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| Rev | Author | Line No. | Line |
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| 2 | pj | 1 | /* @(#)s_log1p.c 5.1 93/09/24 */ |
| 2 | /* |
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| 3 | * ==================================================== |
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| 4 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
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| 5 | * |
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| 6 | * Developed at SunPro, a Sun Microsystems, Inc. business. |
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| 7 | * Permission to use, copy, modify, and distribute this |
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| 8 | * software is freely granted, provided that this notice |
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| 9 | * is preserved. |
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| 10 | * ==================================================== |
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| 11 | */ |
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| 12 | |||
| 13 | #ifndef lint |
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| 14 | static char rcsid[] = "$\Id: s_log1p.c,v 1.2 1995/05/30 05:49:57 rgrimes Exp $"; |
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| 15 | #endif |
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| 16 | |||
| 17 | /* double log1p(double x) |
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| 18 | * |
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| 19 | * Method : |
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| 20 | * 1. Argument Reduction: find k and f such that |
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| 21 | * 1+x = 2^k * (1+f), |
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| 22 | * where sqrt(2)/2 < 1+f < sqrt(2) . |
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| 23 | * |
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| 24 | * Note. If k=0, then f=x is exact. However, if k!=0, then f |
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| 25 | * may not be representable exactly. In that case, a correction |
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| 26 | * term is need. Let u=1+x rounded. Let c = (1+x)-u, then |
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| 27 | * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), |
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| 28 | * and add back the correction term c/u. |
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| 29 | * (Note: when x > 2**53, one can simply return log(x)) |
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| 30 | * |
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| 31 | * 2. Approximation of log1p(f). |
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| 32 | * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) |
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| 33 | * = 2s + 2/3 s**3 + 2/5 s**5 + ....., |
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| 34 | * = 2s + s*R |
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| 35 | * We use a special Reme algorithm on [0,0.1716] to generate |
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| 36 | * a polynomial of degree 14 to approximate R The maximum error |
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| 37 | * of this polynomial approximation is bounded by 2**-58.45. In |
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| 38 | * other words, |
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| 39 | * 2 4 6 8 10 12 14 |
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| 40 | * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s |
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| 41 | * (the values of Lp1 to Lp7 are listed in the program) |
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| 42 | * and |
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| 43 | * | 2 14 | -58.45 |
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| 44 | * | Lp1*s +...+Lp7*s - R(z) | <= 2 |
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| 45 | * | | |
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| 46 | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. |
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| 47 | * In order to guarantee error in log below 1ulp, we compute log |
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| 48 | * by |
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| 49 | * log1p(f) = f - (hfsq - s*(hfsq+R)). |
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| 50 | * |
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| 51 | * 3. Finally, log1p(x) = k*ln2 + log1p(f). |
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| 52 | * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) |
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| 53 | * Here ln2 is split into two floating point number: |
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| 54 | * ln2_hi + ln2_lo, |
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| 55 | * where n*ln2_hi is always exact for |n| < 2000. |
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| 56 | * |
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| 57 | * Special cases: |
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| 58 | * log1p(x) is NaN with signal if x < -1 (including -INF) ; |
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| 59 | * log1p(+INF) is +INF; log1p(-1) is -INF with signal; |
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| 60 | * log1p(NaN) is that NaN with no signal. |
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| 61 | * |
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| 62 | * Accuracy: |
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| 63 | * according to an error analysis, the error is always less than |
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| 64 | * 1 ulp (unit in the last place). |
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| 65 | * |
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| 66 | * Constants: |
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| 67 | * The hexadecimal values are the intended ones for the following |
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| 68 | * constants. The decimal values may be used, provided that the |
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| 69 | * compiler will convert from decimal to binary accurately enough |
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| 70 | * to produce the hexadecimal values shown. |
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| 71 | * |
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| 72 | * Note: Assuming log() return accurate answer, the following |
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| 73 | * algorithm can be used to compute log1p(x) to within a few ULP: |
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| 74 | * |
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| 75 | * u = 1+x; |
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| 76 | * if(u==1.0) return x ; else |
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| 77 | * return log(u)*(x/(u-1.0)); |
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| 78 | * |
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| 79 | * See HP-15C Advanced Functions Handbook, p.193. |
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| 80 | */ |
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| 81 | |||
| 82 | #include "math.h" |
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| 83 | #include "math_private.h" |
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| 84 | |||
| 85 | #ifdef __STDC__ |
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| 86 | static const double |
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| 87 | #else |
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| 88 | static double |
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| 89 | #endif |
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| 90 | ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */ |
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| 91 | ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */ |
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| 92 | two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */ |
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| 93 | Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */ |
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| 94 | Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */ |
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| 95 | Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */ |
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| 96 | Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */ |
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| 97 | Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */ |
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| 98 | Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */ |
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| 99 | Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */ |
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| 100 | |||
| 101 | #ifdef __STDC__ |
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| 102 | static const double zero = 0.0; |
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| 103 | #else |
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| 104 | static double zero = 0.0; |
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| 105 | #endif |
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| 106 | |||
| 107 | #ifdef __STDC__ |
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| 108 | double log1p(double x) |
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| 109 | #else |
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| 110 | double log1p(x) |
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| 111 | double x; |
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| 112 | #endif |
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| 113 | { |
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| 114 | double hfsq,f,c,s,z,R,u; |
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| 115 | int32_t k,hx,hu,ax; |
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| 116 | |||
| 117 | GET_HIGH_WORD(hx,x); |
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| 118 | ax = hx&0x7fffffff; |
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| 119 | |||
| 120 | k = 1; |
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| 121 | if (hx < 0x3FDA827A) { /* x < 0.41422 */ |
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| 122 | if(ax>=0x3ff00000) { /* x <= -1.0 */ |
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| 123 | if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */ |
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| 124 | else return (x-x)/(x-x); /* log1p(x<-1)=NaN */ |
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| 125 | } |
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| 126 | if(ax<0x3e200000) { /* |x| < 2**-29 */ |
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| 127 | if(two54+x>zero /* raise inexact */ |
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| 128 | &&ax<0x3c900000) /* |x| < 2**-54 */ |
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| 129 | return x; |
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| 130 | else |
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| 131 | return x - x*x*0.5; |
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| 132 | } |
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| 133 | if(hx>0||hx<=((int32_t)0xbfd2bec3)) { |
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| 134 | k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */ |
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| 135 | } |
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| 136 | if (hx >= 0x7ff00000) return x+x; |
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| 137 | if(k!=0) { |
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| 138 | if(hx<0x43400000) { |
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| 139 | u = 1.0+x; |
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| 140 | GET_HIGH_WORD(hu,u); |
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| 141 | k = (hu>>20)-1023; |
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| 142 | c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */ |
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| 143 | c /= u; |
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| 144 | } else { |
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| 145 | u = x; |
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| 146 | GET_HIGH_WORD(hu,u); |
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| 147 | k = (hu>>20)-1023; |
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| 148 | c = 0; |
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| 149 | } |
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| 150 | hu &= 0x000fffff; |
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| 151 | if(hu<0x6a09e) { |
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| 152 | SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */ |
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| 153 | } else { |
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| 154 | k += 1; |
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| 155 | SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */ |
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| 156 | hu = (0x00100000-hu)>>2; |
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| 157 | } |
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| 158 | f = u-1.0; |
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| 159 | } |
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| 160 | hfsq=0.5*f*f; |
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| 161 | if(hu==0) { /* |f| < 2**-20 */ |
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| 162 | if(f==zero) if(k==0) return zero; |
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| 163 | else {c += k*ln2_lo; return k*ln2_hi+c;} |
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| 164 | R = hfsq*(1.0-0.66666666666666666*f); |
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| 165 | if(k==0) return f-R; else |
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| 166 | return k*ln2_hi-((R-(k*ln2_lo+c))-f); |
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| 167 | } |
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| 168 | s = f/(2.0+f); |
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| 169 | z = s*s; |
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| 170 | R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))); |
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| 171 | if(k==0) return f-(hfsq-s*(hfsq+R)); else |
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| 172 | return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f); |
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| 173 | } |