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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.
* ====================================================
*/
#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 */
}
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