/* 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.
 * ====================================================
 */

#if !defined(HAVE_ACOSH) || !defined(HAVE_ASINH)
static const double ln2 = 6.93147180559945286227E-01;
static const double two_pow_p28 = 268435456.0; /* 2**28 */
#endif
#if !defined(HAVE_ASINH) || !defined(HAVE_ATANH)
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
#endif
#if !defined(HAVE_ATANH) && !defined(Py_NAN)
static const double zero = 0.0;
#endif


#ifndef HAVE_ACOSH
/* 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));
 }
}
#endif /* HAVE_ACOSH */


#ifndef HAVE_ASINH
/* 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);

}
#endif /* HAVE_ASINH */


#ifndef HAVE_ATANH
/* 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);
}
#endif /* HAVE_ATANH */


#ifndef HAVE_EXPM1
/* 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;
}
#endif /* HAVE_EXPM1 */


/* 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)
{
#ifdef HAVE_LOG1P
 /* Some platforms supply a log1p function but don't respect the sign of
 zero: log1p(-0.0) gives 0.0 instead of the correct result of -0.0.

 To save fiddling with configure tests and platform checks, we handle the
 special case of zero input directly on all platforms.
 */
 if (x == 0.0) {
 return x;
 }
 else {
 return log1p(x);
 }
#else
 /* 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 that 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 /* ifdef HAVE_LOG1P */
}