2014-04-11 Marc Glisse <marc.glisse@inria.fr>
[official-gcc.git] / libgo / go / math / log1p.go
blobc8daaaa1c95ab2293a690a93877cf31439943a82
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.
5 package math
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.
12 // ====================================================
13 // Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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 // ====================================================
22 // double log1p(double x)
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) .
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))
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)).
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.
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.
67 // Accuracy:
68 // according to an error analysis, the error is always less than
69 // 1 ulp (unit in the last place).
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.
77 // Note: Assuming log() return accurate answer, the following
78 // algorithm can be used to compute log1p(x) to within a few ULP:
80 // u = 1+x;
81 // if(u==1.0) return x ; else
82 // return log(u)*(x/(u-1.0));
84 // See HP-15C Advanced Functions Handbook, p.193.
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.
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
96 //extern log1p
97 func libc_log1p(float64) float64
99 func Log1p(x float64) float64 {
100 return libc_log1p(x)
103 func log1p(x float64) float64 {
104 const (
105 Sqrt2M1 = 4.142135623730950488017e-01 // Sqrt(2)-1 = 0x3fda827999fcef34
106 Sqrt2HalfM1 = -2.928932188134524755992e-01 // Sqrt(2)/2-1 = 0xbfd2bec333018866
107 Small = 1.0 / (1 << 29) // 2**-29 = 0x3e20000000000000
108 Tiny = 1.0 / (1 << 54) // 2**-54
109 Two53 = 1 << 53 // 2**53
110 Ln2Hi = 6.93147180369123816490e-01 // 3fe62e42fee00000
111 Ln2Lo = 1.90821492927058770002e-10 // 3dea39ef35793c76
112 Lp1 = 6.666666666666735130e-01 // 3FE5555555555593
113 Lp2 = 3.999999999940941908e-01 // 3FD999999997FA04
114 Lp3 = 2.857142874366239149e-01 // 3FD2492494229359
115 Lp4 = 2.222219843214978396e-01 // 3FCC71C51D8E78AF
116 Lp5 = 1.818357216161805012e-01 // 3FC7466496CB03DE
117 Lp6 = 1.531383769920937332e-01 // 3FC39A09D078C69F
118 Lp7 = 1.479819860511658591e-01 // 3FC2F112DF3E5244
121 // special cases
122 switch {
123 case x < -1 || IsNaN(x): // includes -Inf
124 return NaN()
125 case x == -1:
126 return Inf(-1)
127 case IsInf(x, 1):
128 return Inf(1)
131 absx := x
132 if absx < 0 {
133 absx = -absx
136 var f float64
137 var iu uint64
138 k := 1
139 if absx < Sqrt2M1 { // |x| < Sqrt(2)-1
140 if absx < Small { // |x| < 2**-29
141 if absx < Tiny { // |x| < 2**-54
142 return x
144 return x - x*x*0.5
146 if x > Sqrt2HalfM1 { // Sqrt(2)/2-1 < x
147 // (Sqrt(2)/2-1) < x < (Sqrt(2)-1)
148 k = 0
149 f = x
150 iu = 1
153 var c float64
154 if k != 0 {
155 var u float64
156 if absx < Two53 { // 1<<53
157 u = 1.0 + x
158 iu = Float64bits(u)
159 k = int((iu >> 52) - 1023)
160 if k > 0 {
161 c = 1.0 - (u - x)
162 } else {
163 c = x - (u - 1.0) // correction term
164 c /= u
166 } else {
167 u = x
168 iu = Float64bits(u)
169 k = int((iu >> 52) - 1023)
170 c = 0
172 iu &= 0x000fffffffffffff
173 if iu < 0x0006a09e667f3bcd { // mantissa of Sqrt(2)
174 u = Float64frombits(iu | 0x3ff0000000000000) // normalize u
175 } else {
176 k += 1
177 u = Float64frombits(iu | 0x3fe0000000000000) // normalize u/2
178 iu = (0x0010000000000000 - iu) >> 2
180 f = u - 1.0 // Sqrt(2)/2 < u < Sqrt(2)
182 hfsq := 0.5 * f * f
183 var s, R, z float64
184 if iu == 0 { // |f| < 2**-20
185 if f == 0 {
186 if k == 0 {
187 return 0
188 } else {
189 c += float64(k) * Ln2Lo
190 return float64(k)*Ln2Hi + c
193 R = hfsq * (1.0 - 0.66666666666666666*f) // avoid division
194 if k == 0 {
195 return f - R
197 return float64(k)*Ln2Hi - ((R - (float64(k)*Ln2Lo + c)) - f)
199 s = f / (2.0 + f)
200 z = s * s
201 R = z * (Lp1 + z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))))
202 if k == 0 {
203 return f - (hfsq - s*(hfsq+R))
205 return float64(k)*Ln2Hi - ((hfsq - (s*(hfsq+R) + (float64(k)*Ln2Lo + c))) - f)