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, 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)
  84                 return (j0l(x));
  85         if (n == 1)
  86                 return (j1l(x));
  87         if (x != x)
  88                 return (x+x);
  89         if ((n&1) == 0)
  90                 sgn = 0;                        /* even n */
  91         else
  92                 sgn = signbitl(x);      /* old n  */
  93         x = fabsl(x);
  94         if (x == zero || !finitel(x)) b = zero;
  95         else if ((GENERIC)n <= x) {
  96                                         /*
  97                                          * Safe to use
  98                                          * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
  99                                          */
 100             if (x > 1.0e91L) {
 101                                 /*
 102                                  * x >> n**2
 103                                  *  Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 104                                  *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 105                                  *   Let s=sin(x), c=cos(x),
 106                                  *      xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 107                                  *
 108                                  *         n    sin(xn)*sqt2    cos(xn)*sqt2
 109                                  *      ----------------------------------
 110                                  *         0     s-c             c+s
 111                                  *         1    -s-c            -c+s
 112                                  *         2    -s+c            -c-s
 113                                  *         3     s+c             c-s
 114                                  */
 115                 switch (n&3) {
 116                     case 0: temp =  cosl(x)+sinl(x); break;
 117                     case 1: temp = -cosl(x)+sinl(x); break;
 118                     case 2: temp = -cosl(x)-sinl(x); break;
 119                     case 3: temp =  cosl(x)-sinl(x); break;
 120                 }
 121                 b = invsqrtpi*temp/sqrtl(x);
 122             } else {
 123                         a = j0l(x);
 124                         b = j1l(x);
 125                         for (i = 1; i < n; i++) {
 126                     temp = b;
 127                     b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
 128                     a = temp;
 129                         }
 130             }
 131         } else {
 132             if (x < 1e-17L) {        /* use J(n,x) = 1/n!*(x/2)^n */
 133                 b = powl(0.5L*x, (GENERIC)n);
 134                 if (b != zero) {
 135                     for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
 136                     b = b/a;
 137                 }
 138             } else {
 139                 /* use backward recurrence */
 140                 /*
 141                  *                      x      x^2      x^2
 142                  *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 143                  *                      2n  - 2(n+1) - 2(n+2)
 144                  *
 145                  *                      1      1        1
 146                  *  (for large x)   =  ----  ------   ------   .....
 147                  *                      2n   2(n+1)   2(n+2)
 148                  *                      -- - ------ - ------ -
 149                  *                       x     x         x
 150                  *
 151                  * Let w = 2n/x and h=2/x, then the above quotient
 152                  * is equal to the continued fraction:
 153                  *                  1
 154                  *      = -----------------------
 155                  *                     1
 156                  *         w - -----------------
 157                  *                        1
 158                  *              w+h - ---------
 159                  *                     w+2h - ...
 160                  *
 161                  * To determine how many terms needed, let
 162                  * Q(0) = w, Q(1) = w(w+h) - 1,
 163                  * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 164                  * When Q(k) > 1e4   good for single
 165                  * When Q(k) > 1e9   good for double
 166                  * When Q(k) > 1e17  good for quaduple
 167                  */
 168             /* determin k */
 169                 GENERIC t, v;
 170                 double q0, q1, h, tmp; int k, m;
 171                 w  = (n+n)/(double)x; h = 2.0/(double)x;
 172                 q0 = w;  z = w+h; q1 = w*z - 1.0; k = 1;
 173                 while (q1 < 1.0e17) {
 174                         k += 1; z += h;
 175                         tmp = z*q1 - q0;
 176                         q0 = q1;
 177                         q1 = tmp;
 178                 }
 179                 m = n+n;
 180                 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
 181                 a = t;
 182                 b = one;
 183                 /*
 184                  * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
 185                  * hence, if n*(log(2n/x)) > ...
 186                  *  single 8.8722839355e+01
 187                  *  double 7.09782712893383973096e+02
 188                  *  long double 1.1356523406294143949491931077970765006170e+04
 189                  *  then recurrent value may overflow and the result is
 190                  *  likely underflow to zero
 191                  */
 192                 tmp = n;
 193                 v = two/x;
 194                 tmp = tmp*logl(fabsl(v*tmp));
 195                 if (tmp < 1.1356523406294143949491931077970765e+04L) {
 196                                 for (i = n-1; i > 0; i--) {
 197                         temp = b;
 198                         b = ((i+i)/x)*b - a;
 199                         a = temp;
 200                     }
 201                 } else {
 202                                 for (i = n-1; i > 0; i--) {
 203                         temp = b;
 204                         b = ((i+i)/x)*b - a;
 205                         a = temp;
 206                         if (b > 1e1000L) {
 207                             a /= b;
 208                             t /= b;
 209                             b  = 1.0;
 210                         }
 211                     }
 212                 }
 213                 b = (t*j0l(x)/b);
 214             }
 215         }
 216         if (sgn == 1)
 217                 return (-b);
 218         else
 219                 return (b);
 220 }
 221 
 222 GENERIC ynl(n, x)
 223 int n; GENERIC x; {
 224         int i;
 225         int sign;
 226         GENERIC a, b, temp;
 227 
 228         if (x != x)
 229                 return (x+x);
 230         if (x <= zero) {
 231                 if (x == zero)
 232                         return (-one/zero);
 233                 else
 234                         return (zero/zero);
 235         }
 236         sign = 1;
 237         if (n < 0) {
 238                 n = -n;
 239                 if ((n&1) == 1) sign = -1;
 240         }
 241         if (n == 0)
 242                 return (y0l(x));
 243         if (n == 1)
 244                 return (sign*y1l(x));
 245         if (!finitel(x))
 246                 return (zero);
 247 
 248         if (x > 1.0e91L) {   /* x >> n**2
 249                                     Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 250                                     Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 251                                     Let s = sin(x), c = cos(x),
 252                                         xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
 253 
 254                                            n    sin(xn)*sqt2    cos(xn)*sqt2
 255                                         ----------------------------------
 256                                            0     s-c             c+s
 257                                            1    -s-c            -c+s
 258                                            2    -s+c            -c-s
 259                                            3     s+c             c-s
 260                                  */
 261                 switch (n&3) {
 262                     case 0: temp =  sinl(x)-cosl(x); break;
 263                     case 1: temp = -sinl(x)-cosl(x); break;
 264                     case 2: temp = -sinl(x)+cosl(x); break;
 265                     case 3: temp =  sinl(x)+cosl(x); break;
 266                 }
 267                 b = invsqrtpi*temp/sqrtl(x);
 268         } else {
 269                 a = y0l(x);
 270                 b = y1l(x);
 271                 /*
 272                  * fix 1262058 and take care of non-default rounding
 273                  */
 274                 for (i = 1; i < n; i++) {
 275                         temp = b;
 276                         b *= (GENERIC) (i + i) / x;
 277                         if (b <= -LDBL_MAX)
 278                                 break;
 279                         b -= a;
 280                         a = temp;
 281                 }
 282         }
 283         if (sign > 0)
 284                 return (b);
 285         else
 286                 return (-b);
 287 }