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F(x) = \sqrt{x}
\[
F(x) * F^{(1)} (x) - 0 * F (x) = 1/2
\]
and in the
standard math function differential equation
,
A(x) = 0
,
B(x) = F(x)
,
and
D(x) = 1/2
.
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
,
\[
\begin{array}{rcl}
z^{(0)} & = & \sqrt { x^{(0)} }
\\
e^{(j)}
& = & d^{(j)} + \sum_{k=0}^{j} a^{(j-k)} * z^{(k)}
\\
& = & \left\{ \begin{array}{ll}
1/2 & {\rm if} \; j = 0 \\
0 & {\rm otherwise}
\end{array} \right.
\\
z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} }
\left(
\sum_{k=1}^{j+1} k x^{(k)} e^{(j+1-k)}
- \sum_{k=1}^j k z^{(k)} b^{(j+1-k)}
\right)
\\
& = & \frac{1}{j+1} \frac{1}{ z^{(0)} }
\left(
\frac{j+1}{2} x^{(j+1) }
- \sum_{k=1}^j k z^{(k)} z^{(j+1-k)}
\right)
\end{array}
\]