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TitleUse Newton's method to find the roots of an equation in Visual Basic 6
DescriptionThis example shows how to use Newton's method to find the roots of an equation in Visual Basic 6.
KeywordsNewton's method, equation, root, zero
CategoriesAlgorithms, Graphics
Newton's method calculates the roots of equations. In other words, it finds the values of X for which F(X) = 0.

This program graphs the equation X^3/3 - 2*X + 5. When you click on the graph, it uses Newton's method to find a root of the equation, starting from the X value that you clicked.

' The function.
Private Function F(ByVal X As Double) As Single
    F = X * X * X / 3 - 2 * X * X + 5
End Function

' Draw the background graph.
Private Sub DrawGraph()
Const STEP_SIZE As Single = 0.1
Dim X As Single
Dim Y As Single
Dim y1 As Single
Dim y2 As Single


    ' Draw the axes.
    For X = -10 To 10
        Me.Line (X, -0.5)-Step(0, 1), vbBlue
    Next X
    For Y = -10 To 10
        Me.Line (-0.5, Y)-Step(1, 0), vbBlue
    Next Y
    Me.Line (-10, 0)-Step(20, 0), vbBlue
    Me.Line (0, -10)-Step(0, 20), vbBlue

    ' Draw the curve.
    y2 = F(-10)
    For X = -10 + STEP_SIZE To 10 Step STEP_SIZE
        y1 = y2
        y2 = F(X)
        Me.Line (X - STEP_SIZE, y1)-(X, y2), vbRed
    Next X
End Sub
The way Newton's method works is it starts from an initial guess X0 (given by the point you clicked). It then estimates the difference between that value and root:

    epsilon = -F(x) / dFdx(x)

The method then sets its next guess for x to be the current value plus epsilon (X(n+1) = X(n) + epsilon). It repeats this process until epsilon is smaller than some cutoff value.

' The function's derivative.
Private Function dFdx(ByVal X As Double) As Single
    dFdx = X * X - 4 * X
End Function

' Find the roots by using Newton's method.
Private Sub Form_MouseDown(Button As Integer, Shift As _
     Integer, X As Single,
Y As Single)
Const CUTOFF As Single = 0.0000001
Dim epsilon As Single
Dim iterations As Integer
Dim x0 As Single

    ' Clear previous results.

    x0 = X
    iterations = 0
        iterations = iterations + 1
        Me.Line (x0, -1.5)-Step(0, 3)
        epsilon = -F(x0) / dFdx(x0)
        x0 = x0 + epsilon
    Loop While Abs(epsilon) > CUTOFF
    Me.Line (x0, -3)-Step(0, 6)

    Me.Caption = x0 & " +/-" & epsilon & " in " & _
        iterations & " iterations"
End Sub
Newton's method uses the function's derivative to make its next X value guess. If the curve is relatively flat for the current value of X, then the derivative (which is the slope of the function) gives a value far away from the current one and the method is unstable. In this example, try clicking on a point where X is at or near 0 and see what happens.

For more information on Newton's method, see Eric W. Weisstein's article Newton's Method from MathWorld--A Wolfram Web Resource.

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