/*

 * Copyright (c) 1998, 2001, Oracle and/or its affiliates. All rights reserved.

 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.

 *

 * This code is free software; you can redistribute it and/or modify it

 * under the terms of the GNU General Public License version 2 only, as

 * published by the Free Software Foundation.  Oracle designates this

 * particular file as subject to the "Classpath" exception as provided

 * by Oracle in the LICENSE file that accompanied this code.

 *

 * This code is distributed in the hope that it will be useful, but WITHOUT

 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or

 * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License

 * version 2 for more details (a copy is included in the LICENSE file that

 * accompanied this code).

 *

 * You should have received a copy of the GNU General Public License version

 * 2 along with this work; if not, write to the Free Software Foundation,

 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.

 *

 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA

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 */

/* expm1(x)

 * Returns exp(x)-1, the exponential of x minus 1.

 *

 * Method

 *   1. Argument reduction:

 *      Given x, find r and integer k such that

 *

 *               x = k*ln2 + r,  |r| <= 0.5*ln2 ~ 0.34658

 *

 *      Here a correction term c will be computed to compensate

 *      the error in r when rounded to a floating-point number.

 *

 *   2. Approximating expm1(r) by a special rational function on

 *      the interval [0,0.34658]:

 *      Since

 *          r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 - r^4/360 + ...

 *      we define R1(r*r) by

 *          r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 * R1(r*r)

 *      That is,

 *          R1(r**2) = 6/r *((exp(r)+1)/(exp(r)-1) - 2/r)

 *                   = 6/r * ( 1 + 2.0*(1/(exp(r)-1) - 1/r))

 *                   = 1 - r^2/60 + r^4/2520 - r^6/100800 + ...

 *      We use a special Reme algorithm on [0,0.347] to generate

 *      a polynomial of degree 5 in r*r to approximate R1. The

 *      maximum error of this polynomial approximation is bounded

 *      by 2**-61. In other words,

 *          R1(z) ~ 1.0 + Q1*z + Q2*z**2 + Q3*z**3 + Q4*z**4 + Q5*z**5

 *      where   Q1  =  -1.6666666666666567384E-2,

 *              Q2  =   3.9682539681370365873E-4,

 *              Q3  =  -9.9206344733435987357E-6,

 *              Q4  =   2.5051361420808517002E-7,

 *              Q5  =  -6.2843505682382617102E-9;

 *      (where z=r*r, and the values of Q1 to Q5 are listed below)

 *      with error bounded by

 *          |                  5           |     -61

 *          | 1.0+Q1*z+...+Q5*z   -  R1(z) | <= 2

 *          |                              |

 *

 *      expm1(r) = exp(r)-1 is then computed by the following

 *      specific way which minimize the accumulation rounding error:

 *                             2     3

 *                            r     r    [ 3 - (R1 + R1*r/2)  ]

 *            expm1(r) = r + --- + --- * [--------------------]

 *                            2     2    [ 6 - r*(3 - R1*r/2) ]

 *

 *      To compensate the error in the argument reduction, we use

 *              expm1(r+c) = expm1(r) + c + expm1(r)*c

 *                         ~ expm1(r) + c + r*c

 *      Thus c+r*c will be added in as the correction terms for

 *      expm1(r+c). Now rearrange the term to avoid optimization

 *      screw up:

 *                      (      2                                    2 )

 *                      ({  ( r    [ R1 -  (3 - R1*r/2) ]  )  }    r  )

 *       expm1(r+c)~r - ({r*(--- * [--------------------]-c)-c} - --- )

 *                      ({  ( 2    [ 6 - r*(3 - R1*r/2) ]  )  }    2  )

 *                      (                                             )

 *

 *                 = r - E

 *   3. Scale back to obtain expm1(x):

 *      From step 1, we have

 *         expm1(x) = either 2^k*[expm1(r)+1] - 1

 *                  = or     2^k*[expm1(r) + (1-2^-k)]

 *   4. Implementation notes:

 *      (A). To save one multiplication, we scale the coefficient Qi

 *           to Qi*2^i, and replace z by (x^2)/2.

 *      (B). To achieve maximum accuracy, we compute expm1(x) by

 *        (i)   if x < -56*ln2, return -1.0, (raise inexact if x!=inf)

 *        (ii)  if k=0, return r-E

 *        (iii) if k=-1, return 0.5*(r-E)-0.5

 *        (iv)  if k=1 if r < -0.25, return 2*((r+0.5)- E)

 *                     else          return  1.0+2.0*(r-E);

 *        (v)   if (k<-2||k>56) return 2^k(1-(E-r)) - 1 (or exp(x)-1)

 *        (vi)  if k <= 20, return 2^k((1-2^-k)-(E-r)), else

 *        (vii) return 2^k(1-((E+2^-k)-r))

 *

 * Special cases:

 *      expm1(INF) is INF, expm1(NaN) is NaN;

 *      expm1(-INF) is -1, and

 *      for finite argument, only expm1(0)=0 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 expm1(x) overflow

 *

 * 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.

 */

#include "fdlibm.h"

#ifdef __STDC__

static const double

#else

static double

#endif

one             = 1.0,

huge            = 1.0e+300,

tiny            = 1.0e-300,

o_threshold     = 7.09782712893383973096e+02,/* 0x40862E42, 0xFEFA39EF */

ln2_hi          = 6.93147180369123816490e-01,/* 0x3fe62e42, 0xfee00000 */

ln2_lo          = 1.90821492927058770002e-10,/* 0x3dea39ef, 0x35793c76 */

invln2          = 1.44269504088896338700e+00,/* 0x3ff71547, 0x652b82fe */

        /* scaled coefficients related to expm1 */

Q1  =  -3.33333333333331316428e-02, /* BFA11111 111110F4 */

Q2  =   1.58730158725481460165e-03, /* 3F5A01A0 19FE5585 */

Q3  =  -7.93650757867487942473e-05, /* BF14CE19 9EAADBB7 */

Q4  =   4.00821782732936239552e-06, /* 3ED0CFCA 86E65239 */

Q5  =  -2.01099218183624371326e-07; /* BE8AFDB7 6E09C32D */

#ifdef __STDC__

        double expm1(double x)

#else

        double expm1(x)

        double x;

#endif

{

        double y,hi,lo,c=0,t,e,hxs,hfx,r1;

        int k,xsb;

        unsigned hx;

        hx  = __HI(x);  /* high word of x */

        xsb = hx&0x80000000;            /* sign bit of x */

        if(xsb==0) y=x; else y= -x;     /* y = |x| */

        hx &= 0x7fffffff;               /* high word of |x| */

    /* filter out huge and non-finite argument */

        if(hx >= 0x4043687A) {                  /* if |x|>=56*ln2 */

            if(hx >= 0x40862E42) {              /* if |x|>=709.78... */

                if(hx>=0x7ff00000) {

                    if(((hx&0xfffff)|__LO(x))!=0)

                         return x+x;     /* NaN */

                    else return (xsb==0)? x:-1.0;/* exp(+-inf)={inf,-1} */

                }

                if(x > o_threshold) return huge*huge; /* overflow */

            }

            if(xsb!=0) { /* x < -56*ln2, return -1.0 with inexact */

                if(x+tiny<0.0)          /* raise inexact */

                return tiny-one;        /* return -1 */

            }

        }

    /* argument reduction */

        if(hx > 0x3fd62e42) {           /* if  |x| > 0.5 ln2 */

            if(hx < 0x3FF0A2B2) {       /* and |x| < 1.5 ln2 */

                if(xsb==0)

                    {hi = x - ln2_hi; lo =  ln2_lo;  k =  1;}

                else

                    {hi = x + ln2_hi; lo = -ln2_lo;  k = -1;}

            } else {

                k  = invln2*x+((xsb==0)?0.5:-0.5);

                t  = k;

                hi = x - t*ln2_hi;      /* t*ln2_hi is exact here */

                lo = t*ln2_lo;

            }

            x  = hi - lo;

            c  = (hi-x)-lo;

        }

        else if(hx < 0x3c900000) {      /* when |x|<2**-54, return x */

            t = huge+x; /* return x with inexact flags when x!=0 */

            return x - (t-(huge+x));

        }

        else k = 0;

    /* x is now in primary range */

        hfx = 0.5*x;

        hxs = x*hfx;

        r1 = one+hxs*(Q1+hxs*(Q2+hxs*(Q3+hxs*(Q4+hxs*Q5))));

        t  = 3.0-r1*hfx;

        e  = hxs*((r1-t)/(6.0 - x*t));

        if(k==0) return x - (x*e-hxs);          /* c is 0 */

        else {

            e  = (x*(e-c)-c);

            e -= hxs;

            if(k== -1) return 0.5*(x-e)-0.5;

            if(k==1) {

                if(x < -0.25) return -2.0*(e-(x+0.5));

                else          return  one+2.0*(x-e);

            }

            if (k <= -2 || k>56) {   /* suffice to return exp(x)-1 */

                y = one-(e-x);

                __HI(y) += (k<<20);     /* add k to y's exponent */

                return y-one;

            }

            t = one;

            if(k<20) {

                __HI(t) = 0x3ff00000 - (0x200000>>k);  /* t=1-2^-k */

                y = t-(e-x);

                __HI(y) += (k<<20);     /* add k to y's exponent */

           } else {

                __HI(t)  = ((0x3ff-k)<<20);     /* 2^-k */

                y = x-(e+t);

                y += one;

                __HI(y) += (k<<20);     /* add k to y's exponent */

            }

        }

        return y;

}