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\[
f(x) = 1 + x + x^2 / 2
\]
The corresponding derivative function is
\[
\partial_x f (x) = 1 + x
\]
v_5
in the
exp_2 operation sequence
.
We begin with the function
f_5
where
v_5
is both an argument and the value of the function; i.e.,
\[
\begin{array}{rcl}
f_5 ( v_1 , v_2 , v_3 , v_4 , v_5 ) & = & v_5
\\
\D{f_5}{v_5} & = & 1
\end{array}
\]
All the other partial derivatives of
f_5
are zero.
\[
v_5 = v_2 + v_4
\]
We define the function
f_4 ( v_1 , v_2 , v_3 , v_4 )
as equal to
f_5
except that
v_5
is eliminated using
this operation; i.e.
\[
f_4 =
f_5 [ v_1 , v_2 , v_3 , v_4 , v_5 ( v_2 , v_4 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_4}{v_2}
& = & \D{f_5}{v_2} +
\D{f_5}{v_5} * \D{v_5}{v_2}
& = 1
\\
\D{f_4}{v_4}
& = & \D{f_5}{v_4} +
\D{f_5}{v_5} * \D{v_5}{v_4}
& = 1
\end{array}
\]
All the other partial derivatives of
f_4
are zero.
\[
v_4 = v_3 / 2
\]
We define the function
f_3 ( v_1 , v_2 , v_3 )
as equal to
f_4
except that
v_4
is eliminated using this operation; i.e.,
\[
f_3 =
f_4 [ v_1 , v_2 , v_3 , v_4 ( v_3 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_3}{v_1}
& = & \D{f_4}{v_1}
& = 0
\\
\D{f_3}{v_2}
& = & \D{f_4}{v_2}
& = 1
\\
\D{f_3}{v_3}
& = & \D{f_4}{v_3} +
\D{f_4}{v_4} * \D{v_4}{v_3}
& = 0.5
\end{array}
\]
\[
v_3 = v_1 * v_1
\]
We define the function
f_2 ( v_1 , v_2 )
as equal to
f_3
except that
v_3
is eliminated using this operation; i.e.,
\[
f_2 =
f_3 [ v_1 , v_2 , v_3 ( v_1 ) ]
\]
Note that the value of
v_1
is equal to
x
which is .5 for this evaluation.
It follows that
\[
\begin{array}{rcll}
\D{f_2}{v_1}
& = & \D{f_3}{v_1} +
\D{f_3}{v_3} * \D{v_3}{v_1}
& = 0.5
\\
\D{f_2}{v_2}
& = & \D{f_3}{v_2}
& = 1
\end{array}
\]
\[
v_2 = 1 + v_1
\]
We define the function
f_1 ( v_1 )
as equal to
f_2
except that
v_2
is eliminated using this operation; i.e.,
\[
f_1 =
f_2 [ v_1 , v_2 ( v_1 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_1}{v_1}
& = & \D{f_2}{v_1} +
\D{f_2}{v_2} * \D{v_2}{v_1}
& = 1.5
\end{array}
\]
Note that
v_1
is equal to
x
,
so the derivative of this is the derivative of
the function defined by exp_2.hpp
at
x = .5
.
f_j
that might not be equal to the corresponding
partials of
f_{j+1}
; i.e., the
other partials of
f_j
must be equal to the corresponding
partials of
f_{j+1}
.
x = .1
and we first preform a zero order forward sweep
for the operation sequence used above.
What are the results of a
first order reverse sweep; i.e.,
what are the corresponding derivatives of
f_5 , f_4 , \ldots , f_1
.