src/math/expm1l.c

   1 /* origin: OpenBSD /usr/src/lib/libm/src/ld80/e_expm1l.c */
   2 /*
   3  * Copyright (c) 2008 Stephen L. Moshier <steve@moshier.net>
   4  *
   5  * Permission to use, copy, modify, and distribute this software for any
   6  * purpose with or without fee is hereby granted, provided that the above
   7  * copyright notice and this permission notice appear in all copies.
   8  *
   9  * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
  10  * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
  11  * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
  12  * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
  13  * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
  14  * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
  15  * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
  16  */
  17 /*
  18  *      Exponential function, minus 1
  19  *      Long double precision
  20  *
  21  *
  22  * SYNOPSIS:
  23  *
  24  * long double x, y, expm1l();
  25  *
  26  * y = expm1l( x );
  27  *
  28  *
  29  * DESCRIPTION:
  30  *
  31  * Returns e (2.71828...) raised to the x power, minus 1.
  32  *
  33  * Range reduction is accomplished by separating the argument
  34  * into an integer k and fraction f such that
  35  *
  36  *     x    k  f
  37  *    e  = 2  e.
  38  *
  39  * An expansion x + .5 x^2 + x^3 R(x) approximates exp(f) - 1
  40  * in the basic range [-0.5 ln 2, 0.5 ln 2].
  41  *
  42  *
  43  * ACCURACY:
  44  *
  45  *                      Relative error:
  46  * arithmetic   domain     # trials      peak         rms
  47  *    IEEE    -45,+maxarg   200,000     1.2e-19     2.5e-20
  48  */
  49
  50 #include "libm.h"
  51
  52 #if LDBL_MANT_DIG == 53 && LDBL_MAX_EXP == 1024
  53 long double expm1l(long double x)
  54 {
  55         return expm1(x);
  56 }
  57 #elif LDBL_MANT_DIG == 64 && LDBL_MAX_EXP == 16384
  58
  59 /* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x)
  60    -.5 ln 2  <  x  <  .5 ln 2
  61    Theoretical peak relative error = 3.4e-22  */
  62 static const long double
  63 P0 = -1.586135578666346600772998894928250240826E4L,
  64 P1 =  2.642771505685952966904660652518429479531E3L,
  65 P2 = -3.423199068835684263987132888286791620673E2L,
  66 P3 =  1.800826371455042224581246202420972737840E1L,
  67 P4 = -5.238523121205561042771939008061958820811E-1L,
  68 Q0 = -9.516813471998079611319047060563358064497E4L,
  69 Q1 =  3.964866271411091674556850458227710004570E4L,
  70 Q2 = -7.207678383830091850230366618190187434796E3L,
  71 Q3 =  7.206038318724600171970199625081491823079E2L,
  72 Q4 = -4.002027679107076077238836622982900945173E1L,
  73 /* Q5 = 1.000000000000000000000000000000000000000E0 */
  74 /* C1 + C2 = ln 2 */
  75 C1 = 6.93145751953125E-1L,
  76 C2 = 1.428606820309417232121458176568075500134E-6L,
  77 /* ln 2^-65 */
  78 minarg = -4.5054566736396445112120088E1L,
  79 /* ln 2^16384 */
  80 maxarg = 1.1356523406294143949492E4L;
  81
  82 long double expm1l(long double x)
  83 {
  84         long double px, qx, xx;
  85         int k;
  86
  87         if (isnan(x))
  88                 return x;
  89         if (x > maxarg)
  90                 return x*0x1p16383L; /* overflow, unless x==inf */
  91         if (x == 0.0)
  92                 return x;
  93         if (x < minarg)
  94                 return -1.0;
  95
  96         xx = C1 + C2;
  97         /* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
  98         px = floorl(0.5 + x / xx);
  99         k = px;
 100         /* remainder times ln 2 */
 101         x -= px * C1;
 102         x -= px * C2;
 103
 104         /* Approximate exp(remainder ln 2).*/
 105         px = (((( P4 * x + P3) * x + P2) * x + P1) * x + P0) * x;
 106         qx = (((( x + Q4) * x + Q3) * x + Q2) * x + Q1) * x + Q0;
 107         xx = x * x;
 108         qx = x + (0.5 * xx + xx * px / qx);
 109
 110         /* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2).
 111          We have qx = exp(remainder ln 2) - 1, so
 112          exp(x) - 1  =  2^k (qx + 1) - 1  =  2^k qx + 2^k - 1.  */
 113         px = scalbnl(1.0, k);
 114         x = px * qx + (px - 1.0);
 115         return x;
 116 }
 117 #elif LDBL_MANT_DIG == 113 && LDBL_MAX_EXP == 16384
 118 // TODO: broken implementation to make things compile
 119 long double expm1l(long double x)
 120 {
 121         return expm1(x);
 122 }
 123 #endif