1 /*- 2 * Copyright (c) 1992, 1993 3 * The Regents of the University of California. All rights reserved. 4 * 5 * %sccs.include.redist.c% 6 */ 7 8 #ifndef lint 9 static char sccsid[] = "@(#)jn.c 8.2 (Berkeley) 11/30/93"; 10 #endif /* not lint */ 11 12 /* 13 * 16 December 1992 14 * Minor modifications by Peter McIlroy to adapt non-IEEE architecture. 15 */ 16 17 /* 18 * ==================================================== 19 * Copyright (C) 1992 by Sun Microsystems, Inc. 20 * 21 * Developed at SunPro, a Sun Microsystems, Inc. business. 22 * Permission to use, copy, modify, and distribute this 23 * software is freely granted, provided that this notice 24 * is preserved. 25 * ==================================================== 26 * 27 * ******************* WARNING ******************** 28 * This is an alpha version of SunPro's FDLIBM (Freely 29 * Distributable Math Library) for IEEE double precision 30 * arithmetic. FDLIBM is a basic math library written 31 * in C that runs on machines that conform to IEEE 32 * Standard 754/854. This alpha version is distributed 33 * for testing purpose. Those who use this software 34 * should report any bugs to 35 * 36 * fdlibm-comments@sunpro.eng.sun.com 37 * 38 * -- K.C. Ng, Oct 12, 1992 39 * ************************************************ 40 */ 41 42 /* 43 * jn(int n, double x), yn(int n, double x) 44 * floating point Bessel's function of the 1st and 2nd kind 45 * of order n 46 * 47 * Special cases: 48 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal; 49 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal. 50 * Note 2. About jn(n,x), yn(n,x) 51 * For n=0, j0(x) is called, 52 * for n=1, j1(x) is called, 53 * for n<x, forward recursion us used starting 54 * from values of j0(x) and j1(x). 55 * for n>x, a continued fraction approximation to 56 * j(n,x)/j(n-1,x) is evaluated and then backward 57 * recursion is used starting from a supposed value 58 * for j(n,x). The resulting value of j(0,x) is 59 * compared with the actual value to correct the 60 * supposed value of j(n,x). 61 * 62 * yn(n,x) is similar in all respects, except 63 * that forward recursion is used for all 64 * values of n>1. 65 * 66 */ 67 68 #include <math.h> 69 #include <float.h> 70 #include <errno.h> 71 72 #if defined(vax) || defined(tahoe) 73 #define _IEEE 0 74 #else 75 #define _IEEE 1 76 #define infnan(x) (0.0) 77 #endif 78 79 static double 80 invsqrtpi= 5.641895835477562869480794515607725858441e-0001, 81 two = 2.0, 82 zero = 0.0, 83 one = 1.0; 84 85 double jn(n,x) 86 int n; double x; 87 { 88 int i, sgn; 89 double a, b, temp; 90 double z, w; 91 92 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) 93 * Thus, J(-n,x) = J(n,-x) 94 */ 95 /* if J(n,NaN) is NaN */ 96 if (_IEEE && isnan(x)) return x+x; 97 if (n<0){ 98 n = -n; 99 x = -x; 100 } 101 if (n==0) return(j0(x)); 102 if (n==1) return(j1(x)); 103 sgn = (n&1)&(x < zero); /* even n -- 0, odd n -- sign(x) */ 104 x = fabs(x); 105 if (x == 0 || !finite (x)) /* if x is 0 or inf */ 106 b = zero; 107 else if ((double) n <= x) { 108 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ 109 if (_IEEE && x >= 8.148143905337944345e+090) { 110 /* x >= 2**302 */ 111 /* (x >> n**2) 112 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) 113 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) 114 * Let s=sin(x), c=cos(x), 115 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then 116 * 117 * n sin(xn)*sqt2 cos(xn)*sqt2 118 * ---------------------------------- 119 * 0 s-c c+s 120 * 1 -s-c -c+s 121 * 2 -s+c -c-s 122 * 3 s+c c-s 123 */ 124 switch(n&3) { 125 case 0: temp = cos(x)+sin(x); break; 126 case 1: temp = -cos(x)+sin(x); break; 127 case 2: temp = -cos(x)-sin(x); break; 128 case 3: temp = cos(x)-sin(x); break; 129 } 130 b = invsqrtpi*temp/sqrt(x); 131 } else { 132 a = j0(x); 133 b = j1(x); 134 for(i=1;i<n;i++){ 135 temp = b; 136 b = b*((double)(i+i)/x) - a; /* avoid underflow */ 137 a = temp; 138 } 139 } 140 } else { 141 if (x < 1.86264514923095703125e-009) { /* x < 2**-29 */ 142 /* x is tiny, return the first Taylor expansion of J(n,x) 143 * J(n,x) = 1/n!*(x/2)^n - ... 144 */ 145 if (n > 33) /* underflow */ 146 b = zero; 147 else { 148 temp = x*0.5; b = temp; 149 for (a=one,i=2;i<=n;i++) { 150 a *= (double)i; /* a = n! */ 151 b *= temp; /* b = (x/2)^n */ 152 } 153 b = b/a; 154 } 155 } else { 156 /* use backward recurrence */ 157 /* x x^2 x^2 158 * J(n,x)/J(n-1,x) = ---- ------ ------ ..... 159 * 2n - 2(n+1) - 2(n+2) 160 * 161 * 1 1 1 162 * (for large x) = ---- ------ ------ ..... 163 * 2n 2(n+1) 2(n+2) 164 * -- - ------ - ------ - 165 * x x x 166 * 167 * Let w = 2n/x and h=2/x, then the above quotient 168 * is equal to the continued fraction: 169 * 1 170 * = ----------------------- 171 * 1 172 * w - ----------------- 173 * 1 174 * w+h - --------- 175 * w+2h - ... 176 * 177 * To determine how many terms needed, let 178 * Q(0) = w, Q(1) = w(w+h) - 1, 179 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2), 180 * When Q(k) > 1e4 good for single 181 * When Q(k) > 1e9 good for double 182 * When Q(k) > 1e17 good for quadruple 183 */ 184 /* determine k */ 185 double t,v; 186 double q0,q1,h,tmp; int k,m; 187 w = (n+n)/(double)x; h = 2.0/(double)x; 188 q0 = w; z = w+h; q1 = w*z - 1.0; k=1; 189 while (q1<1.0e9) { 190 k += 1; z += h; 191 tmp = z*q1 - q0; 192 q0 = q1; 193 q1 = tmp; 194 } 195 m = n+n; 196 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t); 197 a = t; 198 b = one; 199 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n) 200 * Hence, if n*(log(2n/x)) > ... 201 * single 8.8722839355e+01 202 * double 7.09782712893383973096e+02 203 * long double 1.1356523406294143949491931077970765006170e+04 204 * then recurrent value may overflow and the result will 205 * likely underflow to zero 206 */ 207 tmp = n; 208 v = two/x; 209 tmp = tmp*log(fabs(v*tmp)); 210 for (i=n-1;i>0;i--){ 211 temp = b; 212 b = ((i+i)/x)*b - a; 213 a = temp; 214 /* scale b to avoid spurious overflow */ 215 # if defined(vax) || defined(tahoe) 216 # define BMAX 1e13 217 # else 218 # define BMAX 1e100 219 # endif /* defined(vax) || defined(tahoe) */ 220 if (b > BMAX) { 221 a /= b; 222 t /= b; 223 b = one; 224 } 225 } 226 b = (t*j0(x)/b); 227 } 228 } 229 return ((sgn == 1) ? -b : b); 230 } 231 double yn(n,x) 232 int n; double x; 233 { 234 int i, sign; 235 double a, b, temp; 236 237 /* Y(n,NaN), Y(n, x < 0) is NaN */ 238 if (x <= 0 || (_IEEE && x != x)) 239 if (_IEEE && x < 0) return zero/zero; 240 else if (x < 0) return (infnan(EDOM)); 241 else if (_IEEE) return -one/zero; 242 else return(infnan(-ERANGE)); 243 else if (!finite(x)) return(0); 244 sign = 1; 245 if (n<0){ 246 n = -n; 247 sign = 1 - ((n&1)<<2); 248 } 249 if (n == 0) return(y0(x)); 250 if (n == 1) return(sign*y1(x)); 251 if(_IEEE && x >= 8.148143905337944345e+090) { /* x > 2**302 */ 252 /* (x >> n**2) 253 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) 254 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) 255 * Let s=sin(x), c=cos(x), 256 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then 257 * 258 * n sin(xn)*sqt2 cos(xn)*sqt2 259 * ---------------------------------- 260 * 0 s-c c+s 261 * 1 -s-c -c+s 262 * 2 -s+c -c-s 263 * 3 s+c c-s 264 */ 265 switch (n&3) { 266 case 0: temp = sin(x)-cos(x); break; 267 case 1: temp = -sin(x)-cos(x); break; 268 case 2: temp = -sin(x)+cos(x); break; 269 case 3: temp = sin(x)+cos(x); break; 270 } 271 b = invsqrtpi*temp/sqrt(x); 272 } else { 273 a = y0(x); 274 b = y1(x); 275 /* quit if b is -inf */ 276 for (i = 1; i < n && !finite(b); i++){ 277 temp = b; 278 b = ((double)(i+i)/x)*b - a; 279 a = temp; 280 } 281 } 282 if (!_IEEE && !finite(b)) 283 return (infnan(-sign * ERANGE)); 284 return ((sign > 0) ? b : -b); 285 } 286