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Mathematical model

  1. Sep 28, 2009 #1
    I have to transform this equation X(j+1)=X(j)+c1*(Y(j)-Y(j-1))+c2(Z(j)-Z(j-1))
    in a differential one.

    I would like to know if is it possible multiply the equation by (1/z), like this:

    (X(j+1)-X(j)) / Z = c1*((Y(j)-Y(j-1))/ Z)+c2((Z(j)-Z(j-1)) / Z)

    Then approximate the difference by derivates:

    dX(j)=c1*dY(j)+c2*dZ(j) with z->0 ??
    dz dz dz
     
  2. jcsd
  3. Sep 28, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What, exactly, is this "z" that you are dividing by? Is it the same as the function Z(j)?
    And why would j+1 go to j as z goes to 0?
     
  4. Sep 28, 2009 #3
    Oh sorry! The "z" that I'm dividing isn't the same Z(j). And j+1=z+delta(z).
     
  5. Sep 29, 2009 #4
    Ok, I'm going to try to write the problem again:
    I have the following equation:

    f(x+h)=f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h)) (1)

    I would like to now if is it possible multiply the equation by (1/h):

    ((f(x+h)-f(x))/ h)= c1*((g(x)-g(x-h))/z)+c2*((h(x)-h(x-h))/z)

    Then approximate the differences by the derivates:

    f '(x)=c1*g'(x)+c2*g'(2).

    Or should I derivate the equation (1) in both sides, like this:

    d[f(x+h)] = d [f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h))]
    dx dx
     
  6. Sep 29, 2009 #5

    You could take the first approach provided the derivative of both g(x) and h(x) exists. This is reminiscent of the Taylor expansion series. Only, your equation does not have the higher order terms in it.
     
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