usr/src/lib/libm/common/LD/jnl.c

   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 __jnl = jnl
  31 #pragma weak __ynl = ynl
  32
  33 /*
  34  * floating point Bessel's function of the 1st and 2nd kind
  35  * of order n: jn(n,x),yn(n,x);
  36  *
  37  * Special cases:
  38  *      y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
  39  *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
  40  * Note 2. About jn(n,x), yn(n,x)
  41  *      For n=0, j0(x) is called,
  42  *      for n=1, j1(x) is called,
  43  *      for n<x, forward recursion us used starting
  44  *      from values of j0(x) and j1(x).
  45  *      for n>x, a continued fraction approximation to
  46  *      j(n,x)/j(n-1,x) is evaluated and then backward
  47  *      recursion is used starting from a supposed value
  48  *      for j(n,x). The resulting value of j(0,x) is
  49  *      compared with the actual value to correct the
  50  *      supposed value of j(n,x).
  51  *
  52  *      yn(n,x) is similar in all respects, except
  53  *      that forward recursion is used for all
  54  *      values of n>1.
  55  *
  56  */
  57
  58 #include "libm.h"
  59 #include "longdouble.h"
  60 #include <float.h>      /* LDBL_MAX */
  61
  62 #define GENERIC long double
  63
  64 static const GENERIC
  65 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
  66 two  = 2.0L,
  67 zero = 0.0L,
  68 one  = 1.0L;
  69
  70 GENERIC
  71 jnl(n, x) int n; GENERIC x; {
  72         int i, sgn;
  73         GENERIC a, b, temp = 0, z, w;
  74
  75         /*
  76          * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
  77          * Thus, J(-n,x) = J(n,-x)
  78          */
  79         if (n < 0) {
  80                 n = -n;
  81                 x = -x;
  82         }
  83         if (n == 0) return (j0l(x));
  84         if (n == 1) return (j1l(x));
  85         if (x != x) return x+x;
  86         if ((n&1) == 0)
  87                 sgn = 0;                        /* even n */
  88         else
  89                 sgn = signbitl(x);      /* old n  */
  90         x = fabsl(x);
  91         if (x == zero || !finitel(x)) b = zero;
  92         else if ((GENERIC)n <= x) {
  93                         /*
  94                          * Safe to use
  95                          * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
  96                          */
  97             if (x > 1.0e91L) {
  98                                 /*
  99                                  * x >> n**2
 100                                  *  Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 101                                  *  Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 102                                  *  Let s=sin(x), c=cos(x),
 103                                  *  xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 104                                  *
 105                                  *         n    sin(xn)*sqt2    cos(xn)*sqt2
 106                                  *      ----------------------------------
 107                                  *         0     s-c             c+s
 108                                  *         1    -s-c            -c+s
 109                                  *         2    -s+c            -c-s
 110                                  *         3     s+c             c-s
 111                                  */
 112                 switch (n&3) {
 113                     case 0: temp =  cosl(x)+sinl(x); break;
 114                     case 1: temp = -cosl(x)+sinl(x); break;
 115                     case 2: temp = -cosl(x)-sinl(x); break;
 116                     case 3: temp =  cosl(x)-sinl(x); break;
 117                 }
 118                 b = invsqrtpi*temp/sqrtl(x);
 119             } else {
 120                         a = j0l(x);
 121                         b = j1l(x);
 122                         for (i = 1; i < n; i++) {
 123                     temp = b;
 124                     b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
 125                     a = temp;
 126                         }
 127             }
 128         } else {
 129             if (x < 1e-17L) {   /* use J(n,x) = 1/n!*(x/2)^n */
 130                 b = powl(0.5L*x, (GENERIC) n);
 131                 if (b != zero) {
 132                     for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
 133                     b = b/a;
 134                 }
 135             } else {
 136                 /*
 137                  * use backward recurrence
 138                  *                      x      x^2      x^2
 139                  *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 140                  *                      2n  - 2(n+1) - 2(n+2)
 141                  *
 142                  *                      1      1        1
 143                  *  (for large x)   =  ----  ------   ------   .....
 144                  *                      2n   2(n+1)   2(n+2)
 145                  *                      -- - ------ - ------ -
 146                  *                       x     x         x
 147                  *
 148                  * Let w = 2n/x and h=2/x, then the above quotient
 149                  * is equal to the continued fraction:
 150                  *                  1
 151                  *      = -----------------------
 152                  *                     1
 153                  *         w - -----------------
 154                  *                        1
 155                  *              w+h - ---------
 156                  *                     w+2h - ...
 157                  *
 158                  * To determine how many terms needed, let
 159                  * Q(0) = w, Q(1) = w(w+h) - 1,
 160                  * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 161                  * When Q(k) > 1e4      good for single
 162                  * When Q(k) > 1e9      good for double
 163                  * When Q(k) > 1e17     good for quaduple
 164                  */
 165             /* determin k */
 166                 GENERIC t, v;
 167                 double q0, q1, h, tmp; int k, m;
 168                 w  = (n+n)/(double)x; h = 2.0/(double)x;
 169                 q0 = w;  z = w+h; q1 = w*z - 1.0; k = 1;
 170                 while (q1 < 1.0e17) {
 171                         k += 1; z += h;
 172                         tmp = z*q1 - q0;
 173                         q0 = q1;
 174                         q1 = tmp;
 175                 }
 176                 m = n+n;
 177                 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
 178                 a = t;
 179                 b = one;
 180                         /*
 181                          * Estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
 182                          * hence, if n*(log(2n/x)) > ...
 183                          * single 8.8722839355e+01
 184                          * double 7.09782712893383973096e+02
 185                          * long double 1.1356523406294143949491931077970765006170e+04
 186                          * then recurrent value may overflow and the result is
 187                          * likely underflow to zero.
 188                          */
 189                 tmp = n;
 190                 v = two/x;
 191                 tmp = tmp*logl(fabsl(v*tmp));
 192                 if (tmp < 1.1356523406294143949491931077970765e+04L) {
 193                                 for (i = n-1; i > 0; i--) {
 194                                 temp = b;
 195                                 b = ((i+i)/x)*b - a;
 196                                 a = temp;
 197                                 }
 198                 } else {
 199                                 for (i = n-1; i > 0; i--) {
 200                                 temp = b;
 201                                 b = ((i+i)/x)*b - a;
 202                                 a = temp;
 203                         if (b > 1e1000L) {
 204                                                 a /= b;
 205                                                 t /= b;
 206                                                 b  = 1.0;
 207                                         }
 208                                 }
 209                 }
 210                         b = (t*j0l(x)/b);
 211             }
 212         }
 213         if (sgn == 1)
 214                 return -b;
 215         else
 216                 return b;
 217 }
 218
 219 GENERIC
 220 ynl(n, x) int n; GENERIC x; {
 221         int i;
 222         int sign;
 223         GENERIC a, b, temp = 0;
 224
 225         if (x != x)
 226                 return x+x;
 227         if (x <= zero) {
 228                 if (x == zero)
 229                         return -one/zero;
 230                 else
 231                         return zero/zero;
 232         }
 233         sign = 1;
 234         if (n < 0) {
 235                 n = -n;
 236                 if ((n&1) == 1) sign = -1;
 237         }
 238         if (n == 0) return (y0l(x));
 239         if (n == 1) return (sign*y1l(x));
 240         if (!finitel(x)) return zero;
 241
 242         if (x > 1.0e91L) {
 243                                 /*
 244                                  * x >> n**2
 245                                  * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 246                                  *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 247                                  *   Let s=sin(x), c=cos(x),
 248                                  * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 249                                  *
 250                                  *         n    sin(xn)*sqt2    cos(xn)*sqt2
 251                                  *      ----------------------------------
 252                                  *         0     s-c             c+s
 253                                  *         1    -s-c            -c+s
 254                                  *         2    -s+c            -c-s
 255                                  *         3     s+c             c-s
 256                                  */
 257                 switch (n&3) {
 258                     case 0: temp =  sinl(x)-cosl(x); break;
 259                     case 1: temp = -sinl(x)-cosl(x); break;
 260                     case 2: temp = -sinl(x)+cosl(x); break;
 261                     case 3: temp =  sinl(x)+cosl(x); break;
 262                 }
 263                 b = invsqrtpi*temp/sqrtl(x);
 264         } else {
 265                 a = y0l(x);
 266                 b = y1l(x);
 267                 /*
 268                  * fix 1262058 and take care of non-default rounding
 269                  */
 270                 for (i = 1; i < n; i++) {
 271                         temp = b;
 272                         b *= (GENERIC) (i + i) / x;
 273                         if (b <= -LDBL_MAX)
 274                                 break;
 275                         b -= a;
 276                         a = temp;
 277                 }
 278         }
 279         if (sign > 0)
 280                 return b;
 281         else
 282                 return -b;
 283 }