xref: /illumos-gate/usr/src/lib/libm/common/Q/jnl.c (revision 685c1a21304711e8d0a21bade51b783487d53044)
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
jnl(int n,GENERIC x)71 jnl(int n, GENERIC x)
72 {
73 	int i, sgn;
74 	GENERIC a, b, temp, z, w;
75 
76 	/*
77 	 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
78 	 * Thus, J(-n,x) = J(n,-x)
79 	 */
80 	if (n < 0) {
81 		n = -n;
82 		x = -x;
83 	}
84 	if (n == 0)
85 		return (j0l(x));
86 	if (n == 1)
87 		return (j1l(x));
88 	if (x != x)
89 		return (x+x);
90 	if ((n&1) == 0)
91 		sgn = 0;			/* even n */
92 	else
93 		sgn = signbitl(x);	/* old n  */
94 	x = fabsl(x);
95 	if (x == zero || !finitel(x)) b = zero;
96 	else if ((GENERIC)n <= x) {
97 					/*
98 					 * Safe to use
99 					 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
100 					 */
101 		if (x > 1.0e91L) {
102 				/*
103 				 * x >> n**2
104 				 *  Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 				 *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
106 				 *   Let s=sin(x), c=cos(x),
107 				 *	xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
108 				 *
109 				 *	   n	sin(xn)*sqt2	cos(xn)*sqt2
110 				 *	----------------------------------
111 				 *	   0	 s-c		 c+s
112 				 *	   1	-s-c		-c+s
113 				 *	   2	-s+c		-c-s
114 				 *	   3	 s+c		 c-s
115 				 */
116 			switch (n&3) {
117 			case 0:
118 				temp =  cosl(x)+sinl(x);
119 				break;
120 			case 1:
121 				temp = -cosl(x)+sinl(x);
122 				break;
123 			case 2:
124 				temp = -cosl(x)-sinl(x);
125 				break;
126 			case 3:
127 				temp =  cosl(x)-sinl(x);
128 				break;
129 			}
130 			b = invsqrtpi*temp/sqrtl(x);
131 		} else {
132 			a = j0l(x);
133 			b = j1l(x);
134 			for (i = 1; i < n; i++) {
135 				temp = b;
136 				/* avoid underflow */
137 				b = b*((GENERIC)(i+i)/x) - a;
138 				a = temp;
139 			}
140 		}
141 	} else {
142 		if (x < 1e-17L) {	/* use J(n,x) = 1/n!*(x/2)^n */
143 			b = powl(0.5L*x, (GENERIC)n);
144 			if (b != zero) {
145 				for (a = one, i = 1; i <= n; i++)
146 					a *= (GENERIC)i;
147 				b = b/a;
148 			}
149 		} else {
150 			/* use backward recurrence */
151 			/* BEGIN CSTYLED */
152 			/*
153 			 *			x      x^2	x^2
154 			 *  J(n,x)/J(n-1,x) =  ----   ------   ------	.....
155 			 *			2n  - 2(n+1) - 2(n+2)
156 			 *
157 			 *			1      1	1
158 			 *  (for large x)   =  ----  ------   ------   .....
159 			 *			2n   2(n+1)   2(n+2)
160 			 *			-- - ------ - ------ -
161 			 *			 x     x	 x
162 			 *
163 			 * Let w = 2n/x and h=2/x, then the above quotient
164 			 * is equal to the continued fraction:
165 			 *		    1
166 			 *	= -----------------------
167 			 *		       1
168 			 *	   w - -----------------
169 			 *			  1
170 			 *		w+h - ---------
171 			 *		       w+2h - ...
172 			 *
173 			 * To determine how many terms needed, let
174 			 * Q(0) = w, Q(1) = w(w+h) - 1,
175 			 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
176 			 * When Q(k) > 1e4	good for single
177 			 * When Q(k) > 1e9	good for double
178 			 * When Q(k) > 1e17	good for quaduple
179 			 */
180 			/* END CSTYLED */
181 			/* determine k */
182 			GENERIC t, v;
183 			double q0, q1, h, tmp;
184 			int k, m;
185 			w  = (n+n)/(double)x;
186 			h = 2.0/(double)x;
187 			q0 = w;
188 			z = w+h;
189 			q1 = w*z - 1.0;
190 			k = 1;
191 			while (q1 < 1.0e17) {
192 				k += 1;
193 				z += h;
194 				tmp = z*q1 - q0;
195 				q0 = q1;
196 				q1 = tmp;
197 			}
198 			m = n+n;
199 			for (t = zero, i = 2*(n+k); i >= m; i -= 2)
200 				t = one/(i/x-t);
201 			a = t;
202 			b = one;
203 			/*
204 			 * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
205 			 * hence, if n*(log(2n/x)) > ...
206 			 *  single:
207 			 *    8.8722839355e+01
208 			 *  double:
209 			 *    7.09782712893383973096e+02
210 			 *  long double:
211 			 *    1.1356523406294143949491931077970765006170e+04
212 			 *  then recurrent value may overflow and the result is
213 			 *  likely underflow to zero
214 			 */
215 			tmp = n;
216 			v = two/x;
217 			tmp = tmp*logl(fabsl(v*tmp));
218 			if (tmp < 1.1356523406294143949491931077970765e+04L) {
219 				for (i = n-1; i > 0; i--) {
220 					temp = b;
221 					b = ((i+i)/x)*b - a;
222 					a = temp;
223 				}
224 			} else {
225 				for (i = n-1; i > 0; i--) {
226 					temp = b;
227 					b = ((i+i)/x)*b - a;
228 					a = temp;
229 					if (b > 1e1000L) {
230 						a /= b;
231 						t /= b;
232 						b  = 1.0;
233 					}
234 				}
235 			}
236 			b = (t*j0l(x)/b);
237 		}
238 	}
239 	if (sgn != 0)
240 		return (-b);
241 	else
242 		return (b);
243 }
244 
245 GENERIC
ynl(int n,GENERIC x)246 ynl(int n, GENERIC x)
247 {
248 	int i;
249 	int sign;
250 	GENERIC a, b, temp;
251 
252 	if (x != x)
253 		return (x+x);
254 	if (x <= zero) {
255 		if (x == zero)
256 			return (-one/zero);
257 		else
258 			return (zero/zero);
259 	}
260 	sign = 1;
261 	if (n < 0) {
262 		n = -n;
263 		if ((n&1) == 1) sign = -1;
264 	}
265 	if (n == 0)
266 		return (y0l(x));
267 	if (n == 1)
268 		return (sign*y1l(x));
269 	if (!finitel(x))
270 		return (zero);
271 
272 	if (x > 1.0e91L) {
273 		/*
274 		 * x >> n**2
275 		 *   Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
276 		 *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
277 		 *   Let s = sin(x), c = cos(x),
278 		 *	xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
279 		 *
280 		 *	   n	sin(xn)*sqt2	cos(xn)*sqt2
281 		 *	----------------------------------
282 		 *	   0	 s-c		 c+s
283 		 *	   1	-s-c		-c+s
284 		 *	   2	-s+c		-c-s
285 		 *	   3	 s+c		 c-s
286 		 */
287 		switch (n&3) {
288 		case 0:
289 			temp =  sinl(x)-cosl(x);
290 			break;
291 		case 1:
292 			temp = -sinl(x)-cosl(x);
293 			break;
294 		case 2:
295 			temp = -sinl(x)+cosl(x);
296 			break;
297 		case 3:
298 			temp =  sinl(x)+cosl(x);
299 			break;
300 		}
301 		b = invsqrtpi*temp/sqrtl(x);
302 	} else {
303 		a = y0l(x);
304 		b = y1l(x);
305 		/*
306 		 * fix 1262058 and take care of non-default rounding
307 		 */
308 		for (i = 1; i < n; i++) {
309 			temp = b;
310 			b *= (GENERIC) (i + i) / x;
311 			if (b <= -LDBL_MAX)
312 				break;
313 			b -= a;
314 			a = temp;
315 		}
316 	}
317 	if (sign > 0)
318 		return (b);
319 	else
320 		return (-b);
321 }
322