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 __jnl = jnl 31 #pragma weak __ynl = ynl 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 "longdouble.h" 60 #include <float.h> /* LDBL_MAX */ 61 62 #define GENERIC long double 63 64 static const GENERIC 65 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L, 66 two = 2.0L, 67 zero = 0.0L, 68 one = 1.0L; 69 70 GENERIC 71 jnl(n, x) int n; GENERIC x; { 72 int i, sgn; 73 GENERIC a, b, temp = 0, z, w; 74 75 /* 76 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) 77 * Thus, J(-n,x) = J(n,-x) 78 */ 79 if (n < 0) { 80 n = -n; 81 x = -x; 82 } 83 if (n == 0) return (j0l(x)); 84 if (n == 1) return (j1l(x)); 85 if (x != x) return x+x; 86 if ((n&1) == 0) 87 sgn = 0; /* even n */ 88 else 89 sgn = signbitl(x); /* old n */ 90 x = fabsl(x); 91 if (x == zero || !finitel(x)) b = zero; 92 else if ((GENERIC)n <= x) { 93 /* 94 * Safe to use 95 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x) 96 */ 97 if (x > 1.0e91L) { 98 /* 99 * x >> n**2 100 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) 101 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) 102 * Let s=sin(x), c=cos(x), 103 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then 104 * 105 * n sin(xn)*sqt2 cos(xn)*sqt2 106 * ---------------------------------- 107 * 0 s-c c+s 108 * 1 -s-c -c+s 109 * 2 -s+c -c-s 110 * 3 s+c c-s 111 */ 112 switch (n&3) { 113 case 0: temp = cosl(x)+sinl(x); break; 114 case 1: temp = -cosl(x)+sinl(x); break; 115 case 2: temp = -cosl(x)-sinl(x); break; 116 case 3: temp = cosl(x)-sinl(x); break; 117 } 118 b = invsqrtpi*temp/sqrtl(x); 119 } else { 120 a = j0l(x); 121 b = j1l(x); 122 for (i = 1; i < n; i++) { 123 temp = b; 124 b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */ 125 a = temp; 126 } 127 } 128 } else { 129 if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */ 130 b = powl(0.5L*x, (GENERIC) n); 131 if (b != zero) { 132 for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i; 133 b = b/a; 134 } 135 } else { 136 /* 137 * use backward recurrence 138 * x x^2 x^2 139 * J(n,x)/J(n-1,x) = ---- ------ ------ ..... 140 * 2n - 2(n+1) - 2(n+2) 141 * 142 * 1 1 1 143 * (for large x) = ---- ------ ------ ..... 144 * 2n 2(n+1) 2(n+2) 145 * -- - ------ - ------ - 146 * x x x 147 * 148 * Let w = 2n/x and h=2/x, then the above quotient 149 * is equal to the continued fraction: 150 * 1 151 * = ----------------------- 152 * 1 153 * w - ----------------- 154 * 1 155 * w+h - --------- 156 * w+2h - ... 157 * 158 * To determine how many terms needed, let 159 * Q(0) = w, Q(1) = w(w+h) - 1, 160 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2), 161 * When Q(k) > 1e4 good for single 162 * When Q(k) > 1e9 good for double 163 * When Q(k) > 1e17 good for quaduple 164 */ 165 /* determin k */ 166 GENERIC t, v; 167 double q0, q1, h, tmp; int k, m; 168 w = (n+n)/(double)x; h = 2.0/(double)x; 169 q0 = w; z = w+h; q1 = w*z - 1.0; k = 1; 170 while (q1 < 1.0e17) { 171 k += 1; z += h; 172 tmp = z*q1 - q0; 173 q0 = q1; 174 q1 = tmp; 175 } 176 m = n+n; 177 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t); 178 a = t; 179 b = one; 180 /* 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*logl(fabsl(v*tmp)); 192 if (tmp < 1.1356523406294143949491931077970765e+04L) { 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 > 1e1000L) { 204 a /= b; 205 t /= b; 206 b = 1.0; 207 } 208 } 209 } 210 b = (t*j0l(x)/b); 211 } 212 } 213 if (sgn == 1) 214 return -b; 215 else 216 return b; 217 } 218 219 GENERIC 220 ynl(n, x) int n; GENERIC x; { 221 int i; 222 int sign; 223 GENERIC a, b, temp = 0; 224 225 if (x != x) 226 return x+x; 227 if (x <= zero) { 228 if (x == zero) 229 return -one/zero; 230 else 231 return zero/zero; 232 } 233 sign = 1; 234 if (n < 0) { 235 n = -n; 236 if ((n&1) == 1) sign = -1; 237 } 238 if (n == 0) return (y0l(x)); 239 if (n == 1) return (sign*y1l(x)); 240 if (!finitel(x)) return zero; 241 242 if (x > 1.0e91L) { 243 /* 244 * x >> n**2 245 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) 246 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) 247 * Let s=sin(x), c=cos(x), 248 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then 249 * 250 * n sin(xn)*sqt2 cos(xn)*sqt2 251 * ---------------------------------- 252 * 0 s-c c+s 253 * 1 -s-c -c+s 254 * 2 -s+c -c-s 255 * 3 s+c c-s 256 */ 257 switch (n&3) { 258 case 0: temp = sinl(x)-cosl(x); break; 259 case 1: temp = -sinl(x)-cosl(x); break; 260 case 2: temp = -sinl(x)+cosl(x); break; 261 case 3: temp = sinl(x)+cosl(x); break; 262 } 263 b = invsqrtpi*temp/sqrtl(x); 264 } else { 265 a = y0l(x); 266 b = y1l(x); 267 /* 268 * fix 1262058 and take care of non-default rounding 269 */ 270 for (i = 1; i < n; i++) { 271 temp = b; 272 b *= (GENERIC) (i + i) / x; 273 if (b <= -LDBL_MAX) 274 break; 275 b -= a; 276 a = temp; 277 } 278 } 279 if (sign > 0) 280 return b; 281 else 282 return -b; 283 } 284