# Homework Help: Exam question help part 2

1. Aug 9, 2008

### franky2727

i posted this on the end of another question i asked but i dont think that anyone is looking at is as it has like 10 replies and i assume most people are assuming that the problem has been solved... which it has

i had a question yesterday entitled exam question help which 2 people were kind enough to work through with me to a point were i now understand how to do this question, as i stated on there yesterday i have another question that is similar but this time it is a bit more complex, could someone please take a look at this one and if possible try to present it in the same way that hallsofivey did yesterday as this was the way that i understood the question.

part (a) which i can do is find the solution to the two dimensional map

Xn+1=Xn -Yn
Yn+1=2Xn +4Yn

with X0=1 and Y0=1

giving me eivgenvalues of 2 and 3 with vectors of (1,-1)T and (1,-2)T respectivly

After which i do the question just the same as the one that was answered for me above and i end up with hopefully the correct answer of Xn=-2 . 3n+3 . 2n

and Yn=4.3n-3.2n

hopefully that bits right. then i have part b of the question which i dont know how to do which is

indicate how the second order map
Xn+1=4xn-3xn-1

can be expressed as a two dimensional map. by concidering the eigenvalues of the associated matrix show that Xn then theres a symbol that i have never seen before, it looks like a 8 on its side but part of the loop is missing on the right hand side then 3n for n large... thanks

2. Aug 9, 2008

### foxjwill

first of all, that symbol, $$\propto$$, means proportional to.

Ok, for your question, try letting $$y_n=x_{n-1}$$, then form a system of linear recurrence relations with

$$\left \{ \begin{matrix} x_{n+1} = \ldots \\ y_{n+1} = \ldots \end{matrix} \right.$$

3. Aug 9, 2008

### konthelion

As for when n is large, take the limit as this means that as $$n \rightarrow \infty$$

of the explicit formula for $$r_{n}$$

$$r_{n} = c_{1} \lambda_{1}^n v_{1}+ c_{2} \lambda_{2} ^n v_{2}$$

where $$\lambda_{1},\lambda_{2}$$ are your eigenvalues, $$v_{1},v_{2}$$ are the corresponding eigenvectors.
Note that: $$r_{n+1}=Ax_{n}$$ with $$r_{0}= \begin{bmatrix} x_{0} \\ y_{0} \end{bmatrix}$$

Basically, after you've found $$r_{n}$$, then simply let $$n = \infty$$, this should give you,I'm not mistaken,the steady-state vector. Then you can simply find $$x_{n}$$ as $$n \rightarrow \infty$$ from this steady-state vector.
========================================

Or you can do it this way:
$$A^{n}= UD^{n}U^{-1}$$.

Notice that $$r_{n+1} = Ar_{n}$$ (I believe I gave a proof of this yesterday)

then

$$r_{\infty} = A^{n}r_{0} = UD^{n}U^{-1}r_{0} = UD^{n}U^{-1} \begin{bmatrix} x_{0} \\ y_{0} \end{bmatrix}$$

Last edited: Aug 9, 2008