0,0 → 1,146 |
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/* @(#)e_log.c 5.1 93/09/24 */ |
/* |
* ==================================================== |
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
* |
* Developed at SunPro, a Sun Microsystems, Inc. business. |
* Permission to use, copy, modify, and distribute this |
* software is freely granted, provided that this notice |
* is preserved. |
* ==================================================== |
*/ |
|
/* __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 "fdlibm.h" |
|
#ifndef _DOUBLE_IS_32BITS |
|
#ifdef __STDC__ |
static const double |
#else |
static double |
#endif |
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 */ |
|
#ifdef __STDC__ |
static const double zero = 0.0; |
#else |
static double zero = 0.0; |
#endif |
|
#ifdef __STDC__ |
double __ieee754_log(double x) |
#else |
double __ieee754_log(x) |
double x; |
#endif |
{ |
double hfsq,f,s,z,R,w,t1,t2,dk; |
__int32_t k,hx,i,j; |
__uint32_t lx; |
|
EXTRACT_WORDS(hx,lx,x); |
|
k=0; |
if (hx < 0x00100000) { /* x < 2**-1022 */ |
if (((hx&0x7fffffff)|lx)==0) |
return -two54/zero; /* 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) { /* |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); |
} |
} |
|
#endif /* defined(_DOUBLE_IS_32BITS) */ |