128-bit long double Bessel functions jn and yn.

1 files changed, 382 insertions, 0 deletions

diff --git a/sysdeps/ieee754/ldbl-128/e_jnl.c b/sysdeps/ieee754/ldbl-128/e_jnl.c
new file mode 100644
index 0000000..a568927
--- /dev/null
+++ b/sysdeps/ieee754/ldbl-128/e_jnl.c
@@ -0,0 +1,382 @@
+/*
+ * ====================================================
+ * 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.
+ * ====================================================
+ */
+
+/* Modifications for 128-bit long double contributed by
+   Stephen L. Moshier <moshier@na-net.ornl.gov>  */
+
+/*
+ * __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"
+
+#ifdef __STDC__
+static const long double
+#else
+static long double
+#endif
+  invsqrtpi = 5.6418958354775628694807945156077258584405E-1L,
+  two = 2.0e0L,
+  one = 1.0e0L,
+  zero = 0.0L;
+
+
+#ifdef __STDC__
+long double
+__ieee754_jnl (int n, long double x)
+#else
+long double
+__ieee754_jnl (n, x)
+     int n;
+     long double x;
+#endif
+{
+  u_int32_t se;
+  int32_t i, ix, sgn;
+  long double a, b, temp, di;
+  long double z, w;
+  ieee854_long_double_shape_type u;
+
+
+  /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
+   * Thus, J(-n,x) = J(n,-x)
+   */
+
+  u.value = x;
+  se = u.parts32.w0;
+  ix = se & 0x7fffffff;
+
+  /* if J(n,NaN) is NaN */
+  if (ix >= 0x7fff0000)
+    {
+      if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
+	return x + x;
+    }
+
+  if (n < 0)
+    {
+      n = -n;
+      x = -x;
+      se ^= 0x80000000;
+    }
+  if (n == 0)
+    return (__ieee754_j0l (x));
+  if (n == 1)
+    return (__ieee754_j1l (x));
+  sgn = (n & 1) & (se >> 31);	/* even n -- 0, odd n -- sign(x) */
+  x = fabsl (x);
+
+  if (x == 0.0L || ix >= 0x7fff0000)	/* if x is 0 or inf */
+    b = zero;
+  else if ((long double) n <= x)
+    {
+      /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
+      if (ix >= 0x412D0000)
+	{			/* x > 2**302 */
+
+	  /* ??? Could use an expansion for large x here.  */
+
+	  /* (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
+	   */
+	  long double s;
+	  long double c;
+	  __sincosl (x, &s, &c);
+	  switch (n & 3)
+	    {
+	    case 0:
+	      temp = c + s;
+	      break;
+	    case 1:
+	      temp = -c + s;
+	      break;
+	    case 2:
+	      temp = -c - s;
+	      break;
+	    case 3:
+	      temp = c - s;
+	      break;
+	    }
+	  b = invsqrtpi * temp / __ieee754_sqrtl (x);
+	}
+      else
+	{
+	  a = __ieee754_j0l (x);
+	  b = __ieee754_j1l (x);
+	  for (i = 1; i < n; i++)
+	    {
+	      temp = b;
+	      b = b * ((long double) (i + i) / x) - a;	/* avoid underflow */
+	      a = temp;
+	    }
+	}
+    }
+  else
+    {
+      if (ix < 0x3fc60000)
+	{			/* x < 2**-57 */
+	  /* x is tiny, return the first Taylor expansion of J(n,x)
+	   * J(n,x) = 1/n!*(x/2)^n  - ...
+	   */
+	  if (n >= 400)		/* underflow, result < 10^-4952 */
+	    b = zero;
+	  else
+	    {
+	      temp = x * 0.5;
+	      b = temp;
+	      for (a = one, i = 2; i <= n; i++)
+		{
+		  a *= (long 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 */
+	  long double t, v;
+	  long double q0, q1, h, tmp;
+	  int32_t k, m;
+	  w = (n + n) / (long double) x;
+	  h = 2.0L / (long double) x;
+	  q0 = w;
+	  z = w + h;
+	  q1 = w * z - 1.0L;
+	  k = 1;
+	  while (q1 < 1.0e17L)
+	    {
+	      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_logl (fabsl (v * tmp));
+
+	  if (tmp < 1.1356523406294143949491931077970765006170e+04L)
+	    {
+	      for (i = n - 1, di = (long double) (i + i); i > 0; i--)
+		{
+		  temp = b;
+		  b *= di;
+		  b = b / x - a;
+		  a = temp;
+		  di -= two;
+		}
+	    }
+	  else
+	    {
+	      for (i = n - 1, di = (long 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 > 1e100L)
+		    {
+		      a /= b;
+		      t /= b;
+		      b = one;
+		    }
+		}
+	    }
+	  b = (t * __ieee754_j0l (x) / b);
+	}
+    }
+  if (sgn == 1)
+    return -b;
+  else
+    return b;
+}
+
+#ifdef __STDC__
+long double
+__ieee754_ynl (int n, long double x)
+#else
+long double
+__ieee754_ynl (n, x)
+     int n;
+     long double x;
+#endif
+{
+  u_int32_t se;
+  int32_t i, ix;
+  int32_t sign;
+  long double a, b, temp;
+  ieee854_long_double_shape_type u;
+
+  u.value = x;
+  se = u.parts32.w0;
+  ix = se & 0x7fffffff;
+
+  /* if Y(n,NaN) is NaN */
+  if (ix >= 0x7fff0000)
+    {
+      if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3)
+	return x + x;
+    }
+  if (x <= 0.0L)
+    {
+      if (x == 0.0L)
+	return -one / zero;
+      if (se & 0x80000000)
+	return zero / zero;
+    }
+  sign = 1;
+  if (n < 0)
+    {
+      n = -n;
+      sign = 1 - ((n & 1) << 1);
+    }
+  if (n == 0)
+    return (__ieee754_y0l (x));
+  if (n == 1)
+    return (sign * __ieee754_y1l (x));
+  if (ix >= 0x7fff0000)
+    return zero;
+  if (ix >= 0x412D0000)
+    {				/* x > 2**302 */
+
+      /* ??? See comment above on the possible futility of this.  */
+
+      /* (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
+       */
+      long double s;
+      long double c;
+      __sincosl (x, &s, &c);
+      switch (n & 3)
+	{
+	case 0:
+	  temp = s - c;
+	  break;
+	case 1:
+	  temp = -s - c;
+	  break;
+	case 2:
+	  temp = -s + c;
+	  break;
+	case 3:
+	  temp = s + c;
+	  break;
+	}
+      b = invsqrtpi * temp / __ieee754_sqrtl (x);
+    }
+  else
+    {
+      a = __ieee754_y0l (x);
+      b = __ieee754_y1l (x);
+      /* quit if b is -inf */
+      u.value = b;
+      se = u.parts32.w0 & 0xffff0000;
+      for (i = 1; i < n && se != 0xffff0000; i++)
+	{
+	  temp = b;
+	  b = ((long double) (i + i) / x) * b - a;
+	  u.value = b;
+	  se = u.parts32.w0 & 0xffff0000;
+	  a = temp;
+	}
+    }
+  if (sign > 0)
+    return b;
+  else
+    return -b;
+}