Can a parameter of a function be NULL? Can a parameter of a function be NULL like in the example below?
$
 f_{AE}(t_x;i_1;i_2;p)= 
  \begin{cases} 
      \hfill \frac{\displaystyle\sum_{k=i_1}^{i_2} R_s(e_{c_k,t_x};a_k)}{p}    \hfill & \text{if $p \neq NULL \land R_s(e_{c_k,t_x};a_k)= p$} \\
      i_2 - i_1    \hfill & \text{if $p = NULL$} \\
  \end{cases}
$
Call option 1:
$f_{AE}(t_1;5;10)$
Call option 2:
$f_{AE}(t_1;5;10;5)$
Or do I need two completly separate functions with different parameters?
thx
 A: The inputs to a function can come from any set you like. There is nothing wrong with saying $p$ belongs to say, the set 
$$\{\text{NULL}, 5,\}$$
But of course, care is needed to make sure you have a well-defined function.
[edit] I see from the comments that you don't just want it to be NULL, you want to completely omit the parameter.
You can do this if you tell your audience that you are doing it, but it is highly non-standard. This is because mathematicians have been taught to curry :) e.g. $A$ is a matrix/linear map, and $A(x)$ is a vector.
A: The short answer is YES, $p$ can take on the "value" NULL for the function you have provided.
For this function, you have allowed $p$ to contribute to the value if and only if it is not NULL. This implies that it has a value in your domain.
If $p$ is NULL, it should not contribute to the value of the function, because it wouldn't make any sense. Consider $$1+NULL = ???,$$ which is meaningless. Most computer languages interpret any math using NULL to equal NULL, so that
$$1+ NULL = NULL.$$
This is accounted for in your function by making the function equal $i_2-i_1$ when $p$ is NULL, which does not depend on $p$.
