--- a/Python/pymath.c	Mon Dec 21 12:50:02 2009 +0000
+++ b/Python/pymath.c	Mon Dec 21 15:27:41 2009 +0000
@@ -77,202 +77,3 @@
     return copysign(y, x);
 }
 #endif /* HAVE_ROUND */
-
-#ifndef HAVE_LOG1P
-#include <float.h>
-
-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 */