| /* Definitions of some C99 math library functions, for those platforms | |
| that don't implement these functions already. */ | |
| #include "Python.h" | |
| #include <float.h> | |
| #include "_math.h" | |
| /* The following copyright notice applies to the original | |
| implementations of acosh, asinh and atanh. */ | |
| /* | |
| * ==================================================== | |
| * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. | |
| * | |
| * Developed at SunPro, a Sun Microsystems, Inc. business. | |
| * Permission to use, copy, modify, and distribute this | |
| * software is freely granted, provided that this notice | |
| * is preserved. | |
| * ==================================================== | |
| */ | |
| static const double ln2 = 6.93147180559945286227E-01; | |
| static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */ | |
| static const double two_pow_p28 = 268435456.0; /* 2**28 */ | |
| static const double zero = 0.0; | |
| /* acosh(x) | |
| * Method : | |
| * Based on | |
| * acosh(x) = log [ x + sqrt(x*x-1) ] | |
| * we have | |
| * acosh(x) := log(x)+ln2, if x is large; else | |
| * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else | |
| * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1. | |
| * | |
| * Special cases: | |
| * acosh(x) is NaN with signal if x<1. | |
| * acosh(NaN) is NaN without signal. | |
| */ | |
| double | |
| _Py_acosh(double x) | |
| { | |
| if (Py_IS_NAN(x)) { | |
| return x+x; | |
| } | |
| if (x < 1.) { /* x < 1; return a signaling NaN */ | |
| errno = EDOM; | |
| #ifdef Py_NAN | |
| return Py_NAN; | |
| #else | |
| return (x-x)/(x-x); | |
| #endif | |
| } | |
| else if (x >= two_pow_p28) { /* x > 2**28 */ | |
| if (Py_IS_INFINITY(x)) { | |
| return x+x; | |
| } | |
| else { | |
| return log(x)+ln2; /* acosh(huge)=log(2x) */ | |
| } | |
| } | |
| else if (x == 1.) { | |
| return 0.0; /* acosh(1) = 0 */ | |
| } | |
| else if (x > 2.) { /* 2 < x < 2**28 */ | |
| double t = x*x; | |
| return log(2.0*x - 1.0 / (x + sqrt(t - 1.0))); | |
| } | |
| else { /* 1 < x <= 2 */ | |
| double t = x - 1.0; | |
| return m_log1p(t + sqrt(2.0*t + t*t)); | |
| } | |
| } | |
| /* asinh(x) | |
| * Method : | |
| * Based on | |
| * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ] | |
| * we have | |
| * asinh(x) := x if 1+x*x=1, | |
| * := sign(x)*(log(x)+ln2)) for large |x|, else | |
| * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else | |
| * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2))) | |
| */ | |
| double | |
| _Py_asinh(double x) | |
| { | |
| double w; | |
| double absx = fabs(x); | |
| if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) { | |
| return x+x; | |
| } | |
| if (absx < two_pow_m28) { /* |x| < 2**-28 */ | |
| return x; /* return x inexact except 0 */ | |
| } | |
| if (absx > two_pow_p28) { /* |x| > 2**28 */ | |
| w = log(absx)+ln2; | |
| } | |
| else if (absx > 2.0) { /* 2 < |x| < 2**28 */ | |
| w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx)); | |
| } | |
| else { /* 2**-28 <= |x| < 2= */ | |
| double t = x*x; | |
| w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t))); | |
| } | |
| return copysign(w, x); | |
| } | |
| /* atanh(x) | |
| * Method : | |
| * 1.Reduced x to positive by atanh(-x) = -atanh(x) | |
| * 2.For x>=0.5 | |
| * 1 2x x | |
| * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------) | |
| * 2 1 - x 1 - x | |
| * | |
| * For x<0.5 | |
| * atanh(x) = 0.5*log1p(2x+2x*x/(1-x)) | |
| * | |
| * Special cases: | |
| * atanh(x) is NaN if |x| >= 1 with signal; | |
| * atanh(NaN) is that NaN with no signal; | |
| * | |
| */ | |
| double | |
| _Py_atanh(double x) | |
| { | |
| double absx; | |
| double t; | |
| if (Py_IS_NAN(x)) { | |
| return x+x; | |
| } | |
| absx = fabs(x); | |
| if (absx >= 1.) { /* |x| >= 1 */ | |
| errno = EDOM; | |
| #ifdef Py_NAN | |
| return Py_NAN; | |
| #else | |
| return x/zero; | |
| #endif | |
| } | |
| if (absx < two_pow_m28) { /* |x| < 2**-28 */ | |
| return x; | |
| } | |
| if (absx < 0.5) { /* |x| < 0.5 */ | |
| t = absx+absx; | |
| t = 0.5 * m_log1p(t + t*absx / (1.0 - absx)); | |
| } | |
| else { /* 0.5 <= |x| <= 1.0 */ | |
| t = 0.5 * m_log1p((absx + absx) / (1.0 - absx)); | |
| } | |
| return copysign(t, x); | |
| } | |
| /* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed | |
| to avoid the significant loss of precision that arises from direct | |
| evaluation of the expression exp(x) - 1, for x near 0. */ | |
| double | |
| _Py_expm1(double x) | |
| { | |
| /* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this | |
| also works fine for infinities and nans. | |
| For smaller x, we can use a method due to Kahan that achieves close to | |
| full accuracy. | |
| */ | |
| if (fabs(x) < 0.7) { | |
| double u; | |
| u = exp(x); | |
| if (u == 1.0) | |
| return x; | |
| else | |
| return (u - 1.0) * x / log(u); | |
| } | |
| else | |
| return exp(x) - 1.0; | |
| } | |
| /* log1p(x) = log(1+x). The log1p function is designed to avoid the | |
| significant loss of precision that arises from direct evaluation when x is | |
| small. */ | |
| double | |
| _Py_log1p(double x) | |
| { | |
| /* For x small, we use the following approach. Let y be the nearest float | |
| to 1+x, then | |
| 1+x = y * (1 - (y-1-x)/y) | |
| so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny, the | |
| second term is well approximated by (y-1-x)/y. If abs(x) >= | |
| DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest | |
| then y-1-x will be exactly representable, and is computed exactly by | |
| (y-1)-x. | |
| If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be | |
| round-to-nearest then this method is slightly dangerous: 1+x could be | |
| rounded up to 1+DBL_EPSILON instead of down to 1, and in that case | |
| y-1-x will not be exactly representable any more and the result can be | |
| off by many ulps. But this is easily fixed: for a floating-point | |
| number |x| < DBL_EPSILON/2., the closest floating-point number to | |
| log(1+x) is exactly x. | |
| */ | |
| double y; | |
| if (fabs(x) < DBL_EPSILON/2.) { | |
| return x; | |
| } | |
| else if (-0.5 <= x && x <= 1.) { | |
| /* WARNING: it's possible than an overeager compiler | |
| will incorrectly optimize the following two lines | |
| to the equivalent of "return log(1.+x)". If this | |
| happens, then results from log1p will be inaccurate | |
| for small x. */ | |
| y = 1.+x; | |
| return log(y)-((y-1.)-x)/y; | |
| } | |
| else { | |
| /* NaNs and infinities should end up here */ | |
| return log(1.+x); | |
| } | |
| } |