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"
62 static const long double
63 invsqrtpi
= 5.6418958354775628694807945156077258584405E-1L,
70 __ieee754_jnl (int n
, long double x
)
74 long double a
, b
, temp
, di
;
76 ieee854_long_double_shape_type u
;
79 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
80 * Thus, J(-n,x) = J(n,-x)
87 /* if J(n,NaN) is NaN */
90 if ((u
.parts32
.w0
& 0xffff) | u
.parts32
.w1
| u
.parts32
.w2
| u
.parts32
.w3
)
101 return (__ieee754_j0l (x
));
103 return (__ieee754_j1l (x
));
104 sgn
= (n
& 1) & (se
>> 31); /* even n -- 0, odd n -- sign(x) */
107 if (x
== 0.0L || ix
>= 0x7fff0000) /* if x is 0 or inf */
109 else if ((long double) n
<= x
)
111 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
112 if (ix
>= 0x412D0000)
115 /* ??? Could use an expansion for large x here. */
118 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
119 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
120 * Let s=sin(x), c=cos(x),
121 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
123 * n sin(xn)*sqt2 cos(xn)*sqt2
124 * ----------------------------------
132 __sincosl (x
, &s
, &c
);
148 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
152 a
= __ieee754_j0l (x
);
153 b
= __ieee754_j1l (x
);
154 for (i
= 1; i
< n
; i
++)
157 b
= b
* ((long double) (i
+ i
) / x
) - a
; /* avoid underflow */
166 /* x is tiny, return the first Taylor expansion of J(n,x)
167 * J(n,x) = 1/n!*(x/2)^n - ...
169 if (n
>= 400) /* underflow, result < 10^-4952 */
175 for (a
= one
, i
= 2; i
<= n
; i
++)
177 a
*= (long double) i
; /* a = n! */
178 b
*= temp
; /* b = (x/2)^n */
185 /* use backward recurrence */
187 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
188 * 2n - 2(n+1) - 2(n+2)
191 * (for large x) = ---- ------ ------ .....
193 * -- - ------ - ------ -
196 * Let w = 2n/x and h=2/x, then the above quotient
197 * is equal to the continued fraction:
199 * = -----------------------
201 * w - -----------------
206 * To determine how many terms needed, let
207 * Q(0) = w, Q(1) = w(w+h) - 1,
208 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
209 * When Q(k) > 1e4 good for single
210 * When Q(k) > 1e9 good for double
211 * When Q(k) > 1e17 good for quadruple
215 long double q0
, q1
, h
, tmp
;
217 w
= (n
+ n
) / (long double) x
;
218 h
= 2.0L / (long double) x
;
232 for (t
= zero
, i
= 2 * (n
+ k
); i
>= m
; i
-= 2)
233 t
= one
/ (i
/ x
- t
);
236 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
237 * Hence, if n*(log(2n/x)) > ...
238 * single 8.8722839355e+01
239 * double 7.09782712893383973096e+02
240 * long double 1.1356523406294143949491931077970765006170e+04
241 * then recurrent value may overflow and the result is
242 * likely underflow to zero
246 tmp
= tmp
* __ieee754_logl (fabsl (v
* tmp
));
248 if (tmp
< 1.1356523406294143949491931077970765006170e+04L)
250 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
261 for (i
= n
- 1, di
= (long double) (i
+ i
); i
> 0; i
--)
268 /* scale b to avoid spurious overflow */
277 /* j0() and j1() suffer enormous loss of precision at and
278 * near zero; however, we know that their zero points never
279 * coincide, so just choose the one further away from zero.
281 z
= __ieee754_j0l (x
);
282 w
= __ieee754_j1l (x
);
283 if (fabsl (z
) >= fabsl (w
))
294 strong_alias (__ieee754_jnl
, __jnl_finite
)
297 __ieee754_ynl (int n
, long double x
)
302 long double a
, b
, temp
;
303 ieee854_long_double_shape_type u
;
307 ix
= se
& 0x7fffffff;
309 /* if Y(n,NaN) is NaN */
310 if (ix
>= 0x7fff0000)
312 if ((u
.parts32
.w0
& 0xffff) | u
.parts32
.w1
| u
.parts32
.w2
| u
.parts32
.w3
)
318 return -HUGE_VALL
+ x
;
320 return zero
/ (zero
* x
);
326 sign
= 1 - ((n
& 1) << 1);
329 return (__ieee754_y0l (x
));
331 return (sign
* __ieee754_y1l (x
));
332 if (ix
>= 0x7fff0000)
334 if (ix
>= 0x412D0000)
337 /* ??? See comment above on the possible futility of this. */
340 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
341 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
342 * Let s=sin(x), c=cos(x),
343 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
345 * n sin(xn)*sqt2 cos(xn)*sqt2
346 * ----------------------------------
354 __sincosl (x
, &s
, &c
);
370 b
= invsqrtpi
* temp
/ __ieee754_sqrtl (x
);
374 a
= __ieee754_y0l (x
);
375 b
= __ieee754_y1l (x
);
376 /* quit if b is -inf */
378 se
= u
.parts32
.w0
& 0xffff0000;
379 for (i
= 1; i
< n
&& se
!= 0xffff0000; i
++)
382 b
= ((long double) (i
+ i
) / x
) * b
- a
;
384 se
= u
.parts32
.w0
& 0xffff0000;
393 strong_alias (__ieee754_ynl
, __ynl_finite
)