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Convergence test

  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    ##P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ... ##
    Determine if the series is convergent or divergent if ## |z| = 2 ##, where, ## z## is a complex number.

    2. Relevant equations
    ##1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}##

    3. The attempt at a solution
    Let, ##z = 2 exp (i \theta)##

    For the first ##N## terms, the summation is,
    ##P_N (z) = \frac{1-(-1)^N exp(iN\theta)}{1+exp(i \theta)}=\frac{1-(-1)^N \cos (N\theta) - i (-1)^N \sin (N \theta)}{1 + exp (i \theta)}##
    As ## N \rightarrow \infty##, ##\cos (N \theta)## and ## \sin (N \theta)## do not converge to a particular value.
    So, I conclude that the series is not convergent for ## |z|=2##
    But the answer says that the series converges to ##\frac{1}{1+\frac{z}{2}}##
     
    Last edited: Jan 28, 2016
  2. jcsd
  3. Jan 28, 2016 #2

    Samy_A

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    The general term of the series is ##a_n=\frac{(-1)^nz^n}{2^n}##.
    Then, for ##|z|=2##, ##|a_n|=|\frac{z^n}{2^n}|=\frac{|z|^n}{2^n}=1##. How can this series converge if the sequence ##a_n## doesn't converge to 0?

    I think you are correct.
     
    Last edited: Jan 28, 2016
  4. Jan 28, 2016 #3
    Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.
     
  5. Jan 28, 2016 #4

    Samy_A

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    I see. They write that it converges for all ##z## with ##|z|=2## except ##z=-2##.
    Let's take ##z=2## then. The series becomes ##1-1+1-1 ...##, so the partial sums alternate between ##0## and ##1##.
    A few lines before this they give an example with the same oscillating series and conclude (correctly in my book) that it doesn't converge:
    limit.jpg

    Call me baffled.
     
  6. Jan 28, 2016 #5

    PeroK

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    ##\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)##
     
  7. Jan 28, 2016 #6

    Samy_A

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    But the book claims that this series also converges for all ##z## with ##|z|=2## except ##z=-2##. This is clearly a highly regarded book, but still, I can't see how the series converges when ##|z|=2##.
     
  8. Jan 28, 2016 #7

    PeroK

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    It doesn't. I thought you'd proved that.
     
  9. Jan 28, 2016 #8

    Samy_A

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    I thought so too, but given the reputation of the book I started to fear I was missing something. Thanks for the second opinion.

    It's not you lacking some important concept here. It surely is a reliable book, but on this point you were right and they made a mistake. Happens to the best.
     
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