Tensor power series

[andydude]

Before we do that though, there are still a few things I would like to clear up. First notation, lets start using Einstein's summation convention (because we need it now), and let

\( \nabla_{\otimes n}F(0)x^{\otimes n}
= \left(\prod_i^n {\nabla}_{(k_i)}\right) \otimes F(0)
\otimes \left(\prod_i^n x^{(k_i)} \right) \)

which allows expressing the power series as:

\( F(X)
= \sum_{n=0}^{\infty} \frac{1}{n!} \nabla_{\otimes n}F(0)X^{\otimes n}
\)

and the Bell "matrix" as: \( B[F] = \nabla_{\otimes j}(F^{\otimes k})(0) \)
and the Vandermonde "vector" as: \( V(X) = X^{\otimes k} \)

Secondly, we need an example, like the Mandelbrot set function:
\( M\left[\begin{tabular}{c}
z \\ c
\end{tabular}\right]
=
\left[\begin{tabular}{c}
z^2 + c \\ c
\end{tabular}\right] \)
we can see that the value at zero is zero:
\( M\left[\begin{tabular}{c}
0 \\ 0
\end{tabular}\right]
=
\left[\begin{tabular}{c}
0 \\ 0
\end{tabular}\right] \)
which means we have a fixed point. Also, the first derivative is:
\( M'\left[\begin{tabular}{c}
0 \\ 0
\end{tabular}\right]
=
\left[\begin{tabular}{cc}
2z & 1 \\
0 & 1 \\
\end{tabular}\right]_{(z,c)=(0,0)}
=
\left[\begin{tabular}{cc}
0 & 1 \\
0 & 1 \\
\end{tabular}\right] \)
and the second derivative is:
\( M''\left[\begin{tabular}{c}
0 \\ 0
\end{tabular}\right]
=
\left[\begin{tabular}{cc}
[2\ 0] & [0\ 0] \\
[0\ 0] & [0\ 0] \\
\end{tabular}\right] \)
and from this it should be clear that the third derivative is zero. Now we can form a power series for this function as follows:
\(
M(X) = \left[\begin{tabular}{cc}
0 & 1 \\
0 & 1 \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ c
\end{tabular}\right]
+
\frac{1}{2!}
\left[\begin{tabular}{cc}
[2\ 0] & [0\ 0] \\
[0\ 0] & [0\ 0] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ c
\end{tabular}\right]^{\otimes 2}
\)

Another example we could try is the logistic map:
\( L\left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]
=
\left[\begin{tabular}{c}
r z (1-z) \\ r
\end{tabular}\right] \)
which has the power series expansion:
\(
L(X) = \left[\begin{tabular}{cc}
0 & 0 \\
0 & 1 \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]
+
\frac{1}{2!}
\left[\begin{tabular}{cc}
[0\ 0] & [1\ 0] \\
[1\ 0] & [0\ 0] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]^{\otimes 2}
+ \frac{1}{3!}
\left[\begin{tabular}{cc}
\left[\begin{tabular}{cc}
0 & 0 \\ -2 & 0
\end{tabular}\right] &
\left[\begin{tabular}{cc}
-2 & 0 \\ 0 & 0
\end{tabular}\right] \\
\left[\begin{tabular}{cc}
-2 & 0 \\ 0 & 0
\end{tabular}\right] &
\left[\begin{tabular}{cc}
0 & 0 \\ 0 & 0
\end{tabular}\right] \\
\end{tabular}\right] \left[\begin{tabular}{c}
z \\ r
\end{tabular}\right]^{\otimes 3}
\)

Now for the fun part. This kind of framework would be general enough to express the exponential factorial as an iterated function! This could be expressed as:
\( E\left[\begin{tabular}{c}
z \\ n
\end{tabular}\right]
=
\left[\begin{tabular}{c}
n^z \\ n + 1
\end{tabular}\right]
\)
and one thing I am curious about is whether regular iteration methods will work on this new power series ring, and whether they could help define the exponential factorial over the real numbers, as opposed to the integers.

Gottfried also brought up a good point. The Bell matrix and Vandermonde vector should be defined first, and well understood, before other iterational things. But this would cause much confusion, because even the Vandermonde vector (V) would not even be a tensor! This is because \( V_1 \) would be a (1,0)-tensor, and \( V_2 \) would be a (2,0)-tensor, and \( V_3 \) would be a (3,0)-tensor, which means whatever container you use to hold these (V) would not have a single rank (tensors must have a rank) and this would have all ranks... very confusing... I must think about this.

Andrew Robbins

PS. I think I forgot a factorial somewhere...
PPS. [update]I fixed the factorials[/update]