So now for more examples, well, how about the last example I gave, the exponential factorial. The function I gave was defined as: \( E(z, n) = (n^z, n+1) \) using tuple-notation as opposed to matrix-notation.
Whatever happens with this tensor power series stuff, it should produce the known power series for a function. So for the 2-variable case (actually, this only applies to one component), the power series should be the same as:
\(
\begin{tabular}{rlll}
E(X) & =
E(0, 1) &
+ \left[\frac{\partial E}{\partial n}\right]_{(z, n)=(0, 1)} (n-1) &
+ \left[\frac{\partial^2 E}{\partial n^2}\right]_{(z, n)=(0, 1)} \frac{(n-1)^2}{2!} \\ &
+ \left[\frac{\partial E}{\partial z}\right]_{(z, n)=(0, 1)} z &
+ \left[\frac{\partial^2 E}{\partial z n}\right]_{(z, n)=(0, 1)} z (n-1) &
+ \left[\frac{\partial^3 E}{\partial z n^2}\right]_{(z, n)=(0, 1)} z \frac{(n-1)^2}{2!} \\ &
+ \left[\frac{\partial^2 E}{\partial z^2}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} &
+ \left[\frac{\partial^3 E}{\partial z^2 n}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} (n-1) &
+ \left[\frac{\partial^4 E}{\partial z^2 n^2}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} \frac{(n-1)^2}{2!} \\ &
+ \cdots
\end{tabular}
\)
First thing to notice is that we can't make a series about (0, 0), because \( 0^0 \) is indeterminate, and the next derivative involves \( 1/0 \) which is infinite. So what we really need is a series about (0, 1) since \( E(0, 1) = (1^0, 1+1) = (1, 2) \) which is finite, but unfortunately, this is not a fixed point, so we can't use regular iteration. That's OK, though, because we're just talking about power series. Using the standard term for Taylor series \( (X - X_0)^n \) where each of the X's are vectors, we find that \( (z, n) - (0, 1) = (z, n-1) \) so the tensor power series will look like:
\(
E(z, n) = (1, 2) + \sum_{k=1}^{\infty} \frac{1}{k!}
\left([\nabla_{\otimes k}E(X)]_{X=(0, 1)} \right)*(z, n-1)^{\otimes k}
\)
where \( (*) \) actually represents k tensor contractions.
The second thing to notice is that using Taylor-Puiseux conversion (from my Abel function post), We can actually find the power series for the top part of the vector function \( n^z \) about (0, 1), which is:
\(
n^z = \sum_{k=0}^{\infty} \frac{(n-1)^k}{k!} \sum_{j=0}^{k} \left[ {k \atop j}\right] z^j
\)
So what do all of the \( [\nabla_{\otimes k}E(X)] \) look like? Well, a simple way to look at it is that each term in the tensor power series is a vector that contains all polynomials of degree k. So, for example, the 2nd term will be a vector that holds all coefficients of \( z^2 \), \( zn \), and \( n^2 \) since these are all polynomials of degree 2. The 3rd term will be a vector that holds all coefficients of \( z^3 \), \( z^2n \), \( zn^2 \), and \( n^3 \), and so on.
Here is the Mathematica code for calculating the multiple-gradients:
and here is the Mathematica code for the tensor power series of the exponential factorial:
To show you exactly how Mathematica evaluates these expressions:
and the next term:
and the next term:
Just to wrap up, and take a step back: although this may seem a bit messy, if this is the only way to convert current (univariate) analytic iteration theory to apply to multi-dimensional dynamical systems, then I think it is worth it.
Besides... we have computers now
Andrew Robbins
Whatever happens with this tensor power series stuff, it should produce the known power series for a function. So for the 2-variable case (actually, this only applies to one component), the power series should be the same as:
\(
\begin{tabular}{rlll}
E(X) & =
E(0, 1) &
+ \left[\frac{\partial E}{\partial n}\right]_{(z, n)=(0, 1)} (n-1) &
+ \left[\frac{\partial^2 E}{\partial n^2}\right]_{(z, n)=(0, 1)} \frac{(n-1)^2}{2!} \\ &
+ \left[\frac{\partial E}{\partial z}\right]_{(z, n)=(0, 1)} z &
+ \left[\frac{\partial^2 E}{\partial z n}\right]_{(z, n)=(0, 1)} z (n-1) &
+ \left[\frac{\partial^3 E}{\partial z n^2}\right]_{(z, n)=(0, 1)} z \frac{(n-1)^2}{2!} \\ &
+ \left[\frac{\partial^2 E}{\partial z^2}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} &
+ \left[\frac{\partial^3 E}{\partial z^2 n}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} (n-1) &
+ \left[\frac{\partial^4 E}{\partial z^2 n^2}\right]_{(z, n)=(0, 1)} \frac{z^2}{2!} \frac{(n-1)^2}{2!} \\ &
+ \cdots
\end{tabular}
\)
First thing to notice is that we can't make a series about (0, 0), because \( 0^0 \) is indeterminate, and the next derivative involves \( 1/0 \) which is infinite. So what we really need is a series about (0, 1) since \( E(0, 1) = (1^0, 1+1) = (1, 2) \) which is finite, but unfortunately, this is not a fixed point, so we can't use regular iteration. That's OK, though, because we're just talking about power series. Using the standard term for Taylor series \( (X - X_0)^n \) where each of the X's are vectors, we find that \( (z, n) - (0, 1) = (z, n-1) \) so the tensor power series will look like:
\(
E(z, n) = (1, 2) + \sum_{k=1}^{\infty} \frac{1}{k!}
\left([\nabla_{\otimes k}E(X)]_{X=(0, 1)} \right)*(z, n-1)^{\otimes k}
\)
where \( (*) \) actually represents k tensor contractions.
The second thing to notice is that using Taylor-Puiseux conversion (from my Abel function post), We can actually find the power series for the top part of the vector function \( n^z \) about (0, 1), which is:
\(
n^z = \sum_{k=0}^{\infty} \frac{(n-1)^k}{k!} \sum_{j=0}^{k} \left[ {k \atop j}\right] z^j
\)
So what do all of the \( [\nabla_{\otimes k}E(X)] \) look like? Well, a simple way to look at it is that each term in the tensor power series is a vector that contains all polynomials of degree k. So, for example, the 2nd term will be a vector that holds all coefficients of \( z^2 \), \( zn \), and \( n^2 \) since these are all polynomials of degree 2. The 3rd term will be a vector that holds all coefficients of \( z^3 \), \( z^2n \), \( zn^2 \), and \( n^3 \), and so on.
Here is the Mathematica code for calculating the multiple-gradients:
Code:
GradientD[expr_List, vars_List, 0] := expr;
GradientD[expr_List, vars_List, 1] := GradientD[expr, vars];
GradientD[expr_List, vars_List, n_] :=
GradientD[GradientD[expr, vars], vars, n - 1];
GradientD[expr_List, vars_List] :=
Map[Function[{f}, Map[Function[{x}, D[f, x]], vars]], expr];Code:
{1, 2} +
{{0, 0},
{0, 1}}.{z, n-1} +
{{{0, 1}, {1, 0}},
{{0, 0}, {0, 0}}}.{z, n-1}.{z, n-1}/2! +
{{{{0, 0}, {0, -1}},
{{0, -1}, {-1, 0}}},
{{{0, 0}, {0, 0}},
{{0, 0}, {0, 0}}}}.{z, n-1}.{z, n-1}.{z, n-1}/3! +
{{{{{0, 0}, {0, 2}},
{{0, 2}, {2, 2}}},
{{{0, 2}, {2, 2}},
{{2, 2}, {2, 0}}}},
{{{{0, 0}, {0, 0}},
{{0, 0}, {0, 0}}},
{{{0, 0}, {0, 0}},
{{0, 0}, {0, 0}}}}}.{z, n-1}.{z, n-1}.{z, n-1}.{z, n-1}/4!To show you exactly how Mathematica evaluates these expressions:
Code:
{{0, 0}, {0, 1}}.{z, n-1}
= {0, n - 1}Code:
{{{0, 1}, {1, 0}}, {{0, 0}, {0, 0}}}.{z, n-1}.{z, n-1}/2!
= {{n - 1, z}, {0, 0}}.{z, n-1}/2!
= {2 (n - 1) z, 0}/2!
= {(n - 1) z, 0}Code:
{{{{0,0},{0,-1}},{{0,-1},{-1,0}}},{{{0,0},{0,0}},{{0,0},{0,0}}}}.{z, n-1}.{z, n-1}.{z, n-1}/3!
= {{{0, 1 - n}, {1 - n, -z}}, {{0, 0}, {0, 0}}}.{z, n-1}.{z, n-1}/3!
= {{-(n - 1)^2, -2 (n - 1) z}, {0, 0}}.{z, n-1}/3!
= {-3 (n - 1)^2 z, 0}/3!
= {-(n - 1)^2 z/2, 0}Just to wrap up, and take a step back: although this may seem a bit messy, if this is the only way to convert current (univariate) analytic iteration theory to apply to multi-dimensional dynamical systems, then I think it is worth it.
Besides... we have computers now
Andrew Robbins
