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394 lines
9.7 KiB
C
394 lines
9.7 KiB
C
/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* Modifications for 128-bit long double are
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Copyright (C) 2001 Stephen L. Moshier <moshier@na-net.ornl.gov>
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and are incorporated herein by permission of the author. The author
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reserves the right to distribute this material elsewhere under different
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copying permissions. These modifications are distributed here under
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the following terms:
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This library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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This library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with this library; if not, see
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<http://www.gnu.org/licenses/>. */
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/*
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* __ieee754_jn(n, x), __ieee754_yn(n, x)
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* floating point Bessel's function of the 1st and 2nd kind
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* of order n
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*
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* Special cases:
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* y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
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* y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
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* Note 2. About jn(n,x), yn(n,x)
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* For n=0, j0(x) is called,
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* for n=1, j1(x) is called,
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* for n<x, forward recursion us used starting
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* from values of j0(x) and j1(x).
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* for n>x, a continued fraction approximation to
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* j(n,x)/j(n-1,x) is evaluated and then backward
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* recursion is used starting from a supposed value
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* for j(n,x). The resulting value of j(0,x) is
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* compared with the actual value to correct the
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* supposed value of j(n,x).
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*
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* yn(n,x) is similar in all respects, except
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* that forward recursion is used for all
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* values of n>1.
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*
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*/
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#include <math.h>
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#include <math_private.h>
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static const long double
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invsqrtpi = 5.6418958354775628694807945156077258584405E-1L,
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two = 2.0e0L,
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one = 1.0e0L,
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zero = 0.0L;
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long double
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__ieee754_jnl (int n, long double x)
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{
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u_int32_t se;
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int32_t i, ix, sgn;
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long double a, b, temp, di;
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long double z, w;
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ieee854_long_double_shape_type u;
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/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
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* Thus, J(-n,x) = J(n,-x)
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*/
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u.value = x;
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se = u.parts32.w0;
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ix = se & 0x7fffffff;
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/* if J(n,NaN) is NaN */
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if (ix >= 0x7fff0000)
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{
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if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
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return x + x;
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}
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if (n < 0)
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{
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n = -n;
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x = -x;
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se ^= 0x80000000;
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}
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if (n == 0)
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return (__ieee754_j0l (x));
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if (n == 1)
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return (__ieee754_j1l (x));
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sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */
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x = fabsl (x);
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if (x == 0.0L || ix >= 0x7fff0000) /* if x is 0 or inf */
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b = zero;
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else if ((long double) n <= x)
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{
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/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
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if (ix >= 0x412D0000)
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{ /* x > 2**302 */
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/* ??? Could use an expansion for large x here. */
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=sin(x), c=cos(x),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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long double s;
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long double c;
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__sincosl (x, &s, &c);
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switch (n & 3)
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{
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case 0:
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temp = c + s;
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break;
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case 1:
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temp = -c + s;
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break;
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case 2:
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temp = -c - s;
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break;
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case 3:
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temp = c - s;
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break;
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}
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b = invsqrtpi * temp / __ieee754_sqrtl (x);
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}
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else
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{
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a = __ieee754_j0l (x);
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b = __ieee754_j1l (x);
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for (i = 1; i < n; i++)
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{
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temp = b;
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b = b * ((long double) (i + i) / x) - a; /* avoid underflow */
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a = temp;
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}
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}
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}
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else
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{
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if (ix < 0x3fc60000)
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{ /* x < 2**-57 */
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/* x is tiny, return the first Taylor expansion of J(n,x)
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* J(n,x) = 1/n!*(x/2)^n - ...
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*/
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if (n >= 400) /* underflow, result < 10^-4952 */
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b = zero;
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else
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{
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temp = x * 0.5;
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b = temp;
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for (a = one, i = 2; i <= n; i++)
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{
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a *= (long double) i; /* a = n! */
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b *= temp; /* b = (x/2)^n */
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}
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b = b / a;
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}
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}
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else
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{
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/* use backward recurrence */
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/* x x^2 x^2
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* J(n,x)/J(n-1,x) = ---- ------ ------ .....
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* 2n - 2(n+1) - 2(n+2)
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*
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* 1 1 1
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* (for large x) = ---- ------ ------ .....
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* 2n 2(n+1) 2(n+2)
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* -- - ------ - ------ -
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* x x x
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*
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* Let w = 2n/x and h=2/x, then the above quotient
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* is equal to the continued fraction:
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* 1
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* = -----------------------
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* 1
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* w - -----------------
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* 1
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* w+h - ---------
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* w+2h - ...
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*
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* To determine how many terms needed, let
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* Q(0) = w, Q(1) = w(w+h) - 1,
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* Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
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* When Q(k) > 1e4 good for single
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* When Q(k) > 1e9 good for double
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* When Q(k) > 1e17 good for quadruple
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*/
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/* determine k */
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long double t, v;
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long double q0, q1, h, tmp;
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int32_t k, m;
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w = (n + n) / (long double) x;
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h = 2.0L / (long double) x;
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q0 = w;
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z = w + h;
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q1 = w * z - 1.0L;
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k = 1;
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while (q1 < 1.0e17L)
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{
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k += 1;
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z += h;
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tmp = z * q1 - q0;
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q0 = q1;
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q1 = tmp;
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}
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m = n + n;
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for (t = zero, i = 2 * (n + k); i >= m; i -= 2)
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t = one / (i / x - t);
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a = t;
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b = one;
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/* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
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* Hence, if n*(log(2n/x)) > ...
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* single 8.8722839355e+01
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* double 7.09782712893383973096e+02
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* long double 1.1356523406294143949491931077970765006170e+04
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* then recurrent value may overflow and the result is
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* likely underflow to zero
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*/
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tmp = n;
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v = two / x;
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tmp = tmp * __ieee754_logl (fabsl (v * tmp));
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if (tmp < 1.1356523406294143949491931077970765006170e+04L)
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{
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for (i = n - 1, di = (long double) (i + i); i > 0; i--)
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{
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temp = b;
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b *= di;
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b = b / x - a;
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a = temp;
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di -= two;
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}
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}
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else
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{
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for (i = n - 1, di = (long double) (i + i); i > 0; i--)
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{
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temp = b;
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b *= di;
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b = b / x - a;
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a = temp;
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di -= two;
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/* scale b to avoid spurious overflow */
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if (b > 1e100L)
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{
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a /= b;
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t /= b;
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b = one;
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}
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}
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}
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/* j0() and j1() suffer enormous loss of precision at and
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* near zero; however, we know that their zero points never
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* coincide, so just choose the one further away from zero.
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*/
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z = __ieee754_j0l (x);
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w = __ieee754_j1l (x);
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if (fabsl (z) >= fabsl (w))
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b = (t * z / b);
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else
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b = (t * w / a);
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}
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}
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if (sgn == 1)
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return -b;
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else
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return b;
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}
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strong_alias (__ieee754_jnl, __jnl_finite)
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long double
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__ieee754_ynl (int n, long double x)
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{
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u_int32_t se;
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int32_t i, ix;
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int32_t sign;
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long double a, b, temp;
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ieee854_long_double_shape_type u;
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u.value = x;
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se = u.parts32.w0;
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ix = se & 0x7fffffff;
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/* if Y(n,NaN) is NaN */
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if (ix >= 0x7fff0000)
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{
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if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
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return x + x;
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}
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if (x <= 0.0L)
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{
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if (x == 0.0L)
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return -HUGE_VALL + x;
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if (se & 0x80000000)
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return zero / (zero * x);
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}
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sign = 1;
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if (n < 0)
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{
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n = -n;
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sign = 1 - ((n & 1) << 1);
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}
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if (n == 0)
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return (__ieee754_y0l (x));
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if (n == 1)
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return (sign * __ieee754_y1l (x));
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if (ix >= 0x7fff0000)
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return zero;
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if (ix >= 0x412D0000)
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{ /* x > 2**302 */
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/* ??? See comment above on the possible futility of this. */
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=sin(x), c=cos(x),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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long double s;
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long double c;
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__sincosl (x, &s, &c);
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switch (n & 3)
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{
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case 0:
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temp = s - c;
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break;
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case 1:
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temp = -s - c;
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break;
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case 2:
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temp = -s + c;
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break;
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case 3:
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temp = s + c;
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break;
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}
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b = invsqrtpi * temp / __ieee754_sqrtl (x);
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}
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else
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{
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a = __ieee754_y0l (x);
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b = __ieee754_y1l (x);
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/* quit if b is -inf */
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u.value = b;
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se = u.parts32.w0 & 0xffff0000;
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for (i = 1; i < n && se != 0xffff0000; i++)
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{
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temp = b;
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b = ((long double) (i + i) / x) * b - a;
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u.value = b;
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se = u.parts32.w0 & 0xffff0000;
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a = temp;
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}
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}
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if (sign > 0)
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return b;
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else
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return -b;
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}
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strong_alias (__ieee754_ynl, __ynl_finite)
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