2 * ====================================================
3 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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
9 * ====================================================
12 /* Modifications for 128-bit long double are
13 Copyright (C) 2001 Stephen L. Moshier <moshier@na-net.ornl.gov>
14 and are incorporated herein by permission of the author. The author
15 reserves the right to distribute this material elsewhere under different
16 copying permissions. These modifications are distributed here under
19 This library is free software; you can redistribute it and/or
20 modify it under the terms of the GNU Lesser General Public
21 License as published by the Free Software Foundation; either
22 version 2.1 of the License, or (at your option) any later version.
24 This library is distributed in the hope that it will be useful,
25 but WITHOUT ANY WARRANTY; without even the implied warranty of
26 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
27 Lesser General Public License for more details.
29 You should have received a copy of the GNU Lesser General Public
30 License along with this library; if not, see
31 <http://www.gnu.org/licenses/>. */
34 * __ieee754_jn(n, x), __ieee754_yn(n, x)
35 * floating point Bessel's function of the 1st and 2nd kind
39 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
40 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
41 * Note 2. About jn(n,x), yn(n,x)
42 * For n=0, j0(x) is called,
43 * for n=1, j1(x) is called,
44 * for n<x, forward recursion us used starting
45 * from values of j0(x) and j1(x).
46 * for n>x, a continued fraction approximation to
47 * j(n,x)/j(n-1,x) is evaluated and then backward
48 * recursion is used starting from a supposed value
49 * for j(n,x). The resulting value of j(0,x) is
50 * compared with the actual value to correct the
51 * supposed value of j(n,x).
53 * yn(n,x) is similar in all respects, except
54 * that forward recursion is used for all
62 #include <math_private.h>
64 static const long double
65 invsqrtpi
= 5.6418958354775628694807945156077258584405E-1L,
72 __ieee754_jnl (int n
, long double x
)
76 long double a
, b
, temp
, di
, ret
;
81 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
82 * Thus, J(-n,x) = J(n,-x)
86 EXTRACT_WORDS (se
, lx
, xhi
);
89 /* if J(n,NaN) is NaN */
92 if (((ix
- 0x7ff00000) | lx
) != 0)
103 return (__ieee754_j0l (x
));
105 return (__ieee754_j1l (x
));
106 sgn
= (n
& 1) & (se
>> 31); /* even n -- 0, odd n -- sign(x) */
110 SET_RESTORE_ROUNDL (FE_TONEAREST
);
111 if (x
== 0.0L || ix
>= 0x7ff00000) /* if x is 0 or inf */
112 return sgn
== 1 ? -zero
: zero
;
113 else if ((long double) n
<= x
)
115 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
116 if (ix
>= 0x52d00000)
119 /* ??? Could use an expansion for large x here. */
122 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
123 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
124 * Let s=sin(x), c=cos(x),
125 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
127 * n sin(xn)*sqt2 cos(xn)*sqt2
128 * ----------------------------------
136 __sincosl (x
, &s
, &c
);
152 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
156 a
= __ieee754_j0l (x
);
157 b
= __ieee754_j1l (x
);
158 for (i
= 1; i
< n
; i
++)
161 b
= b
* ((long double) (i
+ i
) / x
) - a
; /* avoid underflow */
170 /* x is tiny, return the first Taylor expansion of J(n,x)
171 * J(n,x) = 1/n!*(x/2)^n - ...
173 if (n
>= 33) /* underflow, result < 10^-300 */
179 for (a
= one
, i
= 2; i
<= n
; i
++)
181 a
*= (long double) i
; /* a = n! */
182 b
*= temp
; /* b = (x/2)^n */
189 /* use backward recurrence */
191 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
192 * 2n - 2(n+1) - 2(n+2)
195 * (for large x) = ---- ------ ------ .....
197 * -- - ------ - ------ -
200 * Let w = 2n/x and h=2/x, then the above quotient
201 * is equal to the continued fraction:
203 * = -----------------------
205 * w - -----------------
210 * To determine how many terms needed, let
211 * Q(0) = w, Q(1) = w(w+h) - 1,
212 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
213 * When Q(k) > 1e4 good for single
214 * When Q(k) > 1e9 good for double
215 * When Q(k) > 1e17 good for quadruple
219 long double q0
, q1
, h
, tmp
;
221 w
= (n
+ n
) / (long double) x
;
222 h
= 2.0L / (long double) x
;
236 for (t
= zero
, i
= 2 * (n
+ k
); i
>= m
; i
-= 2)
237 t
= one
/ (i
/ x
- t
);
240 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
241 * Hence, if n*(log(2n/x)) > ...
242 * single 8.8722839355e+01
243 * double 7.09782712893383973096e+02
244 * long double 1.1356523406294143949491931077970765006170e+04
245 * then recurrent value may overflow and the result is
246 * likely underflow to zero
250 tmp
= tmp
* __ieee754_logl (fabsl (v
* tmp
));
252 if (tmp
< 1.1356523406294143949491931077970765006170e+04L)
254 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
265 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
272 /* scale b to avoid spurious overflow */
281 /* j0() and j1() suffer enormous loss of precision at and
282 * near zero; however, we know that their zero points never
283 * coincide, so just choose the one further away from zero.
285 z
= __ieee754_j0l (x
);
286 w
= __ieee754_j1l (x
);
287 if (fabsl (z
) >= fabsl (w
))
299 ret
= __copysignl (LDBL_MIN
, ret
) * LDBL_MIN
;
302 strong_alias (__ieee754_jnl
, __jnl_finite
)
305 __ieee754_ynl (int n
, long double x
)
310 long double a
, b
, temp
, ret
;
314 EXTRACT_WORDS (se
, lx
, xhi
);
315 ix
= se
& 0x7fffffff;
317 /* if Y(n,NaN) is NaN */
318 if (ix
>= 0x7ff00000)
320 if (((ix
- 0x7ff00000) | lx
) != 0)
326 return ((n
< 0 && (n
& 1) != 0) ? 1.0L : -1.0L) / 0.0L;
328 return zero
/ (zero
* x
);
334 sign
= 1 - ((n
& 1) << 1);
337 return (__ieee754_y0l (x
));
339 SET_RESTORE_ROUNDL (FE_TONEAREST
);
342 ret
= sign
* __ieee754_y1l (x
);
345 if (ix
>= 0x7ff00000)
347 if (ix
>= 0x52D00000)
350 /* ??? See comment above on the possible futility of this. */
353 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
354 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
355 * Let s=sin(x), c=cos(x),
356 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
358 * n sin(xn)*sqt2 cos(xn)*sqt2
359 * ----------------------------------
367 __sincosl (x
, &s
, &c
);
383 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
387 a
= __ieee754_y0l (x
);
388 b
= __ieee754_y1l (x
);
389 /* quit if b is -inf */
391 GET_HIGH_WORD (se
, xhi
);
393 for (i
= 1; i
< n
&& se
!= 0xfff00000; i
++)
396 b
= ((long double) (i
+ i
) / x
) * b
- a
;
398 GET_HIGH_WORD (se
, xhi
);
403 /* If B is +-Inf, set up errno accordingly. */
405 __set_errno (ERANGE
);
413 ret
= __copysignl (LDBL_MAX
, ret
) * LDBL_MAX
;
416 strong_alias (__ieee754_ynl
, __ynl_finite
)