/* origin: FreeBSD /usr/src/lib/msun/src/e_log.c */
/*
* ====================================================
* 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.
* ====================================================
*/
/* log(x)
* Return the logarithm 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 Remez 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 <math.h>
#include <stdint.h>
static const double
ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */
ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */
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 */
double log(double x)
{
union {double f; uint64_t i;} u = {x};
double_t hfsq,f,s,z,R,w,t1,t2,dk;
uint32_t hx;
int k;
hx = u.i>>32;
k = 0;
if (hx < 0x00100000 || hx>>31) {
if (u.i<<1 == 0)
return -1/(x*x); /* log(+-0)=-inf */
if (hx>>31)
return (x-x)/0.0; /* log(-#) = NaN */
/* subnormal number, scale x up */
k -= 54;
x *= 0x1p54;
u.f = x;
hx = u.i>>32;
} else if (hx >= 0x7ff00000) {
return x;
} else if (hx == 0x3ff00000 && u.i<<32 == 0)
return 0;
/* reduce x into [sqrt(2)/2, sqrt(2)] */
hx += 0x3ff00000 - 0x3fe6a09e;
k += (int)(hx>>20) - 0x3ff;
hx = (hx&0x000fffff) + 0x3fe6a09e;
u.i = (uint64_t)hx<<32 | (u.i&0xffffffff);
x = u.f;
f = x - 1.0;
hfsq = 0.5*f*f;
s = f/(2.0+f);
z = s*s;
w = z*z;
t1 = w*(Lg2+w*(Lg4+w*Lg6));
t2 = z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
R = t2 + t1;
dk = k;
return s*(hfsq+R) + dk*ln2_lo - hfsq + f + dk*ln2_hi;
}
