go acosh 源码

2022-07-15 浏览 (1128)

golang acosh 代码

文件路径:/src/math/acosh.go

/ Copyright 2010 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

/ The original C code, the long comment, and the constants
/ below are from FreeBSD's /usr/src/lib/msun/src/e_acosh.c
/ and came with this notice. The go code is a simplified
/ version of the original C.
/
/ ====================================================
/ Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
/
/ Developed at SunPro, a Sun Microsystems, Inc. business.
/ Permission to use, copy, modify, and distribute this
/ software is freely granted, provided that this notice
/ is preserved.
/ ====================================================
/
/
/ __ieee754_acosh(x)
/ Method :
/	Based on
/	        acosh(x) = log [ x + sqrt(x*x-1) ]
/	we have
/	        acosh(x) := log(x)+ln2,	if x is large; else
/	        acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
/	        acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
/
/ Special cases:
/	acosh(x) is NaN with signal if x<1.
/	acosh(NaN) is NaN without signal.
/

/ Acosh returns the inverse hyperbolic cosine of x.
/
/ Special cases are:
/
/	Acosh(+Inf) = +Inf
/	Acosh(x) = NaN if x < 1
/	Acosh(NaN) = NaN
func Acosh(x float64) float64 {
	if haveArchAcosh {
		return archAcosh(x)
	}
	return acosh(x)
}

func acosh(x float64) float64 {
	const Large = 1 << 28 / 2**28
	/ first case is special case
	switch {
	case x < 1 || IsNaN(x):
		return NaN()
	case x == 1:
		return 0
	case x >= Large:
		return Log(x) + Ln2 / x > 2**28
	case x > 2:
		return Log(2*x - 1/(x+Sqrt(x*x-1))) / 2**28 > x > 2
	}
	t := x - 1
	return Log1p(t + Sqrt(2*t+t*t)) / 2 >= x > 1
}

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