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/* @(#)s_log1p.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.

 * ====================================================

 */

/*

FUNCTION

<<log1p>>, <<log1pf>>---log of <<1 + <[x]>>>

INDEX

        log1p

INDEX

        log1pf

ANSI_SYNOPSIS

        #include <math.h>

        double log1p(double <[x]>);

        float log1pf(float <[x]>);

TRAD_SYNOPSIS

        #include <math.h>

        double log1p(<[x]>)

        double <[x]>;

        float log1pf(<[x]>)

        float <[x]>;

DESCRIPTION

<<log1p>> calculates

@tex

$ln(1+x)$,

@end tex

the natural logarithm of <<1+<[x]>>>.  You can use <<log1p>> rather

than `<<log(1+<[x]>)>>' for greater precision when <[x]> is very

small.

<<log1pf>> calculates the same thing, but accepts and returns

<<float>> values rather than <<double>>.

RETURNS

<<log1p>> returns a <<double>>, the natural log of <<1+<[x]>>>.

<<log1pf>> returns a <<float>>, the natural log of <<1+<[x]>>>.

PORTABILITY

Neither <<log1p>> nor <<log1pf>> is required by ANSI C or by the System V

Interface Definition (Issue 2).

*/

/* double log1p(double x)

 *

 * Method :

 *   1. Argument Reduction: find k and f such that

 *                      1+x = 2^k * (1+f),

 *         where  sqrt(2)/2 < 1+f < sqrt(2) .

 *

 *      Note. If k=0, then f=x is exact. However, if k!=0, then f

 *      may not be representable exactly. In that case, a correction

 *      term is need. Let u=1+x rounded. Let c = (1+x)-u, then

 *      log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u),

 *      and add back the correction term c/u.

 *      (Note: when x > 2**53, one can simply return log(x))

 *

 *   2. Approximation of log1p(f).

 *      Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)

 *               = 2s + 2/3 s**3 + 2/5 s**5 + .....,

 *               = 2s + s*R

 *      We use a special Reme algorithm on [0,0.1716] to generate

 *      a polynomial of degree 14 to approximate R The maximum error

 *      of this polynomial approximation is bounded by 2**-58.45. In

 *      other words,

 *                      2      4      6      8      10      12      14

 *          R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s  +Lp6*s  +Lp7*s

 *      (the values of Lp1 to Lp7 are listed in the program)

 *      and

 *          |      2          14          |     -58.45

 *          | Lp1*s +...+Lp7*s    -  R(z) | <= 2

 *          |                             |

 *      Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.

 *      In order to guarantee error in log below 1ulp, we compute log

 *      by

 *              log1p(f) = f - (hfsq - s*(hfsq+R)).

 *

 *      3. Finally, log1p(x) = k*ln2 + log1p(f).

 *                           = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))

 *         Here ln2 is split into two floating point number:

 *                      ln2_hi + ln2_lo,

 *         where n*ln2_hi is always exact for |n| < 2000.

 *

 * Special cases:

 *      log1p(x) is NaN with signal if x < -1 (including -INF) ;

 *      log1p(+INF) is +INF; log1p(-1) is -INF with signal;

 *      log1p(NaN) is that NaN with no signal.

 *

 * Accuracy:

 *      according to an error analysis, the error is always less than

 *      1 ulp (unit in the last place).

 *

 * 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.

 *

 * Note: Assuming log() return accurate answer, the following

 *       algorithm can be used to compute log1p(x) to within a few ULP:

 *

 *              u = 1+x;

 *              if(u==1.0) return x ; else

 *                         return log(u)*(x/(u-1.0));

 *

 *       See HP-15C Advanced Functions Handbook, p.193.

 */

#include "fdlibm.h"

#ifndef _DOUBLE_IS_32BITS

#ifdef __STDC__

static const double

#else

static double

#endif

ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */

ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */

two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */

Lp1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */

Lp2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */

Lp3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */

Lp4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */

Lp5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */

Lp6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */

Lp7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */

#ifdef __STDC__

static const double zero = 0.0;

#else

static double zero = 0.0;

#endif

#ifdef __STDC__

        double log1p(double x)

#else

        double log1p(x)

        double x;

#endif

{

        double hfsq,f,c,s,z,R,u;

        __int32_t k,hx,hu,ax;

        GET_HIGH_WORD(hx,x);

        ax = hx&0x7fffffff;

        k = 1;

        if (hx < 0x3FDA827A) {                  /* x < 0.41422  */

            if(ax>=0x3ff00000) {                /* x <= -1.0 */

                if(x==-1.0) return -two54/zero; /* log1p(-1)=+inf */

                else return (x-x)/(x-x);        /* log1p(x<-1)=NaN */

            }

            if(ax<0x3e200000) {                 /* |x| < 2**-29 */

                if(two54+x>zero                 /* raise inexact */

                    &&ax<0x3c900000)            /* |x| < 2**-54 */

                    return x;

                else

                    return x - x*x*0.5;

            }

            if(hx>0||hx<=((__int32_t)0xbfd2bec3)) {

                k=0;f=x;hu=1;}   /* -0.2929<x<0.41422 */

        }

        if (hx >= 0x7ff00000) return x+x;

        if(k!=0) {

            if(hx<0x43400000) {

                u  = 1.0+x;

                GET_HIGH_WORD(hu,u);

                k  = (hu>>20)-1023;

                c  = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */

                c /= u;

            } else {

                u  = x;

                GET_HIGH_WORD(hu,u);

                k  = (hu>>20)-1023;

                c  = 0;

            }

            hu &= 0x000fffff;

            if(hu<0x6a09e) {

                SET_HIGH_WORD(u,hu|0x3ff00000); /* normalize u */

            } else {

                k += 1;

                SET_HIGH_WORD(u,hu|0x3fe00000); /* normalize u/2 */

                hu = (0x00100000-hu)>>2;

            }

            f = u-1.0;

        }

        hfsq=0.5*f*f;

        if(hu==0) {      /* |f| < 2**-20 */

          if(f==zero) { if(k==0) return zero;

                      else {c += k*ln2_lo; return k*ln2_hi+c;}}

            R = hfsq*(1.0-0.66666666666666666*f);

            if(k==0) return f-R; else

                     return k*ln2_hi-((R-(k*ln2_lo+c))-f);

        }

        s = f/(2.0+f);

        z = s*s;

        R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7))))));

        if(k==0) return f-(hfsq-s*(hfsq+R)); else

                 return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f);

}

#endif /* _DOUBLE_IS_32BITS */