218 lines
6.1 KiB
C
218 lines
6.1 KiB
C
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/* @(#)s_log1p.c 5.1 93/09/24 */
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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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/*
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FUNCTION
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<<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>
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INDEX
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log1p
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INDEX
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log1pf
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ANSI_SYNOPSIS
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#include <math.h>
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double log1p(double <[x]>);
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float log1pf(float <[x]>);
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TRAD_SYNOPSIS
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#include <math.h>
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double log1p(<[x]>)
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double <[x]>;
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float log1pf(<[x]>)
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float <[x]>;
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DESCRIPTION
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<<log1p>> calculates
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@tex
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$ln(1+x)$,
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@end tex
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the natural logarithm of <<1+<[x]>>>. You can use <<log1p>> rather
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than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
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small.
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<<log1pf>> calculates the same thing, but accepts and returns
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<<float>> values rather than <<double>>.
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RETURNS
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<<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
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<<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
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PORTABILITY
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Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
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Interface Definition (Issue 2).
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*/
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/* double log1p(double 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 log1p(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 Reme 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) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s
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* (the values of Lp1 to Lp7 are listed in the program)
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* and
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* | 2 14 | -58.45
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* | Lp1*s +...+Lp7*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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* log1p(f) = f - (hfsq - s*(hfsq+R)).
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*
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* 3. Finally, log1p(x) = k*ln2 + log1p(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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* 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 "fdlibm.h"
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#ifndef _DOUBLE_IS_32BITS
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#ifdef __STDC__
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static const double
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#else
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static double
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#endif
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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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two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */
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Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
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Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
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Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
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Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
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Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
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Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
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Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
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#ifdef __STDC__
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static const double zero = 0.0;
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#else
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static double zero = 0.0;
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#endif
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#ifdef __STDC__
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double log1p(double x)
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#else
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double log1p(x)
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double x;
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#endif
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{
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double hfsq,f,c,s,z,R,u;
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__int32_t k,hx,hu,ax;
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GET_HIGH_WORD(hx,x);
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ax = hx&0x7fffffff;
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k = 1;
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if (hx < 0x3FDA827A) { /* x < 0.41422 */
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if(ax>=0x3ff00000) { /* x <= -1.0 */
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if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */
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else return (x-x)/(x-x); /* log1p(x<-1)=NaN */
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}
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if(ax<0x3e200000) { /* |x| < 2**-29 */
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if(two54+x>zero /* raise inexact */
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&&ax<0x3c900000) /* |x| < 2**-54 */
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return x;
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else
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return x - x*x*0.5;
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}
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if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
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k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */
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}
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if (hx >= 0x7ff00000) return x+x;
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if(k!=0) {
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if(hx<0x43400000) {
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u = 1.0+x;
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GET_HIGH_WORD(hu,u);
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k = (hu>>20)-1023;
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c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
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c /= u;
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} else {
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u = x;
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GET_HIGH_WORD(hu,u);
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k = (hu>>20)-1023;
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c = 0;
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}
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hu &= 0x000fffff;
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if(hu<0x6a09e) {
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SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */
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} else {
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k += 1;
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SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */
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hu = (0x00100000-hu)>>2;
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}
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f = u-1.0;
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}
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hfsq=0.5*f*f;
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if(hu==0) { /* |f| < 2**-20 */
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if(f==zero) if(k==0) return zero;
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else {c += k*ln2_lo; return k*ln2_hi+c;}
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R = hfsq*(1.0-0.66666666666666666*f);
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if(k==0) return f-R; else
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return k*ln2_hi-((R-(k*ln2_lo+c))-f);
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}
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s = f/(2.0+f);
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z = s*s;
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R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
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if(k==0) return f-(hfsq-s*(hfsq+R)); else
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return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
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}
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#endif /* _DOUBLE_IS_32BITS */
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