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 } |
|