What is the range of $h$?
$f(x)=4x+1$
$g(x)=(x-1)/3$
Let $h=\{f^n(g^m(1)):n,m\in\mathbb{N}\geq0\}$
What is the range of $h$?
Show that $(2\mathbb{N}-1)\subset H$.
... okay I've done a bit more:
I think the closed forms for $g$ and $f$ are:
$g^m(x)=\frac{x}{3^m}-\frac{1}{2}(1-\frac{1}{3^m})$
$f^n(x)=4^nx+\frac{(4^n-1)}{3}$
So
$f^n(g^m(x))=4^n(\frac{x}{3^m}-\frac{1}{2}(1-\frac{1}{3^m}))+\frac{4^n-1}{3}$
$h=4^n(\frac{1}{3^m}-\frac{1}{2}(1-\frac{1}{3^m}))+\frac{4^n-1}{3}$
What is the range of $h$ for $n,m\in\mathbb{N}$ ? Show that $(2\mathbb{N}-1)\subset H$.
Please note that I've added the tags analysis and divergence because i have reason to suspect that the proof that $(2\mathbb{N}-1)\subset H$ may require a proof that one of the functions $f$ or $g$ diverges faster than the other.
If it helps... I think the method required, may be to ask yourself; for any given range of integers from $2^{y-1}$ to $2^y$, what is the least value of $m$ required, to ensure that $h(m,n)$ is exhaustive over $\mathbb{N}$ in that range, for all $n\in \mathbb{N}$.