1 /* Definitions of some C99 math library functions, for those platforms
2    that don't implement these functions already. */
3 
4 #include "Python.h"
5 #include <float.h>
6 #include "_math.h"
7 
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
17  * software is freely granted, provided that this notice
18  * is preserved.
19  * ====================================================
20  */
21 
22 static const double ln2 = 6.93147180559945286227E-01;
23 static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
24 static const double two_pow_p28 = 268435456.0; /* 2**28 */
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 
40 double
_Py_acosh(double x)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         }
58         else {
59             return log(x)+ln2;          /* acosh(huge)=log(2x) */
60         }
61     }
62     else if (x == 1.) {
63         return 0.0;                     /* acosh(1) = 0 */
64     }
65     else if (x > 2.) {                  /* 2 < x < 2**28 */
66         double t = x*x;
67         return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
68     }
69     else {                              /* 1 < x <= 2 */
70         double t = x - 1.0;
71         return m_log1p(t + sqrt(2.0*t + t*t));
72     }
73 }
74 
75 
76 /* asinh(x)
77  * Method :
78  *      Based on
79  *              asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
80  *      we have
81  *      asinh(x) := x  if  1+x*x=1,
82  *               := sign(x)*(log(x)+ln2)) for large |x|, else
83  *               := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
84  *               := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
85  */
86 
87 double
_Py_asinh(double x)88 _Py_asinh(double x)
89 {
90     double w;
91     double absx = fabs(x);
92 
93     if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
94         return x+x;
95     }
96     if (absx < two_pow_m28) {           /* |x| < 2**-28 */
97         return x;                       /* return x inexact except 0 */
98     }
99     if (absx > two_pow_p28) {           /* |x| > 2**28 */
100         w = log(absx)+ln2;
101     }
102     else if (absx > 2.0) {              /* 2 < |x| < 2**28 */
103         w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
104     }
105     else {                              /* 2**-28 <= |x| < 2= */
106         double t = x*x;
107         w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t)));
108     }
109     return copysign(w, x);
110 
111 }
112 
113 /* atanh(x)
114  * Method :
115  *    1.Reduced x to positive by atanh(-x) = -atanh(x)
116  *    2.For x>=0.5
117  *                  1              2x                          x
118  *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)
119  *                  2             1 - x                      1 - x
120  *
121  *      For x<0.5
122  *      atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
123  *
124  * Special cases:
125  *      atanh(x) is NaN if |x| >= 1 with signal;
126  *      atanh(NaN) is that NaN with no signal;
127  *
128  */
129 
130 double
_Py_atanh(double x)131 _Py_atanh(double x)
132 {
133     double absx;
134     double t;
135 
136     if (Py_IS_NAN(x)) {
137         return x+x;
138     }
139     absx = fabs(x);
140     if (absx >= 1.) {                   /* |x| >= 1 */
141         errno = EDOM;
142 #ifdef Py_NAN
143         return Py_NAN;
144 #else
145         return x/0.0;
146 #endif
147     }
148     if (absx < two_pow_m28) {           /* |x| < 2**-28 */
149         return x;
150     }
151     if (absx < 0.5) {                   /* |x| < 0.5 */
152         t = absx+absx;
153         t = 0.5 * m_log1p(t + t*absx / (1.0 - absx));
154     }
155     else {                              /* 0.5 <= |x| <= 1.0 */
156         t = 0.5 * m_log1p((absx + absx) / (1.0 - absx));
157     }
158     return copysign(t, x);
159 }
160 
161 /* Mathematically, expm1(x) = exp(x) - 1.  The expm1 function is designed
162    to avoid the significant loss of precision that arises from direct
163    evaluation of the expression exp(x) - 1, for x near 0. */
164 
165 double
_Py_expm1(double x)166 _Py_expm1(double x)
167 {
168     /* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
169        also works fine for infinities and nans.
170 
171        For smaller x, we can use a method due to Kahan that achieves close to
172        full accuracy.
173     */
174 
175     if (fabs(x) < 0.7) {
176         double u;
177         u = exp(x);
178         if (u == 1.0)
179             return x;
180         else
181             return (u - 1.0) * x / log(u);
182     }
183     else
184         return exp(x) - 1.0;
185 }
186 
187 /* log1p(x) = log(1+x).  The log1p function is designed to avoid the
188    significant loss of precision that arises from direct evaluation when x is
189    small. */
190 
191 #ifdef HAVE_LOG1P
192 
193 double
_Py_log1p(double x)194 _Py_log1p(double x)
195 {
196     /* Some platforms supply a log1p function but don't respect the sign of
197        zero:  log1p(-0.0) gives 0.0 instead of the correct result of -0.0.
198 
199        To save fiddling with configure tests and platform checks, we handle the
200        special case of zero input directly on all platforms.
201     */
202     if (x == 0.0) {
203         return x;
204     }
205     else {
206         return log1p(x);
207     }
208 }
209 
210 #else
211 
212 double
_Py_log1p(double x)213 _Py_log1p(double x)
214 {
215     /* For x small, we use the following approach.  Let y be the nearest float
216        to 1+x, then
217 
218          1+x = y * (1 - (y-1-x)/y)
219 
220        so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny, the
221        second term is well approximated by (y-1-x)/y.  If abs(x) >=
222        DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
223        then y-1-x will be exactly representable, and is computed exactly by
224        (y-1)-x.
225 
226        If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
227        round-to-nearest then this method is slightly dangerous: 1+x could be
228        rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
229        y-1-x will not be exactly representable any more and the result can be
230        off by many ulps.  But this is easily fixed: for a floating-point
231        number |x| < DBL_EPSILON/2., the closest floating-point number to
232        log(1+x) is exactly x.
233     */
234 
235     double y;
236     if (fabs(x) < DBL_EPSILON/2.) {
237         return x;
238     }
239     else if (-0.5 <= x && x <= 1.) {
240         /* WARNING: it's possible than an overeager compiler
241            will incorrectly optimize the following two lines
242            to the equivalent of "return log(1.+x)". If this
243            happens, then results from log1p will be inaccurate
244            for small x. */
245         y = 1.+x;
246         return log(y)-((y-1.)-x)/y;
247     }
248     else {
249         /* NaNs and infinities should end up here */
250         return log(1.+x);
251     }
252 }
253 
254 #endif /* ifdef HAVE_LOG1P */
255