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11175 libm should use signbit() correctly
11188 c99 math macros should return strictly backward compatible values
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--- old/usr/src/lib/libm/common/C/jn.c
+++ new/usr/src/lib/libm/common/C/jn.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 #pragma weak __jn = jn
31 31 #pragma weak __yn = yn
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 <float.h> /* DBL_MIN */
60 60 #include <values.h> /* X_TLOSS */
61 61 #include "xpg6.h" /* __xpg6 */
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62 62
63 63 #define GENERIC double
64 64
65 65 static const GENERIC
66 66 invsqrtpi = 5.641895835477562869480794515607725858441e-0001,
67 67 two = 2.0,
68 68 zero = 0.0,
69 69 one = 1.0;
70 70
71 71 GENERIC
72 -jn(int n, GENERIC x) {
72 +jn(int n, GENERIC x)
73 +{
73 74 int i, sgn;
74 75 GENERIC a, b, temp = 0;
75 76 GENERIC z, w, ox, on;
76 77
77 78 /*
78 79 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
79 80 * Thus, J(-n,x) = J(n,-x)
80 81 */
81 - ox = x; on = (GENERIC)n;
82 + ox = x;
83 + on = (GENERIC)n;
84 +
82 85 if (n < 0) {
83 86 n = -n;
84 87 x = -x;
85 88 }
86 89 if (isnan(x))
87 90 return (x*x); /* + -> * for Cheetah */
88 - if (!((int) _lib_version == libm_ieee ||
89 - (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
90 - if (fabs(x) > X_TLOSS)
91 + if (!((int)_lib_version == libm_ieee ||
92 + (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
93 + if (fabs(x) > X_TLOSS)
91 94 return (_SVID_libm_err(on, ox, 38));
92 95 }
93 96 if (n == 0)
94 97 return (j0(x));
95 98 if (n == 1)
96 99 return (j1(x));
97 100 if ((n&1) == 0)
98 - sgn = 0; /* even n */
101 + sgn = 0; /* even n */
99 102 else
100 103 sgn = signbit(x); /* old n */
101 104 x = fabs(x);
102 105 if (x == zero||!finite(x)) b = zero;
103 106 else if ((GENERIC)n <= x) {
104 107 /*
105 108 * Safe to use
106 109 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
107 110 */
108 - if (x > 1.0e91) {
111 + if (x > 1.0e91) {
109 112 /*
110 113 * x >> n**2
111 114 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
112 115 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
113 116 * Let s=sin(x), c=cos(x),
114 117 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
115 118 *
116 119 * n sin(xn)*sqt2 cos(xn)*sqt2
117 120 * ----------------------------------
118 121 * 0 s-c c+s
119 - * 1 -s-c -c+s
122 + * 1 -s-c -c+s
120 123 * 2 -s+c -c-s
121 124 * 3 s+c c-s
122 125 */
123 - switch (n&3) {
124 - case 0: temp = cos(x)+sin(x); break;
125 - case 1: temp = -cos(x)+sin(x); break;
126 - case 2: temp = -cos(x)-sin(x); break;
127 - case 3: temp = cos(x)-sin(x); break;
128 - }
129 - b = invsqrtpi*temp/sqrt(x);
130 - } else {
126 + switch (n&3) {
127 + case 0:
128 + temp = cos(x)+sin(x);
129 + break;
130 + case 1:
131 + temp = -cos(x)+sin(x);
132 + break;
133 + case 2:
134 + temp = -cos(x)-sin(x);
135 + break;
136 + case 3:
137 + temp = cos(x)-sin(x);
138 + break;
139 + }
140 + b = invsqrtpi*temp/sqrt(x);
141 + } else {
131 142 a = j0(x);
132 143 b = j1(x);
133 144 for (i = 1; i < n; i++) {
134 - temp = b;
135 - b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
136 - a = temp;
145 + temp = b;
146 + /* avoid underflow */
147 + b = b*((GENERIC)(i+i)/x) - a;
148 + a = temp;
137 149 }
138 - }
139 - } else {
140 - if (x < 1e-9) { /* use J(n,x) = 1/n!*(x/2)^n */
141 - b = pow(0.5*x, (GENERIC) n);
142 - if (b != zero) {
143 - for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
144 - b = b/a;
145 - }
146 - } else {
147 - /*
148 - * use backward recurrence
149 - * x x^2 x^2
150 - * J(n,x)/J(n-1,x) = ---- ------ ------ .....
151 - * 2n - 2(n+1) - 2(n+2)
152 - *
153 - * 1 1 1
154 - * (for large x) = ---- ------ ------ .....
155 - * 2n 2(n+1) 2(n+2)
156 - * -- - ------ - ------ -
157 - * x x x
158 - *
159 - * Let w = 2n/x and h = 2/x, then the above quotient
160 - * is equal to the continued fraction:
161 - * 1
162 - * = -----------------------
163 - * 1
164 - * w - -----------------
165 - * 1
166 - * w+h - ---------
167 - * w+2h - ...
168 - *
169 - * To determine how many terms needed, let
170 - * Q(0) = w, Q(1) = w(w+h) - 1,
171 - * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
172 - * When Q(k) > 1e4 good for single
173 - * When Q(k) > 1e9 good for double
174 - * When Q(k) > 1e17 good for quaduple
175 - */
176 - /* determin k */
177 - GENERIC t, v;
178 - double q0, q1, h, tmp; int k, m;
179 - w = (n+n)/(double)x; h = 2.0/(double)x;
180 - q0 = w; z = w + h; q1 = w*z - 1.0; k = 1;
181 - while (q1 < 1.0e9) {
182 - k += 1; z += h;
183 - tmp = z*q1 - q0;
184 - q0 = q1;
185 - q1 = tmp;
186 150 }
187 - m = n+n;
188 - for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
189 - a = t;
190 - b = one;
191 - /*
192 - * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
193 - * hence, if n*(log(2n/x)) > ...
194 - * single 8.8722839355e+01
195 - * double 7.09782712893383973096e+02
196 - * long double 1.1356523406294143949491931077970765006170e+04
197 - * then recurrent value may overflow and the result is
198 - * likely underflow to zero
199 - */
200 - tmp = n;
201 - v = two/x;
202 - tmp = tmp*log(fabs(v*tmp));
203 - if (tmp < 7.09782712893383973096e+02) {
204 - for (i = n-1; i > 0; i--) {
205 - temp = b;
206 - b = ((i+i)/x)*b - a;
207 - a = temp;
208 - }
151 + } else {
152 + if (x < 1e-9) { /* use J(n,x) = 1/n!*(x/2)^n */
153 + b = pow(0.5*x, (GENERIC) n);
154 + if (b != zero) {
155 + for (a = one, i = 1; i <= n; i++)
156 + a *= (GENERIC)i;
157 + b = b/a;
158 + }
209 159 } else {
160 + /*
161 + * use backward recurrence
162 + * x x^2 x^2
163 + * J(n,x)/J(n-1,x) = ---- ------ ------ .....
164 + * 2n - 2(n+1) - 2(n+2)
165 + *
166 + * 1 1 1
167 + * (for large x) = ---- ------ ------ .....
168 + * 2n 2(n+1) 2(n+2)
169 + * -- - ------ - ------ -
170 + * x x x
171 + *
172 + * Let w = 2n/x and h = 2/x, then the above quotient
173 + * is equal to the continued fraction:
174 + * 1
175 + * = -----------------------
176 + * 1
177 + * w - -----------------
178 + * 1
179 + * w+h - ---------
180 + * w+2h - ...
181 + *
182 + * To determine how many terms needed, let
183 + * Q(0) = w, Q(1) = w(w+h) - 1,
184 + * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
185 + * When Q(k) > 1e4 good for single
186 + * When Q(k) > 1e9 good for double
187 + * When Q(k) > 1e17 good for quaduple
188 + */
189 + /* determine k */
190 + GENERIC t, v;
191 + double q0, q1, h, tmp;
192 + int k, m;
193 + w = (n+n)/(double)x;
194 + h = 2.0/(double)x;
195 + q0 = w;
196 + z = w + h;
197 + q1 = w*z - 1.0;
198 + k = 1;
199 +
200 + while (q1 < 1.0e9) {
201 + k += 1;
202 + z += h;
203 + tmp = z*q1 - q0;
204 + q0 = q1;
205 + q1 = tmp;
206 + }
207 + m = n+n;
208 + for (t = zero, i = 2*(n+k); i >= m; i -= 2)
209 + t = one/(i/x-t);
210 + a = t;
211 + b = one;
212 + /*
213 + * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
214 + * hence, if n*(log(2n/x)) > ...
215 + * single 8.8722839355e+01
216 + * double 7.09782712893383973096e+02
217 + * long double 1.1356523406294143949491931077970765006170e+04
218 + * then recurrent value may overflow and the result is
219 + * likely underflow to zero
220 + */
221 + tmp = n;
222 + v = two/x;
223 + tmp = tmp*log(fabs(v*tmp));
224 + if (tmp < 7.09782712893383973096e+02) {
210 225 for (i = n-1; i > 0; i--) {
211 - temp = b;
212 - b = ((i+i)/x)*b - a;
213 - a = temp;
226 + temp = b;
227 + b = ((i+i)/x)*b - a;
228 + a = temp;
229 + }
230 + } else {
231 + for (i = n-1; i > 0; i--) {
232 + temp = b;
233 + b = ((i+i)/x)*b - a;
234 + a = temp;
214 235 if (b > 1e100) {
215 236 a /= b;
216 237 t /= b;
217 238 b = 1.0;
218 239 }
219 240 }
220 - }
241 + }
221 242 b = (t*j0(x)/b);
222 - }
243 + }
223 244 }
224 - if (sgn == 1)
245 + if (sgn != 0)
225 246 return (-b);
226 247 else
227 248 return (b);
228 249 }
229 250
230 251 GENERIC
231 -yn(int n, GENERIC x) {
252 +yn(int n, GENERIC x)
253 +{
232 254 int i;
233 255 int sign;
234 256 GENERIC a, b, temp = 0, ox, on;
235 257
236 - ox = x; on = (GENERIC)n;
258 + ox = x;
259 + on = (GENERIC)n;
237 260 if (isnan(x))
238 261 return (x*x); /* + -> * for Cheetah */
239 262 if (x <= zero) {
240 263 if (x == zero) {
241 264 /* return -one/zero; */
242 265 return (_SVID_libm_err((GENERIC)n, x, 12));
243 266 } else {
244 267 /* return zero/zero; */
245 268 return (_SVID_libm_err((GENERIC)n, x, 13));
246 269 }
247 270 }
248 - if (!((int) _lib_version == libm_ieee ||
249 - (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
250 - if (x > X_TLOSS)
271 + if (!((int)_lib_version == libm_ieee ||
272 + (__xpg6 & _C99SUSv3_math_errexcept) != 0)) {
273 + if (x > X_TLOSS)
251 274 return (_SVID_libm_err(on, ox, 39));
252 275 }
253 276 sign = 1;
254 277 if (n < 0) {
255 278 n = -n;
256 279 if ((n&1) == 1) sign = -1;
257 280 }
258 281 if (n == 0)
259 282 return (y0(x));
260 283 if (n == 1)
261 284 return (sign*y1(x));
262 285 if (!finite(x))
263 286 return (zero);
264 287
265 288 if (x > 1.0e91) {
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266 289 /*
267 290 * x >> n**2
268 291 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
269 292 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
270 293 * Let s = sin(x), c = cos(x),
271 294 * xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
272 295 *
273 296 * n sin(xn)*sqt2 cos(xn)*sqt2
274 297 * ----------------------------------
275 298 * 0 s-c c+s
276 - * 1 -s-c -c+s
299 + * 1 -s-c -c+s
277 300 * 2 -s+c -c-s
278 301 * 3 s+c c-s
279 302 */
280 303 switch (n&3) {
281 - case 0: temp = sin(x)-cos(x); break;
282 - case 1: temp = -sin(x)-cos(x); break;
283 - case 2: temp = -sin(x)+cos(x); break;
284 - case 3: temp = sin(x)+cos(x); break;
304 + case 0:
305 + temp = sin(x)-cos(x);
306 + break;
307 + case 1:
308 + temp = -sin(x)-cos(x);
309 + break;
310 + case 2:
311 + temp = -sin(x)+cos(x);
312 + break;
313 + case 3:
314 + temp = sin(x)+cos(x);
315 + break;
285 316 }
286 317 b = invsqrtpi*temp/sqrt(x);
287 318 } else {
288 319 a = y0(x);
289 320 b = y1(x);
290 321 /*
291 322 * fix 1262058 and take care of non-default rounding
292 323 */
293 324 for (i = 1; i < n; i++) {
294 325 temp = b;
295 326 b *= (GENERIC) (i + i) / x;
296 327 if (b <= -DBL_MAX)
297 328 break;
298 329 b -= a;
299 330 a = temp;
300 331 }
301 332 }
302 333 if (sign > 0)
303 334 return (b);
304 335 else
305 336 return (-b);
306 337 }
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