/*
** 16 December 1992
** Minor modifications by Peter McIlroy to adapt non-IEEE architecture.
*/

/*
** ====================================================
** Copyright (C) 1992 by Sun Microsystems, Inc.
**
** Developed at SunPro, a Sun Microsystems, Inc. business.
** Permission to use, copy, modify, and distribute this
** software is freely granted, provided that this notice 
** is preserved.
** ====================================================
**
** ******************* WARNING ********************
** This is an alpha version of SunPro's FDLIBM (Freely
** Distributable Math Library) for IEEE double precision 
** arithmetic. FDLIBM is a basic math library written
** in C that runs on machines that conform to IEEE 
** Standard 754/854. This alpha version is distributed 
** for testing purpose. Those who use this software 
** should report any bugs to 
**
**		fdlibm-comments@sunpro.eng.sun.com
**
** -- K.C. Ng, Oct 12, 1992
** ************************************************
*/

/*
** jn(int n, double x), yn(int n, double 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.
**	
*/

#define MERLIN_MATH
#include "../merlin.h"

#include <errno.h>

extern double merlin_j0(),merlin_j1(),log(),fabs(),sqrt(),cos(),sin(),merlin_y0(),merlin_y1();
static double invsqrtpi = 5.641895835477562869480794515607725858441e-0001;

double merlin_jn(int n, double x)
{
  int i, sgn;
  double a, b, temp = 0.0;
  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)
  */

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

  if (isnan(x)) return x+x;
  if (n<0)
    {		
      n = -n;
      x = -x;
    }

  if (n==0) return(merlin_j0(x));
  if (n==1) return(merlin_j1(x));
  sgn = (n&1)&(x < 0.0);		/* even n -- 0, odd n -- sign(x) */
  x = fabs(x);
  if (x == 0 || !finite (x)) 	/* if x is 0 or inf */
    b = 0.0;
  else if ((double) n <= x) {
    /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
    if (x >= 8.148143905337944345e+090)
      { /* 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 = merlin_j0(x);
	b = merlin_j1(x);
	for(i=1;i<n;i++)
	  {
	    temp = b;
	    b = b*((double)(i+i)/x) - a; /* avoid underflow */
	    a = temp;
	  }
      }
  }
  else
    {
      if (x < 1.86264514923095703125e-009)
	{ /* 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 = 0.0;
	  else
	    {
	      temp = x*0.5; b = temp;
	      for (a=1.0,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; 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<1.0e9)
	    {
	      k += 1; z += h;
	      tmp = z*q1 - q0;
	      q0 = q1;
	      q1 = tmp;
	    }
	  m = n+n;
	  for(t=0.0, i = 2*(n+k); i>=m; i -= 2) t = 1.0/(i/x-t);
	  a = t;
	  b = 1.0;

	  /*
	  ** 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 will
	  **  likely underflow to zero
	  */

	  tmp = n;
	  v = 2.0/x;
	  tmp = tmp*log(fabs(v*tmp));
	  for (i=n-1;i>0;i--)
	    {
	      temp = b;
	      b = ((i+i)/x)*b - a;
	      a = temp;
	      /* scale b to avoid spurious overflow */

	      if (b > 1e100)
		{
		  a /= b;
		  t /= b;
		  b = 1.0;
		}
	    }
	  b = (t*merlin_j0(x)/b);
	}
    }
  return ((sgn == 1) ? -b : b);
}


double merlin_yn(int n, double x)
{
  int i, sign;
  double a, b, temp = 0.0;

  /* Y(n,NaN), Y(n, x < 0) is NaN */

  if (x <= 0 || (x != x))
    {
      if (x < 0) return 0.0/0.0;
      else       return -1.0/0.0;
    }
  else if (!finite(x)) return(0);
  sign = 1;
  if (n<0)
    {
      n = -n;
      sign = 1 - ((n&1)<<2);
    }
  if (n == 0) return(merlin_y0(x));
  if (n == 1) return(sign*merlin_y1(x));
  if(x >= 8.148143905337944345e+090)
    { /* 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
    {
      a = merlin_y0(x);
      b = merlin_y1(x);

      /* quit if b is -inf */

      for (i = 1; i < n && !finite(b); i++)
	{
	  temp = b;
	  b = ((double)(i+i)/x)*b - a;
	  a = temp;
	}
    }

  return ((sign > 0) ? b : -b);
}
