1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Exam question help part 2

  1. Aug 9, 2008 #1
    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

    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. jcsd
  3. Aug 9, 2008 #2
    first of all, that symbol, [tex]\propto[/tex], means proportional to.

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

    [tex]\left \{ \begin{matrix}
    x_{n+1} = \ldots \\
    y_{n+1} = \ldots
    \end{matrix} \right.
  4. Aug 9, 2008 #3
    As for when n is large, take the limit as this means that as [tex]n \rightarrow \infty[/tex]

    of the explicit formula for [tex]r_{n}[/tex]

    [tex]r_{n} = c_{1} \lambda_{1}^n v_{1}+ c_{2} \lambda_{2} ^n v_{2}[/tex]

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

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

    Or you can do it this way:
    [tex]A^{n}= UD^{n}U^{-1}[/tex].

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


    [tex]r_{\infty} = A^{n}r_{0} = UD^{n}U^{-1}r_{0} = UD^{n}U^{-1} \begin{bmatrix} x_{0} \\ y_{0} \end{bmatrix} [/tex]
    Last edited: Aug 9, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook