Stuck at the inductive step with this proof.
Given the four subfunctions $P_1,P_2,N_1$, and $N_2$ each with subdomain and subrange as shown above, which together make up an iterative piecewise function that we will call $f(x)$. Note that all values in the range of $f$ also appear in the domain of $f$, and vice versa, this guarantees continuous iterations under $f$.
I want to show that the trajectory of all values in the domain of $f(x)$ will converge to a single value, $1$. Part of the tree is shown below.
Base Step: Show that the initial term of each subfunction does converge to 1.
The initial term of $P_1$ is 2 which has trajectory $2→3→-4→-2→-1→1$. This composition is $N_1 ∘N_2∘N_2 ∘P_2∘P_1 (x)=\frac{9}{32}x+\frac{14}{32}$. The initial term of $P_1$ does converge to $1$, so this partial base step is true.
The initial term of $P_2$ is $1$ which has trajectory $1→-1→1$. This composition is $N_1∘P_2 (x)=\frac{3}{4} x+\frac{1}{4}$. The initial term of $P_2$ does converge to $1$, so this partial base step is also true.
The initial term of $N_1$ is $-1$ which has trajectory $-1→1$. This composition is $N_1 (x)=\frac{-1}{2} x+\frac{1}{2}$. The initial term of $N_1$ does converge to $1$, so this partial base step is also true.
The initial term of $N_2$ is $-2$ which has trajectory $-2→-1→1$. This composition is $N_1∘N_2 (x)=\frac{-1}{4} x+\frac{1}{2}$. The initial term of $N_2$ does converge to $1$, so this partial base step is also true.
Since all the partial base steps are true, the base step is true.
The Inductive Step (updated)
Assume that an arbitrary positive integer $x_i$ will converge to $1$ under $f(x)$ after $k$ iterations, that is, there exists a composition $C(x_i)=\frac{3^m}{2^n}x_i+\frac{c}{2^n} =1$ . (This argument will also hold for a negative integer). We need to show that $x_i + 2$ will also converge.
This is as far as I can go.
On a different note however, looking at the digraph, I recognize that all the trajectories flow in one direction. This led me to wonder if once the initial terms converge, wouldn't all subsequent term converge, since iteration of $f$ on any subsequent term is an iteration on the initial term plus some integer. In other words, no number can flow in the other direction to diverge.
Any help is appreciated. Any related resource will also be appreciated.
This isn't what you want to show (it's obvious: the $(k+1)th$ iteration of $x_i$ is $f(1) = -1$, which you already covered in your base case). What you've shown is that if some $x_i$ converges to $1$, then $-1$ converges to $1$, which doesn't actually tell you anything about any points other than those in your base case.
I'd also object to "randomly" - that word has a meaning, and you are not using probabilistic or stochastic methods.