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  1. #1
    Join Date
    Nov 2004

    Exclamation Unanswered: complex maths equation... help needed

    hello people

    hmm have another problem on hand... two maths problem in fact...

    1) suppose i have b= (6.44*10E-9 *c^3/2)/ B

    where b, B is both known... and we need to find unknown value "c" hmm how should we go about doing it .. seems kind of simple but i actually have a hard time doing it....

    2) suppose we have two equation : b = (6.44*10E-9 * c^3/2)/B....(1)

    a = (J *1.54*10E-6*B^2*exp(10.4/sqr(c)) / c...(2)

    suppose a,b and j are known values... and we need to find c and B.... i have problem doing that.. initally i try to find one unknow first .. ie) making B the value we want to find first in terms of c.... so we have

    B = (6.44*10E-9 * c^3/2)/b ... (3)

    then we sub formula (3) into (2) so we have

    a= (J *1.54*10E-6*(((6.44*10E-9 * c^3/2)/b)^2)exp(10.4/sqr(c))/c....(4)

    so with a, J and b known... how are we going to find the value of c since there is going to be exponential part to be solved... hmm i ask around and i was told that by bringing everything to one side and then solve by trial and error.. i would be able to find the unknown values.. but i have been trying and cannot seems to get it...

    if (J *1.54*10E-6*(((6.44*10E-9 * c^3/2)/b)^2)exp(10.4/sqr(c))/c - a =0

    find c..and then B.. how are we going to do it by using a program ????

    hope u guys know wat i am trying to do here and help me out in any way that u can...

    really appreciate the effort...

    thanks in advance...

    Last edited by sportyboy; 02-17-05 at 02:55. Reason: missing parameters

  2. #2
    Join Date
    Jan 2004
    Aberdeen, Scotland
    Iteration, Find a lower bound and an upper bound i.e.
    for equation 1)
    0= (6.44*10E-9 *c^3/2)/ B-b

    find a lower bound for c and an upper bound for C i.e.
    Lower Bound
    (6.44*10E-9 *c^3/2)/ B-b equates < 0
    Upper Bound equates >0
    take the value at the midpoint of this and equate if its above 0 change upper bound
    if its below 0 change lower bound

    then Loop

    here is a v quick example for #1
    try to improve on this and get yourself an answer for #2

  3. #3
    Join Date
    Nov 2004

    help required

    hello david..

    thanks for the quick reply.. but i still have difficulty in solving the iteration... cause i am quite weak in my can u pls help me out here??

    thanks alot for yr help

    regards sportyboy

  4. #4
    Join Date
    Jan 2004
    Aberdeen, Scotland
    it depends how you want to do it,
    are you creating a form for users to input values on or is it a stand alone module,

    im assuming that this is homework so ill just post some ideas on how to go about it,
    I also assume that your using VB,

    youll need a variable for each part of your expression, also for the upper bound and a lower bound
    work out your Value for C by taking the midpoint of your lower bound and your upper bound and adding it to the lower bound

    run it through a do loop testing when your equation gets close to 0 pick a reasonable value to test to i.e. 0.00000000005 reset either upper or lower bound dependant on whether or not you obtain a positive or negative result

    give it a try and post what you get and we can talk about getting Your soloution as good as you need it

    as a quick side note you can evaluate expression 1 by pure algebra and just work it out

  5. #5
    Join Date
    Nov 2004

    Exclamation another few problem.. big ones...

    done with part 1 already...

    hmm actually its my project.. hmm having a lot of problem... well here is another problem that i have face...

    i wan to link VB and matlab together ... using just a click of the button.. already have the interface for it and also the matlab part for plotting has so far been finish... it look something like this///

    V=U( : , 1)'
    I=U( : , 2)'
    while n<N+1

    while n<N+1
    Slope=Slope+( y(1,n)-y(1,n-1) )/( x(1,n)-x(1,n-1) )
    Interception=Interception-( y(1,n)-y(1,n-1) )*x(1,n)/( x(1,n)-x(1,n-1) )+y(1,n)



    Plot_X= [X_Constrain x 0]
    Plot_Y=[Y_Constrain y Interception_Average]
    plot(Plot_X, Plot_Y)
    title('Linear Plot for a&b Value')

    diary M.xls
    M= [exp(Interception_Average) -Slope_Average]
    diary off


    but i have two problems here... 1) i can open only the file that is being place under the xlsread command... i cannot open from a directory and choose the file that i desired...

    2) i search for some reference and then find that by keying in the command...
    Dim x As Long
    x = Shell("E:\MATLAB6p5\bin\win32\matlab.exe", 1)

    i am being able to start the entire matlab program.. but thats not wat i want... cause it will still need to go through whole process... so wat should i do in order to just by a click of a button.. just start matlab and ask user to choose the file that they want and then when press ok .. the plot would be produce... hmm i hope u know wat i am trying to say here.. and of course i do hope that u can help me in any way that u can....

    thanks a lot

  6. #6
    Join Date
    Nov 2004

    Exclamation help me pls

    hello there..

    hmm i have write some code for the problem..but there seems to have a overflow problem...

    the source code look like this..

    Sub Benny()


    Dim alpha As Double
    Dim a As Double
    Dim b As Double
    Dim beta As Double
    Dim funnySign As Double
    Dim row As Integer
    Dim tempStr As String
    Dim temp As Double

    alpha = Range("A1").Value
    a = Range("A2").Value
    b = Range("A3").Value
    beta = 1
    row = 4

    Do While beta < 10 ^ 5
    funnySign = ((b * beta) / (6.44 * (10 ^ 9))) ^ (2 / 3)
    tempStr = Str(funnySign)
    Call formatNumber(tempStr)
    temp = (a - (alpha * 1.54 * (10 ^ -6) * (beta ^ 2) * Exp(10.4 / formatNo)) / formatNo)
    MsgBox temp
    If temp = 0 Then
    Range("A" & row).Value = counter
    row = row + 1
    End If
    beta = beta + 2

    End Sub
    Sub formatNumber(temp As String)
    'up to 3 decimal places
    Dim start As Double
    Dim extract As String

    extract = Mid(temp, 1, 3)
    formatNo = CDbl(extract)

    For counter = 1 To 3
    extract = Mid(temp, counter + 3, 1)
    start = 1
    For times = 1 To counter
    start = start / 10#
    formatNo = formatNo + CDbl(extract) * start
    End Sub

    hmm wat can i do elimate this problem..

    thanks a lot

  7. #7
    Join Date
    Jan 2004
    Aberdeen, Scotland
    are you still trying to solve the equation finding the values that c can hold or are you chart the function now

    formatNo doesnt appear to be a global variable but if it is it will allow for divisions by 0
    im not sure where your overflow is comming in what line does it stop on and what values are you getting

    your second sub is also redundant if your using excel as you can use something like

    to round your number to however many decimal places
    looking at your equation you also cant have negative numbers also
    here is my soloution to part 1 this will need to be heavily altered for part 2

    Sub calcEqn1()
        Dim wks As Worksheet
        Dim a As Double
        Dim b As Double
        Dim beta As Double, betaMin As Double, betaMax As Double
        Dim Gamma As Double 'changed funny sign to gamma
        Dim bolReverse As Boolean
        Dim iterations As Double
        Dim temp As Double
        Set wks = Worksheets("Sheet2")
        a = wks.Range("A2").Value
        b = wks.Range("A3").Value
        beta = 1: betaMin = 1: betaMax = 1
        bolReverse = False
        'need to find upper and lower bounds for beta
        'start with lower bounds
        'do until Gamma is less than 0
            'as the eqaution is a cube root ( values under 1 are not defined (i.e. Imaginary numbers)
            Gamma = (((6.44 * (10 ^ 9) * betaMin ^ (3 / 2))) / b) - a
            If Not bolReverse Then
                If temp <> 0 Then
                    'test if temp < Gamma(Gamma increasing) if so reverse direction of search
                    If temp < Gamma Then bolReverse = True
                End If
                temp = Gamma
                If Gamma > 0 Then betaMin = betaMin - 0.1
                If Gamma > 0 Then betaMin = betaMin + 0.1
            End If
        Loop Until Gamma <= 0
        'next upper bound
            Gamma = Gamma = (((6.44 * (10 ^ 9) * betaMax ^ (3 / 2))) / b) - a
            If Not bolReverse Then
                If Gamma > 0 Then betaMax = betaMax + 0.1
                If Gamma > 0 Then betaMin = betaMax - 0.1
            End If
        Loop Until Gamma >= 0
        'now we have obtained an upper and lower bounds iterate through function
        'we can now calculate beta
        'we want betamin to be the lower value for ease of use
        'so adjust betamin so that its less than betamax
        If betaMin > betaMax Then
            temp = betaMin
            betaMin = betaMax
            betaMax = temp
        End If
            'set up instance of beta as midpoint of betamin and betamax
            beta = betaMin + Abs(betaMax - betaMin) / 2
            'Calculate value of beta
            Gamma = (((6.44 * (10 ^ 9) * beta ^ (3 / 2))) / b) - a
            'check sign of gamma if its < 0 then replace betamin
            If Gamma < 0 Then
                betaMin = beta
                betaMax = beta
            End If
            iterations = iterations + 1
            If iterations > 1000000 Then
                MsgBox "Lots of iterations i give up", vbOKOnly
                Exit Sub
            End If
        Loop Until Abs(Gamma) < 0.000000005
        MsgBox "beta = " & beta & " after " & iterations & " iterations"
    End Sub
    Dont use this as your soloution but see if you can figure out whats going on and write your own (always better for learning how to do things), this uses the principles of iteration i stated above whereas yours just seems to be developing numbers for charting

    also try thinking how you would manually solve the problem and see if you can programatically put these processes down in order (i know how i would do it manually)

    finally is this a maths exercise or is it a programming exercise as i would do the 2 in different ways

  8. #8
    Join Date
    Nov 2004

    help me pls

    this is a programming exercise.. and its really urgent...

    have been trying but still cannot get the correct answer
    there is always a overflow problem...

    wat can i do.. pls help


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