Tensor power series

[andydude]

At first, I thought you wouldn't need the factorials you usually need in normal power series, because all the coefficients in F' and F'' were less than the factorial, but then I realized that there were multiple coefficients in each tensor. So taking the logistic map as an example again:
\(
\left[\begin{tabular}{cc}
[0\ 0] & [1\ 0] \\
[1\ 0] & [0\ 0] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]^{\otimes 2}
= \left[\begin{tabular}{c}
zr + rz \\ 0
\end{tabular}\right]
= \left[\begin{tabular}{c}
2rz \\ 0
\end{tabular}\right]
\)
so
\(
\frac{1}{2!}
\left[\begin{tabular}{cc}
[0\ 0] & [1\ 0] \\
[1\ 0] & [0\ 0] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]^{\otimes 2}
= \left[\begin{tabular}{c}
rz \\ 0
\end{tabular}\right]
\)
as it should, and doing the same thing for the third term:
\(
\frac{1}{3!}
\left[\begin{tabular}{cc}
\left[\begin{tabular}{cc}
0 & 0 \\ -2 & 0
\end{tabular}\right] &
\left[\begin{tabular}{cc}
-2 & 0 \\ 0 & 0
\end{tabular}\right] \\
\left[\begin{tabular}{cc}
-2 & 0 \\ 0 & 0
\end{tabular}\right] &
\left[\begin{tabular}{cc}
0 & 0 \\ 0 & 0
\end{tabular}\right] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]^{\otimes 3}
= \frac{1}{3!}\left[\begin{tabular}{c}
-2z^2r -2zrz -2rz^2 \\ 0
\end{tabular}\right]
= \left[\begin{tabular}{c}
-rz^2 \\ 0
\end{tabular}\right]
\)
as it should.

Andrew Robbins