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Math Help - Uniform Convergence

  1. #1
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    Uniform Convergence

    I'm having trouble with this question:
    Is the $\displaystyle\sum\limits_{n=0}^\infty \frac{n^2 x}{1+{n^4}{x^2}}$
    uniformly convergent over the interval (0,1].
    Now I know that it isn't uniformly convergent, and have started my proof with a contradiction.
    So assuming for contradiction that it is uniformly convergent, we can say that \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall n,m>N |f_n(x) -<br />
f_m(x)|<\epsilon but I don't know where to go from there. Any help?
    Last edited by worc3247; March 23rd 2011 at 12:44 PM.
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  2. #2
    Super Member girdav's Avatar
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    This series is pointwise convergent. We have, if x is positive that \left|\frac{n^2x}{1+n^4x}\right|\leq \left|\frac{n^2x}{n^4x}\right|=\frac 1{n^2} so for all n, \displaystyle\sup_{0<x\leq 1}\left|\frac{n^2x}{1+n^4x}\right|\leq\frac 1{n^2} and the series is normally convergent.
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    Sorry, I can't see how this answers the question?
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  4. #4
    Super Member girdav's Avatar
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    If a series is normally convergent on the set A, then it is uniformly convergent on A.
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  5. #5
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    I thought that was only for closed intervals? So are you saying this is uniformly convergent over (0,1]?
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by worc3247 View Post
    Sorry, I can't see how this answers the question?


    You can use Weierstass M-test.
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  7. #7
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    Please accept my apologies for this, I missed out a squared on the denominator of the series and have only just realised. Really sorry, have corrected the mistake now.
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  8. #8
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    As was mentioned the series is pointwise convergent to 0. Investigate what happens at the point x=\frac{1}{n^2}\in (0,1] for all n!
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Find max f_n(x), by taking the derivative, you'll get x=+/- 1/n^4, x in (0,1] so x=1/n^4

    Placing that x=1/n^4 back to f_n(x). You will get f_n(1/n^4)=1/n^2

    sum 1/n^2 is converges.

    {n^2x}/{1+n^4x^2} < 1/n^2 for all x in (0,1], now the rest comes from Weierstrass test.
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