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Complex series question

  1. Nov 1, 2007 #1
    This is a question from a homework assignment that I turned in today and it's driving me nuts because I don't know if I did it correctly or not. First of all we have these two questions that seem almost identical:

    Question 1
    Show that the following series converges uniformly in the given region.
    [tex]\displaystyle\sum_{n=1}^{\infty}z^{n}[/tex] for [tex]0\le|z|<R[/tex] and [tex]R<1[/tex]


    Question 2
    Show that the sequence:
    [tex]\displaystyle\left\{x^{n}\right\}^{\infty}_{n=1}[/tex]
    converges uniformly inside [tex]0\le|z|<R[/tex] with [tex]R<1[/tex].​

    How I did question 1:

    [tex]\displaystyle\sum_{n=1}^{\infty}z^{n}= z + z^{2} + z^{3} + \cdots + z^{n} + \cdots[/tex]

    Call the nth partial sum of the series [tex]S_{n}[/tex]:

    [tex]S_{n}= z + z^{2} + z^{3} + \cdots + z^{n}[/tex]

    [tex]zS_{n}= z^{2} + z^{3} + \cdots + z^{n} + z^{n+1}[/tex]

    [tex]S_{n}=\frac{z(1 - z^{n})}{1-z}[/tex]

    Now,

    [tex]\displaystyle\sum_{n=1}^{\infty}z^{n}= \displaystyle\lim_{n\to\infty}S_{n}=\displaystyle\lim_{n\to\infty}\frac{z(1 - z^{n})}{1-z}[/tex]

    [tex]=\frac{z}{1-z}[/tex] because [tex]z^{n}\longrightarrow 0[/tex] for [tex]|z|<1[/tex]

    Now we consider:

    [tex]\left|\frac{z}{1-z} -\frac{z(1 - z^{n})}{1-z}\right| = \frac{|z|^{n}}{|1-z|}< \frac{R^{n}}{1-R}[/tex]

    with [tex]\epsilon>0[/tex] we choose N so that N depends on R and [tex]\epsilon[/tex] but not on on z, so that when [tex]n>N_{R,\epsilon}[/tex]

    [tex]\frac{R^{n}}{1-R}< \epsilon[/tex]

    Because we can choose such an N the series converges uniformly.

    For question 2 I was confused, If
    [tex]\displaystyle\sum_{n=1}^{\infty}z^{n}[/tex]
    converges then the sequence [tex]\displaystyle\left\{x^{n}\right\}^{\infty}_{n=1}[/tex]

    must be cauchy, right? So as [tex]n\longrightarrow \infty[/tex] the nth term of the series goes to zero. So I thought I'd need to show that there was a N, not dependent on z such that the nth term would be less than [tex]\epsilon[/tex].

    I found:

    [tex]N = \frac{log \epsilon}{log R} + 1[/tex]

    Is that good enough to show uniform convergence?
     
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 1, 2007 #2

    Dick

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    Science Advisor
    Homework Helper

    I think you are right. But your language is somewhat confusing. "Uniform convergence" is a term to apply to a sequence of functions. In the first case, the sequence of functions is the partial sums of your series, and yes, they converge uniformly to z/(1-z) over the domain |z|<=R<1. In the second case the sequence of functions is f_n=z^n. And they converge uniformly to zero. And, yeah, the second sequence of function are differences of the first sequence of functions, so if the first sequence of functions converges, the second sequence converges to zero. I'm a little confused about what the question is.
     
  4. Nov 1, 2007 #3
    I just didn't know if ai was waaaaaay off base or not about what the 2nd quetions was asking-- and I wanted to know if I did the first one right.

    And I like typing things in latex... that's all! :smile:

    Thanks for responding.
     
  5. Nov 1, 2007 #4

    Dick

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    Science Advisor
    Homework Helper

    You could install a latex compiler and typeset mathematics equations to your hearts content in the privacy of your own home. Anyway, it looks pretty much ok to me.
     
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