I'm reading "Introduction to Set Theory" by Hrbacek and Jech(1999, 3rd ed). On page 47, the recursion theorem is stated as follows.
The recursion theorem: For any set $A$, any $a \in A$, and any function $g : A \times N \to A$, there exists a unique infinite sequence $f : N \to A$ such that
(a) $f_0 = a$;
(b) $f_{n+1} = g(f_n,n)$ for all $n \in N$
The author says that "the proof of the Recursion Theorem consists of devising an explicit definition of $f$. (p. 48)" In case of the function factorial, an explicit definition could be given like this:
$$f_0 = 1 \quad\text{ and }\quad f_m = 1 \times 2 \times \cdots\times (m-1) \times m \;\text{ if }\; m \neq 0 \;\text{ and }\; m \in N \; \cdots \; (1)$$
The problem with this kind of formulation, the author says, is that the ellipsis "$\cdots$" is not precise enough. The solution, according to the book, is to state $f_m$ as the result of a computation.
$$1$$ $$1\times1$$ $$[1\times1]\times2$$ $$[1\times1\times2]\times3$$ $$\vdots$$ $$[1\times1\times2\times\cdots\times(m-1)]\times m$$
Above is an m-step computation t which means that it is a "finite sequence that is of length $m + 1$ where $t_0 = 1$ and $t_{k+1} = t_k \times (k + 1) = g(t_k,k)$ for all $k < m, k \geq 0.$" The rigorous explicit definition of $f$ then is:
$$f_m = t_m$$ where $t$ is an m-step computation (based on $a = 1$ and $g$)
My question is, I don't understand why the ellipsis "$\cdots$" in (1) is not precise enough. What does "precise" mean? How does reformulating it in terms of a computation make it any more "precise"?
When there is a disagreement while playing any game, the participants must be able to get out the 'rule book' to decide how to proceed.
What if a famous mathematician writing out a proof that uses the ellipsis in a much more complex context. Are they following the rules?
Can you see this one "in your own mind":
$a + ar + ar^2 + \ldots + ar^{n-1} = \frac{ a( 1- r^n) } { (1-r) }$
Must we force another person looking at this to see it the way we do, as a
'telescoping cancel/out deal'.