newlib/libm/common/s_log1p.c

   1
   2 /* @(#)s_log1p.c 5.1 93/09/24 */
   3 /*
   4  * ====================================================
   5  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
   6  *
   7  * Developed at SunPro, a Sun Microsystems, Inc. business.
   8  * Permission to use, copy, modify, and distribute this
   9  * software is freely granted, provided that this notice
  10  * is preserved.
  11  * ====================================================
  12  */
  13
  14 /*
  15 FUNCTION
  16 <<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>
  17
  18 INDEX
  19         log1p
  20 INDEX
  21         log1pf
  22
  23 SYNOPSIS
  24         #include <math.h>
  25         double log1p(double <[x]>);
  26         float log1pf(float <[x]>);
  27
  28 DESCRIPTION
  29 <<log1p>> calculates
  30 @tex
  31 $ln(1+x)$,
  32 @end tex
  33 the natural logarithm of <<1+<[x]>>>.  You can use <<log1p>> rather
  34 than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
  35 small.
  36
  37 <<log1pf>> calculates the same thing, but accepts and returns
  38 <<float>> values rather than <<double>>.
  39
  40 RETURNS
  41 <<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
  42 <<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
  43
  44 PORTABILITY
  45 Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
  46 Interface Definition (Issue 2).
  47
  48 */
  49
  50 /* double log1p(double x)
  51  *
  52  * Method :
  53  *   1. Argument Reduction: find k and f such that
  54  *                      1+x = 2^k * (1+f),
  55  *         where  sqrt(2)/2 < 1+f < sqrt(2) .
  56  *
  57  *      Note. If k=0, then f=x is exact. However, if k!=0, then f
  58  *      may not be representable exactly. In that case, a correction
  59  *      term is need. Let u=1+x rounded. Let c = (1+x)-u, then
  60  *      log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
  61  *      and add back the correction term c/u.
  62  *      (Note: when x > 2**53, one can simply return log(x))
  63  *
  64  *   2. Approximation of log1p(f).
  65  *      Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
  66  *               = 2s + 2/3 s**3 + 2/5 s**5 + .....,
  67  *               = 2s + s*R
  68  *      We use a special Remez algorithm on [0,0.1716] to generate
  69  *      a polynomial of degree 14 to approximate R The maximum error
  70  *      of this polynomial approximation is bounded by 2**-58.45. In
  71  *      other words,
  72  *                      2      4      6      8      10      12      14
  73  *          R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s  +Lp6*s  +Lp7*s
  74  *      (the values of Lp1 to Lp7 are listed in the program)
  75  *      and
  76  *          |      2          14          |     -58.45
  77  *          | Lp1*s +...+Lp7*s    -  R(z) | <= 2
  78  *          |                             |
  79  *      Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
  80  *      In order to guarantee error in log below 1ulp, we compute log
  81  *      by
  82  *              log1p(f) = f - (hfsq - s*(hfsq+R)).
  83  *
  84  *      3. Finally, log1p(x) = k*ln2 + log1p(f).
  85  *                           = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
  86  *         Here ln2 is split into two floating point number:
  87  *                      ln2_hi + ln2_lo,
  88  *         where n*ln2_hi is always exact for |n| < 2000.
  89  *
  90  * Special cases:
  91  *      log1p(x) is NaN with signal if x < -1 (including -INF) ;
  92  *      log1p(+INF) is +INF; log1p(-1) is -INF with signal;
  93  *      log1p(NaN) is that NaN with no signal.
  94  *
  95  * Accuracy:
  96  *      according to an error analysis, the error is always less than
  97  *      1 ulp (unit in the last place).
  98  *
  99  * Constants:
 100  * The hexadecimal values are the intended ones for the following
 101  * constants. The decimal values may be used, provided that the
 102  * compiler will convert from decimal to binary accurately enough
 103  * to produce the hexadecimal values shown.
 104  *
 105  * Note: Assuming log() return accurate answer, the following
 106  *       algorithm can be used to compute log1p(x) to within a few ULP:
 107  *
 108  *              u = 1+x;
 109  *              if(u==1.0) return x ; else
 110  *                         return log(u)*(x/(u-1.0));
 111  *
 112  *       See HP-15C Advanced Functions Handbook, p.193.
 113  */
 114
 115 #include "fdlibm.h"
 116 #include "math_config.h"
 117
 118 #ifndef _DOUBLE_IS_32BITS
 119
 120 #ifdef __STDC__
 121 static const double
 122 #else
 123 static double
 124 #endif
 125 ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
 126 ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
 127 two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
 128 Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
 129 Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
 130 Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
 131 Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
 132 Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
 133 Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
 134 Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
 135
 136 #ifdef __STDC__
 137 static const double zero = 0.0;
 138 #else
 139 static double zero = 0.0;
 140 #endif
 141
 142 #ifdef __STDC__
 143         double log1p(double x)
 144 #else
 145         double log1p(x)
 146         double x;
 147 #endif
 148 {
 149         double hfsq,f,c,s,z,R,u;
 150         __int32_t k,hx,hu,ax;
 151
 152         GET_HIGH_WORD(hx,x);
 153         ax = hx&0x7fffffff;
 154
 155         k = 1;
 156         if (hx < 0x3FDA827A) {                  /* x < 0.41422  */
 157             if(ax>=0x3ff00000) {                /* x <= -1.0 */
 158                 if(x==-1.0)
 159                     return __math_divzero (1);  /* log1p(-1)=-inf */
 160                 else
 161                     return __math_invalid (x);  /* log1p(x<-1)=NaN */
 162             }
 163             if(ax<0x3e200000) {                 /* |x| < 2**-29 */
 164                 if(two54+x>zero                 /* raise inexact */
 165                     &&ax<0x3c900000)            /* |x| < 2**-54 */
 166                     return x;
 167                 else
 168                     return x - x*x*0.5;
 169             }
 170             if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
 171                 k=0;f=x;hu=1;}  /* -0.2929<x<0.41422 */
 172         }
 173         if (hx >= 0x7ff00000) return x+x;
 174         if(k!=0) {
 175             if(hx<0x43400000) {
 176                 u  = 1.0+x;
 177                 GET_HIGH_WORD(hu,u);
 178                 k  = (hu>>20)-1023;
 179                 c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
 180                 c /= u;
 181             } else {
 182                 u  = x;
 183                 GET_HIGH_WORD(hu,u);
 184                 k  = (hu>>20)-1023;
 185                 c  = 0;
 186             }
 187             hu &= 0x000fffff;
 188             if(hu<0x6a09e) {
 189                 SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */
 190             } else {
 191                 k += 1;
 192                 SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */
 193                 hu = (0x00100000-hu)>>2;
 194             }
 195             f = u-1.0;
 196         }
 197         hfsq=0.5*f*f;
 198         if(hu==0) {     /* |f| < 2**-20 */
 199           if(f==zero) { if(k==0) return zero;
 200                       else {c += k*ln2_lo; return k*ln2_hi+c;}}
 201             R = hfsq*(1.0-0.66666666666666666*f);
 202             if(k==0) return f-R; else
 203                      return k*ln2_hi-((R-(k*ln2_lo+c))-f);
 204         }
 205         s = f/(2.0+f);
 206         z = s*s;
 207         R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
 208         if(k==0) return f-(hfsq-s*(hfsq+R)); else
 209                  return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
 210 }
 211
 212 #endif /* _DOUBLE_IS_32BITS */