 & 0 & 0 & 0
\end{tabular}\right] \)
gives a non-invertible matrix, so doing the choppy thing gives:
\( \mathbf{C}(\mathbf{B}_x[x + d] - \mathbf{I})\mathbf{D}= \left[\begin{tabular}{ccc}
1 & 0& 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{tabular}\right]\left[\begin{tabular}{ccc}
0 & d & d^2 & d^3 \\
0 & 0 & 2d & 3d^2 \\
0 & 0 & 0 & 3d \\
0 & 0 & 0 & 0
\end{tabular}\right]\left[\begin{tabular}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{tabular}\right] = \left[\begin{tabular}{ccc}
d & d^2 & d^3 \\
0 & 2d & 3d^2 \\
0 & 0 & 3d
\end{tabular}\right] \)
putting this in the matrix equation gives:
\( \left[\begin{tabular}{ccc}
d & d^2 & d^3 \\
0 & 2d & 3d^2 \\
0 & 0 & 3d
\end{tabular}\right]\left[\begin{tabular}{c}
g_1 \\
g_2 \\
g_3
\end{tabular}\right] == \left[\begin{tabular}{c}
1 \\
0 \\
0
\end{tabular}\right] \)
and since the matrix is invertible now, we can multiply both sides by its inverse:
\( \left[\begin{tabular}{c}
g_1 \\
g_2 \\
g_3
\end{tabular}\right] == \left[\begin{tabular}{ccc}
\frac{1}{d} & -\frac{1}{2} & \frac{d}{6} \\
0 & \frac{1}{2d} & -\frac{1}{2} \\
0 & 0 & \frac{1}{3d}
\end{tabular}\right]\left[\begin{tabular}{c}
1 \\
0 \\
0
\end{tabular}\right] = \left[\begin{tabular}{c}
1/d \\
0 \\
0
\end{tabular}\right] \)
This gives the natural Abel function \( A_f(x) = f^{*}(x) = x/d \) (choosing the constant term to be zero) which satisfies the equation \( \frac{x+d}{d} = \frac{x}{d} + 1 \).

I like that example I know you already talked about it, but I wanted to show a matrix-based way of doing the chopping process.

So in general, the matrix equation \( (\mathbf{C}(\mathbf{B}[f] - \mathbf{I})\mathbf{D}) \mathbf{a} = \mathbf{b} \) would give the coefficients \( \mathbf{a} \) of the natural Abel function of f.

Andrew Robbins