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Proof of a limit involving exponentials

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Given [tex]k\in \mathbb{Z} \setminus \{ 0 \}[/tex], prove that [tex]\lim_{n\to \infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i k n \alpha}=0[/tex], for all [tex]\alpha \in \mathbb{R} \setminus \mathbb{Q}[/tex].

    2. Relevant equations



    3. The attempt at a solution

    Well, I had an idea, but I'm not sure how well it works. Even if it did, I'm not sure how to make it rigorous. Basically, my idea was to say that you work your way around the unit circle, and that whenever we come arbitrarily close to hitting the same point again, the sum (between the two points) gets arbitrarily close to zero. Am I on the right track?

    Edit: Alternately, could we use some convergence theorem using counting measure? Also, the limit is supposed to be 0, not 1. I changed it in the code, but it isn't working.
     
  2. jcsd
  3. Oct 27, 2009 #2

    Hurkyl

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    This sounds like a reasonable line of attack.

    Let's idealize for a bit (which may or may not help). For simplicity, let's fix a particular point, say, 1. What would happen if, after some number P of steps you really did return to 1?

    (i.e. what if [itex]\alpha \in \mathbb{Q}[/itex], and [itex]k \alpha \notin \mathbb{Z}[/itex])

    Can you prove the limit goes to zero in this special case?


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  4. Oct 27, 2009 #3
    Ok, I proved it. Thanks a lot for the help.

    (For the curious: You prove it for \alpha k rational. Then you pick r = p/q \in \Q s.t. |\alpha k - p/q | < \epsilon. You do a lot of clever approximation and you get that 1/N * the sum \le (2\epsilon+q)/N, which goes to 0 as N goes to infinity.)
     
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