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Starting to prove by De Moivre's Theorem

  1. May 18, 2008 #1

    rock.freak667

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    [SOLVED] Starting to prove by De Moivre's Theorem

    1. The problem statement, all variables and given/known data

    Use de Moivre's theorem to show that

    [tex]\sum_{n=1} ^{\infty} \frac{sinn\theta}{2^n}=\frac{2^{N+1}sin\theta+sinN\theta-2sin(N+1)\theta}{2^N(5-4cos\theta)}[/tex]

    2. Relevant equations

    [tex](cos\theta + isin\theta)^n=cosn\theta + isinn\theta[/tex]

    3. The attempt at a solution

    Just need a little help on how to start this question. Where would I get the [itex]\frac{sinn\theta}{2^n}[/itex] from?
     
  2. jcsd
  3. May 18, 2008 #2

    Hurkyl

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    As the imaginary part of [itex](cos\theta + isin\theta)^n[/itex], presumably. Oh, I suppose that doesn't work literally.
     
  4. May 19, 2008 #3

    tiny-tim

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    (btw, the ∑ is to N, not to ∞)

    Yes it does … sum [itex]\left(\frac{1}{2}(cos\theta + isin\theta)\right)^n[/itex] from 1 to N, get rid of the complex denominator in the usual way, and take the imaginary part. :smile:
     
  5. May 19, 2008 #4

    rock.freak667

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    But if I expand [itex](cos\theta +isin\theta)^n[/itex] how do I know when to stop since n varies?


    [tex](cos\theta +isin\theta)^n=(isin\theta)^n + ^nC_1(isin\theta)^{n-1}cos\theta +^nC_2(isin\theta)^{n-2}cos^2\theta +...[/tex]
     
  6. May 19, 2008 #5

    Dick

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    Don't expand it! Use deMoivre!
     
  7. May 19, 2008 #6

    rock.freak667

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    But then that would just give me that


    [itex]sin n\theta = Im[(cos\theta +i sin\theta)^n[/itex]

    the left has what I need, the right doesn't seem to be giving me any ideas.
     
  8. May 19, 2008 #7

    Hurkyl

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    Not even after plugging it into the sum and simplifying?
     
  9. May 19, 2008 #8

    rock.freak667

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    Forgive me if I am slow with this, I am very bad with these problems...


    But back to it. [itex]sin n\theta = Im[(cos\theta +i sin\theta)^n][/itex]

    putting that into

    [tex]\sum_{n=1} ^{N} \frac{sinn\theta}{2^n} [/tex]

    I get

    [tex]
    \sum_{n=1} ^{N} \frac{Im[(cos\theta +i sin\theta)^n]}{2^n} =\sum_{n=1} ^{N} (\frac{Im[(cos\theta +i sin\theta)}{2})^n[/tex]

    [tex]\sum_{n=1} ^{N} (\frac{sin\theta}{2})^n[/tex]


    Which is a GP with first term,a=[itex]\frac{sin\theta}{2}[/itex] and common ratio,r=[itex]\frac{sin\theta}{2}[/itex] and |r|<1 ?
     
  10. May 19, 2008 #9

    Dick

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    You can't factor the nth power outside the Im, and you don't want to. The left side is a geometric progression with common ratio exp(i*theta)/2. Sum it first, then take the imaginary part of both sides.
     
  11. May 19, 2008 #10

    rock.freak667

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    Not too clear here.

    I know that [itex]e^{i\theta}=cos\theta +i sin\theta[/itex] so the imaginary part would just be the sine. But if I can't factor the n outside of the Im, how do I get it as a GP?
     
  12. May 19, 2008 #11

    rock.freak667

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    ah I now got what you meant.

    I was to consider the sum from 1 to N of [exp(i*theta)/2]^n and then take the Im part. Well I got everything in the Denominator and two of the terms in the numerator. I am just missing the [itex]2^{N+1}sin\theta[/itex], I have [itex]2^Nsin\theta[/itex]. Shall recheck my algebra in a while.
     
  13. May 19, 2008 #12

    Dick

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    Forget the original series. Sum (exp(i*theta)/2)^n. Then take the imaginary part. The original series will appear on the left side. It will be a surprise.
     
  14. Sep 8, 2010 #13
    Re: [SOLVED] Starting to prove by De Moivre's Theorem

    Can some1 post the whole solution? How to get the 2^N[5-4Cos(theta)]


    How do we get cos???? If we are taking the imaginry part???? Please solve it and post it... thanx in advance
     
  15. Sep 9, 2010 #14
    Re: [SOLVED] Starting to prove by De Moivre's Theorem

    We want to show that
    Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)].
    -------------------------
    Note that e^(iθ) = cos θ + i sin θ.
    So, De Moivre's Theorem yields for any positive integer n
    e^(inθ) = (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ).

    Next, note that
    Σ(n=1 to N) sin(nθ) / 2^n
    = Im [Σ(n=1 to N) e^(inθ) / 2^n], by the first observation
    = Im {Σ(n=1 to N) [e^(iθ)/2]^n}.

    Thus, we need to compute the imaginary part of the finite geometric series
    Σ(n=1 to N) [e^(iθ)/2]^n
    = (e^(iθ)/2) Σ(k=0 to N-1) [e^(iθ)/2]^k, by re-indexing the sum
    = (e^(iθ)/2) (1 - (e^(iθ)/2)^N) / (1 - e^(iθ)/2)
    = e^(iθ) (1 - (e^(iθ)/2)^N) / (2 - e^(iθ))
    = e^(iθ) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))]

    Multiply numerator and denominator by the conjugate (2 - e^(-iθ)):
    ==> e^(iθ)(2 - e^(-iθ)) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))(2 - e^(-iθ))]
    = (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (e^(iθ) + e^(-iθ))]
    = (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (2 cos θ))]
    = (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 4 cos θ)].

    Now, it's a matter of simplifying the numerator.
    (2e^(iθ) - 1) (2^N - e^(iNθ))
    = 2^(N+1) e^(iθ) - 2^N - 2e^(i(N+1)θ) + e^(iNθ)
    = 2^(N+1) (cos θ + i sin θ) - 2^N - 2 (cos(N+1)θ + i sin(N+1)θ) + (cos(Nθ) + i sin(Nθ))
    = [2^(N+1) cos θ - 2^N - 2 cos(N+1)θ + cos(Nθ)] + i [2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ)]

    Its imaginary part is 2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ).

    Therefore, we have obtained
    Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)]
    as required.

    I hope this helps!
     
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