Source

cpython-withatomic / Modules / _math.c

The branch 'legacy-trunk' does not exist.
/* 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);
    }
}