GCC Code Coverage Report
Directory: ./ Exec Total Coverage
File: lib/libm/src/e_jn.c Lines: 0 81 0.0 %
Date: 2017-11-07 Branches: 0 66 0.0 %

Line Branch Exec Source
1
/* @(#)e_jn.c 5.1 93/09/24 */
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
 * jn(n, x), 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<x, forward recursion us used starting
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
#include "math.h"
40
#include "math_private.h"
41
42
static const double
43
invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
44
two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
45
one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
46
47
static const double zero  =  0.00000000000000000000e+00;
48
49
double
50
jn(int n, double x)
51
{
52
	int32_t i,hx,ix,lx, sgn;
53
	double a, b, temp, di;
54
	double z, w;
55
56
    /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
57
     * Thus, J(-n,x) = J(n,-x)
58
     */
59
	EXTRACT_WORDS(hx,lx,x);
60
	ix = 0x7fffffff&hx;
61
    /* if J(n,NaN) is NaN */
62
	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
63
	if(n<0){
64
		n = -n;
65
		x = -x;
66
		hx ^= 0x80000000;
67
	}
68
	if(n==0) return(j0(x));
69
	if(n==1) return(j1(x));
70
	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
71
	x = fabs(x);
72
	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
73
	    b = zero;
74
	else if((double)n<=x) {
75
		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
76
	    if(ix>=0x52D00000) { /* x > 2**302 */
77
    /* (x >> n**2)
78
     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
79
     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
80
     *	    Let s=sin(x), c=cos(x),
81
     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
82
     *
83
     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
84
     *		----------------------------------
85
     *		   0	 s-c		 c+s
86
     *		   1	-s-c 		-c+s
87
     *		   2	-s+c		-c-s
88
     *		   3	 s+c		 c-s
89
     */
90
		switch(n&3) {
91
		    case 0: temp =  cos(x)+sin(x); break;
92
		    case 1: temp = -cos(x)+sin(x); break;
93
		    case 2: temp = -cos(x)-sin(x); break;
94
		    case 3: temp =  cos(x)-sin(x); break;
95
		}
96
		b = invsqrtpi*temp/sqrt(x);
97
	    } else {
98
	        a = j0(x);
99
	        b = j1(x);
100
	        for(i=1;i<n;i++){
101
		    temp = b;
102
		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
103
		    a = temp;
104
	        }
105
	    }
106
	} else {
107
	    if(ix<0x3e100000) {	/* x < 2**-29 */
108
    /* x is tiny, return the first Taylor expansion of J(n,x)
109
     * J(n,x) = 1/n!*(x/2)^n  - ...
110
     */
111
		if(n>33)	/* underflow */
112
		    b = zero;
113
		else {
114
		    temp = x*0.5; b = temp;
115
		    for (a=one,i=2;i<=n;i++) {
116
			a *= (double)i;		/* a = n! */
117
			b *= temp;		/* b = (x/2)^n */
118
		    }
119
		    b = b/a;
120
		}
121
	    } else {
122
		/* use backward recurrence */
123
		/* 			x      x^2      x^2
124
		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
125
		 *			2n  - 2(n+1) - 2(n+2)
126
		 *
127
		 * 			1      1        1
128
		 *  (for large x)   =  ----  ------   ------   .....
129
		 *			2n   2(n+1)   2(n+2)
130
		 *			-- - ------ - ------ -
131
		 *			 x     x         x
132
		 *
133
		 * Let w = 2n/x and h=2/x, then the above quotient
134
		 * is equal to the continued fraction:
135
		 *		    1
136
		 *	= -----------------------
137
		 *		       1
138
		 *	   w - -----------------
139
		 *			  1
140
		 * 	        w+h - ---------
141
		 *		       w+2h - ...
142
		 *
143
		 * To determine how many terms needed, let
144
		 * Q(0) = w, Q(1) = w(w+h) - 1,
145
		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
146
		 * When Q(k) > 1e4	good for single
147
		 * When Q(k) > 1e9	good for double
148
		 * When Q(k) > 1e17	good for quadruple
149
		 */
150
	    /* determine k */
151
		double t,v;
152
		double q0,q1,h,tmp; int32_t k,m;
153
		w  = (n+n)/(double)x; h = 2.0/(double)x;
154
		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
155
		while(q1<1.0e9) {
156
			k += 1; z += h;
157
			tmp = z*q1 - q0;
158
			q0 = q1;
159
			q1 = tmp;
160
		}
161
		m = n+n;
162
		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
163
		a = t;
164
		b = one;
165
		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
166
		 *  Hence, if n*(log(2n/x)) > ...
167
		 *  single 8.8722839355e+01
168
		 *  double 7.09782712893383973096e+02
169
		 *  long double 1.1356523406294143949491931077970765006170e+04
170
		 *  then recurrent value may overflow and the result is
171
		 *  likely underflow to zero
172
		 */
173
		tmp = n;
174
		v = two/x;
175
		tmp = tmp*log(fabs(v*tmp));
176
		if(tmp<7.09782712893383973096e+02) {
177
	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
178
		        temp = b;
179
			b *= di;
180
			b  = b/x - a;
181
		        a = temp;
182
			di -= two;
183
	     	    }
184
		} else {
185
	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
186
		        temp = b;
187
			b *= di;
188
			b  = b/x - a;
189
		        a = temp;
190
			di -= two;
191
		    /* scale b to avoid spurious overflow */
192
			if(b>1e100) {
193
			    a /= b;
194
			    t /= b;
195
			    b  = one;
196
			}
197
	     	    }
198
		}
199
	    	b = (t*j0(x)/b);
200
	    }
201
	}
202
	if(sgn==1) return -b; else return b;
203
}
204
205
double
206
yn(int n, double x)
207
{
208
	int32_t i,hx,ix,lx;
209
	int32_t sign;
210
	double a, b, temp;
211
212
	EXTRACT_WORDS(hx,lx,x);
213
	ix = 0x7fffffff&hx;
214
    /* if Y(n,NaN) is NaN */
215
	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
216
	if((ix|lx)==0) return -one/zero;
217
	if(hx<0) return zero/zero;
218
	sign = 1;
219
	if(n<0){
220
		n = -n;
221
		sign = 1 - ((n&1)<<1);
222
	}
223
	if(n==0) return(y0(x));
224
	if(n==1) return(sign*y1(x));
225
	if(ix==0x7ff00000) return zero;
226
	if(ix>=0x52D00000) { /* x > 2**302 */
227
    /* (x >> n**2)
228
     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
229
     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
230
     *	    Let s=sin(x), c=cos(x),
231
     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
232
     *
233
     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
234
     *		----------------------------------
235
     *		   0	 s-c		 c+s
236
     *		   1	-s-c 		-c+s
237
     *		   2	-s+c		-c-s
238
     *		   3	 s+c		 c-s
239
     */
240
		switch(n&3) {
241
		    case 0: temp =  sin(x)-cos(x); break;
242
		    case 1: temp = -sin(x)-cos(x); break;
243
		    case 2: temp = -sin(x)+cos(x); break;
244
		    case 3: temp =  sin(x)+cos(x); break;
245
		}
246
		b = invsqrtpi*temp/sqrt(x);
247
	} else {
248
	    u_int32_t high;
249
	    a = y0(x);
250
	    b = y1(x);
251
	/* quit if b is -inf */
252
	    GET_HIGH_WORD(high,b);
253
	    for(i=1;i<n&&high!=0xfff00000;i++){
254
		temp = b;
255
		b = ((double)(i+i)/x)*b - a;
256
		GET_HIGH_WORD(high,b);
257
		a = temp;
258
	    }
259
	}
260
	if(sign>0) return b; else return -b;
261
}