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 /*
15 FUNCTION
16 <<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>
17
18 INDEX
19 log1p
20 INDEX
21 log1pf
22
23 SYNOPSIS
24 #include <math.h>
25 double log1p(double <[x]>);
26 float log1pf(float <[x]>);
27
28 DESCRIPTION
29 <<log1p>> calculates
30 @tex
31 $ln(1+x)$,
32 @end tex
33 the natural logarithm of <<1+<[x]>>>. You can use <<log1p>> rather
34 than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very
35 small.
36
37 <<log1pf>> calculates the same thing, but accepts and returns
38 <<float>> values rather than <<double>>.
39
40 RETURNS
41 <<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.
42 <<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.
43
44 PORTABILITY
45 Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V
46 Interface 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 Remez 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 #include "math_config.h"
117
118 #ifdef _NEED_FLOAT64
119
120 static const __float64
121 ln2_hi = _F_64(6.93147180369123816490e-01), /* 3fe62e42 fee00000 */
122 ln2_lo = _F_64(1.90821492927058770002e-10), /* 3dea39ef 35793c76 */
123 two54 = _F_64(1.80143985094819840000e+16), /* 43500000 00000000 */
124 Lp1 = _F_64(6.666666666666735130e-01), /* 3FE55555 55555593 */
125 Lp2 = _F_64(3.999999999940941908e-01), /* 3FD99999 9997FA04 */
126 Lp3 = _F_64(2.857142874366239149e-01), /* 3FD24924 94229359 */
127 Lp4 = _F_64(2.222219843214978396e-01), /* 3FCC71C5 1D8E78AF */
128 Lp5 = _F_64(1.818357216161805012e-01), /* 3FC74664 96CB03DE */
129 Lp6 = _F_64(1.531383769920937332e-01), /* 3FC39A09 D078C69F */
130 Lp7 = _F_64(1.479819860511658591e-01); /* 3FC2F112 DF3E5244 */
131
132 static const __float64 zero = _F_64(0.0);
133
134 __float64
log1p64(__float64 x)135 log1p64(__float64 x)
136 {
137 __float64 hfsq,f,c=0,s,z,R,u;
138 __int32_t k,hx,hu,ax;
139
140 GET_HIGH_WORD(hx,x);
141 ax = hx&0x7fffffff;
142
143 k = 1;
144 if (hx < 0x3FDA827A) { /* x < 0.41422 */
145 if(ax>=0x3ff00000) { /* x <= -1.0 */
146 if(x==_F_64(-1.0))
147 return __math_divzero (1); /* log1p(-1)=-inf */
148 else
149 return __math_invalid (x); /* log1p(x<-1)=NaN */
150 }
151 if(ax<0x3e200000) { /* |x| < 2**-29 */
152 if(two54+x>zero /* raise inexact */
153 &&ax<0x3c900000) /* |x| < 2**-54 */
154 return x;
155 else
156 return x - x*x*_F_64(0.5);
157 }
158 if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {
159 k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */
160 }
161 if (hx >= 0x7ff00000) return x+x;
162 if(k!=0) {
163 if(hx<0x43400000) {
164 u = _F_64(1.0)+x;
165 GET_HIGH_WORD(hu,u);
166 k = (hu>>20)-1023;
167 c = (k>0)? _F_64(1.0)-(u-x):x-(u-_F_64(1.0));/* correction term */
168 c /= u;
169 } else {
170 u = x;
171 GET_HIGH_WORD(hu,u);
172 k = (hu>>20)-1023;
173 c = 0;
174 }
175 hu &= 0x000fffff;
176 if(hu<0x6a09e) {
177 SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */
178 } else {
179 k += 1;
180 SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */
181 hu = (0x00100000-hu)>>2;
182 }
183 f = u - _F_64(1.0);
184 }
185 hfsq=_F_64(0.5)*f*f;
186 if(hu==0) { /* |f| < 2**-20 */
187 if(f==zero) { if(k==0) return zero;
188 else {c += k*ln2_lo; return k*ln2_hi+c;}}
189 R = hfsq*(_F_64(1.0)-_F_64(0.66666666666666666)*f);
190 if(k==0) return f-R; else
191 return k*ln2_hi-((R-(k*ln2_lo+c))-f);
192 }
193 s = f/(_F_64(2.0)+f);
194 z = s*s;
195 R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));
196 if(k==0) return f-(hfsq-s*(hfsq+R)); else
197 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);
198 }
199
200 _MATH_ALIAS_d_d(log1p)
201
202 #endif /* _NEED_FLOAT64 */
203
