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Complex Induction

  1. Feb 2, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Prove by induction [tex]cos(6\theta)+isin(6\theta)=(cos\theta +isin\theta )^6[/tex] for all [tex]\theta[/tex]

    2. Relevant equations
    by DeMoivres theorem, [tex](cos\theta +isin\theta )^n=cos(n\theta)+isin(n\theta)[/tex]

    3. The attempt at a solution
    Well when I first looked at this problem I thought there was an error in the question because from what I know, induction has a flaw in that it can only prove for integers, not ALL values.

    Anyway, so I need to prove:

    [tex]cos6(k+1)+isin6(k+1)=(cos(k+1)+isin(k+1)) ^6[/tex]

    Any help would be appreciated :smile:
     
  2. jcsd
  3. Feb 2, 2009 #2

    Mark44

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    I think you can use Euler's formula here: e^(ia) = cos(a) + i sin(a). You also need the fact that e^(ab) = (e^a)^b.
     
  4. Feb 3, 2009 #3

    Mentallic

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    So
    [tex]cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{6i})^{k+1}=e^{6i}.(e^{6i})^k[/tex]

    Sorry but I'm unsure what to do from here. And please correct me if I'm wrong.
     
  5. Feb 3, 2009 #4

    Mark44

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    Try this instead.
    [tex]cos6(k+1)+isin6(k+1)=e^{i6(k+1)}=(e^{i(k + 1)})^{6}[/tex]

    Now you're almost there...
     
  6. Feb 3, 2009 #5

    gabbagabbahey

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    I think you may be misinterpreting the question here. I assume that you are supposed to start by showing [itex]cos(n\theta)+isin(n\theta)=(cos\theta +isin\theta )^n[/itex] for all [itex]\theta[/itex], and some positive integer [itex]n[/itex] (I'd use n=1 as your starting point) by expanding [itex](cos\theta +isin\theta )^k[/itex] and using the multiple angle trig identities. Then assuming it is true for [itex]n=k[/itex] show that it is true for [itex]n=k+1[/itex] and hence by induction that it is true for all n, including [itex]n=6[/itex]

    Mark's method isn't really a proof by induction, since you are assuming that Euler's formula holds for all [itex]\theta[/itex].
     
  7. Feb 3, 2009 #6

    Mark44

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    It seemed odd to me, too, but I'm just going by what mentallic wrote in the first post in this thread:
    and
     
  8. Feb 3, 2009 #7

    gabbagabbahey

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    It is a poorly worded question; they may as well have just asked to prove by induction that [itex](x)^6=x^6[/itex] for all x.
     
  9. Feb 4, 2009 #8

    Mentallic

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    Well this is the only question I got wrong in that test (besides a couple sloppy errors). I was unsure how I would prove for all [tex]\theta[/tex] without using DeMoivres theorem, and at the same time, if I were to assume the theorem true, then... I have pretty much the same thing as what you said gabbagabbahey:
    So I instead went ahead and proved DeMoivres theorem [tex]cos(n\theta)+isin(n\theta)=(cos\theta+isin\theta)^n[/tex]
    And thus, true for n=6 and I probably cheated here when I finished with the statement, hence true for all [tex]\theta[/tex].

    Sorry Mark, but I was never taught Euler's method. We rather used the mod-arg form [tex]rcis\theta[/tex]. But I have heard that there is a clear relationship between these 2 forms, so if we could convert to the mod-arg form, I could proceed :smile:

    If it is poorly worded, what should the question have been instead? Because this is exactly what we needed to answer.
     
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