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 Tensor power series andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/22/2008, 04:11 AM (This post was last modified: 05/22/2008, 04:14 AM by andydude.) 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: 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];and here is the Mathematica code for the tensor power series of the exponential factorial: 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}and the next term: 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}and the next term: 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 « Next Oldest | Next Newest »

 Messages In This Thread Tensor power series - by andydude - 05/13/2008, 07:58 AM RE: Tensor power series - by andydude - 05/13/2008, 07:59 AM RE: Tensor power series - by andydude - 05/13/2008, 08:11 AM RE: Tensor power series - by andydude - 05/14/2008, 06:18 AM RE: Tensor power series - by Gottfried - 05/20/2008, 08:39 PM RE: Tensor power series - by andydude - 05/22/2008, 12:58 AM RE: Tensor power series - by andydude - 05/22/2008, 04:11 AM RE: Tensor power series - by andydude - 05/22/2008, 04:36 AM RE: Tensor power series - by bo198214 - 05/24/2008, 10:10 AM RE: Tensor power series - by andydude - 06/04/2008, 08:08 AM

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