lib/sdk/crt/math/ieee754/jn_yn.c

   1 /* @(#)e_jn.c 5.1 93/09/24 */
   2 /*
   3  * ====================================================
   4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
   5  *
   6  * Developed at SunPro, 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 #if defined(LIBM_SCCS) && !defined(lint)
  14 static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
  15 #endif
  16
  17 /*
  18  * __ieee754_jn(n, x), __ieee754_yn(n, x)
  19  * floating point Bessel's function of the 1st and 2nd kind
  20  * of order n
  21  *
  22  * Special cases:
  23  *      y0(0)=y1(0)=yn(n,0) = -inf with overflow signal;
  24  *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
  25  * Note 2. About jn(n,x), yn(n,x)
  26  *      For n=0, j0(x) is called,
  27  *      for n=1, j1(x) is called,
  28  *      for n<x, forward recursion us used starting
  29  *      from values of j0(x) and j1(x).
  30  *      for n>x, a continued fraction approximation to
  31  *      j(n,x)/j(n-1,x) is evaluated and then backward
  32  *      recursion is used starting from a supposed value
  33  *      for j(n,x). The resulting value of j(0,x) is
  34  *      compared with the actual value to correct the
  35  *      supposed value of j(n,x).
  36  *
  37  *      yn(n,x) is similar in all respects, except
  38  *      that forward recursion is used for all
  39  *      values of n>1.
  40  *
  41  */
  42
  43 #include "math.h"
  44 #include "ieee754.h"
  45
  46 #ifdef __STDC__
  47 static const double
  48 #else
  49 static double
  50 #endif
  51 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
  52 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
  53 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
  54
  55 #ifdef __STDC__
  56 static const double zero  =  0.00000000000000000000e+00;
  57 #else
  58 static double zero  =  0.00000000000000000000e+00;
  59 #endif
  60
  61 #ifdef __STDC__
  62         double __ieee754_jn(int n, double x)
  63 #else
  64         double __ieee754_jn(n,x)
  65         int n; double x;
  66 #endif
  67 {
  68         int32_t i,hx,ix,lx, sgn;
  69         double a, b, temp, di;
  70         double z, w;
  71
  72     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
  73      * Thus, J(-n,x) = J(n,-x)
  74      */
  75         EXTRACT_WORDS(hx,lx,x);
  76         ix = 0x7fffffff&hx;
  77     /* if J(n,NaN) is NaN */
  78         if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
  79         if(n<0){
  80                 n = -n;
  81                 x = -x;
  82                 hx ^= 0x80000000;
  83         }
  84         if(n==0) return(__ieee754_j0(x));
  85         if(n==1) return(__ieee754_j1(x));
  86         sgn = (n&1)&(hx>>31);   /* even n -- 0, odd n -- sign(x) */
  87         x = fabs(x);
  88         if((ix|lx)==0||ix>=0x7ff00000)  /* if x is 0 or inf */
  89             b = zero;
  90         else if((double)n<=x) {
  91                 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
  92             if(ix>=0x52D00000) { /* x > 2**302 */
  93     /* (x >> n**2)
  94      *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
  95      *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
  96      *      Let s=sin(x), c=cos(x),
  97      *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
  98      *
  99      *             n    sin(xn)*sqt2    cos(xn)*sqt2
 100      *          ----------------------------------
 101      *             0     s-c             c+s
 102      *             1    -s-c            -c+s
 103      *             2    -s+c            -c-s
 104      *             3     s+c             c-s
 105      */
 106                 double s;
 107                 double c;
 108                 __sincos (x, &s, &c);
 109                 switch(n&3) {
 110                     case 0: temp =  c + s; break;
 111                     case 1: temp = -c + s; break;
 112                     case 2: temp = -c - s; break;
 113                     case 3: temp =  c - s; break;
 114                 }
 115                 b = invsqrtpi*temp/__ieee754_sqrt(x);
 116             } else {
 117                 a = __ieee754_j0(x);
 118                 b = __ieee754_j1(x);
 119                 for(i=1;i<n;i++){
 120                     temp = b;
 121                     b = b*((double)(i+i)/x) - a; /* avoid underflow */
 122                     a = temp;
 123                 }
 124             }
 125         } else {
 126             if(ix<0x3e100000) { /* x < 2**-29 */
 127     /* x is tiny, return the first Taylor expansion of J(n,x)
 128      * J(n,x) = 1/n!*(x/2)^n  - ...
 129      */
 130                 if(n>33)        /* underflow */
 131                     b = zero;
 132                 else {
 133                     temp = x*0.5; b = temp;
 134                     for (a=one,i=2;i<=n;i++) {
 135                         a *= (double)i;         /* a = n! */
 136                         b *= temp;              /* b = (x/2)^n */
 137                     }
 138                     b = b/a;
 139                 }
 140             } else {
 141                 /* use backward recurrence */
 142                 /*                      x      x^2      x^2
 143                  *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
 144                  *                      2n  - 2(n+1) - 2(n+2)
 145                  *
 146                  *                      1      1        1
 147                  *  (for large x)   =  ----  ------   ------   .....
 148                  *                      2n   2(n+1)   2(n+2)
 149                  *                      -- - ------ - ------ -
 150                  *                       x     x         x
 151                  *
 152                  * Let w = 2n/x and h=2/x, then the above quotient
 153                  * is equal to the continued fraction:
 154                  *                  1
 155                  *      = -----------------------
 156                  *                     1
 157                  *         w - -----------------
 158                  *                        1
 159                  *              w+h - ---------
 160                  *                     w+2h - ...
 161                  *
 162                  * To determine how many terms needed, let
 163                  * Q(0) = w, Q(1) = w(w+h) - 1,
 164                  * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
 165                  * When Q(k) > 1e4      good for single
 166                  * When Q(k) > 1e9      good for double
 167                  * When Q(k) > 1e17     good for quadruple
 168                  */
 169             /* determine k */
 170                 double t,v;
 171                 double q0,q1,h,tmp; int32_t k,m;
 172                 w  = (n+n)/(double)x; h = 2.0/(double)x;
 173                 q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
 174                 while(q1<1.0e9) {
 175                         k += 1; z += h;
 176                         tmp = z*q1 - q0;
 177                         q0 = q1;
 178                         q1 = tmp;
 179                 }
 180                 m = n+n;
 181                 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
 182                 a = t;
 183                 b = one;
 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*__ieee754_log(fabs(v*tmp));
 195                 if(tmp<7.09782712893383973096e+02) {
 196                     for(i=n-1,di=(double)(i+i);i>0;i--){
 197                         temp = b;
 198                         b *= di;
 199                         b  = b/x - a;
 200                         a = temp;
 201                         di -= two;
 202                     }
 203                 } else {
 204                     for(i=n-1,di=(double)(i+i);i>0;i--){
 205                         temp = b;
 206                         b *= di;
 207                         b  = b/x - a;
 208                         a = temp;
 209                         di -= two;
 210                     /* scale b to avoid spurious overflow */
 211                         if(b>1e100) {
 212                             a /= b;
 213                             t /= b;
 214                             b  = one;
 215                         }
 216                     }
 217                 }
 218                 b = (t*__ieee754_j0(x)/b);
 219             }
 220         }
 221         if(sgn==1) return -b; else return b;
 222 }
 223
 224 #ifdef __STDC__
 225         double __ieee754_yn(int n, double x)
 226 #else
 227         double __ieee754_yn(n,x)
 228         int n; double x;
 229 #endif
 230 {
 231         int32_t i,hx,ix,lx;
 232         int32_t sign;
 233         double a, b, temp;
 234
 235         EXTRACT_WORDS(hx,lx,x);
 236         ix = 0x7fffffff&hx;
 237     /* if Y(n,NaN) is NaN */
 238         if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
 239         if((ix|lx)==0) return -HUGE_VAL+x; /* -inf and overflow exception.  */;
 240         if(hx<0) return zero/(zero*x);
 241         sign = 1;
 242         if(n<0){
 243                 n = -n;
 244                 sign = 1 - ((n&1)<<1);
 245         }
 246         if(n==0) return(__ieee754_y0(x));
 247         if(n==1) return(sign*__ieee754_y1(x));
 248         if(ix==0x7ff00000) return zero;
 249         if(ix>=0x52D00000) { /* x > 2**302 */
 250     /* (x >> n**2)
 251      *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 252      *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
 253      *      Let s=sin(x), c=cos(x),
 254      *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
 255      *
 256      *             n    sin(xn)*sqt2    cos(xn)*sqt2
 257      *          ----------------------------------
 258      *             0     s-c             c+s
 259      *             1    -s-c            -c+s
 260      *             2    -s+c            -c-s
 261      *             3     s+c             c-s
 262      */
 263                 double c;
 264                 double s;
 265                 __sincos (x, &s, &c);
 266                 switch(n&3) {
 267                     case 0: temp =  s - c; break;
 268                     case 1: temp = -s - c; break;
 269                     case 2: temp = -s + c; break;
 270                     case 3: temp =  s + c; break;
 271                 }
 272                 b = invsqrtpi*temp/__ieee754_sqrt(x);
 273         } else {
 274             u_int32_t high;
 275             a = __ieee754_y0(x);
 276             b = __ieee754_y1(x);
 277         /* quit if b is -inf */
 278             GET_HIGH_WORD(high,b);
 279             for(i=1;i<n&&high!=0xfff00000;i++){
 280                 temp = b;
 281                 b = ((double)(i+i)/x)*b - a;
 282                 GET_HIGH_WORD(high,b);
 283                 a = temp;
 284             }
 285         }
 286         if(sign>0) return b; else return -b;
 287 }