Examples of partial functions in which the domain is not known? I was reading this, it mentions about a kind of function in which the exact domain is not known. The only example given is this one - and I'm not really sure I understood it. I got curious about it:


*

*What are examples of these functions? (beyond the one I suggested as an example)

*What would be needed to know their exact domain?


I got curious about it because I'm an undergraduate student, every function I've seen had a known domain. The idea of not having a domain is completely alien to me. Notice that It's notvery clear what exact domain is, I have in mind only the discrepancy of domain/image. 
 A: Consider the partial function which takes as input a Turing machine and returns as output its halting time if it has one. The exact domain of this function, namely the set of Turing machines which have halting times, is in a pretty strong sense impossible to know because the halting problem is undecidable. 
A: The Collatz conjecture provides a nice example:
$$
f(n) = \begin{cases}
1 & \text{ for } n = 1, \\
f(n/2) & \text{ for } n \text{ even}, \\
f(3n+1) & \text{ for } n \text{ odd},
\end{cases}
$$
for any positive $n \in \mathbb{N}$. The conjecture is (equivalent to) that the domain of $f$ is $\mathbb{N}\setminus\{0\}$, but we don't know it. However, if it is true, then $f \equiv (n \mapsto 1)$.
I hope this helps $\ddot\smile$
A: You can build such a function whenever you have a function $f$ whose zeros are unknown. For example, set $$
  f(z) = \frac{1}{\zeta(z)}
$$
where $\zeta(z)$ is the Riemann zeta function. It is conjectured, but not proven, that for all zeros of $\zeta(z)$, either $z = -2n$, $n \in \mathbb{N}$ (these are called the trivial zeros) or $\textrm{Re} z = \frac{1}{2}$ (this is what is unknown). It has been proved that all non-trivial zeros lie in the strip $0 < \textrm{Re} z < 1$, so we currently know that the domain of $f$ is at least $$
  \textrm{dom } f = \{z \in \mathbb{C} \mid -\tfrac{1}{2}z \notin \mathbb{N} \textrm{ and } \textrm{Re } z \notin (0,1)\} \text{.}
$$
If the riemann hypothesis turns out to be true, the domain can be enlarged to $$
  \textrm{dom } f = \{z \in \mathbb{C} \mid -\tfrac{1}{2}z \notin \mathbb{N} \textrm{ and } \textrm{Re } z \neq \tfrac{1}{2}\} \text{.}
$$
A: Suppose $f\subset A\times B$. Then $f$ is a partial function if and only if for all $x\in A$ and all $y_1,y_2\in B$, if $(x,y_1)\in f$ and $(x,y_2)\in f$ then $y_1=y_2$. (Unlike others, this definition does not specify a domain for $f$. Implicitly, the domain is some subset of $A$.)
You can still use the functional notation $f(x)=y$, but you cannot assume that $f(x)$ is defined for every $x\in A$.
Examples:
$f= \{(2,9), (5, 7)\}\subset N^2$ is a partial function on $N$ where $f(2)=9$ and $f(5)=7$.
$g= \{(2,9), (2, 7)\}\subset N^2$ is not a partial function on $N$ since $2$ is mapped to different values.
$h= \{(9,2), (7,2)\}\subset N^2$ is a partial function on $N$  even though different elements of $N$ are mapped to the same value, where $h(9)=2$ and $h(7)=2$.
