Derivatives
Given @(@
X(t)
@)@, we define the function
@[@
Z(t) = \R{erf}[ X(t) ]
@]@
It follows that
@[@
\begin{array}{rcl}
\R{erf}^{(1)} ( u ) & = & ( 2 / \sqrt{\pi} ) \exp \left( - u^2 \right)
\\
Z^{(1)} (t) & = & \R{erf}^{(1)} [ X(t) ] X^{(1)} (t) = Y(t) X^{(1)} (t)
\end{array}
@]@
where we define the function
@[@
Y(t) = \frac{2}{ \sqrt{\pi} } \exp \left[ - X(t)^2 \right]
@]@
Taylor Coefficients Recursion
Suppose that we are given the Taylor coefficients
up to order @(@
j
@)@ for the function @(@
X(t)
@)@ and @(@
Y(t)
@)@.
We need a formula that computes the coefficient of order @(@
j
@)@
for @(@
Z(t)
@)@.
Using the equation above for @(@
Z^{(1)} (t)
@)@, we have
@[@
\begin{array}{rcl}
\sum_{k=1}^j k z^{(k)} t^{k-1}
& = &
\left[ \sum_{k=0}^j y^{(k)} t^k \right]
\left[ \sum_{k=1}^j k x^{(k)} t^{k-1} \right]
+
o( t^{j-1} )
\end{array}
@]@
Setting the coefficients of @(@
t^{j-1}
@)@ equal, we have
@[@
\begin{array}{rcl}
j z^{(j)}
=
\sum_{k=1}^j k x^{(k)} y^{(j-k)}
\\
z^{(j)}
=
\frac{1}{j} \sum_{k=1}^j k x^{(k)} y^{(j-k)}
\end{array}
@]@
Input File: omh/appendix/theory/erf_forward.omh