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Uniform convergence of a series of functions

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove:
    (1) the series
    [tex]\sum_{n=0}^\infty (-1)^n x^n (1-x)[/tex]
    converges absolutely and uniformly on the interval [0,1]

    (2) the series
    [tex]\sum_{n=0}^\infty x^n (1-x)[/tex]
    converges absolutely and uniformly on the interval [0,1]

    3. The attempt at a solution
    I have shown, by induction, that the limiting function of the second series is 1 - xn+1, which goes to 1. Thus the series of functions converges (absolutely, since all values are positive) but is 0 at x = 1, so thus not continuous. Therefore, the convergence of the second series is not uniform. However, this also shows that the first series converges absolutely.

    Where I am stuck is with uniform convergence of the first series. Using partial sums I was able to show that the series converges to (1-x)/(1+x), but how do I show this is uniform? I don't think it's enough to say that the limiting function is continuous in the given interval.

    Can anyone tell me if I'm on the right track or what I can use to prove uniform convergence? Thanks.
     
  2. jcsd
  3. Aug 5, 2008 #2
    Let [tex]\epsilon \in \mathbb{R}[/tex]. Show that for all [tex]\epsilon > 1[/tex], [tex]\sum_n \left ( {-1 \over \epsilon} \right ) (\epsilon x)^n (1-x)[/tex] satisfies Abel's Uniform Convergence Test.
     
  4. Aug 6, 2008 #3
    I'm afraid I don't quite understand your restatement of series. Is it missing a power of n on the first term?

    If that's the case, then I believe I need to show the following:
    1. [tex]{( \epsilon x)^n } [/tex] is uniformly bounded and eventually monotonic.
    2. [tex] \sum_{n=0}^\infty {\left(\frac{1}{\epsilon}\right)}^n (1-x) [/tex] converges uniformly

    Is this correct? Thanks.
     
  5. Aug 8, 2008 #4
    Let me try this:

    I claim:
    [tex]\sum_{n=0}^N (-1)^n x^n (1-x) = (1-x) \frac{1 - (-x)^{N+1}}{1+x}[/tex]
    [tex]\sum_{n=0}^{\infty} (-1)^n x^n (1-x) = \frac{1-x}{1+x}[/tex]

    If the first converges uniformly to the second, then as n goes to infinity:
    [tex]\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| \rightarrow 0[/tex]

    This is true since:
    [tex]\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| =
    \left| \frac{(1-x)}{1+x} (-1) (-x)^{N+1}\right| = (x)^{N+1} \left| \frac{(1-x)}{1+x}\right| \rightarrow 0 (1) = 0[/tex]

    Hence the series converges uniformly. Can anyone validate this or pick out a problem with it? Thanks.
     
  6. Aug 13, 2008 #5
    Not to bump this thread but I think I have a better solution now:
    [tex]\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| = \frac{1 - x}{1 + x} \left| -(-x)^{N+1} \right| = \frac{1 - x}{1 + x}(x)^{N+1} \leq (1-x)x^{N+1}[/tex]

    Keeping in mind that we are on the interval [0,1], we can take the derivative of the last function and, setting it equal to 0, find the the function has a maximum value at x=(N+1)/(N+2). Thus,

    [tex]\left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| \leq (1-x)x^{N+1} \leq (1- \frac{N+1}{N+2})(\frac{N+1}{N+2})^{N+1} \leq \frac{1}{N+2}[/tex]

    Then this is less than any arbitrary epsilon if
    [tex]N > \frac{1}{\epsilon} - 2[/tex]

    Since N does not depend on the choice of x, this convergence is uniform.

    I think this is the right idea but I'd appreciate any comments or validations. Thanks.
     
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