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Math Help - rate of convergence

  1. #1
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    rate of convergence

    i've been stuck on this proof for hours:

    suppose 0<p<q, a_n=a+O(n^(-p)) and b_n=b+O(n^(-q)) where O is the order of convergence as n approaches infinite. i need to prove that a_n*b_n=ab+O(n^(-p)), in another words that this function converges to ab with order of convergence n^(-p).

    by definition, a sequence a_n converges to a with rate of convergence O(b_n) if there exists k such that abs (a_n-a)<=k*abs(b_n).
    is it possible to prove (or even true) that abs (a_n*b_n-ab) <= abs (a_n-a)* abs (b_n-b)? this would allow us to find such a k.

    thanks in advance for any help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by squarerootof2 View Post
    i've been stuck on this proof for hours:

    suppose 0<p<q, a_n=a+O(n^(-p)) and b_n=b+O(n^(-q)) where O is the order of convergence as n approaches infinite. i need to prove that a_n*b_n=ab+O(n^(-p)), in another words that this function converges to ab with order of convergence n^(-p).

    by definition, a sequence a_n converges to a with rate of convergence O(b_n) if there exists k such that abs (a_n-a)<=k*abs(b_n).
    is it possible to prove (or even true) that abs (a_n*b_n-ab) <= abs (a_n-a)* abs (b_n-b)? this would allow us to find such a k.

    thanks in advance for any help!
    The point is that given the properties of big-O that:

    a_n b_n = ab +aO(n^{-q})+bO(n^{-p})+ O(n^{-q})O(n^{-p})

    and as 0<p<q we are left with only the O(n^{-p}) term, so:

    a_n b_n = ab +aO(n^{-p})=ab+O(n^{-p}).

    Now if you are expected to prove this from the definition of big-O notation that is a different problem.

    RonL
    Last edited by CaptainBlack; June 30th 2008 at 02:54 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The point is that given the properties of big-O that:

    a_n b_n = ab +aO(n^{-q})+bO(n^{-p})+ O(n^{-q})O(n^{-p})

    and as 0<p<q we are left with only the O(n^{-p}) term, so:

    a_n b_n = ab +aO(n^{-q})=ab+O(n^{-p}).

    Now if you are expected to prove this from the definition of big-O notation that is a different problem.

    RonL
    i'm not convinced of your second statement. why would 0<p<q condition imply that we have a_n*b_n=ab+aO(n^(-q))=ab+O(n^(-p))?
    Last edited by CaptainBlack; June 30th 2008 at 02:55 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by squarerootof2 View Post
    i'm not convinced of your second statement. why would 0<p<q condition imply that we have a_n*b_n=ab+aO(n^(-q))=ab+O(n^(-p))?

    There is a typo now corrected
    RonL
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