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Weierstrass M-Test and Absolutely Uniformly Convergence

  1. Mar 2, 2008 #1
    I'm posting this again because the other was plagued with errors in the first post. My fault and I apologize. I'll do a better job this time, I hope.

    1. The problem statement, all variables and given/known data
    Let f(n,x) = [tex]
    \sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}
    [/tex]
    a) Test for absolutely convergence on [0,1]
    b) Test for uniformly convergence on [0,1]
    c) Is [tex]
    \sum\limits_{n = 1}^{\inf } {|f(n,x)|}
    [/tex] uniformly convergent on [0,1]?



    2. Relevant equations
    N/A


    3. The attempt at a solution
    Alright, a) is simple actually, a simple application of the root/ratio test will do, but here comes the catch.

    I know that b) is true, since a power series is uniformly convergent inside its radius of convergence and the endpoint in this case is trivial. It just seems, however, that with a comparison with a geometric series, I can use the Weierstrass M-Test to prove all 3 at once.

    For any 'x' in [0,1), I can choose an a such that 0 <= x <= a <1 by the density of R. Therefore
    [tex]
    |\sum\limits_{n = 1}^{\inf } {f(n,x)} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}
    [/tex]
    and for x = 1 and any a > 0
    [tex]
    |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}[/tex]

    I have just chosen the function to be uniformally cauchy on [0,1) U {1} (have I?) by a basic comparison test with a series which is uniformly cauchy on (0,1) and by considering the end point separately.
    But it also seems that the Weierstrass M-Test gives me "Absolutely Uniformly Convergence" (that is, the absolute value of the function is also uniformly convergent) for free. Am I right? Is that always gonna hold true, or is this a property of power series?
     
  2. jcsd
  3. Mar 2, 2008 #2
    Alternatively, could I also just make a few arguments reasoning from (a).

    I know the series is absolutely convergent on [0,1]. So its absolute value will be uniformly convergent inside its radius of convergence. I proceed to test the endpoint x = 1, which turns it into a series of zeros. So I have just shown its absolute value to be uniformly convergent on [0,1) U {1}.
    If the absolute value is uniformly convergent, than the series itself must be uniformly convergent.

    Is this an acceptable way to proceed?
     
  4. Mar 2, 2008 #3
    Please, I don't like constantly bumping my threads, but I'm really struggling to understand this concept. =/
     
  5. Mar 2, 2008 #4
    Yeah, I'm pretty sure the Weierstrass M-test tests for uniform convergence by essentially testing for absolute uniform convergence...
     
  6. Mar 3, 2008 #5
    Thank you many times for replying.

    Okay, so is my proof mathematically sound?

    I also have a quick follow up though regarding the order/flow of the argument.

    In my argument, I essentially "fix" my 'x' and choose my 'a'. But it's also true that between 'a' and 1, I can find an 'x'. Now, I'm certain I can write a more rigorous argument showing that the limit of 'a' is 1, but I'm wondering if that's an overkill, since my whole purpose in bringing in teh Weierstrass M Test was to simplify things (as well as solve them, since it's the only method I know).
     
  7. Mar 3, 2008 #6
    Well, I believe that to use the Weierstrass M-test you have to find a bound that works for all x on whatever set you're testing convergence in, not just a given x in the set. That is, your bound a is allowed to be a function of n, but it cannot be a function of x.
     
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