/****************************************************************************
 *
 * $Source: /usr/local/cvsroot/gccsdk/unixlib/source/math/jn.c,v $
 * $Date: 2003/12/29 19:02:38 $
 * $Revision: 1.4 $
 * $State: Exp $
 * $Author: peter $
 *
 ***************************************************************************/

#ifdef EMBED_RCSID
static const char rcs_id[] = "$Id: jn.c,v 1.4 2003/12/29 19:02:38 peter Exp $";
#endif

/* @(#)e_jn.c 5.1 93/09/24 */
/*
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * 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.
 * ====================================================
 */

/*
 * __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.
 *
 */

#undef __UNIXLIB_INTERNALS

#include <math.h>
#include <unixlib/math.h>
#include <unixlib/types.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;

double
jn (int n, double x)
{
  __int32_t i, hx, ix, lx, sgn;
  double a, b, temp = 0.0, 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 | ((__uint32_t) (lx | -lx)) >> 31) > 0x7ff00000)
    return x + x;
  if (n < 0)
    {
      n = -n;
      x = -x;
      hx ^= 0x80000000;
    }
  if (n == 0)
    return (j0 (x));
  if (n == 1)
    return (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 = j0 (x);
	  b = 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 * 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 * j0 (x) / b);
	}
    }
  if (sgn == 1)
    return -b;
  else
    return b;
}

double
yn (int n, double x)
{
  __int32_t i, hx, ix, lx;
  __int32_t sign;
  double a, b, temp = 0.0;

  EXTRACT_WORDS (hx, lx, x);
  ix = 0x7fffffff & hx;
  /* if Y(n,NaN) is NaN */
  if ((ix | ((__uint32_t) (lx | -lx)) >> 31) > 0x7ff00000)
    return x + x;
  if ((ix | lx) == 0)
    return -one / zero;
  if (hx < 0)
    return zero / zero;
  sign = 1;
  if (n < 0)
    {
      n = -n;
      sign = 1 - ((n & 1) << 1);
    }
  if (n == 0)
    return (y0 (x));
  if (n == 1)
    return (sign * 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
    {
      __uint32_t high;
      a = y0 (x);
      b = 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;
}
