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@(@\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 .
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{array}{rcl} {\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{array} @]@

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
  1. Using the definition of @(@ k( x, \varepsilon ) @)@ 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_eps.hpp