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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.

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 ). $




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, $f^{n+1} (x)$.

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.)

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

Notice that for $alpha>0$:

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

The opposite arguments can be made around $-2$. Since $x_1in[-2,-2]$, your iteration is restricted to this range

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 $x_n = 2cos(2 pi f_n)$. Then

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

So it follows that $cos(2pi f_{N+n}) = cos(2pi cdot 3^n f_N)$. 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

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

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.