xref: /freebsd/lib/msun/src/e_jn.c (revision b1f9167f94059fd55c630891d359bcff987bd7eb)
1 
2 /* @(#)e_jn.c 1.4 95/01/18 */
3 /*
4  * ====================================================
5  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6  *
7  * Developed at SunSoft, a Sun Microsystems, Inc. business.
8  * Permission to use, copy, modify, and distribute this
9  * software is freely granted, provided that this notice
10  * is preserved.
11  * ====================================================
12  */
13 
14 #include <sys/cdefs.h>
15 __FBSDID("$FreeBSD$");
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 "math_private.h"
45 
46 static const double
47 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
48 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
49 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
50 
51 static const double zero  =  0.00000000000000000000e+00;
52 
53 double
54 __ieee754_jn(int n, double x)
55 {
56 	int32_t i,hx,ix,lx, sgn;
57 	double a, b, temp, di;
58 	double z, w;
59 
60     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
61      * Thus, J(-n,x) = J(n,-x)
62      */
63 	EXTRACT_WORDS(hx,lx,x);
64 	ix = 0x7fffffff&hx;
65     /* if J(n,NaN) is NaN */
66 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
67 	if(n<0){
68 		n = -n;
69 		x = -x;
70 		hx ^= 0x80000000;
71 	}
72 	if(n==0) return(__ieee754_j0(x));
73 	if(n==1) return(__ieee754_j1(x));
74 	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
75 	x = fabs(x);
76 	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
77 	    b = zero;
78 	else if((double)n<=x) {
79 		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
80 	    if(ix>=0x52D00000) { /* x > 2**302 */
81     /* (x >> n**2)
82      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
83      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
84      *	    Let s=sin(x), c=cos(x),
85      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
86      *
87      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
88      *		----------------------------------
89      *		   0	 s-c		 c+s
90      *		   1	-s-c 		-c+s
91      *		   2	-s+c		-c-s
92      *		   3	 s+c		 c-s
93      */
94 		switch(n&3) {
95 		    case 0: temp =  cos(x)+sin(x); break;
96 		    case 1: temp = -cos(x)+sin(x); break;
97 		    case 2: temp = -cos(x)-sin(x); break;
98 		    case 3: temp =  cos(x)-sin(x); break;
99 		}
100 		b = invsqrtpi*temp/sqrt(x);
101 	    } else {
102 	        a = __ieee754_j0(x);
103 	        b = __ieee754_j1(x);
104 	        for(i=1;i<n;i++){
105 		    temp = b;
106 		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
107 		    a = temp;
108 	        }
109 	    }
110 	} else {
111 	    if(ix<0x3e100000) {	/* x < 2**-29 */
112     /* x is tiny, return the first Taylor expansion of J(n,x)
113      * J(n,x) = 1/n!*(x/2)^n  - ...
114      */
115 		if(n>33)	/* underflow */
116 		    b = zero;
117 		else {
118 		    temp = x*0.5; b = temp;
119 		    for (a=one,i=2;i<=n;i++) {
120 			a *= (double)i;		/* a = n! */
121 			b *= temp;		/* b = (x/2)^n */
122 		    }
123 		    b = b/a;
124 		}
125 	    } else {
126 		/* use backward recurrence */
127 		/* 			x      x^2      x^2
128 		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
129 		 *			2n  - 2(n+1) - 2(n+2)
130 		 *
131 		 * 			1      1        1
132 		 *  (for large x)   =  ----  ------   ------   .....
133 		 *			2n   2(n+1)   2(n+2)
134 		 *			-- - ------ - ------ -
135 		 *			 x     x         x
136 		 *
137 		 * Let w = 2n/x and h=2/x, then the above quotient
138 		 * is equal to the continued fraction:
139 		 *		    1
140 		 *	= -----------------------
141 		 *		       1
142 		 *	   w - -----------------
143 		 *			  1
144 		 * 	        w+h - ---------
145 		 *		       w+2h - ...
146 		 *
147 		 * To determine how many terms needed, let
148 		 * Q(0) = w, Q(1) = w(w+h) - 1,
149 		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
150 		 * When Q(k) > 1e4	good for single
151 		 * When Q(k) > 1e9	good for double
152 		 * When Q(k) > 1e17	good for quadruple
153 		 */
154 	    /* determine k */
155 		double t,v;
156 		double q0,q1,h,tmp; int32_t k,m;
157 		w  = (n+n)/(double)x; h = 2.0/(double)x;
158 		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
159 		while(q1<1.0e9) {
160 			k += 1; z += h;
161 			tmp = z*q1 - q0;
162 			q0 = q1;
163 			q1 = tmp;
164 		}
165 		m = n+n;
166 		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
167 		a = t;
168 		b = one;
169 		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
170 		 *  Hence, if n*(log(2n/x)) > ...
171 		 *  single 8.8722839355e+01
172 		 *  double 7.09782712893383973096e+02
173 		 *  long double 1.1356523406294143949491931077970765006170e+04
174 		 *  then recurrent value may overflow and the result is
175 		 *  likely underflow to zero
176 		 */
177 		tmp = n;
178 		v = two/x;
179 		tmp = tmp*__ieee754_log(fabs(v*tmp));
180 		if(tmp<7.09782712893383973096e+02) {
181 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
182 		        temp = b;
183 			b *= di;
184 			b  = b/x - a;
185 		        a = temp;
186 			di -= two;
187 	     	    }
188 		} else {
189 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
190 		        temp = b;
191 			b *= di;
192 			b  = b/x - a;
193 		        a = temp;
194 			di -= two;
195 		    /* scale b to avoid spurious overflow */
196 			if(b>1e100) {
197 			    a /= b;
198 			    t /= b;
199 			    b  = one;
200 			}
201 	     	    }
202 		}
203 		z = __ieee754_j0(x);
204 		w = __ieee754_j1(x);
205 		if (fabs(z) >= fabs(w))
206 		    b = (t*z/b);
207 		else
208 		    b = (t*w/a);
209 	    }
210 	}
211 	if(sgn==1) return -b; else return b;
212 }
213 
214 double
215 __ieee754_yn(int n, double x)
216 {
217 	int32_t i,hx,ix,lx;
218 	int32_t sign;
219 	double a, b, temp;
220 
221 	EXTRACT_WORDS(hx,lx,x);
222 	ix = 0x7fffffff&hx;
223     /* if Y(n,NaN) is NaN */
224 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
225 	if((ix|lx)==0) return -one/zero;
226 	if(hx<0) return zero/zero;
227 	sign = 1;
228 	if(n<0){
229 		n = -n;
230 		sign = 1 - ((n&1)<<1);
231 	}
232 	if(n==0) return(__ieee754_y0(x));
233 	if(n==1) return(sign*__ieee754_y1(x));
234 	if(ix==0x7ff00000) return zero;
235 	if(ix>=0x52D00000) { /* x > 2**302 */
236     /* (x >> n**2)
237      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
238      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
239      *	    Let s=sin(x), c=cos(x),
240      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
241      *
242      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
243      *		----------------------------------
244      *		   0	 s-c		 c+s
245      *		   1	-s-c 		-c+s
246      *		   2	-s+c		-c-s
247      *		   3	 s+c		 c-s
248      */
249 		switch(n&3) {
250 		    case 0: temp =  sin(x)-cos(x); break;
251 		    case 1: temp = -sin(x)-cos(x); break;
252 		    case 2: temp = -sin(x)+cos(x); break;
253 		    case 3: temp =  sin(x)+cos(x); break;
254 		}
255 		b = invsqrtpi*temp/sqrt(x);
256 	} else {
257 	    u_int32_t high;
258 	    a = __ieee754_y0(x);
259 	    b = __ieee754_y1(x);
260 	/* quit if b is -inf */
261 	    GET_HIGH_WORD(high,b);
262 	    for(i=1;i<n&&high!=0xfff00000;i++){
263 		temp = b;
264 		b = ((double)(i+i)/x)*b - a;
265 		GET_HIGH_WORD(high,b);
266 		a = temp;
267 	    }
268 	}
269 	if(sign>0) return b; else return -b;
270 }
271