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\(\newcommand{\W}[1]{ \; #1 \; }\)
\(\newcommand{\R}[1]{ {\rm #1} }\)
\(\newcommand{\B}[1]{ {\bf #1} }\)
\(\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }\)
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\(\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }\)
pow_forward
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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
\(b^j = x^j\) , \(d^j = 0\) ,
\[\begin{split}a^{(j)} = \left\{ \begin{array}{ll}
y & \R{if} \; j = 0
\\
0 & \R{otherwise}
\end{array} \right.\end{split}\]
Taylor Coefficients Recursion
z^(0)
\[z^{(0)} = F ( x^{(0)} )\]
e^(j)
\begin{eqnarray}
e^{(j)} & = & d^{(j)} + \sum_{k=0}^j a^{(j-k)} * z^{(k)}
\\
e^{(j)} & = & y * z^{(j)}
\end{eqnarray}
z^j
For \(j = 0, \ldots , p-1\)
\begin{eqnarray}
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{eqnarray}
For \(j = 1, \ldots , p\)
\begin{eqnarray}
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{eqnarray}