BroydenFletcherGoldfarbShanno Class

public class BroydenFletcherGoldfarbShanno : BaseGradientOptimizationMethod, 
	IGradientOptimizationMethod, IOptimizationMethod, IOptimizationMethod<double[], double>, 
	IGradientOptimizationMethod<double[], double>, IFunctionOptimizationMethod<double[], double>, 
	IOptimizationMethod<BroydenFletcherGoldfarbShannoStatus>, IOptimizationMethod<double[], double, BroydenFletcherGoldfarbShannoStatus>

Public Class BroydenFletcherGoldfarbShanno
	Inherits BaseGradientOptimizationMethod
	Implements IGradientOptimizationMethod, IOptimizationMethod, IOptimizationMethod(Of Double(), Double), 
	IGradientOptimizationMethod(Of Double(), Double), IFunctionOptimizationMethod(Of Double(), Double), 
	IOptimizationMethod(Of BroydenFletcherGoldfarbShannoStatus), IOptimizationMethod(Of Double(), Double, BroydenFletcherGoldfarbShannoStatus)

// Suppose we would like to find the minimum of the function
// 
//   f(x,y)  =  -exp{-(x-1)²} - exp{-(y-2)²/2}
// 

// First we need write down the function either as a named
// method, an anonymous method or as a lambda function:

Func<double[], double> f = (x) =>
    -Math.Exp(-Math.Pow(x[0] - 1, 2)) - Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2));

// Now, we need to write its gradient, which is just the
// vector of first partial derivatives del_f / del_x, as:
// 
//   g(x,y)  =  { del f / del x, del f / del y }
// 

Func<double[], double[]> g = (x) => new double[] 
{
    // df/dx = {-2 e^(-    (x-1)^2) (x-1)}
    2 * Math.Exp(-Math.Pow(x[0] - 1, 2)) * (x[0] - 1),

    // df/dy = {-  e^(-1/2 (y-2)^2) (y-2)}
    Math.Exp(-0.5 * Math.Pow(x[1] - 2, 2)) * (x[1] - 2)
};

// Finally, we can create the L-BFGS solver, passing the functions as arguments
var lbfgs = new BroydenFletcherGoldfarbShanno(numberOfVariables: 2, function: f, gradient: g);

// And then minimize the function:
bool success = lbfgs.Minimize();
double minValue = lbfgs.Value;
double[] solution = lbfgs.Solution;

// The resultant minimum value should be -2, and the solution
// vector should be { 1.0, 2.0 }. The answer can be checked on
// Wolfram Alpha by clicking the following the link:

// http://www.wolframalpha.com/input/?i=maximize+%28exp%28-%28x-1%29%C2%B2%29+%2B+exp%28-%28y-2%29%C2%B2%2F2%29%29