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t
(see Section 10.2 of
Evaluating Derivatives
).
Here and below
@(@
X(t)
@)@, @(@
Y(t)
@)@, and
Z(t)
are scalar valued functions
and the corresponding p
-th order Taylor coefficients row vectors are
@(@
x
@)@, @(@
y
@)@ and @(@
z
@)@; i.e.,
@[@
\begin{array}{lcr}
X(t) & = & x^{(0)} + x^{(1)} * t + \cdots + x^{(p)} * t^p + o( t^p ) \\
Y(t) & = & y^{(0)} + y^{(1)} * t + \cdots + y^{(p)} * t^p + o( t^p ) \\
Z(t) & = & z^{(0)} + z^{(1)} * t + \cdots + z^{(p)} * t^p + o( t^p )
\end{array}
@]@
For the purposes of this section, we are given @(@
x
@)@ and @(@
y
@)@
and need to determine @(@
z
@)@.
p
-th order Taylor coefficient row vectors for
@(@
A [ X (t) ]
@)@, @(@
B [ X (t) ]
@)@ and @(@
D [ X (t) ]
@)@
respectively.
We assume that these coefficients are known functions of @(@
x
@)@,
the p
-th order Taylor coefficients for @(@
X(t)
@)@.
p
-th order Taylor coefficient row vector for @(@
Z(t)
@)@,
in terms of these other known coefficients.
It follows from the formulas above that
@[@
\begin{array}{rcl}
Z^{(1)} (t)
& = & F^{(1)} [ X(t) ] * X^{(1)} (t)
\\
B[ X(t) ] * Z^{(1)} (t)
& = & \{ D[ X(t) ] + A[ X(t) ] * Z(t) \} * X^{(1)} (t)
\\
B[ X(t) ] * Z^{(1)} (t) & = & E(t) * X^{(1)} (t)
\end{array}
@]@
where we define
@[@
E(t) = D[X(t)] + A[X(t)] * Z(t)
@]@
We can compute the value of @(@
z^{(0)}
@)@ using the formula
@[@
z^{(0)} = F ( x^{(0)} )
@]@
Suppose by induction (on @(@
j
@)@) that we are given the
Taylor coefficients of @(@
E(t)
@)@ up to order @(@
j-1
@)@; i.e.,
@(@
e^{(k)}
@)@ for @(@
k = 0 , \ldots , j-1
@)@
and the coefficients
@(@
z^{(k)}
@)@ for @(@
k = 0 , \ldots , j
@)@.
We can compute @(@
e^{(j)}
@)@ using the formula
@[@
e^{(j)} = d^{(j)} + \sum_{k=0}^j a^{(j-k)} * z^{(k)}
@]@
We need to complete the induction by finding formulas for @(@
z^{(j+1)}
@)@.
It follows from the definition of @(@
E(t)
@)@ that
@[@
\left( \sum_{k=0}^j b^{(k)} * t^k \right)
*
\left( \sum_{k=1}^{j+1} k z^{(k)} * t^{k-1} \right)
=
\left( \sum_{k=0}^j e^{(k)} * t^k \right)
*
\left( \sum_{k=1}^{j+1} k x^{(k)} * t^{k-1} \right)
+
o( t^p )
@]@
Setting the left and right side coefficients of @(@
t^j
@)@ equal,
and using the formula for
multiplication
,
we obtain
@[@
\begin{array}{rcl}
\sum_{k=0}^j b^{(k)} (j+1-k) z^{(j+1-k)}
& = &
\sum_{k=0}^j e^{(k)} (j+1-k) x^{(j+1-k)}
\\
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(
\sum_{k=1}^{j+1} k x^{(k)} e^{(j+1-k)}
- \sum_{k=1}^j k z^{(k)} b^{(j+1-k)}
\right)
\end{array}
@]@
This completes the induction that computes @(@
e^{(j)}
@)@
and @(@
z^{(j+1)}
@)@.
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