Homework Help: Second Order Homogeneous Recurrence Relation

1. Oct 13, 2011

nobahar

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

This means that $t^{n} = At^{n-1} + Bt^{n-2}$.

Using n = 2, you end up with the quadratic $t^{2} – At – B = 0$, 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 [$a_{n} = Aa_{n-1} + Ba_{n-2}$]?”

To answer this, Epp multiplies the quadratic by $t^{n-2}$.

Here’s my issue: I don’t understand the point of the first part. Why not just start with the quadratic and multiply by $t^{n-2}$? This shows that the sequence $1, t^{1}, t^{2}, t^{3},…,t^{n},….$ satisfies $a_{n} = Aa_{n-1} + Ba_{n-2}$ (since $t^{n} = At^{n-1} + Bt^{n-2}$ is derived from the quadratic by multiplying by $t^{n-2}$) 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.