1/*
2 * Copyright (c) 1998, 2003, Oracle and/or its affiliates. All rights reserved.
3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
4 *
5 * This code is free software; you can redistribute it and/or modify it
6 * under the terms of the GNU General Public License version 2 only, as
7 * published by the Free Software Foundation. Oracle designates this
8 * particular file as subject to the "Classpath" exception as provided
9 * by Oracle in the LICENSE file that accompanied this code.
10 *
11 * This code is distributed in the hope that it will be useful, but WITHOUT
12 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
13 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
14 * version 2 for more details (a copy is included in the LICENSE file that
15 * accompanied this code).
16 *
17 * You should have received a copy of the GNU General Public License version
18 * 2 along with this work; if not, write to the Free Software Foundation,
19 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
20 *
21 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
22 * or visit www.oracle.com if you need additional information or have any
23 * questions.
24 */
25
26/* double log1p(double x)
27 *
28 * Method :
29 * 1. Argument Reduction: find k and f such that
30 * 1+x = 2^k * (1+f),
31 * where sqrt(2)/2 < 1+f < sqrt(2) .
32 *
33 * Note. If k=0, then f=x is exact. However, if k!=0, then f
34 * may not be representable exactly. In that case, a correction
35 * term is need. Let u=1+x rounded. Let c = (1+x)-u, then
36 * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
37 * and add back the correction term c/u.
38 * (Note: when x > 2**53, one can simply return log(x))
39 *
40 * 2. Approximation of log1p(f).
41 * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
42 * = 2s + 2/3 s**3 + 2/5 s**5 + .....,
43 * = 2s + s*R
44 * We use a special Reme algorithm on [0,0.1716] to generate
45 * a polynomial of degree 14 to approximate R The maximum error
46 * of this polynomial approximation is bounded by 2**-58.45. In
47 * other words,
48 * 2 4 6 8 10 12 14
49 * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s
50 * (the values of Lp1 to Lp7 are listed in the program)
51 * and
52 * | 2 14 | -58.45
53 * | Lp1*s +...+Lp7*s - R(z) | <= 2
54 * | |
55 * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
56 * In order to guarantee error in log below 1ulp, we compute log
57 * by
58 * log1p(f) = f - (hfsq - s*(hfsq+R)).
59 *
60 * 3. Finally, log1p(x) = k*ln2 + log1p(f).
61 * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
62 * Here ln2 is split into two floating point number:
63 * ln2_hi + ln2_lo,
64 * where n*ln2_hi is always exact for |n| < 2000.
65 *
66 * Special cases:
67 * log1p(x) is NaN with signal if x < -1 (including -INF) ;
68 * log1p(+INF) is +INF; log1p(-1) is -INF with signal;
69 * log1p(NaN) is that NaN with no signal.
70 *
71 * Accuracy:
72 * according to an error analysis, the error is always less than
73 * 1 ulp (unit in the last place).
74 *
75 * Constants:
76 * The hexadecimal values are the intended ones for the following
77 * constants. The decimal values may be used, provided that the
78 * compiler will convert from decimal to binary accurately enough
79 * to produce the hexadecimal values shown.
80 *
81 * Note: Assuming log() return accurate answer, the following
82 * algorithm can be used to compute log1p(x) to within a few ULP:
83 *
84 * u = 1+x;
85 * if(u==1.0) return x ; else
86 * return log(u)*(x/(u-1.0));
87 *
88 * See HP-15C Advanced Functions Handbook, p.193.
89 */
90
91#include "fdlibm.h"
92
93#ifdef __STDC__
94static const double
95#else
96static double
97#endif
98ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */
99ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */
100two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */
101Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
102Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
103Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
104Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
105Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
106Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
107Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
108
109static double zero = 0.0;
110
111#ifdef __STDC__
112 double log1p(double x)
113#else
114 double log1p(x)
115 double x;
116#endif
117{
118 double hfsq,f=0,c=0,s,z,R,u;
119 int k,hx,hu=0,ax;
120
121 hx = __HI(x); /* high word of x */
122 ax = hx&0x7fffffff;
123
124 k = 1;
125 if (hx < 0x3FDA827A) { /* x < 0.41422 */
126 if(ax>=0x3ff00000) { /* x <= -1.0 */
127 /*
128 * Added redundant test against hx to work around VC++
129 * code generation problem.
130 */
131 if(x==-1.0 && (hx==0xbff00000)) /* log1p(-1)=-inf */
132 return -two54/zero;
133 else
134 return (x-x)/(x-x); /* log1p(x<-1)=NaN */
135 }
136 if(ax<0x3e200000) { /* |x| < 2**-29 */
137 if(two54+x>zero /* raise inexact */
138 &&ax<0x3c900000) /* |x| < 2**-54 */
139 return x;
140 else
141 return x - x*x*0.5;
142 }
143 if(hx>0||hx<=((int)0xbfd2bec3)) {
144 k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */
145 }
146 if (hx >= 0x7ff00000) return x+x;
147 if(k!=0) {
148 if(hx<0x43400000) {
149 u = 1.0+x;
150 hu = __HI(u); /* high word of u */
151 k = (hu>>20)-1023;
152 c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */
153 c /= u;
154 } else {
155 u = x;
156 hu = __HI(u); /* high word of u */
157 k = (hu>>20)-1023;
158 c = 0;
159 }
160 hu &= 0x000fffff;
161 if(hu<0x6a09e) {
162 __HI(u) = hu|0x3ff00000; /* normalize u */
163 } else {
164 k += 1;
165 __HI(u) = hu|0x3fe00000; /* normalize u/2 */
166 hu = (0x00100000-hu)>>2;
167 }
168 f = u-1.0;
169 }
170 hfsq=0.5*f*f;
171 if(hu==0) { /* |f| < 2**-20 */
172 if(f==zero) { if(k==0) return zero;
173 else {c += k*ln2_lo; return k*ln2_hi+c;}}
174 R = hfsq*(1.0-0.66666666666666666*f);
175 if(k==0) return f-R; else
176 return k*ln2_hi-((R-(k*ln2_lo+c))-f);
177 }
178 s = f/(2.0+f);
179 z = s*s;
180 R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
181 if(k==0) return f-(hfsq-s*(hfsq+R)); else
182 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
183}
184