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Partial limits of cos(pi*n/3)

  1. Nov 10, 2012 #1
    Hi,
    Trying to find all partial limits of cos(pi*n/3), I separated it into:
    a_3k -> -1
    a_6k -> 1

    Is this a valid approach? Are there any other partial limits?
     
  2. jcsd
  3. Nov 10, 2012 #2

    HallsofIvy

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    Essentially you are looking at n= 3 (mod 6) and n= 0 (mod 6), which is a very good idea, but those two are not the only possibilities. Suppose n is equal 1 (mod 6). That is, n= 1+ 6k. In that case, [itex]\pi n/3= \pi/3+ 2\pi k[/itex]. Now, [itex]cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)[/itex] so [itex]cos(\pi/3+ 2\pi k)= cos(\pi/3)cos(2\pi k)+ sun(\pi/3)sin(2\pi k)= cos(\pi/3)[/itex] for all k.

    Similarly, look at n= 2+ 6k, n= 4+ 6k, and n= 5+ 6k.
     
    Last edited: Nov 11, 2012
  4. Nov 10, 2012 #3
    Thanks! Hence this sequence has only four partial limits. Is that correct?
     
  5. Nov 10, 2012 #4
    And what about the sequence ncos(pi*n/4). I believe it has only two partial limits, converging to +-infinity. Is that true?
     
  6. Nov 10, 2012 #5

    haruspex

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    In all sequences cos(n*pi/m), constant integer m, each value that the sequence can take occurs infinitely often. So every value is the limit of a subsequence. So, yes, for m=3 there are 4 values.
    For n*cos(n*pi/4), what are the possible values in the sequence?
     
  7. Nov 10, 2012 #6
    Would it be incorrect to say that n*cos(n*pi/4) has merely two partial limits (+-infinity) as cos(n*pi/4) is bounded whereas n isn't?
     
  8. Nov 10, 2012 #7

    haruspex

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    I could be wrong, but I wouldn't have thought infinity qualified as the limit of a sequence. It isn't a number, so you can't converge to it.
    OTOH, there is one finite limit for this sequence. What are all the possible values of cos(n*pi/4) ?
     
  9. Nov 10, 2012 #8
    Infinity, plus and minus, could well serve as PL's in this case, as thus the question is formulated in the textbook. In any case, the possible values are +-1/sqrt(2), 0 and 1. But that multiplied by infinity would either be infinity or, in the case of 0, undetermined, would it not?
     
  10. Nov 11, 2012 #9

    haruspex

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    No, you're not multiplying cos(n*pi/4) by infinity, you're taking a limit. In particular, what is the limit of n*cos(n*pi/4) for the subsequence n = 4k+2?
     
  11. Nov 11, 2012 #10
    I still don't see why it wouldn't be infinity. Could you help me see it?
     
  12. Nov 11, 2012 #11

    haruspex

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    (4k+2)cos((4k+2)pi/4) = (4k+2)cos(pi*k+pi/2) = 0 for all k. Therefore the sequence converges to 0.
     
  13. Nov 11, 2012 #12
    Thank you. Are there are any other partial limits?
     
  14. Nov 11, 2012 #13

    vela

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    You tell us.
     
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