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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 ! + ⋯ We define k ( x , ε ) as the smallest non-negative integer such that ε ≥ x k / k ! ; i.e., k ( x , ε ) = min { k ∈ Z + | ε ≥ x k / k ! } The mathematical form for our approximation of the exponential function is exp _ eps ( x , ε ) = { 1 exp _ eps ( - x , ε ) if x < 0 1 + x 1 / 1 ! + ⋯ + x k ( x , ε ) / k ( x , ε ) ! otherwise
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 ε 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. It returns true for success and false for failure.

Exercises
  1. Using the definition of k ( x , ε ) above, what is the value of k ( .5 , 1 ) , k ( .5 , .1 ) , and k ( .5 , .01 ) ?
  2. 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 ?
  3. 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 ?
Input File: introduction/exp_apx/exp_eps.hpp