Take the 2-minute tour ×
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.

I am considering the sequence $a_{n+1}=a_n^2-1$ and I want to examine for what values of $a_1\in\mathbb{R}$ the sequence converges. I know that if it converges it converges to $\frac{1\pm\sqrt{5}}{2}$. Any hints on how to approach this problem?

share|improve this question
1  
To get an idea of what the answer would be, you could draw the graphs of $y=x$ and $y=x^2-1$ and then use cobwebbing with different values of $a_1$. –  user84413 Aug 7 at 19:31

1 Answer 1

If $f(x)=x^2-1$, then $|f^{\prime}(x)|=|2x|>1$ for $x=\frac{1\pm\sqrt{5}}{2}$, so both equilibrium points are locally unstable. Therefore the sequence will only converge for values of $a_1$ which are eventually mapped to

$\phi_{1}=\frac{1+\sqrt{5}}{2}\;$ or $\;\;\phi_{2}=\frac{1-\sqrt{5}}{2}$ by $f(x)$,

such as $\;\;\pm\phi_{1}, \pm\phi_{2}, \pm\sqrt{\phi_{1}}, \pm\sqrt{1+\phi_{1}}, \pm\sqrt{1+\sqrt{1+\phi_{1}}},\cdots$

share|improve this answer
1  
This is not true. The two equilibrium points are unstable only rule out the case $a_n$ approach them closer and closer as a limit. It didn't rule out the case at some $n$, $a_n$ lands on one of these point exactly and then trap at that point forever. For an example, the sequence start with $a_1 = \frac{\sqrt{5}-1}{2}$ or $\sqrt{1 + \frac{\sqrt{5}-1}{2}}, \ldots$ converges to $\frac{1-\sqrt{5}}{2}$. –  achille hui Aug 7 at 22:56
    
@user84413 What do you mean that the equilibrium points are locally unstable? Unfortunately I am not familiar with this. –  user150391 Aug 8 at 7:14
    
@achille hui Thanks for your correction; I realized this last night after I had posted my answer. –  user84413 Aug 8 at 15:58
    
This means that if $a_1$ is a value near one of the two possible limits, the sequence will not in general approach the limit, but will diverge instead. (Please see @achille hui's comment.) –  user84413 Aug 8 at 16:03

Your Answer

 
discard

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.