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 * ====================================================
13 #if defined(LIBM_SCCS) && !defined(lint)
14 static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
18 * __ieee754_jn(n, x), __ieee754_yn(n, x)
19 * floating point Bessel's function of the 1st and 2nd kind
23 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
24 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
25 * Note 2. About jn(n,x), yn(n,x)
26 * For n=0, j0(x) is called,
27 * for n=1, j1(x) is called,
28 * for n<x, forward recursion us used starting
29 * from values of j0(x) and j1(x).
30 * for n>x, a continued fraction approximation to
31 * j(n,x)/j(n-1,x) is evaluated and then backward
32 * recursion is used starting from a supposed value
33 * for j(n,x). The resulting value of j(0,x) is
34 * compared with the actual value to correct the
35 * supposed value of j(n,x).
37 * yn(n,x) is similar in all respects, except
38 * that forward recursion is used for all
44 #include "math_private.h"
52 invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
53 two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
54 one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
57 static const double zero = 0.00000000000000000000e+00;
59 static double zero = 0.00000000000000000000e+00;
63 double attribute_hidden __ieee754_jn(int n, double x)
65 double attribute_hidden __ieee754_jn(n,x)
69 int32_t i,hx,ix,lx, sgn;
70 double a, b, temp=0, di;
73 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
74 * Thus, J(-n,x) = J(n,-x)
76 EXTRACT_WORDS(hx,lx,x);
78 /* if J(n,NaN) is NaN */
79 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
85 if(n==0) return(__ieee754_j0(x));
86 if(n==1) return(__ieee754_j1(x));
87 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
89 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
91 else if((double)n<=x) {
92 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
93 if(ix>=0x52D00000) { /* x > 2**302 */
95 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
96 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
97 * Let s=sin(x), c=cos(x),
98 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
100 * n sin(xn)*sqt2 cos(xn)*sqt2
101 * ----------------------------------
108 case 0: temp = cos(x)+sin(x); break;
109 case 1: temp = -cos(x)+sin(x); break;
110 case 2: temp = -cos(x)-sin(x); break;
111 case 3: temp = cos(x)-sin(x); break;
113 b = invsqrtpi*temp/sqrt(x);
119 b = b*((double)(i+i)/x) - a; /* avoid underflow */
124 if(ix<0x3e100000) { /* x < 2**-29 */
125 /* x is tiny, return the first Taylor expansion of J(n,x)
126 * J(n,x) = 1/n!*(x/2)^n - ...
128 if(n>33) /* underflow */
131 temp = x*0.5; b = temp;
132 for (a=one,i=2;i<=n;i++) {
133 a *= (double)i; /* a = n! */
134 b *= temp; /* b = (x/2)^n */
139 /* use backward recurrence */
141 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
142 * 2n - 2(n+1) - 2(n+2)
145 * (for large x) = ---- ------ ------ .....
147 * -- - ------ - ------ -
150 * Let w = 2n/x and h=2/x, then the above quotient
151 * is equal to the continued fraction:
153 * = -----------------------
155 * w - -----------------
160 * To determine how many terms needed, let
161 * Q(0) = w, Q(1) = w(w+h) - 1,
162 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
163 * When Q(k) > 1e4 good for single
164 * When Q(k) > 1e9 good for double
165 * When Q(k) > 1e17 good for quadruple
169 double q0,q1,h,tmp; int32_t k,m;
170 w = (n+n)/(double)x; h = 2.0/(double)x;
171 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
179 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
182 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
183 * Hence, if n*(log(2n/x)) > ...
184 * single 8.8722839355e+01
185 * double 7.09782712893383973096e+02
186 * long double 1.1356523406294143949491931077970765006170e+04
187 * then recurrent value may overflow and the result is
188 * likely underflow to zero
192 tmp = tmp*__ieee754_log(fabs(v*tmp));
193 if(tmp<7.09782712893383973096e+02) {
194 for(i=n-1,di=(double)(i+i);i>0;i--){
202 for(i=n-1,di=(double)(i+i);i>0;i--){
208 /* scale b to avoid spurious overflow */
216 b = (t*__ieee754_j0(x)/b);
219 if(sgn==1) return -b; else return b;
223 double attribute_hidden __ieee754_yn(int n, double x)
225 double attribute_hidden __ieee754_yn(n,x)
233 EXTRACT_WORDS(hx,lx,x);
235 /* if Y(n,NaN) is NaN */
236 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
237 if((ix|lx)==0) return -one/zero;
238 if(hx<0) return zero/zero;
242 sign = 1 - ((n&1)<<1);
244 if(n==0) return(__ieee754_y0(x));
245 if(n==1) return(sign*__ieee754_y1(x));
246 if(ix==0x7ff00000) return zero;
247 if(ix>=0x52D00000) { /* x > 2**302 */
249 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
250 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
251 * Let s=sin(x), c=cos(x),
252 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
254 * n sin(xn)*sqt2 cos(xn)*sqt2
255 * ----------------------------------
262 case 0: temp = sin(x)-cos(x); break;
263 case 1: temp = -sin(x)-cos(x); break;
264 case 2: temp = -sin(x)+cos(x); break;
265 case 3: temp = sin(x)+cos(x); break;
267 b = invsqrtpi*temp/sqrt(x);
272 /* quit if b is -inf */
273 GET_HIGH_WORD(high,b);
274 for(i=1;i<n&&high!=0xfff00000;i++){
276 b = ((double)(i+i)/x)*b - a;
277 GET_HIGH_WORD(high,b);
281 if(sign>0) return b; else return -b;