1 /*
   2  * CDDL HEADER START
   3  *
   4  * The contents of this file are subject to the terms of the
   5  * Common Development and Distribution License (the "License").
   6  * You may not use this file except in compliance with the License.
   7  *
   8  * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
   9  * or http://www.opensolaris.org/os/licensing.
  10  * See the License for the specific language governing permissions
  11  * and limitations under the License.
  12  *
  13  * When distributing Covered Code, include this CDDL HEADER in each
  14  * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
  15  * If applicable, add the following below this CDDL HEADER, with the
  16  * fields enclosed by brackets "[]" replaced with your own identifying
  17  * information: Portions Copyright [yyyy] [name of copyright owner]
  18  *
  19  * CDDL HEADER END
  20  */
  21 
  22 /*
  23  * Copyright 2011 Nexenta Systems, Inc.  All rights reserved.
  24  */
  25 /*
  26  * Copyright 2006 Sun Microsystems, Inc.  All rights reserved.
  27  * Use is subject to license terms.
  28  */
  29 
  30 #pragma weak expm1 = __expm1
  31 
  32 /* INDENT OFF */
  33 /*
  34  * expm1(x)
  35  * Returns exp(x)-1, the exponential of x minus 1.
  36  *
  37  * Method
  38  *   1. Arugment reduction:
  39  *      Given x, find r and integer k such that
  40  *
  41  *               x = k*ln2 + r,  |r| <= 0.5*ln2 ~ 0.34658
  42  *
  43  *      Here a correction term c will be computed to compensate
  44  *      the error in r when rounded to a floating-point number.
  45  *
  46  *   2. Approximating expm1(r) by a special rational function on
  47  *      the interval [0,0.34658]:
  48  *      Since
  49  *          r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 - r^4/360 + ...
  50  *      we define R1(r*r) by
  51  *          r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 * R1(r*r)
  52  *      That is,
  53  *          R1(r**2) = 6/r *((exp(r)+1)/(exp(r)-1) - 2/r)
  54  *                   = 6/r * ( 1 + 2.0*(1/(exp(r)-1) - 1/r))
  55  *                   = 1 - r^2/60 + r^4/2520 - r^6/100800 + ...
  56  *      We use a special Reme algorithm on [0,0.347] to generate
  57  *      a polynomial of degree 5 in r*r to approximate R1. The
  58  *      maximum error of this polynomial approximation is bounded
  59  *      by 2**-61. In other words,
  60  *          R1(z) ~ 1.0 + Q1*z + Q2*z**2 + Q3*z**3 + Q4*z**4 + Q5*z**5
  61  *      where   Q1  =  -1.6666666666666567384E-2,
  62  *              Q2  =   3.9682539681370365873E-4,
  63  *              Q3  =  -9.9206344733435987357E-6,
  64  *              Q4  =   2.5051361420808517002E-7,
  65  *              Q5  =  -6.2843505682382617102E-9;
  66  *      (where z=r*r, and the values of Q1 to Q5 are listed below)
  67  *      with error bounded by
  68  *          |                  5           |     -61
  69  *          | 1.0+Q1*z+...+Q5*z   -  R1(z) | <= 2
  70  *          |                              |
  71  *
  72  *      expm1(r) = exp(r)-1 is then computed by the following
  73  *      specific way which minimize the accumulation rounding error:
  74  *                             2     3
  75  *                            r     r    [ 3 - (R1 + R1*r/2)  ]
  76  *            expm1(r) = r + --- + --- * [--------------------]
  77  *                            2     2    [ 6 - r*(3 - R1*r/2) ]
  78  *
  79  *      To compensate the error in the argument reduction, we use
  80  *              expm1(r+c) = expm1(r) + c + expm1(r)*c
  81  *                         ~ expm1(r) + c + r*c
  82  *      Thus c+r*c will be added in as the correction terms for
  83  *      expm1(r+c). Now rearrange the term to avoid optimization
  84  *      screw up:
  85  *                      (      2                                    2 )
  86  *                      ({  ( r    [ R1 -  (3 - R1*r/2) ]  )  }    r  )
  87  *       expm1(r+c)~r - ({r*(--- * [--------------------]-c)-c} - --- )
  88  *                      ({  ( 2    [ 6 - r*(3 - R1*r/2) ]  )  }    2  )
  89  *                      (                                             )
  90  *
  91  *                 = r - E
  92  *   3. Scale back to obtain expm1(x):
  93  *      From step 1, we have
  94  *         expm1(x) = either 2^k*[expm1(r)+1] - 1
  95  *                  = or     2^k*[expm1(r) + (1-2^-k)]
  96  *   4. Implementation notes:
  97  *      (A). To save one multiplication, we scale the coefficient Qi
  98  *           to Qi*2^i, and replace z by (x^2)/2.
  99  *      (B). To achieve maximum accuracy, we compute expm1(x) by
 100  *        (i)   if x < -56*ln2, return -1.0, (raise inexact if x != inf)
 101  *        (ii)  if k=0, return r-E
 102  *        (iii) if k=-1, return 0.5*(r-E)-0.5
 103  *        (iv)  if k=1 if r < -0.25, return 2*((r+0.5)- E)
 104  *                                      else         return  1.0+2.0*(r-E);
 105  *        (v)   if (k<-2||k>56) return 2^k(1-(E-r)) - 1 (or exp(x)-1)
 106  *        (vi)  if k <= 20, return 2^k((1-2^-k)-(E-r)), else
 107  *        (vii) return 2^k(1-((E+2^-k)-r))
 108  *
 109  * Special cases:
 110  *      expm1(INF) is INF, expm1(NaN) is NaN;
 111  *      expm1(-INF) is -1, and
 112  *      for finite argument, only expm1(0)=0 is exact.
 113  *
 114  * Accuracy:
 115  *      according to an error analysis, the error is always less than
 116  *      1 ulp (unit in the last place).
 117  *
 118  * Misc. info.
 119  *      For IEEE double
 120  *          if x >  7.09782712893383973096e+02 then expm1(x) overflow
 121  *
 122  * Constants:
 123  * The hexadecimal values are the intended ones for the following
 124  * constants. The decimal values may be used, provided that the
 125  * compiler will convert from decimal to binary accurately enough
 126  * to produce the hexadecimal values shown.
 127  */
 128 /* INDENT ON */
 129 
 130 #include "libm_synonyms.h"      /* __expm1 */
 131 #include "libm_macros.h"
 132 #include <math.h>
 133 
 134 static const double xxx[] = {
 135 /* one */                1.0,
 136 /* huge */               1.0e+300,
 137 /* tiny */               1.0e-300,
 138 /* o_threshold */        7.09782712893383973096e+02,    /* 40862E42 FEFA39EF */
 139 /* ln2_hi */             6.93147180369123816490e-01,    /* 3FE62E42 FEE00000 */
 140 /* ln2_lo */             1.90821492927058770002e-10,    /* 3DEA39EF 35793C76 */
 141 /* invln2 */             1.44269504088896338700e+00,    /* 3FF71547 652B82FE */
 142 /* scaled coefficients related to expm1 */
 143 /* Q1 */                -3.33333333333331316428e-02,    /* BFA11111 111110F4 */
 144 /* Q2 */                 1.58730158725481460165e-03,    /* 3F5A01A0 19FE5585 */
 145 /* Q3 */                -7.93650757867487942473e-05,    /* BF14CE19 9EAADBB7 */
 146 /* Q4 */                 4.00821782732936239552e-06,    /* 3ED0CFCA 86E65239 */
 147 /* Q5 */                -2.01099218183624371326e-07     /* BE8AFDB7 6E09C32D */
 148 };
 149 #define one             xxx[0]
 150 #define huge            xxx[1]
 151 #define tiny            xxx[2]
 152 #define o_threshold     xxx[3]
 153 #define ln2_hi          xxx[4]
 154 #define ln2_lo          xxx[5]
 155 #define invln2          xxx[6]
 156 #define Q1              xxx[7]
 157 #define Q2              xxx[8]
 158 #define Q3              xxx[9]
 159 #define Q4              xxx[10]
 160 #define Q5              xxx[11]
 161 
 162 double
 163 expm1(double x) {
 164         double y, hi, lo, c = 0.0L, t, e, hxs, hfx, r1;
 165         int k, xsb;
 166         unsigned hx;
 167 
 168         hx = ((unsigned *) &x)[HIWORD];             /* high word of x */
 169         xsb = hx & 0x80000000;                      /* sign bit of x */
 170         if (xsb == 0)
 171                 y = x;
 172         else
 173                 y = -x;                         /* y = |x| */
 174         hx &= 0x7fffffff;                   /* high word of |x| */
 175 
 176         /* filter out huge and non-finite argument */
 177         /* for example exp(38)-1 is approximately 3.1855932e+16 */
 178         if (hx >= 0x4043687A) {
 179                 /* if |x|>=56*ln2 (~38.8162...) */
 180                 if (hx >= 0x40862E42) {              /* if |x|>=709.78... -> inf */
 181                         if (hx >= 0x7ff00000) {
 182                                 if (((hx & 0xfffff) | ((int *) &x)[LOWORD])
 183                                         != 0)
 184                                         return (x * x); /* + -> * for Cheetah */
 185                                 else
 186                                         /* exp(+-inf)={inf,-1} */
 187                                         return (xsb == 0 ? x : -1.0);
 188                         }
 189                         if (x > o_threshold)
 190                                 return (huge * huge);   /* overflow */
 191                 }
 192                 if (xsb != 0) {         /* x < -56*ln2, return -1.0 w/inexact */
 193                         if (x + tiny < 0.0)          /* raise inexact */
 194                                 return (tiny - one);    /* return -1 */
 195                 }
 196         }
 197 
 198         /* argument reduction */
 199         if (hx > 0x3fd62e42) {                       /* if  |x| > 0.5 ln2 */
 200                 if (hx < 0x3FF0A2B2) {               /* and |x| < 1.5 ln2 */
 201                         if (xsb == 0) {         /* positive number */
 202                                 hi = x - ln2_hi;
 203                                 lo = ln2_lo;
 204                                 k = 1;
 205                         } else {
 206                                 /* negative number */
 207                                 hi = x + ln2_hi;
 208                                 lo = -ln2_lo;
 209                                 k = -1;
 210                         }
 211                 } else {
 212                         /* |x| > 1.5 ln2 */
 213                         k = (int) (invln2 * x + (xsb == 0 ? 0.5 : -0.5));
 214                         t = k;
 215                         hi = x - t * ln2_hi;    /* t*ln2_hi is exact here */
 216                         lo = t * ln2_lo;
 217                 }
 218                 x = hi - lo;
 219                 c = (hi - x) - lo; /* still at |x| > 0.5 ln2 */
 220         } else if (hx < 0x3c900000) {
 221                 /* when |x|<2**-54, return x */
 222                 t = huge + x;           /* return x w/inexact when x != 0 */
 223                 return (x - (t - (huge + x)));
 224         } else
 225                 /* |x| <= 0.5 ln2 */
 226                 k = 0;
 227 
 228         /* x is now in primary range */
 229         hfx = 0.5 * x;
 230         hxs = x * hfx;
 231         r1 = one + hxs * (Q1 + hxs * (Q2 + hxs * (Q3 + hxs * (Q4 + hxs * Q5))));
 232         t = 3.0 - r1 * hfx;
 233         e = hxs * ((r1 - t) / (6.0 - x * t));
 234         if (k == 0) /* |x| <= 0.5 ln2 */
 235                 return (x - (x * e - hxs));
 236         else {          /* |x| > 0.5 ln2 */
 237                 e = (x * (e - c) - c);
 238                 e -= hxs;
 239                 if (k == -1)
 240                         return (0.5 * (x - e) - 0.5);
 241                 if (k == 1) {
 242                         if (x < -0.25)
 243                                 return (-2.0 * (e - (x + 0.5)));
 244                         else
 245                                 return (one + 2.0 * (x - e));
 246                 }
 247                 if (k <= -2 || k > 56) {  /* suffice to return exp(x)-1 */
 248                         y = one - (e - x);
 249                         ((int *) &y)[HIWORD] += k << 20;
 250                         return (y - one);
 251                 }
 252                 t = one;
 253                 if (k < 20) {
 254                         ((int *) &t)[HIWORD] = 0x3ff00000 - (0x200000 >> k);
 255                                                         /* t = 1 - 2^-k */
 256                         y = t - (e - x);
 257                         ((int *) &y)[HIWORD] += k << 20;
 258                 } else {
 259                         ((int *) &t)[HIWORD] = (0x3ff - k) << 20; /* 2^-k */
 260                         y = x - (e + t);
 261                         y += one;
 262                         ((int *) &y)[HIWORD] += k << 20;
 263                 }
 264         }
 265         return (y);
 266 }