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  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. ANSI_SYNOPSIS
  24.         #include <math.h>
  25.         double log1p(double <[x]>);
  26.         float log1pf(float <[x]>);
  27.  
  28. TRAD_SYNOPSIS
  29.         #include <math.h>
  30.         double log1p(<[x]>)
  31.         double <[x]>;
  32.  
  33.         float log1pf(<[x]>)
  34.         float <[x]>;
  35.  
  36. DESCRIPTION
  37. <<log1p>> calculates
  38. @tex
  39. $ln(1+x)$,
  40. @end tex
  41. the natural logarithm of <<1+<[x]>>>.  You can use <<log1p>> rather
  42. than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
  43. small.
  44.  
  45. <<log1pf>> calculates the same thing, but accepts and returns
  46. <<float>> values rather than <<double>>.
  47.  
  48. RETURNS
  49. <<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
  50. <<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
  51.  
  52. PORTABILITY
  53. Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
  54. Interface Definition (Issue 2).
  55.  
  56. */
  57.  
  58. /* double log1p(double x)
  59.  *
  60.  * Method :                  
  61.  *   1. Argument Reduction: find k and f such that
  62.  *                      1+x = 2^k * (1+f),
  63.  *         where  sqrt(2)/2 < 1+f < sqrt(2) .
  64.  *
  65.  *      Note. If k=0, then f=x is exact. However, if k!=0, then f
  66.  *      may not be representable exactly. In that case, a correction
  67.  *      term is need. Let u=1+x rounded. Let c = (1+x)-u, then
  68.  *      log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
  69.  *      and add back the correction term c/u.
  70.  *      (Note: when x > 2**53, one can simply return log(x))
  71.  *
  72.  *   2. Approximation of log1p(f).
  73.  *      Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
  74.  *               = 2s + 2/3 s**3 + 2/5 s**5 + .....,
  75.  *               = 2s + s*R
  76.  *      We use a special Reme algorithm on [0,0.1716] to generate
  77.  *      a polynomial of degree 14 to approximate R The maximum error
  78.  *      of this polynomial approximation is bounded by 2**-58.45. In
  79.  *      other words,
  80.  *                      2      4      6      8      10      12      14
  81.  *          R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s  +Lp6*s  +Lp7*s
  82.  *      (the values of Lp1 to Lp7 are listed in the program)
  83.  *      and
  84.  *          |      2          14          |     -58.45
  85.  *          | Lp1*s +...+Lp7*s    -  R(z) | <= 2
  86.  *          |                             |
  87.  *      Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
  88.  *      In order to guarantee error in log below 1ulp, we compute log
  89.  *      by
  90.  *              log1p(f) = f - (hfsq - s*(hfsq+R)).
  91.  *     
  92.  *      3. Finally, log1p(x) = k*ln2 + log1p(f).  
  93.  *                           = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
  94.  *         Here ln2 is split into two floating point number:
  95.  *                      ln2_hi + ln2_lo,
  96.  *         where n*ln2_hi is always exact for |n| < 2000.
  97.  *
  98.  * Special cases:
  99.  *      log1p(x) is NaN with signal if x < -1 (including -INF) ;
  100.  *      log1p(+INF) is +INF; log1p(-1) is -INF with signal;
  101.  *      log1p(NaN) is that NaN with no signal.
  102.  *
  103.  * Accuracy:
  104.  *      according to an error analysis, the error is always less than
  105.  *      1 ulp (unit in the last place).
  106.  *
  107.  * Constants:
  108.  * The hexadecimal values are the intended ones for the following
  109.  * constants. The decimal values may be used, provided that the
  110.  * compiler will convert from decimal to binary accurately enough
  111.  * to produce the hexadecimal values shown.
  112.  *
  113.  * Note: Assuming log() return accurate answer, the following
  114.  *       algorithm can be used to compute log1p(x) to within a few ULP:
  115.  *     
  116.  *              u = 1+x;
  117.  *              if(u==1.0) return x ; else
  118.  *                         return log(u)*(x/(u-1.0));
  119.  *
  120.  *       See HP-15C Advanced Functions Handbook, p.193.
  121.  */
  122.  
  123. #include "fdlibm.h"
  124.  
  125. #ifndef _DOUBLE_IS_32BITS
  126.  
  127. #ifdef __STDC__
  128. static const double
  129. #else
  130. static double
  131. #endif
  132. ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
  133. ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
  134. two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
  135. Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
  136. Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
  137. Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
  138. Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
  139. Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
  140. Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
  141. Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
  142.  
  143. #ifdef __STDC__
  144. static const double zero = 0.0;
  145. #else
  146. static double zero = 0.0;
  147. #endif
  148.  
  149. #ifdef __STDC__
  150.         double log1p(double x)
  151. #else
  152.         double log1p(x)
  153.         double x;
  154. #endif
  155. {
  156.         double hfsq,f,c,s,z,R,u;
  157.         __int32_t k,hx,hu,ax;
  158.  
  159.         GET_HIGH_WORD(hx,x);
  160.         ax = hx&0x7fffffff;
  161.  
  162.         k = 1;
  163.         if (hx < 0x3FDA827A) {                  /* x < 0.41422  */
  164.             if(ax>=0x3ff00000) {                /* x <= -1.0 */
  165.                 if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */
  166.                 else return (x-x)/(x-x);        /* log1p(x<-1)=NaN */
  167.             }
  168.             if(ax<0x3e200000) {                 /* |x| < 2**-29 */
  169.                 if(two54+x>zero                 /* raise inexact */
  170.                     &&ax<0x3c900000)            /* |x| < 2**-54 */
  171.                     return x;
  172.                 else
  173.                     return x - x*x*0.5;
  174.             }
  175.             if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
  176.                 k=0;f=x;hu=1;}  /* -0.2929<x<0.41422 */
  177.         }
  178.         if (hx >= 0x7ff00000) return x+x;
  179.         if(k!=0) {
  180.             if(hx<0x43400000) {
  181.                 u  = 1.0+x;
  182.                 GET_HIGH_WORD(hu,u);
  183.                 k  = (hu>>20)-1023;
  184.                 c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
  185.                 c /= u;
  186.             } else {
  187.                 u  = x;
  188.                 GET_HIGH_WORD(hu,u);
  189.                 k  = (hu>>20)-1023;
  190.                 c  = 0;
  191.             }
  192.             hu &= 0x000fffff;
  193.             if(hu<0x6a09e) {
  194.                 SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */
  195.             } else {
  196.                 k += 1;
  197.                 SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */
  198.                 hu = (0x00100000-hu)>>2;
  199.             }
  200.             f = u-1.0;
  201.         }
  202.         hfsq=0.5*f*f;
  203.         if(hu==0) {     /* |f| < 2**-20 */
  204.           if(f==zero) { if(k==0) return zero;  
  205.                       else {c += k*ln2_lo; return k*ln2_hi+c;}}
  206.             R = hfsq*(1.0-0.66666666666666666*f);
  207.             if(k==0) return f-R; else
  208.                      return k*ln2_hi-((R-(k*ln2_lo+c))-f);
  209.         }
  210.         s = f/(2.0+f);
  211.         z = s*s;
  212.         R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
  213.         if(k==0) return f-(hfsq-s*(hfsq+R)); else
  214.                  return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
  215. }
  216.  
  217. #endif /* _DOUBLE_IS_32BITS */
  218.