I am learning how $Z$-Transforms work, but I have no encountered a situation in which the bound does not account for any signal.
Take for example the following discrete time signal: \begin{cases} x(-2)=4\\ x(-1)=0\\ x(0)=2\\ x(1)=4\\ x(2)=6\\ x(3)=8\\ x(4)=-2\\ \end{cases} I need to find the $Z$-transform for $x(n)=0$ for $n$ from $-\infty$ to $-3$ and also for $n$ from $5$ to $\infty$.
My answer for this would be $0$ for both cases.
Is this correct?
For finding the $Z$ transform $\sum_{n=-\infty}^{\infty}x(n)z^{-n}$ you need the value of a sequence $x(n)~\forall n\in\Bbb Z$. So those "two cases" are actually the description of the same sequence.
The $Z$-transform is given by $x(-2)z^2+x(-1)z^1+\cdot\cdot\cdot+x(4)z^{-4}=4z^2+2+4/z+6/z^2+8/z^3-2/z^4$ since the remaining $x(i)$'s are all zero.