1 // Copyright 2010 The Go Authors. All rights reserved. 2 // Use of this source code is governed by a BSD-style 3 // license that can be found in the LICENSE file. 4 5 package math 6 7 // The original C code, the long comment, and the constants 8 // below are from FreeBSD's /usr/src/lib/msun/src/s_log1p.c 9 // and came with this notice. The go code is a simplified 10 // version of the original C. 11 // 12 // ==================================================== 13 // Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. 14 // 15 // Developed at SunPro, a Sun Microsystems, Inc. business. 16 // Permission to use, copy, modify, and distribute this 17 // software is freely granted, provided that this notice 18 // is preserved. 19 // ==================================================== 20 // 21 // 22 // double log1p(double x) 23 // 24 // Method : 25 // 1. Argument Reduction: find k and f such that 26 // 1+x = 2**k * (1+f), 27 // where sqrt(2)/2 < 1+f < sqrt(2) . 28 // 29 // Note. If k=0, then f=x is exact. However, if k!=0, then f 30 // may not be representable exactly. In that case, a correction 31 // term is need. Let u=1+x rounded. Let c = (1+x)-u, then 32 // log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), 33 // and add back the correction term c/u. 34 // (Note: when x > 2**53, one can simply return log(x)) 35 // 36 // 2. Approximation of log1p(f). 37 // Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) 38 // = 2s + 2/3 s**3 + 2/5 s**5 + ....., 39 // = 2s + s*R 40 // We use a special Reme algorithm on [0,0.1716] to generate 41 // a polynomial of degree 14 to approximate R The maximum error 42 // of this polynomial approximation is bounded by 2**-58.45. In 43 // other words, 44 // 2 4 6 8 10 12 14 45 // R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s 46 // (the values of Lp1 to Lp7 are listed in the program) 47 // and 48 // | 2 14 | -58.45 49 // | Lp1*s +...+Lp7*s - R(z) | <= 2 50 // | | 51 // Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. 52 // In order to guarantee error in log below 1ulp, we compute log 53 // by 54 // log1p(f) = f - (hfsq - s*(hfsq+R)). 55 // 56 // 3. Finally, log1p(x) = k*ln2 + log1p(f). 57 // = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) 58 // Here ln2 is split into two floating point number: 59 // ln2_hi + ln2_lo, 60 // where n*ln2_hi is always exact for |n| < 2000. 61 // 62 // Special cases: 63 // log1p(x) is NaN with signal if x < -1 (including -INF) ; 64 // log1p(+INF) is +INF; log1p(-1) is -INF with signal; 65 // log1p(NaN) is that NaN with no signal. 66 // 67 // Accuracy: 68 // according to an error analysis, the error is always less than 69 // 1 ulp (unit in the last place). 70 // 71 // Constants: 72 // The hexadecimal values are the intended ones for the following 73 // constants. The decimal values may be used, provided that the 74 // compiler will convert from decimal to binary accurately enough 75 // to produce the hexadecimal values shown. 76 // 77 // Note: Assuming log() return accurate answer, the following 78 // algorithm can be used to compute log1p(x) to within a few ULP: 79 // 80 // u = 1+x; 81 // if(u==1.0) return x ; else 82 // return log(u)*(x/(u-1.0)); 83 // 84 // See HP-15C Advanced Functions Handbook, p.193. 85 86 // Log1p returns the natural logarithm of 1 plus its argument x. 87 // It is more accurate than Log(1 + x) when x is near zero. 88 // 89 // Special cases are: 90 // Log1p(+Inf) = +Inf 91 // Log1p(0) = 0 92 // Log1p(-1) = -Inf 93 // Log1p(x < -1) = NaN 94 // Log1p(NaN) = NaN 95 func Log1p(x float64) float64 96 97 func log1p(x float64) float64 { 98 const ( 99 Sqrt2M1 = 4.142135623730950488017e-01 // Sqrt(2)-1 = 0x3fda827999fcef34 100 Sqrt2HalfM1 = -2.928932188134524755992e-01 // Sqrt(2)/2-1 = 0xbfd2bec333018866 101 Small = 1.0 / (1 << 29) // 2**-29 = 0x3e20000000000000 102 Tiny = 1.0 / (1 << 54) // 2**-54 103 Two53 = 1 << 53 // 2**53 104 Ln2Hi = 6.93147180369123816490e-01 // 3fe62e42fee00000 105 Ln2Lo = 1.90821492927058770002e-10 // 3dea39ef35793c76 106 Lp1 = 6.666666666666735130e-01 // 3FE5555555555593 107 Lp2 = 3.999999999940941908e-01 // 3FD999999997FA04 108 Lp3 = 2.857142874366239149e-01 // 3FD2492494229359 109 Lp4 = 2.222219843214978396e-01 // 3FCC71C51D8E78AF 110 Lp5 = 1.818357216161805012e-01 // 3FC7466496CB03DE 111 Lp6 = 1.531383769920937332e-01 // 3FC39A09D078C69F 112 Lp7 = 1.479819860511658591e-01 // 3FC2F112DF3E5244 113 ) 114 115 // special cases 116 switch { 117 case x < -1 || IsNaN(x): // includes -Inf 118 return NaN() 119 case x == -1: 120 return Inf(-1) 121 case IsInf(x, 1): 122 return Inf(1) 123 } 124 125 absx := x 126 if absx < 0 { 127 absx = -absx 128 } 129 130 var f float64 131 var iu uint64 132 k := 1 133 if absx < Sqrt2M1 { // |x| < Sqrt(2)-1 134 if absx < Small { // |x| < 2**-29 135 if absx < Tiny { // |x| < 2**-54 136 return x 137 } 138 return x - x*x*0.5 139 } 140 if x > Sqrt2HalfM1 { // Sqrt(2)/2-1 < x 141 // (Sqrt(2)/2-1) < x < (Sqrt(2)-1) 142 k = 0 143 f = x 144 iu = 1 145 } 146 } 147 var c float64 148 if k != 0 { 149 var u float64 150 if absx < Two53 { // 1<<53 151 u = 1.0 + x 152 iu = Float64bits(u) 153 k = int((iu >> 52) - 1023) 154 if k > 0 { 155 c = 1.0 - (u - x) 156 } else { 157 c = x - (u - 1.0) // correction term 158 c /= u 159 } 160 } else { 161 u = x 162 iu = Float64bits(u) 163 k = int((iu >> 52) - 1023) 164 c = 0 165 } 166 iu &= 0x000fffffffffffff 167 if iu < 0x0006a09e667f3bcd { // mantissa of Sqrt(2) 168 u = Float64frombits(iu | 0x3ff0000000000000) // normalize u 169 } else { 170 k++ 171 u = Float64frombits(iu | 0x3fe0000000000000) // normalize u/2 172 iu = (0x0010000000000000 - iu) >> 2 173 } 174 f = u - 1.0 // Sqrt(2)/2 < u < Sqrt(2) 175 } 176 hfsq := 0.5 * f * f 177 var s, R, z float64 178 if iu == 0 { // |f| < 2**-20 179 if f == 0 { 180 if k == 0 { 181 return 0 182 } 183 c += float64(k) * Ln2Lo 184 return float64(k)*Ln2Hi + c 185 } 186 R = hfsq * (1.0 - 0.66666666666666666*f) // avoid division 187 if k == 0 { 188 return f - R 189 } 190 return float64(k)*Ln2Hi - ((R - (float64(k)*Ln2Lo + c)) - f) 191 } 192 s = f / (2.0 + f) 193 z = s * s 194 R = z * (Lp1 + z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))) 195 if k == 0 { 196 return f - (hfsq - s*(hfsq+R)) 197 } 198 return float64(k)*Ln2Hi - ((hfsq - (s*(hfsq+R) + (float64(k)*Ln2Lo + c))) - f) 199 } 200
