/* 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;
}