POK
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00001 /* 00002 * POK header 00003 * 00004 * The following file is a part of the POK project. Any modification should 00005 * made according to the POK licence. You CANNOT use this file or a part of 00006 * this file is this part of a file for your own project 00007 * 00008 * For more information on the POK licence, please see our LICENCE FILE 00009 * 00010 * Please follow the coding guidelines described in doc/CODING_GUIDELINES 00011 * 00012 * Copyright (c) 2007-2009 POK team 00013 * 00014 * Created by julien on Fri Jan 30 14:41:34 2009 00015 */ 00016 00017 /* @(#)s_log1p.c 5.1 93/09/24 */ 00018 /* 00019 * ==================================================== 00020 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. 00021 * 00022 * Developed at SunPro, a Sun Microsystems, Inc. business. 00023 * Permission to use, copy, modify, and distribute this 00024 * software is freely granted, provided that this notice 00025 * is preserved. 00026 * ==================================================== 00027 */ 00028 00029 /* double log1p(double x) 00030 * 00031 * Method : 00032 * 1. Argument Reduction: find k and f such that 00033 * 1+x = 2^k * (1+f), 00034 * where sqrt(2)/2 < 1+f < sqrt(2) . 00035 * 00036 * Note. If k=0, then f=x is exact. However, if k!=0, then f 00037 * may not be representable exactly. In that case, a correction 00038 * term is need. Let u=1+x rounded. Let c = (1+x)-u, then 00039 * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), 00040 * and add back the correction term c/u. 00041 * (Note: when x > 2**53, one can simply return log(x)) 00042 * 00043 * 2. Approximation of log1p(f). 00044 * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) 00045 * = 2s + 2/3 s**3 + 2/5 s**5 + ....., 00046 * = 2s + s*R 00047 * We use a special Reme algorithm on [0,0.1716] to generate 00048 * a polynomial of degree 14 to approximate R The maximum error 00049 * of this polynomial approximation is bounded by 2**-58.45. In 00050 * other words, 00051 * 2 4 6 8 10 12 14 00052 * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s 00053 * (the values of Lp1 to Lp7 are listed in the program) 00054 * and 00055 * | 2 14 | -58.45 00056 * | Lp1*s +...+Lp7*s - R(z) | <= 2 00057 * | | 00058 * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. 00059 * In order to guarantee error in log below 1ulp, we compute log 00060 * by 00061 * log1p(f) = f - (hfsq - s*(hfsq+R)). 00062 * 00063 * 3. Finally, log1p(x) = k*ln2 + log1p(f). 00064 * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) 00065 * Here ln2 is split into two floating point number: 00066 * ln2_hi + ln2_lo, 00067 * where n*ln2_hi is always exact for |n| < 2000. 00068 * 00069 * Special cases: 00070 * log1p(x) is NaN with signal if x < -1 (including -INF) ; 00071 * log1p(+INF) is +INF; log1p(-1) is -INF with signal; 00072 * log1p(NaN) is that NaN with no signal. 00073 * 00074 * Accuracy: 00075 * according to an error analysis, the error is always less than 00076 * 1 ulp (unit in the last place). 00077 * 00078 * Constants: 00079 * The hexadecimal values are the intended ones for the following 00080 * constants. The decimal values may be used, provided that the 00081 * compiler will convert from decimal to binary accurately enough 00082 * to produce the hexadecimal values shown. 00083 * 00084 * Note: Assuming log() return accurate answer, the following 00085 * algorithm can be used to compute log1p(x) to within a few ULP: 00086 * 00087 * u = 1+x; 00088 * if(u==1.0) return x ; else 00089 * return log(u)*(x/(u-1.0)); 00090 * 00091 * See HP-15C Advanced Functions Handbook, p.193. 00092 */ 00093 00094 #ifdef POK_NEEDS_LIBMATH 00095 00096 #include <types.h> 00097 #include "math_private.h" 00098 00099 static const double 00100 ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */ 00101 ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */ 00102 two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */ 00103 Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */ 00104 Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */ 00105 Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */ 00106 Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */ 00107 Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */ 00108 Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */ 00109 Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */ 00110 00111 static const double zero = 0.0; 00112 00113 double 00114 log1p(double x) 00115 { 00116 double hfsq,f,c,s,z,R,u; 00117 int32_t k,hx,hu,ax; 00118 00119 f = c = 0; 00120 hu = 0; 00121 GET_HIGH_WORD(hx,x); 00122 ax = hx&0x7fffffff; 00123 00124 k = 1; 00125 if (hx < 0x3FDA827A) { /* x < 0.41422 */ 00126 if(ax>=0x3ff00000) { /* x <= -1.0 */ 00127 if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */ 00128 else return (x-x)/(x-x); /* log1p(x<-1)=NaN */ 00129 } 00130 if(ax<0x3e200000) { /* |x| < 2**-29 */ 00131 if(two54+x>zero /* raise inexact */ 00132 &&ax<0x3c900000) /* |x| < 2**-54 */ 00133 return x; 00134 else 00135 return x - x*x*0.5; 00136 } 00137 if(hx>0||hx<=((int32_t)0xbfd2bec3)) { 00138 k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */ 00139 } 00140 if (hx >= 0x7ff00000) return x+x; 00141 if(k!=0) { 00142 if(hx<0x43400000) { 00143 u = 1.0+x; 00144 GET_HIGH_WORD(hu,u); 00145 k = (hu>>20)-1023; 00146 c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */ 00147 c /= u; 00148 } else { 00149 u = x; 00150 GET_HIGH_WORD(hu,u); 00151 k = (hu>>20)-1023; 00152 c = 0; 00153 } 00154 hu &= 0x000fffff; 00155 if(hu<0x6a09e) { 00156 SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */ 00157 } else { 00158 k += 1; 00159 SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */ 00160 hu = (0x00100000-hu)>>2; 00161 } 00162 f = u-1.0; 00163 } 00164 hfsq=0.5*f*f; 00165 if(hu==0) { /* |f| < 2**-20 */ 00166 if(f==zero) { if(k==0) return zero; 00167 else {c += k*ln2_lo; return k*ln2_hi+c;} 00168 } 00169 R = hfsq*(1.0-0.66666666666666666*f); 00170 if(k==0) return f-R; else 00171 return k*ln2_hi-((R-(k*ln2_lo+c))-f); 00172 } 00173 s = f/(2.0+f); 00174 z = s*s; 00175 R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))); 00176 if(k==0) return f-(hfsq-s*(hfsq+R)); else 00177 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f); 00178 } 00179 #endif 00180