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Math Help - Recurrence Relation

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    Recurrence Relation

    Show that the sequence {a_{n}} is a solution of the recurrence relation a_{n}=-3a_{n-1}+4a_{n-1} if a_{n}=0

    So far, I plugged in a_{n}=0 to get a_{n}=-3(0)+4(0)=0 to show the sequence is a solution, but I think I'm missing a step. How can I figure out what a_{n-1} and a_{n-2} are?
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    Quote Originally Posted by jennifer1004 View Post
    Show that the sequence {a_{n}} is a solution of the recurrence relation a_{n}=-3a_{n-1}+4a_{n-1} if a_{n}=0

    So far, I plugged in a_{n}=0 to get a_{n}=-3(0)+4(0)=0 to show the sequence is a solution, but I think I'm missing a step. How can I figure out what a_{n-1} and a_{n-2} are?
    From what it looks like the two terms on the RHS can be added together to give

    a_n = a_{n-1}.

    Is your right hand side right or is it

    a_{n}=-3a_{n-1}+4a_{n-2}?
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    You're right. I'm sorry about that. it is 4a_{n-2}.

    The edited question is:

    Show that the sequence is a solution of the recurrence relation a_{n}=-3a_{n-1}+4a_{n-2}
    Last edited by mr fantastic; January 13th 2009 at 03:12 PM. Reason: Merged and edited from a duplicate thread containing the correction
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  4. #4
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    Quote Originally Posted by jennifer1004 View Post
    You're right. I'm sorry about that. it is 4a_{n-2}.
    Typically given

    a_{n}=-3a_{n-1}+4a_{n-2} for n \ge 2

    you are given a_0\; \text{and}\; a_1, then the recurrance relationship gives the rest of the a_n 's. Is your problem different?
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    In other problems I was given the first two cases, but not this one. The entire question is:

    Show that the sequence { a_n} is a solution of the recurrence relation a_{n}=-3a_{n-1}+4a_{n-2} if a_{n}=0
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    Quote Originally Posted by jennifer1004 View Post
    In other problems I was given the first two cases, but not this one. The entire question is:

    Show that the sequence { a_n} is a solution of the recurrence relation a_{n}=-3a_{n-1}+4a_{n-2} if a_{n}=0
    Well, you got me. The a_n = 0 is bothersome as it specifies nothing of the value of n. Of course the sequence work for a_n=0 for all n.

    For what it's worth. The recursion equation has the exact solution

    a_n = c_1 + c_2(-4)^n for arbitrary constants c_1,\;\;c_2

    Sorry I couldn't have been of more help.
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    Thanks

    Thank you for your help. I was stumped as well because there does not seem to be enough information to work with. I have more of these questions, but hopefully I can post one that will make sense. Thanks again for trying to make sense out of nothing!
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