/*
 * CDDL HEADER START
 *
 * The contents of this file are subject to the terms of the
 * Common Development and Distribution License (the "License").
 * You may not use this file except in compliance with the License.
 *
 * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
 * or http://www.opensolaris.org/os/licensing.
 * See the License for the specific language governing permissions
 * and limitations under the License.
 *
 * When distributing Covered Code, include this CDDL HEADER in each
 * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
 * If applicable, add the following below this CDDL HEADER, with the
 * fields enclosed by brackets "[]" replaced with your own identifying
 * information: Portions Copyright [yyyy] [name of copyright owner]
 *
 * CDDL HEADER END
 */

/*
 * Copyright 2011 Nexenta Systems, Inc.  All rights reserved.
 */
/*
 * Copyright 2006 Sun Microsystems, Inc.  All rights reserved.
 * Use is subject to license terms.
 */

#pragma weak __jnl = jnl
#pragma weak __ynl = ynl

/*
 * floating point Bessel's function of the 1st and 2nd kind
 * of order n: jn(n,x),yn(n,x);
 *
 * Special cases:
 *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
 *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
 * Note 2. About jn(n,x), yn(n,x)
 *	For n=0, j0(x) is called,
 *	for n=1, j1(x) is called,
 *	for n<x, forward recursion us used starting
 *	from values of j0(x) and j1(x).
 *	for n>x, a continued fraction approximation to
 *	j(n,x)/j(n-1,x) is evaluated and then backward
 *	recursion is used starting from a supposed value
 *	for j(n,x). The resulting value of j(0,x) is
 *	compared with the actual value to correct the
 *	supposed value of j(n,x).
 *
 *	yn(n,x) is similar in all respects, except
 *	that forward recursion is used for all
 *	values of n>1.
 *
 */

#include "libm.h"
#include "longdouble.h"
#include <float.h>	/* LDBL_MAX */

#define	GENERIC long double

static const GENERIC
invsqrtpi = 5.641895835477562869480794515607725858441e-0001L,
two  = 2.0L,
zero = 0.0L,
one  = 1.0L;

GENERIC
jnl(n, x) int n; GENERIC x; {
	int i, sgn;
	GENERIC a, b, temp = 0, z, w;

	/*
	 * J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
	 * Thus, J(-n,x) = J(n,-x)
	 */
	if (n < 0) {
		n = -n;
		x = -x;
	}
	if (n == 0) return (j0l(x));
	if (n == 1) return (j1l(x));
	if (x != x) return x+x;
	if ((n&1) == 0)
		sgn = 0; 			/* even n */
	else
		sgn = signbitl(x);	/* old n  */
	x = fabsl(x);
	if (x == zero || !finitel(x)) b = zero;
	else if ((GENERIC)n <= x) {
			/*
			 * Safe to use
			 * J(n+1,x)=2n/x *J(n,x)-J(n-1,x)
			 */
	    if (x > 1.0e91L) {
				/*
				 * x >> n**2
				 *  Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
				 *  Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
				 *  Let s=sin(x), c=cos(x),
				 *  xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
				 *
				 *	   n	sin(xn)*sqt2	cos(xn)*sqt2
				 *	----------------------------------
				 *	   0	 s-c		 c+s
				 *	   1	-s-c 		-c+s
				 *	   2	-s+c		-c-s
				 *	   3	 s+c		 c-s
				 */
		switch (n&3) {
		    case 0: temp =  cosl(x)+sinl(x); break;
		    case 1: temp = -cosl(x)+sinl(x); break;
		    case 2: temp = -cosl(x)-sinl(x); break;
		    case 3: temp =  cosl(x)-sinl(x); break;
		}
		b = invsqrtpi*temp/sqrtl(x);
	    } else {
			a = j0l(x);
			b = j1l(x);
			for (i = 1; i < n; i++) {
		    temp = b;
		    b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */
		    a = temp;
			}
	    }
	} else {
	    if (x < 1e-17L) {	/* use J(n,x) = 1/n!*(x/2)^n */
		b = powl(0.5L*x, (GENERIC) n);
		if (b != zero) {
		    for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i;
		    b = b/a;
		}
	    } else {
		/*
		 * use backward recurrence
		 * 			x      x^2      x^2
		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
		 *			2n  - 2(n+1) - 2(n+2)
		 *
		 * 			1      1        1
		 *  (for large x)   =  ----  ------   ------   .....
		 *			2n   2(n+1)   2(n+2)
		 *			-- - ------ - ------ -
		 *			 x     x         x
		 *
		 * Let w = 2n/x and h=2/x, then the above quotient
		 * is equal to the continued fraction:
		 *		    1
		 *	= -----------------------
		 *		       1
		 *	   w - -----------------
		 *			  1
		 * 	        w+h - ---------
		 *		       w+2h - ...
		 *
		 * To determine how many terms needed, let
		 * Q(0) = w, Q(1) = w(w+h) - 1,
		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
		 * When Q(k) > 1e4	good for single
		 * When Q(k) > 1e9	good for double
		 * When Q(k) > 1e17	good for quaduple
		 */
	    /* determin k */
		GENERIC t, v;
		double q0, q1, h, tmp; int k, m;
		w  = (n+n)/(double)x; h = 2.0/(double)x;
		q0 = w;  z = w+h; q1 = w*z - 1.0; k = 1;
		while (q1 < 1.0e17) {
			k += 1; z += h;
			tmp = z*q1 - q0;
			q0 = q1;
			q1 = tmp;
		}
		m = n+n;
		for (t = zero, i = 2*(n+k); i >= m; i -= 2) t = one/(i/x-t);
		a = t;
		b = one;
			/*
			 * Estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
			 * hence, if n*(log(2n/x)) > ...
			 * single 8.8722839355e+01
			 * double 7.09782712893383973096e+02
			 * long double 1.1356523406294143949491931077970765006170e+04
			 * then recurrent value may overflow and the result is
			 * likely underflow to zero.
			 */
		tmp = n;
		v = two/x;
		tmp = tmp*logl(fabsl(v*tmp));
		if (tmp < 1.1356523406294143949491931077970765e+04L) {
				for (i = n-1; i > 0; i--) {
				temp = b;
				b = ((i+i)/x)*b - a;
				a = temp;
				}
		} else {
				for (i = n-1; i > 0; i--) {
				temp = b;
				b = ((i+i)/x)*b - a;
				a = temp;
			if (b > 1e1000L) {
						a /= b;
						t /= b;
						b  = 1.0;
					}
				}
		}
			b = (t*j0l(x)/b);
	    }
	}
	if (sgn == 1)
		return -b;
	else
		return b;
}

GENERIC
ynl(n, x) int n; GENERIC x; {
	int i;
	int sign;
	GENERIC a, b, temp = 0;

	if (x != x)
		return x+x;
	if (x <= zero) {
		if (x == zero)
			return -one/zero;
		else
			return zero/zero;
	}
	sign = 1;
	if (n < 0) {
		n = -n;
		if ((n&1) == 1) sign = -1;
	}
	if (n == 0) return (y0l(x));
	if (n == 1) return (sign*y1l(x));
	if (!finitel(x)) return zero;

	if (x > 1.0e91L) {
				/*
				 * x >> n**2
				 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
				 *   Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
				 *   Let s=sin(x), c=cos(x),
				 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
				 *
				 *	   n	sin(xn)*sqt2	cos(xn)*sqt2
				 *	----------------------------------
				 * 	   0	 s-c		 c+s
				 *	   1	-s-c 		-c+s
				 * 	   2	-s+c		-c-s
				 *	   3	 s+c		 c-s
				 */
		switch (n&3) {
		    case 0: temp =  sinl(x)-cosl(x); break;
		    case 1: temp = -sinl(x)-cosl(x); break;
		    case 2: temp = -sinl(x)+cosl(x); break;
		    case 3: temp =  sinl(x)+cosl(x); break;
		}
		b = invsqrtpi*temp/sqrtl(x);
	} else {
		a = y0l(x);
		b = y1l(x);
		/*
		 * fix 1262058 and take care of non-default rounding
		 */
		for (i = 1; i < n; i++) {
			temp = b;
			b *= (GENERIC) (i + i) / x;
			if (b <= -LDBL_MAX)
				break;
			b -= a;
			a = temp;
		}
	}
	if (sign > 0)
		return b;
	else
		return -b;
}