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4acf9396 GCI |
1 | /* @(#)s_log1p.c 5.1 93/09/24 */ |
2 | /* | |
3 | * ==================================================== | |
4 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. | |
5 | * | |
6 | * Developed at SunPro, a Sun Microsystems, Inc. business. | |
7 | * Permission to use, copy, modify, and distribute this | |
8 | * software is freely granted, provided that this notice | |
9 | * is preserved. | |
10 | * ==================================================== | |
11 | */ | |
12 | ||
13 | #ifndef lint | |
14 | static char rcsid[] = "$Id: s_log1p.c,v 1.4 1994/03/03 17:04:40 jtc Exp $"; | |
15 | #endif | |
16 | ||
17 | /* double log1p(double x) | |
18 | * | |
19 | * Method : | |
20 | * 1. Argument Reduction: find k and f such that | |
21 | * 1+x = 2^k * (1+f), | |
22 | * where sqrt(2)/2 < 1+f < sqrt(2) . | |
23 | * | |
24 | * Note. If k=0, then f=x is exact. However, if k!=0, then f | |
25 | * may not be representable exactly. In that case, a correction | |
26 | * term is need. Let u=1+x rounded. Let c = (1+x)-u, then | |
27 | * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), | |
28 | * and add back the correction term c/u. | |
29 | * (Note: when x > 2**53, one can simply return log(x)) | |
30 | * | |
31 | * 2. Approximation of log1p(f). | |
32 | * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) | |
33 | * = 2s + 2/3 s**3 + 2/5 s**5 + ....., | |
34 | * = 2s + s*R | |
35 | * We use a special Reme algorithm on [0,0.1716] to generate | |
36 | * a polynomial of degree 14 to approximate R The maximum error | |
37 | * of this polynomial approximation is bounded by 2**-58.45. In | |
38 | * other words, | |
39 | * 2 4 6 8 10 12 14 | |
40 | * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s | |
41 | * (the values of Lp1 to Lp7 are listed in the program) | |
42 | * and | |
43 | * | 2 14 | -58.45 | |
44 | * | Lp1*s +...+Lp7*s - R(z) | <= 2 | |
45 | * | | | |
46 | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. | |
47 | * In order to guarantee error in log below 1ulp, we compute log | |
48 | * by | |
49 | * log1p(f) = f - (hfsq - s*(hfsq+R)). | |
50 | * | |
51 | * 3. Finally, log1p(x) = k*ln2 + log1p(f). | |
52 | * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) | |
53 | * Here ln2 is split into two floating point number: | |
54 | * ln2_hi + ln2_lo, | |
55 | * where n*ln2_hi is always exact for |n| < 2000. | |
56 | * | |
57 | * Special cases: | |
58 | * log1p(x) is NaN with signal if x < -1 (including -INF) ; | |
59 | * log1p(+INF) is +INF; log1p(-1) is -INF with signal; | |
60 | * log1p(NaN) is that NaN with no signal. | |
61 | * | |
62 | * Accuracy: | |
63 | * according to an error analysis, the error is always less than | |
64 | * 1 ulp (unit in the last place). | |
65 | * | |
66 | * Constants: | |
67 | * The hexadecimal values are the intended ones for the following | |
68 | * constants. The decimal values may be used, provided that the | |
69 | * compiler will convert from decimal to binary accurately enough | |
70 | * to produce the hexadecimal values shown. | |
71 | * | |
72 | * Note: Assuming log() return accurate answer, the following | |
73 | * algorithm can be used to compute log1p(x) to within a few ULP: | |
74 | * | |
75 | * u = 1+x; | |
76 | * if(u==1.0) return x ; else | |
77 | * return log(u)*(x/(u-1.0)); | |
78 | * | |
79 | * See HP-15C Advanced Functions Handbook, p.193. | |
80 | */ | |
81 | ||
82 | #include "math.h" | |
83 | #include <machine/endian.h> | |
84 | ||
85 | #if BYTE_ORDER == LITTLE_ENDIAN | |
86 | #define n0 1 | |
87 | #else | |
88 | #define n0 0 | |
89 | #endif | |
90 | ||
91 | #ifdef __STDC__ | |
92 | static const double | |
93 | #else | |
94 | static double | |
95 | #endif | |
96 | ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */ | |
97 | ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */ | |
98 | two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */ | |
99 | Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */ | |
100 | Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */ | |
101 | Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */ | |
102 | Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */ | |
103 | Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */ | |
104 | Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */ | |
105 | Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */ | |
106 | ||
107 | static double zero = 0.0; | |
108 | ||
109 | #ifdef __STDC__ | |
110 | double log1p(double x) | |
111 | #else | |
112 | double log1p(x) | |
113 | double x; | |
114 | #endif | |
115 | { | |
116 | double hfsq,f,c,s,z,R,u; | |
117 | int k,hx,hu,ax; | |
118 | ||
119 | hx = *(n0+(int*)&x); /* high word of x */ | |
120 | ax = hx&0x7fffffff; | |
121 | ||
122 | k = 1; | |
123 | if (hx < 0x3FDA827A) { /* x < 0.41422 */ | |
124 | if(ax>=0x3ff00000) { /* x <= -1.0 */ | |
125 | if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */ | |
126 | else return (x-x)/(x-x); /* log1p(x<-1)=NaN */ | |
127 | } | |
128 | if(ax<0x3e200000) { /* |x| < 2**-29 */ | |
129 | if(two54+x>zero /* raise inexact */ | |
130 | &&ax<0x3c900000) /* |x| < 2**-54 */ | |
131 | return x; | |
132 | else | |
133 | return x - x*x*0.5; | |
134 | } | |
135 | if(hx>0||hx<=((int)0xbfd2bec3)) { | |
136 | k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */ | |
137 | } | |
138 | if (hx >= 0x7ff00000) return x+x; | |
139 | if(k!=0) { | |
140 | if(hx<0x43400000) { | |
141 | u = 1.0+x; | |
142 | hu = *(n0+(int*)&u); /* high word of u */ | |
143 | k = (hu>>20)-1023; | |
144 | c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */ | |
145 | c /= u; | |
146 | } else { | |
147 | u = x; | |
148 | hu = *(n0+(int*)&u); /* high word of u */ | |
149 | k = (hu>>20)-1023; | |
150 | c = 0; | |
151 | } | |
152 | hu &= 0x000fffff; | |
153 | if(hu<0x6a09e) { | |
154 | *(n0+(int*)&u) = hu|0x3ff00000; /* normalize u */ | |
155 | } else { | |
156 | k += 1; | |
157 | *(n0+(int*)&u) = hu|0x3fe00000; /* normalize u/2 */ | |
158 | hu = (0x00100000-hu)>>2; | |
159 | } | |
160 | f = u-1.0; | |
161 | } | |
162 | hfsq=0.5*f*f; | |
163 | if(hu==0) { /* |f| < 2**-20 */ | |
164 | if(f==zero) if(k==0) return zero; | |
165 | else {c += k*ln2_lo; return k*ln2_hi+c;} | |
166 | R = hfsq*(1.0-0.66666666666666666*f); | |
167 | if(k==0) return f-R; else | |
168 | return k*ln2_hi-((R-(k*ln2_lo+c))-f); | |
169 | } | |
170 | s = f/(2.0+f); | |
171 | z = s*s; | |
172 | R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))); | |
173 | if(k==0) return f-(hfsq-s*(hfsq+R)); else | |
174 | return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f); | |
175 | } |