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#include "Python.h"
#ifndef HAVE_HYPOT
double hypot(double x, double y)
{
double yx;
x = fabs(x);
y = fabs(y);
if (x < y) {
double temp = x;
x = y;
y = temp;
}
if (x == 0.)
return 0.;
else {
yx = y/x;
return x*sqrt(1.+yx*yx);
}
}
#endif /* HAVE_HYPOT */
#ifndef HAVE_COPYSIGN
static double
copysign(double x, double y)
{
/* use atan2 to distinguish -0. from 0. */
if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
return fabs(x);
} else {
return -fabs(x);
}
}
#endif /* HAVE_COPYSIGN */
#ifndef HAVE_LOG1P
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 */
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