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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 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#define __ieee754_j0 j0
47#define __ieee754_y0 y0
48#define __ieee754_j1 j1
49#define __ieee754_y1 y1
50#define __ieee754_jn jn
51#define __ieee754_yn yn
52#define __ieee754_log log
53
54#ifdef __STDC__
55static const double
56#else
57static double
58#endif
59invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
60two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
61one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
62
63#ifdef __STDC__
64static const double zero = 0.00000000000000000000e+00;
65#else
66static double zero = 0.00000000000000000000e+00;
67#endif
68
69#ifdef __STDC__
70 double __ieee754_jn(int n, double x)
71#else
72 double __ieee754_jn(n,x)
73 int n; double x;
74#endif
75{
76 int32_t i,hx,ix,lx, sgn;
77 double a, b, temp, di;
78 double z, w;
79
80 temp = 0.0;
81 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
82 * Thus, J(-n,x) = J(n,-x)
83 */
84 EXTRACT_WORDS(hx,lx,x);
85 ix = 0x7fffffff&hx;
86 /* if J(n,NaN) is NaN */
87 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
88 if(n<0){
89 n = -n;
90 x = -x;
91 hx ^= 0x80000000;
92 }
93 if(n==0) return(__ieee754_j0(x));
94 if(n==1) return(__ieee754_j1(x));
95 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
96 x = fabs(x);
97 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
98 b = zero;
99 else if((double)n<=x) {
100 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
101 if(ix>=0x52D00000) { /* x > 2**302 */
102 /* (x >> n**2)
103 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
104 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 * Let s=sin(x), c=cos(x),
106 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
107 *
108 * n sin(xn)*sqt2 cos(xn)*sqt2
109 * ----------------------------------
110 * 0 s-c c+s
111 * 1 -s-c -c+s
112 * 2 -s+c -c-s
113 * 3 s+c c-s
114 */
115 switch(n&3) {
116 case 0: temp = cos(x)+sin(x); break;
117 case 1: temp = -cos(x)+sin(x); break;
118 case 2: temp = -cos(x)-sin(x); break;
119 case 3: temp = cos(x)-sin(x); break;
120 }
121 b = invsqrtpi*temp/sqrt(x);
122 } else {
123 a = __ieee754_j0(x);
124 b = __ieee754_j1(x);
125 for(i=1;i<n;i++){
126 temp = b;
127 b = b*((double)(i+i)/x) - a; /* avoid underflow */
128 a = temp;
129 }
130 }
131 } else {
132 if(ix<0x3e100000) { /* x < 2**-29 */
133 /* x is tiny, return the first Taylor expansion of J(n,x)
134 * J(n,x) = 1/n!*(x/2)^n - ...
135 */
136 if(n>33) /* underflow */
137 b = zero;
138 else {
139 temp = x*0.5; b = temp;
140 for (a=one,i=2;i<=n;i++) {
141 a *= (double)i; /* a = n! */
142 b *= temp; /* b = (x/2)^n */
143 }
144 b = b/a;
145 }
146 } else {
147 /* use backward recurrence */
148 /* x x^2 x^2
149 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
150 * 2n - 2(n+1) - 2(n+2)
151 *
152 * 1 1 1
153 * (for large x) = ---- ------ ------ .....
154 * 2n 2(n+1) 2(n+2)
155 * -- - ------ - ------ -
156 * x x x
157 *
158 * Let w = 2n/x and h=2/x, then the above quotient
159 * is equal to the continued fraction:
160 * 1
161 * = -----------------------
162 * 1
163 * w - -----------------
164 * 1
165 * w+h - ---------
166 * w+2h - ...
167 *
168 * To determine how many terms needed, let
169 * Q(0) = w, Q(1) = w(w+h) - 1,
170 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
171 * When Q(k) > 1e4 good for single
172 * When Q(k) > 1e9 good for double
173 * When Q(k) > 1e17 good for quadruple
174 */
175 /* determine k */
176 double t,v;
177 double q0,q1,h,tmp; int32_t k,m;
178 w = (n+n)/(double)x; h = 2.0/(double)x;
179 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
180 while(q1<1.0e9) {
181 k += 1; z += h;
182 tmp = z*q1 - q0;
183 q0 = q1;
184 q1 = tmp;
185 }
186 m = n+n;
187 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
188 a = t;
189 b = one;
190 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
191 * Hence, if n*(log(2n/x)) > ...
192 * single 8.8722839355e+01
193 * double 7.09782712893383973096e+02
194 * long double 1.1356523406294143949491931077970765006170e+04
195 * then recurrent value may overflow and the result is
196 * likely underflow to zero
197 */
198 tmp = n;
199 v = two/x;
200 tmp = tmp*__ieee754_log(fabs(v*tmp));
201 if(tmp<7.09782712893383973096e+02) {
202 for(i=n-1,di=(double)(i+i);i>0;i--){
203 temp = b;
204 b *= di;
205 b = b/x - a;
206 a = temp;
207 di -= two;
208 }
209 } else {
210 for(i=n-1,di=(double)(i+i);i>0;i--){
211 temp = b;
212 b *= di;
213 b = b/x - a;
214 a = temp;
215 di -= two;
216 /* scale b to avoid spurious overflow */
217 if(b>1e100) {
218 a /= b;
219 t /= b;
220 b = one;
221 }
222 }
223 }
224 b = (t*__ieee754_j0(x)/b);
225 }
226 }
227 if(sgn==1) return -b; else return b;
228}
229
230#ifdef __STDC__
231 double __ieee754_yn(int n, double x)
232#else
233 double __ieee754_yn(n,x)
234 int n; double x;
235#endif
236{
237 int32_t i,hx,ix,lx;
238 int32_t sign;
239 double a, b, temp;
240
241 temp = 0.0;
242 EXTRACT_WORDS(hx,lx,x);
243 ix = 0x7fffffff&hx;
244 /* if Y(n,NaN) is NaN */
245 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
246 if((ix|lx)==0) return -one/zero;
247 if(hx<0) return zero/zero;
248 sign = 1;
249 if(n<0){
250 n = -n;
251 sign = 1 - ((n&1)<<1);
252 }
253 if(n==0) return(__ieee754_y0(x));
254 if(n==1) return(sign*__ieee754_y1(x));
255 if(ix==0x7ff00000) return zero;
256 if(ix>=0x52D00000) { /* x > 2**302 */
257 /* (x >> n**2)
258 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
259 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
260 * Let s=sin(x), c=cos(x),
261 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
262 *
263 * n sin(xn)*sqt2 cos(xn)*sqt2
264 * ----------------------------------
265 * 0 s-c c+s
266 * 1 -s-c -c+s
267 * 2 -s+c -c-s
268 * 3 s+c c-s
269 */
270 switch(n&3) {
271 case 0: temp = sin(x)-cos(x); break;
272 case 1: temp = -sin(x)-cos(x); break;
273 case 2: temp = -sin(x)+cos(x); break;
274 case 3: temp = sin(x)+cos(x); break;
275 }
276 b = invsqrtpi*temp/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}