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 jn = __jn
  31 #pragma weak yn = __yn
  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 <float.h>        /* DBL_MIN */
  60 #include <values.h>       /* X_TLOSS */
  61 #include "xpg6.h"       /* __xpg6 */
  62 
  63 #define GENERIC double
  64 
  65 static const GENERIC
  66         invsqrtpi = 5.641895835477562869480794515607725858441e-0001,
  67         two     = 2.0,
  68         zero    = 0.0,
  69         one     = 1.0;
  70 
  71 GENERIC
  72 jn(int n, GENERIC x) {
  73         int i, sgn;
  74         GENERIC a, b, temp = 0;
  75         GENERIC z, w, ox, on;
  76 
  77     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
  78      * Thus, J(-n,x) = J(n,-x)
  79      */
  80         ox = x; on = (GENERIC)n;
  81         if(n<0){
  82                 n = -n;
  83                 x = -x;
  84         }
  85         if(isnan(x)) return x*x;        /* + -> * for Cheetah */
  86         if (!((int) _lib_version == libm_ieee ||
  87                 (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
  88             if(fabs(x) > X_TLOSS) return _SVID_libm_err(on,ox,38);
  89         }
  90         if(n==0) return(j0(x));
  91         if(n==1) return(j1(x));
  92         if((n&1)==0)
  93                 sgn=0;                  /* even n */
  94         else
  95                 sgn = signbit(x);       /* old n  */
  96         x = fabs(x);
  97         if(x == zero||!finite(x)) b = zero;
  98         else if((GENERIC)n<=x) {     /* Safe to use
  99                                            J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
 100                                          */
 101             if(x>1.0e91) {   /* 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 =  cos(x)+sin(x); break;
 116                     case 1: temp = -cos(x)+sin(x); break;
 117                     case 2: temp = -cos(x)-sin(x); break;
 118                     case 3: temp =  cos(x)-sin(x); break;
 119                 }
 120                 b = invsqrtpi*temp/sqrt(x);
 121             } else {
 122                 a = j0(x);
 123                 b = j1(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-9) {     /* use J(n,x) = 1/n!*(x/2)^n */
 132                 b = pow(0.5*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                 /* use backward recurrence */
 139                 /*                      x      x^2      x^2
 140                  *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 141                  *                      2n  - 2(n+1) - 2(n+2)
 142                  *
 143                  *                      1      1        1
 144                  *  (for large x)   =  ----  ------   ------   .....
 145                  *                      2n   2(n+1)   2(n+2)
 146                  *                      -- - ------ - ------ -
 147                  *                       x     x         x
 148                  *
 149                  * Let w = 2n/x and h=2/x, then the above quotient
 150                  * is equal to the continued fraction:
 151                  *                  1
 152                  *      = -----------------------
 153                  *                     1
 154                  *         w - -----------------
 155                  *                        1
 156                  *              w+h - ---------
 157                  *                     w+2h - ...
 158                  *
 159                  * To determine how many terms needed, let
 160                  * Q(0) = w, Q(1) = w(w+h) - 1,
 161                  * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 162                  * When Q(k) > 1e4   good for single
 163                  * When Q(k) > 1e9   good for double
 164                  * When Q(k) > 1e17  good for quaduple
 165                  */
 166             /* determin k */
 167                 GENERIC t,v;
 168                 double q0,q1,h,tmp; int k,m;
 169                 w  = (n+n)/(double)x; h = 2.0/(double)x;
 170                 q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
 171                 while(q1<1.0e9) {
 172                         k += 1; z += h;
 173                         tmp = z*q1 - q0;
 174                         q0 = q1;
 175                         q1 = tmp;
 176                 }
 177                 m = n+n;
 178                 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
 179                 a = t;
 180                 b = one;
 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*log(fabs(v*tmp));
 192                 if(tmp<7.09782712893383973096e+02) {
 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>1e100) {
 204                             a /= b;
 205                             t /= b;
 206                             b  = 1.0;
 207                         }
 208                     }
 209                 }
 210                 b = (t*j0(x)/b);
 211             }
 212         }
 213         if(sgn==1) return -b; else return b;
 214 }
 215 
 216 GENERIC
 217 yn(int n, GENERIC x) {
 218         int i;
 219         int sign;
 220         GENERIC a, b, temp = 0, ox, on;
 221 
 222         ox = x; on = (GENERIC)n;
 223         if(isnan(x)) return x*x;        /* + -> * for Cheetah */
 224         if (x <= zero) {
 225                 if(x==zero) {
 226                         /* return -one/zero; */
 227                         return _SVID_libm_err((GENERIC)n,x,12);
 228                 } else {
 229                         /* return zero/zero; */
 230                         return _SVID_libm_err((GENERIC)n,x,13);
 231                 }
 232         }
 233         if (!((int) _lib_version == libm_ieee ||
 234                 (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
 235             if(x > X_TLOSS) return _SVID_libm_err(on,ox,39);
 236         }
 237         sign = 1;
 238         if(n<0){
 239                 n = -n;
 240                 if((n&1) == 1) sign = -1;
 241         }
 242         if(n==0) return(y0(x));
 243         if(n==1) return(sign*y1(x));
 244         if(!finite(x)) return zero;
 245 
 246         if(x>1.0e91) {       /* 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 =  sin(x)-cos(x); break;
 261                     case 1: temp = -sin(x)-cos(x); break;
 262                     case 2: temp = -sin(x)+cos(x); break;
 263                     case 3: temp =  sin(x)+cos(x); break;
 264                 }
 265                 b = invsqrtpi*temp/sqrt(x);
 266         } else {
 267                 a = y0(x);
 268                 b = y1(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 <= -DBL_MAX)
 276                                 break;
 277                         b -= a;
 278                         a = temp;
 279                 }
 280         }
 281         if(sign>0) return b; else return -b;
 282 }