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f(u, v)
where
u \in B^n
we use the notation
\[
\D{f}{u} (u, v) = \left[ \D{f}{u_1} (u, v) , \cdots , \D{f}{u_n} (u, v) \right]
\]
F : B^n \rightarrow B^m
,
a matrix of Taylor coefficients
x \in B^{n \times p}
,
and a weight vector
w \in B^m
.
We define the functions
X : B \times B^{n \times p} \rightarrow B^n
,
W : B \times B^{n \times p} \rightarrow B
, and
W_j : B^{n \times p} \rightarrow B
by
\[
\begin{array}{rcl}
X(t , x) & = & x^{(0)} + x^{(1)} t + \cdots + x^{(p-1)} t^{p-1}
\\
W(t, x) & = & w_0 F_0 [X(t, x)] + \cdots + w_{m-1} F_{m-1} [X(t, x)]
\\
W_j (x) & = & \frac{1}{j!} \Dpow{j}{t} W(0, x)
\end{array}
\]
where
x^{(j)}
is the j-th column of
x \in B^{n \times p}
.
The theorem below implies that
\[
\D{ W_j }{ x^{(i)} } (x) = \D{ W_{j-i} }{ x^{(0)} } (x)
\]
A general reverse sweep
calculates the values
\[
\D{ W_{p-1} }{ x^{(i)} } (x) \hspace{1cm} (i = 0 , \ldots , p-1)
\]
But the return values for a reverse sweep are specified
in terms of the more useful values
\[
\D{ W_j }{ x^{(0)} } (x) \hspace{1cm} (j = 0 , \ldots , p-1)
\]
F : B^n \rightarrow B^m
is a
p
times
continuously differentiable function.
Define the functions
Z : B \times B^{n \times p} \rightarrow B^n
,
Y : B \times B^{n \times p }\rightarrow B^m
,
and
y^{(j)} : B^{n \times p }\rightarrow B^m
by
\[
\begin{array}{rcl}
Z(t, x) & = & x^{(0)} + x^{(1)} t + \cdots + x^{(p-1)} t^{p-1}
\\
Y(t, x) & = & F [ Z(t, x) ]
\\
y^{(j)} (x) & = & \frac{1}{j !} \Dpow{j}{t} Y(0, x)
\end{array}
\]
where
x^{(j)}
denotes the j-th column of
x \in B^{n \times p}
.
It follows that
for all
i, j
such that
i \leq j < p
,
\[
\begin{array}{rcl}
\D{ y^{(j)} }{ x^{(i)} } (x) & = & \D{ y^{(j-i)} }{ x^{(0)} } (x)
\end{array}
\]
\[
\begin{array}{rclr}
\D{ y^{(j)} }{ x^{(i)} } (x)
& = &
\frac{1}{j ! } \D{ }{ x^{(i)} }
\left[ \Dpow{j}{t} (F \circ Z) (t, x) \right]_{t=0}
\\
& = &
\frac{1}{j ! } \left[ \Dpow{j}{t}
\D{ }{ x^{(i)} } (F \circ Z) (t, x)
\right]_{t=0}
\\
& = &
\frac{1}{j ! } \left\{
\Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
\right\}_{t=0}
\end{array}
\]
For
k > i
, the k-th
partial of
t^i
with respect to
t
is zero.
Thus, the partial with respect to
t
is given by
\[
\begin{array}{rcl}
\Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
& = &
\sum_{k=0}^i
\left( \begin{array}{c} j \\ k \end{array} \right)
\frac{ i ! }{ (i - k) ! } t^{i-k} \;
\Dpow{j-k}{t} ( F^{(1)} \circ Z ) (t, x)
\\
\left\{
\Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
\right\}_{t=0}
& = &
\left( \begin{array}{c} j \\ i \end{array} \right)
i ! \Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
\\
& = &
\frac{ j ! }{ (j - i) ! }
\Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
\\
\D{ y^{(j)} }{ x^{(i)} } (x)
& = &
\frac{ 1 }{ (j - i) ! }
\Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
\end{array}
\]
Applying this formula to the case where
j
is replaced by
j - i
and
i
is replaced by zero,
we obtain
\[
\D{ y^{(j-i)} }{ x^{(0)} } (x)
=
\frac{ 1 }{ (j - i) ! }
\Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
=
\D{ y^{(j)} }{ x^{(i)} } (x)
\]
which completes the proof