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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
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