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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/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
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 __jnl = jnl
31 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 */
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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 -jnl(n, x) int n; GENERIC x; {
71 +jnl(int n, GENERIC x)
72 +{
72 73 int i, sgn;
73 74 GENERIC a, b, temp, z, w;
74 75
75 76 /*
76 77 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
77 78 * Thus, J(-n,x) = J(n,-x)
78 79 */
79 80 if (n < 0) {
80 81 n = -n;
81 82 x = -x;
82 83 }
83 84 if (n == 0)
84 85 return (j0l(x));
85 86 if (n == 1)
86 87 return (j1l(x));
87 88 if (x != x)
88 89 return (x+x);
89 90 if ((n&1) == 0)
90 - sgn = 0; /* even n */
91 + sgn = 0; /* even n */
91 92 else
92 93 sgn = signbitl(x); /* old n */
93 94 x = fabsl(x);
94 95 if (x == zero || !finitel(x)) b = zero;
95 96 else if ((GENERIC)n <= x) {
96 97 /*
97 98 * Safe to use
98 99 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
99 100 */
100 - if (x > 1.0e91L) {
101 + if (x > 1.0e91L) {
101 102 /*
102 103 * x >> n**2
103 104 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
104 105 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 106 * Let s=sin(x), c=cos(x),
106 107 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
107 108 *
108 109 * n sin(xn)*sqt2 cos(xn)*sqt2
109 110 * ----------------------------------
110 111 * 0 s-c c+s
111 - * 1 -s-c -c+s
112 + * 1 -s-c -c+s
112 113 * 2 -s+c -c-s
113 114 * 3 s+c c-s
114 115 */
115 - switch (n&3) {
116 - case 0: temp = cosl(x)+sinl(x); break;
117 - case 1: temp = -cosl(x)+sinl(x); break;
118 - case 2: temp = -cosl(x)-sinl(x); break;
119 - case 3: temp = cosl(x)-sinl(x); break;
120 - }
121 - b = invsqrtpi*temp/sqrtl(x);
122 - } else {
116 + switch (n&3) {
117 + case 0:
118 + temp = cosl(x)+sinl(x);
119 + break;
120 + case 1:
121 + temp = -cosl(x)+sinl(x);
122 + break;
123 + case 2:
124 + temp = -cosl(x)-sinl(x);
125 + break;
126 + case 3:
127 + temp = cosl(x)-sinl(x);
128 + break;
129 + }
130 + b = invsqrtpi*temp/sqrtl(x);
131 + } else {
123 132 a = j0l(x);
124 133 b = j1l(x);
125 134 for (i = 1; i < n; i++) {
126 - temp = b;
127 - b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
128 - a = temp;
135 + temp = b;
136 + /* avoid underflow */
137 + b = b*((GENERIC)(i+i)/x) - a;
138 + a = temp;
129 139 }
130 - }
131 - } else {
132 - if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */
133 - b = powl(0.5L*x, (GENERIC)n);
134 - if (b != zero) {
135 - for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
136 - b = b/a;
137 - }
138 - } else {
139 - /* use backward recurrence */
140 - /*
141 - * x x^2 x^2
142 - * J(n,x)/J(n-1,x) = ---- ------ ------ .....
143 - * 2n - 2(n+1) - 2(n+2)
144 - *
145 - * 1 1 1
146 - * (for large x) = ---- ------ ------ .....
147 - * 2n 2(n+1) 2(n+2)
148 - * -- - ------ - ------ -
149 - * x x x
150 - *
151 - * Let w = 2n/x and h=2/x, then the above quotient
152 - * is equal to the continued fraction:
153 - * 1
154 - * = -----------------------
155 - * 1
156 - * w - -----------------
157 - * 1
158 - * w+h - ---------
159 - * w+2h - ...
160 - *
161 - * To determine how many terms needed, let
162 - * Q(0) = w, Q(1) = w(w+h) - 1,
163 - * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
164 - * When Q(k) > 1e4 good for single
165 - * When Q(k) > 1e9 good for double
166 - * When Q(k) > 1e17 good for quaduple
167 - */
168 - /* determin k */
169 - GENERIC t, v;
170 - double q0, q1, h, tmp; int k, m;
171 - w = (n+n)/(double)x; h = 2.0/(double)x;
172 - q0 = w; z = w+h; q1 = w*z - 1.0; k = 1;
173 - while (q1 < 1.0e17) {
174 - k += 1; z += h;
175 - tmp = z*q1 - q0;
176 - q0 = q1;
177 - q1 = tmp;
178 140 }
179 - m = n+n;
180 - for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
181 - a = t;
182 - b = one;
183 - /*
184 - * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
185 - * hence, if n*(log(2n/x)) > ...
186 - * single 8.8722839355e+01
187 - * double 7.09782712893383973096e+02
188 - * long double 1.1356523406294143949491931077970765006170e+04
189 - * then recurrent value may overflow and the result is
190 - * likely underflow to zero
191 - */
192 - tmp = n;
193 - v = two/x;
194 - tmp = tmp*logl(fabsl(v*tmp));
195 - if (tmp < 1.1356523406294143949491931077970765e+04L) {
196 - for (i = n-1; i > 0; i--) {
197 - temp = b;
198 - b = ((i+i)/x)*b - a;
199 - a = temp;
200 - }
141 + } else {
142 + if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */
143 + b = powl(0.5L*x, (GENERIC)n);
144 + if (b != zero) {
145 + for (a = one, i = 1; i <= n; i++)
146 + a *= (GENERIC)i;
147 + b = b/a;
148 + }
201 149 } else {
150 + /* use backward recurrence */
151 + /*
152 + * x x^2 x^2
153 + * J(n,x)/J(n-1,x) = ---- ------ ------ .....
154 + * 2n - 2(n+1) - 2(n+2)
155 + *
156 + * 1 1 1
157 + * (for large x) = ---- ------ ------ .....
158 + * 2n 2(n+1) 2(n+2)
159 + * -- - ------ - ------ -
160 + * x x x
161 + *
162 + * Let w = 2n/x and h=2/x, then the above quotient
163 + * is equal to the continued fraction:
164 + * 1
165 + * = -----------------------
166 + * 1
167 + * w - -----------------
168 + * 1
169 + * w+h - ---------
170 + * w+2h - ...
171 + *
172 + * To determine how many terms needed, let
173 + * Q(0) = w, Q(1) = w(w+h) - 1,
174 + * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
175 + * When Q(k) > 1e4 good for single
176 + * When Q(k) > 1e9 good for double
177 + * When Q(k) > 1e17 good for quaduple
178 + */
179 + /* determine k */
180 + GENERIC t, v;
181 + double q0, q1, h, tmp;
182 + int k, m;
183 + w = (n+n)/(double)x;
184 + h = 2.0/(double)x;
185 + q0 = w;
186 + z = w+h;
187 + q1 = w*z - 1.0;
188 + k = 1;
189 + while (q1 < 1.0e17) {
190 + k += 1;
191 + z += h;
192 + tmp = z*q1 - q0;
193 + q0 = q1;
194 + q1 = tmp;
195 + }
196 + m = n+n;
197 + for (t = zero, i = 2*(n+k); i >= m; i -= 2)
198 + t = one/(i/x-t);
199 + a = t;
200 + b = one;
201 + /*
202 + * estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
203 + * hence, if n*(log(2n/x)) > ...
204 + * single 8.8722839355e+01
205 + * double 7.09782712893383973096e+02
206 + * long double 1.1356523406294143949491931077970765006170e+04
207 + * then recurrent value may overflow and the result is
208 + * likely underflow to zero
209 + */
210 + tmp = n;
211 + v = two/x;
212 + tmp = tmp*logl(fabsl(v*tmp));
213 + if (tmp < 1.1356523406294143949491931077970765e+04L) {
214 + for (i = n-1; i > 0; i--) {
215 + temp = b;
216 + b = ((i+i)/x)*b - a;
217 + a = temp;
218 + }
219 + } else {
202 220 for (i = n-1; i > 0; i--) {
203 - temp = b;
204 - b = ((i+i)/x)*b - a;
205 - a = temp;
206 - if (b > 1e1000L) {
207 - a /= b;
208 - t /= b;
209 - b = 1.0;
210 - }
211 - }
221 + temp = b;
222 + b = ((i+i)/x)*b - a;
223 + a = temp;
224 + if (b > 1e1000L) {
225 + a /= b;
226 + t /= b;
227 + b = 1.0;
228 + }
229 + }
230 + }
231 + b = (t*j0l(x)/b);
212 232 }
213 - b = (t*j0l(x)/b);
214 - }
215 233 }
216 - if (sgn == 1)
234 + if (sgn != 0)
217 235 return (-b);
218 236 else
219 237 return (b);
220 238 }
221 239
222 -GENERIC ynl(n, x)
223 -int n; GENERIC x; {
240 +GENERIC
241 +ynl(int n, GENERIC x)
242 +{
224 243 int i;
225 244 int sign;
226 245 GENERIC a, b, temp;
227 246
228 247 if (x != x)
229 248 return (x+x);
230 249 if (x <= zero) {
231 250 if (x == zero)
232 251 return (-one/zero);
233 252 else
234 253 return (zero/zero);
235 254 }
236 255 sign = 1;
237 256 if (n < 0) {
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238 257 n = -n;
239 258 if ((n&1) == 1) sign = -1;
240 259 }
241 260 if (n == 0)
242 261 return (y0l(x));
243 262 if (n == 1)
244 263 return (sign*y1l(x));
245 264 if (!finitel(x))
246 265 return (zero);
247 266
248 - if (x > 1.0e91L) { /* x >> n**2
249 - Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
250 - Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
251 - Let s = sin(x), c = cos(x),
252 - xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
253 -
254 - n sin(xn)*sqt2 cos(xn)*sqt2
255 - ----------------------------------
256 - 0 s-c c+s
257 - 1 -s-c -c+s
258 - 2 -s+c -c-s
259 - 3 s+c c-s
260 - */
267 + if (x > 1.0e91L) {
268 + /*
269 + * x >> n**2
270 + * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
271 + * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
272 + * Let s = sin(x), c = cos(x),
273 + * xn = x-(2n+1)*pi/4, sqt2 = sqrt(2), then
274 + *
275 + * n sin(xn)*sqt2 cos(xn)*sqt2
276 + * ----------------------------------
277 + * 0 s-c c+s
278 + * 1 -s-c -c+s
279 + * 2 -s+c -c-s
280 + * 3 s+c c-s
281 + */
261 282 switch (n&3) {
262 - case 0: temp = sinl(x)-cosl(x); break;
263 - case 1: temp = -sinl(x)-cosl(x); break;
264 - case 2: temp = -sinl(x)+cosl(x); break;
265 - case 3: temp = sinl(x)+cosl(x); break;
283 + case 0:
284 + temp = sinl(x)-cosl(x);
285 + break;
286 + case 1:
287 + temp = -sinl(x)-cosl(x);
288 + break;
289 + case 2:
290 + temp = -sinl(x)+cosl(x);
291 + break;
292 + case 3:
293 + temp = sinl(x)+cosl(x);
294 + break;
266 295 }
267 296 b = invsqrtpi*temp/sqrtl(x);
268 297 } else {
269 298 a = y0l(x);
270 299 b = y1l(x);
271 300 /*
272 301 * fix 1262058 and take care of non-default rounding
273 302 */
274 303 for (i = 1; i < n; i++) {
275 304 temp = b;
276 305 b *= (GENERIC) (i + i) / x;
277 306 if (b <= -LDBL_MAX)
278 307 break;
279 308 b -= a;
280 309 a = temp;
281 310 }
282 311 }
283 312 if (sign > 0)
284 313 return (b);
285 314 else
286 315 return (-b);
287 316 }
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