xref: /dragonfly/contrib/openbsd_libm/src/e_jn.c (revision 9348a738)
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 /*
14  * jn(n, x), yn(n, x)
15  * floating point Bessel's function of the 1st and 2nd kind
16  * of order n
17  *
18  * Special cases:
19  *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
20  *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
21  * Note 2. About jn(n,x), yn(n,x)
22  *	For n=0, j0(x) is called,
23  *	for n=1, j1(x) is called,
24  *	for n<x, forward recursion us used starting
25  *	from values of j0(x) and j1(x).
26  *	for n>x, a continued fraction approximation to
27  *	j(n,x)/j(n-1,x) is evaluated and then backward
28  *	recursion is used starting from a supposed value
29  *	for j(n,x). The resulting value of j(0,x) is
30  *	compared with the actual value to correct the
31  *	supposed value of j(n,x).
32  *
33  *	yn(n,x) is similar in all respects, except
34  *	that forward recursion is used for all
35  *	values of n>1.
36  *
37  */
38 
39 #include "math.h"
40 #include "math_private.h"
41 
42 static const double
43 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
44 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
45 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
46 
47 static const double zero  =  0.00000000000000000000e+00;
48 
49 double
50 jn(int n, double x)
51 {
52 	int32_t i,hx,ix,lx, sgn;
53 	double a, b, temp, di;
54 	double z, w;
55 
56     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
57      * Thus, J(-n,x) = J(n,-x)
58      */
59 	EXTRACT_WORDS(hx,lx,x);
60 	ix = 0x7fffffff&hx;
61     /* if J(n,NaN) is NaN */
62 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
63 	if(n<0){
64 		n = -n;
65 		x = -x;
66 		hx ^= 0x80000000;
67 	}
68 	if(n==0) return(j0(x));
69 	if(n==1) return(j1(x));
70 	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
71 	x = fabs(x);
72 	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
73 	    b = zero;
74 	else if((double)n<=x) {
75 		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
76 	    if(ix>=0x52D00000) { /* x > 2**302 */
77     /* (x >> n**2)
78      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
79      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
80      *	    Let s=sin(x), c=cos(x),
81      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
82      *
83      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
84      *		----------------------------------
85      *		   0	 s-c		 c+s
86      *		   1	-s-c 		-c+s
87      *		   2	-s+c		-c-s
88      *		   3	 s+c		 c-s
89      */
90 		switch(n&3) {
91 		    case 0: temp =  cos(x)+sin(x); break;
92 		    case 1: temp = -cos(x)+sin(x); break;
93 		    case 2: temp = -cos(x)-sin(x); break;
94 		    case 3: temp =  cos(x)-sin(x); break;
95 		}
96 		b = invsqrtpi*temp/sqrt(x);
97 	    } else {
98 	        a = j0(x);
99 	        b = j1(x);
100 	        for(i=1;i<n;i++){
101 		    temp = b;
102 		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
103 		    a = temp;
104 	        }
105 	    }
106 	} else {
107 	    if(ix<0x3e100000) {	/* x < 2**-29 */
108     /* x is tiny, return the first Taylor expansion of J(n,x)
109      * J(n,x) = 1/n!*(x/2)^n  - ...
110      */
111 		if(n>33)	/* underflow */
112 		    b = zero;
113 		else {
114 		    temp = x*0.5; b = temp;
115 		    for (a=one,i=2;i<=n;i++) {
116 			a *= (double)i;		/* a = n! */
117 			b *= temp;		/* b = (x/2)^n */
118 		    }
119 		    b = b/a;
120 		}
121 	    } else {
122 		/* use backward recurrence */
123 		/* 			x      x^2      x^2
124 		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
125 		 *			2n  - 2(n+1) - 2(n+2)
126 		 *
127 		 * 			1      1        1
128 		 *  (for large x)   =  ----  ------   ------   .....
129 		 *			2n   2(n+1)   2(n+2)
130 		 *			-- - ------ - ------ -
131 		 *			 x     x         x
132 		 *
133 		 * Let w = 2n/x and h=2/x, then the above quotient
134 		 * is equal to the continued fraction:
135 		 *		    1
136 		 *	= -----------------------
137 		 *		       1
138 		 *	   w - -----------------
139 		 *			  1
140 		 * 	        w+h - ---------
141 		 *		       w+2h - ...
142 		 *
143 		 * To determine how many terms needed, let
144 		 * Q(0) = w, Q(1) = w(w+h) - 1,
145 		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
146 		 * When Q(k) > 1e4	good for single
147 		 * When Q(k) > 1e9	good for double
148 		 * When Q(k) > 1e17	good for quadruple
149 		 */
150 	    /* determine k */
151 		double t,v;
152 		double q0,q1,h,tmp; int32_t k,m;
153 		w  = (n+n)/(double)x; h = 2.0/(double)x;
154 		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
155 		while(q1<1.0e9) {
156 			k += 1; z += h;
157 			tmp = z*q1 - q0;
158 			q0 = q1;
159 			q1 = tmp;
160 		}
161 		m = n+n;
162 		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
163 		a = t;
164 		b = one;
165 		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
166 		 *  Hence, if n*(log(2n/x)) > ...
167 		 *  single 8.8722839355e+01
168 		 *  double 7.09782712893383973096e+02
169 		 *  long double 1.1356523406294143949491931077970765006170e+04
170 		 *  then recurrent value may overflow and the result is
171 		 *  likely underflow to zero
172 		 */
173 		tmp = n;
174 		v = two/x;
175 		tmp = tmp*log(fabs(v*tmp));
176 		if(tmp<7.09782712893383973096e+02) {
177 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
178 		        temp = b;
179 			b *= di;
180 			b  = b/x - a;
181 		        a = temp;
182 			di -= two;
183 	     	    }
184 		} else {
185 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
186 		        temp = b;
187 			b *= di;
188 			b  = b/x - a;
189 		        a = temp;
190 			di -= two;
191 		    /* scale b to avoid spurious overflow */
192 			if(b>1e100) {
193 			    a /= b;
194 			    t /= b;
195 			    b  = one;
196 			}
197 	     	    }
198 		}
199 	    	b = (t*j0(x)/b);
200 	    }
201 	}
202 	if(sgn==1) return -b; else return b;
203 }
204 
205 double
206 yn(int n, double x)
207 {
208 	int32_t i,hx,ix,lx;
209 	int32_t sign;
210 	double a, b, temp;
211 
212 	EXTRACT_WORDS(hx,lx,x);
213 	ix = 0x7fffffff&hx;
214     /* if Y(n,NaN) is NaN */
215 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
216 	if((ix|lx)==0) return -one/zero;
217 	if(hx<0) return zero/zero;
218 	sign = 1;
219 	if(n<0){
220 		n = -n;
221 		sign = 1 - ((n&1)<<1);
222 	}
223 	if(n==0) return(y0(x));
224 	if(n==1) return(sign*y1(x));
225 	if(ix==0x7ff00000) return zero;
226 	if(ix>=0x52D00000) { /* x > 2**302 */
227     /* (x >> n**2)
228      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
229      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
230      *	    Let s=sin(x), c=cos(x),
231      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
232      *
233      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
234      *		----------------------------------
235      *		   0	 s-c		 c+s
236      *		   1	-s-c 		-c+s
237      *		   2	-s+c		-c-s
238      *		   3	 s+c		 c-s
239      */
240 		switch(n&3) {
241 		    case 0: temp =  sin(x)-cos(x); break;
242 		    case 1: temp = -sin(x)-cos(x); break;
243 		    case 2: temp = -sin(x)+cos(x); break;
244 		    case 3: temp =  sin(x)+cos(x); break;
245 		}
246 		b = invsqrtpi*temp/sqrt(x);
247 	} else {
248 	    u_int32_t high;
249 	    a = y0(x);
250 	    b = y1(x);
251 	/* quit if b is -inf */
252 	    GET_HIGH_WORD(high,b);
253 	    for(i=1;i<n&&high!=0xfff00000;i++){
254 		temp = b;
255 		b = ((double)(i+i)/x)*b - a;
256 		GET_HIGH_WORD(high,b);
257 		a = temp;
258 	    }
259 	}
260 	if(sign>0) return b; else return -b;
261 }
262