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Conjecture Regarding rotation of a set by a sequence of rational angles.

  1. Aug 2, 2012 #1
    Conjecture Regarding Rotation of a Set by a Sequence of Angles.

    Consider the following sequence, where the elements are rational numbers mulriplied by [itex]\pi[/itex]:
    [itex] (\alpha_{i}) = \hspace{2 mm}\pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/32,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/16,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/8,\hspace{2 mm} \pi/4,\hspace{2 mm} 3\pi/64,\hspace{2 mm} \pi/4,\hspace{2 mm} \cdots[/itex]

    Let [itex]K \subset ℝ^{2}[/itex] be a compact set. Also let [itex]R_{\alpha_{i}}[/itex] denote the rotation by [itex]\alpha_{i}[/itex].

    Suppose [itex]R_{\alpha_{i}}K = \hspace{2 mm} K[/itex] for each [itex]\alpha_{i} \in (\alpha_{i})[/itex].

    Question: Is it true that for all [itex]\theta \in [0, 2\pi)[/itex] [itex]R_{\theta}K = \hspace{2 mm} K[/itex].

    If instead we had the sequence [itex](n\alpha)[/itex] where [itex]\alpha[/itex] is an irrational number, it is trivial that the conjecture holds. This is trivial due to the following fact from the study of continued fractions:
    Given any real number on a circle, it can be approximated arbitrarily close by multiples of an irrational number.
    But if [itex]\alpha[/itex] is a rational number this doesn't hold since after a finite number of rotations you will get back to where you started from. However in the question above we don't have rotations by a fixed rational number and the answer is not immediate!
    Last edited: Aug 2, 2012
  2. jcsd
  3. Aug 2, 2012 #2


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    There are some things not making sense to me in this question.
    Most obviously, the sequence clearly consists of irrationals, so I guess you mean rational multiples of pi. Secondly, I don't see any significance in showing it as a sequence, including repeats. It seems to be used only as a set - so why the repeats?
    That said, if I've understood the question...
    K is closed under rotations by 3nπ.2-m, for all positive integers m, n.
    Given θ, let xi be the bits of the binary fraction expressing θ/3π. From this you can construct a sequence of rotations converging on θ.
  4. Aug 3, 2012 #3
    That's exactly right. Thank you for pointing that out. I corrected the question.

    The reason for the repeats is the following:
    I'm preforming a symmetrization on the set K and the algorithm is such that it produces the above sequence.

    I got confused myself because once I got the sequence, I thought the rule is that I must follow the sequence to get arbitrary close to a θ. But you're absolutely right. Once I show [itex] R_{\alpha_{i}}K = K[/itex] for each [itex] \alpha_{i}[/itex], I'm done. This is because of the role that "n" is playing in your solution.

    When I get a little too excited I need someone to check I'm not doing something stupid. Thanks for the comment!
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