1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
|
# ifndef CPPAD_INTRODUCTION_EXP_EPS_HPP
# define CPPAD_INTRODUCTION_EXP_EPS_HPP
// SPDX-License-Identifier: EPL-2.0 OR GPL-2.0-or-later
// SPDX-FileCopyrightText: Bradley M. Bell <bradbell@seanet.com>
// SPDX-FileContributor: 2003-24 Bradley M. Bell
// ----------------------------------------------------------------------------
/*
{xrst_begin exp_eps}
{xrst_spell
apx
yyyymmdd
}
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 :ref:`exp_2-name` for a simpler example).
Mathematical Function
*********************
The exponential function can be defined by
.. math::
\exp (x) = 1 + x^1 / 1 ! + x^2 / 2 ! + \cdots
We define :math:`k ( x, \varepsilon )` as the smallest
non-negative integer such that :math:`\varepsilon \geq x^k / k !`; i.e.,
.. math::
k( x, \varepsilon ) =
\min \{ k \in {\rm Z}_+ \; | \; \varepsilon \geq x^k / k ! \}
The mathematical form for our approximation of the exponential function is
.. math::
:nowrap:
\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
:ref:`installation<Install-name>` 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 :math:`\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`` :
.. list-table::
:widths: auto
* - **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 :math:`u * v`
* - *u* / *v*
- *Type*
- result is value of :math:`u / v`
* - *u* + *v*
- *Type*
- result is value of :math:`u + v`
* - ``-`` *u*
- *Type*
- result is value of :math:`- u`
{xrst_toc_hidden
introduction/exp_eps.xrst
introduction/exp_eps_cppad.cpp
}
Implementation
**************
The file :ref:`exp_eps.hpp-name`
contains a C++ implementation of this function.
Test
****
The file :ref:`exp_eps.cpp-name`
contains a test of this implementation.
Exercises
*********
#. Using the definition of :math:`k( x, \varepsilon )` above,
what is the value of
:math:`k(.5, 1)`, :math:`k(.5, .1)`, and :math:`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 :ref:`exp_eps.hpp-name` ?
#. Continuing the previous exercise,
what is the value assigned to
``k`` , ``temp`` , ``term`` , and ``sum``
the second time through the ``while`` loop in :ref:`exp_eps.hpp-name` ?
{xrst_end exp_eps}
-----------------------------------------------------------------------------
*/
// BEGIN C++
template <class Type>
Type exp_eps(const Type &x, const Type &epsilon)
{ // abs_x = |x|
Type abs_x = x;
if( Type(0) > x )
abs_x = - x;
// initialize
int k = 0; // initial order
Type term = 1.; // term = |x|^k / k !
Type sum = term; // initial sum
while(term > epsilon)
{ k = k + 1; // order for next term
Type temp = term * abs_x; // term = |x|^k / (k-1)!
term = temp / Type(k); // term = |x|^k / k !
sum = sum + term; // sum = 1 + ... + |x|^k / k !
}
// In the case where x is negative, use exp(x) = 1 / exp(-|x|)
if( Type(0) > x )
sum = Type(1) / sum;
return sum;
}
// END C++
# endif
|
