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
& 0xfffff) | u
.parts32
.w1
102 | (u
.parts32
.w2
& 0x7fffffff) | u
.parts32
.w3
)
113 return (__ieee754_j0l (x
));
115 return (__ieee754_j1l (x
));
116 sgn
= (n
& 1) & (se
>> 31); /* even n -- 0, odd n -- sign(x) */
119 if (x
== 0.0L || ix
>= 0x7ff00000) /* if x is 0 or inf */
121 else if ((long double) n
<= x
)
123 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
124 if (ix
>= 0x52d00000)
127 /* ??? Could use an expansion for large x here. */
130 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
131 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
132 * Let s=sin(x), c=cos(x),
133 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
135 * n sin(xn)*sqt2 cos(xn)*sqt2
136 * ----------------------------------
144 __sincosl (x
, &s
, &c
);
160 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
164 a
= __ieee754_j0l (x
);
165 b
= __ieee754_j1l (x
);
166 for (i
= 1; i
< n
; i
++)
169 b
= b
* ((long double) (i
+ i
) / x
) - a
; /* avoid underflow */
178 /* x is tiny, return the first Taylor expansion of J(n,x)
179 * J(n,x) = 1/n!*(x/2)^n - ...
181 if (n
>= 33) /* underflow, result < 10^-300 */
187 for (a
= one
, i
= 2; i
<= n
; i
++)
189 a
*= (long double) i
; /* a = n! */
190 b
*= temp
; /* b = (x/2)^n */
197 /* use backward recurrence */
199 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
200 * 2n - 2(n+1) - 2(n+2)
203 * (for large x) = ---- ------ ------ .....
205 * -- - ------ - ------ -
208 * Let w = 2n/x and h=2/x, then the above quotient
209 * is equal to the continued fraction:
211 * = -----------------------
213 * w - -----------------
218 * To determine how many terms needed, let
219 * Q(0) = w, Q(1) = w(w+h) - 1,
220 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
221 * When Q(k) > 1e4 good for single
222 * When Q(k) > 1e9 good for double
223 * When Q(k) > 1e17 good for quadruple
227 long double q0
, q1
, h
, tmp
;
229 w
= (n
+ n
) / (long double) x
;
230 h
= 2.0L / (long double) x
;
244 for (t
= zero
, i
= 2 * (n
+ k
); i
>= m
; i
-= 2)
245 t
= one
/ (i
/ x
- t
);
248 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
249 * Hence, if n*(log(2n/x)) > ...
250 * single 8.8722839355e+01
251 * double 7.09782712893383973096e+02
252 * long double 1.1356523406294143949491931077970765006170e+04
253 * then recurrent value may overflow and the result is
254 * likely underflow to zero
258 tmp
= tmp
* __ieee754_logl (fabsl (v
* tmp
));
260 if (tmp
< 1.1356523406294143949491931077970765006170e+04L)
262 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
273 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
280 /* scale b to avoid spurious overflow */
289 /* j0() and j1() suffer enormous loss of precision at and
290 * near zero; however, we know that their zero points never
291 * coincide, so just choose the one further away from zero.
293 z
= __ieee754_j0l (x
);
294 w
= __ieee754_j1l (x
);
295 if (fabsl (z
) >= fabsl (w
))
309 __ieee754_ynl (int n
, long double x
)
320 long double a
, b
, temp
;
321 ieee854_long_double_shape_type u
;
325 ix
= se
& 0x7fffffff;
327 /* if Y(n,NaN) is NaN */
328 if (ix
>= 0x7ff00000)
330 if ((u
.parts32
.w0
& 0xfffff) | u
.parts32
.w1
331 | (u
.parts32
.w2
& 0x7fffffff) | u
.parts32
.w3
)
337 return -HUGE_VALL
+ x
;
339 return zero
/ (zero
* x
);
345 sign
= 1 - ((n
& 1) << 1);
348 return (__ieee754_y0l (x
));
350 return (sign
* __ieee754_y1l (x
));
351 if (ix
>= 0x7ff00000)
353 if (ix
>= 0x52D00000)
356 /* ??? See comment above on the possible futility of this. */
359 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
360 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
361 * Let s=sin(x), c=cos(x),
362 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
364 * n sin(xn)*sqt2 cos(xn)*sqt2
365 * ----------------------------------
373 __sincosl (x
, &s
, &c
);
389 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
393 a
= __ieee754_y0l (x
);
394 b
= __ieee754_y1l (x
);
395 /* quit if b is -inf */
397 se
= u
.parts32
.w0
& 0xfff00000;
398 for (i
= 1; i
< n
&& se
!= 0xfff00000; i
++)
401 b
= ((long double) (i
+ i
) / x
) * b
- a
;
403 se
= u
.parts32
.w0
& 0xfff00000;