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$\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }$
Inverse Cosine and Hyperbolic Cosine Forward Mode Theory

Derivatives
$$\begin{array}{rcl} \R{acos}^{(1)} (x) & = & - 1 / \sqrt{ 1 - x * x } \\ \R{acosh}^{(1)} (x) & = & + 1 / \sqrt{ x * x - 1} \end{array}$$If $F(x)$ is $\R{acos} (x)$ or $\R{acosh} (x)$ the corresponding derivative satisfies the equation $$\sqrt{ \mp ( x * x - 1 ) } * F^{(1)} (x) - 0 * F (u) = \mp 1$$ and in the standard math function differential equation , $A(x) = 0$, $B(x) = \sqrt{ \mp( x * x - 1 ) }$, and $D(x) = \mp 1$. We use $a$, $b$, $d$ and $z$ to denote the Taylor coefficients for $A [ X (t) ]$, $B [ X (t) ]$, $D [ X (t) ]$, and $F [ X(t) ]$ respectively.

Taylor Coefficients Recursion
We define $Q(x) = \mp ( x * x - 1 )$ and let $q$ be the corresponding Taylor coefficients for $Q[ X(t) ]$. It follows that $$q^{(j)} = \left\{ \begin{array}{ll} \mp ( x^{(0)} * x^{(0)} - 1 ) & {\rm if} \; j = 0 \\ \mp \sum_{k=0}^j x^{(k)} x^{(j-k)} & {\rm otherwise} \end{array} \right.$$ It follows that $B[ X(t) ] = \sqrt{ Q[ X(t) ] }$ and from the equations for the square root that for $j = 0 , 1, \ldots$, $$\begin{array}{rcl} b^{(0)} & = & \sqrt{ q^{(0)} } \\ b^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( \frac{j+1}{2} q^{(j+1) } - \sum_{k=1}^j k b^{(k)} b^{(j+1-k)} \right) \end{array}$$ It now follows from the general Taylor coefficients recursion formula that for $j = 0 , 1, \ldots$, $$\begin{array}{rcl} z^{(0)} & = & F ( x^{(0)} ) \\ e^{(j)} & = & d^{(j)} + \sum_{k=0}^{j} a^{(j-k)} * z^{(k)} \\ & = & \left\{ \begin{array}{ll} \mp 1 & {\rm if} \; j = 0 \\ 0 & {\rm otherwise} \end{array} \right. \\ z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( \sum_{k=0}^j e^{(k)} (j+1-k) x^{(j+1-k)} - \sum_{k=1}^j b^{(k)} (j+1-k) z^{(j+1-k)} \right) \\ z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( \mp (j+1) x^{(j+1)} - \sum_{k=1}^j k z^{(k)} b^{(j+1-k)} \right) \end{array}$$
Input File: omh/appendix/theory/acos_forward.omh