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Math Help - determine the order of convergence of steffensen's method

  1. #1
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    determine the order of convergence of steffensen's method

    Hey guys. This numerical analysis section is really throwing me. I don't quite know how to handle this order of convergence stuff. Here's just one example of a problem that's giving me trouble:

    Let F(x)=x-\frac{f(x)}{g(x)}, where g(x)=\frac{f(x+f(x))-f(x)}{f(x)} and f is smooth.

    Suppose a sequence \{x_n\} satisfies x_{n+1}=F(x_n) and converges to r, where r is a simple zero of f. We define "order of convergence" to be the largest real number q such that

    \lim_{n\to\infty}\frac{|x_{n+1}-r|}{|x_n-r|^q}

    exists and is nonzero. Determine the order of convergence.
    Any help would be much appreciated !
    Last edited by hatsoff; September 27th 2011 at 08:59 AM.
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  2. #2
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    Re: determine the order of convergence of steffensen's method

    Quote Originally Posted by hatsoff View Post
    Hey guys. This numerical analysis section is really throwing me. I don't quite know how to handle this order of convergence stuff. Here's just one example of a problem that's giving me trouble:



    Any help would be much appreciated !
    I'm going to assume f\in C^1(\mathbb{R}) (I think a locally bounded derivative would be enough).

    Obviously your limit is 0 when q<0. Since g(r)=f'(r)\neq 0 (I'm assuming this is your definition of simple zero) F'(r) exists, F(r)=r so if we put q>1 we get

    \frac{|F(x_n)-F(r)|}{|x_n-r|^q}=\frac{|F(x_n)-F(r)|}{|x_n-r|}|x_n-r|^{1-q}

    and the first term goes to |F'(r)| while the second blows up so the limit doesn't exist. If q=1 in the above, calculting F'(r) we arrive at the conclusion that the limit in this case is |F'(r)|=2.

    Now if 0<q<1 we get, using the triangle inequality first, the fact that f is locally Lipschitz continous and that f(r)=0

    \frac{|F(x_n)-r|}{|x_n-r|^{q}} \leq |x_n-r|^{1-q} + \frac{1}{|g(x_n)|} \frac{ |f(x_n)|}{|x_n-r|^q} \leq |x_n-r|^{1-q}\left(1+\frac{1}{|g(x_n)|}\right)\rightarrow0

    So in fact q=1 is the only value for which the limit isn't zero or infinite.
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  3. #3
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    Re: determine the order of convergence of steffensen's method

    Thanks, but I get F'(r)=0, not |F'(r)|=2.

    Oh, and I forgot to say f is smooth.

    I guess I could maybe try computing F''(r) and using Taylor's series, but to compute F''(r) would take perhaps an hour or more---assuming I didn't make any elementary mistakes. Perhaps there is a more efficient method of solution ... ?
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  4. #4
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    Re: determine the order of convergence of steffensen's method

    Yes, I misread a sign, but the argument still works: The limit is then 0 for q\leq 1 and infinity for q>1. Have you asked a professor for clarification? Maybe there's something we're missing.
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