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 = 0, 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) return (j0l(x));
84 if (n == 1) return (j1l(x));
85 if (x != x) return x+x;
86 if ((n&1) == 0)
87 sgn = 0; /* even n */
88 else
89 sgn = signbitl(x); /* old n */
90 x = fabsl(x);
91 if (x == zero || !finitel(x)) b = zero;
92 else if ((GENERIC)n <= x) {
93 /*
94 * Safe to use
95 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
96 */
97 if (x > 1.0e91L) {
98 /*
99 * x >> n**2
100 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
101 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
102 * Let s=sin(x), c=cos(x),
103 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
104 *
105 * n sin(xn)*sqt2 cos(xn)*sqt2
106 * ----------------------------------
107 * 0 s-c c+s
108 * 1 -s-c -c+s
109 * 2 -s+c -c-s
110 * 3 s+c c-s
111 */
112 switch (n&3) {
113 case 0: temp = cosl(x)+sinl(x); break;
114 case 1: temp = -cosl(x)+sinl(x); break;
115 case 2: temp = -cosl(x)-sinl(x); break;
116 case 3: temp = cosl(x)-sinl(x); break;
117 }
118 b = invsqrtpi*temp/sqrtl(x);
119 } else {
120 a = j0l(x);
121 b = j1l(x);
122 for (i = 1; i < n; i++) {
123 temp = b;
124 b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
125 a = temp;
126 }
127 }
128 } else {
129 if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */
130 b = powl(0.5L*x, (GENERIC) n);
131 if (b != zero) {
132 for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
133 b = b/a;
134 }
135 } else {
136 /*
137 * use backward recurrence
138 * x x^2 x^2
139 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
140 * 2n - 2(n+1) - 2(n+2)
141 *
142 * 1 1 1
143 * (for large x) = ---- ------ ------ .....
144 * 2n 2(n+1) 2(n+2)
145 * -- - ------ - ------ -
146 * x x x
147 *
148 * Let w = 2n/x and h=2/x, then the above quotient
149 * is equal to the continued fraction:
150 * 1
151 * = -----------------------
152 * 1
153 * w - -----------------
154 * 1
155 * w+h - ---------
156 * w+2h - ...
157 *
158 * To determine how many terms needed, let
159 * Q(0) = w, Q(1) = w(w+h) - 1,
160 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
161 * When Q(k) > 1e4 good for single
162 * When Q(k) > 1e9 good for double
163 * When Q(k) > 1e17 good for quaduple
164 */
165 /* determin k */
166 GENERIC t, v;
167 double q0, q1, h, tmp; int k, m;
168 w = (n+n)/(double)x; h = 2.0/(double)x;
169 q0 = w; z = w+h; q1 = w*z - 1.0; k = 1;
170 while (q1 < 1.0e17) {
171 k += 1; z += h;
172 tmp = z*q1 - q0;
173 q0 = q1;
174 q1 = tmp;
175 }
176 m = n+n;
177 for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
178 a = t;
179 b = one;
180 /*
181 * Estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
182 * hence, if n*(log(2n/x)) > ...
183 * single 8.8722839355e+01
184 * double 7.09782712893383973096e+02
185 * long double 1.1356523406294143949491931077970765006170e+04
186 * then recurrent value may overflow and the result is
187 * likely underflow to zero.
188 */
189 tmp = n;
190 v = two/x;
191 tmp = tmp*logl(fabsl(v*tmp));
192 if (tmp < 1.1356523406294143949491931077970765e+04L) {
193 for (i = n-1; i > 0; i--) {
194 temp = b;
195 b = ((i+i)/x)*b - a;
196 a = temp;
197 }
198 } else {
199 for (i = n-1; i > 0; i--) {
200 temp = b;
201 b = ((i+i)/x)*b - a;
202 a = temp;
203 if (b > 1e1000L) {
204 a /= b;
205 t /= b;
206 b = 1.0;
207 }
208 }
209 }
210 b = (t*j0l(x)/b);
211 }
212 }
213 if (sgn == 1)
214 return -b;
215 else
216 return b;
217 }
218
219 GENERIC
220 ynl(n, x) int n; GENERIC x; {
221 int i;
222 int sign;
223 GENERIC a, b, temp = 0;
224
225 if (x != x)
226 return x+x;
227 if (x <= zero) {
228 if (x == zero)
229 return -one/zero;
230 else
231 return zero/zero;
232 }
233 sign = 1;
234 if (n < 0) {
235 n = -n;
236 if ((n&1) == 1) sign = -1;
237 }
238 if (n == 0) return (y0l(x));
239 if (n == 1) return (sign*y1l(x));
240 if (!finitel(x)) return zero;
241
242 if (x > 1.0e91L) {
243 /*
244 * x >> n**2
245 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
246 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
247 * Let s=sin(x), c=cos(x),
248 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
249 *
250 * n sin(xn)*sqt2 cos(xn)*sqt2
251 * ----------------------------------
252 * 0 s-c c+s
253 * 1 -s-c -c+s
254 * 2 -s+c -c-s
255 * 3 s+c c-s
256 */
257 switch (n&3) {
258 case 0: temp = sinl(x)-cosl(x); break;
259 case 1: temp = -sinl(x)-cosl(x); break;
260 case 2: temp = -sinl(x)+cosl(x); break;
261 case 3: temp = sinl(x)+cosl(x); break;
262 }
263 b = invsqrtpi*temp/sqrtl(x);
264 } else {
265 a = y0l(x);
266 b = y1l(x);
267 /*
268 * fix 1262058 and take care of non-default rounding
269 */
270 for (i = 1; i < n; i++) {
271 temp = b;
272 b *= (GENERIC) (i + i) / x;
273 if (b <= -LDBL_MAX)
274 break;
275 b -= a;
276 a = temp;
277 }
278 }
279 if (sign > 0)
280 return b;
281 else
282 return -b;
283 }
|
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 = 0, 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
93 sgn = signbitl(x); /* old n */
94 x = fabsl(x);
95 if (x == zero || !finitel(x)) b = zero;
96 else if ((GENERIC)n <= x) {
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 /*
151 * use backward recurrence
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 = 0;
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)
259 sign = -1;
260 }
261 if (n == 0)
262 return (y0l(x));
263 if (n == 1)
264 return (sign*y1l(x));
265 if (!finitel(x))
266 return (zero);
267
268 if (x > 1.0e91L) {
269 /*
270 * x >> n**2
271 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
272 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
273 * Let s=sin(x), c=cos(x),
274 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
275 *
276 * n sin(xn)*sqt2 cos(xn)*sqt2
277 * ----------------------------------
278 * 0 s-c c+s
279 * 1 -s-c -c+s
280 * 2 -s+c -c-s
281 * 3 s+c c-s
282 */
283 switch (n&3) {
284 case 0:
285 temp = sinl(x)-cosl(x);
286 break;
287 case 1:
288 temp = -sinl(x)-cosl(x);
289 break;
290 case 2:
291 temp = -sinl(x)+cosl(x);
292 break;
293 case 3:
294 temp = sinl(x)+cosl(x);
295 break;
296 }
297 b = invsqrtpi*temp/sqrtl(x);
298 } else {
299 a = y0l(x);
300 b = y1l(x);
301 /*
302 * fix 1262058 and take care of non-default rounding
303 */
304 for (i = 1; i < n; i++) {
305 temp = b;
306 b *= (GENERIC) (i + i) / x;
307 if (b <= -LDBL_MAX)
308 break;
309 b -= a;
310 a = temp;
311 }
312 }
313 if (sign > 0)
314 return (b);
315 else
316 return (-b);
317 }
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