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Recurrence Relations

  1. Jun 21, 2011 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    [tex]a_n+3a_{n-1}-10a_{n-2}=2^n[/tex]


    3. The attempt at a solution
    I missed the lectures that addressed how to solve these kinds of problems, and while studying my recommended text book it only went as far as solving recurrence relations that are equal to 0 as opposed to 2n. I understand how to solve the simple recurrence relations but I have no clue as to what to do with these.

    Also, if you may, can you please explain any special cases that I should be looking for?

    For example,

    [tex]a_n+2a_{n-1}+a_{n-2}=0[/tex] would need to be handled differently because of the double root associated with it.
     
  2. jcsd
  3. Jun 21, 2011 #2
    This should be transformable into a homogenous recurrence though it might be a bit messy. Consider 2n+1-2n = 2n
     
  4. Jun 21, 2011 #3

    Mentallic

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    Sorry but I'm going to need a little more than that, this is my first time answering questions of this form.
     
  5. Jun 21, 2011 #4
    [itex]a_{n} + 3a_{n-1} - 10a_{n-2} - (a_{n-1} + 3a_{n-2} - 10a_{n-3}) = 2^{n} - 2^{n-1} = 2^{n-1}[/itex]

    also [itex]2^{n-1} = a_{n-1} + 3a_{n-2} - 10a_{n-3}[/itex]

    so [itex]a_{n} + 2a_{n-1} -13a_{n-2} + 10a_{n-3} = a_{n-1} + 3a_{n-2} - 10a_{n-3}[/itex]
    or [itex]a_{n} + a_{n-1} - 16a_{n-2} + 20a_{n-3} = 0[/itex]

    This is a homogenous recurrence and should be solvable.
     
  6. Jun 21, 2011 #5

    Mentallic

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    Oh thank you so much! I hadn't actually looked at the exponent of 2n in that way. This was very helpful :smile:
     
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