For what values $x_0$, does the sequence $x_{n+1} = x_n^2 - \dfrac{x_n}{2}$ converge? This is the question:
Let $x_n$ be a sequence of real numbers defined by $x_{n+1} = x_n^2 - \dfrac{x_n}{2}$, with $n \geqslant 0$.
For what values $x_0$, does this sequence converge? And it converges to what?
My first idea on how to solve this was to determine for what values the sequence decreases. So, $x_n \geqslant x_{n+1}$ implies $x_n \geqslant x_n^2 - \dfrac{x_n}{2}$, and from that we get that the sequence decreases if $x_n \in [0,\frac{3}{2}]$.
Testing some values, we see that the sequence does converge on $x_0 = 0$, $x_n = 1$, $x_n = \frac{1}{2}$, $x_n = \frac{3}{2}$, for the values $0$, $0$, $0$, and $\frac{3}{2}$ respectively.
All done, I don't have any clues on how to proceed. Any help will be welcome.
 A: First, you should find the fixed points of
$$f(x)=x^2-\frac{x}{2}$$
by $f(x)=x$. By solving this, you will get $x=0$ or $x=1.5$.
Now study this criterion about the convergence of the fixed point method.
A: As it was already mentioned, the only possible limits are the fixed points of $f(x)=x^2-\frac x2$, i.e. $x=0$ and $x=\frac 32$.

*

*If $x_0>\frac 32$, since $f$ is increasing in that region, the sequence will be increasing, hence divergent (it cannot converge to any $x^*>\frac 32$.


*Similarly, since $f(x)>\frac 32$ for $x< -1$, If $x_0< -1$ the sequence will also diverge. (taking $x_0<-1$ implies that $x_1> \frac 32$)


*When $x_0 = -1$ or $x_0= \frac 32$ the sequence converges to $\frac 32$. The second because $\frac 32$ is a fixed point and the first because $f(-1)=\frac 32$.


*For the remaining $x_0$, the sequence converges to $0$. You can see this by showing that the fixed point theorem conditions are met in some smaller set, for instance $I=[-\frac 18, \frac 18]$, and using the monotonicity of $f$, argue for other values of $x_0 \in (-1,\frac 32)\setminus I$, the sequence terms eventually fall into $I$.
A: 
Start from the $x-$axis for $x_0$ and follow the arrows to get $x_n$.
CASE I: $x_0\in\left(-\infty.-1\right)\cup\left(\frac32,+\infty\right), x_n$ diverges because $x^2-\frac x2>x$.
CASE II: $x_0\in\left\{-\frac32.-1\right\}, x_n$ converges to $x_n=\frac32$.
CASE III: $x_0\in\left(-1,\frac32\right), x_n$ spirals inwards to $x_n=0$ because of monotonicity of $y=x^2-\color{red}{\frac 12}x$. Had it been $x_{n+1}=x_n^2-\color{red}{2}x_n, x_n$ would've spiralled outwards. The grey part is drawn just for comparison. You may neglect it altogether.
A: Here is the result:
$$ \lim_{n\to\infty} x_n = \begin{cases}
0, & \text{if } x_0 \in (-1, \frac{3}{2}); \\
\frac{3}{2}, & \text{if } x_0 \in \{-1, \frac{3}{2}\}; \\
+\infty, & \text{if } x_0 \in (-\infty, -1)\cup(\frac{3}{2},\infty). \tag{*}
\end{cases} $$
To analyze this behavior, define $f(x) = x^2 - \frac{x}{2}$.
Case 1. Suppose that $x_0 \in (-1,\frac{3}{2})$. Then $x_1 = f(x_0) \in (-\frac{1}{2},\frac{3}{2})$. Moreover,
$$|f(x)| \leq |x| \qquad\text{for}\qquad x \in (-\tfrac{1}{2},\tfrac{3}{2})$$
and the equality holds if and only if $x=0$. Therefore $|x_n|$ converges to $0$ in this case.
Case 2. Suppose that $x_0 \in \{-1, \frac{3}{2}\}$. Then $x_n = \frac{3}{2}$ for all $n\geq 1$.
Case 3. If $x_0 \in (-\infty, -1)\cup(\frac{3}{2},\infty)$, then $x_1 \in (\frac{3}{2}, \infty)$. Also,
$$ f(x) > x \qquad \text{for} \qquad x > \tfrac{3}{2}. $$
So it follows that $(x_n)_{n\geq 1}$ is strictly increasing. Finally, $(x_n)$ cannot be bounded, for otherwise $(x_n)$ converges to some point which must be a fixed point of $f$, contradicting the fact that the only fixed points of $f$ are $0$ and $\frac{3}{2}$.
