Taylor polynomial. System of equations for the coefficients.

[marraco]

P*A = A*Bb

I think that we are speaking of different things.

Obviously, there should be a way to demonstrate the equivalence of both, because they are trying to solve the same problem; looking for the same solution.

But as I understand, the Carleman matrix A only contains powers of a_i coefficients, yet if you look at the red side, it cannot be written as a matrix product A*Bb, because it needs to have products of a_i coefficients (like \( a_1^3.a_3^2.a_5^8.a_... \)). Maybe it is a power of A.Bb, or something like A^Bb?

The Pascal matrix on the blue side is the exponential of a much simpler matrix

\(
\exp
\left (
\left [
\begin{matrix}
. & 1 & . & . & . & . & . \\
. & . & 2 & . & . & . & . \\
. & . & . & 3 & . & . & . \\
. & . & . & . & 4 & . & . \\
. & . & . & . & . & 5 & . \\
. & . & . & . & . & . & 6 \\
. & . & . & . & . & . & .
\end{matrix}
\right ]
\right )
=
\left [
\begin{matrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
. & 1 & 2 & 3 & 4 & 5 & 6 \\
. & . & 1 & 3 & 6 & 10 & 15 \\
. & . & . & 1 & 4 & 10 & 20 \\
. & . & . & . & 1 & 5 & 15 \\
. & . & . & . & . & 1 & 6 \\
. & . & . & . & . & . & 1
\end{matrix}
\right ] \)

Maybe the equation can be greatly simplified by taking a logarithm of both sides.