src/math/log1p.c

   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 }