1/* @(#)e_jn.c 5.1 93/09/24 */
2/*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunPro, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
9 * is preserved.
10 * ====================================================
11 */
12
13#if defined(LIBM_SCCS) && !defined(lint)
14static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
15#endif
16
17/*
18 * __ieee754_jn(n, x), __ieee754_yn(n, x)
19 * floating point Bessel's function of the 1st and 2nd kind
20 * of order n
21 *
22 * Special cases:
23 *	y0(0)=y1(0)=yn(n,0) = -inf with overflow signal;
24 *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
25 * Note 2. About jn(n,x), yn(n,x)
26 *	For n=0, j0(x) is called,
27 *	for n=1, j1(x) is called,
28 *	for n<x, forward recursion us used starting
29 *	from values of j0(x) and j1(x).
30 *	for n>x, a continued fraction approximation to
31 *	j(n,x)/j(n-1,x) is evaluated and then backward
32 *	recursion is used starting from a supposed value
33 *	for j(n,x). The resulting value of j(0,x) is
34 *	compared with the actual value to correct the
35 *	supposed value of j(n,x).
36 *
37 *	yn(n,x) is similar in all respects, except
38 *	that forward recursion is used for all
39 *	values of n>1.
40 *
41 */
42
43#include "math.h"
44#include "math_private.h"
45
46#ifdef __STDC__
47static const double
48#else
49static double
50#endif
51invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
52two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
53one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
54
55#ifdef __STDC__
56static const double zero  =  0.00000000000000000000e+00;
57#else
58static double zero  =  0.00000000000000000000e+00;
59#endif
60
61#ifdef __STDC__
62	double __ieee754_jn(int n, double x)
63#else
64	double __ieee754_jn(n,x)
65	int n; double x;
66#endif
67{
68	int32_t i,hx,ix,lx, sgn;
69	double a, b, temp, di;
70	double z, w;
71
72    /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
73     * Thus, J(-n,x) = J(n,-x)
74     */
75	EXTRACT_WORDS(hx,lx,x);
76	ix = 0x7fffffff&hx;
77    /* if J(n,NaN) is NaN */
78	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
79	if(n<0){
80		n = -n;
81		x = -x;
82		hx ^= 0x80000000;
83	}
84	if(n==0) return(__ieee754_j0(x));
85	if(n==1) return(__ieee754_j1(x));
86	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
87	x = fabs(x);
88	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
89	    b = zero;
90	else if((double)n<=x) {
91		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
92	    if(ix>=0x52D00000) { /* x > 2**302 */
93    /* (x >> n**2)
94     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
95     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
96     *	    Let s=sin(x), c=cos(x),
97     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
98     *
99     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
100     *		----------------------------------
101     *		   0	 s-c		 c+s
102     *		   1	-s-c 		-c+s
103     *		   2	-s+c		-c-s
104     *		   3	 s+c		 c-s
105     */
106		double s;
107		double c;
108		__sincos (x, &s, &c);
109		switch(n&3) {
110		    case 0: temp =  c + s; break;
111		    case 1: temp = -c + s; break;
112		    case 2: temp = -c - s; break;
113		    case 3: temp =  c - s; break;
114		}
115		b = invsqrtpi*temp/__ieee754_sqrt(x);
116	    } else {
117	        a = __ieee754_j0(x);
118	        b = __ieee754_j1(x);
119	        for(i=1;i<n;i++){
120		    temp = b;
121		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
122		    a = temp;
123	        }
124	    }
125	} else {
126	    if(ix<0x3e100000) {	/* x < 2**-29 */
127    /* x is tiny, return the first Taylor expansion of J(n,x)
128     * J(n,x) = 1/n!*(x/2)^n  - ...
129     */
130		if(n>33)	/* underflow */
131		    b = zero;
132		else {
133		    temp = x*0.5; b = temp;
134		    for (a=one,i=2;i<=n;i++) {
135			a *= (double)i;		/* a = n! */
136			b *= temp;		/* b = (x/2)^n */
137		    }
138		    b = b/a;
139		}
140	    } else {
141		/* use backward recurrence */
142		/* 			x      x^2      x^2
143		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
144		 *			2n  - 2(n+1) - 2(n+2)
145		 *
146		 * 			1      1        1
147		 *  (for large x)   =  ----  ------   ------   .....
148		 *			2n   2(n+1)   2(n+2)
149		 *			-- - ------ - ------ -
150		 *			 x     x         x
151		 *
152		 * Let w = 2n/x and h=2/x, then the above quotient
153		 * is equal to the continued fraction:
154		 *		    1
155		 *	= -----------------------
156		 *		       1
157		 *	   w - -----------------
158		 *			  1
159		 * 	        w+h - ---------
160		 *		       w+2h - ...
161		 *
162		 * To determine how many terms needed, let
163		 * Q(0) = w, Q(1) = w(w+h) - 1,
164		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
165		 * When Q(k) > 1e4	good for single
166		 * When Q(k) > 1e9	good for double
167		 * When Q(k) > 1e17	good for quadruple
168		 */
169	    /* determine k */
170		double t,v;
171		double q0,q1,h,tmp; int32_t k,m;
172		w  = (n+n)/(double)x; h = 2.0/(double)x;
173		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
174		while(q1<1.0e9) {
175			k += 1; z += h;
176			tmp = z*q1 - q0;
177			q0 = q1;
178			q1 = tmp;
179		}
180		m = n+n;
181		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
182		a = t;
183		b = one;
184		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
185		 *  Hence, if n*(log(2n/x)) > ...
186		 *  single 8.8722839355e+01
187		 *  double 7.09782712893383973096e+02
188		 *  long double 1.1356523406294143949491931077970765006170e+04
189		 *  then recurrent value may overflow and the result is
190		 *  likely underflow to zero
191		 */
192		tmp = n;
193		v = two/x;
194		tmp = tmp*__ieee754_log(fabs(v*tmp));
195		if(tmp<7.09782712893383973096e+02) {
196	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
197		        temp = b;
198			b *= di;
199			b  = b/x - a;
200		        a = temp;
201			di -= two;
202	     	    }
203		} else {
204	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
205		        temp = b;
206			b *= di;
207			b  = b/x - a;
208		        a = temp;
209			di -= two;
210		    /* scale b to avoid spurious overflow */
211			if(b>1e100) {
212			    a /= b;
213			    t /= b;
214			    b  = one;
215			}
216	     	    }
217		}
218	    	b = (t*__ieee754_j0(x)/b);
219	    }
220	}
221	if(sgn==1) return -b; else return b;
222}
223
224#ifdef __STDC__
225	double __ieee754_yn(int n, double x)
226#else
227	double __ieee754_yn(n,x)
228	int n; double x;
229#endif
230{
231	int32_t i,hx,ix,lx;
232	int32_t sign;
233	double a, b, temp;
234
235	EXTRACT_WORDS(hx,lx,x);
236	ix = 0x7fffffff&hx;
237    /* if Y(n,NaN) is NaN */
238	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
239	if((ix|lx)==0) return -HUGE_VAL+x; /* -inf and overflow exception.  */;
240#if (defined(__BEOS__) || defined(__HAIKU__))
241	if(hx<0) return -HUGE_VAL+x;
242#else
243	if(hx<0) return zero/(zero*x);
244#endif
245	sign = 1;
246	if(n<0){
247		n = -n;
248		sign = 1 - ((n&1)<<1);
249	}
250	if(n==0) return(__ieee754_y0(x));
251	if(n==1) return(sign*__ieee754_y1(x));
252	if(ix==0x7ff00000) return zero;
253	if(ix>=0x52D00000) { /* x > 2**302 */
254    /* (x >> n**2)
255     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
256     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
257     *	    Let s=sin(x), c=cos(x),
258     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
259     *
260     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
261     *		----------------------------------
262     *		   0	 s-c		 c+s
263     *		   1	-s-c 		-c+s
264     *		   2	-s+c		-c-s
265     *		   3	 s+c		 c-s
266     */
267		double c;
268		double s;
269		__sincos (x, &s, &c);
270		switch(n&3) {
271		    case 0: temp =  s - c; break;
272		    case 1: temp = -s - c; break;
273		    case 2: temp = -s + c; break;
274		    case 3: temp =  s + c; break;
275		}
276		b = invsqrtpi*temp/__ieee754_sqrt(x);
277	} else {
278	    u_int32_t high;
279	    a = __ieee754_y0(x);
280	    b = __ieee754_y1(x);
281	/* quit if b is -inf */
282	    GET_HIGH_WORD(high,b);
283	    for(i=1;i<n&&high!=0xfff00000;i++){
284		temp = b;
285		b = ((double)(i+i)/x)*b - a;
286		GET_HIGH_WORD(high,b);
287		a = temp;
288	    }
289	}
290	if(sign>0) return b; else return -b;
291}
292