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5fea2ccc77
The tree's pretty inconsistent about single space vs double space after a period in documentation. Make it consistently a single space, per earlier decisions. This means contributors won't be confused by misleading precedence. This CL doesn't use go/doc to parse. It only addresses // comments. It was generated with: $ perl -i -npe 's,^(\s*// .+[a-z]\.) +([A-Z]),$1 $2,' $(git grep -l -E '^\s*//(.+\.) +([A-Z])') $ go test go/doc -update Change-Id: Iccdb99c37c797ef1f804a94b22ba5ee4b500c4f7 Reviewed-on: https://go-review.googlesource.com/20022 Reviewed-by: Rob Pike <r@golang.org> Reviewed-by: Dave Day <djd@golang.org> Run-TryBot: Brad Fitzpatrick <bradfitz@golang.org> TryBot-Result: Gobot Gobot <gobot@golang.org>
201 lines
6.3 KiB
Go
201 lines
6.3 KiB
Go
// Copyright 2010 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package math
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// The original C code, the long comment, and the constants
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// below are from FreeBSD's /usr/src/lib/msun/src/s_log1p.c
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// and came with this notice. The go code is a simplified
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// version of the original C.
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//
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// ====================================================
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// Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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//
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// Developed at SunPro, a Sun Microsystems, Inc. business.
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// Permission to use, copy, modify, and distribute this
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// software is freely granted, provided that this notice
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// is preserved.
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// ====================================================
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//
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//
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// double log1p(double x)
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//
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// Method :
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// 1. Argument Reduction: find k and f such that
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// 1+x = 2**k * (1+f),
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// where sqrt(2)/2 < 1+f < sqrt(2) .
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//
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// Note. If k=0, then f=x is exact. However, if k!=0, then f
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// may not be representable exactly. In that case, a correction
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// term is need. Let u=1+x rounded. Let c = (1+x)-u, then
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// log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),
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// and add back the correction term c/u.
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// (Note: when x > 2**53, one can simply return log(x))
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//
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// 2. Approximation of log1p(f).
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// Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
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// = 2s + 2/3 s**3 + 2/5 s**5 + .....,
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// = 2s + s*R
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// We use a special Reme algorithm on [0,0.1716] to generate
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// a polynomial of degree 14 to approximate R The maximum error
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// of this polynomial approximation is bounded by 2**-58.45. In
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// other words,
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// 2 4 6 8 10 12 14
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// R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s
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// (the values of Lp1 to Lp7 are listed in the program)
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// and
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// | 2 14 | -58.45
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// | Lp1*s +...+Lp7*s - R(z) | <= 2
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// | |
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// Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
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// In order to guarantee error in log below 1ulp, we compute log
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// by
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// log1p(f) = f - (hfsq - s*(hfsq+R)).
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//
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// 3. Finally, log1p(x) = k*ln2 + log1p(f).
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// = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
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// Here ln2 is split into two floating point number:
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// ln2_hi + ln2_lo,
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// where n*ln2_hi is always exact for |n| < 2000.
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//
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// Special cases:
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// log1p(x) is NaN with signal if x < -1 (including -INF) ;
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// log1p(+INF) is +INF; log1p(-1) is -INF with signal;
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// log1p(NaN) is that NaN with no signal.
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//
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// Accuracy:
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// according to an error analysis, the error is always less than
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// 1 ulp (unit in the last place).
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//
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// Constants:
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// The hexadecimal values are the intended ones for the following
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// constants. The decimal values may be used, provided that the
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// compiler will convert from decimal to binary accurately enough
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// to produce the hexadecimal values shown.
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//
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// Note: Assuming log() return accurate answer, the following
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// algorithm can be used to compute log1p(x) to within a few ULP:
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//
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// u = 1+x;
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// if(u==1.0) return x ; else
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// return log(u)*(x/(u-1.0));
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//
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// See HP-15C Advanced Functions Handbook, p.193.
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// Log1p returns the natural logarithm of 1 plus its argument x.
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// It is more accurate than Log(1 + x) when x is near zero.
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//
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// Special cases are:
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// Log1p(+Inf) = +Inf
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// Log1p(±0) = ±0
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// Log1p(-1) = -Inf
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// Log1p(x < -1) = NaN
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// Log1p(NaN) = NaN
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func Log1p(x float64) float64
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func log1p(x float64) float64 {
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const (
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Sqrt2M1 = 4.142135623730950488017e-01 // Sqrt(2)-1 = 0x3fda827999fcef34
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Sqrt2HalfM1 = -2.928932188134524755992e-01 // Sqrt(2)/2-1 = 0xbfd2bec333018866
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Small = 1.0 / (1 << 29) // 2**-29 = 0x3e20000000000000
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Tiny = 1.0 / (1 << 54) // 2**-54
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Two53 = 1 << 53 // 2**53
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Ln2Hi = 6.93147180369123816490e-01 // 3fe62e42fee00000
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Ln2Lo = 1.90821492927058770002e-10 // 3dea39ef35793c76
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Lp1 = 6.666666666666735130e-01 // 3FE5555555555593
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Lp2 = 3.999999999940941908e-01 // 3FD999999997FA04
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Lp3 = 2.857142874366239149e-01 // 3FD2492494229359
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Lp4 = 2.222219843214978396e-01 // 3FCC71C51D8E78AF
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Lp5 = 1.818357216161805012e-01 // 3FC7466496CB03DE
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Lp6 = 1.531383769920937332e-01 // 3FC39A09D078C69F
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Lp7 = 1.479819860511658591e-01 // 3FC2F112DF3E5244
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)
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// special cases
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switch {
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case x < -1 || IsNaN(x): // includes -Inf
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return NaN()
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case x == -1:
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return Inf(-1)
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case IsInf(x, 1):
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return Inf(1)
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}
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absx := x
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if absx < 0 {
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absx = -absx
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}
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var f float64
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var iu uint64
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k := 1
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if absx < Sqrt2M1 { // |x| < Sqrt(2)-1
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if absx < Small { // |x| < 2**-29
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if absx < Tiny { // |x| < 2**-54
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return x
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}
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return x - x*x*0.5
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}
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if x > Sqrt2HalfM1 { // Sqrt(2)/2-1 < x
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// (Sqrt(2)/2-1) < x < (Sqrt(2)-1)
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k = 0
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f = x
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iu = 1
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}
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}
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var c float64
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if k != 0 {
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var u float64
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if absx < Two53 { // 1<<53
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u = 1.0 + x
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iu = Float64bits(u)
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k = int((iu >> 52) - 1023)
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if k > 0 {
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c = 1.0 - (u - x)
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} else {
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c = x - (u - 1.0) // correction term
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c /= u
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}
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} else {
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u = x
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iu = Float64bits(u)
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k = int((iu >> 52) - 1023)
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c = 0
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}
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iu &= 0x000fffffffffffff
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if iu < 0x0006a09e667f3bcd { // mantissa of Sqrt(2)
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u = Float64frombits(iu | 0x3ff0000000000000) // normalize u
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} else {
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k += 1
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u = Float64frombits(iu | 0x3fe0000000000000) // normalize u/2
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iu = (0x0010000000000000 - iu) >> 2
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}
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f = u - 1.0 // Sqrt(2)/2 < u < Sqrt(2)
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}
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hfsq := 0.5 * f * f
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var s, R, z float64
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if iu == 0 { // |f| < 2**-20
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if f == 0 {
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if k == 0 {
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return 0
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} else {
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c += float64(k) * Ln2Lo
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return float64(k)*Ln2Hi + c
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}
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}
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R = hfsq * (1.0 - 0.66666666666666666*f) // avoid division
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if k == 0 {
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return f - R
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}
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return float64(k)*Ln2Hi - ((R - (float64(k)*Ln2Lo + c)) - f)
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}
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s = f / (2.0 + f)
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z = s * s
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R = z * (Lp1 + z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))))
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if k == 0 {
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return f - (hfsq - s*(hfsq+R))
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}
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return float64(k)*Ln2Hi - ((hfsq - (s*(hfsq+R) + (float64(k)*Ln2Lo + c))) - f)
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}
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