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@(@\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }@)@
Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory

Derivatives
@[@ \begin{array}{rcl} \tan^{(1)} ( u ) & = & [ \cos (u)^2 + \sin (u)^2 ] / \cos (u)^2 \\ & = & 1 + \tan (u)^2 \\ \tanh^{(1)} ( u ) & = & [ \cosh (u)^2 - \sinh (u)^2 ] / \cosh (u)^2 \\ & = & 1 - \tanh (u)^2 \end{array} @]@If @(@ F(u) @)@ is @(@ \tan (u) @)@ or @(@ \tanh (u) @)@ the corresponding derivative is given by @[@ F^{(1)} (u) = 1 \pm F(u)^2 @]@ Given @(@ X(t) @)@, we define the function @(@ Z(t) = F[ X(t) ] @)@. It follows that @[@ Z^{(1)} (t) = F^{(1)} [ X(t) ] X^{(1)} (t) = [ 1 \pm Y(t) ] X^{(1)} (t) @]@ where we define the function @(@ Y(t) = Z(t)^2 @)@.

Taylor Coefficients Recursion
Suppose that we are given the Taylor coefficients up to order @(@ j @)@ for the function @(@ X(t) @)@ and up to order @(@ j-1 @)@ for the functions @(@ Y(t) @)@ and @(@ Z(t) @)@. We need a formula that computes the coefficient of order @(@ j @)@ for @(@ Y(t) @)@ and @(@ Z(t) @)@. Using the equation above for @(@ Z^{(1)} (t) @)@, we have @[@ \begin{array}{rcl} \sum_{k=1}^j k z^{(k)} t^{k-1} & = & \sum_{k=1}^j k x^{(k)} t^{k-1} \pm \left[ \sum_{k=0}^{j-1} y^{(k)} t^k \right] \left[ \sum_{k=1}^j k x^{(k)} t^{k-1} \right] + o( t^{j-1} ) \end{array} @]@ Setting the coefficients of @(@ t^{j-1} @)@ equal, we have @[@ \begin{array}{rcl} j z^{(j)} = j x^{(j)} \pm \sum_{k=1}^j k x^{(k)} y^{(j-k)} \\ z^{(j)} = x^{(j)} \pm \frac{1}{j} \sum_{k=1}^j k x^{(k)} y^{(j-k)} \end{array} @]@ Once we have computed @(@ z^{(j)} @)@, we can compute @(@ y^{(j)} @)@ as follows: @[@ y^{(j)} = \sum_{k=0}^j z^{(k)} z^{(j-k)} @]@
Input File: omh/appendix/theory/tan_forward.omh