123 lines
3.8 KiB
C
123 lines
3.8 KiB
C
/* origin: FreeBSD /usr/src/lib/msun/src/s_log1p.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 SunPro, 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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/* double log1p(double x)
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* Return the natural logarithm of 1+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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* 1+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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* Note. If k=0, then f=x is exact. However, if k!=0, then f
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* may not be representable exactly. In that case, a correction
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* term is need. Let u=1+x rounded. Let c = (1+x)-u, then
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* log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
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* and add back the correction term c/u.
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* (Note: when x > 2**53, one can simply return log(x))
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*
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* 2. Approximation of log(1+f): See log.c
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*
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* 3. Finally, log1p(x) = k*ln2 + log(1+f) + c/u. See log.c
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*
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* Special cases:
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* log1p(x) is NaN with signal if x < -1 (including -INF) ;
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* log1p(+INF) is +INF; log1p(-1) is -INF with signal;
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* log1p(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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* Note: Assuming log() return accurate answer, the following
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* algorithm can be used to compute log1p(x) to within a few ULP:
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*
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* u = 1+x;
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* if(u==1.0) return x ; else
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* return log(u)*(x/(u-1.0));
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*
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* See HP-15C Advanced Functions Handbook, p.193.
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*/
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#include "libc/math/libm.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 log1p(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,c,s,z,R,w,t1,t2,dk;
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uint32_t hx,hu;
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int k;
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hx = u.i>>32;
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k = 1;
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if (hx < 0x3fda827a || hx>>31) { /* 1+x < sqrt(2)+ */
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if (hx >= 0xbff00000) { /* x <= -1.0 */
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if (x == -1)
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return x/0.0; /* log1p(-1) = -inf */
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return (x-x)/0.0; /* log1p(x<-1) = NaN */
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}
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if (hx<<1 < 0x3ca00000<<1) { /* |x| < 2**-53 */
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/* underflow if subnormal */
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if ((hx&0x7ff00000) == 0)
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FORCE_EVAL((float)x);
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return x;
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}
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if (hx <= 0xbfd2bec4) { /* sqrt(2)/2- <= 1+x < sqrt(2)+ */
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k = 0;
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c = 0;
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f = x;
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}
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} else if (hx >= 0x7ff00000)
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return x;
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if (k) {
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u.f = 1 + x;
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hu = u.i>>32;
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hu += 0x3ff00000 - 0x3fe6a09e;
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k = (int)(hu>>20) - 0x3ff;
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/* correction term ~ log(1+x)-log(u), avoid underflow in c/u */
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if (k < 54) {
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c = k >= 2 ? 1-(u.f-x) : x-(u.f-1);
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c /= u.f;
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} else
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c = 0;
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/* reduce u into [sqrt(2)/2, sqrt(2)] */
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hu = (hu&0x000fffff) + 0x3fe6a09e;
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u.i = (uint64_t)hu<<32 | (u.i&0xffffffff);
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f = u.f - 1;
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}
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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+c) - hfsq + f + dk*ln2_hi;
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}
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