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Undetermined Coefficients of Difference Equations

  1. Mar 10, 2010 #1
    I am working on a section on undetermined coefficients of a difference. I'm really trying to understand the material, but am struggling.
    [tex]y_{n+1}[/tex]-2[tex]y_{n}[/tex]=2
    I got as far as saying for the homogenous equation y=[tex]c_{1}[/tex][tex](-1)^{n}[/tex].
    I'm assuming the guess next would be A, but I'm confused.
    Maybe if someone could help me with some examples on how to do this using the 2 or something like 2^n or cos(n). I'm trying to learn something something new on my own and it's hard.
     
  2. jcsd
  3. Mar 11, 2010 #2

    tiny-tim

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    Hi kathrynag! :smile:

    Do you mean yn+1 - 2yn = 2 ? :confused:

    If so, the homogenous equation for this recurrence relation is yn+1 - 2yn = 0, and C(-1)n is not a solution, is it?

    For the particular solution, try a polynomial of lowest possible degree (start with zero!)
     
  4. Mar 11, 2010 #3
    Yeah that's what i meant. I tried letting the polynomial =A.
    Then A(n+1)+A=2
    An+2A=2
    This is where I get stuck.
    Now if I had a number like n^2 on the end polynomial would be An^2+Bn+C
    An^2(n+1)+Cn(n+1)+Cn+C+An^2+Bn+C=n^2
    An^3+n^2(2A+C)+n(2C+B)+(2C)=n^2
    I guess I get stuck at the end each time. Say I had cosn. Then polynomial would be Acosn+Bsinn.
    Acos(n+1)+Bsin(n+1)+Acosn+Bsin(n)=cosn

    Maybe I'm doing something wrong somewhere.
     
  5. Mar 11, 2010 #4

    tiny-tim

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    Hi kathrynag! :smile:
    Now you're mystifying me …

    where did A(n+1) come from? :confused:

    If the polynomial is A, then yn+1 - 2yn is -A …

    carry on from there. :smile:
     
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