xref: /netbsd-src/lib/libm/src/e_jn.c (revision ae1bfcddc410612bc8c58b807e1830becb69a24c)
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 #ifndef lint
14 static char rcsid[] = "$Id: e_jn.c,v 1.4 1994/03/03 17:04:16 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 division by zero 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 <machine/endian.h>
45 
46 #if BYTE_ORDER == LITTLE_ENDIAN
47 #define n0	1
48 #else
49 #define n0	0
50 #endif
51 
52 #ifdef __STDC__
53 static const double
54 #else
55 static double
56 #endif
57 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
58 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
59 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
60 
61 static double zero  =  0.00000000000000000000e+00;
62 
63 #ifdef __STDC__
64 	double __ieee754_jn(int n, double x)
65 #else
66 	double __ieee754_jn(n,x)
67 	int n; double x;
68 #endif
69 {
70 	int i,hx,ix,lx, sgn;
71 	double a, b, temp, di;
72 	double z, w;
73 
74     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
75      * Thus, J(-n,x) = J(n,-x)
76      */
77 	hx = *(n0+(int*)&x);
78 	ix = 0x7fffffff&hx;
79 	lx = *(1-n0+(int*)&x);
80     /* if J(n,NaN) is NaN */
81 	if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
82 	if(n<0){
83 		n = -n;
84 		x = -x;
85 		hx ^= 0x80000000;
86 	}
87 	if(n==0) return(__ieee754_j0(x));
88 	if(n==1) return(__ieee754_j1(x));
89 	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
90 	x = fabs(x);
91 	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
92 	    b = zero;
93 	else if((double)n<=x) {
94 		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
95 	    if(ix>=0x52D00000) { /* x > 2**302 */
96     /* (x >> n**2)
97      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
98      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
99      *	    Let s=sin(x), c=cos(x),
100      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
101      *
102      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
103      *		----------------------------------
104      *		   0	 s-c		 c+s
105      *		   1	-s-c 		-c+s
106      *		   2	-s+c		-c-s
107      *		   3	 s+c		 c-s
108      */
109 		switch(n&3) {
110 		    case 0: temp =  cos(x)+sin(x); break;
111 		    case 1: temp = -cos(x)+sin(x); break;
112 		    case 2: temp = -cos(x)-sin(x); break;
113 		    case 3: temp =  cos(x)-sin(x); break;
114 		}
115 		b = invsqrtpi*temp/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; int 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 	int i,hx,ix,lx;
232 	int sign;
233 	double a, b, temp;
234 
235 	hx = *(n0+(int*)&x);
236 	ix = 0x7fffffff&hx;
237 	lx = *(1-n0+(int*)&x);
238     /* if Y(n,NaN) is NaN */
239 	if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
240 	if((ix|lx)==0) return -one/zero;
241 	if(hx<0) return zero/zero;
242 	sign = 1;
243 	if(n<0){
244 		n = -n;
245 		sign = 1 - ((n&1)<<2);
246 	}
247 	if(n==0) return(__ieee754_y0(x));
248 	if(n==1) return(sign*__ieee754_y1(x));
249 	if(ix==0x7ff00000) return zero;
250 	if(ix>=0x52D00000) { /* x > 2**302 */
251     /* (x >> n**2)
252      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
253      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
254      *	    Let s=sin(x), c=cos(x),
255      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
256      *
257      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
258      *		----------------------------------
259      *		   0	 s-c		 c+s
260      *		   1	-s-c 		-c+s
261      *		   2	-s+c		-c-s
262      *		   3	 s+c		 c-s
263      */
264 		switch(n&3) {
265 		    case 0: temp =  sin(x)-cos(x); break;
266 		    case 1: temp = -sin(x)-cos(x); break;
267 		    case 2: temp = -sin(x)+cos(x); break;
268 		    case 3: temp =  sin(x)+cos(x); break;
269 		}
270 		b = invsqrtpi*temp/sqrt(x);
271 	} else {
272 	    a = __ieee754_y0(x);
273 	    b = __ieee754_y1(x);
274 	/* quit if b is -inf */
275 	    for(i=1;i<n&&(*(n0+(int*)&b)!=0xfff00000);i++){
276 		temp = b;
277 		b = ((double)(i+i)/x)*b - a;
278 		a = temp;
279 	    }
280 	}
281 	if(sign>0) return b; else return -b;
282 }
283