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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}} }@)@ This is cppad-20221105 documentation. Here is a link to its current documentation .
Power Function Forward Mode Theory
We consider the operation @(@ F(x) = x^y @)@ where @(@ x @)@ is a variable and @(@ y @)@ is a parameter.

Derivatives
The corresponding derivative satisfies the equation @[@ x * F^{(1)} (x) - y F(x) = 0 @]@ This is the standard math function differential equation , where @(@ A(x) = y @)@, @(@ B(x) = x @)@, and @(@ D(x) = 0 @)@. 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. It follows that @(@ b^j = x^j @)@, @(@ d^j = 0 @)@, @[@ a^{(j)} = \left\{ \begin{array}{ll} y & \R{if} \; j = 0 \\ 0 & \R{otherwise} \end{array} \right. @]@

Taylor Coefficients Recursion

z^(0)
@[@ z^{(0)} = F ( x^{(0)} ) @]@
e^(j)
@[@ \begin{array}{rcl} e^{(j)} & = & d^{(j)} + \sum_{k=0}^j a^{(j-k)} * z^{(k)} \\ e^{(j)} & = & y * z^{(j)} \end{array} @]@
z^j
For @(@ j = 0, \ldots , p-1 @)@ @[@ \begin{array}{rcl} z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } \left( \sum_{k=1}^{j+1} k x^{(k)} e^{(j+1-k)} - \sum_{k=1}^j k z^{(k)} b^{(j+1-k)} \right) \\ & = & \frac{1}{j+1} \frac{1}{ x^{(0)} } \left( y \sum_{k=1}^{j+1} k x^{(k)} z^{(j+1-k)} - \sum_{k=1}^j k z^{(k)} x^{(j+1-k)} \right) \\ & = & \frac{1}{j+1} \frac{1}{ x^{(0)} } \left( y (j+1) x^{(j+1)} z^{(0)} + \sum_{k=1}^j k ( y x^{(k)} z^{(j+1-k)} - z^{(k)} x^{(j+1-k)} ) \right) \\ & = & y z^{(0)} x^{(j+1)} / x^{(0)} + \frac{1}{j+1} \frac{1}{ x^{(0)} } \sum_{k=1}^j k ( y x^{(k)} z^{(j+1-k)} - z^{(k)} x^{(j+1-k)} ) \end{array} @]@ For @(@ j = 1, \ldots , p @)@ @[@ \begin{array}{rcl} z^{(j)} & = & \left. \left( y z^{(0)} x^{(j)} + \frac{1}{j} \sum_{k=1}^{j-1} k ( y x^{(k)} z^{(j-k)} - z^{(k)} x^{(j-k)} ) \right) \right/ x^{(0)} \end{array} @]@
Input File: omh/theory/pow_forward.omh