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 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 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/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 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 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 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
