Compute limit of the sequence $x_n$ given by $x_{n+2}=-\frac{1}{2}(x_{n+1}-x_n^2)^2+x_n^4$ Let $(x_n)$ be a real sequence such that $x_0=a\in\mathbb{R},x_1=b\in\mathbb{R},x_{n+2}=-\dfrac{1}{2}\left(x_{n+1}-x_{n}^2\right)^2+x_{n}^4\;\forall n\in\mathbb{N} $ and $|x_n|\leq \dfrac{3}{4},\forall n\in\mathbb{N}$. The sequence $x_n$ is convergence or not and compute $\lim x_n$ (if exists)?
I think we could use $\inf$ and $\sup$. But I am not sure
 A: As Greg Martin stated in comment, the limit satisfies the equation $\displaystyle L=-\frac{1}{2}(L-L^2)^2+L^4$, not hard to find(by rearranging and factoring the equation) the solutions of the equation are $0,1,-1,-2$.
Because $\displaystyle |x_n|\leq \frac{3}{4}$ the sequence cannot converge to $1, -1, -2$. The sequence can only converge to $0$. It remains to prove the sequence indeed converges to 0, which can be done by estimating an descending upper bound of $|x_n|$
To do the estimation, we start by proving a lemma, for real numbers $\alpha$ and $\beta$ we have $|\frac{1}{2}\alpha^2+\alpha\beta-\frac{1}{2}\beta^2|\leq \frac{\sqrt2}{2}(\alpha^2+\beta^2)$.
To prove it, let $\gamma^2=\alpha^2+\beta^2$, we can parametrize $\alpha$ and $\beta$ to get $\alpha=\gamma \cos(\theta),\beta=\gamma \sin(\theta)$, which means $$\frac{1}{2}\alpha^2+\alpha\beta-\frac{1}{2}\beta^2=\frac{\gamma^2}{2}(\cos(2\theta)+sin(2\theta))=\frac{\sqrt2\gamma^2}{2}(sin(2\theta+\pi/4))$$ and therefore the inequality.
Back to the question, if $|x_n|\leq c$ and $|x_{n+1}|\leq c$, $\frac{3}{4}\geq c\geq 0$, by the lemma(set $\alpha=x_n^2$,$\beta=x_{n+1}$) $|x_{n+2}|=|-\frac{1}{2}(x_{n+1}-x_n^2)^2+x_n^4|=|\frac{1}{2}x_n^4+x_n^2x_{n+1}-\frac{1}{2}x_{n+1}^2|\leq|\frac{\sqrt2}{2}(x_n^4+x_{n+1}^2)|\leq\frac{\sqrt2}{2}(c^4+c^2)=\frac{\sqrt2}{2}(c^3+c)c\leq \frac{\sqrt2}{2}((\frac{3}{4})^3+(\frac{3}{4}))c<0.83c$
Therefore we get $|x_1|\leq\frac{3}{4},|x_2|\leq\frac{3}{4},|x_3|\leq(0.83)\frac{3}{4},|x_4|\leq(0.83)\frac{3}{4},|x_5|\leq(0.83)^2\frac{3}{4}......|x_{2n}|\leq(0.83)^{n-1}\frac{3}{4}$, where $(0.83)^n \rightarrow 0$ as $n \rightarrow \infty$, by comparison we conclude the original sequence converges to 0.
