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..
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...
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?