/// <summary>
/// Computes the given Bernoulli number.
/// </summary>
/// <param name="n">The index of the Bernoulli number to compute, which must be non-negative.</param>
/// <returns>The Bernoulli number B<sub>n</sub>.</returns>
/// <exception cref="ArgumentOutOfRangeException"><paramref name="n"/> is negative.</exception>
/// <remarks>
/// <para>B<sub>n</sub> vanishes for all odd n except n=1. For n about 260 or larger, B<sub>n</sub> overflows a double.</para>
/// </remarks>
/// <seealso href="http://en.wikipedia.org/wiki/Bernoulli_number"/>
/// <seealso href="http://mathworld.wolfram.com/BernoulliNumber.html"/>
public static double BernoulliNumber(int n)
{
if (n < 0)
{
throw new ArgumentOutOfRangeException(nameof(n));
}
// B_1 is the only odd Bernoulli number.
if (n == 1)
{
return(-1.0 / 2.0);
}
// For all other odd arguments, return zero.
if (n % 2 != 0)
{
return(0.0);
}
// If the argument is small enough, look up the answer in our stored array.
int m = n / 2;
if (m < Bernoulli.Length)
{
return(Bernoulli[m]);
}
// Otherwise, use the relationship with the Riemann zeta function to get the Bernoulli number.
// Since this is only done for large n, it would probably be faster to just sum the zeta series explicitly here.
double B = 2.0 * AdvancedMath.RiemannZeta(n) / AdvancedMath.PowOverGammaPlusOne(Global.TwoPI, n);
if (m % 2 == 0)
{
B = -B;
}
return(B);
}
