Revision control
Copy as Markdown
Other Tools
/* 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.
*/
const LN2_HI: f64 = 6.93147180369123816490e-01; /* 3fe62e42 fee00000 */
const LN2_LO: f64 = 1.90821492927058770002e-10; /* 3dea39ef 35793c76 */
const LG1: f64 = 6.666666666666735130e-01; /* 3FE55555 55555593 */
const LG2: f64 = 3.999999999940941908e-01; /* 3FD99999 9997FA04 */
const LG3: f64 = 2.857142874366239149e-01; /* 3FD24924 94229359 */
const LG4: f64 = 2.222219843214978396e-01; /* 3FCC71C5 1D8E78AF */
const LG5: f64 = 1.818357216161805012e-01; /* 3FC74664 96CB03DE */
const LG6: f64 = 1.531383769920937332e-01; /* 3FC39A09 D078C69F */
const LG7: f64 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
#[cfg_attr(all(test, assert_no_panic), no_panic::no_panic)]
pub fn log(mut x: f64) -> f64 {
let x1p54 = f64::from_bits(0x4350000000000000); // 0x1p54 === 2 ^ 54
let mut ui = x.to_bits();
let mut hx: u32 = (ui >> 32) as u32;
let mut k: i32 = 0;
if (hx < 0x00100000) || ((hx >> 31) != 0) {
/* x < 2**-126 */
if ui << 1 == 0 {
return -1. / (x * x); /* log(+-0)=-inf */
}
if hx >> 31 != 0 {
return (x - x) / 0.0; /* log(-#) = NaN */
}
/* subnormal number, scale x up */
k -= 54;
x *= x1p54;
ui = x.to_bits();
hx = (ui >> 32) as u32;
} else if hx >= 0x7ff00000 {
return x;
} else if hx == 0x3ff00000 && ui << 32 == 0 {
return 0.;
}
/* reduce x into [sqrt(2)/2, sqrt(2)] */
hx += 0x3ff00000 - 0x3fe6a09e;
k += ((hx >> 20) as i32) - 0x3ff;
hx = (hx & 0x000fffff) + 0x3fe6a09e;
ui = ((hx as u64) << 32) | (ui & 0xffffffff);
x = f64::from_bits(ui);
let f: f64 = x - 1.0;
let hfsq: f64 = 0.5 * f * f;
let s: f64 = f / (2.0 + f);
let z: f64 = s * s;
let w: f64 = z * z;
let t1: f64 = w * (LG2 + w * (LG4 + w * LG6));
let t2: f64 = z * (LG1 + w * (LG3 + w * (LG5 + w * LG7)));
let r: f64 = t2 + t1;
let dk: f64 = k as f64;
s * (hfsq + r) + dk * LN2_LO - hfsq + f + dk * LN2_HI
}