compiler/stdc/math/e_atanh.c

   1
   2 /* @(#)e_atanh.c 1.3 95/01/18 */
   3 /*
   4  * ====================================================
   5  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
   6  *
   7  * Developed at SunSoft, a Sun Microsystems, Inc. business.
   8  * Permission to use, copy, modify, and distribute this
   9  * software is freely granted, provided that this notice
  10  * is preserved.
  11  * ====================================================
  12  *
  13  */
  14
  15 #ifndef lint
  16 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_atanh.c,v 1.7 2005/02/04 18:26:05 das Exp $";
  17 #endif
  18
  19 /* __ieee754_atanh(x)
  20  * Method :
  21  *    1.Reduced x to positive by atanh(-x) = -atanh(x)
  22  *    2.For x>=0.5
  23  *                  1              2x                          x
  24  *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
  25  *                  2             1 - x                      1 - x
  26  *
  27  *      For x<0.5
  28  *      atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
  29  *
  30  * Special cases:
  31  *      atanh(x) is NaN if |x| > 1 with signal;
  32  *      atanh(NaN) is that NaN with no signal;
  33  *      atanh(+-1) is +-INF with signal.
  34  *
  35  */
  36
  37 #include "math.h"
  38 #include "math_private.h"
  39
  40 static const double one = 1.0, huge = 1e300;
  41 static const double zero = 0.0;
  42
  43 double
  44 __ieee754_atanh(double x)
  45 {
  46         double t;
  47         int32_t hx,ix;
  48         uint32_t lx;
  49         EXTRACT_WORDS(hx,lx,x);
  50         ix = hx&0x7fffffff;
  51         if ((ix|((lx|(-lx))>>31))>0x3ff00000) /* |x|>1 */
  52             return (x-x)/(x-x);
  53         if(ix==0x3ff00000)
  54             return x/zero;
  55         if(ix<0x3e300000&&(huge+x)>zero) return x;      /* x<2**-28 */
  56         SET_HIGH_WORD(x,ix);
  57         if(ix<0x3fe00000) {             /* x < 0.5 */
  58             t = x+x;
  59             t = 0.5*log1p(t+t*x/(one-x));
  60         } else
  61             t = 0.5*log1p((x+x)/(one-x));
  62         if(hx>=0) return t; else return -t;
  63 }