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Complex no. query

  1. Oct 30, 2005 #1
    the first two parts of this question are easy to prove:

    1 + cosy/2+ cos(2y)/4 + cos(3y)/8 ......... =(4-2cosy)/(5-4cosy)

    this can be done using infinite geometric series, and taking the real part.

    Then cos(4y) =8({(cosy)^4} -{(cosy)^2}) + 1

    which can be done using double angle formulae.

    Now we need to solve

    x^4 - x^2 + 1/16 = 0

    This can be done easily using complex no's, but I'm not sure how to do it using the previous results. I've tried setting cos(4y) = 1/16, and then letting x=cosy, but I end up with some horrible expression involving cos(0.25*arcos(1/16)) which is obviously not what I'm supposed to get.

    Thanks
     
  2. jcsd
  3. Oct 31, 2005 #2

    CarlB

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    Try putting [tex]\cos(4y) = 1/2[/tex].

    Carl
     
  4. Oct 31, 2005 #3

    Hurkyl

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    You notice that:

    (cos y)^4 - (cos y)^2 + (1/8)(1 - cos 4y) = 0?
     
  5. Nov 1, 2005 #4
    thanks v. much. i was probably suffering some sort of miniature brain death by putting cosy = 1/16 (instead of cosy = 1/2!) The trig solution and the completing the square solutions match up as well, so it all work out.
     
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