Prove that if $x<0$ then $F_\mu^n(x)\rightarrow -\infty$ as $n\rightarrow \infty$, where $F_\mu=\mu x(1-x)$.

Question

Logistic map is given as $$x_{n+1}=\mu x_n(1-x_n)$$ Let $F_\mu=\mu x(1-x)$.

Therefore, for $\mu>1$, prove that if $x<0$ then $F_\mu^n(x)\rightarrow -\infty$ as $n\rightarrow \infty$.

Answer 1

Fix some $\mu>1$. Since $x_0<0$, $\mu\left(1-x_0\right)>t_0$ for some $t_0>1$. Note that $$μ(1−x_n)=\mu(1-x_{n-1})+\mu x_{n-1}^2>μ(1−x_{n−1})>t_0,$$ for every $n$, in particular $\{x_n\}$ is decreasing and $x_n<t_0^nx_0$ for every $n$. Since $t^nx_0\to-\infty$, this proves that $\left\{ x_{n}\right\} $ is strictly decreasing with no lower bound. (This argument is in @Jean-Claude Arbaut's comment).