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Uniform convergence

  1. May 11, 2005 #1
    Hi,

    I don't know how to analyse uniform convergence/local uniform convergence for this series of functions:

    [tex]
    \sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}
    [/tex]

    Then
    [tex]
    f_{n}^{'} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}
    [/tex]
    [tex]
    f_{n}^{'} = 0 \Leftrightarrow x = \frac{1}{3^{n}\left(\frac{\pi}{2} + k\pi\right)}
    [/tex]

    I tried to prove the convergence using the theorem about change of sum and derivation - if I showed that
    [tex]
    \sum_{n=1}^{\infty} f_{n}^{'} \rightrightarrows^{loc} on (a,b)
    [/tex]

    I could tell that also
    [tex]
    \sum_{n=1}^{\infty} f_{n} \rightrightarrows^{loc} on (a,b)
    [/tex]

    Maybe I could use Dirichlet/Abel to show the convergence, but I'm not sure I have met the preconditions. Anyway, if I wanted to show directly convergence of the original series using the most straightforward criterion - Weierstrass, could I write it this way: ?

    Maximum for [itex]f_{n}[/itex] is in [itex]x = \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi\right)}[/itex]. Then
    [tex]
    S_{n} := \sup_{x \in (0,\infty)} \left|2^{n}\sin \frac{1}{3^{n}x}\right| = 2^{n}
    [/tex]

    But
    [tex]
    \sum_{n=1}^{\infty} S_{n} = \infty
    [/tex]
    so I haven't shown the convergence. Anyway, as we can see, x at which our function has the maximum is moving to zero as n grows. So, what if I chose another interval for x, let's say [itex]x \in (\epsilon, \infty), \epsilon > 0[/itex]? Then

    [tex]
    \exists n_0 \in \mathbb{N} \mbox{ such that } \forall n \ge n_0\ \ \ \ \ \ \
    \frac{1}{3^{n}\left(\frac{\pi}{2} + 2k\pi}\right)} < \epsilon
    [/tex]

    I feel this is maybe good idea, but don't know how to pull it into finish or how to use this idea...

    Thank you for your help.
     
    Last edited: May 11, 2005
  2. jcsd
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