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  1. #1
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    uniformly continuous

    How do i prove that x*sin(x) is not uniformly continuous?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by math8553 View Post
    How do i prove that x*sin(x) is not uniformly continuous?
    x \sin(x)-(x+\delta) \sin(x+\delta)=x \delta \cos(x) + xO(\delta^2)

    and so for \delta small enough and x \ne n \pi/2, \ n\in \mathbb{Z} there is a k>0 independent of x such that:

    x \sin(x)-(x+\delta) \sin(x+\delta)<kx \delta \cos(x)

    So however small we make |\delta|>0 there is an x which makes |x \sin(x)-(x+\delta) \sin(x+\delta)| as large as we like.

    Note: The O(\delta^2) term can be made independent of x as it stands in place of a remainder term of a Taylor series which is the product of a constant \kappa, \delta^2 and a circular trig function at a point in [x,x+\delta] all of which which is bounded by |\kappa \delta^2|. If the reader needs the detail they can develop it them selves.

    CB
    Last edited by CaptainBlack; March 15th 2010 at 03:01 AM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by math8553 View Post
    How do i prove that x*sin(x) is not uniformly continuous?
    I think CB assumed you meant uniformly continuous on \mathbb{R}. You need to specify these things. This function is actually uniformly continuous on any bounded subspace of \mathbb{R}
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Drexel28 View Post
    I think CB assumed you meant uniformly continuous on \mathbb{R}. You need to specify these things. This function is actually uniformly continuous on any bounded subspace of \mathbb{R}
    That is what i assumed and for the reason you give.

    (I had expected some complaints rather than thanks for my demonstration as it is not a conventional way of doing these things, with the real detail left to the reader to confirm).

    CB
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    T
    with the real detail left to the reader to confirm).
    Yes, but that is the way it should be.

    Plus. New and strange methods are my favorite!
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