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Math Help - Convergence of Series of Functions

  1. #1
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    Convergence of Series of Functions

    Hello!

    I am working on what seems to be a very "concrete" exercise in showing convergence.

    on the interval [0,1], show that the functions converge to the function f(x) point-wise, and determine if they converge uniformly.

    a) \{f_n(x)\}_{n=1}^{\infty} = \{\frac{2x}{1+nx}\} where f(x) = 0

    b) \{f_n(x)\}_{n=1}^{\infty} = \{\frac{cos(nx)}{\sqrt{n}}\} where f(x) = 0

    c) \{f_n(x)\}_{n=1}^{\infty} = \{\frac{n^3x}{1+n^4x}\} where f(x) = 0

    What I've got so far...

    So, to show a function is convergent point-wise, we just try to show that the NUMBER sequence \{f_n(x)\} converges to f(x) = 0 so, \forall \epsilon > 0, \exists N such that n > N implies f_n(x) - 0 < \epsilon

    Here is where I get confused. Do we define an epsilon in terms of n?! Which would depend on the series of functions in all 3 cases?
    And for uniform convergence, we are seeking a similar result but with an epsilon which is independent of n?

    Any help/guidance to set up the question further appreciated!!
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  2. #2
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    I assume this are sequences of functions as opposed to series. In that case, regarding your questions, for pointwise convergence we just need an N dependent on x and n. For uniform convergence we need an N dependent only on n (The epsilon is arbitrary so it makes no sense that it would depend on something). For example in the second one we have the bound 0\leq \frac{|\cos(nx)|}{\sqrt{n}} \leq \frac{1}{\sqrt{n}} from which one sees that one can make the sequence of functions arbitrarily arbtrarily close to 0 independently of any point.
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  3. #3
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    Yes... of course! Sorry, sequences of functions, not series. Thank you for pointing that out.

    Also, of course, I was trying to find epsilon as a function of N but that's ridiculous, I'm not sure how my mind slipped on such basic epsilon-delta arguments. I'll blame it on an algebra OD.

    Thanks!
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