11175 libm should use signbit() correctly 11188 c99 math macros should return strictly backward compatible values
51 *
52 * yn(n,x) is similar in all respects, except
53 * that forward recursion is used for all
54 * values of n>1.
55 *
56 */
57
58 #include "libm.h"
59 #include "longdouble.h"
60 #include <float.h> /* LDBL_MAX */
61
62 #define GENERIC long double
63
64 static const GENERIC
65 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
66 two = 2.0L,
67 zero = 0.0L,
68 one = 1.0L;
69
70 GENERIC
71 jnl(n, x) int n; GENERIC x; {
72 int i, sgn;
73 GENERIC a, b, temp, z, w;
74
75 /*
76 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
77 * Thus, J(-n,x) = J(n,-x)
78 */
79 if (n < 0) {
80 n = -n;
81 x = -x;
82 }
83 if (n == 0)
84 return (j0l(x));
85 if (n == 1)
86 return (j1l(x));
87 if (x != x)
88 return (x+x);
89 if ((n&1) == 0)
90 sgn = 0; /* even n */
91 else
96 /*
97 * Safe to use
98 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
99 */
100 if (x > 1.0e91L) {
101 /*
102 * x >> n**2
103 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
104 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 * Let s=sin(x), c=cos(x),
106 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
107 *
108 * n sin(xn)*sqt2 cos(xn)*sqt2
109 * ----------------------------------
110 * 0 s-c c+s
111 * 1 -s-c -c+s
112 * 2 -s+c -c-s
113 * 3 s+c c-s
114 */
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 {
123 a = j0l(x);
124 b = j1l(x);
125 for (i = 1; i < n; i++) {
126 temp = b;
127 b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
128 a = temp;
129 }
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 }
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 }
201 } else {
202 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 }
212 }
213 b = (t*j0l(x)/b);
214 }
215 }
216 if (sgn == 1)
217 return (-b);
218 else
219 return (b);
220 }
221
222 GENERIC ynl(n, x)
223 int n; GENERIC x; {
224 int i;
225 int sign;
226 GENERIC a, b, temp;
227
228 if (x != x)
229 return (x+x);
230 if (x <= zero) {
231 if (x == zero)
232 return (-one/zero);
233 else
234 return (zero/zero);
235 }
236 sign = 1;
237 if (n < 0) {
238 n = -n;
239 if ((n&1) == 1) sign = -1;
240 }
241 if (n == 0)
242 return (y0l(x));
243 if (n == 1)
244 return (sign*y1l(x));
245 if (!finitel(x))
246 return (zero);
247
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 */
261 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;
266 }
267 b = invsqrtpi*temp/sqrtl(x);
268 } else {
269 a = y0l(x);
270 b = y1l(x);
271 /*
272 * fix 1262058 and take care of non-default rounding
273 */
274 for (i = 1; i < n; i++) {
275 temp = b;
276 b *= (GENERIC) (i + i) / x;
277 if (b <= -LDBL_MAX)
278 break;
279 b -= a;
280 a = temp;
281 }
282 }
283 if (sign > 0)
284 return (b);
285 else
| 51 * 52 * yn(n,x) is similar in all respects, except 53 * that forward recursion is used for all 54 * values of n>1. 55 * 56 */ 57 58 #include "libm.h" 59 #include "longdouble.h" 60 #include <float.h> /* LDBL_MAX */ 61 62 #define GENERIC long double 63 64 static const GENERIC 65 invsqrtpi = 5.641895835477562869480794515607725858441e-0001L, 66 two = 2.0L, 67 zero = 0.0L, 68 one = 1.0L; 69 70 GENERIC 71 jnl(int n, GENERIC x) 72 { 73 int i, sgn; 74 GENERIC a, b, temp, z, w; 75 76 /* 77 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) 78 * Thus, J(-n,x) = J(n,-x) 79 */ 80 if (n < 0) { 81 n = -n; 82 x = -x; 83 } 84 if (n == 0) 85 return (j0l(x)); 86 if (n == 1) 87 return (j1l(x)); 88 if (x != x) 89 return (x+x); 90 if ((n&1) == 0) 91 sgn = 0; /* even n */ 92 else
97 /*
98 * Safe to use
99 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
100 */
101 if (x > 1.0e91L) {
102 /*
103 * x >> n**2
104 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
105 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
106 * Let s=sin(x), c=cos(x),
107 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
108 *
109 * n sin(xn)*sqt2 cos(xn)*sqt2
110 * ----------------------------------
111 * 0 s-c c+s
112 * 1 -s-c -c+s
113 * 2 -s+c -c-s
114 * 3 s+c c-s
115 */
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 {
132 a = j0l(x);
133 b = j1l(x);
134 for (i = 1; i < n; i++) {
135 temp = b;
136 /* avoid underflow */
137 b = b*((GENERIC)(i+i)/x) - a;
138 a = temp;
139 }
140 }
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 }
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 {
220 for (i = n-1; i > 0; i--) {
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);
232 }
233 }
234 if (sgn != 0)
235 return (-b);
236 else
237 return (b);
238 }
239
240 GENERIC
241 ynl(int n, GENERIC x)
242 {
243 int i;
244 int sign;
245 GENERIC a, b, temp;
246
247 if (x != x)
248 return (x+x);
249 if (x <= zero) {
250 if (x == zero)
251 return (-one/zero);
252 else
253 return (zero/zero);
254 }
255 sign = 1;
256 if (n < 0) {
257 n = -n;
258 if ((n&1) == 1) sign = -1;
259 }
260 if (n == 0)
261 return (y0l(x));
262 if (n == 1)
263 return (sign*y1l(x));
264 if (!finitel(x))
265 return (zero);
266
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 */
282 switch (n&3) {
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;
295 }
296 b = invsqrtpi*temp/sqrtl(x);
297 } else {
298 a = y0l(x);
299 b = y1l(x);
300 /*
301 * fix 1262058 and take care of non-default rounding
302 */
303 for (i = 1; i < n; i++) {
304 temp = b;
305 b *= (GENERIC) (i + i) / x;
306 if (b <= -LDBL_MAX)
307 break;
308 b -= a;
309 a = temp;
310 }
311 }
312 if (sign > 0)
313 return (b);
314 else
|