First Order Expansion
We define @(@
x(t)
@)@ near @(@
t = 0
@)@ by the first order expansion
@[@
x(t) = x^{(0)} + x^{(1)} * t
@]@
it follows that @(@
x^{(0)}
@)@ is the zero,
and @(@
x^{(1)}
@)@ the first,
order derivative of @(@
x(t)
@)@
at @(@
t = 0
@)@.
Purpose
In general, a first order forward sweep is given the
zero order derivative
for all of the variables in an operation sequence,
and the first order derivatives for the independent variables.
It uses these to compute the first order derivatives,
and thereby obtain the first order expansion,
for all the other variables in the operation sequence.
Mathematical Form
Suppose that we use the algorithm exp_2.hpp
to compute
@[@
f(x) = 1 + x + x^2 / 2
@]@
The corresponding derivative function is
@[@
\partial_x f (x) = 1 + x
@]@
An algorithmic differentiation package
does not operate on the mathematical form of the function,
or its derivative,
but rather on the
operation sequence
for the for the algorithm that is used to evaluate the function.
Operation Sequence
We consider the case where exp_2.hpp
is executed with
@(@
x = .5
@)@.
The corresponding operation sequence and
zero order forward mode values
(see zero order sweep
)
are inputs and are used by a first order forward sweep.
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
@)@ so that differentiation
with respect to @(@
t
@)@,
at @(@
t = 0
@)@, is the same as 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_5^{(1)} =
\left[ \D{v_5}{t} \right]_{t=0} =
\left[ \D{}{t} f ( x^{(0)} + x^{(1)} t ) \right]_{t=0}
\\
& = &
f^{(1)} ( x^{(0)} ) * x^{(1)} =
f^{(1)} ( x^{(0)} )
\end{array}
@]@
(We have used the fact that @(@
x^{(1)} = 1
@)@.)
Verification
The file exp_2_for1.cpp
contains a routine
which verifies the values computed above.
It returns true for success and false for failure.
Which statement in the routine defined by exp_2_for1.cpp
uses
the values that are calculated by the routine
defined by exp_2_for0.cpp
?
Suppose that @(@
x = .1
@)@,
what are the results of a zero and first order forward sweep for
the operation sequence above;
i.e., what are the corresponding values for
@(@
v_1^{(0)}, v_2^{(0)}, \cdots , v_5^{(0)}
@)@ and
@(@
v_1^{(1)}, v_2^{(1)}, \cdots , v_5^{(1)}
@)@ ?
Create a modified version of exp_2_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_2_for1.cpp
.