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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.,
\[
\begin{array}{lcr}
X(t) & = & x^{(0)} + x^{(1)} * t + \cdots + x^{(p)} * t^p + O( t^{p+1} ) \\
Y(t) & = & y^{(0)} + y^{(1)} * t + \cdots + y^{(p)} * t^p + O( t^{p+1} ) \\
Z(t) & = & z^{(0)} + z^{(1)} * t + \cdots + z^{(p)} * t^p + O( t^{p+1} )
\end{array}
\]
For the purposes of this discussion,
we are given the p-th order Taylor coefficient row vectors
x
,
y
, and
z
.
In addition, we are given the partial derivatives of a scalar valued function
\[
G ( z^{(j)} , \ldots , z^{(0)}, x, y)
\]
We need to compute the partial derivatives of the scalar valued function
\[
H ( z^{(j-1)} , \ldots , z^{(0)}, x, y) =
G ( z^{(j)}, z^{(j-1)} , \ldots , z^{(0)}, x , y )
\]
where
z^{(j)}
is expressed as a function of the
j-1-th order Taylor coefficient row
vector for
Z
and the vectors
x
,
y
; i.e.,
z^{(j)}
above is a shorthand for
\[
z^{(j)} ( z^{(j-1)} , \ldots , z^{(0)}, x, y )
\]
If we do not provide a formula for
a partial derivative of
H
, then that partial derivative
has the same value as for the function
G
.
\[
z^{(j)} = x^{(j)} + y^{(j)}
\]
If follows that for
k = 0 , \ldots , j
and
l = 0 , \ldots , j-1
\[
\begin{array}{rcl}
\D{H}{ x^{(k)} } & = &
\D{G}{ x^{(k)} } + \D{G}{ z^{(k)} } \\
\\
\D{H}{ y^{(k)} } & = &
\D{G}{ y^{(k)} } + \D{G}{ z^{(k)} }
\\
\D{H}{ z^{(l)} } & = & \D{G}{ z^{(l)} }
\end{array}
\]
\[
z^{(j)} = x^{(j)} - y^{(j)}
\]
If follows that for
k = 0 , \ldots , j
\[
\begin{array}{rcl}
\D{H}{ x^{(k)} } & = &
\D{G}{ x^{(k)} } - \D{G}{ z^{(k)} } \\
\\
\D{H}{ y^{(k)} } & = &
\D{G}{ y^{(k)} } - \D{G}{ z^{(k)} }
\end{array}
\]
\[
z^{(j)} = \sum_{k=0}^j x^{(j-k)} * y^{(k)}
\]
If follows that for
k = 0 , \ldots , j
and
l = 0 , \ldots , j-1
\[
\begin{array}{rcl}
\D{H}{ x^{(j-k)} } & = &
\D{G}{ x^{(j-k)} } +
\sum_{k=0}^j \D{G}{ z^{(j)} } y^{(k)}
\\
\D{H}{ y^{(k)} } & = &
\D{G}{ y^{(k)} } +
\sum_{k=0}^j \D{G}{ z^{(j)} } x^{(j-k)}
\end{array}
\]
\[
z^{(j)} =
\frac{1}{y^{(0)}}
\left(
x^{(j)} - \sum_{k=1}^j z^{(j-k)} y^{(k)}
\right)
\]
If follows that for
k = 1 , \ldots , j
\[
\begin{array}{rcl}
\D{H}{ x^{(j)} } & = &
\D{G}{ x^{(j)} } + \D{G}{ z^{(j)} } \frac{1}{y^{(0)}}
\\
\D{H}{ z^{(j-k)} } & = &
\D{G}{ z^{(j-k)} } - \D{G}{ z^{(j)} } \frac{1}{y^{(0)}} y^{(k)}
\\
\D{H}{ y^{(k)} } & = &
\D{G}{ y^{(k)} } - \D{G}{ z^{(j)} } \frac{1}{y^{(0)}} z^{(j-k)}
\\
\D{H}{ y^{(0)} } & = &
\D{G}{ y^{(0)} } - \D{G}{ z^{(j)} } \frac{1}{y^{(0)}} \frac{1}{y^{(0)}}
\left(
x^{(j)} - \sum_{k=1}^j z^{(j-k)} y^{(k)}
\right)
\\
& = &
\D{G}{ y^{(0)} } - \D{G}{ z^{(j)} } \frac{1}{y^{(0)}} z^{(j)}
\end{array}
\]
\[
G ( z^{(j)} , \ldots , z^{(0)}, x)
\]
We need to compute the partial derivatives of the scalar valued function
\[
H ( z^{(j-1)} , \ldots , z^{(0)}, x) =
G ( z^{(j)}, z^{(j-1)} , \ldots , z^{(0)}, x)
\]
where
z^{(j)}
is expressed as a function of the
j-1-th order Taylor coefficient row
vector for
Z
and the vector
x
; i.e.,
z^{(j)}
above is a shorthand for
\[
z^{(j)} ( z^{(j-1)} , \ldots , z^{(0)}, x )
\]