exp_eps¶

An Epsilon Accurate Exponential Approximation¶

Syntax¶

# include "exp_eps.hpp"

y = exp_eps ( x , epsilon )

Purpose¶

This is a an example algorithm that is used to demonstrate how Algorithmic Differentiation works with loops and boolean decision variables (see exp_2 for a simpler example).

Mathematical Function¶

The exponential function can be defined by

\[\exp (x) = 1 + x^1 / 1 ! + x^2 / 2 ! + \cdots\]

We define \(k ( x, \varepsilon )\) as the smallest non-negative integer such that \(\varepsilon \geq x^k / k !\); i.e.,

\[k( x, \varepsilon ) = \min \{ k \in {\rm Z}_+ \; | \; \varepsilon \geq x^k / k ! \}\]

The mathematical form for our approximation of the exponential function is

\begin{eqnarray} {\rm exp\_eps} (x , \varepsilon ) & = & \left\{ \begin{array}{ll} \frac{1}{ {\rm exp\_eps} (-x , \varepsilon ) } & {\rm if} \; x < 0 \\ 1 + x^1 / 1 ! + \cdots + x^{k( x, \varepsilon)} / k( x, \varepsilon ) ! & {\rm otherwise} \end{array} \right. \end{eqnarray}

include¶

The include command in the syntax is relative to

cppad- yyyymmdd / introduction/exp_apx

where cppad- yyyymmdd is the distribution directory created during the beginning steps of the installation of CppAD.

x¶

The argument x has prototype

const Type & x

(see Type below). It specifies the point at which to evaluate the approximation for the exponential function.

epsilon¶

The argument epsilon has prototype

const Type & epsilon

It specifies the accuracy with which to approximate the exponential function value; i.e., it is the value of \(\varepsilon\) in the exponential function approximation defined above.

y¶

The result y has prototype

Type y

It is the value of the exponential function approximation defined above.

Type¶

If u and v are Type objects and i is an int :

Operation	Result Type	Description
Type ( i )	Type	object with value equal to i
Type u = v	Type	construct u with value equal to v
u > v	`bool`	true, if u greater than v , an false otherwise
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 \(u + v\)
`-` u	Type	result is value of \(- u\)

Implementation¶

The file exp_eps.hpp contains a C++ implementation of this function.

Test¶

The file exp_eps.cpp contains a test of this implementation.

Exercises¶

Using the definition of \(k( x, \varepsilon )\) above, what is the value of \(k(.5, 1)\), \(k(.5, .1)\), and \(k(.5, .01)\) ?
Suppose that we make the following call to exp_eps :
```
double x       = 1.;
double epsilon = .01;
double y = exp_eps(x, epsilon);
```
What is the value assigned to k , temp , term , and sum the first time through the while loop in exp_eps.hpp ?
Continuing the previous exercise, what is the value assigned to k , temp , term , and sum the second time through the while loop in exp_eps.hpp ?