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5261 libm should stop using synonyms.h
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--- old/usr/src/lib/libm/common/Q/jnl.c
+++ new/usr/src/lib/libm/common/Q/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
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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 -#pragma weak jnl = __jnl
31 -#pragma weak ynl = __ynl
30 +#pragma weak __jnl = jnl
31 +#pragma weak __ynl = ynl
32 32
33 33 /*
34 34 * floating point Bessel's function of the 1st and 2nd kind
35 35 * of order n: jn(n,x),yn(n,x);
36 36 *
37 37 * Special cases:
38 38 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
39 39 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
40 40 * Note 2. About jn(n,x), yn(n,x)
41 41 * For n=0, j0(x) is called,
42 42 * for n=1, j1(x) is called,
43 43 * for n<x, forward recursion us used starting
44 44 * from values of j0(x) and j1(x).
45 45 * for n>x, a continued fraction approximation to
46 46 * j(n,x)/j(n-1,x) is evaluated and then backward
47 47 * recursion is used starting from a supposed value
48 48 * for j(n,x). The resulting value of j(0,x) is
49 49 * compared with the actual value to correct the
50 50 * supposed value of j(n,x).
51 51 *
52 52 * yn(n,x) is similar in all respects, except
53 53 * that forward recursion is used for all
54 54 * values of n>1.
55 55 *
56 56 */
57 57
58 58 #include "libm.h"
59 59 #include "longdouble.h"
60 60 #include <float.h> /* LDBL_MAX */
61 61
62 62 #define GENERIC long double
63 63
64 64 static const GENERIC
65 65 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
66 66 two = 2.0L,
67 67 zero = 0.0L,
68 68 one = 1.0L;
69 69
70 70 GENERIC
71 71 jnl(n, x) int n; GENERIC x; {
72 72 int i, sgn;
73 73 GENERIC a, b, temp, z, w;
74 74
75 75 /*
76 76 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
77 77 * Thus, J(-n,x) = J(n,-x)
78 78 */
79 79 if (n < 0) {
80 80 n = -n;
81 81 x = -x;
82 82 }
83 83 if (n == 0)
84 84 return (j0l(x));
85 85 if (n == 1)
86 86 return (j1l(x));
87 87 if (x != x)
88 88 return (x+x);
89 89 if ((n&1) == 0)
90 90 sgn = 0; /* even n */
91 91 else
92 92 sgn = signbitl(x); /* old n */
93 93 x = fabsl(x);
94 94 if (x == zero || !finitel(x)) b = zero;
95 95 else if ((GENERIC)n <= x) {
96 96 /*
97 97 * Safe to use
98 98 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
99 99 */
100 100 if (x > 1.0e91L) {
101 101 /*
102 102 * x >> n**2
103 103 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
104 104 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 105 * Let s=sin(x), c=cos(x),
106 106 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
107 107 *
108 108 * n sin(xn)*sqt2 cos(xn)*sqt2
109 109 * ----------------------------------
110 110 * 0 s-c c+s
111 111 * 1 -s-c -c+s
112 112 * 2 -s+c -c-s
113 113 * 3 s+c c-s
114 114 */
115 115 switch (n&3) {
116 116 case 0: temp = cosl(x)+sinl(x); break;
117 117 case 1: temp = -cosl(x)+sinl(x); break;
118 118 case 2: temp = -cosl(x)-sinl(x); break;
119 119 case 3: temp = cosl(x)-sinl(x); break;
120 120 }
121 121 b = invsqrtpi*temp/sqrtl(x);
122 122 } else {
123 123 a = j0l(x);
124 124 b = j1l(x);
125 125 for (i = 1; i < n; i++) {
126 126 temp = b;
127 127 b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
128 128 a = temp;
129 129 }
130 130 }
131 131 } else {
132 132 if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */
133 133 b = powl(0.5L*x, (GENERIC)n);
134 134 if (b != zero) {
135 135 for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
136 136 b = b/a;
137 137 }
138 138 } else {
139 139 /* use backward recurrence */
140 140 /*
141 141 * x x^2 x^2
142 142 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
143 143 * 2n - 2(n+1) - 2(n+2)
144 144 *
145 145 * 1 1 1
146 146 * (for large x) = ---- ------ ------ .....
147 147 * 2n 2(n+1) 2(n+2)
148 148 * -- - ------ - ------ -
149 149 * x x x
150 150 *
151 151 * Let w = 2n/x and h=2/x, then the above quotient
152 152 * is equal to the continued fraction:
153 153 * 1
154 154 * = -----------------------
155 155 * 1
156 156 * w - -----------------
157 157 * 1
158 158 * w+h - ---------
159 159 * w+2h - ...
160 160 *
161 161 * To determine how many terms needed, let
162 162 * Q(0) = w, Q(1) = w(w+h) - 1,
163 163 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
164 164 * When Q(k) > 1e4 good for single
165 165 * When Q(k) > 1e9 good for double
166 166 * When Q(k) > 1e17 good for quaduple
167 167 */
168 168 /* determin k */
169 169 GENERIC t, v;
170 170 double q0, q1, h, tmp; int k, m;
171 171 w = (n+n)/(double)x; h = 2.0/(double)x;
172 172 q0 = w; z = w+h; q1 = w*z - 1.0; k = 1;
173 173 while (q1 < 1.0e17) {
174 174 k += 1; z += h;
175 175 tmp = z*q1 - q0;
176 176 q0 = q1;
177 177 q1 = tmp;
178 178 }
179 179 m = n+n;
180 180 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
181 181 a = t;
182 182 b = one;
183 183 /*
184 184 * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
185 185 * hence, if n*(log(2n/x)) > ...
186 186 * single 8.8722839355e+01
187 187 * double 7.09782712893383973096e+02
188 188 * long double 1.1356523406294143949491931077970765006170e+04
189 189 * then recurrent value may overflow and the result is
190 190 * likely underflow to zero
191 191 */
192 192 tmp = n;
193 193 v = two/x;
194 194 tmp = tmp*logl(fabsl(v*tmp));
195 195 if (tmp < 1.1356523406294143949491931077970765e+04L) {
196 196 for (i = n-1; i > 0; i--) {
197 197 temp = b;
198 198 b = ((i+i)/x)*b - a;
199 199 a = temp;
200 200 }
201 201 } else {
202 202 for (i = n-1; i > 0; i--) {
203 203 temp = b;
204 204 b = ((i+i)/x)*b - a;
205 205 a = temp;
206 206 if (b > 1e1000L) {
207 207 a /= b;
208 208 t /= b;
209 209 b = 1.0;
210 210 }
211 211 }
212 212 }
213 213 b = (t*j0l(x)/b);
214 214 }
215 215 }
216 216 if (sgn == 1)
217 217 return (-b);
218 218 else
219 219 return (b);
220 220 }
221 221
222 222 GENERIC ynl(n, x)
223 223 int n; GENERIC x; {
224 224 int i;
225 225 int sign;
226 226 GENERIC a, b, temp;
227 227
228 228 if (x != x)
229 229 return (x+x);
230 230 if (x <= zero) {
231 231 if (x == zero)
232 232 return (-one/zero);
233 233 else
234 234 return (zero/zero);
235 235 }
236 236 sign = 1;
237 237 if (n < 0) {
238 238 n = -n;
239 239 if ((n&1) == 1) sign = -1;
240 240 }
241 241 if (n == 0)
242 242 return (y0l(x));
243 243 if (n == 1)
244 244 return (sign*y1l(x));
245 245 if (!finitel(x))
246 246 return (zero);
247 247
248 248 if (x > 1.0e91L) { /* x >> n**2
249 249 Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
250 250 Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
251 251 Let s = sin(x), c = cos(x),
252 252 xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
253 253
254 254 n sin(xn)*sqt2 cos(xn)*sqt2
255 255 ----------------------------------
256 256 0 s-c c+s
257 257 1 -s-c -c+s
258 258 2 -s+c -c-s
259 259 3 s+c c-s
260 260 */
261 261 switch (n&3) {
262 262 case 0: temp = sinl(x)-cosl(x); break;
263 263 case 1: temp = -sinl(x)-cosl(x); break;
264 264 case 2: temp = -sinl(x)+cosl(x); break;
265 265 case 3: temp = sinl(x)+cosl(x); break;
266 266 }
267 267 b = invsqrtpi*temp/sqrtl(x);
268 268 } else {
269 269 a = y0l(x);
270 270 b = y1l(x);
271 271 /*
272 272 * fix 1262058 and take care of non-default rounding
273 273 */
274 274 for (i = 1; i < n; i++) {
275 275 temp = b;
276 276 b *= (GENERIC) (i + i) / x;
277 277 if (b <= -LDBL_MAX)
278 278 break;
279 279 b -= a;
280 280 a = temp;
281 281 }
282 282 }
283 283 if (sign > 0)
284 284 return (b);
285 285 else
286 286 return (-b);
287 287 }
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