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...

Please help..(Worried)

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- Jan 11th 2011, 11:03 AMsilverblue22Solving 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...

Please help..(Worried) - Jan 11th 2011, 11:05 AMsnowtea
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. - Jan 11th 2011, 11:18 AMsilverblue22
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... - Jan 11th 2011, 12:28 PMsnowtea
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? - Jan 11th 2011, 12:45 PMsilverblue22
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? - Jan 11th 2011, 02:05 PMchisigma