text: 
  /* @(#)s_log1p.c 5.1 93/09/24 */
  /*
   * ====================================================
   * 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.
   * ====================================================
   */
  
  /*
  FUNCTION
  <<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>
  
  INDEX
  	log1p
  INDEX
  	log1pf
  
  ANSI_SYNOPSIS
  	#include <math.h>
  	double log1p(double <[x]>);
  	float log1pf(float <[x]>);
  
  TRAD_SYNOPSIS
  	#include <math.h>
  	double log1p(<[x]>)
  	double <[x]>;
  
  	float log1pf(<[x]>)
  	float <[x]>;
  
  DESCRIPTION
  <<log1p>> calculates 
  @tex
  $ln(1+x)$, 
  @end tex
  the natural logarithm of <<1+<[x]>>>.  You can use <<log1p>> rather
  than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
  small.
  
  <<log1pf>> calculates the same thing, but accepts and returns
  <<float>> values rather than <<double>>.
  
  RETURNS
  <<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
  <<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
  
  PORTABILITY
  Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
  Interface Definition (Issue 2).
  
  */
  
  /* double log1p(double x)
   *
   * Method :                  
   *   1. Argument Reduction: find k and f such that 
   *			1+x = 2^k * (1+f), 
   *	   where  sqrt(2)/2 < 1+f < sqrt(2) .
   *
   *      Note. If k=0, then f=x is exact. However, if k!=0, then f
   *	may not be representable exactly. In that case, a correction
   *	term is need. Let u=1+x rounded. Let c = (1+x)-u, then
   *	log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
   *	and add back the correction term c/u.
   *	(Note: when x > 2**53, one can simply return log(x))
   *
   *   2. Approximation of log1p(f).
   *	Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
   *		 = 2s + 2/3 s**3 + 2/5 s**5 + .....,
   *	     	 = 2s + s*R
   *      We use a special Reme algorithm on [0,0.1716] to generate 
   * 	a polynomial of degree 14 to approximate R The maximum error 
   *	of this polynomial approximation is bounded by 2**-58.45. In
   *	other words,
   *		        2      4      6      8      10      12      14
   *	    R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s  +Lp6*s  +Lp7*s
   *  	(the values of Lp1 to Lp7 are listed in the program)
   *	and
   *	    |      2          14          |     -58.45
   *	    | Lp1*s +...+Lp7*s    -  R(z) | <= 2 
   *	    |                             |
   *	Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
   *	In order to guarantee error in log below 1ulp, we compute log
   *	by
   *		log1p(f) = f - (hfsq - s*(hfsq+R)).
   *	
   *	3. Finally, log1p(x) = k*ln2 + log1p(f).  
   *		 	     = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
   *	   Here ln2 is split into two floating point number: 
   *			ln2_hi + ln2_lo,
   *	   where n*ln2_hi is always exact for |n| < 2000.
   *
   * Special cases:
   *	log1p(x) is NaN with signal if x < -1 (including -INF) ; 
   *	log1p(+INF) is +INF; log1p(-1) is -INF with signal;
   *	log1p(NaN) is that NaN with no signal.
   *
   * Accuracy:
   *	according to an error analysis, the error is always less than
   *	1 ulp (unit in the last place).
   *
   * Constants:
   * The hexadecimal values are the intended ones for the following 
   * constants. The decimal values may be used, provided that the 
   * compiler will convert from decimal to binary accurately enough 
   * to produce the hexadecimal values shown.
   *
   * Note: Assuming log() return accurate answer, the following
   * 	 algorithm can be used to compute log1p(x) to within a few ULP:
   *	
   *		u = 1+x;
   *		if(u==1.0) return x ; else
   *			   return log(u)*(x/(u-1.0));
   *
   *	 See HP-15C Advanced Functions Handbook, p.193.
   */
  
  #include "fdlibm.h"
  
  #ifndef _DOUBLE_IS_32BITS
  
  #ifdef __STDC__
  static const double
  #else
  static double
  #endif
  ln2_hi  =  6.93147180369123816490e-01,	/* 3fe62e42 fee00000 */
  ln2_lo  =  1.90821492927058770002e-10,	/* 3dea39ef 35793c76 */
  two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
  Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
  Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
  Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
  Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
  Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
  Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
  Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
  
  #ifdef __STDC__
  static const double zero = 0.0;
  #else
  static double zero = 0.0;
  #endif
  
  #ifdef __STDC__
  	double log1p(double x)
  #else
  	double log1p(x)
  	double x;
  #endif
  {
  	double hfsq,f,c,s,z,R,u;
  	__int32_t k,hx,hu,ax;
  
  	GET_HIGH_WORD(hx,x);
  	ax = hx&0x7fffffff;
  
  	k = 1;
  	if (hx < 0x3FDA827A) {			/* x < 0.41422  */
  	    if(ax>=0x3ff00000) {		/* x <= -1.0 */
  		if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */
  		else return (x-x)/(x-x);	/* log1p(x<-1)=NaN */
  	    }
  	    if(ax<0x3e200000) {			/* |x| < 2**-29 */
  		if(two54+x>zero			/* raise inexact */
  	            &&ax<0x3c900000) 		/* |x| < 2**-54 */
  		    return x;
  		else
  		    return x - x*x*0.5;
  	    }
  	    if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
  		k=0;f=x;hu=1;}	/* -0.2929<x<0.41422 */
  	} 
  	if (hx >= 0x7ff00000) return x+x;
  	if(k!=0) {
  	    if(hx<0x43400000) {
  		u  = 1.0+x; 
  		GET_HIGH_WORD(hu,u);
  	        k  = (hu>>20)-1023;
  	        c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
  		c /= u;
  	    } else {
  		u  = x;
  		GET_HIGH_WORD(hu,u);
  	        k  = (hu>>20)-1023;
  		c  = 0;
  	    }
  	    hu &= 0x000fffff;
  	    if(hu<0x6a09e) {
  	        SET_HIGH_WORD(u,hu|0x3ff00000);	/* normalize u */
  	    } else {
  	        k += 1; 
  		SET_HIGH_WORD(u,hu|0x3fe00000);	/* normalize u/2 */
  	        hu = (0x00100000-hu)>>2;
  	    }
  	    f = u-1.0;
  	}
  	hfsq=0.5*f*f;
  	if(hu==0) {	/* |f| < 2**-20 */
            if(f==zero) { if(k==0) return zero;  
                        else {c += k*ln2_lo; return k*ln2_hi+c;}}
  	    R = hfsq*(1.0-0.66666666666666666*f);
  	    if(k==0) return f-R; else
  	    	     return k*ln2_hi-((R-(k*ln2_lo+c))-f);
  	}
   	s = f/(2.0+f); 
  	z = s*s;
  	R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
  	if(k==0) return f-(hfsq-s*(hfsq+R)); else
  		 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
  }
  
  #endif /* _DOUBLE_IS_32BITS */
  