Super Powers

Calculus Level pending

Let ${\left( a_n\right)}_{n \in \mathbb{N} }$ be a sequence defined, recursively, by:

$\begin{cases} a_1=x \\ \forall n\in \mathbb{N}\backslash\{1\} , & a_{n+1}=x^{a_n}\\ \end{cases}$

For some $x \in \mathbb{R}^{+}$ .

Let $x\in \mathbb{R}^{+}$ be such that $\lim a_n=2$

(these means, intuitively, that $x^{x^{x^{x^{x^{ ... }}}}} =2$ ).

Find the value of $\lfloor x^6+x^4+x^2+1 \rfloor$ .

1 solution

Paulo Guilherme Santos
Jun 28, 2015

Let ${\left( a_n \right)}_{n \in \mathbb{N} }$ be such that

$\begin{cases} a_1=x \\ \forall n \in \mathbb{N}, & a_{n+1}=x^{a_n} \\ \end{cases}$

For some $x \in \mathbb{R}^{+}$ such that

$\lim a_n=2$ .

We have that

$\log_x \left(2 \right)=2 \implies 2=x^2 \implies x=\sqrt2$ .

This means that

$\lfloor x^6+x^4+x^2+1 \rfloor= \lfloor {\left(\sqrt2 \right)}^6+{\left(\sqrt2 \right)}^4+{\left(\sqrt2 \right)}^2+1 \rfloor=\lfloor 15 \rfloor$ .