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128-bit long double Bessel functions jn and yn.
[thirdparty/glibc.git] / sysdeps / ieee754 / ldbl-128 / e_jnl.c
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1/*
2 * ====================================================
3 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
4 *
5 * Developed at SunPro, a Sun Microsystems, Inc. business.
6 * Permission to use, copy, modify, and distribute this
7 * software is freely granted, provided that this notice
8 * is preserved.
9 * ====================================================
10 */
11
12/* Modifications for 128-bit long double contributed by
13 Stephen L. Moshier <moshier@na-net.ornl.gov> */
14
15/*
16 * __ieee754_jn(n, x), __ieee754_yn(n, x)
17 * floating point Bessel's function of the 1st and 2nd kind
18 * of order n
19 *
20 * Special cases:
21 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
22 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
23 * Note 2. About jn(n,x), yn(n,x)
24 * For n=0, j0(x) is called,
25 * for n=1, j1(x) is called,
26 * for n<x, forward recursion us used starting
27 * from values of j0(x) and j1(x).
28 * for n>x, a continued fraction approximation to
29 * j(n,x)/j(n-1,x) is evaluated and then backward
30 * recursion is used starting from a supposed value
31 * for j(n,x). The resulting value of j(0,x) is
32 * compared with the actual value to correct the
33 * supposed value of j(n,x).
34 *
35 * yn(n,x) is similar in all respects, except
36 * that forward recursion is used for all
37 * values of n>1.
38 *
39 */
40
41#include "math.h"
42#include "math_private.h"
43
44#ifdef __STDC__
45static const long double
46#else
47static long double
48#endif
49 invsqrtpi = 5.6418958354775628694807945156077258584405E-1L,
50 two = 2.0e0L,
51 one = 1.0e0L,
52 zero = 0.0L;
53
54
55#ifdef __STDC__
56long double
57__ieee754_jnl (int n, long double x)
58#else
59long double
60__ieee754_jnl (n, x)
61 int n;
62 long double x;
63#endif
64{
65 u_int32_t se;
66 int32_t i, ix, sgn;
67 long double a, b, temp, di;
68 long double z, w;
69 ieee854_long_double_shape_type u;
70
71
72 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
73 * Thus, J(-n,x) = J(n,-x)
74 */
75
76 u.value = x;
77 se = u.parts32.w0;
78 ix = se & 0x7fffffff;
79
80 /* if J(n,NaN) is NaN */
81 if (ix >= 0x7fff0000)
82 {
83 if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
84 return x + x;
85 }
86
87 if (n < 0)
88 {
89 n = -n;
90 x = -x;
91 se ^= 0x80000000;
92 }
93 if (n == 0)
94 return (__ieee754_j0l (x));
95 if (n == 1)
96 return (__ieee754_j1l (x));
97 sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */
98 x = fabsl (x);
99
100 if (x == 0.0L || ix >= 0x7fff0000) /* if x is 0 or inf */
101 b = zero;
102 else if ((long double) n <= x)
103 {
104 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
105 if (ix >= 0x412D0000)
106 { /* x > 2**302 */
107
108 /* ??? Could use an expansion for large x here. */
109
110 /* (x >> n**2)
111 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
112 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
113 * Let s=sin(x), c=cos(x),
114 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
115 *
116 * n sin(xn)*sqt2 cos(xn)*sqt2
117 * ----------------------------------
118 * 0 s-c c+s
119 * 1 -s-c -c+s
120 * 2 -s+c -c-s
121 * 3 s+c c-s
122 */
123 long double s;
124 long double c;
125 __sincosl (x, &s, &c);
126 switch (n & 3)
127 {
128 case 0:
129 temp = c + s;
130 break;
131 case 1:
132 temp = -c + s;
133 break;
134 case 2:
135 temp = -c - s;
136 break;
137 case 3:
138 temp = c - s;
139 break;
140 }
141 b = invsqrtpi * temp / __ieee754_sqrtl (x);
142 }
143 else
144 {
145 a = __ieee754_j0l (x);
146 b = __ieee754_j1l (x);
147 for (i = 1; i < n; i++)
148 {
149 temp = b;
150 b = b * ((long double) (i + i) / x) - a; /* avoid underflow */
151 a = temp;
152 }
153 }
154 }
155 else
156 {
157 if (ix < 0x3fc60000)
158 { /* x < 2**-57 */
159 /* x is tiny, return the first Taylor expansion of J(n,x)
160 * J(n,x) = 1/n!*(x/2)^n - ...
161 */
162 if (n >= 400) /* underflow, result < 10^-4952 */
163 b = zero;
164 else
165 {
166 temp = x * 0.5;
167 b = temp;
168 for (a = one, i = 2; i <= n; i++)
169 {
170 a *= (long double) i; /* a = n! */
171 b *= temp; /* b = (x/2)^n */
172 }
173 b = b / a;
174 }
175 }
176 else
177 {
178 /* use backward recurrence */
179 /* x x^2 x^2
180 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
181 * 2n - 2(n+1) - 2(n+2)
182 *
183 * 1 1 1
184 * (for large x) = ---- ------ ------ .....
185 * 2n 2(n+1) 2(n+2)
186 * -- - ------ - ------ -
187 * x x x
188 *
189 * Let w = 2n/x and h=2/x, then the above quotient
190 * is equal to the continued fraction:
191 * 1
192 * = -----------------------
193 * 1
194 * w - -----------------
195 * 1
196 * w+h - ---------
197 * w+2h - ...
198 *
199 * To determine how many terms needed, let
200 * Q(0) = w, Q(1) = w(w+h) - 1,
201 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
202 * When Q(k) > 1e4 good for single
203 * When Q(k) > 1e9 good for double
204 * When Q(k) > 1e17 good for quadruple
205 */
206 /* determine k */
207 long double t, v;
208 long double q0, q1, h, tmp;
209 int32_t k, m;
210 w = (n + n) / (long double) x;
211 h = 2.0L / (long double) x;
212 q0 = w;
213 z = w + h;
214 q1 = w * z - 1.0L;
215 k = 1;
216 while (q1 < 1.0e17L)
217 {
218 k += 1;
219 z += h;
220 tmp = z * q1 - q0;
221 q0 = q1;
222 q1 = tmp;
223 }
224 m = n + n;
225 for (t = zero, i = 2 * (n + k); i >= m; i -= 2)
226 t = one / (i / x - t);
227 a = t;
228 b = one;
229 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
230 * Hence, if n*(log(2n/x)) > ...
231 * single 8.8722839355e+01
232 * double 7.09782712893383973096e+02
233 * long double 1.1356523406294143949491931077970765006170e+04
234 * then recurrent value may overflow and the result is
235 * likely underflow to zero
236 */
237 tmp = n;
238 v = two / x;
239 tmp = tmp * __ieee754_logl (fabsl (v * tmp));
240
241 if (tmp < 1.1356523406294143949491931077970765006170e+04L)
242 {
243 for (i = n - 1, di = (long double) (i + i); i > 0; i--)
244 {
245 temp = b;
246 b *= di;
247 b = b / x - a;
248 a = temp;
249 di -= two;
250 }
251 }
252 else
253 {
254 for (i = n - 1, di = (long double) (i + i); i > 0; i--)
255 {
256 temp = b;
257 b *= di;
258 b = b / x - a;
259 a = temp;
260 di -= two;
261 /* scale b to avoid spurious overflow */
262 if (b > 1e100L)
263 {
264 a /= b;
265 t /= b;
266 b = one;
267 }
268 }
269 }
270 b = (t * __ieee754_j0l (x) / b);
271 }
272 }
273 if (sgn == 1)
274 return -b;
275 else
276 return b;
277}
278
279#ifdef __STDC__
280long double
281__ieee754_ynl (int n, long double x)
282#else
283long double
284__ieee754_ynl (n, x)
285 int n;
286 long double x;
287#endif
288{
289 u_int32_t se;
290 int32_t i, ix;
291 int32_t sign;
292 long double a, b, temp;
293 ieee854_long_double_shape_type u;
294
295 u.value = x;
296 se = u.parts32.w0;
297 ix = se & 0x7fffffff;
298
299 /* if Y(n,NaN) is NaN */
300 if (ix >= 0x7fff0000)
301 {
302 if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
303 return x + x;
304 }
305 if (x <= 0.0L)
306 {
307 if (x == 0.0L)
308 return -one / zero;
309 if (se & 0x80000000)
310 return zero / zero;
311 }
312 sign = 1;
313 if (n < 0)
314 {
315 n = -n;
316 sign = 1 - ((n & 1) << 1);
317 }
318 if (n == 0)
319 return (__ieee754_y0l (x));
320 if (n == 1)
321 return (sign * __ieee754_y1l (x));
322 if (ix >= 0x7fff0000)
323 return zero;
324 if (ix >= 0x412D0000)
325 { /* x > 2**302 */
326
327 /* ??? See comment above on the possible futility of this. */
328
329 /* (x >> n**2)
330 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
331 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
332 * Let s=sin(x), c=cos(x),
333 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
334 *
335 * n sin(xn)*sqt2 cos(xn)*sqt2
336 * ----------------------------------
337 * 0 s-c c+s
338 * 1 -s-c -c+s
339 * 2 -s+c -c-s
340 * 3 s+c c-s
341 */
342 long double s;
343 long double c;
344 __sincosl (x, &s, &c);
345 switch (n & 3)
346 {
347 case 0:
348 temp = s - c;
349 break;
350 case 1:
351 temp = -s - c;
352 break;
353 case 2:
354 temp = -s + c;
355 break;
356 case 3:
357 temp = s + c;
358 break;
359 }
360 b = invsqrtpi * temp / __ieee754_sqrtl (x);
361 }
362 else
363 {
364 a = __ieee754_y0l (x);
365 b = __ieee754_y1l (x);
366 /* quit if b is -inf */
367 u.value = b;
368 se = u.parts32.w0 & 0xffff0000;
369 for (i = 1; i < n && se != 0xffff0000; i++)
370 {
371 temp = b;
372 b = ((long double) (i + i) / x) * b - a;
373 u.value = b;
374 se = u.parts32.w0 & 0xffff0000;
375 a = temp;
376 }
377 }
378 if (sign > 0)
379 return b;
380 else
381 return -b;
382}