log1p.c raw

   1  /* origin: FreeBSD /usr/src/lib/msun/src/s_log1p.c */
   2  /*
   3   * ====================================================
   4   * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
   5   *
   6   * Developed at SunPro, a Sun Microsystems, Inc. business.
   7   * Permission to use, copy, modify, and distribute this
   8   * software is freely granted, provided that this notice
   9   * is preserved.
  10   * ====================================================
  11   */
  12  /* double log1p(double x)
  13   * Return the natural logarithm of 1+x.
  14   *
  15   * Method :
  16   *   1. Argument Reduction: find k and f such that
  17   *                      1+x = 2^k * (1+f),
  18   *         where  sqrt(2)/2 < 1+f < sqrt(2) .
  19   *
  20   *      Note. If k=0, then f=x is exact. However, if k!=0, then f
  21   *      may not be representable exactly. In that case, a correction
  22   *      term is need. Let u=1+x rounded. Let c = (1+x)-u, then
  23   *      log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
  24   *      and add back the correction term c/u.
  25   *      (Note: when x > 2**53, one can simply return log(x))
  26   *
  27   *   2. Approximation of log(1+f): See log.c
  28   *
  29   *   3. Finally, log1p(x) = k*ln2 + log(1+f) + c/u. See log.c
  30   *
  31   * Special cases:
  32   *      log1p(x) is NaN with signal if x < -1 (including -INF) ;
  33   *      log1p(+INF) is +INF; log1p(-1) is -INF with signal;
  34   *      log1p(NaN) is that NaN with no signal.
  35   *
  36   * Accuracy:
  37   *      according to an error analysis, the error is always less than
  38   *      1 ulp (unit in the last place).
  39   *
  40   * Constants:
  41   * The hexadecimal values are the intended ones for the following
  42   * constants. The decimal values may be used, provided that the
  43   * compiler will convert from decimal to binary accurately enough
  44   * to produce the hexadecimal values shown.
  45   *
  46   * Note: Assuming log() return accurate answer, the following
  47   *       algorithm can be used to compute log1p(x) to within a few ULP:
  48   *
  49   *              u = 1+x;
  50   *              if(u==1.0) return x ; else
  51   *                         return log(u)*(x/(u-1.0));
  52   *
  53   *       See HP-15C Advanced Functions Handbook, p.193.
  54   */
  55  
  56  #include "libm.h"
  57  
  58  static const double
  59  ln2_hi = 6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
  60  ln2_lo = 1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
  61  Lg1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
  62  Lg2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
  63  Lg3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
  64  Lg4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
  65  Lg5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
  66  Lg6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
  67  Lg7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
  68  
  69  double log1p(double x)
  70  {
  71  	union {double f; uint64_t i;} u = {x};
  72  	double_t hfsq,f,c,s,z,R,w,t1,t2,dk;
  73  	uint32_t hx,hu;
  74  	int k;
  75  
  76  	hx = u.i>>32;
  77  	k = 1;
  78  	if (hx < 0x3fda827a || hx>>31) {  /* 1+x < sqrt(2)+ */
  79  		if (hx >= 0xbff00000) {  /* x <= -1.0 */
  80  			if (x == -1)
  81  				return x/0.0; /* log1p(-1) = -inf */
  82  			return (x-x)/0.0;     /* log1p(x<-1) = NaN */
  83  		}
  84  		if (hx<<1 < 0x3ca00000<<1) {  /* |x| < 2**-53 */
  85  			/* underflow if subnormal */
  86  			if ((hx&0x7ff00000) == 0)
  87  				FORCE_EVAL((float)x);
  88  			return x;
  89  		}
  90  		if (hx <= 0xbfd2bec4) {  /* sqrt(2)/2- <= 1+x < sqrt(2)+ */
  91  			k = 0;
  92  			c = 0;
  93  			f = x;
  94  		}
  95  	} else if (hx >= 0x7ff00000)
  96  		return x;
  97  	if (k) {
  98  		u.f = 1 + x;
  99  		hu = u.i>>32;
 100  		hu += 0x3ff00000 - 0x3fe6a09e;
 101  		k = (int)(hu>>20) - 0x3ff;
 102  		/* correction term ~ log(1+x)-log(u), avoid underflow in c/u */
 103  		if (k < 54) {
 104  			c = k >= 2 ? 1-(u.f-x) : x-(u.f-1);
 105  			c /= u.f;
 106  		} else
 107  			c = 0;
 108  		/* reduce u into [sqrt(2)/2, sqrt(2)] */
 109  		hu = (hu&0x000fffff) + 0x3fe6a09e;
 110  		u.i = (uint64_t)hu<<32 | (u.i&0xffffffff);
 111  		f = u.f - 1;
 112  	}
 113  	hfsq = 0.5*f*f;
 114  	s = f/(2.0+f);
 115  	z = s*s;
 116  	w = z*z;
 117  	t1 = w*(Lg2+w*(Lg4+w*Lg6));
 118  	t2 = z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
 119  	R = t2 + t1;
 120  	dk = k;
 121  	return s*(hfsq+R) + (dk*ln2_lo+c) - hfsq + f + dk*ln2_hi;
 122  }
 123