Quelle e_log.cpp Sprache: C

/* @(#)e_log.c 1.3 95/01/18 */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/

//#include <sys/cdefs.h>
//__FBSDID("$FreeBSD$");

/* __ieee754_log(x)
* Return the logrithm of x
*
* Method :
*   1. Argument Reduction: find k and f such that
* x = 2^k * (1+f),
*    where  sqrt(2)/2 < 1+f < sqrt(2) .
*
*   2. Approximation of log(1+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) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s  +Lg6*s  +Lg7*s
*   (the values of Lg1 to Lg7 are listed in the program)
* and
*     |      2          14          |     -58.45
*     | Lg1*s +...+Lg7*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
* log(1+f) = f - s*(f - R) (if f is not too large)
* log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
*
* 3. Finally,  log(x) = k*ln2 + log(1+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:
* log(x) is NaN with signal if x < 0 (including -INF) ;
* log(+INF) is +INF; log(0) is -INF with signal;
* log(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.
*/

#include <float.h>

#include "math_private.h"

static const double
ln2_hi  =  6.93147180369123816490e-01, /* 3fe62e42 fee00000 */
ln2_lo  =  1.90821492927058770002e-10, /* 3dea39ef 35793c76 */
two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
Lg1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
Lg2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
Lg3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
Lg4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
Lg5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
Lg6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
Lg7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */

static const double zero   =  0.0;
static volatile double vzero = 0.0;

double
__ieee754_log(double x)
{
double hfsq,f,s,z,R,w,t1,t2,dk;
int32_t k,hx,i,j;
u_int32_t lx;

EXTRACT_WORDS(hx,lx,x);

k=0;
if (hx < 0x00100000) {   /* x < 2**-1022  */
     if (((hx&0x7fffffff)|lx)==0)
  return -two54/vzero;  /* log(+-0)=-inf */
     if (hx<0) return (x-x)/zero; /* log(-#) = NaN */
     k -= 54; x *= two54; /* subnormal number, scale up x */
     GET_HIGH_WORD(hx,x);
}
if (hx >= 0x7ff00000) return x+x;
k += (hx>>20)-1023;
hx &= 0x000fffff;
i = (hx+0x95f64)&0x100000;
SET_HIGH_WORD(x,hx|(i^0x3ff00000)); /* normalize x or x/2 */
k += (i>>20);
f = x-1.0;
if((0x000fffff&(2+hx))<3) { /* -2**-20 <= f < 2**-20 */
     if(f==zero) {
  if(k==0) {
      return zero;
  } else {
      dk=(double)k;
      return dk*ln2_hi+dk*ln2_lo;
  }
     }
     R = f*f*(0.5-0.33333333333333333*f);
     if(k==0) return f-R; else {dk=(double)k;
           return dk*ln2_hi-((R-dk*ln2_lo)-f);}
}
  s = f/(2.0+f);
dk = (double)k;
z = s*s;
i = hx-0x6147a;
w = z*z;
j = 0x6b851-hx;
t1= w*(Lg2+w*(Lg4+w*Lg6));
t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
i |= j;
R = t2+t1;
if(i>0) {
     hfsq=0.5*f*f;
     if(k==0) return f-(hfsq-s*(hfsq+R)); else
       return dk*ln2_hi-((hfsq-(s*(hfsq+R)+dk*ln2_lo))-f);
} else {
     if(k==0) return f-s*(f-R); else
       return dk*ln2_hi-((s*(f-R)-dk*ln2_lo)-f);
}
}

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