\(\newcommand{\W}[1]{ \; #1 \; }\) \(\newcommand{\R}[1]{ {\rm #1} }\) \(\newcommand{\B}[1]{ {\bf #1} }\) \(\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }\) \(\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }\) \(\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }\) \(\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }\)

exp_2

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Second Order Exponential Approximation

Syntax

# include "exp_2.hpp"

y = exp_2 ( x )

Purpose

This is a simple example algorithm that is used to demonstrate Algorithmic Differentiation (see exp_eps for a more complex example).

Mathematical Form

The exponential function can be defined by

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

The second order approximation for the exponential function is

\[{\rm exp\_2} (x) = 1 + x + x^2 / 2\]

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 second order exponential approximation.

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	construct object with value equal to i
Type u = v	Type	construct u with value equal to 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 \(u + v\)

Contents

Name	Title
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

Implementation

The file exp_2.hpp contains a C++ implementation of this function.

Test

The file exp_2.cpp contains a test of this implementation.

Exercises

1. Suppose that we make the call

   double x = .1;
   double y = exp_2(x);
   

   What is the value assigned to v1 , v2 , … ,``v5`` in exp_2.hpp ?

2. Extend the routine 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).

3. Suppose that we make the call

   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 ) ?