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Unif. conv. of fx=Σc_n*x^n on |x|<h implies unif. conv. on |x|<=h?

  1. Dec 1, 2007 #1
    [SOLVED] unif. conv. of fx=Σc_n*x^n on |x|<h implies unif. conv. on |x|<=h???

    my whole question is on the title..uniform convergence of power series..

    The answer is yes..how can I prove that? help me lol

    and I`d like to know..the existence of limit..

    [1 3 5 ''' (2n-1)]/[2 4 6 ''' (2n)] -> ???? as n goes infinity..

    (sorry for my bad Eng..)
     
  2. jcsd
  3. Dec 1, 2007 #2

    EnumaElish

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    Suppose there is no unif. conv. on the closure (|x| < h). What does that imply about the behavior of the function as |x| ---> h (versus at |x| = h)?
     
  4. Dec 1, 2007 #3
    I liked this problem because I hadn't thought about it before and my initial thought was to apply Abel's theorem, but it turned out not to be the thing to do here.


    Fact 1) Suppose f_n -> f uniformly on [a,b), and f_n(b) -> f(b).
    prove: f_n -> f uniformly on [a,b].


    Fact 2) If f_n is a sequence of continuous functions on [a,b] that converges
    uniformly to f(x) on [a,b), then lim f_n(b) exists.

    proof:

    Fix e > 0.
    By uniform convergence, choose N such that such that |f_n(x)-f_m(x)| < e for n,m > N, x in [a,b).
    By continuity, |f_n(b)-f_m(b)| = lim |f_n(x)-f_m(x)| <= e.
    Hence f_n(b) is Cauchy, thus converges.


    Applying 1),2) above, you get:

    If f_n is a sequence of continuous functions on [a,b] such that f_n -> f(x) uniformly on [a,b),
    then f_n -> f(x) uniformly on [a,b]. (Where in the hypothesis, it is not assumed lim f_n(b) exists.)

    (*note: f(b) is defined to be lim f_n(b), in the conclusion of the proof.)
     
    Last edited: Dec 2, 2007
  5. Dec 1, 2007 #4
    thx much..:)

    but after showing that lim f_n(b) exists(f_n Cauchy seq. at x=b)..

    isn`t it necessary to show that the limit tends to f(b)=Σc_n b^n??

    so |f_n(b)-f(b)| -> 0 as n-> infinity..???
     
  6. Dec 2, 2007 #5
    You have [tex]f_n(b) = \sum_{i=0}^n c_i b^i[/tex] is a Cauchy sequence. By definition, [tex]\sum_{i=0}^\infty c_i b^i[/tex] is the limit of this sequence. So no, there's nothing to show.

    In the above proof, I didn't assume lim f_n(b) existed. Once I showed it existed, of course we have to define f(b) = lim f_n(b). But in the case of power series, that is precisely what the definition of f(x) is.
     
    Last edited: Dec 2, 2007
  7. Dec 6, 2007 #6
    Got it~! thx very much for saving me!
     
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