src/math/jn.c

   1 /* origin: FreeBSD /usr/src/lib/msun/src/e_jn.c */
   2 /*
   3  * ====================================================
   4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
   5  *
   6  * Developed at SunSoft, a Sun Microsystems, Inc. business.
   7  * Permission to use, copy, modify, and distribute this
   8  * software is freely granted, provided that this notice
   9  * is preserved.
  10  * ====================================================
  11  */
  12 /*
  13  * jn(n, x), yn(n, x)
  14  * floating point Bessel's function of the 1st and 2nd kind
  15  * of order n
  16  *
  17  * Special cases:
  18  *      y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
  19  *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
  20  * Note 2. About jn(n,x), yn(n,x)
  21  *      For n=0, j0(x) is called,
  22  *      for n=1, j1(x) is called,
  23  *      for n<=x, forward recursion is used starting
  24  *      from values of j0(x) and j1(x).
  25  *      for n>x, a continued fraction approximation to
  26  *      j(n,x)/j(n-1,x) is evaluated and then backward
  27  *      recursion is used starting from a supposed value
  28  *      for j(n,x). The resulting value of j(0,x) is
  29  *      compared with the actual value to correct the
  30  *      supposed value of j(n,x).
  31  *
  32  *      yn(n,x) is similar in all respects, except
  33  *      that forward recursion is used for all
  34  *      values of n>1.
  35  */
  36
  37 #include "libm.h"
  38
  39 static const double invsqrtpi = 5.64189583547756279280e-01; /* 0x3FE20DD7, 0x50429B6D */
  40
  41 double jn(int n, double x)
  42 {
  43         uint32_t ix, lx;
  44         int nm1, i, sign;
  45         double a, b, temp;
  46
  47         EXTRACT_WORDS(ix, lx, x);
  48         sign = ix>>31;
  49         ix &= 0x7fffffff;
  50
  51         if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
  52                 return x;
  53
  54         /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
  55          * Thus, J(-n,x) = J(n,-x)
  56          */
  57         /* nm1 = |n|-1 is used instead of |n| to handle n==INT_MIN */
  58         if (n == 0)
  59                 return j0(x);
  60         if (n < 0) {
  61                 nm1 = -(n+1);
  62                 x = -x;
  63                 sign ^= 1;
  64         } else
  65                 nm1 = n-1;
  66         if (nm1 == 0)
  67                 return j1(x);
  68
  69         sign &= n;  /* even n: 0, odd n: signbit(x) */
  70         x = fabs(x);
  71         if ((ix|lx) == 0 || ix == 0x7ff00000)  /* if x is 0 or inf */
  72                 b = 0.0;
  73         else if (nm1 < x) {
  74                 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
  75                 if (ix >= 0x52d00000) { /* x > 2**302 */
  76                         /* (x >> n**2)
  77                          *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
  78                          *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
  79                          *      Let s=sin(x), c=cos(x),
  80                          *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
  81                          *
  82                          *             n    sin(xn)*sqt2    cos(xn)*sqt2
  83                          *          ----------------------------------
  84                          *             0     s-c             c+s
  85                          *             1    -s-c            -c+s
  86                          *             2    -s+c            -c-s
  87                          *             3     s+c             c-s
  88                          */
  89                         switch(nm1&3) {
  90                         case 0: temp = -cos(x)+sin(x); break;
  91                         case 1: temp = -cos(x)-sin(x); break;
  92                         case 2: temp =  cos(x)-sin(x); break;
  93                         default:
  94                         case 3: temp =  cos(x)+sin(x); break;
  95                         }
  96                         b = invsqrtpi*temp/sqrt(x);
  97                 } else {
  98                         a = j0(x);
  99                         b = j1(x);
 100                         for (i=0; i<nm1; ) {
 101                                 i++;
 102                                 temp = b;
 103                                 b = b*(2.0*i/x) - a; /* avoid underflow */
 104                                 a = temp;
 105                         }
 106                 }
 107         } else {
 108                 if (ix < 0x3e100000) { /* x < 2**-29 */
 109                         /* x is tiny, return the first Taylor expansion of J(n,x)
 110                          * J(n,x) = 1/n!*(x/2)^n  - ...
 111                          */
 112                         if (nm1 > 32)  /* underflow */
 113                                 b = 0.0;
 114                         else {
 115                                 temp = x*0.5;
 116                                 b = temp;
 117                                 a = 1.0;
 118                                 for (i=2; i<=nm1+1; i++) {
 119                                         a *= (double)i; /* a = n! */
 120                                         b *= temp;      /* b = (x/2)^n */
 121                                 }
 122                                 b = b/a;
 123                         }
 124                 } else {
 125                         /* use backward recurrence */
 126                         /*                      x      x^2      x^2
 127                          *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 128                          *                      2n  - 2(n+1) - 2(n+2)
 129                          *
 130                          *                      1      1        1
 131                          *  (for large x)   =  ----  ------   ------   .....
 132                          *                      2n   2(n+1)   2(n+2)
 133                          *                      -- - ------ - ------ -
 134                          *                       x     x         x
 135                          *
 136                          * Let w = 2n/x and h=2/x, then the above quotient
 137                          * is equal to the continued fraction:
 138                          *                  1
 139                          *      = -----------------------
 140                          *                     1
 141                          *         w - -----------------
 142                          *                        1
 143                          *              w+h - ---------
 144                          *                     w+2h - ...
 145                          *
 146                          * To determine how many terms needed, let
 147                          * Q(0) = w, Q(1) = w(w+h) - 1,
 148                          * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 149                          * When Q(k) > 1e4      good for single
 150                          * When Q(k) > 1e9      good for double
 151                          * When Q(k) > 1e17     good for quadruple
 152                          */
 153                         /* determine k */
 154                         double t,q0,q1,w,h,z,tmp,nf;
 155                         int k;
 156
 157                         nf = nm1 + 1.0;
 158                         w = 2*nf/x;
 159                         h = 2/x;
 160                         z = w+h;
 161                         q0 = w;
 162                         q1 = w*z - 1.0;
 163                         k = 1;
 164                         while (q1 < 1.0e9) {
 165                                 k += 1;
 166                                 z += h;
 167                                 tmp = z*q1 - q0;
 168                                 q0 = q1;
 169                                 q1 = tmp;
 170                         }
 171                         for (t=0.0, i=k; i>=0; i--)
 172                                 t = 1/(2*(i+nf)/x - t);
 173                         a = t;
 174                         b = 1.0;
 175                         /*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
 176                          *  Hence, if n*(log(2n/x)) > ...
 177                          *  single 8.8722839355e+01
 178                          *  double 7.09782712893383973096e+02
 179                          *  long double 1.1356523406294143949491931077970765006170e+04
 180                          *  then recurrent value may overflow and the result is
 181                          *  likely underflow to zero
 182                          */
 183                         tmp = nf*log(fabs(w));
 184                         if (tmp < 7.09782712893383973096e+02) {
 185                                 for (i=nm1; i>0; i--) {
 186                                         temp = b;
 187                                         b = b*(2.0*i)/x - a;
 188                                         a = temp;
 189                                 }
 190                         } else {
 191                                 for (i=nm1; i>0; i--) {
 192                                         temp = b;
 193                                         b = b*(2.0*i)/x - a;
 194                                         a = temp;
 195                                         /* scale b to avoid spurious overflow */
 196                                         if (b > 0x1p500) {
 197                                                 a /= b;
 198                                                 t /= b;
 199                                                 b  = 1.0;
 200                                         }
 201                                 }
 202                         }
 203                         z = j0(x);
 204                         w = j1(x);
 205                         if (fabs(z) >= fabs(w))
 206                                 b = t*z/b;
 207                         else
 208                                 b = t*w/a;
 209                 }
 210         }
 211         return sign ? -b : b;
 212 }
 213
 214
 215 double yn(int n, double x)
 216 {
 217         uint32_t ix, lx, ib;
 218         int nm1, sign, i;
 219         double a, b, temp;
 220
 221         EXTRACT_WORDS(ix, lx, x);
 222         sign = ix>>31;
 223         ix &= 0x7fffffff;
 224
 225         if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
 226                 return x;
 227         if (sign && (ix|lx)!=0) /* x < 0 */
 228                 return 0/0.0;
 229         if (ix == 0x7ff00000)
 230                 return 0.0;
 231
 232         if (n == 0)
 233                 return y0(x);
 234         if (n < 0) {
 235                 nm1 = -(n+1);
 236                 sign = n&1;
 237         } else {
 238                 nm1 = n-1;
 239                 sign = 0;
 240         }
 241         if (nm1 == 0)
 242                 return sign ? -y1(x) : y1(x);
 243
 244         if (ix >= 0x52d00000) { /* x > 2**302 */
 245                 /* (x >> n**2)
 246                  *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 247                  *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 248                  *      Let s=sin(x), c=cos(x),
 249                  *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 250                  *
 251                  *             n    sin(xn)*sqt2    cos(xn)*sqt2
 252                  *          ----------------------------------
 253                  *             0     s-c             c+s
 254                  *             1    -s-c            -c+s
 255                  *             2    -s+c            -c-s
 256                  *             3     s+c             c-s
 257                  */
 258                 switch(nm1&3) {
 259                 case 0: temp = -sin(x)-cos(x); break;
 260                 case 1: temp = -sin(x)+cos(x); break;
 261                 case 2: temp =  sin(x)+cos(x); break;
 262                 default:
 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                 /* quit if b is -inf */
 270                 GET_HIGH_WORD(ib, b);
 271                 for (i=0; i<nm1 && ib!=0xfff00000; ){
 272                         i++;
 273                         temp = b;
 274                         b = (2.0*i/x)*b - a;
 275                         GET_HIGH_WORD(ib, b);
 276                         a = temp;
 277                 }
 278         }
 279         return sign ? -b : b;
 280 }