log1p.c

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/*-
 * Copyright (c) 2014-2018 Carsten Sonne Larsen <cs@innolan.net>
 * All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 *
 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR
 * IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
 * OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
 * IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT,
 * INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
 * DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
 * THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
 * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF
 * THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
 *
 * Project homepage:
 * https://amath.innolan.net
 *
 * The original source code can be obtained from:
 * http://www.netlib.org/fdlibm/s_log1p.c
 * 
 * =================================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * Developed at SunSoft, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * is preserved.
 * =================================================================
 */

#include "prim.h"

/* 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 Remes 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.
 */

static const double
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 */

static double zero = 0.0;

double log1p(double x)
{
    double hfsq,f,c,s,z,R,u;
    int32_t k,hx,hu,ax;

    f = 0.0;
    c = 0.0;
    hu = 0;

    GET_HIGH_WORD(hx,x); /* high word of 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<=((int)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); /* high word of 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); /* high word of 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);
}