No longer places function number in subclass box.
[cake.git] / compiler / mlib / e_jn.c
blob057bd045c1f7692e6ac3c42dcfafc14e16a572e4
2 /* @(#)e_jn.c 1.4 95/01/18 */
3 /*
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
7 * Developed at SunSoft, a Sun Microsystems, Inc. business.
8 * Permission to use, copy, modify, and distribute this
9 * software is freely granted, provided that this notice
10 * is preserved.
11 * ====================================================
14 #ifndef lint
15 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_jn.c,v 1.9 2005/02/04 18:26:06 das Exp $";
16 #endif
19 * __ieee754_jn(n, x), __ieee754_yn(n, x)
20 * floating point Bessel's function of the 1st and 2nd kind
21 * of order n
23 * Special cases:
24 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
25 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
26 * Note 2. About jn(n,x), yn(n,x)
27 * For n=0, j0(x) is called,
28 * for n=1, j1(x) is called,
29 * for n<x, forward recursion us used starting
30 * from values of j0(x) and j1(x).
31 * for n>x, a continued fraction approximation to
32 * j(n,x)/j(n-1,x) is evaluated and then backward
33 * recursion is used starting from a supposed value
34 * for j(n,x). The resulting value of j(0,x) is
35 * compared with the actual value to correct the
36 * supposed value of j(n,x).
38 * yn(n,x) is similar in all respects, except
39 * that forward recursion is used for all
40 * values of n>1.
44 #include "math.h"
45 #include "math_private.h"
47 static const double
48 invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
49 two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
50 one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
52 static const double zero = 0.00000000000000000000e+00;
54 double
55 __ieee754_jn(int n, double x)
57 int32_t i,hx,ix,lx, sgn;
58 double a, b, temp, di;
59 double z, w;
61 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
62 * Thus, J(-n,x) = J(n,-x)
64 EXTRACT_WORDS(hx,lx,x);
65 ix = 0x7fffffff&hx;
66 /* if J(n,NaN) is NaN */
67 if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
68 if(n<0){
69 n = -n;
70 x = -x;
71 hx ^= 0x80000000;
73 if(n==0) return(__ieee754_j0(x));
74 if(n==1) return(__ieee754_j1(x));
75 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
76 x = fabs(x);
77 if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
78 b = zero;
79 else if((double)n<=x) {
80 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
81 if(ix>=0x52D00000) { /* x > 2**302 */
82 /* (x >> n**2)
83 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
84 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
85 * Let s=sin(x), c=cos(x),
86 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
88 * n sin(xn)*sqt2 cos(xn)*sqt2
89 * ----------------------------------
90 * 0 s-c c+s
91 * 1 -s-c -c+s
92 * 2 -s+c -c-s
93 * 3 s+c c-s
95 switch(n&3) {
96 case 0: temp = cos(x)+sin(x); break;
97 case 1: temp = -cos(x)+sin(x); break;
98 case 2: temp = -cos(x)-sin(x); break;
99 case 3: temp = cos(x)-sin(x); break;
101 b = invsqrtpi*temp/sqrt(x);
102 } else {
103 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;
111 } else {
112 if(ix<0x3e100000) { /* x < 2**-29 */
113 /* x is tiny, return the first Taylor expansion of J(n,x)
114 * J(n,x) = 1/n!*(x/2)^n - ...
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 */
124 b = b/a;
126 } else {
127 /* use backward recurrence */
128 /* x x^2 x^2
129 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
130 * 2n - 2(n+1) - 2(n+2)
132 * 1 1 1
133 * (for large x) = ---- ------ ------ .....
134 * 2n 2(n+1) 2(n+2)
135 * -- - ------ - ------ -
136 * x x x
138 * Let w = 2n/x and h=2/x, then the above quotient
139 * is equal to the continued fraction:
141 * = -----------------------
143 * w - -----------------
145 * w+h - ---------
146 * w+2h - ...
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),
151 * When Q(k) > 1e4 good for single
152 * When Q(k) > 1e9 good for double
153 * When Q(k) > 1e17 good for quadruple
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;
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
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;
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;
204 b = (t*__ieee754_j0(x)/b);
207 if(sgn==1) return -b; else return b;
210 double
211 __ieee754_yn(int n, double x)
213 int32_t i,hx,ix,lx;
214 int32_t sign;
215 double a, b, temp;
217 EXTRACT_WORDS(hx,lx,x);
218 ix = 0x7fffffff&hx;
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;
223 sign = 1;
224 if(n<0){
225 n = -n;
226 sign = 1 - ((n&1)<<1);
228 if(n==0) return(__ieee754_y0(x));
229 if(n==1) return(sign*__ieee754_y1(x));
230 if(ix==0x7ff00000) return zero;
231 if(ix>=0x52D00000) { /* x > 2**302 */
232 /* (x >> n**2)
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 * ----------------------------------
240 * 0 s-c c+s
241 * 1 -s-c -c+s
242 * 2 -s+c -c-s
243 * 3 s+c c-s
245 switch(n&3) {
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);
252 } else {
253 u_int32_t high;
254 a = __ieee754_y0(x);
255 b = __ieee754_y1(x);
256 /* quit if b is -inf */
257 GET_HIGH_WORD(high,b);
258 for(i=1;i<n&&high!=0xfff00000;i++){
259 temp = b;
260 b = ((double)(i+i)/x)*b - a;
261 GET_HIGH_WORD(high,b);
262 a = temp;
265 if(sign>0) return b; else return -b;