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Sum of a geometric series of complex numbers

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Given an integer n and an angle θ let
    Sn(θ) = ∑(eikθ) from k=-n to k=n

    And show that this sum = sinα / sinβ

    2. Relevant equations
    Sum from 0 to n of xk is (xk+1-1)/(x-1)

    3. The attempt at a solution

    The series can be rewritten by taking out a factor of e-iθ as
    e-iθ∑(e)k from 0 to 2n.

    = (einθ-1)/(e-1)
    Subsequent attempts at rearrangement, e.g. trying to realize the denominator by multiplying by e-iθ-1 have not got me any closer to the answer. Not sure what to do from here...
     
  2. jcsd
  3. Feb 25, 2015 #2
    Are you absolutely sure you printed the problem correctly? I ask since I get the following,

    [itex]
    S_n (\theta) = \sum \limits _{k=-n}^{n} e^{ik\theta}
    [/itex]

    [itex]
    = 1 + \sum \limits _{k=1}^{n} \left( e^{ik\theta} + e^{-ik\theta} \right)
    [/itex]

    [itex]
    = 1 + 2 \sum \limits _{k=1}^{n} \cos (k \theta )
    [/itex]

    Where I see no possible way for rewriting in terms of what you have there. Are [itex] \alpha, \beta [/itex] constants?
     
  4. Feb 25, 2015 #3
    ##e^{ik\theta}=cosk\theta+isink\theta##
    Split the sumation from -n to -1, then from 1 to n. For k=0, you can simply substitute.
    now again split the summation. On for summing cos terms and other for sin terms.
    do you notice that the angle inside the cosine and sine terms are in AP?
    That is:Σcoskθ from -n to -1 = ##cos(-n\theta )+cos(-n\theta +\theta )+.....+cos(-n\theta + (n-1)\theta )##
    You have a simple formula for evaluating its sum:
    Sin(a)+sin(a+b)+....+sin(a+(n-1)b)= try deriving it....
     
    Last edited: Feb 25, 2015
  5. Feb 25, 2015 #4

    Fredrik

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    Not sure what you did here.
    $$(e^{i\theta})^k =(e^{-i\theta}e^{2i\theta})^k = e^{-ik\theta} e^{2ik\theta}.$$
     
    Last edited: Feb 25, 2015
  6. Feb 25, 2015 #5
    Reasoning as follows:
    Series is z-n+z-n+1...+zn
    So if I take out a factor of z-n, as in divide by it, then that becomes the powers are from 0 to 2n, and the geometric sum to n formula can be used.

    Although come to think of it, it isn't entirely clear that that's allowed... does it break the rules?
     
  7. Feb 25, 2015 #6

    Fredrik

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    Thanks for the clarification. Yes, this you can do. I thought you were taking something that depends on k outside of the sum, but I see now that you're not.
     
  8. Feb 25, 2015 #7
    So evaluate 4 summations? When writing out the coskθ expansion, don't think I follow. Why isn't there a cos(-θ) term? That sounds like it has a really obvious answer... Sorry if it's a silly question!
     
  9. Feb 25, 2015 #8
    They look completely different. But I can't find why mine is wrong, so maybe they're the same? But where are the sin terms going to come from?Oh, and I don't think alpha and beta are constant, they depend on n and/or theta according to the question!
    I'll give it a go with this expression, see if I can get there. Thanks :)

    Edit: definitely different. What kind of series is that, anyway?!
     
    Last edited: Feb 25, 2015
  10. Feb 25, 2015 #9
    It was a typo. Its edited now
     
  11. Feb 25, 2015 #10

    Ray Vickson

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    If ##S_0 = \sum_{k=1}^n e^{i k \theta}##, then ##S = 1 + 2 \text{Re}(S_0)##, where ##\text{Re}## denotes the "real part". Use the geometric sum formula to get ##S_0##, then re-write it to extract the real and imaginary parts.
     
  12. Feb 25, 2015 #11
    What does S denote? I am a little bit lost. And should one part, either real or imaginary, be zero? Because the whole series is meant to sum to just one term?
     
  13. Feb 25, 2015 #12

    Ray Vickson

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    My ##S## is your ##S_n(\theta)##.

    In your original sum, for every ##+k## term there is also a corresponding ##-k## term, so the imaginary parts cancel out when you perform the whole sum. The reason this happens is that the ##+k## and ##-k## terms taken together produce
    [tex] e^{ik\theta} + e^{-ik\theta} = 2 \cos(k \theta) = 2\, \text{Re}(e^{i k \theta}).[/tex]
    Therefore, it suffices to sum the real parts for ##k## from 1 to ##n##. What makes this do-able is the fact that the sum of the real parts equals the real part of the sum. Thus, you can sum first, then take the real part; that will allow you to perform calculations like ##\sum_{k=1}^n \cos(k \theta)## by using geometric sums.
     
  14. Feb 26, 2015 #13
    I have another approach, which almost solved it. I think something is wrong with problem itself, or perhaps som eother information is missing. However, will give my seconds try, and perhaps you'll find, some mistake I made and solve it... Anyways,
    [itex]
    S_n(\theta) = \sum \limits _{k=-n}^n e^{ik\theta}
    [/itex]

    adding [itex] e^0 - 1 = 0 [/itex] to the sum enables us to write,

    [itex]
    = \left( \sum \limits _{k=0}^n e^{+ik\theta}\right) + \left( \sum \limits _{k=0}^n e^{-ik\theta}\right) - 1
    [/itex]

    Then using the final sum formula for a geometrical series as you yourself posted,

    [itex]
    = \frac{e^{+i(n+1)\theta}-1}{e^{+i\theta}-1} + \frac{e^{-i(n+1)\theta}-1}{e^{-i\theta}-1} - 1
    [/itex]

    Make all fractions have the same denominator

    [itex]
    = -\frac{e^{+i(n+1)\theta}-1}{1-e^{i\theta}} + \frac{e^{-i(n+1)\theta + i\theta}-e^{i\theta}}{1-e^{i\theta}} - \frac{1-e^{i\theta}}{1-e^{i\theta}}
    [/itex]

    Rearrange,

    [itex]
    = \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta} + 1 + e^{-i(n+1)\theta + i\theta}-e^{i\theta} - 1+e^{i\theta}\right)
    [/itex]

    Some stuff goes away,

    [itex]
    = \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta}+ e^{-i(n+1)\theta}\cdot e^{i\theta}\right)
    [/itex]

    In which WolframAlpha.com simplifies to,

    [itex]
    = \cos(n \theta) + \frac{\cos(\theta / 2)}{\sin(\theta / 2)}\cdot \sin(n \theta)
    [/itex]


    As said, this is as far as I come. I suggest skim over my calculations and se if you find any faults.. If not, there is something wrong with the problem, or with the geometric series sum formula..

    Good luck!
     
    Last edited: Feb 26, 2015
  15. Feb 26, 2015 #14
    Pretty sure all that is right! If you keep going, make a common denominator, then it simplifies down to sin/sin! What a horrible question. Thanks for your help :)
     
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