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Need an equation for the following (this isn't coursework but was told to place here)

  1. Jul 21, 2009 #1
    Can anyone figure this out... searching for an algebraic equation (not a student)

    I'm trying to unravel some equations in a spreadsheet and as such I need to find a non-time-series, algebraic equation for the following data expressed for finding C:

    A + B always equals 10 (i.e. A1 + B1 = 10; A2 + B2 = 10; ...)

    when:

    A0 = 0... B0 = 10 .... C0 = 10

    A1 = 1 ... B1 = 9 ... C1 = 4

    A2 = 2 ... B2 = 8 ... C2 = 2

    A3 = 3 ... B3 = 7 ... C3 = 0.83 repeating 3

    A4 = 4 ... B4 = 6 ... C4 = 0.2

    A5 = 5 ... B5 = 5 ... C5 = 0

    A6 = 6 ... B6 = 4 ... C6 = 0.2

    A7 = 7 ... B7 = 3 ... C7 = 0.83 repeating 3

    A8 = 8 ... B8 = 2 ... C8 = 2

    A9 = 9 ... B9 = 1 ... C9 = 4

    A10 = 10 ... B10 = 0 ... C10 = 10

    an eqational attempts would be something along the lines of:

    (A - B) / (A + B) = C

    (A - B) / (A times B) = C

    (A - B)^2 / (A + B) = C

    Note: C3 = C7 = 0.83 = 5 / 6 (serves as a decent clue as it displays both a numerator and a denominator)

    once again, I am not seeking a time series equation...
     
    Last edited: Jul 21, 2009
  2. jcsd
  3. Jul 23, 2009 #2

    Dick

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    Re: Need an equation for the following (this isn't coursework but was told to place h

    What's a 'not time series equation'? Do you want an exact polynomial through those points or what?
     
  4. Jul 24, 2009 #3

    HallsofIvy

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    Re: Can anyone figure this out... searching for an algebraic equation (not a student)

    You are seeking an equation. "Time" is not a mathematical concept so "time series" is not relevant here.

    Since x+ y= 10, y is fixed once you know x so you can just look for a function of x giving those values. I notice this is symmertric about x= 5 so I would look for f(x)= a(x-5)2n Taking f(0)= a(5)2n= 10 and f(1)= a(4)2n= 4 we can divide the first equation by the second and get
    [tex]\frac{5^{2n}}{4^{2n}}= \left(\frac{5}{4}\right)^{2n}= \frac{10}{4}= \frac{5}{2}[/itex]
    Taking log of both sides, 2n log(5/4)= log(5/2) or 2n(0.096910)=0.39794 so 2n= 4.106. Let's just call that "4". Then, from f(0)= a(5)2n= a(54)= 125a= 10, a= 10/125= 0.08.

    See if f(x,y)= 0.08 x4 fits the other values.
     
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