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Converegence Problem for forward and backward propogation

  1. Sep 13, 2012 #1
    The function is J(n-1)+J(n+1)=2nJ(n)

    I need to prove that it diverges in forward direction but converges in backward direction.
    I am unable to find any method, kindly suggest.
     
  2. jcsd
  3. Sep 13, 2012 #2
    Are you sure that you typed the function correctly?
     
  4. Sep 13, 2012 #3
    Yes.

    for forward direction we can re-write it this way..
    J(n)=2*(n-1)*J(n-1)-J(n-2)

    and for backward propogation...
    J(n)=2*(n+1)*J(n+1)-J(n+2)
     
  5. Sep 13, 2012 #4
  6. Sep 13, 2012 #5

    Ray Vickson

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    You need to show your work.

    RGV
     
  7. Sep 13, 2012 #6
    The typical approach shown in text is to show that if the solution of such function truns out to be in form of
    [itex]C.λ^n[/itex] then we can say that such recursive function diverges, I have tried the same approach for the problem but the λ I am getting turns out to be function of n.

    I can scan and upload my work if you wish.
     
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