/home/jaouen/pok_official/pok/trunk/libpok/libm/e_acosh.c

/*
 *                               POK header
 * 
 * The following file is a part of the POK project. Any modification should
 * made according to the POK licence. You CANNOT use this file or a part of
 * this file is this part of a file for your own project
 *
 * For more information on the POK licence, please see our LICENCE FILE
 *
 * Please follow the coding guidelines described in doc/CODING_GUIDELINES
 *
 *                                      Copyright (c) 2007-2009 POK team 
 *
 * Created by julien on Fri Jan 30 14:41:34 2009 
 */

/* @(#)e_acosh.c 5.1 93/09/24 */
/*
 * ====================================================
 * 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.
 */

#ifdef POK_NEEDS_LIBMATH

#include <libm.h>
#include "math_private.h"

static const double
one     = 1.0,
ln2     = 6.93147180559945286227e-01;  /* 0x3FE62E42, 0xFEFA39EF */

double
__ieee754_acosh(double x)
{
        double t;
        int32_t hx;
        uint32_t lx;
        EXTRACT_WORDS(hx,lx,x);
        if(hx<0x3ff00000) {             /* x < 1 */
            return (x-x)/(x-x);
        } else if(hx >=0x41b00000) {    /* x > 2**28 */
            if(hx >=0x7ff00000) {       /* x is inf of NaN */
                return x+x;
            } else
                return __ieee754_log(x)+ln2;    /* acosh(huge)=log(2x) */
        } else if(((hx-0x3ff00000)|lx)==0) {
            return 0.0;                 /* acosh(1) = 0 */
        } else if (hx > 0x40000000) {   /* 2**28 > x > 2 */
            t=x*x;
            return __ieee754_log(2.0*x-one/(x+__ieee754_sqrt(t-one)));
        } else {                        /* 1<x<2 */
            t = x-one;
            return log1p(t+sqrt(2.0*t+t*t));
        }
}

#endif