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Uniform convergens and continuity on R

  1. May 10, 2006 #1
    Hello people,

    I'm tasked with showing the following:

    given the series [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}[/tex]

    (1) show that it converges Uniformly [tex]f_n(x) :\mathbb{R} \rightarrow \mathbb{R}[/tex].

    (2) Next show the function

    [tex]f(x) = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2} [/tex]

    is continious on [tex]\mathbb{R}[/tex]

    (1) Suppose [tex]f_n = \frac{1}{x^2 + n^2}[/tex],

    Then [tex]f_n[/tex] is uniformly convergens if

    [tex]sup _{x \in \mathbb{R}} |f_n(x) - x|[/tex]. Now

    [tex]_{sup} _{x \in \mathbb{R}} |f_n(x) - x| = |\frac{1}{x^2 - n^2} - x| =
    _{sup} _{x \in \mathbb{R}} \frac{1}{x^2 - n^2}[/tex]

    The deriative of f_n(x) is non-negative on [tex]\mathbb{R}[/tex], so its increasing and is hence maximumized at [tex]x = \mathbb{R}[/tex]. So the supremum is [tex]1/n^2[/tex]. This does tend to zero as [tex]n \rightarrow \infty[/tex]. So therefore it converge Uniformly.

    Am I on the right track here? If yes any hints on how to prove the continuety ?

    I know that its something to do with:

    \integral_{1} ^{\infty} 1/x^2 + n^2 dx = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}

    Sincerely Hummingbird25
    Last edited: May 10, 2006
  2. jcsd
  3. May 10, 2006 #2

    matt grime

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    Science Advisor
    Homework Helper

    you've not defined f_n

    You start an 'if' statement and then do not complete it with a condition about anything.
  4. May 10, 2006 #3
    Hi :)

    f_n: \mathbb{R} \rightarrow \mathb{R}

    f_n(x) should have said f_n

    There is one sup to much sorry.


    Last edited: May 10, 2006
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