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Looking for General solution for a difference equation

  1. Mar 1, 2010 #1


    I know I can set At+1=At=A and solve for a special solution.

    What would be a general solution?

    I am not taking a course in Difference Equation, and this is not my homework but I encounter a similar question and I reduce it to this form.

    Last edited: Mar 1, 2010
  2. jcsd
  3. Mar 1, 2010 #2
    Or One step further, how to solve this one?


    A_t means A with t subscipt

    I don't think it is as easy as it looks.
    Last edited: Mar 1, 2010
  4. Mar 2, 2010 #3


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    It doesn't even look like anything. :tongue:

    Is your second one meant to be

    (1+r)At+1-At+At+1*At=r ?
  5. Mar 2, 2010 #4
    Yes That is what I mean. And it is just algebra manipulation of the original equation
  6. Mar 4, 2010 #5


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    Your starting form is

    [tex]A_{t+1} = \frac{A_t + r}{A_t + r + 1},[/tex]
    which is perhaps more convenient to write as

    [tex]A_{t+1} = 1 - \frac{1}{A_t+r+1}[/tex]

    What you could to do try to solve it is find A_2 in terms of A_1, plug that into A_3 to get it in terms of A_1, and so on, to see if you can guess what A_t for general t will look like. (I don't know if this one will turn out nice. I did the first couple of steps myself, and it looks like you may want to look into http://en.wikipedia.org/wiki/Continued_fraction (go to the page on periodic continued fractions too). There may be nice solutions for specific values of r or A_1. (Of course, A_1 will be irrelevant as t -> infinity)
    Last edited by a moderator: Apr 24, 2017
  7. Mar 4, 2010 #6
    Specific solution for A_t is easy to solve

    Set A_t=A=A_t+1

    We can solve that r=A/(1-A) or the other way, A=(1/2)(-r+(-)(r^2+4r)^1/2

    But I still find it very hard to find a general solution.
  8. Mar 5, 2010 #7
    A general solution is hard to find.
    But you can get some idea about the asymptotic behaviour when the sequence is approaching equilibrium A.

    So assume your solution is


    Where the sequence p goes to zero as k increases. At the limit of large k, you can expand the RHS to a taylor series. Because we want a linear easy solution, we will only care about the first order. So:


    Let A+r+1=b to make it look nicer. Also 1-1/b=A. So:




    So at large k's, the series will be:


    This will be a real approximation only if abs(1-A)<1 (the solution is really stable), and that depends on what value of A you have chose, and what is r.
  9. Mar 5, 2010 #8
    How did u expand it?

  10. Mar 11, 2010 #9
    This is a geometrical series:


    setting q-> (-q) the terms will occur with alternating signs so the linear approximation will be

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