\newcommand{\W}[1]{ \; #1 \; }
\newcommand{\R}[1]{ {\rm #1} }
\newcommand{\B}[1]{ {\bf #1} }
\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }
\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }
\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }
\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }This is cppad-20221105 documentation. Here is a link to its
current documentation
.
exp_eps: First Order Forward Sweep
First Order Expansion
We define
x(t)
and
\varepsilon(t) ]
near
t = 0
by the first order expansions
\begin{array}{rcl}
x(t) & = & x^{(0)} + x^{(1)} * t
\\
\varepsilon(t) & = & \varepsilon^{(0)} + \varepsilon^{(1)} * t
\end{array}
It follows that
x^{(0)}
(
\varepsilon^{(0)}
) is the zero,
and
x^{(1)}
(
\varepsilon^{(1)}
) the first,
order derivative of
x(t)
at
t = 0
(
\varepsilon (t)
)
at
t = 0
.
Mathematical Form
Suppose that we use the algorithm exp_eps.hpp
to compute
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 derivative with respect to
x
,
and the value of the derivative, are
\partial_x f ( x , \varepsilon ) = 1 + x = 1.5
Index
The Index column contains the index in the operation sequence
of the corresponding atomic operation.
A Forward sweep starts with the first operation
and ends with the last.
Operation
The Operation column contains the
mathematical function corresponding to each atomic operation in the sequence.
Zero Order
The Zero Order column contains the zero order derivatives
for the corresponding variable in the operation sequence
(see zero order sweep
).
Derivative
The Derivative column contains the
mathematical function corresponding to the derivative
with respect to
t
,
at
t = 0
, for each variable in the sequence.
First Order
The First Order column contains the first order derivatives
for the corresponding variable in the operation sequence; i.e.,
v_j (t) = v_j^{(0)} + v_j^{(1)} t
We use
x^{(1)} = 1
and
\varepsilon^{(1)} = 0
,
so that differentiation with respect to
t
,
at
t = 0
,
is the same partial differentiation with respect to
x
at
x = x^{(0)}
.
Return Value
The derivative of the return value for this case is
\begin{array}{rcl}
1.5
& = &
v_7^{(1)} =
\left[ \D{v_7}{t} \right]_{t=0} =
\left[ \D{}{t} f( x^{(0)} + x^{(1)} * t , \varepsilon^{(0)} ) \right]_{t=0}
\\
& = &
\partial_x f ( x^{(0)} , \varepsilon^{(0)} ) * x^{(1)} =
\partial_x f ( x^{(0)} , \varepsilon^{(0)} )
\end{array}
(We have used the fact that
x^{(1)} = 1
and
\varepsilon^{(1)} = 0
.)
Suppose that
x = .1
,
what are the results of a zero and first order forward mode sweep for
the operation sequence above;
i.e., what are the corresponding values for
v_1^{(0)}, v_2^{(0)}, \cdots , v_7^{(0)}
and
v_1^{(1)}, v_2^{(1)}, \cdots , v_7^{(1)}
?
Create a modified version of exp_eps_for1.cpp
that verifies
the derivative values from the previous exercise.
Also create and run a main program that reports the result
of calling the modified version of
exp_eps_for1.cpp
.
Suppose that
x = .1
and
\epsilon = .2
,
what is the operation sequence corresponding to
exp_eps(x, epsilon)
