What range of values for $q\in[0, 1]$ imply the convergence of $x_{n+1} = 1+qx_n^2$

Question

Let: $$ x_1 = a\\ 0<a<1\\ x_{n+1} = 1+qx_n^2\\ n\in \Bbb N $$ For what values of $q\in[0, 1]$ the sequence $\{x_n\}$ is convergent?

I've started with the following. We are given that $a\in(0,1)$, then: $$ x_1 = a\\ x_2 = 1+qa^2 $$

Obviously since $a<1$: $$ x_1 < x_2 $$ Suppose $\forall n\in\Bbb N: x_{n+1}>x_n$: $$ x_{n+1} > x_n \\ x_{n+1}^2 > x_n^2 \\ qx_{n+1}^2 > qx_n^2 \\ 1+qx_{n+1}^2 > 1+qx_n^2 \\ x_{n+2} > x_{n+1} $$ The induction shows $x_n$ is monotonically increasing. A monotonic sequence is convergent when it's bounded. Suppose $x_n$ is convergent. Let's try to find the fixed point of the recurrence: $$ x = 1 + qx^2\\ qx^2 - x + 1 = 0 \\ x = \frac{1\pm \sqrt{1-4q}}{2q} $$

For a fixed point to exist (at least in $\Bbb R$) we need: $$ 1-4q \ge 0 \\ q\le {1\over 4} $$ Or given initial conditions for $q$: $$ 0\le q\le {1\over 4} $$

However the solution above is missing upper bound for $x_n$ which I was not able to find. What would be the way to finish the problem? (In the answer section $q$ is indeed given in $q\in\left[0, {1\over 4}\right]$, but i still need to justify that)

Answer 1

• $(x_n)$converges for $0\le q\le \frac{1}{4}$:
  
  Let $\alpha=\frac{1+\sqrt{1-4q}}{2q}$ be the larger root of the equation $qt^2-t+1=0$. Since $\frac{1}{\alpha}$ is the other root, we find that $\alpha \ge 1\ge x_0$. We also find that $$ \alpha \ge x_n \implies \alpha =q\alpha^2 +1\ge qx_n^2 +1=x_{n+1}. $$ By induction, we have $\alpha \ge x_n$ for all $n$. Combining with the fact (in the OP) that $(x_n)$ is increasing, there exists $$ l=\lim_{n\to \infty} x_n \le \alpha $$ by monotone convergence theorem.
  
  
• $(x_n)$ diverges (to $\infty$) for $q>\frac{1}{4}$:
  
  We find that $$ qt^2 -t+1 =q\left(t-\frac{1}{2q}\right)^2+1-\frac{1}{4q}\ge 1-\frac{1}{4q}=c>0. $$ This implies $$ x_{n+1} =qx_n^2+1 \ge x_n +c,\quad\forall n. $$ Therefore, by induction we have $$ x_n \ge x_0 +nc \stackrel{n\to\infty}\longrightarrow \infty. $$ Summing up, $(x_n)$ converges if and only if $q\in [0,\frac{1}{4}].$

Answer 2

Assume $0\le q\le1/4$ and let $$ x^*=\frac{1-\sqrt{1-4\,q}}{2\,q}=\frac{2}{1+\sqrt{1-4\,q}}. $$ $x^*$ is a fixed point and $x^*>1$ (in particular, $a<x^*$.) We have also that $$ x<1+q\,x^2,\quad 0<x<x^*. $$ Since $0<a<x^*$, it follows that $0<x_n<x^*$ for all $n\in\Bbb N$.