1 /* @(#)e_jn.c 5.1 93/09/24 */
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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
10 * ====================================================
12 * $NetBSD: e_jn.c,v 1.12 2002/05/26 22:01:50 wiz Exp $
13 * $DragonFly: src/lib/libm/src/e_jn.c,v 1.1 2005/07/26 21:15:20 joerg Exp $
18 * floating point Bessel's function of the 1st and 2nd kind
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).
36 * yn(n,x) is similar in all respects, except
37 * that forward recursion is used for all
43 #include "math_private.h"
46 invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
47 two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
48 one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
50 static const double zero = 0.00000000000000000000e+00;
55 int32_t i,hx,ix,lx, sgn;
56 double a, b, temp, di;
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)
63 EXTRACT_WORDS(hx,lx,x);
65 /* if J(n,NaN) is NaN */
66 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
72 if(n==0) return(j0(x));
73 if(n==1) return(j1(x));
74 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
76 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
78 else if((double)n<=x) {
79 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
80 if(ix>=0x52D00000) { /* x > 2**302 */
82 * 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)
84 * Let s=sin(x), c=cos(x),
85 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
87 * n sin(xn)*sqt2 cos(xn)*sqt2
88 * ----------------------------------
95 case 0: temp = cos(x)+sin(x); break;
96 case 1: temp = -cos(x)+sin(x); break;
97 case 2: temp = -cos(x)-sin(x); break;
98 case 3: temp = cos(x)-sin(x); break;
100 b = invsqrtpi*temp/sqrt(x);
106 b = b*((double)(i+i)/x) - a; /* avoid underflow */
111 if(ix<0x3e100000) { /* x < 2**-29 */
112 /* x is tiny, return the first Taylor expansion of J(n,x)
113 * J(n,x) = 1/n!*(x/2)^n - ...
115 if(n>33) /* underflow */
118 temp = x*0.5; b = temp;
119 for (a=one,i=2;i<=n;i++) {
120 a *= (double)i; /* a = n! */
121 b *= temp; /* b = (x/2)^n */
126 /* use backward recurrence */
128 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
129 * 2n - 2(n+1) - 2(n+2)
132 * (for large x) = ---- ------ ------ .....
134 * -- - ------ - ------ -
137 * Let w = 2n/x and h=2/x, then the above quotient
138 * is equal to the continued fraction:
140 * = -----------------------
142 * w - -----------------
147 * To determine how many terms needed, let
148 * Q(0) = w, Q(1) = w(w+h) - 1,
149 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
150 * When Q(k) > 1e4 good for single
151 * When Q(k) > 1e9 good for double
152 * When Q(k) > 1e17 good for quadruple
156 double q0,q1,h,tmp; int32_t k,m;
157 w = (n+n)/(double)x; h = 2.0/(double)x;
158 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
166 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
169 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
170 * Hence, if n*(log(2n/x)) > ...
171 * single 8.8722839355e+01
172 * double 7.09782712893383973096e+02
173 * long double 1.1356523406294143949491931077970765006170e+04
174 * then recurrent value may overflow and the result is
175 * likely underflow to zero
179 tmp = tmp*log(fabs(v*tmp));
180 if(tmp<7.09782712893383973096e+02) {
181 for(i=n-1,di=(double)(i+i);i>0;i--){
189 for(i=n-1,di=(double)(i+i);i>0;i--){
195 /* scale b to avoid spurious overflow */
206 if(sgn==1) return -b; else return b;
217 EXTRACT_WORDS(hx,lx,x);
219 /* if Y(n,NaN) is NaN */
220 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
221 if((ix|lx)==0) return -one/zero;
222 if(hx<0) return zero/zero;
226 sign = 1 - ((n&1)<<1);
228 if(n==0) return(y0(x));
229 if(n==1) return(sign*y1(x));
230 if(ix==0x7ff00000) return zero;
231 if(ix>=0x52D00000) { /* x > 2**302 */
233 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
234 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
235 * Let s=sin(x), c=cos(x),
236 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
238 * n sin(xn)*sqt2 cos(xn)*sqt2
239 * ----------------------------------
246 case 0: temp = sin(x)-cos(x); break;
247 case 1: temp = -sin(x)-cos(x); break;
248 case 2: temp = -sin(x)+cos(x); break;
249 case 3: temp = sin(x)+cos(x); break;
251 b = invsqrtpi*temp/sqrt(x);
256 /* quit if b is -inf */
257 GET_HIGH_WORD(high,b);
258 for(i=1;i<n&&high!=0xfff00000;i++){
260 b = ((double)(i+i)/x)*b - a;
261 GET_HIGH_WORD(high,b);
265 if(sign>0) return b; else return -b;