Suppose we start with a rational number $a_0$, and define $a_{n+1}=2a_n^2-1$ for $n\geq 0$. For what $a_0$ will it be the case that $a_i=a_j$ for some $i\neq j$?

We can start with something like $a_0=1$, then $a_1=1$ so $a_0=a_1$.

If $a_0=0$, we get $0, -1, 1, 1, \ldots$

Likewise if $a_0=-1$, we get $-1,1,1,\ldots$.

But how can we find all $a_0$?

  • $\begingroup$ Well, for example for $\;a_0= 1\implies a_1=a_0\;$ ... $\endgroup$ – DonAntonio Nov 12 '13 at 19:37

Let $a_{0} = \cos \theta$. Then it is easy to check that $a_{n} = \cos (2^{n}\theta)$. So if $a_{i} = a_{j}$ for some $i \neq j$, then we must have

\begin{align*} \cos(2^{i}\theta) = \cos(2^{j}\theta) &\quad \Longleftrightarrow \quad 2^{i}\theta = 2n\pi \pm 2^{j}\theta, \quad n \in \Bbb{Z} \\ &\quad \Longleftrightarrow \quad \theta = \frac{2n\pi}{2^{i} \pm 2^{j}}, \quad n \in \Bbb{Z} \end{align*}

Thus the problem reduces to find the condition of $(i, j, n, \pm)$ such that

$$ \cos \left( \frac{2n\pi}{2^{i} \pm 2^{j}} \right) \in \Bbb{Q}. $$

Referring to this posting, this is possible if and only if

\begin{align*} \theta \equiv 0, \pm \frac{\pi}{3}, \pm \frac{\pi}{2}, \pm \frac{2\pi}{3} \pmod{2\pi} \end{align*}

This corresponds to $a_{0} \in \{0, \pm \frac{1}{2}, \pm 1 \}$.

  • $\begingroup$ Why did you decide that $\;|a_0|\le 1\;$ , @sos440 ? This follows from my answer, of course, but I can't see any reason why in yours... $\endgroup$ – DonAntonio Nov 13 '13 at 4:26
  • $\begingroup$ @DonAntonio, This solution works if $\theta$ is allowed to have complex values. But a simpler argument is as follow: If $|a_{0}| > 1$ and $a_{0} = \cosh \varphi$, then $\a_{n} = \cosh (2^{n}\varphi)$. $\endgroup$ – Sangchul Lee Nov 13 '13 at 14:10

In general

$$a_{n+1}:=2a_n^2-1=2a_m^2-1=:a_m\iff a_n=\pm a_m$$

If we choose $\;n\;$ to be the minimal index s.t. $\;a_{n+1}=a_{m+1}\;$ , for some $\;m\neq n\;$ , the above means that

$$a_n=-a_m\iff 2a_{n-1}^2-1=-2a_{m-1}^2+1\iff a_{n-1}^2+a_{m-1}^2=1\ldots$$

Try to take it from here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.