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Math Help - Uniformly continuous #2

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    MHF Contributor Also sprach Zarathustra's Avatar
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    Uniformly continuous #2

    Prove that f(x)=sin(x)/x + cos(sqrt(x)) + 10x is uniformly continuous in her domain.
    Last edited by mr fantastic; January 3rd 2010 at 10:00 PM. Reason: Re-titled
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    It's easy to prove that for differentiable functions it is equivalent:

    1) f has a bounded derivative
    2) f is Lipschitz

    Since Lipschitz \Rightarrow unif. cont. we only have to prove f has bounded derivative at infinity which follows almost immediately so f is unif. cont. in any interval of the form [a, \infty ) where a>0, but (I believe, I didn't really check it) that it is not unif. cont. in (0, \infty ) since the sequence 1/n seems (!) to map to an unbounded sequence.
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    Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

    Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.
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    Quote Originally Posted by Enrique2 View Post
    Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

    Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.
    Well, continous in [0,1] since otherwise the square root is not defined.

    Also, yeah I started confusing f and f', so yeah f is unif cont in [0,a] too.
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    Quote Originally Posted by Enrique2 View Post
    Deining f(0)=2 (because of the limit of sinx/x when x goes to 0), you have that f is continuous in [-1,1] and then uniformly continuous (no problem of course removing 0 from the domain of definition as it seems that you are asked).

    Outside this interval you can check that it is Lipschitz as Jose27 suggests, using mean value.
    Note that this function is not defined on [-1,0) because \cos\left(\sqrt{x}\right) is not defined.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    And without derivatives?
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    I'm sorry but being analysis questions I think we're allowed to invoke some of stuff from Calculus I, that is what Jose27 just said: bounded derivative implies Lipschitz.

    Why making these things harder than they look like.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!
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    Yes, clearly as Plato and Jose27 have noticed my former argument is not correct, we only have defined the function on positive numbers. The argument works in [0,1] instead of [-1,1] .

    The argument that I believe krizalid is suggesting is not so trivial. The bounded derivative argument only works easily outside an interval [0,a] with a>0. In [0,a] you can use that sin(\sqrt{x})=O( \sqrt{x}) when x approaches 0, but it is not for me easier than using Heine Cantor in a bounded interval and observig that the derivative is bounded outside, that is the correct point to solve a lot of these kind of problems. It was not my idea at all being "too much sophisticated", simply I didn't see the argument in [0,a], that is nice!.
    Last edited by Enrique2; January 3rd 2010 at 03:21 PM.
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    Quote Originally Posted by Also sprach Zarathustra View Post
    The tools for this questions not include derivatives. Lipschitz yes! Derivatives no!

    Sorry! Without the derivatives, proving Lipschitz is harder. Then I believe you will be allowed to use Heine Cantor in [0,a], and making calculations outside, but I don't see at the moment an immediate way for suggesting you how to do these calculations in a simple way, sorry again!
    Last edited by Enrique2; January 3rd 2010 at 02:57 PM. Reason: Heine Cantor and not Borel!!!
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