/// <summary>
/// Returns K{n}(x) for integer order
/// </summary>
/// <param name="n"></param>
/// <param name="x"></param>
/// <returns></returns>
public static double KN(int n, double x)
{
if (x < 0)
{
Policies.ReportDomainError("BesselK(v: {0}, x: {1}): Requires x >= 0 for real result", n, x);
return(double.NaN);
}
if (x == 0)
{
return(double.PositiveInfinity);
}
// even function
// K{-n}(z) = K{n}(z)
if (n < 0)
{
n = -n;
}
if (n == 0)
{
return(K0(x));
}
if (n == 1)
{
return(K1(x));
}
double v = n;
// Hankel is fast, and reasonably accurate, saving us from many recurrences.
if (x >= HankelAsym.IKMinX(v))
{
return(HankelAsym.K(v, x));
}
// the uniform expansion is here as a last resort
// to limit the number of recurrences, but it is less accurate.
if (UniformAsym.IsIKAvailable(v, x))
{
return(UniformAsym.K(v, x));
}
// Since K{v}(x) has a (e^-x)/sqrt(x) multiplier
// using recurrence can underflow too quickly for large x,
// so, use a scaled version
double result;
if (x > 1)
{
double prev = K0(x, true);
double current = K1(x, true);
// for large v and x this number can get very large
// maximum observed K(1000,10) = 2^6211
var(Kv, _, binaryScale) = Recurrence.ForwardK_B(1, x, n - 1, current, prev);
// Compute: value * 2^(binaryScale) * e^-x
if (x < -DoubleX.MinLogValue)
{
DoubleX exs = DoubleX.Ldexp(DoubleX.Exp(-x), binaryScale);
result = Math2.Ldexp(Kv * exs.Mantissa, exs.Exponent);
}
else
{
result = Math.Exp(-x + Math.Log(Kv) + binaryScale * Constants.Ln2);
}
}
else
{
double prev = K0(x);
double current = K1(x);
result = Recurrence.ForwardK(1, x, n - 1, current, prev).Kvpn;
}
return(result);
}
