xref: /titanic_50/usr/src/lib/libm/common/LD/jnl.c (revision ddc0e0b53c661f6e439e3b7072b3ef353eadb4af)
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(n,x)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
ynl(n,x)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