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
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
|
/* Definitions of some C99 math library functions, for those platforms
that don't implement these functions already. */
#include "Python.h"
#include <float.h>
/* The following copyright notice applies to the original
implementations of acosh, asinh and atanh. */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
static const double ln2 = 6.93147180559945286227E-01;
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
static const double two_pow_p28 = 268435456.0; /* 2**28 */
static const double zero = 0.0;
/* acosh(x)
* Method :
* Based on
* acosh(x) = log [ x + sqrt(x*x-1) ]
* we have
* acosh(x) := log(x)+ln2, if x is large; else
* acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
* acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
*
* Special cases:
* acosh(x) is NaN with signal if x<1.
* acosh(NaN) is NaN without signal.
*/
double
_Py_acosh(double x)
{
if (Py_IS_NAN(x)) {
return x+x;
}
if (x < 1.) { /* x < 1; return a signaling NaN */
errno = EDOM;
#ifdef Py_NAN
return Py_NAN;
#else
return (x-x)/(x-x);
#endif
}
else if (x >= two_pow_p28) { /* x > 2**28 */
if (Py_IS_INFINITY(x)) {
return x+x;
} else {
return log(x)+ln2; /* acosh(huge)=log(2x) */
}
}
else if (x == 1.) {
return 0.0; /* acosh(1) = 0 */
}
else if (x > 2.) { /* 2 < x < 2**28 */
double t = x*x;
return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
}
else { /* 1 < x <= 2 */
double t = x - 1.0;
return log1p(t + sqrt(2.0*t + t*t));
}
}
/* asinh(x)
* Method :
* Based on
* asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
* we have
* asinh(x) := x if 1+x*x=1,
* := sign(x)*(log(x)+ln2)) for large |x|, else
* := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
* := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
*/
double
_Py_asinh(double x)
{
double w;
double absx = fabs(x);
if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
return x+x;
}
if (absx < two_pow_m28) { /* |x| < 2**-28 */
return x; /* return x inexact except 0 */
}
if (absx > two_pow_p28) { /* |x| > 2**28 */
w = log(absx)+ln2;
}
else if (absx > 2.0) { /* 2 < |x| < 2**28 */
w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
}
else { /* 2**-28 <= |x| < 2= */
double t = x*x;
w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
}
return copysign(w, x);
}
/* atanh(x)
* Method :
* 1.Reduced x to positive by atanh(-x) = -atanh(x)
* 2.For x>=0.5
* 1 2x x
* atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
* 2 1 - x 1 - x
*
* For x<0.5
* atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
*
* Special cases:
* atanh(x) is NaN if |x| >= 1 with signal;
* atanh(NaN) is that NaN with no signal;
*
*/
double
_Py_atanh(double x)
{
double absx;
double t;
if (Py_IS_NAN(x)) {
return x+x;
}
absx = fabs(x);
if (absx >= 1.) { /* |x| >= 1 */
errno = EDOM;
#ifdef Py_NAN
return Py_NAN;
#else
return x/zero;
#endif
}
if (absx < two_pow_m28) { /* |x| < 2**-28 */
return x;
}
if (absx < 0.5) { /* |x| < 0.5 */
t = absx+absx;
t = 0.5 * log1p(t + t*absx / (1.0 - absx));
}
else { /* 0.5 <= |x| <= 1.0 */
t = 0.5 * log1p((absx + absx) / (1.0 - absx));
}
return copysign(t, x);
}
/* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed
to avoid the significant loss of precision that arises from direct
evaluation of the expression exp(x) - 1, for x near 0. */
double
_Py_expm1(double x)
{
/* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
also works fine for infinities and nans.
For smaller x, we can use a method due to Kahan that achieves close to
full accuracy.
*/
if (fabs(x) < 0.7) {
double u;
u = exp(x);
if (u == 1.0)
return x;
else
return (u - 1.0) * x / log(u);
}
else
return exp(x) - 1.0;
}
/* log1p(x) = log(1+x). The log1p function is designed to avoid the
significant loss of precision that arises from direct evaluation when x is
small. */
double
_Py_log1p(double x)
{
/* For x small, we use the following approach. Let y be the nearest float
to 1+x, then
1+x = y * (1 - (y-1-x)/y)
so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny, the
second term is well approximated by (y-1-x)/y. If abs(x) >=
DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
then y-1-x will be exactly representable, and is computed exactly by
(y-1)-x.
If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
round-to-nearest then this method is slightly dangerous: 1+x could be
rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
y-1-x will not be exactly representable any more and the result can be
off by many ulps. But this is easily fixed: for a floating-point
number |x| < DBL_EPSILON/2., the closest floating-point number to
log(1+x) is exactly x.
*/
double y;
if (fabs(x) < DBL_EPSILON/2.) {
return x;
} else if (-0.5 <= x && x <= 1.) {
/* WARNING: it's possible than an overeager compiler
will incorrectly optimize the following two lines
to the equivalent of "return log(1.+x)". If this
happens, then results from log1p will be inaccurate
for small x. */
y = 1.+x;
return log(y)-((y-1.)-x)/y;
} else {
/* NaNs and infinities should end up here */
return log(1.+x);
}
}
|
