1
2 /*
3 * Copyright (c) 1998, 2001, Oracle and/or its affiliates. All rights reserved.
4 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
5 *
6 * This code is free software; you can redistribute it and/or modify it
7 * under the terms of the GNU General Public License version 2 only, as
8 * published by the Free Software Foundation. Oracle designates this
9 * particular file as subject to the "Classpath" exception as provided
10 * by Oracle in the LICENSE file that accompanied this code.
11 *
12 * This code is distributed in the hope that it will be useful, but WITHOUT
13 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
14 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
15 * version 2 for more details (a copy is included in the LICENSE file that
16 * accompanied this code).
17 *
18 * You should have received a copy of the GNU General Public License version
19 * 2 along with this work; if not, write to the Free Software Foundation,
20 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
21 *
22 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
23 * or visit www.oracle.com if you need additional information or have any
24 * questions.
25 */
26
27 /*
28 * __ieee754_jn(n, x), __ieee754_yn(n, x)
29 * floating point Bessel's function of the 1st and 2nd kind
30 * of order n
31 *
32 * Special cases:
33 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
34 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
35 * Note 2. About jn(n,x), yn(n,x)
36 * For n=0, j0(x) is called,
37 * for n=1, j1(x) is called,
38 * for n<x, forward recursion us used starting
39 * from values of j0(x) and j1(x).
40 * for n>x, a continued fraction approximation to
41 * j(n,x)/j(n-1,x) is evaluated and then backward
42 * recursion is used starting from a supposed value
43 * for j(n,x). The resulting value of j(0,x) is
44 * compared with the actual value to correct the
45 * supposed value of j(n,x).
46 *
47 * yn(n,x) is similar in all respects, except
48 * that forward recursion is used for all
49 * values of n>1.
50 *
51 */
52
53 #include "fdlibm.h"
54
55 #ifdef __STDC__
56 static const double
57 #else
58 static double
59 #endif
60 invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
61 two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
62 one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
63
64 static double zero = 0.00000000000000000000e+00;
65
66 #ifdef __STDC__
67 double __ieee754_jn(int n, double x)
68 #else
69 double __ieee754_jn(n,x)
70 int n; double x;
71 #endif
72 {
73 int i,hx,ix,lx, sgn;
74 double a, b, temp = 0, di;
75 double z, w;
76
77 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
78 * Thus, J(-n,x) = J(n,-x)
79 */
80 hx = __HI(x);
81 ix = 0x7fffffff&hx;
82 lx = __LO(x);
83 /* if J(n,NaN) is NaN */
84 if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
85 if(n<0){
86 n = -n;
87 x = -x;
88 hx ^= 0x80000000;
89 }
90 if(n==0) return(__ieee754_j0(x));
91 if(n==1) return(__ieee754_j1(x));
92 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
93 x = fabs(x);
94 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
95 b = zero;
96 else if((double)n<=x) {
97 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
98 if(ix>=0x52D00000) { /* x > 2**302 */
99 /* (x >> n**2)
100 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
101 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
102 * Let s=sin(x), c=cos(x),
103 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
104 *
105 * n sin(xn)*sqt2 cos(xn)*sqt2
106 * ----------------------------------
107 * 0 s-c c+s
108 * 1 -s-c -c+s
109 * 2 -s+c -c-s
110 * 3 s+c c-s
111 */
112 switch(n&3) {
113 case 0: temp = cos(x)+sin(x); break;
114 case 1: temp = -cos(x)+sin(x); break;
115 case 2: temp = -cos(x)-sin(x); break;
116 case 3: temp = cos(x)-sin(x); break;
117 }
118 b = invsqrtpi*temp/sqrt(x);
119 } else {
120 a = __ieee754_j0(x);
121 b = __ieee754_j1(x);
122 for(i=1;i<n;i++){
123 temp = b;
124 b = b*((double)(i+i)/x) - a; /* avoid underflow */
125 a = temp;
126 }
127 }
128 } else {
129 if(ix<0x3e100000) { /* x < 2**-29 */
130 /* x is tiny, return the first Taylor expansion of J(n,x)
131 * J(n,x) = 1/n!*(x/2)^n - ...
132 */
133 if(n>33) /* underflow */
134 b = zero;
135 else {
136 temp = x*0.5; b = temp;
137 for (a=one,i=2;i<=n;i++) {
138 a *= (double)i; /* a = n! */
139 b *= temp; /* b = (x/2)^n */
140 }
141 b = b/a;
142 }
143 } else {
144 /* use backward recurrence */
145 /* x x^2 x^2
146 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
147 * 2n - 2(n+1) - 2(n+2)
148 *
149 * 1 1 1
150 * (for large x) = ---- ------ ------ .....
151 * 2n 2(n+1) 2(n+2)
152 * -- - ------ - ------ -
153 * x x x
154 *
155 * Let w = 2n/x and h=2/x, then the above quotient
156 * is equal to the continued fraction:
157 * 1
158 * = -----------------------
159 * 1
160 * w - -----------------
161 * 1
162 * w+h - ---------
163 * w+2h - ...
164 *
165 * To determine how many terms needed, let
166 * Q(0) = w, Q(1) = w(w+h) - 1,
167 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
168 * When Q(k) > 1e4 good for single
169 * When Q(k) > 1e9 good for double
170 * When Q(k) > 1e17 good for quadruple
171 */
172 /* determine k */
173 double t,v;
174 double q0,q1,h,tmp; int k,m;
175 w = (n+n)/(double)x; h = 2.0/(double)x;
176 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
177 while(q1<1.0e9) {
178 k += 1; z += h;
179 tmp = z*q1 - q0;
180 q0 = q1;
181 q1 = tmp;
182 }
183 m = n+n;
184 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
185 a = t;
186 b = one;
187 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
188 * Hence, if n*(log(2n/x)) > ...
189 * single 8.8722839355e+01
190 * double 7.09782712893383973096e+02
191 * long double 1.1356523406294143949491931077970765006170e+04
192 * then recurrent value may overflow and the result is
193 * likely underflow to zero
194 */
195 tmp = n;
196 v = two/x;
197 tmp = tmp*__ieee754_log(fabs(v*tmp));
198 if(tmp<7.09782712893383973096e+02) {
199 for(i=n-1,di=(double)(i+i);i>0;i--){
200 temp = b;
201 b *= di;
202 b = b/x - a;
203 a = temp;
204 di -= two;
205 }
206 } else {
207 for(i=n-1,di=(double)(i+i);i>0;i--){
208 temp = b;
209 b *= di;
210 b = b/x - a;
211 a = temp;
212 di -= two;
213 /* scale b to avoid spurious overflow */
214 if(b>1e100) {
215 a /= b;
216 t /= b;
217 b = one;
218 }
219 }
220 }
221 b = (t*__ieee754_j0(x)/b);
222 }
223 }
224 if(sgn==1) return -b; else return b;
225 }
226
227 #ifdef __STDC__
228 double __ieee754_yn(int n, double x)
229 #else
230 double __ieee754_yn(n,x)
231 int n; double x;
232 #endif
233 {
234 int i,hx,ix,lx;
235 int sign;
236 double a, b, temp = 0;
237
238 hx = __HI(x);
239 ix = 0x7fffffff&hx;
240 lx = __LO(x);
241 /* if Y(n,NaN) is NaN */
242 if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
243 if((ix|lx)==0) return -one/zero;
244 if(hx<0) return zero/zero;
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 switch(n&3) {
268 case 0: temp = sin(x)-cos(x); break;
269 case 1: temp = -sin(x)-cos(x); break;
270 case 2: temp = -sin(x)+cos(x); break;
271 case 3: temp = sin(x)+cos(x); break;
272 }
273 b = invsqrtpi*temp/sqrt(x);
274 } else {
275 a = __ieee754_y0(x);
276 b = __ieee754_y1(x);
277 /* quit if b is -inf */
278 for(i=1;i<n&&(__HI(b) != 0xfff00000);i++){
279 temp = b;
280 b = ((double)(i+i)/x)*b - a;
281 a = temp;
282 }
283 }
284 if(sign>0) return b; else return -b;
285 }