lines 6-117 of file: xrst/theory/pow_forward.xrst {xrst_begin pow_forward} Power Function Forward Mode Theory ################################## We consider the operation :math:`F(x) = x^y` where :math:`x` is a variable and :math:`y` is a parameter. Derivatives *********** The corresponding derivative satisfies the equation .. math:: x * F^{(1)} (x) - y F(x) = 0 This is the :ref:`standard math function differential equation` , where :math:`A(x) = y`, :math:`B(x) = x`, and :math:`D(x) = 0`. We use :math:`a`, :math:`b`, :math:`d`, and :math:`z` to denote the Taylor coefficients for :math:`A [ X (t) ]`, :math:`B [ X (t) ]`, :math:`D [ X (t) ]`, and :math:`F [ X(t) ]` respectively. It follows that :math:`b^j = x^j`, :math:`d^j = 0`, .. math:: a^{(j)} = \left\{ \begin{array}{ll} y & \R{if} \; j = 0 \\ 0 & \R{otherwise} \end{array} \right. Taylor Coefficients Recursion ***************************** z^(0) ===== .. math:: z^{(0)} = F ( x^{(0)} ) e^(j) ===== .. math:: :nowrap: \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 :math:`j = 0, \ldots , p-1` .. math:: :nowrap: \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 :math:`j = 1, \ldots , p` .. math:: :nowrap: \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} {xrst_end pow_forward}