1 /*
2 * ====================================================
3 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
4 *
5 * Developed at SunPro, a Sun Microsystems, Inc. business.
6 * Permission to use, copy, modify, and distribute this
7 * software is freely granted, provided that this notice
8 * is preserved.
9 * ====================================================
10 */
11
12 /*
13 * __ieee754_jn(n, x), __ieee754_yn(n, x)
14 * floating point Bessel's function of the 1st and 2nd kind
15 * of order n
16 *
17 * Special cases:
18 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
19 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
20 * Note 2. About jn(n,x), yn(n,x)
21 * For n=0, j0(x) is called,
22 * for n=1, j1(x) is called,
23 * for n<x, forward recursion us used starting
24 * from values of j0(x) and j1(x).
25 * for n>x, a continued fraction approximation to
26 * j(n,x)/j(n-1,x) is evaluated and then backward
27 * recursion is used starting from a supposed value
28 * for j(n,x). The resulting value of j(0,x) is
29 * compared with the actual value to correct the
30 * supposed value of j(n,x).
31 *
32 * yn(n,x) is similar in all respects, except
33 * that forward recursion is used for all
34 * values of n>1.
35 *
36 */
37
38 #include "math.h"
39 #include "math_private.h"
40
41 static const double
42 invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
43 two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
44 one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
45
46 static const double zero = 0.00000000000000000000e+00;
47
__ieee754_jn(int n,double x)48 double __ieee754_jn(int n, double x)
49 {
50 int32_t i,hx,ix,lx, sgn;
51 double a, b, temp=0, di;
52 double z, w;
53
54 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
55 * Thus, J(-n,x) = J(n,-x)
56 */
57 EXTRACT_WORDS(hx,lx,x);
58 ix = 0x7fffffff&hx;
59 /* if J(n,NaN) is NaN */
60 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
61 if(n<0){
62 n = -n;
63 x = -x;
64 hx ^= 0x80000000;
65 }
66 if(n==0) return(__ieee754_j0(x));
67 if(n==1) return(__ieee754_j1(x));
68 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
69 x = fabs(x);
70 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
71 b = zero;
72 else if((double)n<=x) {
73 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
74 if(ix>=0x52D00000) { /* x > 2**302 */
75 /* (x >> n**2)
76 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
77 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
78 * Let s=sin(x), c=cos(x),
79 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
80 *
81 * n sin(xn)*sqt2 cos(xn)*sqt2
82 * ----------------------------------
83 * 0 s-c c+s
84 * 1 -s-c -c+s
85 * 2 -s+c -c-s
86 * 3 s+c c-s
87 */
88 switch(n&3) {
89 case 0: temp = cos(x)+sin(x); break;
90 case 1: temp = -cos(x)+sin(x); break;
91 case 2: temp = -cos(x)-sin(x); break;
92 case 3: temp = cos(x)-sin(x); break;
93 }
94 b = invsqrtpi*temp/sqrt(x);
95 } else {
96 a = __ieee754_j0(x);
97 b = __ieee754_j1(x);
98 for(i=1;i<n;i++){
99 temp = b;
100 b = b*((double)(i+i)/x) - a; /* avoid underflow */
101 a = temp;
102 }
103 }
104 } else {
105 if(ix<0x3e100000) { /* x < 2**-29 */
106 /* x is tiny, return the first Taylor expansion of J(n,x)
107 * J(n,x) = 1/n!*(x/2)^n - ...
108 */
109 if(n>33) /* underflow */
110 b = zero;
111 else {
112 temp = x*0.5; b = temp;
113 for (a=one,i=2;i<=n;i++) {
114 a *= (double)i; /* a = n! */
115 b *= temp; /* b = (x/2)^n */
116 }
117 b = b/a;
118 }
119 } else {
120 /* use backward recurrence */
121 /* x x^2 x^2
122 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
123 * 2n - 2(n+1) - 2(n+2)
124 *
125 * 1 1 1
126 * (for large x) = ---- ------ ------ .....
127 * 2n 2(n+1) 2(n+2)
128 * -- - ------ - ------ -
129 * x x x
130 *
131 * Let w = 2n/x and h=2/x, then the above quotient
132 * is equal to the continued fraction:
133 * 1
134 * = -----------------------
135 * 1
136 * w - -----------------
137 * 1
138 * w+h - ---------
139 * w+2h - ...
140 *
141 * To determine how many terms needed, let
142 * Q(0) = w, Q(1) = w(w+h) - 1,
143 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
144 * When Q(k) > 1e4 good for single
145 * When Q(k) > 1e9 good for double
146 * When Q(k) > 1e17 good for quadruple
147 */
148 /* determine k */
149 double t,v;
150 double q0,q1,h,tmp; int32_t k,m;
151 w = (n+n)/(double)x; h = 2.0/(double)x;
152 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
153 while(q1<1.0e9) {
154 k += 1; z += h;
155 tmp = z*q1 - q0;
156 q0 = q1;
157 q1 = tmp;
158 }
159 m = n+n;
160 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
161 a = t;
162 b = one;
163 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
164 * Hence, if n*(log(2n/x)) > ...
165 * single 8.8722839355e+01
166 * double 7.09782712893383973096e+02
167 * long double 1.1356523406294143949491931077970765006170e+04
168 * then recurrent value may overflow and the result is
169 * likely underflow to zero
170 */
171 tmp = n;
172 v = two/x;
173 tmp = tmp*__ieee754_log(fabs(v*tmp));
174 if(tmp<7.09782712893383973096e+02) {
175 for(i=n-1,di=(double)(i+i);i>0;i--){
176 temp = b;
177 b *= di;
178 b = b/x - a;
179 a = temp;
180 di -= two;
181 }
182 } else {
183 for(i=n-1,di=(double)(i+i);i>0;i--){
184 temp = b;
185 b *= di;
186 b = b/x - a;
187 a = temp;
188 di -= two;
189 /* scale b to avoid spurious overflow */
190 if(b>1e100) {
191 a /= b;
192 t /= b;
193 b = one;
194 }
195 }
196 }
197 b = (t*__ieee754_j0(x)/b);
198 }
199 }
200 if(sgn==1) return -b; else return b;
201 }
202
__ieee754_yn(int n,double x)203 double __ieee754_yn(int n, double x)
204 {
205 int32_t i,hx,ix,lx;
206 int32_t sign;
207 double a, b, temp=0;
208
209 EXTRACT_WORDS(hx,lx,x);
210 ix = 0x7fffffff&hx;
211 /* if Y(n,NaN) is NaN */
212 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
213 if((ix|lx)==0) return -one/zero;
214 if(hx<0) return zero/zero;
215 sign = 1;
216 if(n<0){
217 n = -n;
218 sign = 1 - ((n&1)<<1);
219 }
220 if(n==0) return(__ieee754_y0(x));
221 if(n==1) return(sign*__ieee754_y1(x));
222 if(ix==0x7ff00000) return zero;
223 if(ix>=0x52D00000) { /* x > 2**302 */
224 /* (x >> n**2)
225 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
226 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
227 * Let s=sin(x), c=cos(x),
228 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
229 *
230 * n sin(xn)*sqt2 cos(xn)*sqt2
231 * ----------------------------------
232 * 0 s-c c+s
233 * 1 -s-c -c+s
234 * 2 -s+c -c-s
235 * 3 s+c c-s
236 */
237 switch(n&3) {
238 case 0: temp = sin(x)-cos(x); break;
239 case 1: temp = -sin(x)-cos(x); break;
240 case 2: temp = -sin(x)+cos(x); break;
241 case 3: temp = sin(x)+cos(x); break;
242 }
243 b = invsqrtpi*temp/sqrt(x);
244 } else {
245 u_int32_t high;
246 a = __ieee754_y0(x);
247 b = __ieee754_y1(x);
248 /* quit if b is -inf */
249 GET_HIGH_WORD(high,b);
250 for(i=1;i<n&&high!=0xfff00000;i++){
251 temp = b;
252 b = ((double)(i+i)/x)*b - a;
253 GET_HIGH_WORD(high,b);
254 a = temp;
255 }
256 }
257 if(sign>0) return b; else return -b;
258 }
259