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Looking for General solution for a difference equation

  1. Mar 1, 2010 #1
    At+1=(At+r)/(At+r+1)

    A1=constant

    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.

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

    (1+r)A_t+1-A_t+A_t+1*A_t=r

    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

    epenguin

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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

    Mute

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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

    [tex]A_{k}=A+p_{k}[/tex]

    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:

    [tex]A+p_{k+1}=1-\frac{1}{A+r+1}\frac{1}{1+\frac{p_{k}}{A+r+1}}=1-\frac{1}{A+r+1}(1-\frac{p_{k}}{A+r+1})[/tex]

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

    [tex]A+p_{k+1}=A+(1-A)^{2}p_{k}[/tex]

    Or

    [tex]p_{k+1}-(1-A)^{2}p_{k}=0[/tex]

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

    [tex]A_{k}=A+(1-A)^{2k}[/tex]

    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:

    [tex]\frac{1}{1-q}=1+q+q^{2}+q^{3}+...[/tex]

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

    [tex]\frac{1}{1+q}=1-q[/tex]
     
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