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--- old/usr/src/lib/libm/common/LD/jnl.c
+++ new/usr/src/lib/libm/common/LD/jnl.c
1 1 /*
2 2 * CDDL HEADER START
3 3 *
4 4 * The contents of this file are subject to the terms of the
5 5 * Common Development and Distribution License (the "License").
6 6 * You may not use this file except in compliance with the License.
7 7 *
8 8 * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
9 9 * or http://www.opensolaris.org/os/licensing.
10 10 * See the License for the specific language governing permissions
11 11 * and limitations under the License.
12 12 *
13 13 * When distributing Covered Code, include this CDDL HEADER in each
14 14 * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
15 15 * If applicable, add the following below this CDDL HEADER, with the
16 16 * fields enclosed by brackets "[]" replaced with your own identifying
17 17 * information: Portions Copyright [yyyy] [name of copyright owner]
18 18 *
19 19 * CDDL HEADER END
20 20 */
21 21
22 22 /*
23 23 * Copyright 2011 Nexenta Systems, Inc. All rights reserved.
24 24 */
25 25 /*
26 26 * Copyright 2006 Sun Microsystems, Inc. All rights reserved.
27 27 * Use is subject to license terms.
28 28 */
29 29
30 30 #if defined(ELFOBJ)
31 31 #pragma weak jnl = __jnl
32 32 #pragma weak ynl = __ynl
33 33 #endif
34 34
35 35 /*
36 36 * floating point Bessel's function of the 1st and 2nd kind
37 37 * of order n: jn(n,x),yn(n,x);
38 38 *
39 39 * Special cases:
40 40 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
41 41 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
42 42 * Note 2. About jn(n,x), yn(n,x)
43 43 * For n=0, j0(x) is called,
44 44 * for n=1, j1(x) is called,
45 45 * for n<x, forward recursion us used starting
46 46 * from values of j0(x) and j1(x).
47 47 * for n>x, a continued fraction approximation to
48 48 * j(n,x)/j(n-1,x) is evaluated and then backward
49 49 * recursion is used starting from a supposed value
50 50 * for j(n,x). The resulting value of j(0,x) is
51 51 * compared with the actual value to correct the
52 52 * supposed value of j(n,x).
53 53 *
54 54 * yn(n,x) is similar in all respects, except
55 55 * that forward recursion is used for all
56 56 * values of n>1.
57 57 *
58 58 */
59 59
60 60 #include "libm.h"
61 61 #include "longdouble.h"
62 62 #include <float.h> /* LDBL_MAX */
63 63
64 64 #define GENERIC long double
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65 65
66 66 static const GENERIC
67 67 invsqrtpi= 5.641895835477562869480794515607725858441e-0001L,
68 68 two = 2.0L,
69 69 zero = 0.0L,
70 70 one = 1.0L;
71 71
72 72 GENERIC
73 73 jnl(n,x) int n; GENERIC x;{
74 74 int i, sgn;
75 - GENERIC a, b, temp, z, w;
75 + GENERIC a, b, temp = 0, z, w;
76 76
77 77 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
78 78 * Thus, J(-n,x) = J(n,-x)
79 79 */
80 80 if(n<0){
81 81 n = -n;
82 82 x = -x;
83 83 }
84 84 if(n==0) return(j0l(x));
85 85 if(n==1) return(j1l(x));
86 86 if(x!=x) return x+x;
87 87 if((n&1)==0)
88 88 sgn=0; /* even n */
89 89 else
90 90 sgn = signbitl(x); /* old n */
91 91 x = fabsl(x);
92 92 if(x == zero||!finitel(x)) b = zero;
93 93 else if((GENERIC)n<=x) { /* Safe to use
94 94 J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
95 95 */
96 96 if(x>1.0e91L) { /* x >> n**2
97 97 Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
98 98 Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
99 99 Let s=sin(x), c=cos(x),
100 100 xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
101 101
102 102 n sin(xn)*sqt2 cos(xn)*sqt2
103 103 ----------------------------------
104 104 0 s-c c+s
105 105 1 -s-c -c+s
106 106 2 -s+c -c-s
107 107 3 s+c c-s
108 108 */
109 109 switch(n&3) {
110 110 case 0: temp = cosl(x)+sinl(x); break;
111 111 case 1: temp = -cosl(x)+sinl(x); break;
112 112 case 2: temp = -cosl(x)-sinl(x); break;
113 113 case 3: temp = cosl(x)-sinl(x); break;
114 114 }
115 115 b = invsqrtpi*temp/sqrtl(x);
116 116 } else {
117 117 a = j0l(x);
118 118 b = j1l(x);
119 119 for(i=1;i<n;i++){
120 120 temp = b;
121 121 b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
122 122 a = temp;
123 123 }
124 124 }
125 125 } else {
126 126 if(x<1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */
127 127 b = powl(0.5L*x,(GENERIC) n);
128 128 if (b!=zero) {
129 129 for(a=one,i=1;i<=n;i++) a *= (GENERIC)i;
130 130 b = b/a;
131 131 }
132 132 } else {
133 133 /* use backward recurrence */
134 134 /* x x^2 x^2
135 135 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
136 136 * 2n - 2(n+1) - 2(n+2)
137 137 *
138 138 * 1 1 1
139 139 * (for large x) = ---- ------ ------ .....
140 140 * 2n 2(n+1) 2(n+2)
141 141 * -- - ------ - ------ -
142 142 * x x x
143 143 *
144 144 * Let w = 2n/x and h=2/x, then the above quotient
145 145 * is equal to the continued fraction:
146 146 * 1
147 147 * = -----------------------
148 148 * 1
149 149 * w - -----------------
150 150 * 1
151 151 * w+h - ---------
152 152 * w+2h - ...
153 153 *
154 154 * To determine how many terms needed, let
155 155 * Q(0) = w, Q(1) = w(w+h) - 1,
156 156 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
157 157 * When Q(k) > 1e4 good for single
158 158 * When Q(k) > 1e9 good for double
159 159 * When Q(k) > 1e17 good for quaduple
160 160 */
161 161 /* determin k */
162 162 GENERIC t,v;
163 163 double q0,q1,h,tmp; int k,m;
164 164 w = (n+n)/(double)x; h = 2.0/(double)x;
165 165 q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
166 166 while(q1<1.0e17) {
167 167 k += 1; z += h;
168 168 tmp = z*q1 - q0;
169 169 q0 = q1;
170 170 q1 = tmp;
171 171 }
172 172 m = n+n;
173 173 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
174 174 a = t;
175 175 b = one;
176 176 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
177 177 hence, if n*(log(2n/x)) > ...
178 178 single 8.8722839355e+01
179 179 double 7.09782712893383973096e+02
180 180 long double 1.1356523406294143949491931077970765006170e+04
181 181 then recurrent value may overflow and the result is
182 182 likely underflow to zero
183 183 */
184 184 tmp = n;
185 185 v = two/x;
186 186 tmp = tmp*logl(fabsl(v*tmp));
187 187 if(tmp<1.1356523406294143949491931077970765e+04L) {
188 188 for(i=n-1;i>0;i--){
189 189 temp = b;
190 190 b = ((i+i)/x)*b - a;
191 191 a = temp;
192 192 }
193 193 } else {
194 194 for(i=n-1;i>0;i--){
195 195 temp = b;
196 196 b = ((i+i)/x)*b - a;
197 197 a = temp;
198 198 if(b>1e1000L) {
199 199 a /= b;
200 200 t /= b;
201 201 b = 1.0;
202 202 }
203 203 }
204 204 }
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205 205 b = (t*j0l(x)/b);
206 206 }
207 207 }
208 208 if(sgn==1) return -b; else return b;
209 209 }
210 210
211 211 GENERIC ynl(n,x)
212 212 int n; GENERIC x;{
213 213 int i;
214 214 int sign;
215 - GENERIC a, b, temp;
215 + GENERIC a, b, temp = 0;
216 216
217 - if(x!=x) return x+x;
218 - if (x <= zero)
217 + if(x!=x)
218 + return x+x;
219 + if (x <= zero) {
219 220 if(x==zero)
220 221 return -one/zero;
221 222 else
222 223 return zero/zero;
224 + }
223 225 sign = 1;
224 226 if(n<0){
225 227 n = -n;
226 228 if((n&1) == 1) sign = -1;
227 229 }
228 230 if(n==0) return(y0l(x));
229 231 if(n==1) return(sign*y1l(x));
230 232 if(!finitel(x)) return zero;
231 233
232 234 if(x>1.0e91L) { /* x >> n**2
233 235 Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
234 236 Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
235 237 Let s=sin(x), c=cos(x),
236 238 xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
237 239
238 240 n sin(xn)*sqt2 cos(xn)*sqt2
239 241 ----------------------------------
240 242 0 s-c c+s
241 243 1 -s-c -c+s
242 244 2 -s+c -c-s
243 245 3 s+c c-s
244 246 */
245 247 switch(n&3) {
246 248 case 0: temp = sinl(x)-cosl(x); break;
247 249 case 1: temp = -sinl(x)-cosl(x); break;
248 250 case 2: temp = -sinl(x)+cosl(x); break;
249 251 case 3: temp = sinl(x)+cosl(x); break;
250 252 }
251 253 b = invsqrtpi*temp/sqrtl(x);
252 254 } else {
253 255 a = y0l(x);
254 256 b = y1l(x);
255 257 /*
256 258 * fix 1262058 and take care of non-default rounding
257 259 */
258 260 for (i = 1; i < n; i++) {
259 261 temp = b;
260 262 b *= (GENERIC) (i + i) / x;
261 263 if (b <= -LDBL_MAX)
262 264 break;
263 265 b -= a;
264 266 a = temp;
265 267 }
266 268 }
267 269 if(sign>0) return b; else return -b;
268 270 }
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