Mathematical function differentiation with C#?

Question

I see that I can declare a function with (say)

public double Function(double parameter)

but what if I do want to take the derivative of tha

Answer 1

You can't calculate the exact derivative of a function using a computer program (unless you're doing symbolic math... but that's another, way more complicated, topic).

There are several approaches to computing a numerical derivative of a function. The simplest is the centered three-point method:

• Take a small number h
• Evaluate [f(x+h) - f(x-h)] / 2h
• Voilà, an approximation of f'(x), with only two function evaluations

Another approach is the centered five-point method:

• Take a small number h
• Evaluate [f(x-2h) - 8f(x-h) + 8f(x+h) - f(x+2h)] / 12h
• Voilà, a better approximation of f'(x), but it requires more function evaluations

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Another topic is how to implement this using C#. First, you need a delegate that represents a function that maps a subset of the real numbers onto a another subset of the real numbers:

delegate double RealFunction(double arg);

Then, you need a routing that evaluates the derivative:

public double h = 10e-6; // I'm not sure if this is valid C#, I'm used to C++

static double Derivative(RealFunction f, double arg)
{
    double h2 = h*2;
    return (f(x-h2) - 8*f(x-h) + 8*f(x+h) - f(x+h2)) / (h2*6);
}

If you want an object-oriented implementation, you should create the following classes:

interface IFunction
{
    // Since operator () can't be overloaded, we'll use this trick.
    double this[double arg] { get; }
}

class Function : IFunction
{
    RealFunction func;

    public Function(RealFunction func)
    { this.func = func; }

    public double this[double arg]
    { get { return func(arg); } }
}

class Derivative : IFunction
{
    IFunction func;
    public static double h = 10e-6;

    public Derivative(IFunction func)
    { this.func = func; }

    public double this[double arg]
    {
        get
        {
            double h2 = h*2;
            return (
                func[arg - h2] - func[arg + h2] +
                ( func[arg + h]  - func[arg - h] ) * 8
                ) / (h2 * 6);
        }
    }
}