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Second Order Homogeneous Recurrence Relation

  1. Oct 13, 2011 #1
    Hello!

    I was reading the proof (I think it constitutes a proof) for second order homogeneous recursive relations from the book Discrete Mathematics with Applications by S. Epp, and it seems, to me at least, to be excessive; which suggests that I don’t understand the proof.

    It goes something like the following (the proof is here: http://books.google.co.uk/books?id=...er homogeneous recurrence relations&f=false):

    Firstly, there is the recurrence relation [itex]a_{n} = Aa_{n-1} + Ba_{n-2}[/itex], then the book states that “uppose that for some number t… the sequence 1, t^{1}, t^{2}, t^{3},…,t^{n},….
    satisfies [[itex]a_{n} = Aa_{n-1} + Ba_{n-2}[/itex]].”

    This means that [itex]t^{n} = At^{n-1} + Bt^{n-2}[/itex].

    Using n = 2, you end up with the quadratic [itex]t^{2} – At – B = 0[/itex], from which you can derive the values for t (the roots of the quadratic).

    The book then states: “Now work backward. Suppose t is any number that satisfies [the quadratic]. Does the sequence 1, t^{1}, t^{2}, t^{3},…,t^{n},…. satisfy [[itex]a_{n} = Aa_{n-1} + Ba_{n-2}[/itex]]?”

    To answer this, Epp multiplies the quadratic by [itex]t^{n-2}[/itex].

    Here’s my issue: I don’t understand the point of the first part. Why not just start with the quadratic and multiply by [itex]t^{n-2}[/itex]? This shows that the sequence [itex]1, t^{1}, t^{2}, t^{3},…,t^{n},….[/itex] satisfies [itex]a_{n} = Aa_{n-1} + Ba_{n-2}[/itex] (since [itex]t^{n} = At^{n-1} + Bt^{n-2}[/itex] is derived from the quadratic by multiplying by [itex]t^{n-2}[/itex]) when t is equal to the roots of the quadratic. Why suppose that there is a sequence 1, t, t2,… that satisfies the recurrence relation then work towards the quadratic formula? I don’t see what it demonstrates? Why not just start with the quadratic and show such a relation exists, and work from there?

    Any help appreciated.
     
  2. jcsd
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