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Sum of Complex geometric series

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Use cos ( n * x) = (z ^ n + z ^ -n)/2 to express

    cos x + cos 3x + cos 5x + ... + cos([2n -1]x)

    as a geometric series in terms of z. Hence find this sum in terms of x

    2. Relevant equations

    3. The attempt at a solution

    (z + z^-1)/2 + (z^3 + z^-3)/2 + ... + (z^2n-1 + z-(2n-1))

    I want to use the sum of a geometric series, but I can't work out what the common ratio of the terms is. If someone could point me in the right direction it would be greatly appreciated. Thank you very much in advance for you your help.

    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2


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    Split it into two series. The +n's make one series and the -n's make another.
  4. Nov 4, 2007 #3
    Thank you very much for your reply.

    After an hour of frustrating and sadly unfruitful algebraic manipulation, I am still no closer to the answer of sin(2nx) / 2sin(x).

    Any further help would be greatly appreciated.
  5. Nov 4, 2007 #4


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    It is a pain in the neck, isn't it? I finally got it. And I can't say it took me much less than an hour. I'm assuming the z you are talking about is exp(ix), right? After your series summation for the powers of z, you should get (z-z^(2n+1))/(1-z^2), right? With an analogous result for z^(-1). Once you've done that, turn it back into sin's and cos's of multiples of x using deMoivre. Now you combine the two fractions back into one. If you do that correctly you should find all of the imaginary parts cancel. Now you just have to smash stuff back together using stuff like cos(a+/-b)=cos(a)cos(b)+/-sin(a)sin(b). It finally does work. I wish I could give you an easier method...
  6. Nov 5, 2007 #5

    Gib Z

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    Forget about substituting that ugly expression, just Remember [tex]\cos (nx) = R( e^{inx})[/tex] Where R(t) denotes the Real part of t.
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