Higher Dimensional Numerical Integration in Visual Basic QuickStart Sample
Illustrates numerical integration of functions in higher dimensions using classes in the Extreme.Mathematics.Calculus namespace in Visual Basic.
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Option Infer On
' The numerical integration classes reside in the
' Extreme.Mathematics.Calculus namespace.
Imports Extreme.Mathematics.Calculus
' Function delegates reside in the Extreme.Mathematics
' namespace.
Imports Extreme.Mathematics
Imports Extreme.Mathematics.Algorithms
' Illustrates numerical integration in higher dimensions using
' classes in the Extreme.Mathematics.Calculus namespace of Extreme Numerics.NET.
Module NDIntegration
Sub Main()
' The license is verified at runtime. We're using
' a demo license here. For more information, see
' https://numerics.net/trial-key
Extreme.License.Verify("Demo license")
'
' Two-dimensional integration
'
' The function we are integrating must be
' provided as a Func(Of Double, Double, Double) delegate.
' For more information about this delegate, see the
' FunctionDelegates QuickStart sample.
'
' The functions used in this sample are defined at
' the end of this file.
Dim f1 As Func(Of Double, Double, Double) = AddressOf Integrand1
Dim f2 As Func(Of Double, Double, Double) = AddressOf Integrand2
Dim f3 As Func(Of Double, Double, Double) = AddressOf Integrand3
' Variable to hold the result:
Dim result As Double
' The AdaptiveIntegrator2D class is the most efficient
' 2D integrator in most cases. It uses an adaptive algorithm.
' Construct an instance of the integrator class:
Dim integrator1 As New AdaptiveIntegrator2D()
' An example of setting the integrand and bounds through properties
' is given below. Here, we put the integrand and the bounds
' of the integration region directly in the call to Integrate,
' which performs the calculation:
integrator1.Integrate(f1, 0, 1, 0, 1)
Console.WriteLine("4 / (1 + 2x + 2y) on [0,1] * [0,1]")
Console.WriteLine(" Value: {0:F15}", integrator1.Result)
Console.WriteLine(" Exact value: {0:F15} = Ln(3125 / 729)", Math.Log(3125.0 / 729.0))
' To see whether the algorithm ended normally,
' inspect the Status property:
Console.WriteLine(" Status: {0}",
integrator1.Status)
Console.WriteLine(" Estimated error: {0}",
integrator1.EstimatedError)
Console.WriteLine(" Iterations: {0}",
integrator1.IterationsNeeded)
Console.WriteLine(" Function evaluations: {0}",
integrator1.EvaluationsNeeded)
' Another integrator uses repeated 1-dimensional
' integration:
Dim integrator2 As New Repeated1DIntegrator2D()
' You can set the order of integration, as well as
' the integration rules for the X and the Y direction:
integrator2.InitialDirection = Repeated1DIntegratorDirection.X
' You can set the integrand and the bounds of the integration region
' by setting properties of the integrator object:
integrator2.Integrand = f2
integrator2.XLowerBound = 0.0
integrator2.XUpperBound = 1.0
integrator2.YLowerBound = 0.0
integrator2.YUpperBound = 1.0
result = integrator2.Integrate()
Console.WriteLine("Pi^2 / 4 Sin(Pi x) Sin(Pi y) on [0,1] * [0,1]")
Console.WriteLine(" Value: {0:F15}", integrator2.Result)
Console.WriteLine(" Exact value: {0:F15}", 1.0)
' To see whether the algorithm ended normally,
' inspect the Status property:
Console.WriteLine(" Status: {0}",
integrator2.Status)
Console.WriteLine(" Estimated error: {0}",
integrator2.EstimatedError)
Console.WriteLine(" Iterations: {0}",
integrator2.IterationsNeeded)
Console.WriteLine(" Function evaluations: {0}",
integrator2.EvaluationsNeeded)
'
' Integration over arbitrary regions
'
' The repeated 1D integrator can also be used to compute
' integrals over arbitrary regions. To do this, you need to
' supply function that return the lower bound and upper bound
' of the region as a function of x.
' Here, we integrate x^2 * y^2 over the unit disk.
integrator2.LowerBoundFunction = AddressOf DiskLowerBound
integrator2.UpperBoundFunction = AddressOf DiskUpperBound
integrator2.XLowerBound = -1.0
integrator2.XUpperBound = 1.0
integrator2.Integrand = f3
result = integrator2.Integrate()
Console.WriteLine("x^2 * y^2 on the unit disk")
Console.WriteLine(" Value: {0:F15}", integrator2.Result)
Console.WriteLine(" Exact value: {0:F15} = Pi / 24", Math.PI / 24)
' To see whether the algorithm ended normally,
' inspect the Status property:
Console.WriteLine(" Status: {0}",
integrator2.Status)
Console.WriteLine(" Estimated error: {0}",
integrator2.EstimatedError)
Console.WriteLine(" Iterations: {0}",
integrator2.IterationsNeeded)
Console.WriteLine(" Function evaluations: {0}",
integrator2.EvaluationsNeeded)
Console.Write("Press Enter key to exit...")
Console.ReadLine()
End Sub
' <summary>
' Function to integrate over [0, 1] * [0, 1]
' </summary>
Function Integrand1(x As Double, y As Double) As Double
Return 4 / (1 + 2 * x + 2 * y)
End Function
' <summary>
' Function to integrate over [0, 1] * [0, 1]
' </summary>
Function Integrand2(x As Double, y As Double) As Double
Return Math.PI * Math.PI / 4 * Math.Sin(Math.PI * x) * Math.Sin(Math.PI * y)
End Function
' <summary>
' Lower bound of a unit disk as a function of x.
' </summary>
Function DiskLowerBound(x As Double) As Double
x = Math.Abs(x)
If (x >= 1) Then
Return 0
End If
Return -Math.Sqrt(1 - x * x)
End Function
' <summary>
' Upper bound of a unit disk as a function of x.
' </summary>
Function DiskUpperBound(x As Double) As Double
x = Math.Abs(x)
If (x >= 1) Then
Return 0
End If
Return Math.Sqrt(1 - x * x)
End Function
' <summary>
' Function to integrate over the unit disc.
' </summary>
Function Integrand3(x As Double, y As Double) As Double
Return x * x * y * y
End Function
End Module