4
  1. Determine the sequences $a_0 , a_1 , a_2 ,\dots$ which satisfy all of the following conditions:

a) $a_{n+1} = 2a_n^2 − 1$ for every integer $n ≥ 0,$

b) $a_0$ is a rational number and

c) $a_i =a_j$ for some $i,j$ with $i \neq j$.

You can see it clearer here: http://www.bmoc.maths.org/home/bmo1-2009.pdf

I have managed to work out the modulus of $a_0$ must be equal to or below $1$. Indeed the $1,-1,1/2,-1/2$ and $0$ all work. I'm not familiar with sequences questions so I'm not really sure what I'm looking for in terms of a proof. I think I may also worked out that if we call $a_0$ as $a/b$ then $a+b$ and $2b$ must be square numbers since the rest of the rest of the terms can be put into the form $a/b$ (obviously not the same a and b), but also in the form $2a^2-b^2/b^2$ (again not the same a and b, sorry if I'm confusing people), since the denominator plus the numerator will be in the form $2a^2$, I thought that only $a_0$ can re-appear and so its $a+b$ and $2b$ must be square, but I'm not sure if I made a mistake, because it doesn't seem to work with some values.

Thanks in advance for any contributions.

Arpan
  • 2,633
MadChickenMan
  • 793

1 Answers1

3

You have $a_{n+1}=f(a_n)$ where $f(x)=2x^2-1$. Since $f(x)>x$ whenever $x>1$, we must have $a_n\leq 1$ for every $n$. Since $f(x)>1$ whenever $x<-1$, we must have $a_n\geq -1$ for every $n$. So we have $a_n\in [-1,1]$ for every $n$. In particular, $a_0\in [-1,1]$, so there is a $\theta\in{\mathbb R}$ such that $a_0=\cos(\theta)$. By elementary trigonometry and induction, one deduces $a_n=\cos(2^n\theta)$ for every $n$.

The condition $a_i=a_j,i\neq j$ then implies that $(2^j-2^i)\theta$ is an integer multiple of $2\pi$, so that $\theta$ is a rational multiple of $\pi$.

So $\frac{\theta}{\pi}$ and $\cos(\theta)$ are both rational. It is well known that in this case, $\theta$ must be an integer multiple of $\frac{\pi}{2}$ or $\frac{\pi}{3}$ (see, for example, the last paragraph in Quiaochu’s answer here ).

The corresponding values for $a_0$ are $0,\pm \frac{1}{2},\pm 1$.

Community
  • 1
Ewan Delanoy
  • 61,600
  • Very clever, I would never have thought to use trigonometry. One last thing though, does what you say in the second to last paragraph mean a0 must be those values or just some a. – MadChickenMan Apr 05 '15 at 12:53
  • Never mind, I've figured it out now. Do you think I would be expected to know the statement in the second to last paragraph though. I'm only really familiar with basic trig, is this fact well know enough so that it would be expected of 16-18 year olds to know. – MadChickenMan Apr 05 '15 at 13:01
  • @MadChickenMan You're right, there’s probably a more complicated solution that avoids the rational cosine stuff. On the other hand, the rational cosine stuff is more natural at a more advanced level. – Ewan Delanoy Apr 05 '15 at 14:17