kopia lustrzana https://github.com/micropython/micropython
119 wiersze
3.9 KiB
C
119 wiersze
3.9 KiB
C
/* origin: FreeBSD /usr/src/lib/msun/src/e_log.c */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunSoft, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* log(x)
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* Return the logarithm of x
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*
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* Method :
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* 1. Argument Reduction: find k and f such that
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* x = 2^k * (1+f),
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* where sqrt(2)/2 < 1+f < sqrt(2) .
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*
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* 2. Approximation of log(1+f).
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* Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
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* = 2s + 2/3 s**3 + 2/5 s**5 + .....,
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* = 2s + s*R
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* We use a special Remez algorithm on [0,0.1716] to generate
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* a polynomial of degree 14 to approximate R The maximum error
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* of this polynomial approximation is bounded by 2**-58.45. In
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* other words,
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* 2 4 6 8 10 12 14
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* R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s +Lg6*s +Lg7*s
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* (the values of Lg1 to Lg7 are listed in the program)
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* and
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* | 2 14 | -58.45
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* | Lg1*s +...+Lg7*s - R(z) | <= 2
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* | |
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* Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
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* In order to guarantee error in log below 1ulp, we compute log
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* by
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* log(1+f) = f - s*(f - R) (if f is not too large)
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* log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
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*
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* 3. Finally, log(x) = k*ln2 + log(1+f).
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* = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
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* Here ln2 is split into two floating point number:
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* ln2_hi + ln2_lo,
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* where n*ln2_hi is always exact for |n| < 2000.
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*
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* Special cases:
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* log(x) is NaN with signal if x < 0 (including -INF) ;
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* log(+INF) is +INF; log(0) is -INF with signal;
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* log(NaN) is that NaN with no signal.
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*
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* Accuracy:
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* according to an error analysis, the error is always less than
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* 1 ulp (unit in the last place).
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*
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* Constants:
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* The hexadecimal values are the intended ones for the following
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* constants. The decimal values may be used, provided that the
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* compiler will convert from decimal to binary accurately enough
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* to produce the hexadecimal values shown.
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*/
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#include <math.h>
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#include <stdint.h>
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static const double
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ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */
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ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */
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Lg1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
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Lg2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
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Lg3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
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Lg4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
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Lg5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
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Lg6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
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Lg7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
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double log(double x)
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{
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union {double f; uint64_t i;} u = {x};
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double_t hfsq,f,s,z,R,w,t1,t2,dk;
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uint32_t hx;
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int k;
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hx = u.i>>32;
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k = 0;
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if (hx < 0x00100000 || hx>>31) {
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if (u.i<<1 == 0)
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return -1/(x*x); /* log(+-0)=-inf */
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if (hx>>31)
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return (x-x)/0.0; /* log(-#) = NaN */
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/* subnormal number, scale x up */
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k -= 54;
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x *= 0x1p54;
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u.f = x;
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hx = u.i>>32;
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} else if (hx >= 0x7ff00000) {
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return x;
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} else if (hx == 0x3ff00000 && u.i<<32 == 0)
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return 0;
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/* reduce x into [sqrt(2)/2, sqrt(2)] */
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hx += 0x3ff00000 - 0x3fe6a09e;
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k += (int)(hx>>20) - 0x3ff;
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hx = (hx&0x000fffff) + 0x3fe6a09e;
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u.i = (uint64_t)hx<<32 | (u.i&0xffffffff);
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x = u.f;
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f = x - 1.0;
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hfsq = 0.5*f*f;
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s = f/(2.0+f);
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z = s*s;
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w = z*z;
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t1 = w*(Lg2+w*(Lg4+w*Lg6));
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t2 = z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
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R = t2 + t1;
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dk = k;
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return s*(hfsq+R) + dk*ln2_lo - hfsq + f + dk*ln2_hi;
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}
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