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Mark Dickinson664b5112009-12-16 20:23:42 +00001/* Definitions of some C99 math library functions, for those platforms
2 that don't implement these functions already. */
3
Mark Dickinsonf3718592009-12-21 15:27:41 +00004#include "Python.h"
Mark Dickinson664b5112009-12-16 20:23:42 +00005#include <float.h>
Mark Dickinson82ebc1f2009-12-21 15:42:00 +00006#include "_math.h"
Mark Dickinsonf3718592009-12-21 15:27:41 +00007
8/* The following copyright notice applies to the original
9 implementations of acosh, asinh and atanh. */
10
11/*
12 * ====================================================
13 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
14 *
15 * Developed at SunPro, a Sun Microsystems, Inc. business.
16 * Permission to use, copy, modify, and distribute this
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000017 * software is freely granted, provided that this notice
Mark Dickinsonf3718592009-12-21 15:27:41 +000018 * is preserved.
19 * ====================================================
20 */
21
22static const double ln2 = 6.93147180559945286227E-01;
23static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
24static const double two_pow_p28 = 268435456.0; /* 2**28 */
25static const double zero = 0.0;
26
27/* acosh(x)
28 * Method :
29 * Based on
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000030 * acosh(x) = log [ x + sqrt(x*x-1) ]
Mark Dickinsonf3718592009-12-21 15:27:41 +000031 * we have
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000032 * acosh(x) := log(x)+ln2, if x is large; else
33 * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
34 * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
Mark Dickinsonf3718592009-12-21 15:27:41 +000035 *
36 * Special cases:
37 * acosh(x) is NaN with signal if x<1.
38 * acosh(NaN) is NaN without signal.
39 */
40
41double
42_Py_acosh(double x)
43{
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000044 if (Py_IS_NAN(x)) {
45 return x+x;
46 }
47 if (x < 1.) { /* x < 1; return a signaling NaN */
48 errno = EDOM;
Mark Dickinsonf3718592009-12-21 15:27:41 +000049#ifdef Py_NAN
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000050 return Py_NAN;
Mark Dickinsonf3718592009-12-21 15:27:41 +000051#else
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000052 return (x-x)/(x-x);
Mark Dickinsonf3718592009-12-21 15:27:41 +000053#endif
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000054 }
55 else if (x >= two_pow_p28) { /* x > 2**28 */
56 if (Py_IS_INFINITY(x)) {
57 return x+x;
Mark Dickinson6f493b72010-07-06 15:00:40 +000058 }
59 else {
Mark Dickinson6c3bcb72010-07-05 20:14:26 +000060 return log(x)+ln2; /* acosh(huge)=log(2x) */
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000061 }
62 }
63 else if (x == 1.) {
Mark Dickinson6c3bcb72010-07-05 20:14:26 +000064 return 0.0; /* acosh(1) = 0 */
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000065 }
Mark Dickinson6c3bcb72010-07-05 20:14:26 +000066 else if (x > 2.) { /* 2 < x < 2**28 */
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000067 double t = x*x;
68 return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
69 }
70 else { /* 1 < x <= 2 */
71 double t = x - 1.0;
72 return m_log1p(t + sqrt(2.0*t + t*t));
73 }
Mark Dickinsonf3718592009-12-21 15:27:41 +000074}
75
76
77/* asinh(x)
78 * Method :
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000079 * Based on
80 * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
81 * we have
82 * asinh(x) := x if 1+x*x=1,
83 * := sign(x)*(log(x)+ln2)) for large |x|, else
84 * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
85 * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
Mark Dickinsonf3718592009-12-21 15:27:41 +000086 */
87
88double
89_Py_asinh(double x)
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000090{
91 double w;
92 double absx = fabs(x);
Mark Dickinsonf3718592009-12-21 15:27:41 +000093
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000094 if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
95 return x+x;
96 }
97 if (absx < two_pow_m28) { /* |x| < 2**-28 */
Mark Dickinson6c3bcb72010-07-05 20:14:26 +000098 return x; /* return x inexact except 0 */
Antoine Pitrouf95a1b32010-05-09 15:52:27 +000099 }
100 if (absx > two_pow_p28) { /* |x| > 2**28 */
101 w = log(absx)+ln2;
102 }
103 else if (absx > 2.0) { /* 2 < |x| < 2**28 */
104 w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
105 }
106 else { /* 2**-28 <= |x| < 2= */
107 double t = x*x;
108 w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t)));
109 }
110 return copysign(w, x);
111
Mark Dickinsonf3718592009-12-21 15:27:41 +0000112}
113
114/* atanh(x)
115 * Method :
116 * 1.Reduced x to positive by atanh(-x) = -atanh(x)
117 * 2.For x>=0.5
Mark Dickinson6c3bcb72010-07-05 20:14:26 +0000118 * 1 2x x
119 * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)
120 * 2 1 - x 1 - x
Mark Dickinsonf3718592009-12-21 15:27:41 +0000121 *
122 * For x<0.5
123 * atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
124 *
125 * Special cases:
126 * atanh(x) is NaN if |x| >= 1 with signal;
127 * atanh(NaN) is that NaN with no signal;
128 *
129 */
130
131double
132_Py_atanh(double x)
133{
Antoine Pitrouf95a1b32010-05-09 15:52:27 +0000134 double absx;
135 double t;
Mark Dickinsonf3718592009-12-21 15:27:41 +0000136
Antoine Pitrouf95a1b32010-05-09 15:52:27 +0000137 if (Py_IS_NAN(x)) {
138 return x+x;
139 }
140 absx = fabs(x);
141 if (absx >= 1.) { /* |x| >= 1 */
142 errno = EDOM;
Mark Dickinsonf3718592009-12-21 15:27:41 +0000143#ifdef Py_NAN
Antoine Pitrouf95a1b32010-05-09 15:52:27 +0000144 return Py_NAN;
Mark Dickinsonf3718592009-12-21 15:27:41 +0000145#else
Antoine Pitrouf95a1b32010-05-09 15:52:27 +0000146 return x/zero;
Mark Dickinsonf3718592009-12-21 15:27:41 +0000147#endif
Antoine Pitrouf95a1b32010-05-09 15:52:27 +0000148 }
149 if (absx < two_pow_m28) { /* |x| < 2**-28 */
150 return x;
151 }
152 if (absx < 0.5) { /* |x| < 0.5 */
153 t = absx+absx;
154 t = 0.5 * m_log1p(t + t*absx / (1.0 - absx));
155 }
156 else { /* 0.5 <= |x| <= 1.0 */
157 t = 0.5 * m_log1p((absx + absx) / (1.0 - absx));
158 }
159 return copysign(t, x);
Mark Dickinsonf3718592009-12-21 15:27:41 +0000160}
Mark Dickinson664b5112009-12-16 20:23:42 +0000161
162/* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed
163 to avoid the significant loss of precision that arises from direct
164 evaluation of the expression exp(x) - 1, for x near 0. */
165
166double
167_Py_expm1(double x)
168{
169 /* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
170 also works fine for infinities and nans.
171
172 For smaller x, we can use a method due to Kahan that achieves close to
173 full accuracy.
174 */
175
176 if (fabs(x) < 0.7) {
Mark Dickinson6f493b72010-07-06 15:00:40 +0000177 double u;
178 u = exp(x);
179 if (u == 1.0)
180 return x;
181 else
182 return (u - 1.0) * x / log(u);
Mark Dickinson664b5112009-12-16 20:23:42 +0000183 }
184 else
Mark Dickinson6f493b72010-07-06 15:00:40 +0000185 return exp(x) - 1.0;
Mark Dickinson664b5112009-12-16 20:23:42 +0000186}
Mark Dickinsonf3718592009-12-21 15:27:41 +0000187
188/* log1p(x) = log(1+x). The log1p function is designed to avoid the
189 significant loss of precision that arises from direct evaluation when x is
190 small. */
191
Mark Dickinson05d79e92012-08-18 12:24:30 +0100192#ifdef HAVE_LOG1P
193
194double
195_Py_log1p(double x)
196{
197 /* Some platforms supply a log1p function but don't respect the sign of
198 zero: log1p(-0.0) gives 0.0 instead of the correct result of -0.0.
199
200 To save fiddling with configure tests and platform checks, we handle the
201 special case of zero input directly on all platforms.
202 */
203 if (x == 0.0) {
204 return x;
205 }
206 else {
207 return log1p(x);
208 }
209}
210
211#else
212
Mark Dickinsonf3718592009-12-21 15:27:41 +0000213double
214_Py_log1p(double x)
215{
216 /* For x small, we use the following approach. Let y be the nearest float
217 to 1+x, then
218
Mark Dickinson6c3bcb72010-07-05 20:14:26 +0000219 1+x = y * (1 - (y-1-x)/y)
Mark Dickinsonf3718592009-12-21 15:27:41 +0000220
221 so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny, the
222 second term is well approximated by (y-1-x)/y. If abs(x) >=
223 DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
224 then y-1-x will be exactly representable, and is computed exactly by
225 (y-1)-x.
226
227 If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
228 round-to-nearest then this method is slightly dangerous: 1+x could be
229 rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
230 y-1-x will not be exactly representable any more and the result can be
231 off by many ulps. But this is easily fixed: for a floating-point
232 number |x| < DBL_EPSILON/2., the closest floating-point number to
233 log(1+x) is exactly x.
234 */
235
236 double y;
237 if (fabs(x) < DBL_EPSILON/2.) {
Mark Dickinson6f493b72010-07-06 15:00:40 +0000238 return x;
239 }
240 else if (-0.5 <= x && x <= 1.) {
241 /* WARNING: it's possible than an overeager compiler
242 will incorrectly optimize the following two lines
243 to the equivalent of "return log(1.+x)". If this
244 happens, then results from log1p will be inaccurate
245 for small x. */
246 y = 1.+x;
247 return log(y)-((y-1.)-x)/y;
248 }
249 else {
250 /* NaNs and infinities should end up here */
251 return log(1.+x);
Mark Dickinsonf3718592009-12-21 15:27:41 +0000252 }
253}
Mark Dickinson05d79e92012-08-18 12:24:30 +0100254
255#endif /* ifdef HAVE_LOG1P */