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Answer:
Here is a VBA user-defined function (UDF) that implements the Secant method:
Function Secant(X0 As Double, X1 As Double) As Double
' Returns the root of a function of the form F(x) = 0
' using the Secant method.
' X1 is a first guess at the value of x that solves the equation
' X0 is a "previous" value not equal to X1.
' This function assumes there is an external function named FS that
' represents the function whose root is to be solved
Dim X As Double 'the current guess for root being sought
Dim Xold As Double 'previous guess for root being sought
Dim DeltaX As Double
Dim Iter As Integer 'iteration counter
Const Tol = 0.00000001 'convergence tolerance
Xold = X0
X = X1
'permit a maximum of 100 iterations
For Iter = 1 To 100
DeltaX = (X - Xold) / (1 - FS(Xold) / FS(X))
X = X - DeltaX
If Abs(DeltaX) < Tol Then GoTo Solution
Next Iter
MsgBox "No root found", vbExclamation, "Secant result"
Solution:
Secant = X
End Function
As you can see, you must provide a function FS that is the function you desire the root of. I used the following function for my test of Secant:
Function FS(X As Double) As Double
'Example function cubic equation
FS = X^3 - X - 1
End Function
and I used Secant to solve this by entering it into a cell like this:
=Secant(1.4,1.3)
where 1.4 and 1.3 are the two "previous" (guess) values for x. I got a result of 1.324718, which is
Explanation:
i just saw it ^^"v