Is there a pattern to determine which angles will create a cyclic sequence?

[e(ho0n3]

[SOLVED] Cyclic Sequence of Angles

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

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

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

 

[Dick]

Yes. Any rational multiple of pi creates a cyclic sequence. Why are you insecure about this?

 

[e(ho0n3]

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?

 

[Dick]

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.

 

[e(ho0n3]

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.