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Using complex exponentials to prove 1+acostheta

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Use complex exponentials to prove 1 + acos(theta) + a^2cos(2theta) + a^3cos(3theta)... = (1 - acos(theta))/(1 - 2acos(theta) + a^2)


    2. Relevant equations

    euler's e^itheta/2 +e^-itheta/2=2cos(2theta)

    3. The attempt at a solution

    a^(n)cos(ntheta) = e^nitheta = e^-nitheta

    from there i got the series

    (a^n(e^itheta)^n)/2

    now from here I think I setup the summation formula but this is where I get stuck. Any help is greatly apprecited.
     
  2. jcsd
  3. Sep 6, 2012 #2
    Is |a| < 1?
     
  4. Sep 6, 2012 #3
    Yes, a is a real constant and |a| < 1. sorry about that
     
  5. Sep 6, 2012 #4
    [tex]
    a^n \cos n\theta = a^n\frac {e^{in\theta} + e^{-in\theta}} {2}
    = \frac {a^ne^{in\theta} + a^ne^{-in\theta}} {2}
    = \frac {p^n + q^n} {2}
    \\ p = (\ln a)e^{i\theta}, \ |p| < 1
    \\ q = (\ln a)e^{-i\theta}, \ |q| < 1
    [/tex]

    What is the sum of [itex]p^n[/itex] and [itex]q^n[/itex]?
     
  6. Sep 6, 2012 #5
    That makes sense. That will then give me a real and an imaginary result of which I take the real I believe. Also, I apologize for the typo on the first post.
     
  7. Sep 7, 2012 #6
    Well, you can take the real part, but the sum is real anyway.
     
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