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Math Help - A perhaps silly question about > 0

  1. #1
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    A perhaps silly question about > 0

    Now and then I see domains defined as in this example:
    1)  x \geq \epsilon > 0
    for any epsilon greater than zero, no matter how small.

    I would be tempted to believe that the above cannot be much different from
    2)  x > 0
    as in my understanding that already excludes the case x=0.
    Yet, there must be some subtle reason why some author prefers to explicitly write it as in 1)

    Can it be just a matter of conventions ? or am I missing some fundamental definition ?
    Thanks
    Luca
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Good question. Obviously, this will depend on the actual issue, but I can give an example of why that might happen.

    Very often, a function will converge uniformly everywhere except in a neighbourhood of some point, say 0, where perhaps the function is undefined and around which it does not behave well. Then we cannot say it converges uniformly for x>0, but we can say that it converges uniformly for x\geq\epsilon>0.
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  3. #3
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    The geometric series might be a good example

    Thanks Bruno,
    I believe I have finally got it.
    And perhaps, the geometric series might be a good example.

    Let us consider the geometric series

    \sum_{n=1}^\infty z^n

    z being a complex number. It is known that said series is convergent within |z| < 1.
    However, for claiming in addition that the convergence is uniform it is probably more correct to define a disk of uniform convergence as

    |z| \leq 1-\epsilon < 1

    for any \epsilon > 0, no matter how small.

    Exactly as you suggested, in such case |z| < 1 would not be sufficient for uniform convergence, as it would result in the need to add more and more terms to get sufficient accuracy as we choose |z| closer and closer to 1. Whereas chosing a fixed value for \epsilon > 0 would allow us to find an index N_{\delta}, function only of a \delta>0, at which we could stop adding the terms z^n so that

    \left|\sum_{n=1}^{N_{\delta}} z^n -\frac{1}{1-z}\right|<\delta

    is valid for any z in the domain |z| \leq 1-\epsilon < 1

    Thanks for the tip.

    Luca
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