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 5package 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 95func Log1p(x float64) float64 96 97func 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 // correction term 155 if k > 0 { 156 c = 1.0 - (u - x) 157 } else { 158 c = x - (u - 1.0) 159 } 160 c /= u 161 } else { 162 u = x 163 iu = Float64bits(u) 164 k = int((iu >> 52) - 1023) 165 c = 0 166 } 167 iu &= 0x000fffffffffffff 168 if iu < 0x0006a09e667f3bcd { // mantissa of Sqrt(2) 169 u = Float64frombits(iu | 0x3ff0000000000000) // normalize u 170 } else { 171 k++ 172 u = Float64frombits(iu | 0x3fe0000000000000) // normalize u/2 173 iu = (0x0010000000000000 - iu) >> 2 174 } 175 f = u - 1.0 // Sqrt(2)/2 < u < Sqrt(2) 176 } 177 hfsq := 0.5 * f * f 178 var s, R, z float64 179 if iu == 0 { // |f| < 2**-20 180 if f == 0 { 181 if k == 0 { 182 return 0 183 } 184 c += float64(k) * Ln2Lo 185 return float64(k)*Ln2Hi + c 186 } 187 R = hfsq * (1.0 - 0.66666666666666666*f) // avoid division 188 if k == 0 { 189 return f - R 190 } 191 return float64(k)*Ln2Hi - ((R - (float64(k)*Ln2Lo + c)) - f) 192 } 193 s = f / (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 { 197 return f - (hfsq - s*(hfsq+R)) 198 } 199 return float64(k)*Ln2Hi - ((hfsq - (s*(hfsq+R) + (float64(k)*Ln2Lo + c))) - f) 200} 201