source: trunk/libs/newlib/src/newlib/libm/common/s_log1p.c @ 503

Last change on this file since 503 was 444, checked in by satin@…, 6 years ago

add newlib,libalmos-mkh, restructure shared_syscalls.h and mini-libc
File size: 6.0 KB
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1
2/* @(#)s_log1p.c 5.1 93/09/24 */
3/*
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 *
7 * Developed at SunPro, a Sun Microsystems, Inc. business.
8 * Permission to use, copy, modify, and distribute this
9 * software is freely granted, provided that this notice
10 * is preserved.
11 * ====================================================
12 */
13
14/*
15FUNCTION
16<<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>
17
18INDEX
19        log1p
20INDEX
21        log1pf
22
23SYNOPSIS
24        #include <math.h>
25        double log1p(double <[x]>);
26        float log1pf(float <[x]>);
27
28DESCRIPTION
29<<log1p>> calculates
30@tex
31$ln(1+x)$,
32@end tex
33the natural logarithm of <<1+<[x]>>>.  You can use <<log1p>> rather
34than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
35small.
36
37<<log1pf>> calculates the same thing, but accepts and returns
38<<float>> values rather than <<double>>.
39
40RETURNS
41<<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
42<<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
43
44PORTABILITY
45Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
46Interface Definition (Issue 2).
47
48*/
49
50/* double log1p(double x)
51 *
52 * Method :                 
53 *   1. Argument Reduction: find k and f such that
54 *                      1+x = 2^k * (1+f),
55 *         where  sqrt(2)/2 < 1+f < sqrt(2) .
56 *
57 *      Note. If k=0, then f=x is exact. However, if k!=0, then f
58 *      may not be representable exactly. In that case, a correction
59 *      term is need. Let u=1+x rounded. Let c = (1+x)-u, then
60 *      log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
61 *      and add back the correction term c/u.
62 *      (Note: when x > 2**53, one can simply return log(x))
63 *
64 *   2. Approximation of log1p(f).
65 *      Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
66 *               = 2s + 2/3 s**3 + 2/5 s**5 + .....,
67 *               = 2s + s*R
68 *      We use a special Reme algorithm on [0,0.1716] to generate
69 *      a polynomial of degree 14 to approximate R The maximum error
70 *      of this polynomial approximation is bounded by 2**-58.45. In
71 *      other words,
72 *                      2      4      6      8      10      12      14
73 *          R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s  +Lp6*s  +Lp7*s
74 *      (the values of Lp1 to Lp7 are listed in the program)
75 *      and
76 *          |      2          14          |     -58.45
77 *          | Lp1*s +...+Lp7*s    -  R(z) | <= 2
78 *          |                             |
79 *      Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
80 *      In order to guarantee error in log below 1ulp, we compute log
81 *      by
82 *              log1p(f) = f - (hfsq - s*(hfsq+R)).
83 *     
84 *      3. Finally, log1p(x) = k*ln2 + log1p(f). 
85 *                           = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
86 *         Here ln2 is split into two floating point number:
87 *                      ln2_hi + ln2_lo,
88 *         where n*ln2_hi is always exact for |n| < 2000.
89 *
90 * Special cases:
91 *      log1p(x) is NaN with signal if x < -1 (including -INF) ;
92 *      log1p(+INF) is +INF; log1p(-1) is -INF with signal;
93 *      log1p(NaN) is that NaN with no signal.
94 *
95 * Accuracy:
96 *      according to an error analysis, the error is always less than
97 *      1 ulp (unit in the last place).
98 *
99 * Constants:
100 * The hexadecimal values are the intended ones for the following
101 * constants. The decimal values may be used, provided that the
102 * compiler will convert from decimal to binary accurately enough
103 * to produce the hexadecimal values shown.
104 *
105 * Note: Assuming log() return accurate answer, the following
106 *       algorithm can be used to compute log1p(x) to within a few ULP:
107 *     
108 *              u = 1+x;
109 *              if(u==1.0) return x ; else
110 *                         return log(u)*(x/(u-1.0));
111 *
112 *       See HP-15C Advanced Functions Handbook, p.193.
113 */
114
115#include "fdlibm.h"
116
117#ifndef _DOUBLE_IS_32BITS
118
119#ifdef __STDC__
120static const double
121#else
122static double
123#endif
124ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
125ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
126two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
127Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
128Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
129Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
130Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
131Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
132Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
133Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */
134
135#ifdef __STDC__
136static const double zero = 0.0;
137#else
138static double zero = 0.0;
139#endif
140
141#ifdef __STDC__
142        double log1p(double x)
143#else
144        double log1p(x)
145        double x;
146#endif
147{
148        double hfsq,f,c,s,z,R,u;
149        __int32_t k,hx,hu,ax;
150
151        GET_HIGH_WORD(hx,x);
152        ax = hx&0x7fffffff;
153
154        k = 1;
155        if (hx < 0x3FDA827A) {                  /* x < 0.41422  */
156            if(ax>=0x3ff00000) {                /* x <= -1.0 */
157                if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */
158                else return (x-x)/(x-x);        /* log1p(x<-1)=NaN */
159            }
160            if(ax<0x3e200000) {                 /* |x| < 2**-29 */
161                if(two54+x>zero                 /* raise inexact */
162                    &&ax<0x3c900000)            /* |x| < 2**-54 */
163                    return x;
164                else
165                    return x - x*x*0.5;
166            }
167            if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
168                k=0;f=x;hu=1;}  /* -0.2929<x<0.41422 */
169        } 
170        if (hx >= 0x7ff00000) return x+x;
171        if(k!=0) {
172            if(hx<0x43400000) {
173                u  = 1.0+x; 
174                GET_HIGH_WORD(hu,u);
175                k  = (hu>>20)-1023;
176                c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
177                c /= u;
178            } else {
179                u  = x;
180                GET_HIGH_WORD(hu,u);
181                k  = (hu>>20)-1023;
182                c  = 0;
183            }
184            hu &= 0x000fffff;
185            if(hu<0x6a09e) {
186                SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */
187            } else {
188                k += 1; 
189                SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */
190                hu = (0x00100000-hu)>>2;
191            }
192            f = u-1.0;
193        }
194        hfsq=0.5*f*f;
195        if(hu==0) {     /* |f| < 2**-20 */
196          if(f==zero) { if(k==0) return zero; 
197                      else {c += k*ln2_lo; return k*ln2_hi+c;}}
198            R = hfsq*(1.0-0.66666666666666666*f);
199            if(k==0) return f-R; else
200                     return k*ln2_hi-((R-(k*ln2_lo+c))-f);
201        }
202        s = f/(2.0+f); 
203        z = s*s;
204        R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
205        if(k==0) return f-(hfsq-s*(hfsq+R)); else
206                 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
207}
208
209#endif /* _DOUBLE_IS_32BITS */
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