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Homework Help: Pointwise & Unifrom Convergence

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all intervals on which the sequence of functions
    fn(x) = x2n / [n + x2n], n≥1, converges uniformly.

    2. Relevant equations
    3. The attempt at a solution
    I think we'll need to compute the pointwise limit first.
    For x=0 or x=1, the pointwise limit is 0.
    But what is its pointwise limit for x>0?

    Any help is appreciated!
     
    Last edited: Apr 13, 2010
  2. jcsd
  3. Apr 13, 2010 #2

    Gib Z

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    Thing thing to do would be to consider the cases (0,1), x=1 and x>1 separately.
     
  4. Apr 13, 2010 #3
    OK, I think for all of the cases x=0,x=1,0<x<1, the pointwise limit would be 0.

    But how can we compute the pointwise limit for x>1??
     
  5. Apr 13, 2010 #4

    Gib Z

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    Write [tex]\frac{x^{2n}}{x^{2n}+n} =\frac{x^{2n}+n - n}{x^{2n}+n} = 1 - \frac{n}{n+ x^{2n}}[/tex]

    On the right hand side, the second term is certainly always less than [tex]\frac{n}{x^{2n}}[/tex]. You have a linear term divided by an exponential term, so what is the limit of that term?
     
  6. Apr 13, 2010 #5
    n/x2n ->0 for x>1

    x2n / [n + x2n] ≥ 1 - n/x2n

    1 - n/x2n->1, but still how can we compute the limit of x2n / [n + x2n]?
     
    Last edited: Apr 13, 2010
  7. Apr 13, 2010 #6

    Gib Z

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    I'm not sure you know where all the inequalities are pointing at the moment. Can you see how

    [tex]0 < \frac{n}{n+x^{2n} } < \frac{n}{x^{2n}}[/tex]

    You know what the term on the far right approaches, so what must the middle term approach? And so then what is the limit of the function if |x|> 1 ?
     
    Last edited: Apr 13, 2010
  8. Apr 13, 2010 #7
    n/(n+x2n < n/x2n
    => -n/(n+x2n > -n/x2n
    => 1 -n/(n+x2n > 1 -n/x2n

    Thus, x2n / [n + x2n] > 1 - n/x2n

    1 - n/x2n -> 1

    But still this doesn't say what x2n / [n + x2n] would converge to...
     
  9. Apr 13, 2010 #8

    Gib Z

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    Okay, just read post 6 again and tell me what the term on the far right converges to. And so, what do you conclude that the middle term goes to?
     
  10. Apr 13, 2010 #9
    That middle term goes to 0, so fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise. Now how can we determine uniform convergence. I know that for any interval [a,b] containing 1 or -1 since it's discontinuous there, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1?
     
    Last edited: Apr 13, 2010
  11. Apr 13, 2010 #10

    Gib Z

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    Come on, try something! and Show me what you have done, and where you get stuck.

    Of course you have to apply a definition of uniform convergence and see if the function satisfies it on some intervals.
     
  12. Apr 13, 2010 #11
    fn(x) = x2n / [n + x2n]

    For the pointwise limit, I got fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise.

    So I know that for any interval [a,b] containing 1 or -1, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1??
    This is what I got, but I'm stuck...
    fn(x) <= b2n / [n + a2n] < b2n / [n + 1]
    How can we prove that the RHS ->0 ?

    Please help!
    Thanks!
     
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