Since I am not educated enough in the field of mathematical analysis of integer sequences, I am wondering if there is a difference between a sequence converging towards zero and a sequence entering an repeating cycle of $2,1$ or $4,2,1$ as in for example the Collatz functions? And is convergence towards 0 or convergence towards 1 in the same thing?
I read from wiki that "Leaving aside the cycle $0$ → $0$ which cannot be entered from outside". This is true for the most known Collatz functions, but what if one of these functions instead converges towards zero instead of 1 or the repeating cycles? Would it be any different in regards to finding a proof or analysing the iterated function? Or would such a function have any consequence on the conjecture itself?
I know zero could just be a symbol like any other integer, so I dont know how one would approach finding out if such a function converges? (since it may show zero for numbers up to $2^{10}$ instead of $2,1$ cycle).
To clarify the question a bit more:
Let $X$ be a set and $f:X\rightarrow X$ be a function.
Is proving convergence of $f^n(x) = 1\rightarrow4\rightarrow2\rightarrow1 ...$ any different from proving: $f^n(x) \rightarrow 0.$ ?
I dont know but is proving convergence towards $1$ or the cycle more difficult than a function that converges towards zero alone?
Also if $f(0) = 0$ is a fixedpoint of a Collatz function, what would that mean?
The number $x_0$ converges under iteration of $f$ to the finite cycle $1, 4, 2$ if and only if $x_0$ converges under iteration of $f^3$ to one of $1$, $4$ or $2$, where $f^3(x)=f(f(f(x)))$. This is a standard trick in dynamics to reduce the study of convergence of cycles to the study of convergence to fixed points (note that $1, 2$ and $4$ are all fixed points of $f^3$). In that sense, in any dynamical problem, the study of convergence to cycles is generally not any more difficult than the study of convergence to fixed points.
For the Collatz conjecture we want to prove that all starting points end in the cycle $1, 4, 2$, which is equivalent to proving that all starting points end at $1$, $4$ or $2$ under the iteration of $f^3$.