/*--------------------------------------------------------------------

 *© Portions copyright (c) 2006 Nokia Corporation.  All rights reserved.

 *--------------------------------------------------------------------

*/

/* @(#)e_jn.c 1.4 95/01/18 */

/*

 * ====================================================

 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.

 *

 * Developed at SunSoft, a Sun Microsystems, Inc. business.

 * Permission to use, copy, modify, and distribute this

 * software is freely granted, provided that this notice 

 * is preserved.

 * ====================================================

 */

#ifndef __SYMBIAN32__

#ifndef lint

static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_jn.c,v 1.9 2005/02/04 18:26:06 das Exp $";

#endif

#endif //__SYMBIAN32__

/*

 * __ieee754_jn(n, x), __ieee754_yn(n, x)

 * floating point Bessel's function of the 1st and 2nd kind

 * of order n

 *          

 * 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 <math.h>

#include "math_private.h"

static const double

invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */

two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */

one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */

static const double zero  =  0.00000000000000000000e+00;

EXPORT_C double

__ieee754_jn(int n, double x)

{

	int32_t i,hx,ix,lx, sgn;

	double a, b, temp, di;

	double 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)

     */

	EXTRACT_WORDS(hx,lx,x);

	ix = 0x7fffffff&hx;

    /* if J(n,NaN) is NaN */

	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;

	if(n<0){		

		n = -n;

		x = -x;

		hx ^= 0x80000000;

	}

	if(n==0) return(__ieee754_j0(x));

	if(n==1) return(__ieee754_j1(x));

	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */

	x = fabs(x);

	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */

	    b = zero;

	else if((double)n<=x) {   

		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */

	    if(ix>=0x52D00000) { /* x > 2**302 */

    /* (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 =  cos(x)+sin(x); break;

		    case 1: temp = -cos(x)+sin(x); break;

		    case 2: temp = -cos(x)-sin(x); break;

		    case 3: temp =  cos(x)-sin(x); break;

		}

		b = invsqrtpi*temp/sqrt(x);

	    } else {	

	        a = __ieee754_j0(x);

	        b = __ieee754_j1(x);

	        for(i=1;i<n;i++){

		    temp = b;

		    b = b*((double)(i+i)/x) - a; /* avoid underflow */

		    a = temp;

	        }

	    }

	} else {

	    if(ix<0x3e100000) {	/* x < 2**-29 */

    /* x is tiny, return the first Taylor expansion of J(n,x) 

     * J(n,x) = 1/n!*(x/2)^n  - ...

     */

		if(n>33)	/* underflow */

		    b = zero;

		else {

		    temp = x*0.5; b = temp;

		    for (a=one,i=2;i<=n;i++) {

			a *= (double)i;		/* a = n! */

			b *= temp;		/* b = (x/2)^n */

		    }

		    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 quadruple 

		 */

	    /* determine k */

		double t,v;

		double q0,q1,h,tmp; int32_t 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<1.0e9) {

			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*__ieee754_log(fabs(v*tmp));

		if(tmp<7.09782712893383973096e+02) {

	    	    for(i=n-1,di=(double)(i+i);i>0;i--){

		        temp = b;

			b *= di;

			b  = b/x - a;

		        a = temp;

			di -= two;

	     	    }

		} else {

	    	    for(i=n-1,di=(double)(i+i);i>0;i--){

		        temp = b;

			b *= di;

			b  = b/x - a;

		        a = temp;

			di -= two;

		    /* scale b to avoid spurious overflow */

			if(b>1e100) {

			    a /= b;

			    t /= b;

			    b  = one;

			}

	     	    }

		}

	    	b = (t*__ieee754_j0(x)/b);

	    }

	}

	if(sgn==1) return -b; else return b;

}

EXPORT_C double

__ieee754_yn(int n, double x)

{

	int32_t i,hx,ix,lx;

	int32_t sign;

	double a, b, temp;

	EXTRACT_WORDS(hx,lx,x);

	ix = 0x7fffffff&hx;

    /* if Y(n,NaN) is NaN */

	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;

	if((ix|lx)==0) return -one/zero;

	#ifdef __SYMBIAN32__

	if(hx<0) return -one/zero;

	#else

	if(hx<0) return zero/zero;

	#endif //__SYMBIAN32__

	sign = 1;

	if(n<0){

		n = -n;

		sign = 1 - ((n&1)<<1);

	}

	if(n==0) return(__ieee754_y0(x));

	if(n==1) return(sign*__ieee754_y1(x));

	if(ix==0x7ff00000) return zero;

	if(ix>=0x52D00000) { /* x > 2**302 */

    /* (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 =  sin(x)-cos(x); break;

		    case 1: temp = -sin(x)-cos(x); break;

		    case 2: temp = -sin(x)+cos(x); break;

		    case 3: temp =  sin(x)+cos(x); break;

		}

		b = invsqrtpi*temp/sqrt(x);

	} else {

	    u_int32_t high;

	    a = __ieee754_y0(x);

	    b = __ieee754_y1(x);

	/* quit if b is -inf */

	    GET_HIGH_WORD(high,b);

	    for(i=1;i<n&&high!=0xfff00000;i++){

		temp = b;

		b = ((double)(i+i)/x)*b - a;

		GET_HIGH_WORD(high,b);

		a = temp;

	    }

	}

	if(sign>0) return b; else return -b;

}