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Math Help - uniform convergence on (1,a)

  1. #1
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    uniform convergence on (1,a)

    The function f_n=\frac{x^n}{x^n+1} converges uniformly to f(x)=1 on the interval (1,a).

    Proof.

    taking x=a

    as |\frac{a^n}{a^n+1}-1|<\epsilon\rightarrow \frac{1}{1+a^n}<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n

    So can I take N={ln(\frac{1-\epsilon}{\epsilon})} and then I have that:

    \forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon

    is this correct or is it wildly off- uniform convergence is kind of killing me

    thanks for any help
    Last edited by hmmmm; October 15th 2011 at 12:02 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: uniform convergence on (1,a)

    Quote Originally Posted by hmmmm View Post
    The function f_n=\frac{x^n}{x^n+1} converges uniformly to f(x)=1 on the interval (1,a).

    Proof.

    taking x=a

    as |\frac{a^n}{a^n+1}-1|<\epsilon\rightarrow \frac{1}{1+a^n}<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n

    So can I take N={ln(\frac{1-\epsilon}{\epsilon})} and then I have that:

    \forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon

    is this correct or is it wildly off- uniform convergence is kind of killing me

    thanks for any help
    I don't quite understand what you're doing here, why are you choosing this a? Are you secretly plugging this in because it's the max, and you just omitted this? Give us more words, explaining your choices etc. and then we can help you decide of what you did is correct. A quick check: you should be proving that for every \varepsilon> there exists N\in\mathbb{N} such that \|f_n(x)-f(x)\|_\infty<\varepsilon where \|\cdot\|_\infty is the infinity norm. This 'looks' like what you're doing, but please fill in more holes.
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  3. #3
    Senior Member
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    Re: uniform convergence on (1,a)

    The function f_n=\frac{x^n}{x^n+1} converges uniformly to f(x)=1 on the interval (1,a).

    Proof.

    taking x=a as this will be the max of the f_n(x) so if the max is less then epsilon then f_n(x) will be for all other values of x.

    as |\frac{a^n}{a^n+1}-1|<|\frac{a^n}{a^n+1}-1|<\epsilon

    and as \frac{a^n}{a^n+1}-1=\frac{-1}{1+a^n}

    and so:

    |\frac{1}{1+a^n}|<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n

    So can I take N={ln(\frac{1-\epsilon}{\epsilon})} and then I have that:

    \forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon

    I think this is a bit better? Sorry about it being a bit of a mess I am a little confused by this. Thanks very much for the help

    thanks for any help
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