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 } 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 } 233
