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
59 #include "quadmath-imp.h"
61 static const __float128
62 invsqrtpi = 5.6418958354775628694807945156077258584405E-1Q,
69 jnq (int n, __float128 x)
73 __float128 a, b, temp, di;
78 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
79 * Thus, J(-n,x) = J(n,-x)
86 /* if J(n,NaN) is NaN */
89 if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3)
103 sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */
106 if (x == 0.0Q || ix >= 0x7fff0000) /* if x is 0 or inf */
108 else if ((__float128) n <= x)
110 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
111 if (ix >= 0x412D0000)
114 /* ??? Could use an expansion for large x here. */
117 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
118 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
119 * Let s=sin(x), c=cos(x),
120 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
122 * n sin(xn)*sqt2 cos(xn)*sqt2
123 * ----------------------------------
147 b = invsqrtpi * temp / sqrtq (x);
153 for (i = 1; i < n; i++)
156 b = b * ((__float128) (i + i) / x) - a; /* avoid underflow */
165 /* x is tiny, return the first Taylor expansion of J(n,x)
166 * J(n,x) = 1/n!*(x/2)^n - ...
168 if (n >= 400) /* underflow, result < 10^-4952 */
174 for (a = one, i = 2; i <= n; i++)
176 a *= (__float128) i; /* a = n! */
177 b *= temp; /* b = (x/2)^n */
184 /* use backward recurrence */
186 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
187 * 2n - 2(n+1) - 2(n+2)
190 * (for large x) = ---- ------ ------ .....
192 * -- - ------ - ------ -
195 * Let w = 2n/x and h=2/x, then the above quotient
196 * is equal to the continued fraction:
198 * = -----------------------
200 * w - -----------------
205 * To determine how many terms needed, let
206 * Q(0) = w, Q(1) = w(w+h) - 1,
207 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
208 * When Q(k) > 1e4 good for single
209 * When Q(k) > 1e9 good for double
210 * When Q(k) > 1e17 good for quadruple
214 __float128 q0, q1, h, tmp;
216 w = (n + n) / (__float128) x;
217 h = 2.0Q / (__float128) x;
231 for (t = zero, i = 2 * (n + k); i >= m; i -= 2)
232 t = one / (i / x - t);
235 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
236 * Hence, if n*(log(2n/x)) > ...
237 * single 8.8722839355e+01
238 * double 7.09782712893383973096e+02
239 * __float128 1.1356523406294143949491931077970765006170e+04
240 * then recurrent value may overflow and the result is
241 * likely underflow to zero
245 tmp = tmp * logq (fabsq (v * tmp));
247 if (tmp < 1.1356523406294143949491931077970765006170e+04Q)
249 for (i = n - 1, di = (__float128) (i + i); i > 0; i--)
260 for (i = n - 1, di = (__float128) (i + i); i > 0; i--)
267 /* scale b to avoid spurious overflow */
276 b = (t * j0q (x) / b);
286 ynq (int n, __float128 x)
291 __float128 a, b, temp;
296 ix = se & 0x7fffffff;
298 /* if Y(n,NaN) is NaN */
299 if (ix >= 0x7fff0000)
301 if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3)
307 return -HUGE_VALQ + x;
309 return zero / (zero * x);
315 sign = 1 - ((n & 1) << 1);
320 return (sign * y1q (x));
321 if (ix >= 0x7fff0000)
323 if (ix >= 0x412D0000)
326 /* ??? See comment above on the possible futility of this. */
329 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
330 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
331 * Let s=sin(x), c=cos(x),
332 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
334 * n sin(xn)*sqt2 cos(xn)*sqt2
335 * ----------------------------------
359 b = invsqrtpi * temp / sqrtq (x);
365 /* quit if b is -inf */
367 se = u.words32.w0 & 0xffff0000;
368 for (i = 1; i < n && se != 0xffff0000; i++)
371 b = ((__float128) (i + i) / x) * b - a;
373 se = u.words32.w0 & 0xffff0000;