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, write to the Free Software
31 Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */
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
60 #include "math_private.h"
63 static const long double
67 invsqrtpi
= 5.6418958354775628694807945156077258584405E-1L,
75 __ieee754_jnl (int n
, long double x
)
85 long double a
, b
, temp
, di
;
87 ieee854_long_double_shape_type u
;
90 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
91 * Thus, J(-n,x) = J(n,-x)
98 /* if J(n,NaN) is NaN */
101 if ((u
.parts32
.w0
& 0xffff) | u
.parts32
.w1
| u
.parts32
.w2
| u
.parts32
.w3
)
112 return (__ieee754_j0l (x
));
114 return (__ieee754_j1l (x
));
115 sgn
= (n
& 1) & (se
>> 31); /* even n -- 0, odd n -- sign(x) */
118 if (x
== 0.0L || ix
>= 0x7fff0000) /* if x is 0 or inf */
120 else if ((long double) n
<= x
)
122 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
123 if (ix
>= 0x412D0000)
126 /* ??? Could use an expansion for large x here. */
129 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
130 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
131 * Let s=sin(x), c=cos(x),
132 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
134 * n sin(xn)*sqt2 cos(xn)*sqt2
135 * ----------------------------------
143 __sincosl (x
, &s
, &c
);
159 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
163 a
= __ieee754_j0l (x
);
164 b
= __ieee754_j1l (x
);
165 for (i
= 1; i
< n
; i
++)
168 b
= b
* ((long double) (i
+ i
) / x
) - a
; /* avoid underflow */
177 /* x is tiny, return the first Taylor expansion of J(n,x)
178 * J(n,x) = 1/n!*(x/2)^n - ...
180 if (n
>= 400) /* underflow, result < 10^-4952 */
186 for (a
= one
, i
= 2; i
<= n
; i
++)
188 a
*= (long double) i
; /* a = n! */
189 b
*= temp
; /* b = (x/2)^n */
196 /* use backward recurrence */
198 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
199 * 2n - 2(n+1) - 2(n+2)
202 * (for large x) = ---- ------ ------ .....
204 * -- - ------ - ------ -
207 * Let w = 2n/x and h=2/x, then the above quotient
208 * is equal to the continued fraction:
210 * = -----------------------
212 * w - -----------------
217 * To determine how many terms needed, let
218 * Q(0) = w, Q(1) = w(w+h) - 1,
219 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
220 * When Q(k) > 1e4 good for single
221 * When Q(k) > 1e9 good for double
222 * When Q(k) > 1e17 good for quadruple
226 long double q0
, q1
, h
, tmp
;
228 w
= (n
+ n
) / (long double) x
;
229 h
= 2.0L / (long double) x
;
243 for (t
= zero
, i
= 2 * (n
+ k
); i
>= m
; i
-= 2)
244 t
= one
/ (i
/ x
- t
);
247 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
248 * Hence, if n*(log(2n/x)) > ...
249 * single 8.8722839355e+01
250 * double 7.09782712893383973096e+02
251 * long double 1.1356523406294143949491931077970765006170e+04
252 * then recurrent value may overflow and the result is
253 * likely underflow to zero
257 tmp
= tmp
* __ieee754_logl (fabsl (v
* tmp
));
259 if (tmp
< 1.1356523406294143949491931077970765006170e+04L)
261 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
272 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
279 /* scale b to avoid spurious overflow */
288 b
= (t
* __ieee754_j0l (x
) / b
);
299 __ieee754_ynl (int n
, long double x
)
310 long double a
, b
, temp
;
311 ieee854_long_double_shape_type u
;
315 ix
= se
& 0x7fffffff;
317 /* if Y(n,NaN) is NaN */
318 if (ix
>= 0x7fff0000)
320 if ((u
.parts32
.w0
& 0xffff) | u
.parts32
.w1
| u
.parts32
.w2
| u
.parts32
.w3
)
326 return -HUGE_VALL
+ x
;
328 return zero
/ (zero
* x
);
334 sign
= 1 - ((n
& 1) << 1);
337 return (__ieee754_y0l (x
));
339 return (sign
* __ieee754_y1l (x
));
340 if (ix
>= 0x7fff0000)
342 if (ix
>= 0x412D0000)
345 /* ??? See comment above on the possible futility of this. */
348 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
349 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
350 * Let s=sin(x), c=cos(x),
351 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
353 * n sin(xn)*sqt2 cos(xn)*sqt2
354 * ----------------------------------
362 __sincosl (x
, &s
, &c
);
378 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
382 a
= __ieee754_y0l (x
);
383 b
= __ieee754_y1l (x
);
384 /* quit if b is -inf */
386 se
= u
.parts32
.w0
& 0xffff0000;
387 for (i
= 1; i
< n
&& se
!= 0xffff0000; i
++)
390 b
= ((long double) (i
+ i
) / x
) * b
- a
;
392 se
= u
.parts32
.w0
& 0xffff0000;