1. Not finding help here? Sign up for a free 30min 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!

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]


    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]


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

    Here I'm stock.

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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor
    Homework Helper

    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 }
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Rate of Convergence
  1. Rates of convergence (Replies: 8)

  2. Convergence rate (Replies: 1)

  3. Rate of convergeance (Replies: 2)