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The Android Open Source Project1dc9e472009-03-03 19:28:35 -08001/* @(#)e_jn.c 1.4 95/01/18 */
2/*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
Elliott Hughesab528072018-07-24 00:01:52 +00008 * software is freely granted, provided that this notice
The Android Open Source Project1dc9e472009-03-03 19:28:35 -08009 * is preserved.
10 * ====================================================
11 */
12
Elliott Hughesa0ee0782013-01-30 19:06:37 -080013#include <sys/cdefs.h>
Elliott Hughesff49a3c2019-10-24 18:13:32 -070014__FBSDID("$FreeBSD: head/lib/msun/src/e_jn.c 343953 2019-02-10 08:46:07Z peterj $");
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080015
16/*
17 * __ieee754_jn(n, x), __ieee754_yn(n, x)
18 * floating point Bessel's function of the 1st and 2nd kind
19 * of order n
Elliott Hughesab528072018-07-24 00:01:52 +000020 *
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080021 * Special cases:
22 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
23 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
24 * Note 2. About jn(n,x), yn(n,x)
25 * For n=0, j0(x) is called,
26 * for n=1, j1(x) is called,
27 * for n<x, forward recursion us used starting
28 * from values of j0(x) and j1(x).
29 * for n>x, a continued fraction approximation to
30 * j(n,x)/j(n-1,x) is evaluated and then backward
31 * recursion is used starting from a supposed value
32 * for j(n,x). The resulting value of j(0,x) is
33 * compared with the actual value to correct the
34 * supposed value of j(n,x).
35 *
36 * yn(n,x) is similar in all respects, except
37 * that forward recursion is used for all
38 * values of n>1.
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080039 */
40
41#include "math.h"
42#include "math_private.h"
43
Elliott Hughes8cff2f92015-08-28 20:21:43 -070044static const volatile double vone = 1, vzero = 0;
45
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080046static const double
47invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
48two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
49one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
50
51static const double zero = 0.00000000000000000000e+00;
52
53double
54__ieee754_jn(int n, double x)
55{
56 int32_t i,hx,ix,lx, sgn;
Elliott Hughesff49a3c2019-10-24 18:13:32 -070057 double a, b, c, s, temp, di;
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080058 double z, w;
59
60 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
61 * Thus, J(-n,x) = J(n,-x)
62 */
63 EXTRACT_WORDS(hx,lx,x);
64 ix = 0x7fffffff&hx;
65 /* if J(n,NaN) is NaN */
66 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
Elliott Hughesab528072018-07-24 00:01:52 +000067 if(n<0){
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080068 n = -n;
69 x = -x;
70 hx ^= 0x80000000;
71 }
72 if(n==0) return(__ieee754_j0(x));
73 if(n==1) return(__ieee754_j1(x));
74 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
75 x = fabs(x);
76 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
77 b = zero;
Elliott Hughesab528072018-07-24 00:01:52 +000078 else if((double)n<=x) {
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080079 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
80 if(ix>=0x52D00000) { /* x > 2**302 */
Elliott Hughesab528072018-07-24 00:01:52 +000081 /* (x >> n**2)
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080082 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
83 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
Elliott Hughesab528072018-07-24 00:01:52 +000084 * Let s=sin(x), c=cos(x),
85 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2), then
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080086 *
87 * n sin(xn)*sqt2 cos(xn)*sqt2
88 * ----------------------------------
89 * 0 s-c c+s
90 * 1 -s-c -c+s
91 * 2 -s+c -c-s
92 * 3 s+c c-s
93 */
Elliott Hughesff49a3c2019-10-24 18:13:32 -070094 sincos(x, &s, &c);
The Android Open Source Project1dc9e472009-03-03 19:28:35 -080095 switch(n&3) {
Elliott Hughesff49a3c2019-10-24 18:13:32 -070096 case 0: temp = c+s; break;
97 case 1: temp = -c+s; break;
98 case 2: temp = -c-s; break;
99 case 3: temp = c-s; break;
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800100 }
101 b = invsqrtpi*temp/sqrt(x);
Elliott Hughesab528072018-07-24 00:01:52 +0000102 } else {
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800103 a = __ieee754_j0(x);
104 b = __ieee754_j1(x);
105 for(i=1;i<n;i++){
106 temp = b;
107 b = b*((double)(i+i)/x) - a; /* avoid underflow */
108 a = temp;
109 }
110 }
111 } else {
112 if(ix<0x3e100000) { /* x < 2**-29 */
Elliott Hughesab528072018-07-24 00:01:52 +0000113 /* x is tiny, return the first Taylor expansion of J(n,x)
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800114 * J(n,x) = 1/n!*(x/2)^n - ...
115 */
116 if(n>33) /* underflow */
117 b = zero;
118 else {
119 temp = x*0.5; b = temp;
120 for (a=one,i=2;i<=n;i++) {
121 a *= (double)i; /* a = n! */
122 b *= temp; /* b = (x/2)^n */
123 }
124 b = b/a;
125 }
126 } else {
127 /* use backward recurrence */
Elliott Hughesab528072018-07-24 00:01:52 +0000128 /* x x^2 x^2
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800129 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
130 * 2n - 2(n+1) - 2(n+2)
131 *
Elliott Hughesab528072018-07-24 00:01:52 +0000132 * 1 1 1
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800133 * (for large x) = ---- ------ ------ .....
134 * 2n 2(n+1) 2(n+2)
Elliott Hughesab528072018-07-24 00:01:52 +0000135 * -- - ------ - ------ -
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800136 * x x x
137 *
138 * Let w = 2n/x and h=2/x, then the above quotient
139 * is equal to the continued fraction:
140 * 1
141 * = -----------------------
142 * 1
143 * w - -----------------
144 * 1
145 * w+h - ---------
146 * w+2h - ...
147 *
148 * To determine how many terms needed, let
149 * Q(0) = w, Q(1) = w(w+h) - 1,
150 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
Elliott Hughesab528072018-07-24 00:01:52 +0000151 * When Q(k) > 1e4 good for single
152 * When Q(k) > 1e9 good for double
153 * When Q(k) > 1e17 good for quadruple
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800154 */
155 /* determine k */
156 double t,v;
157 double q0,q1,h,tmp; int32_t k,m;
158 w = (n+n)/(double)x; h = 2.0/(double)x;
159 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
160 while(q1<1.0e9) {
161 k += 1; z += h;
162 tmp = z*q1 - q0;
163 q0 = q1;
164 q1 = tmp;
165 }
166 m = n+n;
167 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
168 a = t;
169 b = one;
170 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
171 * Hence, if n*(log(2n/x)) > ...
172 * single 8.8722839355e+01
173 * double 7.09782712893383973096e+02
174 * long double 1.1356523406294143949491931077970765006170e+04
175 * then recurrent value may overflow and the result is
176 * likely underflow to zero
177 */
178 tmp = n;
179 v = two/x;
180 tmp = tmp*__ieee754_log(fabs(v*tmp));
181 if(tmp<7.09782712893383973096e+02) {
182 for(i=n-1,di=(double)(i+i);i>0;i--){
183 temp = b;
184 b *= di;
185 b = b/x - a;
186 a = temp;
187 di -= two;
188 }
189 } else {
190 for(i=n-1,di=(double)(i+i);i>0;i--){
191 temp = b;
192 b *= di;
193 b = b/x - a;
194 a = temp;
195 di -= two;
196 /* scale b to avoid spurious overflow */
197 if(b>1e100) {
198 a /= b;
199 t /= b;
200 b = one;
201 }
202 }
203 }
Elliott Hughesa0ee0782013-01-30 19:06:37 -0800204 z = __ieee754_j0(x);
205 w = __ieee754_j1(x);
206 if (fabs(z) >= fabs(w))
207 b = (t*z/b);
208 else
209 b = (t*w/a);
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800210 }
211 }
212 if(sgn==1) return -b; else return b;
213}
214
215double
216__ieee754_yn(int n, double x)
217{
218 int32_t i,hx,ix,lx;
219 int32_t sign;
Elliott Hughesff49a3c2019-10-24 18:13:32 -0700220 double a, b, c, s, temp;
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800221
222 EXTRACT_WORDS(hx,lx,x);
223 ix = 0x7fffffff&hx;
Elliott Hughes8cff2f92015-08-28 20:21:43 -0700224 /* yn(n,NaN) = NaN */
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800225 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
Elliott Hughes8cff2f92015-08-28 20:21:43 -0700226 /* yn(n,+-0) = -inf and raise divide-by-zero exception. */
227 if((ix|lx)==0) return -one/vzero;
228 /* yn(n,x<0) = NaN and raise invalid exception. */
229 if(hx<0) return vzero/vzero;
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800230 sign = 1;
231 if(n<0){
232 n = -n;
233 sign = 1 - ((n&1)<<1);
234 }
235 if(n==0) return(__ieee754_y0(x));
236 if(n==1) return(sign*__ieee754_y1(x));
237 if(ix==0x7ff00000) return zero;
238 if(ix>=0x52D00000) { /* x > 2**302 */
Elliott Hughesab528072018-07-24 00:01:52 +0000239 /* (x >> n**2)
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800240 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
241 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
Elliott Hughesab528072018-07-24 00:01:52 +0000242 * Let s=sin(x), c=cos(x),
243 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2), then
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800244 *
245 * n sin(xn)*sqt2 cos(xn)*sqt2
246 * ----------------------------------
247 * 0 s-c c+s
248 * 1 -s-c -c+s
249 * 2 -s+c -c-s
250 * 3 s+c c-s
251 */
Elliott Hughesff49a3c2019-10-24 18:13:32 -0700252 sincos(x, &s, &c);
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800253 switch(n&3) {
Elliott Hughesff49a3c2019-10-24 18:13:32 -0700254 case 0: temp = s-c; break;
255 case 1: temp = -s-c; break;
256 case 2: temp = -s+c; break;
257 case 3: temp = s+c; break;
The Android Open Source Project1dc9e472009-03-03 19:28:35 -0800258 }
259 b = invsqrtpi*temp/sqrt(x);
260 } else {
261 u_int32_t high;
262 a = __ieee754_y0(x);
263 b = __ieee754_y1(x);
264 /* quit if b is -inf */
265 GET_HIGH_WORD(high,b);
266 for(i=1;i<n&&high!=0xfff00000;i++){
267 temp = b;
268 b = ((double)(i+i)/x)*b - a;
269 GET_HIGH_WORD(high,b);
270 a = temp;
271 }
272 }
273 if(sign>0) return b; else return -b;
274}