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exp_eps(
x,
epsilon)
with x is equal to .5
and epsilon is equal to .2.
For this case, the mathematical function for the operation sequence
corresponding to exp_eps
is
\[
f ( x , \varepsilon ) = 1 + x + x^2 / 2
\]
The corresponding partial derivatives,
and the value of the derivatives, are
\[
\begin{array}{rcl}
\partial_x f ( x , \varepsilon ) & = & 1 + x = 1.5
\\
\partial_\varepsilon f ( x , \varepsilon ) & = & 0
\end{array}
\]
\varepsilon
is an independent variable,
it could included as an argument to all of the
f_j
functions below.
The result would be that all the partials with respect to
\varepsilon
would be zero and hence we drop it to simplify
the presentation.
v_7
.
We begin with the function
f_7
where
v_7
is both an argument and the value of the function; i.e.,
\[
\begin{array}{rcl}
f_7 ( v_1 , v_2 , v_3 , v_4 , v_5 , v_6 , v_7 ) & = & v_7
\\
\D{f_7}{v_7} & = & 1
\end{array}
\]
All the other partial derivatives of
f_7
are zero.
\[
v_7 = v_4 + v_6
\]
We define the function
f_6 ( v_1 , v_2 , v_3 , v_4 , v_5 , v_6 )
as equal to
f_7
except that
v_7
is eliminated using
this operation; i.e.
\[
f_6 =
f_7 [ v_1 , v_2 , v_3 , v_4 , v_5 , v_6 , v_7 ( v_4 , v_6 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_6}{v_4}
& = & \D{f_7}{v_4} +
\D{f_7}{v_7} * \D{v_7}{v_4}
& = 1
\\
\D{f_6}{v_6}
& = & \D{f_7}{v_6} +
\D{f_7}{v_7} * \D{v_7}{v_6}
& = 1
\end{array}
\]
All the other partial derivatives of
f_6
are zero.
\[
v_6 = v_5 / 2
\]
We define the function
f_5 ( v_1 , v_2 , v_3 , v_4 , v_5 )
as equal to
f_6
except that
v_6
is eliminated using this operation; i.e.,
\[
f_5 =
f_6 [ v_1 , v_2 , v_3 , v_4 , v_5 , v_6 ( v_5 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_5}{v_4}
& = & \D{f_6}{v_4}
& = 1
\\
\D{f_5}{v_5}
& = & \D{f_6}{v_5} +
\D{f_6}{v_6} * \D{v_6}{v_5}
& = 0.5
\end{array}
\]
All the other partial derivatives of
f_5
are zero.
\[
v_5 = v_3 * v_1
\]
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.,
\[
f_4 =
f_5 [ v_1 , v_2 , v_3 , v_4 , v_5 ( v_3 , v_1 ) ]
\]
Given the information from the forward sweep, we have
v_3 = 0.5
and
v_1 = 0.5
.
It follows that
\[
\begin{array}{rcll}
\D{f_4}{v_1}
& = & \D{f_5}{v_1} +
\D{f_5}{v_5} * \D{v_5}{v_1}
& = 0.25
\\
\D{f_4}{v_2} & = & \D{f_5}{v_2} & = 0
\\
\D{f_4}{v_3}
& = & \D{f_5}{v_3} +
\D{f_5}{v_5} * \D{v_5}{v_3}
& = 0.25
\\
\D{f_4}{v_4}
& = & \D{f_5}{v_4}
& = 1
\end{array}
\]
\[
v_4 = 1 + v_3
\]
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
\[
\begin{array}{rcll}
\D{f_3}{v_1}
& = & \D{f_4}{v_1}
& = 0.25
\\
\D{f_3}{v_2} & = & \D{f_4}{v_2} & = 0
\\
\D{f_3}{v_3}
& = & \D{f_4}{v_3} +
\D{f_4}{v_4} * \D{v_4}{v_3}
& = 1.25
\end{array}
\]
\[
v_3 = v_2 / 1
\]
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.,
\[
f_2 =
f_4 [ v_1 , v_2 , v_3 ( v_2 ) ]
\]
It follows that
\[
\begin{array}{rcll}
\D{f_2}{v_1}
& = & \D{f_3}{v_1}
& = 0.25
\\
\D{f_2}{v_2} & = & \D{f_3}{v_2} +
\D{f_3}{v_3} * \D{v_3}{v_2}
& = 1.25
\end{array}
\]
\[
v_2 = 1 * v_1
\]
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 exp_eps(
x,
epsilon)
at x equal to .5 and epsilon equal .2 is
1.5 in the x direction and zero in the epsilon direction.
We also note that
forward
forward mode gave the
same result for the partial in the x direction.
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 mode sweep
for the operation sequence used above (in reverse order).
What are the results of a
first order reverse mode sweep; i.e.,
what are the corresponding values for
\D{f_j}{v_k}
for all
j, k
such that
\D{f_j}{v_k} \neq 0
.