Subversion Repositories Kolibri OS

Rev

Rev 4872 | Details | Compare with Previous | Last modification | View Log | RSS feed

Rev Author Line No. Line
4349 Serge 1
 
2
/*
3
 * ====================================================
4
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5
 *
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
9
 * is preserved.
10
 * ====================================================
11
 */
12
13
 
14
 * __ieee754_jn(n, x), __ieee754_yn(n, x)
15
 * floating point Bessel's function of the 1st and 2nd kind
16
 * of order n
17
 *
18
 * Special cases:
19
 *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
20
 *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
21
 * Note 2. About jn(n,x), yn(n,x)
22
 *	For n=0, j0(x) is called,
23
 *	for n=1, j1(x) is called,
24
 *	for n
25
 *	from values of j0(x) and j1(x).
26
 *	for n>x, a continued fraction approximation to
27
 *	j(n,x)/j(n-1,x) is evaluated and then backward
28
 *	recursion is used starting from a supposed value
29
 *	for j(n,x). The resulting value of j(0,x) is
30
 *	compared with the actual value to correct the
31
 *	supposed value of j(n,x).
32
 *
33
 *	yn(n,x) is similar in all respects, except
34
 *	that forward recursion is used for all
35
 *	values of n>1.
36
 *
37
 */
38
39
 
40
41
 
42
43
 
44
static const double
45
#else
46
static double
47
#endif
48
invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
49
two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
50
one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
51
52
 
53
static const double zero  =  0.00000000000000000000e+00;
54
#else
55
static double zero  =  0.00000000000000000000e+00;
56
#endif
57
58
 
59
	double __ieee754_jn(int n, double x)
60
#else
61
	double __ieee754_jn(n,x)
62
	int n; double x;
63
#endif
64
{
65
	__int32_t i,hx,ix,lx, sgn;
66
	double a, b, temp, di;
67
	double z, w;
68
69
 
70
     * Thus, J(-n,x) = J(n,-x)
71
     */
72
	EXTRACT_WORDS(hx,lx,x);
73
	ix = 0x7fffffff&hx;
74
    /* if J(n,NaN) is NaN */
75
	if((ix|((__uint32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
76
	if(n<0){
77
		n = -n;
78
		x = -x;
79
		hx ^= 0x80000000;
80
	}
81
	if(n==0) return(__ieee754_j0(x));
82
	if(n==1) return(__ieee754_j1(x));
83
	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
84
	x = fabs(x);
85
	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
86
	    b = zero;
87
	else if((double)n<=x) {
88
		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
89
	    if(ix>=0x52D00000) { /* x > 2**302 */
90
    /* (x >> n**2)
91
     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
92
     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
93
     *	    Let s=sin(x), c=cos(x),
94
     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
95
     *
96
     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
97
     *		----------------------------------
98
     *		   0	 s-c		 c+s
99
     *		   1	-s-c 		-c+s
100
     *		   2	-s+c		-c-s
101
     *		   3	 s+c		 c-s
102
     */
103
		switch(n&3) {
104
		    case 0: temp =  cos(x)+sin(x); break;
105
		    case 1: temp = -cos(x)+sin(x); break;
106
		    case 2: temp = -cos(x)-sin(x); break;
107
		    case 3: temp =  cos(x)-sin(x); break;
108
		}
109
		b = invsqrtpi*temp/__ieee754_sqrt(x);
110
	    } else {
111
	        a = __ieee754_j0(x);
112
	        b = __ieee754_j1(x);
113
	        for(i=1;i
114
		    temp = b;
115
		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
116
		    a = temp;
117
	        }
118
	    }
119
	} else {
120
	    if(ix<0x3e100000) {	/* x < 2**-29 */
121
    /* x is tiny, return the first Taylor expansion of J(n,x)
122
     * J(n,x) = 1/n!*(x/2)^n  - ...
123
     */
124
		if(n>33)	/* underflow */
125
		    b = zero;
126
		else {
127
		    temp = x*0.5; b = temp;
128
		    for (a=one,i=2;i<=n;i++) {
129
			a *= (double)i;		/* a = n! */
130
			b *= temp;		/* b = (x/2)^n */
131
		    }
132
		    b = b/a;
133
		}
134
	    } else {
135
		/* use backward recurrence */
136
		/* 			x      x^2      x^2
137
		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
138
		 *			2n  - 2(n+1) - 2(n+2)
139
		 *
140
		 * 			1      1        1
141
		 *  (for large x)   =  ----  ------   ------   .....
142
		 *			2n   2(n+1)   2(n+2)
143
		 *			-- - ------ - ------ -
144
		 *			 x     x         x
145
		 *
146
		 * Let w = 2n/x and h=2/x, then the above quotient
147
		 * is equal to the continued fraction:
148
		 *		    1
149
		 *	= -----------------------
150
		 *		       1
151
		 *	   w - -----------------
152
		 *			  1
153
		 * 	        w+h - ---------
154
		 *		       w+2h - ...
155
		 *
156
		 * To determine how many terms needed, let
157
		 * Q(0) = w, Q(1) = w(w+h) - 1,
158
		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
159
		 * When Q(k) > 1e4	good for single
160
		 * When Q(k) > 1e9	good for double
161
		 * When Q(k) > 1e17	good for quadruple
162
		 */
163
	    /* determine k */
164
		double t,v;
165
		double q0,q1,h,tmp; __int32_t k,m;
166
		w  = (n+n)/(double)x; h = 2.0/(double)x;
167
		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
168
		while(q1<1.0e9) {
169
			k += 1; z += h;
170
			tmp = z*q1 - q0;
171
			q0 = q1;
172
			q1 = tmp;
173
		}
174
		m = n+n;
175
		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
176
		a = t;
177
		b = one;
178
		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
179
		 *  Hence, if n*(log(2n/x)) > ...
180
		 *  single 8.8722839355e+01
181
		 *  double 7.09782712893383973096e+02
182
		 *  long double 1.1356523406294143949491931077970765006170e+04
183
		 *  then recurrent value may overflow and the result is
184
		 *  likely underflow to zero
185
		 */
186
		tmp = n;
187
		v = two/x;
188
		tmp = tmp*__ieee754_log(fabs(v*tmp));
189
		if(tmp<7.09782712893383973096e+02) {
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
	     	    }
197
		} else {
198
	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
199
		        temp = b;
200
			b *= di;
201
			b  = b/x - a;
202
		        a = temp;
203
			di -= two;
204
		    /* scale b to avoid spurious overflow */
205
			if(b>1e100) {
206
			    a /= b;
207
			    t /= b;
208
			    b  = one;
209
			}
210
	     	    }
211
		}
212
	    	b = (t*__ieee754_j0(x)/b);
213
	    }
214
	}
215
	if(sgn==1) return -b; else return b;
216
}
217
218
 
219
	double __ieee754_yn(int n, double x)
220
#else
221
	double __ieee754_yn(n,x)
222
	int n; double x;
223
#endif
224
{
225
	__int32_t i,hx,ix,lx;
226
	__int32_t sign;
227
	double a, b, temp;
228
229
 
230
	ix = 0x7fffffff&hx;
231
    /* if Y(n,NaN) is NaN */
232
	if((ix|((__uint32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
233
	if((ix|lx)==0) return -one/zero;
234
	if(hx<0) return zero/zero;
235
	sign = 1;
236
	if(n<0){
237
		n = -n;
238
		sign = 1 - ((n&1)<<1);
239
	}
240
	if(n==0) return(__ieee754_y0(x));
241
	if(n==1) return(sign*__ieee754_y1(x));
242
	if(ix==0x7ff00000) return zero;
243
	if(ix>=0x52D00000) { /* x > 2**302 */
244
    /* (x >> n**2)
245
     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
246
     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
247
     *	    Let s=sin(x), c=cos(x),
248
     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
249
     *
250
     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
251
     *		----------------------------------
252
     *		   0	 s-c		 c+s
253
     *		   1	-s-c 		-c+s
254
     *		   2	-s+c		-c-s
255
     *		   3	 s+c		 c-s
256
     */
257
		switch(n&3) {
258
		    case 0: temp =  sin(x)-cos(x); break;
259
		    case 1: temp = -sin(x)-cos(x); break;
260
		    case 2: temp = -sin(x)+cos(x); break;
261
		    case 3: temp =  sin(x)+cos(x); break;
262
		}
263
		b = invsqrtpi*temp/__ieee754_sqrt(x);
264
	} else {
265
	    __uint32_t high;
266
	    a = __ieee754_y0(x);
267
	    b = __ieee754_y1(x);
268
	/* quit if b is -inf */
269
	    GET_HIGH_WORD(high,b);
270
	    for(i=1;i
271
		temp = b;
272
		b = ((double)(i+i)/x)*b - a;
273
		GET_HIGH_WORD(high,b);
274
		a = temp;
275
	    }
276
	}
277
	if(sign>0) return b; else return -b;
278
}
279
280
 
281
>