On convergence of a recursively defined sequence For $\alpha \in \mathbb{R}$, define the sequence $\{x_n \}$ by $x_1 = \alpha$, and $x_{n+1} = x_n^2 - 1 $. It is true that if the sequence converges, then it must converge to $ (1 \pm \sqrt{5} ) /2$. Find all values of $\alpha$ s.t. the sequence converges to $ (1 + \sqrt{5} ) /2$. 
My main line of attack that I kept returning to was to try to express $x_n$ as a function of $\alpha$, so as to determine what properties $\alpha$ must have for the desired convergence. But the sequence becomes very unweilding as an explicit function of $\alpha$. I've also discovered that convergence will fail for values of $\alpha \in \{1, -1,  0, \}$, but also have trouble determining the all the bad values of $\alpha$. Hints appreciated.    
 A: Let $\displaystyle \varphi = \frac{1 + \sqrt{5}}{2}$
I get that $\displaystyle \varphi$ and $\displaystyle -\varphi$ are the only two starting values where the sequence converges to $\displaystyle \varphi$.
Consider
$x_{n+1} - x_n = (x_n + \frac{1}{\varphi})(x_n - \varphi)$
Now 
If $x_n \gt \varphi$, then $x_{n+1} \gt x_n$. So if the sequence ever goes greater that $\varphi$, it will keep increasing.
If $-\frac{1}{\varphi} \lt x_n \lt \varphi$, then $x_{n+1} \lt x_n$.
If $x_n \lt -\frac{1}{\varphi}$ then $x_{n+1} \gt x_n$.
If the sequence ever goes below $\varphi$ (except for $-\varphi$), then the sequence is "pulled" towards the left or right of $\varphi$, but never towards it (basically always away from it).
Also notice that, there is no starting value for which $x_{n+1} = -\varphi$, as $-\varphi = c^2 -1$  has no real solutions.
Also the only value for $x_n$ which makes $x_{n+1} = \varphi$, is $x_n = \pm \varphi$.
A: There are only two values such that $x_n$ converges to $\alpha = \frac{1 +\sqrt{5}}{2}$ wich are $\alpha$ and $- \alpha$ . 
$f(\alpha + h ) = \alpha + h.f'(\alpha) + o(h)$ so there exists $\epsilon$ such that for $|h|<\epsilon$
$$\alpha + h.\frac{(f'(\alpha) +1 )}{2} \leq f(\alpha + h) \leq \alpha + h.(f'(\alpha) +1 )$$
since $f'(\alpha) > 1$ if $0 <|x_n - \alpha| < \epsilon$ there exists $k \geq n$ such that $| x_k - \alpha| \geq \epsilon$ so the only way $x_n$ converges to $\alpha$ is that $(x_n)$ is eventually constant. This happens only for $x_0 =\alpha$ or $-\alpha$
A: $\bf Hint:$ If $(x_n)$ converges then $\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n=\alpha$.
A: An alternate way to illustrate this is to note that the limits are the fixed points of $f(x) = x^2 -1$. A fixed point $x$ is attracting iff $|f'(x)|<1$. 
The fixed points in this problem are solutions of $x^2 - x - 1 = 0$, and $|f'(x)|= |2x|\ge 1$ for both
of those roots, hence they are repelling fixed points. Therefore any other seed value will not lead to a convergent sequence.
