## Python-2.7.3/Modules/_math.c

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```  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  * ====================================================
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 static const double zero = 0.0;
26
27 /* acosh(x)
28  * Method :
29  *      Based on
30  *            acosh(x) = log [ x + sqrt(x*x-1) ]
31  *      we have
32  *            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.
35  *
36  * Special cases:
37  *      acosh(x) is NaN with signal if x<1.
38  *      acosh(NaN) is NaN without signal.
39  */
40
41 double
42 _Py_acosh(double x)
43 {
44     if (Py_IS_NAN(x)) {
45         return x+x;
46     }
47     if (x < 1.) {                       /* x < 1;  return a signaling NaN */
48         errno = EDOM;
49 #ifdef Py_NAN
50         return Py_NAN;
51 #else
52         return (x-x)/(x-x);
53 #endif
54     }
55     else if (x >= two_pow_p28) {        /* x > 2**28 */
56         if (Py_IS_INFINITY(x)) {
57             return x+x;
58         }
59         else {
60             return log(x)+ln2;          /* acosh(huge)=log(2x) */
61         }
62     }
63     else if (x == 1.) {
64         return 0.0;                     /* acosh(1) = 0 */
65     }
66     else if (x > 2.) {                  /* 2 < x < 2**28 */
67         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     }
74 }
75
76
77 /* asinh(x)
78  * Method :
79  *      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)))
86  */
87
88 double
89 _Py_asinh(double x)
90 {
91     double w;
92     double absx = fabs(x);
93
94     if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
95         return x+x;
96     }
97     if (absx < two_pow_m28) {           /* |x| < 2**-28 */
98         return x;                       /* return x inexact except 0 */
99     }
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
112 }
113
114 /* atanh(x)
115  * Method :
116  *    1.Reduced x to positive by atanh(-x) = -atanh(x)
117  *    2.For x>=0.5
118  *                  1              2x                          x
119  *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)
120  *                  2             1 - x                      1 - x
121  *
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
131 double
132 _Py_atanh(double x)
133 {
134     double absx;
135     double t;
136
137     if (Py_IS_NAN(x)) {
138         return x+x;
139     }
140     absx = fabs(x);
141     if (absx >= 1.) {                   /* |x| >= 1 */
142         errno = EDOM;
143 #ifdef Py_NAN
144         return Py_NAN;
145 #else
146         return x/zero;
147 #endif
148     }
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);
160 }
161
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
166 double
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) {
177         double u;
178         u = exp(x);
179         if (u == 1.0)
180             return x;
181         else
182             return (u - 1.0) * x / log(u);
183     }
184     else
185         return exp(x) - 1.0;
186 }
187
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
192 double
193 _Py_log1p(double x)
194 {
195     /* For x small, we use the following approach.  Let y be the nearest float
196        to 1+x, then
197
198          1+x = y * (1 - (y-1-x)/y)
199
200        so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny, the
201        second term is well approximated by (y-1-x)/y.  If abs(x) >=
202        DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
203        then y-1-x will be exactly representable, and is computed exactly by
204        (y-1)-x.
205
206        If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
207        round-to-nearest then this method is slightly dangerous: 1+x could be
208        rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
209        y-1-x will not be exactly representable any more and the result can be
210        off by many ulps.  But this is easily fixed: for a floating-point
211        number |x| < DBL_EPSILON/2., the closest floating-point number to
212        log(1+x) is exactly x.
213     */
214
215     double y;
216     if (fabs(x) < DBL_EPSILON/2.) {
217         return x;
218     }
219     else if (-0.5 <= x && x <= 1.) {
220         /* WARNING: it's possible than an overeager compiler
221            will incorrectly optimize the following two lines
222            to the equivalent of "return log(1.+x)". If this
223            happens, then results from log1p will be inaccurate
224            for small x. */
225         y = 1.+x;
226         return log(y)-((y-1.)-x)/y;
227     }
228     else {
229         /* NaNs and infinities should end up here */
230         return log(1.+x);
231     }
232 }
```