//
// ====================================================
// 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.
// ====================================================
///
// __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).
//
var ln2_hi = 6.93147180369123816490e-01; // 3fe62e42 fee00000
var ln2_lo = 1.90821492927058770002e-10; // 3dea39ef 35793c76
var two54 = 1.80143985094819840000e+16; // 43500000 00000000
var Lg1 = 6.666666666666735130e-01; // 3FE55555 55555593
var Lg2 = 3.999999999940941908e-01; // 3FD99999 9997FA04
var Lg3 = 2.857142874366239149e-01; // 3FD24924 94229359
var Lg4 = 2.222219843214978396e-01; // 3FCC71C5 1D8E78AF
var Lg5 = 1.818357216161805012e-01; // 3FC74664 96CB03DE
var Lg6 = 1.531383769920937332e-01; // 3FC39A09 D078C69F
var Lg7 = 1.479819860511658591e-01; // 3FC2F112 DF3E5244
function log(x) {
//console.log("arg = " + x);
var hx = _DoubleHi(x);
var lx = _DoubleLo(x);
var k = 0;
if (hx < 0x00100000) { // x < 2**-1022
if (((hx & 0x7fffffff) | lx) == 0)
return -Infinity; // log(+-0)=-inf
if (hx<0) return NaN; // log(-#) = NaN
k -= 54;
x *= two54; // subnormal number, scale up x
hx = _DoubleHi(x); // high word of x
}
// x is infinity or NaN, so return infinity or NaN, respectively.
if (hx >= 0x7ff00000)
return x+x;
k += (hx >> 20) - 1023;
hx &= 0x000fffff;
var i = (hx + 0x95f64) & 0x100000;
//__HI(x) = hx|(i^0x3ff00000);
// normalize x or x/2
x = _ConstructDouble(hx|(i ^ 0x3ff00000), lx);
k += (i >> 20);
var f = x - 1.0;
//console.log("x = " + x + ", k = " + k + ", f = " + f);
var R;
// Maybe replace this test with Math.abs(f) < 2^-20?
if ((0x000fffff & (2+hx)) < 3) { // |f| < 2**-20
// I think the only way f = 0 is if x is a power of 2.
if (f==0) {
if (k==0)
return 0;
else {
var dk = k;
return dk * ln2_hi + dk * ln2_lo;
}
}
R = f * f * (0.5 - 0.33333333333333333 * f);
if (k == 0)
return f - R;
else {
var dk = k;
return dk * ln2_hi - ((R - dk * ln2_lo) - f);
}
}
var s = f / (2.0 + f);
var dk = k;
var z = s * s;
// I think this computation of i, j is figuring out if f is not
// too large.
i = hx - 0x6147a;
var j = 0x6b851 - hx;
var w = z * z;
//console.log("hx = " + hx + ", i = " + i + ", j = " + j);
var t1= w * (Lg2 + w * (Lg4 + w * Lg6));
var t2= z * (Lg1 + w * (Lg3 + w * (Lg5 + w * Lg7)));
i |= j;
R = t2 + t1;
if (i > 0) {
// This appears to be handling the "better accuracy" case
// given at the end of item 2, in the Method section above.
//console.log("i = " + i + ", k = " + k);
var 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 {
// This looks like the "f is not too large" case given at the
// end of item 2, in the Method section above.
//console.log("i = " + i + ", k = " + k);
if (k == 0)
return f - s * (f - R);
else
return dk * ln2_hi - ((s * (f - R) - dk * ln2_lo) - f);
}
}
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