# Thread: Solving a recurrence relation.

1. ## Solving a recurrence relation.

Xn+2=6Xn+1-9Xn

Where the first 5 terms of the sequence are: 0,1,6,27,108

I have an example of what the lecturer did...and tried to do th same for this homework question but got it wrong...

2. First of all, you seemed to have typed the problem wrong. All of the terms have Xn? Or maybe you forgot parenthesis?
Why don't you show us what you did, and we can point out where you went wrong.

3. No they all have Xn in them..

Ok...em I don't know how to do all the symbols and stuff on this so hopefully it'll be ok to read..
I'm gona use % instead of lambda

I let Xn+2-6Xn+1+9Xn=0
Let Un=%n
(%)^2-6(%)+9=0
Roots at %+/- =3

So a(%)^n-b(%)+c=0

Then, just following his example I said general solution is of the form
Un=%+n, Un=z%+n (those +n's are subscript, and z is alpha in the notes)

From that i said Un=z%+n + j%-n (where z,j are alpha and beta)
And that's all not right...

4. First of all, look at the Latex tutorial in the Latex subforum to learn Latex. You can also double click on any formula to see how it is written in Latex.

$\displaystyle x_{n+2} - 6x_{n+1} + 9x_n = 0$

Characteristic equation is:
$\displaystyle \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0$

You actually have a double root $\displaystyle \lambda=3$.

How does the solution look like when you have a double root?

5. Oh, cool, I'll have a look... Em we didn't have to do a graph or anything.

I could pt up the example he did in the lecture and see if that's easier? Might be easier to explain with the whole thing there?

6. Originally Posted by silverblue22
Xn+2=6Xn+1-9Xn

... where the first 5 terms of the sequence are: 0,1,6,27,108...
The problem is correctly extablished giving the first two terms of sequence $\displaystyle x_{0}$ and $\displaystyle x_{1}$... more terms are of course redundant...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$