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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}