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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 Dickinsonf3718592009-12-21 15:27:41 +00006
7/* The following copyright notice applies to the original
8 implementations of acosh, asinh and atanh. */
9
10/*
11 * ====================================================
12 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
13 *
14 * Developed at SunPro, a Sun Microsystems, Inc. business.
15 * Permission to use, copy, modify, and distribute this
16 * software is freely granted, provided that this notice
17 * is preserved.
18 * ====================================================
19 */
20
21static const double ln2 = 6.93147180559945286227E-01;
22static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
23static const double two_pow_p28 = 268435456.0; /* 2**28 */
24static const double zero = 0.0;
25
26/* acosh(x)
27 * Method :
28 * Based on
29 * acosh(x) = log [ x + sqrt(x*x-1) ]
30 * we have
31 * acosh(x) := log(x)+ln2, if x is large; else
32 * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
33 * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
34 *
35 * Special cases:
36 * acosh(x) is NaN with signal if x<1.
37 * acosh(NaN) is NaN without signal.
38 */
39
40double
41_Py_acosh(double x)
42{
43 if (Py_IS_NAN(x)) {
44 return x+x;
45 }
46 if (x < 1.) { /* x < 1; return a signaling NaN */
47 errno = EDOM;
48#ifdef Py_NAN
49 return Py_NAN;
50#else
51 return (x-x)/(x-x);
52#endif
53 }
54 else if (x >= two_pow_p28) { /* x > 2**28 */
55 if (Py_IS_INFINITY(x)) {
56 return x+x;
57 } else {
58 return log(x)+ln2; /* acosh(huge)=log(2x) */
59 }
60 }
61 else if (x == 1.) {
62 return 0.0; /* acosh(1) = 0 */
63 }
64 else if (x > 2.) { /* 2 < x < 2**28 */
65 double t = x*x;
66 return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
67 }
68 else { /* 1 < x <= 2 */
69 double t = x - 1.0;
70 return log1p(t + sqrt(2.0*t + t*t));
71 }
72}
73
74
75/* asinh(x)
76 * Method :
77 * Based on
78 * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
79 * we have
80 * asinh(x) := x if 1+x*x=1,
81 * := sign(x)*(log(x)+ln2)) for large |x|, else
82 * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
83 * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
84 */
85
86double
87_Py_asinh(double x)
88{
89 double w;
90 double absx = fabs(x);
91
92 if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
93 return x+x;
94 }
95 if (absx < two_pow_m28) { /* |x| < 2**-28 */
96 return x; /* return x inexact except 0 */
97 }
98 if (absx > two_pow_p28) { /* |x| > 2**28 */
99 w = log(absx)+ln2;
100 }
101 else if (absx > 2.0) { /* 2 < |x| < 2**28 */
102 w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
103 }
104 else { /* 2**-28 <= |x| < 2= */
105 double t = x*x;
106 w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
107 }
108 return copysign(w, x);
109
110}
111
112/* atanh(x)
113 * Method :
114 * 1.Reduced x to positive by atanh(-x) = -atanh(x)
115 * 2.For x>=0.5
116 * 1 2x x
117 * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
118 * 2 1 - x 1 - x
119 *
120 * For x<0.5
121 * atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
122 *
123 * Special cases:
124 * atanh(x) is NaN if |x| >= 1 with signal;
125 * atanh(NaN) is that NaN with no signal;
126 *
127 */
128
129double
130_Py_atanh(double x)
131{
132 double absx;
133 double t;
134
135 if (Py_IS_NAN(x)) {
136 return x+x;
137 }
138 absx = fabs(x);
139 if (absx >= 1.) { /* |x| >= 1 */
140 errno = EDOM;
141#ifdef Py_NAN
142 return Py_NAN;
143#else
144 return x/zero;
145#endif
146 }
147 if (absx < two_pow_m28) { /* |x| < 2**-28 */
148 return x;
149 }
150 if (absx < 0.5) { /* |x| < 0.5 */
151 t = absx+absx;
152 t = 0.5 * log1p(t + t*absx / (1.0 - absx));
153 }
154 else { /* 0.5 <= |x| <= 1.0 */
155 t = 0.5 * log1p((absx + absx) / (1.0 - absx));
156 }
157 return copysign(t, x);
158}
Mark Dickinson664b5112009-12-16 20:23:42 +0000159
160/* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed
161 to avoid the significant loss of precision that arises from direct
162 evaluation of the expression exp(x) - 1, for x near 0. */
163
164double
165_Py_expm1(double x)
166{
167 /* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
168 also works fine for infinities and nans.
169
170 For smaller x, we can use a method due to Kahan that achieves close to
171 full accuracy.
172 */
173
174 if (fabs(x) < 0.7) {
175 double u;
176 u = exp(x);
177 if (u == 1.0)
178 return x;
179 else
180 return (u - 1.0) * x / log(u);
181 }
182 else
183 return exp(x) - 1.0;
184}
Mark Dickinsonf3718592009-12-21 15:27:41 +0000185
186/* log1p(x) = log(1+x). The log1p function is designed to avoid the
187 significant loss of precision that arises from direct evaluation when x is
188 small. */
189
190double
191_Py_log1p(double x)
192{
193 /* For x small, we use the following approach. Let y be the nearest float
194 to 1+x, then
195
196 1+x = y * (1 - (y-1-x)/y)
197
198 so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny, the
199 second term is well approximated by (y-1-x)/y. If abs(x) >=
200 DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
201 then y-1-x will be exactly representable, and is computed exactly by
202 (y-1)-x.
203
204 If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
205 round-to-nearest then this method is slightly dangerous: 1+x could be
206 rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
207 y-1-x will not be exactly representable any more and the result can be
208 off by many ulps. But this is easily fixed: for a floating-point
209 number |x| < DBL_EPSILON/2., the closest floating-point number to
210 log(1+x) is exactly x.
211 */
212
213 double y;
214 if (fabs(x) < DBL_EPSILON/2.) {
215 return x;
216 } else if (-0.5 <= x && x <= 1.) {
217 /* WARNING: it's possible than an overeager compiler
218 will incorrectly optimize the following two lines
219 to the equivalent of "return log(1.+x)". If this
220 happens, then results from log1p will be inaccurate
221 for small x. */
222 y = 1.+x;
223 return log(y)-((y-1.)-x)/y;
224 } else {
225 /* NaNs and infinities should end up here */
226 return log(1.+x);
227 }
228}