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/* 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);
}
}
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