xref: /titanic_50/usr/src/lib/libm/common/LD/jnl.c (revision 89621fe174cf95ae903df6ceab605bf24d696ac3)
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 #if defined(ELFOBJ)
31 #pragma weak jnl = __jnl
32 #pragma weak ynl = __ynl
33 #endif
34 
35 /*
36  * floating point Bessel's function of the 1st and 2nd kind
37  * of order n: jn(n,x),yn(n,x);
38  *
39  * Special cases:
40  *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
41  *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
42  * Note 2. About jn(n,x), yn(n,x)
43  *	For n=0, j0(x) is called,
44  *	for n=1, j1(x) is called,
45  *	for n<x, forward recursion us used starting
46  *	from values of j0(x) and j1(x).
47  *	for n>x, a continued fraction approximation to
48  *	j(n,x)/j(n-1,x) is evaluated and then backward
49  *	recursion is used starting from a supposed value
50  *	for j(n,x). The resulting value of j(0,x) is
51  *	compared with the actual value to correct the
52  *	supposed value of j(n,x).
53  *
54  *	yn(n,x) is similar in all respects, except
55  *	that forward recursion is used for all
56  *	values of n>1.
57  *
58  */
59 
60 #include "libm.h"
61 #include "longdouble.h"
62 #include <float.h>	/* LDBL_MAX */
63 
64 #define	GENERIC long double
65 
66 static const GENERIC
67 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
68 two  = 2.0L,
69 zero = 0.0L,
70 one  = 1.0L;
71 
72 GENERIC
73 jnl(n, x) int n; GENERIC x; {
74 	int i, sgn;
75 	GENERIC a, b, temp = 0, z, w;
76 
77 	/*
78 	 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
79 	 * Thus, J(-n,x) = J(n,-x)
80 	 */
81 	if (n < 0) {
82 		n = -n;
83 		x = -x;
84 	}
85 	if (n == 0) return (j0l(x));
86 	if (n == 1) return (j1l(x));
87 	if (x != x) return x+x;
88 	if ((n&1) == 0)
89 		sgn = 0; 			/* even n */
90 	else
91 		sgn = signbitl(x);	/* old n  */
92 	x = fabsl(x);
93 	if (x == zero || !finitel(x)) b = zero;
94 	else if ((GENERIC)n <= x) {
95 			/*
96 			 * Safe to use
97 			 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
98 			 */
99 	    if (x > 1.0e91L) {
100 				/*
101 				 * 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 =  cosl(x)+sinl(x); break;
116 		    case 1: temp = -cosl(x)+sinl(x); break;
117 		    case 2: temp = -cosl(x)-sinl(x); break;
118 		    case 3: temp =  cosl(x)-sinl(x); break;
119 		}
120 		b = invsqrtpi*temp/sqrtl(x);
121 	    } else {
122 			a = j0l(x);
123 			b = j1l(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-17L) {	/* use J(n,x) = 1/n!*(x/2)^n */
132 		b = powl(0.5L*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 		/*
139 		 * use backward recurrence
140 		 * 			x      x^2      x^2
141 		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
142 		 *			2n  - 2(n+1) - 2(n+2)
143 		 *
144 		 * 			1      1        1
145 		 *  (for large x)   =  ----  ------   ------   .....
146 		 *			2n   2(n+1)   2(n+2)
147 		 *			-- - ------ - ------ -
148 		 *			 x     x         x
149 		 *
150 		 * Let w = 2n/x and h=2/x, then the above quotient
151 		 * is equal to the continued fraction:
152 		 *		    1
153 		 *	= -----------------------
154 		 *		       1
155 		 *	   w - -----------------
156 		 *			  1
157 		 * 	        w+h - ---------
158 		 *		       w+2h - ...
159 		 *
160 		 * To determine how many terms needed, let
161 		 * Q(0) = w, Q(1) = w(w+h) - 1,
162 		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
163 		 * When Q(k) > 1e4	good for single
164 		 * When Q(k) > 1e9	good for double
165 		 * When Q(k) > 1e17	good for quaduple
166 		 */
167 	    /* determin k */
168 		GENERIC t, v;
169 		double q0, q1, h, tmp; int k, m;
170 		w  = (n+n)/(double)x; h = 2.0/(double)x;
171 		q0 = w;  z = w+h; q1 = w*z - 1.0; k = 1;
172 		while (q1 < 1.0e17) {
173 			k += 1; z += h;
174 			tmp = z*q1 - q0;
175 			q0 = q1;
176 			q1 = tmp;
177 		}
178 		m = n+n;
179 		for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
180 		a = t;
181 		b = one;
182 			/*
183 			 * Estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
184 			 * hence, if n*(log(2n/x)) > ...
185 			 * single 8.8722839355e+01
186 			 * double 7.09782712893383973096e+02
187 			 * long double 1.1356523406294143949491931077970765006170e+04
188 			 * then recurrent value may overflow and the result is
189 			 * likely underflow to zero.
190 			 */
191 		tmp = n;
192 		v = two/x;
193 		tmp = tmp*logl(fabsl(v*tmp));
194 		if (tmp < 1.1356523406294143949491931077970765e+04L) {
195 				for (i = n-1; i > 0; i--) {
196 				temp = b;
197 				b = ((i+i)/x)*b - a;
198 				a = temp;
199 				}
200 		} else {
201 				for (i = n-1; i > 0; i--) {
202 				temp = b;
203 				b = ((i+i)/x)*b - a;
204 				a = temp;
205 			if (b > 1e1000L) {
206 						a /= b;
207 						t /= b;
208 						b  = 1.0;
209 					}
210 				}
211 		}
212 			b = (t*j0l(x)/b);
213 	    }
214 	}
215 	if (sgn == 1)
216 		return -b;
217 	else
218 		return b;
219 }
220 
221 GENERIC
222 ynl(n, x) int n; GENERIC x; {
223 	int i;
224 	int sign;
225 	GENERIC a, b, temp = 0;
226 
227 	if (x != x)
228 		return x+x;
229 	if (x <= zero) {
230 		if (x == zero)
231 			return -one/zero;
232 		else
233 			return zero/zero;
234 	}
235 	sign = 1;
236 	if (n < 0) {
237 		n = -n;
238 		if ((n&1) == 1) sign = -1;
239 	}
240 	if (n == 0) return (y0l(x));
241 	if (n == 1) return (sign*y1l(x));
242 	if (!finitel(x)) return zero;
243 
244 	if (x > 1.0e91L) {
245 				/*
246 				 * 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 =  sinl(x)-cosl(x); break;
261 		    case 1: temp = -sinl(x)-cosl(x); break;
262 		    case 2: temp = -sinl(x)+cosl(x); break;
263 		    case 3: temp =  sinl(x)+cosl(x); break;
264 		}
265 		b = invsqrtpi*temp/sqrtl(x);
266 	} else {
267 		a = y0l(x);
268 		b = y1l(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 <= -LDBL_MAX)
276 				break;
277 			b -= a;
278 			a = temp;
279 		}
280 	}
281 	if (sign > 0)
282 		return b;
283 	else
284 		return -b;
285 }
286