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 #if defined(ELFOBJ)
  31 #pragma weak jnl = __jnl
  32 #pragma weak ynl = __ynl
  33 #endif
  34 
  35 /*
  36  * floating point Bessel's function of the 1st and 2nd kind
  37  * of order n: jn(n,x),yn(n,x);
  38  *
  39  * Special cases:
  40  *      y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
  41  *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
  42  * Note 2. About jn(n,x), yn(n,x)
  43  *      For n=0, j0(x) is called,
  44  *      for n=1, j1(x) is called,
  45  *      for n<x, forward recursion us used starting
  46  *      from values of j0(x) and j1(x).
  47  *      for n>x, a continued fraction approximation to
  48  *      j(n,x)/j(n-1,x) is evaluated and then backward
  49  *      recursion is used starting from a supposed value
  50  *      for j(n,x). The resulting value of j(0,x) is
  51  *      compared with the actual value to correct the
  52  *      supposed value of j(n,x).
  53  *
  54  *      yn(n,x) is similar in all respects, except
  55  *      that forward recursion is used for all
  56  *      values of n>1.
  57  *
  58  */
  59 
  60 #include "libm.h"
  61 #include "longdouble.h"
  62 #include <float.h>        /* LDBL_MAX */
  63 
  64 #define GENERIC long double
  65 
  66 static const GENERIC
  67 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
  68 two  = 2.0L,
  69 zero = 0.0L,
  70 one  = 1.0L;
  71 
  72 GENERIC
  73 jnl(n, x) int n; GENERIC x; {
  74         int i, sgn;
  75         GENERIC a, b, temp = 0, z, w;
  76 
  77         /*
  78          * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
  79          * Thus, J(-n,x) = J(n,-x)
  80          */
  81         if (n < 0) {
  82                 n = -n;
  83                 x = -x;
  84         }
  85         if (n == 0) return (j0l(x));
  86         if (n == 1) return (j1l(x));
  87         if (x != x) return x+x;
  88         if ((n&1) == 0)
  89                 sgn = 0;                        /* even n */
  90         else
  91                 sgn = signbitl(x);      /* old n  */
  92         x = fabsl(x);
  93         if (x == zero || !finitel(x)) b = zero;
  94         else if ((GENERIC)n <= x) {
  95                         /*
  96                          * Safe to use
  97                          * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
  98                          */
  99             if (x > 1.0e91L) {
 100                                 /*
 101                                  * x >> n**2
 102                                  *  Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 103                                  *  Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 104                                  *  Let s=sin(x), c=cos(x),
 105                                  *  xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 106                                  *
 107                                  *         n    sin(xn)*sqt2    cos(xn)*sqt2
 108                                  *      ----------------------------------
 109                                  *         0     s-c             c+s
 110                                  *         1    -s-c            -c+s
 111                                  *         2    -s+c            -c-s
 112                                  *         3     s+c             c-s
 113                                  */
 114                 switch (n&3) {
 115                     case 0: temp =  cosl(x)+sinl(x); break;
 116                     case 1: temp = -cosl(x)+sinl(x); break;
 117                     case 2: temp = -cosl(x)-sinl(x); break;
 118                     case 3: temp =  cosl(x)-sinl(x); break;
 119                 }
 120                 b = invsqrtpi*temp/sqrtl(x);
 121             } else {
 122                         a = j0l(x);
 123                         b = j1l(x);
 124                         for (i = 1; i < n; i++) {
 125                     temp = b;
 126                     b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
 127                     a = temp;
 128                         }
 129             }
 130         } else {
 131             if (x < 1e-17L) {        /* use J(n,x) = 1/n!*(x/2)^n */
 132                 b = powl(0.5L*x, (GENERIC) n);
 133                 if (b != zero) {
 134                     for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
 135                     b = b/a;
 136                 }
 137             } else {
 138                 /*
 139                  * use backward recurrence
 140                  *                      x      x^2      x^2
 141                  *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 142                  *                      2n  - 2(n+1) - 2(n+2)
 143                  *
 144                  *                      1      1        1
 145                  *  (for large x)   =  ----  ------   ------   .....
 146                  *                      2n   2(n+1)   2(n+2)
 147                  *                      -- - ------ - ------ -
 148                  *                       x     x         x
 149                  *
 150                  * Let w = 2n/x and h=2/x, then the above quotient
 151                  * is equal to the continued fraction:
 152                  *                  1
 153                  *      = -----------------------
 154                  *                     1
 155                  *         w - -----------------
 156                  *                        1
 157                  *              w+h - ---------
 158                  *                     w+2h - ...
 159                  *
 160                  * To determine how many terms needed, let
 161                  * Q(0) = w, Q(1) = w(w+h) - 1,
 162                  * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 163                  * When Q(k) > 1e4   good for single
 164                  * When Q(k) > 1e9   good for double
 165                  * When Q(k) > 1e17  good for quaduple
 166                  */
 167             /* determin k */
 168                 GENERIC t, v;
 169                 double q0, q1, h, tmp; int k, m;
 170                 w  = (n+n)/(double)x; h = 2.0/(double)x;
 171                 q0 = w;  z = w+h; q1 = w*z - 1.0; k = 1;
 172                 while (q1 < 1.0e17) {
 173                         k += 1; z += h;
 174                         tmp = z*q1 - q0;
 175                         q0 = q1;
 176                         q1 = tmp;
 177                 }
 178                 m = n+n;
 179                 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
 180                 a = t;
 181                 b = one;
 182                         /*
 183                          * Estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
 184                          * hence, if n*(log(2n/x)) > ...
 185                          * single 8.8722839355e+01
 186                          * double 7.09782712893383973096e+02
 187                          * long double 1.1356523406294143949491931077970765006170e+04
 188                          * then recurrent value may overflow and the result is
 189                          * likely underflow to zero.
 190                          */
 191                 tmp = n;
 192                 v = two/x;
 193                 tmp = tmp*logl(fabsl(v*tmp));
 194                 if (tmp < 1.1356523406294143949491931077970765e+04L) {
 195                                 for (i = n-1; i > 0; i--) {
 196                                 temp = b;
 197                                 b = ((i+i)/x)*b - a;
 198                                 a = temp;
 199                                 }
 200                 } else {
 201                                 for (i = n-1; i > 0; i--) {
 202                                 temp = b;
 203                                 b = ((i+i)/x)*b - a;
 204                                 a = temp;
 205                         if (b > 1e1000L) {
 206                                                 a /= b;
 207                                                 t /= b;
 208                                                 b  = 1.0;
 209                                         }
 210                                 }
 211                 }
 212                         b = (t*j0l(x)/b);
 213             }
 214         }
 215         if (sgn == 1)
 216                 return -b;
 217         else
 218                 return b;
 219 }
 220 
 221 GENERIC
 222 ynl(n, x) int n; GENERIC x; {
 223         int i;
 224         int sign;
 225         GENERIC a, b, temp = 0;
 226 
 227         if (x != x)
 228                 return x+x;
 229         if (x <= zero) {
 230                 if (x == zero)
 231                         return -one/zero;
 232                 else
 233                         return zero/zero;
 234         }
 235         sign = 1;
 236         if (n < 0) {
 237                 n = -n;
 238                 if ((n&1) == 1) sign = -1;
 239         }
 240         if (n == 0) return (y0l(x));
 241         if (n == 1) return (sign*y1l(x));
 242         if (!finitel(x)) return zero;
 243 
 244         if (x > 1.0e91L) {
 245                                 /*
 246                                  * x >> n**2
 247                                  * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 248                                  *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 249                                  *   Let s=sin(x), c=cos(x),
 250                                  * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 251                                  *
 252                                  *         n    sin(xn)*sqt2    cos(xn)*sqt2
 253                                  *      ----------------------------------
 254                                  *         0     s-c             c+s
 255                                  *         1    -s-c            -c+s
 256                                  *         2    -s+c            -c-s
 257                                  *         3     s+c             c-s
 258                                  */
 259                 switch (n&3) {
 260                     case 0: temp =  sinl(x)-cosl(x); break;
 261                     case 1: temp = -sinl(x)-cosl(x); break;
 262                     case 2: temp = -sinl(x)+cosl(x); break;
 263                     case 3: temp =  sinl(x)+cosl(x); break;
 264                 }
 265                 b = invsqrtpi*temp/sqrtl(x);
 266         } else {
 267                 a = y0l(x);
 268                 b = y1l(x);
 269                 /*
 270                  * fix 1262058 and take care of non-default rounding
 271                  */
 272                 for (i = 1; i < n; i++) {
 273                         temp = b;
 274                         b *= (GENERIC) (i + i) / x;
 275                         if (b <= -LDBL_MAX)
 276                                 break;
 277                         b -= a;
 278                         a = temp;
 279                 }
 280         }
 281         if (sign > 0)
 282                 return b;
 283         else
 284                 return -b;
 285 }