Question

我正在尝试在Excel上的VBA中创建一个使用正割方法收敛到单个根的代码。 x0是第一个猜测，x1是第二个猜测。出于某种原因，每次我将Secant_it输入Excel单元格时，我都会#VALUE!。我知道问题不在于Secant，因为当我将其输入单元格时，我得到第一个值的正确数字。我在这里做错了什么？

Function Secant(x0, x1)
    Dim x_new As Double
    x_new = x1 - f(x1) * (x1 - x0) / (f(x1) - f(x0))
    Secant = x_new
End Function

Function Secant_it(x0, x1)
    Dim x_new As Double
    Dim j As Integer
    j = 1
    x_new = Secant(x0, x1)
    While j < 21
        x0 = x1
        x1 = x_new
        x_new = Secant(x0, x1)
        j = j + 1
    Wend
        Secant_it = x_new
End Function

Answer 1

我将您的功能更改为

Function Secant(x0, x1)
    Dim x_new As Double
    Dim a As Double, b As Double

    a = f(x1)
    b = f(x0)

    x_new = x1 - a * (x1 - x0) / (a - b)
    Secant = x_new
End Function

Function Secant_it(x0, x1)
    Dim x_new As Double
    Dim j As Integer

    j = 1
    x_new = Secant(x0, x1)
    While j < 21
        x0 = x1
        x1 = x_new
        x_new = Secant(x0, x1)
        Debug.Print x_new
        j = j + 1
    Wend
        Secant_it = x_new
End Function

Function f(x)
    f = (Exp(x / 10) * Cos(x)) - (x ^ 2)
End Function

我注意到您收到了#VALUE!，因为该函数在j = 9之后中断，因为a中的b和Secant变为0。因此，0除以Secant函数会导致错误。

出于测试目的，我采用了=Secant_it(2,3)

Answer 2

你的收敛测试导致溢出，因为在经过一定次数的迭代后，x0和x1变得非常接近，使得f（x1）-f（x0）太小。

您应该将收敛测试更改为检查x0和x1之间的差异是否低于所需的精度。

Function Secant_it(ByVal x0 As Double, ByVal x1 As Double)
    Dim x_new As Double
    Dim j As Integer
    j = 1
    x_new = Secant(x0, x1)
    'While j < 21
     While Abs(x1 - x0) > 0.000001 ' or any precision you desire
        x0 = x1
        x1 = x_new
        x_new = Secant(x0, x1)
        j = j + 1
    Wend
        Secant_it = x_new
End Function

Secant_it方法给出#VALUE！错误

2 个答案: