Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have problems with:
Let $a_1\in\mathbb{R}$ and $a_{n+1}=a_n^2-1$ for $\forall n\in\mathbb{N}$.

Prove, that if $|a_1|\leqslant \dfrac{1+\sqrt{5}}{2} $ then $a_n$ is limited, and otherwise,
if $|a_1|>\dfrac{1+\sqrt{5}}{2}$ then $a_n$ diverges to $+\infty$.

So I assumed that $a_{n+1}-a_n>0$. Then for sure $a_{n}\geqslant a_n^2-1$.

If we want $a_n$ to be increasing then it must lie between $\dfrac{1-\sqrt{5}}{2}$ and $\dfrac{1+\sqrt{5}}{2}$.

This is a place where problems appears. I am able to calculate it (to that point of course), but I don't fully understand what am I doing. Can somebody help me? Thanks in advance!

share|cite|improve this question
$a_{n+1} - a_n > 0$ is equivalent to $a_n < a_n^2 - 1$. – WimC Nov 14 '12 at 20:39
up vote 1 down vote accepted

The recurrence is $$a_{n+1}=a_n^2-1$$ but, because $a_{n+1}$ does not care about the sign of $a_n$, we can instead analyse the boundedness/divergence of the non-negative sequence $|a_{n+1}|=|a_n|^2-1$.

Let us consider the $z$ for which $z^2-1 \le z$, rewriting this and solving the quadratic we get $(z - \varphi)(z - \bar \varphi) \le 0$ which can only happen when $\bar \varphi \le z \le \varphi$. This implies about our sequence that if $|a_n| \le \varphi$, $|a_{n+1}| \le \varphi$ (note, we can't say that $|a_{n+1}| \le |a_n|$).

We have found an interval $[0,\varphi]$ on which the recurrence is bounded. Let us consider $a_n > \varphi$, since $\varphi$ satisfies $z^2 - 1 = z$ let $a_1 = \varphi + r$ (some positive $r$) and observe $a_{n+1} = \varphi + (2 \varphi r + r^2)$ which is clearly bigger than $a_n$ (since both $r$ and $\varphi$ are positive). Therefore if $a_1 \in (\varphi,\infty)$ the sequence diverges.

share|cite|improve this answer

Hint: Have a look at the function $f(x)=x^2-x-1$ and try to find its zero using the iteration method which is basicaly is your problem $x_{n+1}=x^2_n - 1.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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