Mark Dickinson | 664b511 | 2009-12-16 20:23:42 +0000 | [diff] [blame] | 1 | /* Definitions of some C99 math library functions, for those platforms |
| 2 | that don't implement these functions already. */ |
| 3 | |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 4 | #include "Python.h" |
Mark Dickinson | 664b511 | 2009-12-16 20:23:42 +0000 | [diff] [blame] | 5 | #include <float.h> |
Mark Dickinson | 82ebc1f | 2009-12-21 15:42:00 +0000 | [diff] [blame] | 6 | #include "_math.h" |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 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; |
Mark Dickinson | 82ebc1f | 2009-12-21 15:42:00 +0000 | [diff] [blame] | 71 | return m_log1p(t + sqrt(2.0*t + t*t)); |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 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; |
Mark Dickinson | 82ebc1f | 2009-12-21 15:42:00 +0000 | [diff] [blame] | 107 | w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t))); |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 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; |
Mark Dickinson | 82ebc1f | 2009-12-21 15:42:00 +0000 | [diff] [blame] | 153 | t = 0.5 * m_log1p(t + t*absx / (1.0 - absx)); |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 154 | } |
| 155 | else { /* 0.5 <= |x| <= 1.0 */ |
Mark Dickinson | 82ebc1f | 2009-12-21 15:42:00 +0000 | [diff] [blame] | 156 | t = 0.5 * m_log1p((absx + absx) / (1.0 - absx)); |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 157 | } |
| 158 | return copysign(t, x); |
| 159 | } |
Mark Dickinson | 664b511 | 2009-12-16 20:23:42 +0000 | [diff] [blame] | 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 | } |
Mark Dickinson | f371859 | 2009-12-21 15:27:41 +0000 | [diff] [blame] | 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 | } |