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Homework Help: Rate of Convergence

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the Rate of Convergence of [tex] \alpha = \frac{2*n^{2}+n+1}{n^{2}-3} [/tex]

    n=1,2,3,...,...

    2. Relevant equations

    [tex]lim n->\infty=\alpha _{n}[/tex]

    [tex] |\alpha-\alpha _{n} |\leq K*|\beta n|[/tex]

    3. The attempt at a solution

    I found the limit of alpha [tex]\alpha _{n}= 2[/tex]

    Then,

    [tex] |\frac{2*n^{2}+n+1}{n^{2}-3 -2}|=\frac{n+7}{|n^{2}-3|}[/tex]

    Here I'm stock.

    -Link
     
    Last edited: Oct 13, 2007
  2. jcsd
  3. Oct 13, 2007 #2

    morphism

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    What do you mean by rate of convergence?
     
  4. Oct 14, 2007 #3

    CompuChip

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    Heh, you misplaced a bracket, obviously you meant
    [tex] |\frac{2*n^{2}+n+1}{n^{2}-3} -2|=\frac{n+7}{|n^{2}-3|}[/tex]

    I also wonder what you mean by rate of convergence, but taking the "simple" definition on this Wikipedia page I think you want to start out by filling in
    [tex]\frac{a_{n+1} - 2}{a_n - 2}
    = \frac{ \frac{2n^2+n+1}{n^2-3} - 2 }{ \frac{2*n^{2}+n+1}{n^{2}-3} - 2 }
    [/tex]
    and work it out as you did above, then take the limit
    [tex]\lim_{n \to \infty} \frac{a_{n+1} - 2}{a_n - 2}[/tex].

    I don't know what definition you use though.
     
  5. Oct 14, 2007 #4
    Yes I misplaced a bracket, thanks compuchip.

    Rate of convergence definition.

    Suppose [tex] \left \{\beta _{n} \right\}}^{\infty}_{n=1}[/tex] is a sequence known to converge to zero, and [tex] \left\{\alpha _{n} \right\} ^{\infty}_{n=1}[/tex] converges to a number [tex]\alpha [/tex]. If a positive constant K ecists with

    [tex]| \alpha _{n} - \alpha| \leq K|\beta _{n}| [/tex], for large n,

    then we way that [tex] \left\{\alpha _{n} \right\} ^{\infty}_{n=1}[/tex] converges to [tex]\alpha[/tex] with rate of convergence [tex] O( \beta _{n})[/tex]. It is idndicated by writing [tex]\alpha _{n}=\alpha + O( \beta _{n}) [/tex].

    Obtained from "Numerical Analysis 8th ed", by Burden and Faires.
     
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