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https://github.com/Sneed-Group/Poodletooth-iLand
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255 lines
7 KiB
C
255 lines
7 KiB
C
/* Definitions of some C99 math library functions, for those platforms
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that don't implement these functions already. */
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#include "Python.h"
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#include <float.h>
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#include "_math.h"
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/* The following copyright notice applies to the original
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implementations of acosh, asinh and atanh. */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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static const double ln2 = 6.93147180559945286227E-01;
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static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
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static const double two_pow_p28 = 268435456.0; /* 2**28 */
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static const double zero = 0.0;
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/* acosh(x)
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* Method :
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* Based on
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* acosh(x) = log [ x + sqrt(x*x-1) ]
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* we have
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* acosh(x) := log(x)+ln2, if x is large; else
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* acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
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* acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
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*
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* Special cases:
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* acosh(x) is NaN with signal if x<1.
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* acosh(NaN) is NaN without signal.
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*/
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double
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_Py_acosh(double x)
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{
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if (Py_IS_NAN(x)) {
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return x+x;
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}
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if (x < 1.) { /* x < 1; return a signaling NaN */
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errno = EDOM;
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#ifdef Py_NAN
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return Py_NAN;
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#else
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return (x-x)/(x-x);
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#endif
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}
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else if (x >= two_pow_p28) { /* x > 2**28 */
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if (Py_IS_INFINITY(x)) {
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return x+x;
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}
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else {
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return log(x)+ln2; /* acosh(huge)=log(2x) */
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}
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}
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else if (x == 1.) {
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return 0.0; /* acosh(1) = 0 */
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}
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else if (x > 2.) { /* 2 < x < 2**28 */
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double t = x*x;
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return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
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}
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else { /* 1 < x <= 2 */
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double t = x - 1.0;
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return m_log1p(t + sqrt(2.0*t + t*t));
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}
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}
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/* asinh(x)
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* Method :
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* Based on
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* asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
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* we have
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* asinh(x) := x if 1+x*x=1,
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* := sign(x)*(log(x)+ln2)) for large |x|, else
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* := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
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* := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
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*/
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double
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_Py_asinh(double x)
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{
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double w;
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double absx = fabs(x);
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if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
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return x+x;
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}
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if (absx < two_pow_m28) { /* |x| < 2**-28 */
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return x; /* return x inexact except 0 */
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}
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if (absx > two_pow_p28) { /* |x| > 2**28 */
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w = log(absx)+ln2;
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}
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else if (absx > 2.0) { /* 2 < |x| < 2**28 */
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w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
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}
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else { /* 2**-28 <= |x| < 2= */
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double t = x*x;
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w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t)));
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}
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return copysign(w, x);
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}
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/* atanh(x)
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* Method :
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* 1.Reduced x to positive by atanh(-x) = -atanh(x)
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* 2.For x>=0.5
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* 1 2x x
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* atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)
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* 2 1 - x 1 - x
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*
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* For x<0.5
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* atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
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*
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* Special cases:
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* atanh(x) is NaN if |x| >= 1 with signal;
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* atanh(NaN) is that NaN with no signal;
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*
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*/
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double
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_Py_atanh(double x)
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{
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double absx;
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double t;
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if (Py_IS_NAN(x)) {
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return x+x;
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}
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absx = fabs(x);
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if (absx >= 1.) { /* |x| >= 1 */
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errno = EDOM;
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#ifdef Py_NAN
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return Py_NAN;
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#else
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return x/zero;
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#endif
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}
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if (absx < two_pow_m28) { /* |x| < 2**-28 */
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return x;
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}
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if (absx < 0.5) { /* |x| < 0.5 */
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t = absx+absx;
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t = 0.5 * m_log1p(t + t*absx / (1.0 - absx));
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}
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else { /* 0.5 <= |x| <= 1.0 */
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t = 0.5 * m_log1p((absx + absx) / (1.0 - absx));
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}
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return copysign(t, x);
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}
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/* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed
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to avoid the significant loss of precision that arises from direct
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evaluation of the expression exp(x) - 1, for x near 0. */
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double
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_Py_expm1(double x)
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{
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/* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
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also works fine for infinities and nans.
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For smaller x, we can use a method due to Kahan that achieves close to
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full accuracy.
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*/
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if (fabs(x) < 0.7) {
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double u;
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u = exp(x);
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if (u == 1.0)
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return x;
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else
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return (u - 1.0) * x / log(u);
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}
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else
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return exp(x) - 1.0;
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}
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/* log1p(x) = log(1+x). The log1p function is designed to avoid the
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significant loss of precision that arises from direct evaluation when x is
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small. */
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#ifdef HAVE_LOG1P
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double
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_Py_log1p(double x)
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{
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/* Some platforms supply a log1p function but don't respect the sign of
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zero: log1p(-0.0) gives 0.0 instead of the correct result of -0.0.
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To save fiddling with configure tests and platform checks, we handle the
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special case of zero input directly on all platforms.
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*/
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if (x == 0.0) {
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return x;
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}
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else {
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return log1p(x);
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}
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}
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#else
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double
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_Py_log1p(double x)
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{
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/* For x small, we use the following approach. Let y be the nearest float
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to 1+x, then
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1+x = y * (1 - (y-1-x)/y)
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so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny, the
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second term is well approximated by (y-1-x)/y. If abs(x) >=
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DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
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then y-1-x will be exactly representable, and is computed exactly by
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(y-1)-x.
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If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
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round-to-nearest then this method is slightly dangerous: 1+x could be
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rounded up to 1+DBL_EPSILON instead of down to 1, and in that case
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y-1-x will not be exactly representable any more and the result can be
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off by many ulps. But this is easily fixed: for a floating-point
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number |x| < DBL_EPSILON/2., the closest floating-point number to
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log(1+x) is exactly x.
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*/
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double y;
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if (fabs(x) < DBL_EPSILON/2.) {
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return x;
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}
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else if (-0.5 <= x && x <= 1.) {
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/* WARNING: it's possible than an overeager compiler
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will incorrectly optimize the following two lines
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to the equivalent of "return log(1.+x)". If this
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happens, then results from log1p will be inaccurate
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for small x. */
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y = 1.+x;
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return log(y)-((y-1.)-x)/y;
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}
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else {
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/* NaNs and infinities should end up here */
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return log(1.+x);
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}
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}
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#endif /* ifdef HAVE_LOG1P */
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