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Convergence of series of the form (r^n)cos(nx)

  1. Aug 31, 2008 #1
    Show that

    [tex] \sum^{\infty}_{n=0} p^{n}cos\left(n\theta\right) = \frac{1-pcos\theta}{1-2pcos\theta+p^{2}} [/tex]

    if [tex] \left|p\right|<1[/tex].

    Looking at this series, I see that p will approach zero as n approaches infinity, while the series oscillates because of the cosine term. The convergence is easy to see, but not the function that it converges to. The series is reminiscent of a geometric series, or possibly a Fourier series, but the cosine term really complicates things for me. I tried looking for similar problems on the web, but only found examples that apply the tests of convergence rather than determining this function.
     
  2. jcsd
  3. Aug 31, 2008 #2

    tiny-tim

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    Hi buffordboy23! :smile:

    erm … if p is real, then [tex] p^{n}cos\left(n\theta\right) = \Re(p^ne^{in\theta}) [/tex] :wink:
     
  4. Aug 31, 2008 #3
    tiny-tim,

    I am still clueless. I have thought of this option but still cannot proceed. I have tried various expansions and equivalent expressions for the initial cosine term--such as cos nx = [exp(nx) + exp(-nx)]/2--and the equivalent form that you suggest. Unfortunately, my text is rather terse on the convergence of an infinite series, as it only discusses the Cauchy-criterion and Weierstrauss condition--methods to show convergence, but not the function that it is convergent to--and the text is also devoid of example problems. Is there a specific name of the concept to be used in this application?
     
  5. Aug 31, 2008 #4

    tiny-tim

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    Hi buffordboy23! :smile:

    Can you sum [tex]\sum_{n = 0}^\infty p^n[/tex] (for |p| < 1) ? :smile:
     
  6. Aug 31, 2008 #5
    Yes, this is just a geometric series, where the constant term is 1, so the sum of this series is

    [tex]\frac{1}{1-p}[/tex]

    EDIT: I never saw anything before with the form cos(nx), and could not find any examples anywhere either.
     
  7. Aug 31, 2008 #6

    tiny-tim

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    ok … now apply that to [tex] \sum_{n=0}^\infty p^ne^{in\theta} [/tex] :wink:
     
  8. Aug 31, 2008 #7
    I am still confused; I don't ever recall seeing the summation of a complex exponential. I am thinking that I can somehow split the terms so that I have two summation terms, which then will lead to the product of two separate terms.
     
  9. Aug 31, 2008 #8

    tiny-tim

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    [tex] \sum_{n=0}^\infty p^ne^{in\theta}\ =\ \sum_{n=0}^\infty \left(pe^{i\theta}\right)^n[/tex]
     
  10. Aug 31, 2008 #9
    Yes...I see it clearly now. The funny thing is that I had this form at one moment, but failed to recognize its obscure relation to the geometric series.

    Thanks tiny-tim for your time and assistance.
     
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