Numerical Differentiation

The FunctionMath class contains a number of static (Shared) methods for approximating the derivative of a function. Finite differences are used to approximate the derivative. An adaptive method finds the best choice of finite difference, and estimate the error. You can also create a delegate that evaluates the derivative function of an arbitrary function numerically.

All methods listed in this section are defined as extension methods. This means that in most languages they can be called as if they were instance methods of their first argument. The examples illustrate how this is done.

The Derivative method takes two or three parameters. The first argument is a Func<T, TResult> that specifies the function to differentiate. The second argumentspecifies the point at which to evaluate the derivative. The following example evaluates the numerical derivative of the cosine function at x = 1.

Console.WriteLine("Numerical derivative of cos(x):");
double result = FunctionMath.Derivative(Math.Cos, 1.0);
Console.WriteLine("  Result = {0}", result);
Console.WriteLine("  Actual = {0}", -Math.Sin(1));

Console.WriteLine("Numerical derivative of cos(x):")
Dim result As Double = FunctionMath.Derivative(AddressOf Math.Cos, 1.0)
Console.WriteLine("  Result = { 0 }", result)
Console.WriteLine("  Actual = { 0 }", -Math.Sin(1))

No code example is currently available or this language may not be supported.

Console.WriteLine("Numerical derivative of cos(x):")
let result = FunctionMath.Derivative(Func<_,_>(Math.Cos), 1.0)
Console.WriteLine("  Result = {0}", result)
Console.WriteLine("  Actual = {0}", -Math.Sin(1.0))
//

The optional third parameter, of type DifferencesDirection, specifies the kind of finite differences to use in the calculation. The possible values for this parameter are summarized in table 3.

Passing DifferencesDirection, or omitting the third parameter invokes the algorithm using central differences. The target function is evaluated on both sides of the target point.

In some cases this may not work. When the target function is either undefined, the algorithm will break down. If the target function contains a discontinuity, the result will be wrong. In these cases, you should use either forward or backward differences.

Passing DifferencesDirection to Derivative invokes the algorithm using backward differences. The target function is evaluated only to the left of the target point and at the target point, not to the right of the target point. Passing DifferencesDirection to Derivative invokes the algorithm using forward differences. The target function is evaluated only to the right of the target point and at the target point, not to the left of the target point.

The BackwardDerivative, CentralDerivative and ForwardDerivative methods call these respective algorithms directly. They optionally take an out parameter, in which the estimated absolute error is returned.

It may be useful to be able to use the numerical derivative as a function that can be evaluated at arbitrary points. This can be done using the GetNumericalDifferentiator method. This method returns a Func<T, TResult> delegate that represents the derivative of the argument passed to the method. By default, central differences are used. To obtain forward or backward derivatives, use the GetForwardDifferentiator and GetBackwardDifferentiator methods, respectively.

Func<double, double> f = Math.Cos;
Func<double, double> df = FunctionMath.GetNumericalDifferentiator(f);
// Pass both to an equation solver:
double root = FunctionMath.FindZero(f, df, 1.0);

Dim f As Func(Of Double, Double) = AddressOf Math.Cos
Dim df As Func(Of Double, Double) = f.GetNumericalDifferentiator()
' Pass both to an equation solver
Dim root As Double = FunctionMath.FindZero(f, df, 1.0)

No code example is currently available or this language may not be supported.

let f = Func<_,_>(Math.Cos);
let df = f.GetNumericalDifferentiator()
// Pass both to an equation solver:
let root = FunctionMath.FindZero(f, df, 1.0)
//

Using a numerical derivative for an algorithm that requires it, as in the example above, is usually not a good idea. Most such algorithms have alternatives that do not use a derivative, but may be more expensive in terms of function evaluations. Still, numerical differentiation is relatively expensive, costing at least 5 evaluations of the original function. This is an important consideration when deciding which algorithm to use.

In addition to simple derivatives of functions of one variable, numerical approximations to the gradient of a multivariate function, or the Jacobian of a set of multivariate functions can also be computed. The GetNumericalGradient method returns a delegate that evaluates the gradient of its argument. The GetNumericalJacobian

Numerical differentiation is an inherently unstable process. There is really no way to avoid subtracting two numbers that are very close, resulting in very significant round-off error. The best we can hope for is a relative accuracy of roughly half the machine precision (SqrtEpsilon).

Reference

S.D. Conte and Carl de Boor, Elementary Numerical Analysis: An Algorithmic Approach, McGraw-Hill, 1972.