# 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.

• Sorry, what do you mean by converging? Do you mean the terms of the sequence repeat? Commented Jul 29, 2020 at 13:46
• @A-levelStudent mathworld.wolfram.com/…. Commented Jul 29, 2020 at 15:26
• @K.defaoite thanks for that, I've never encountered convergence with a recurrence relation :) Commented Jul 29, 2020 at 15:55

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 $$x=1.5$$.

• I don't understand why I can't add the link properly... Commented Jul 29, 2020 at 13:24
• The fixed point theorem gives you only sufficient conditions for convergence, and so it falls i bit short when it comes to deciding for which initial approximations does the sequence converge. Commented Jul 29, 2020 at 13:35
• Fixed the link, hope it's OK @VIVID Commented Jul 29, 2020 at 13:47
• @grand_chat Thank you Commented Jul 29, 2020 at 15:09

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 $$I$$.

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.

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 $$\frac{3}{2}$$.