B4J Question How to solve nonlinear equation in B4J?

[daiweisc]

In B4J, there is a nonlinear equation to solve?
How to solve nonlinear equation in B4J?

 

[MicroDrie]

Perhaps it is useful to be a little more concrete about what the nonlinear equation problem exactly is, how the result should be presented and what you have or have not achieved yourself. Claus's xChart Class and b4xlib can display very slick graphical results.

 

[daiweisc]

Thanks for reply. The below is my qustion :

λ is a unknown number. From this formula, it is necessary to calculate λ in B4J.

 

[derez]

run a loop where lamda is changed in small increments (not including lamda <= 0):
res = 1/sqrt(lamda) - 2*logarithm(sqrt(lamda)-0.8))
when res is close enough to zero - log lamda and exit.

 

[angel_]

You can use the newton-raphson algorithm but you need to know the derivative and an initial approximation.

If the derived function is correct it would be something like this, you could also do it with a for and set a maximum number of iterations

Sub GetLambda(Re As Double) As Double
    Dim y   As Double         'f(x) = 0
    Dim x0  As Double = 1    'initial approach
    Dim x1  As Double
    Dim dx  As Double         'derivative function
    Dim Precision As Double = 0.0001
    Dim Check As Boolean = False
    
    Do While Check = False
        y = 2 * x0 * Logarithm(Re * Sqrt(x0) - 0.8, cE) - 1
        
        dx = 2 * Logarithm(Re * Sqrt(x0) - 0.8, cE) + 2 * x0 * Re * 1 / 2 * Power(x0, - 1 / 2) / (Re * Sqrt(x0) - 0.8)
    
        x1 = x0 - y / dx
    
        If Abs(x1 - x0) < Precision Then
            Return x1
        Else
            x0 = x1
        End If
    Loop
End Sub

 

[derez]

Dim lamda As Double
Dim res As Double
Dim p As Double
For lamda = 3.6 To 10 Step 0.0001
    p = Sqrt(lamda)
    res = 1/p - 2*Logarithm(p-0.8,cE)
    Log("lamda = " & lamda & "    res =  " & res)
    If Abs(res) <= 0.00001 Then Exit
Next

...
lamda = 4.29630000000342 res = 8.778585356983282E-5
lamda = 4.29640000000342 res = 4.426572195592726E-5
lamda = 4.2965000000034195 res = 7.469458777054072E-7

 

[Sandman]

I'm curious, why is it important to solve this using B4J? I imagine it's very simple to solve using Wolfram Alpha, for instance.

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

 

[daiweisc]

Thanks for your reply.
And another question:
How to show Double Lamda by Scientific notation, keep four decimal digits？

 

[William Lancee]

    Dim greek() As Int = Array As Int(0x03B1,0x03B2,0x03B3,0x03B4,0x03B5,0x03B6,0x03B7,0x03B8,0x03B9,0x03BA,0x03BB,0x03BC,0x03BD,0x03BE,0x03BF)
    Log(Chr(greek(10)) & " = " & NumberFormat2(lamda, 0, 4, 4, False))

λ = 4.2965

 

[William Lancee]

Or

    Dim greek As String =   "αβγδεζηθικλμνξοπρστυφχψωΑΒΓΔΕΖΗΘΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ"
    Dim english As String = "abgdezhziklmvxoprstufcywABGDEZHIKLMVXOPRSTUFCYW"
    Log(greek.charAt(english.IndexOf("l")) & " = " & NumberFormat2(lamda, 1, 4, 4, False))

 

[daiweisc]

Thanks for reply.
Another question, why the below code is wrong . The first line has a wave line :

Private Sub    GetLambda(Reynolds As Double) As Double
    Private y   As Double               'f(x) = 0
    Private x0  As Double = 0.0473    'initial approach
    Private x1  As Double
    Private dx  As Double         'derivative function
    Private Precision As Double = 0.0001
    Private LowerBound As Double = 1E-8
    Private MaxIterations As Int = 100
    Private Check As Boolean = False

    Do While Check = False
        Private count = 1 As Int
        ' Let x0 , x1 = Sqrt(Lambda) , simplify iteration formula
        y = 2 * x0 * Logarithm(Reynolds * x0 - 0.8, 10) - 1
        dx = 2 * Logarithm(Reynolds * x0 - 0.8, 10) + 2 * x0 * Reynolds / ((Reynolds * x0 - 0.8) * Logarithm(10 , cE))
        x1 = x0 - y / dx
        If ( Abs(x1 - x0) < Precision Or y = LowerBound ) Then
            Verify = True         
            Return x1
        Else
            x0 = x1
        End If
        If ( dx = 0 Or count > MaxIterations) Then
            Verify = False
            Return x1
        End If
        count=count+1
    Loop
End Sub

 

[daiweisc]

 

[klaus]

You should have a message on top of the Logs tab saying :

It is a warning, which reminds you that not all paths return a value.
But in your case it is OK, you can ignore it.

 

[daiweisc]

Thanks for reply.

 

[daiweisc]

Can the Sub function return two values?
Is the syntax of the 1 , 20 , 26 line codes correct ?

La , Verify = GetLambda(Re)

Private Sub    GetLambda(Reynolds As Double) As Double
    Private y   As Double               'f(x) = 0
    Private x0  As Double = 0.0276    'initial approach
    Private x1  As Double
    Private dx  As Double         'derivative function
    Private Precision As Double = 0.0001
    Private LowerBound As Double = 1E-8
    Private MaxIterations As Int = 100
    Private Check As Boolean = False

    Do While Check = False
        Private count = 1 As Int
        y = 2 * x0 * Logarithm(Reynolds * Sqrt(x0) - 0.8, 10) - 1
        dx = 2 * Logarithm(Reynolds * Sqrt(x0) - 0.8, 10) + Sqrt(x0) * Reynolds / ((Reynolds * Sqrt(x0) - 0.8) * Logarithm(10 , cE))
        x1 = x0 - y / dx
        If ( Abs(x1 - x0) < Precision Or y = LowerBound ) Then
            Verify = True          
            Return x1 , Verify
        Else
            x0 = x1
        End If
        If ( dx = 0 Or count > MaxIterations) Then
            Verify = False
            Return x1 , Verify
        End If
        count=count+1
    Loop
End Sub

 

[rbghongade]

Return multiple values as an array. The code in your previous post won't work.

 

[Brian Dean]

Return multiple values as an array.

Another alternative which I think is more flexible and better in some situations is to use an object ...

  Type result (item1 as int, item2 as double)
 
  . . .
 
  Sub Calculate(arg1 as double) As result
    Dim r as result
    . . .
    r.item1 = firstValue
    . . .
   
    r.item2 = secondValue
    . . .
    return r
  End Sub

 

[daiweisc]

Thanks for your reply.