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F(x) = \arctan(x)
\[
(1 + x * x ) * F^{(1)} (x) - 0 * F (x) = 1
\]
and in the
standard math function differential equation
,
A(x) = 0
,
B(x) = 1 + x * x
,
and
D(x) = 1
.
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 now follows from the general
Taylor coefficients recursion formula
that for
j = 0 , 1, \ldots
,
\[
\begin{array}{rcl}
z^{(0)} & = & \arctan ( x^{(0)} )
\\
b^{(j)}
& = & \left\{ \begin{array}{ll}
1 + x^{(0)} * x^{(0)} & {\rm if} \; j = 0 \\
\sum_{k=0}^j x^{(k)} x^{(j-k)} & {\rm otherwise}
\end{array} \right.
\\
e^{(j)}
& = & d^{(j)} + \sum_{k=0}^{j} a^{(j-k)} * z^{(k)}
\\
& = & \left\{ \begin{array}{ll}
1 & {\rm if} \; j = 0 \\
0 & {\rm otherwise}
\end{array} \right.
\\
z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} }
\left(
\sum_{k=0}^j e^{(k)} (j+1-k) x^{(j+1-k)}
- \sum_{k=1}^j b^{(k)} (j+1-k) z^{(j+1-k)}
\right)
\\
z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} }
\left(
(j+1) x^{(j+1)}
- \sum_{k=1}^j k z^{(k)} b^{(j+1-k)}
\right)
\end{array}
\]