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Thread: Recurrence Relation

  1. #1
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    Recurrence Relation

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

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

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

    $\displaystyle a_n = a_{n-1}$.

    Is your right hand side right or is it

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

    The edited question is:

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

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

    you are given $\displaystyle a_0\; \text{and}\; a_1$, then the recurrance relationship gives the rest of the $\displaystyle a_n 's$. 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 {$\displaystyle a_n$} is a solution of the recurrence relation $\displaystyle a_{n}=-3a_{n-1}+4a_{n-2}$ if $\displaystyle a_{n}=0$
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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 {$\displaystyle a_n$} is a solution of the recurrence relation $\displaystyle a_{n}=-3a_{n-1}+4a_{n-2}$ if $\displaystyle a_{n}=0$
    Well, you got me. The $\displaystyle a_n = 0$ is bothersome as it specifies nothing of the value of n. Of course the sequence work for $\displaystyle a_n=0$ for all n.

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

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

    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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