How to prove that $x_{n+1}=g(x_n)=x_n^2+1/4\to\infty$ as $n$ increases?

Question

Let $g(x)=x^{2}+1/4$. If $g^{n}(x)=g(g(...g(x)))$ ($n$ times) how to prove for $|x|>1/2$ then $g^{n}(x)\to\infty$ as $n$ increases without bounded?

Answer 1

It is sufficient to notice that $g(g(x))-g(x) = x^4-\frac{x^2}{2}+\frac{1}{16}$, and the latter function is strictly positive whenever $|x|\ne \frac 12$. Therefore, the sequence $x_n$ is strictly increasing.

Suppose that this sequence is bounded, then the upper boundary is equal to the limit of this sequence, which we will denote as $x_*$. Thus, $g(x_*) = x_*$, hence $x_* = \pm \frac 12$, which is impossible (we already know that $x_*>x_2>\frac 12$).

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as @YvesDaoust noticed, the growth of the sequence follows from the fact that $g(x)>|x|$ whenever $|x|>\frac 12$.

Answer 2

Note that $g(x)=x^2+1/4\geq x$ and equality holds iff $x=1/2$.

This means that if $x>1/2$ then the sequence $g^{n}(x)$ is strictly increasing.

Let $L=\sup_n g^{n}(x)=\lim_{n\to+\infty}g^{n}(x)$. If $L$ is finite then, $g^{n+1}(x)=g(g^n(x))$ implies that $L=g(L)$, that is $L=1/2$. Contradiction.

Answer 3

Set $x=\frac12+a$, $a>0$. Then $g(x)=\frac12+a+a^2$.

Playing around with some recursion steps leads to the hypothesis $$x_n=g^n(x)\ge\frac12+a+na^2,$$ which is obviously true for $x_0=\frac12+a$. Induction step: \begin{align} x_{n+1}=g(x_n) &\ge\frac12+(a+na^2)+(a+na^2)^2 \\ &\ge \frac12+a+(n+1)a^2 \end{align} which leads to unbounded growth of the recursive sequence.

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One could even find $x_n\ge\frac12+a+na^2+n(n-1)a^3$ for faster growth, but that is not necessary to prove the claim.

Answer 4

I would like to offer a direct proof. First I prove that the relation is increasing. That is $i>n \Rightarrow g_i(x)\geq g_n(x)$. This is quite straightforward one merely needs to prove that $g(x)>x$ which comes from the fact that \begin{equation} g(x)-x = x^2 -x+1/4=(x-1/2)^2>0 \end{equation} Next I prove that $\forall y \exists n:g_n(x)>y$:

Consider $x>1/2\Rightarrow x=\delta+1/2$ \begin{eqnarray} g(\delta+1/2+n\delta^2)&=&n\delta^{3}+n^2\delta^4+(n+1)\delta^2+\delta+1/2\geq(n+1)\delta^2+\delta+1/2\\ \Rightarrow g_n(1/2+\delta)&\geq&n\delta^2+\delta+1/2\\ \Rightarrow n &=& \lceil\frac{y-x}{(x-1/2)^2} \rceil+1 \end{eqnarray} gives such an $n$.

We have thus shown that the iterations of this function are increasing and unbounded.