diff options
Diffstat (limited to 'Python/pymath.c')
| -rw-r--r-- | Python/pymath.c | 199 |
1 files changed, 0 insertions, 199 deletions
diff --git a/Python/pymath.c b/Python/pymath.c index db2920c..83105f2 100644 --- a/Python/pymath.c +++ b/Python/pymath.c @@ -77,202 +77,3 @@ round(double x) return copysign(y, x); } #endif /* HAVE_ROUND */ - -#ifndef HAVE_LOG1P -#include <float.h> - -double -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); - } -} -#endif /* HAVE_LOG1P */ - -/* - * ==================================================== - * 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; - -/* 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))) - */ - -#ifndef HAVE_ASINH -double -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 = log1p(absx + t / (1.0 + sqrt(1.0 + t))); - } - return copysign(w, x); - -} -#endif /* HAVE_ASINH */ - -/* 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. - */ - -#ifndef HAVE_ACOSH -double -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 log1p(t + sqrt(2.0*t + t*t)); - } -} -#endif /* HAVE_ACOSH */ - -/* 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; - * - */ - -#ifndef HAVE_ATANH -double -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 * log1p(t + t*absx / (1.0 - absx)); - } - else { /* 0.5 <= |x| <= 1.0 */ - t = 0.5 * log1p((absx + absx) / (1.0 - absx)); - } - return copysign(t, x); -} -#endif /* HAVE_ATANH */ |