Modules/_math.c

   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 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                 } 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
  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
 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/zero;
 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
 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 double
 192 _Py_log1p(double x)
 193 {
 194     /* For x small, we use the following approach.  Let y be the nearest float
 195        to 1+x, then
 196
 197           1+x = y * (1 - (y-1-x)/y)
 198
 199        so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny, the
 200        second term is well approximated by (y-1-x)/y.  If abs(x) >=
 201        DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
 202        then y-1-x will be exactly representable, and is computed exactly by
 203        (y-1)-x.
 204
 205        If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
 206        round-to-nearest then this method is slightly dangerous: 1+x could be
 207        rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
 208        y-1-x will not be exactly representable any more and the result can be
 209        off by many ulps.  But this is easily fixed: for a floating-point
 210        number |x| < DBL_EPSILON/2., the closest floating-point number to
 211        log(1+x) is exactly x.
 212     */
 213
 214     double y;
 215     if (fabs(x) < DBL_EPSILON/2.) {
 216         return x;
 217     } else if (-0.5 <= x && x <= 1.) {
 218         /* WARNING: it's possible than an overeager compiler
 219            will incorrectly optimize the following two lines
 220            to the equivalent of "return log(1.+x)". If this
 221            happens, then results from log1p will be inaccurate
 222            for small x. */
 223         y = 1.+x;
 224         return log(y)-((y-1.)-x)/y;
 225     } else {
 226         /* NaNs and infinities should end up here */
 227         return log(1.+x);
 228     }
 229 }