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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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