Find all possible values of $a$ for which the sequence $x_n$ converges

Question

$$x_{n+1} =\frac{x_n^2}{2} + \frac{a}{2} $$ $$x_0 = 0$$ $$x_n \rightarrow x$$ => $$x = \frac{x^2}{2} + \frac{a}{2}$$ $$x = 1 \pm \sqrt{1-a}$$ I have already checked that for a positive value of a the sequence converges only if it is not greater than $1$. But there were difficulties with the negative

positive

drawn two graphs

blue $f$: $x_{n+1} = f(x_n)$

black $g: x_{n+2} = g(x_n)$

noticed that when $a<-3$ $x_n$ seems to diverge

$$\{0,g(0),g(g(0)),\ldots \}= \{x_0, x_2,x_4,\ldots \}$$

$$\{a/2,g(a/2),g(g(a/2),\ldots\} = \{x_1,x_3,x_4,\ldots\}$$

https://www.desmos.com/calculator/t8m2lb9f9q

necessary condition $-3 \le a \le 1$

Answer 1

Consider the discrete derivative

$$x_{n+1}-x_n = \frac{(x_n+1)^2+(a-1)}{2}$$

The sequence can only converge to points where the discrete derivative is $0$ (called equilibrium or stationary points in a dynamical systems setting) which implies $a\leq 1$. However at $a=1$, the equilibrium point is not a stable point as the discrete derivative is positive on both sides.

For $a<1$, regardless of positive or negative since $a=0$ is not a bifurcation point of this system, just as you found the equilibrium points are

$$x_{\pm} = -1 \pm \sqrt{1-a}$$

However since the discrete derivative is an upward facing parabola, the positive root is an unstable equilibrium point (small perturbations away from the positive root tend to move further away) and the negative root is a stable equilibrium point (small perturbations away from the negative root tend to move closer back). This means we have the following convergences for the limit

$$\lim_{n\to\infty}x_n = \begin{cases}x_- & x_0 < x_+ \\ x_+ & x_0 = x_+ \\ \infty & x_0 > x_+\end{cases}$$

Edit: This answer has still not been fixed yet, but I am leaving it up to not remove the positive discussion it generated in the comments that improved my understanding of the problem and has several of OP's attempts to work at a solution.

Answer 2

necessary and sufficient condition $0 \le a\le 1$

($a<0$) necessary condition => $-3 \le a < 0$

$-3 \le a \le 0 \space \space $$\sqrt{-a} \le x_n \le 0 \space \space and \space\space x_{n}>x_{n-1} => x_{n+1}<x_n$

$x_n = x_{N-1} + \Sigma_{k=N}^{n} (x_k - x_{k-1})$

and it suffices to prove the convergence of the series $x_{n+1}-x_n = y(t) (x_n-x_{n-1}) \space \space -3<a<0 => |y(t)| <q<1$

when $-3 < a< 0$ can choose $|\frac{a-1}{4}|<q<1$ and $|x_{n+1} -x_n| < q|x_{n}-x_{n-1}|$ at some $n>N-1$

$|x_n| \le |x_{N-1}| + \Sigma_{k=N}^{n} |x_k - x_{k-1}|< |x_{N-1}| + |x_N-x_{N-1}| \Sigma_{k=N}^n q^{k-N} $ that is, the sequence converges

It remains to prove the convergence for a=-3