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# include "exp_2.hpp"
y = exp_2(
x)
\[
\exp (x) = 1 + x^1 / 1 ! + x^2 / 2 ! + \cdots
\]
The second order approximation for the exponential function is
\[
{\rm exp\_2} (x) = 1 + x + x^2 / 2
\]
cppad-
yy-
mm-
dd/introduction/exp_apx
where cppad-
yy-
mm-
dd is the distribution directory
created during the beginning steps of the
installation
of CppAD.
const
Type &
x
(see Type below).
It specifies the point at which to evaluate the
approximation for the second order exponential approximation.
Type
y
It is the value of the exponential function
approximation defined above.
int
:
Operation | Result Type | Description |
Type( i)
| Type | object with value equal to i |
u = v
| Type | new u (and result) is value of v |
u * v
| Type |
result is value of
u * v
|
u / v
| Type |
result is value of
u / v
|
u + v
| Type |
result is value of
u + v
|
exp_2.hpp | exp_2: Implementation |
exp_2.cpp | exp_2: Test |
exp_2_for0 | exp_2: Operation Sequence and Zero Order Forward Mode |
exp_2_for1 | exp_2: First Order Forward Mode |
exp_2_rev1 | exp_2: First Order Reverse Mode |
exp_2_for2 | exp_2: Second Order Forward Mode |
exp_2_rev2 | exp_2: Second Order Reverse Mode |
exp_2_cppad | exp_2: CppAD Forward and Reverse Sweeps |
double x = .1;
double y = exp_2(x);
What is the value assigned to
v1
, v2
, ... ,v5
in exp_2.hpp
?
exp_2.hpp
to
a routine exp_3.hpp
that computes
\[
1 + x^2 / 2 ! + x^3 / 3 !
\]
Do this in a way that only assigns one value to each variable
(as exp_2
does).
double x = .5;
double y = exp_3(x);
using exp_3
created in the previous problem.
What is the value assigned to the new variables in exp_3
(variables that are in exp_3
and not in exp_2
) ?