infinitely many negative and infinitely many positive numbers

Question

Suppose that $$x_1=\frac{1}{4}, \ x_{n+1}=x_{n}^3-3x_n.$$

Show that the sequence has infinitely many negative and infinitely many positive numbers.

My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $\sqrt{3}$. I want to show that the sequence cannot escape some interval.

Answer 1

Let $f(x) = x^3 - 3x$ and explore, for which $x$ one has $$ |f'(x)| < 1. $$

Answer 2

Notice that for $\alpha>0$:

$$(2+\alpha)^3-3(2+\alpha)=2+9\alpha+6\alpha^2+\alpha^3>2+\alpha$$ while for $\beta\in[0,2]$: $$(2-\beta)^3-3(2-\beta)=2-9\beta+6\beta^2-\beta^3<2-\beta$$

The opposite arguments can be made around $-2$. Since $x_1\in[-2,-2]$, your iteration is restricted to this range

Answer 3

You have essentially the right idea. Here are some hints to help you complete your proof.

Let $f(x) = x^3 - 3x $.

1. Show that $x_n \in ( -2, 2 ). $
   In particular, we have to show that $x_n \neq \pm 1$. See comments below.

2. Show that if $ x \in (0, \sqrt{3})$, then $f(x) < 0 $.

3. Show that if $x \in ( \sqrt{3} , 2 )$, then $ 0 < f(x) < x$.
   This tells us that the values will decrease. However, do they decrease enough?

4. Show that if $ x \in ( \sqrt{3} ,2 )$, then there eixsts an $n$ such that $ f^n(x) < \sqrt{3}$.
   This tells us that the values decrease enough to force a negative value, $f^{n+1} (x)$.

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Note: There are multiple ways of doing 4. If you are stuck, consider $ \frac{ 2 - f(x) } { 2 - x }$. This tells you how quickly you're moving away from 2 (and hence will be less than $\sqrt{3}$ eventually.)

Answer 4

The desired claim follows from the following two observations:

Claim. If $x_n \in \{-1/4, 1/4\}$, then $x_n \neq 0$ for all $n \geq 1$.

Proof. Let $p_1 = \operatorname{sign}(x_1)$ and $p_{n+1} = p_n^3 - 3p_n 4^{2 \cdot 3^{n-1}}$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^{3^{n-1}}$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.

Claim 2. If either $x_n \geq 0$ for any sufficiently large $n$ or $x_n \leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.

Proof. Write $x_n = 2\cos(2 \pi f_n)$. Then

$$ \cos(2 \pi f_{n+1}) = \frac{x_{n+1}}{2} = \frac{x_n^3 - 3x_n}{2} = 4\cos^3(2 \pi f_n) - 3\cos(2 \pi f_n) = \cos(2 \pi \cdot 3f_n). $$

So it follows that $\cos(2\pi f_{N+n}) = \cos(2\pi \cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n \geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_{N+n} \geq 0$ for all $n \geq 0$. Then each $3^n f_N$ must avoid the sets $(\frac{1}{4}, \frac{3}{4}) + \mathbb{Z}$. So it follows that

$$ f_N \in \mathbb{R} \setminus \bigcup_{n=0}^{\infty} \bigcup_{k\in\mathbb{Z}} \left( \frac{4k+1}{4 \cdot 3^n}, \frac{4k+3}{4 \cdot 3^n} \right) = \left\{ k \pm \frac{1}{4\cdot 3^n} : n \geq 0 \right\}. $$

This implies that $3^{n_0} f_N = \pm 1$ for some $n_0 \geq 0$, and hence $x_{N+n} = 0$ for all $n \geq n_0$. This proves the desired claim.