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Cyclic Sequence of Angles

  1. Feb 20, 2008 #1
    [SOLVED] Cyclic Sequence of Angles

    Fix an angle [itex]\theta[/itex]. Let n be a positive integer and define [itex]\theta_n = n\theta \bmod 2\pi[/itex].

    The sequence [itex]\theta_1, \theta_2, \ldots[/itex] is cyclic if if it starts repeating itself at some point, i.e. the sequence has the form [itex]\theta_1, \ldots, \theta_k, \theta_1 \ldots[/itex].

    What I would like to find out is: For which angles [itex]\theta[/itex] is the sequence [itex]\{\theta_n\}[/itex] cyclic? If for some integer m > 1, [itex]\theta_1 = \theta_m \equiv \theta = m\theta[/itex], then [itex]m\theta = \theta + 2\pi x[/itex] for some non-negative integer x. Solving for [itex]\theta[/itex], I get [itex]\theta = 2\pi x / (m - 1)[/itex]. So it seems that any rational multiple of [itex]\pi[/itex] will create a cyclic sequence. Is this correct?
  2. jcsd
  3. Feb 20, 2008 #2


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    Yes. Any rational multiple of pi creates a cyclic sequence. Why are you insecure about this?
  4. Feb 21, 2008 #3
    This all began when I started contemplating about the limit as n approaches infinity of zn, z being complex with |z| < 1. If I represent z as r(cos t + isin t), zn = rn(cos nt + isin nt). If {nt} is cyclic, then I could break up the sine and cosine terms, multiply through by rn and apply the limit on each term. Each term goes to 0 because r < 1 so the limit is 0. Right?
  5. Feb 21, 2008 #4


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    Yes, but you don't have to worry about 'cyclic' |cos nt+i*sin nt| is bounded, since |cos|<=1 and |sin|<=1. Regardless of the arguments. So if you multiply by r^n with r<1, the result certainly goes to 0.
  6. Feb 21, 2008 #5
    That makes sense. So for any complex z with |z| < 1, zn goes to 0 as n goes to infinity. I began feeling paranoid about this when I was trying to compute the limit of nzn as n goes to infinity. I rewrote this as n/z-n and applied l'Hopital's rule to get -zn/log z. I wasn't sure about the limit of zn here, but now I am. Thanks.
    Last edited: Feb 21, 2008
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