$Id: atan_forward.omh 3682 2015-05-08 21:59:05Z bradbell $ // BEGIN SHORT COPYRIGHT /* -------------------------------------------------------------------------- CppAD: C++ Algorithmic Differentiation: Copyright (C) 2003-15 Bradley M. Bell CppAD is distributed under multiple licenses. This distribution is under the terms of the Eclipse Public License Version 1.0. A copy of this license is included in the COPYING file of this distribution. Please visit http://www.coin-or.org/CppAD/ for information on other licenses. -------------------------------------------------------------------------- */ // END SHORT COPYRIGHT $begin atan_forward$$ $spell atan atanh Taylor $$ $section Inverse Tangent and Hyperbolic Tangent Forward Mode Theory$$ $mindex atan, atanh$$ $head Derivatives$$ $latex \[ \begin{array}{rcl} \R{atan}^{(1)} (x) & = & 1 / ( 1 + x * x ) \\ \R{atanh}^{(1)} (x) & = & 1 / ( 1 - x * x ) \end{array} \] $$ If $latex F(x)$$ is $latex \R{atan} (x)$$ or $latex \R{atanh} (x)$$, the corresponding derivative satisfies the equation $latex \[ (1 \pm x * x ) * F^{(1)} (x) - 0 * F (x) = 1 \] $$ and in the $cref/standard math function differential equation /ForwardTheory /Standard Math Functions /Differential Equation /$$, $latex A(x) = 0$$, $latex B(x) = 1 \pm x * x $$, and $latex D(x) = 1$$. We use $latex a$$, $latex b$$, $latex d$$ and $latex z$$ to denote the Taylor coefficients for $latex A [ X (t) ] $$, $latex B [ X (t) ]$$, $latex D [ X (t) ] $$, and $latex F [ X(t) ] $$ respectively. $head Taylor Coefficients Recursion$$ For $latex j = 0 , 1, \ldots$$, $latex \[ \begin{array}{rcl} z^{(0)} & = & F( x^{(0)} ) \\ b^{(j)} & = & \left\{ \begin{array}{ll} 1 \pm x^{(0)} * x^{(0)} & {\rm if} \; j = 0 \\ \pm \sum_{k=0}^j x^{(k)} x^{(j-k)} & {\rm otherwise} \end{array} \right. \\ e^{(j)} & = & d^{(j)} + \sum_{k=0}^{j} a^{(j-k)} * z^{(k)} \\ & = & \left\{ \begin{array}{ll} 1 & {\rm if} \; j = 0 \\ 0 & {\rm otherwise} \end{array} \right. \\ z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( \sum_{k=0}^j e^{(k)} (j+1-k) x^{(j+1-k)} - \sum_{k=1}^j b^{(k)} (j+1-k) z^{(j+1-k)} \right) \\ z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( (j+1) x^{(j+1)} - \sum_{k=1}^j k z^{(k)} b^{(j+1-k)} \right) \end{array} \] $$ $end