golang exp 代码
文件路径：/src/math/exp.go
/ Copyright 2009 The Go Authors. All rights reserved.
/ Use of this source code is governed by a BSD-style
/ license that can be found in the LICENSE file.
package math
/ Exp returns e**x, the base-e exponential of x.
/
/ Special cases are:
/
/ Exp(+Inf) = +Inf
/ Exp(NaN) = NaN
/
/ Very large values overflow to 0 or +Inf.
/ Very small values underflow to 1.
func Exp(x float64) float64 {
if haveArchExp {
return archExp(x)
}
return exp(x)
}
/ The original C code, the long comment, and the constants
/ below are from FreeBSD's /usr/src/lib/msun/src/e_exp.c
/ and came with this notice. The go code is a simplified
/ version of the original C.
/
/ ====================================================
/ Copyright (C) 2004 by Sun Microsystems, Inc. All rights reserved.
/
/ Permission to use, copy, modify, and distribute this
/ software is freely granted, provided that this notice
/ is preserved.
/ ====================================================
/
/
/ exp(x)
/ Returns the exponential of x.
/
/ Method
/ 1. Argument reduction:
/ Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
/ Given x, find r and integer k such that
/
/ x = k*ln2 + r, |r| <= 0.5*ln2.
/
/ Here r will be represented as r = hi-lo for better
/ accuracy.
/
/ 2. Approximation of exp(r) by a special rational function on
/ the interval [0,0.34658]:
/ Write
/ R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
/ We use a special Remez algorithm on [0,0.34658] to generate
/ a polynomial of degree 5 to approximate R. The maximum error
/ of this polynomial approximation is bounded by 2**-59. In
/ other words,
/ R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5
/ (where z=r*r, and the values of P1 to P5 are listed below)
/ and
/ | 5 | -59
/ | 2.0+P1*z+...+P5*z - R(z) | <= 2
/ | |
/ The computation of exp(r) thus becomes
/ 2*r
/ exp(r) = 1 + -------
/ R - r
/ r*R1(r)
/ = 1 + r + ----------- (for better accuracy)
/ 2 - R1(r)
/ where
/ 2 4 10
/ R1(r) = r - (P1*r + P2*r + ... + P5*r ).
/
/ 3. Scale back to obtain exp(x):
/ From step 1, we have
/ exp(x) = 2**k * exp(r)
/
/ Special cases:
/ exp(INF) is INF, exp(NaN) is NaN;
/ exp(-INF) is 0, and
/ for finite argument, only exp(0)=1 is exact.
/
/ Accuracy:
/ according to an error analysis, the error is always less than
/ 1 ulp (unit in the last place).
/
/ Misc. info.
/ For IEEE double
/ if x > 7.09782712893383973096e+02 then exp(x) overflow
/ if x < -7.45133219101941108420e+02 then exp(x) underflow
/
/ Constants:
/ The hexadecimal values are the intended ones for the following
/ constants. The decimal values may be used, provided that the
/ compiler will convert from decimal to binary accurately enough
/ to produce the hexadecimal values shown.
func exp(x float64) float64 {
const (
Ln2Hi = 6.93147180369123816490e-01
Ln2Lo = 1.90821492927058770002e-10
Log2e = 1.44269504088896338700e+00
Overflow = 7.09782712893383973096e+02
Underflow = -7.45133219101941108420e+02
NearZero = 1.0 / (1 << 28) / 2**-28
)
/ special cases
switch {
case IsNaN(x) || IsInf(x, 1):
return x
case IsInf(x, -1):
return 0
case x > Overflow:
return Inf(1)
case x < Underflow:
return 0
case -NearZero < x && x < NearZero:
return 1 + x
}
/ reduce; computed as r = hi - lo for extra precision.
var k int
switch {
case x < 0:
k = int(Log2e*x - 0.5)
case x > 0:
k = int(Log2e*x + 0.5)
}
hi := x - float64(k)*Ln2Hi
lo := float64(k) * Ln2Lo
/ compute
return expmulti(hi, lo, k)
}
/ Exp2 returns 2**x, the base-2 exponential of x.
/
/ Special cases are the same as Exp.
func Exp2(x float64) float64 {
if haveArchExp2 {
return archExp2(x)
}
return exp2(x)
}
func exp2(x float64) float64 {
const (
Ln2Hi = 6.93147180369123816490e-01
Ln2Lo = 1.90821492927058770002e-10
Overflow = 1.0239999999999999e+03
Underflow = -1.0740e+03
)
/ special cases
switch {
case IsNaN(x) || IsInf(x, 1):
return x
case IsInf(x, -1):
return 0
case x > Overflow:
return Inf(1)
case x < Underflow:
return 0
}
/ argument reduction; x = r×lg(e) + k with |r| ≤ ln(2)/2.
/ computed as r = hi - lo for extra precision.
var k int
switch {
case x > 0:
k = int(x + 0.5)
case x < 0:
k = int(x - 0.5)
}
t := x - float64(k)
hi := t * Ln2Hi
lo := -t * Ln2Lo
/ compute
return expmulti(hi, lo, k)
}
/ exp1 returns e**r × 2**k where r = hi - lo and |r| ≤ ln(2)/2.
func expmulti(hi, lo float64, k int) float64 {
const (
P1 = 1.66666666666666657415e-01 /* 0x3FC55555; 0x55555555 */
P2 = -2.77777777770155933842e-03 /* 0xBF66C16C; 0x16BEBD93 */
P3 = 6.61375632143793436117e-05 /* 0x3F11566A; 0xAF25DE2C */
P4 = -1.65339022054652515390e-06 /* 0xBEBBBD41; 0xC5D26BF1 */
P5 = 4.13813679705723846039e-08 /* 0x3E663769; 0x72BEA4D0 */
)
r := hi - lo
t := r * r
c := r - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))))
y := 1 - ((lo - (r*c)/(2-c)) - hi)
/ TODO(rsc): make sure Ldexp can handle boundary k
return Ldexp(y, k)
}