Convergence of $x_{n+1}=x_n^2+\dfrac{\alpha^2}{16}$

Consider the sequence $(x_n) _{n\in N}$ defined by $x_0=\dfrac{1}{2}$, and $x_{n+1}=x_n^2+\dfrac{\alpha^2}{16}$, $\alpha>0$. For what values of $\alpha$ does this sequence converge? diverge? Find $\lim_{n\rightarrow \infty}x_n$ for $\alpha^2=3$ Any form of directive will help

Solution 1:

Hint: if $(x_n)_{n\in \mathbb N}$ converges to $l$, then $l = l^2+\alpha^2/16$.

Solution 2:

First notice that $x_{n}\geq 0\,\forall n\in\mathbb{N}$.

Now $x_{n+1}-x_{n}=(x_{n}-x_{n-1})(x_{n}+x_{n-1})$ which is $\geq$ or $\leq$ $0$ as $x_{n}$ is increasing or decreasing. Inductively it boils down to whether $x_{1}>x_{0}$ or $x_{1}<x_{0}$. Then analyzing boundedness for different values of $\alpha$ gives you convergence.

Then you assume limit is $l$ .

Then $l=l^{2}+\frac{\alpha^{2}}{16}$. Solving this quadratic gives you the limit. And using monotonic increasing or decreasing and the bounds you get to conclude which root of this quadratic is actually the limit.