I just came across this post, but I can't quite follow it.
What is $(A(z))^t$ and why does it hold that $(A(z))^t = \sum_{n \geq 0}c_nz^n$. Further, I think that $(A(z))^t * z = (A(z))$, but how does this help me to find a functional equation for $(A(z))$?
I'm new to the topic of generating functions and just tried to find some examples here that I can use to get familiar with it, sorry if I miss something obvious.
Any help is appreciated!
The generating function of a sequence $a_n$ is the polynomial $A(z) = \sum_{n\geq 0} a_n z^n$. This polinomial is formal, you don't need to interpret it as a function.
The expression $(A(z))^t$ means you are multiplying polynomial $A(z)$ $t$ times. Since $A(z)$ has infinitely many terms, this might be confusing at first, but multiplication is carried out as you would do with finite polynomials:
\begin{equation} (a_0 + a_1 z + a_2 z^2 + \ldots )^2 = a_0^2 + (a_0 a_1 + a_1 a_0)z + (a_0 a_2 + a_1 a_1 + a_2 a_0) z^2 + \ldots \end{equation}
Notice that the coeficient of $z^n$ in $A(z)^2$ is $\sum_{k=0}^n a_k a_{n-k}$. When computing the $t$-th power, we have an analogous result. The coeficient $c_n$ of $z^n$ in $A(z)^t$ is given by
\begin{equation} c_n = \sum_{n_1 + n_2 + \ldots + n_t = n} a_{n_1} a_{n_2} \ldots a_{n_t} \end{equation}
as is written in equation (3) of the previous post. The important feature of generating functions is that you can translate some relations between terms of the sequence $a_n$ into relations between its generating function $A(z)$. In equation (1) of the previous post it was proved one of this relations. He proved \begin{equation} (A(z))^t = \sum_{n \geq 0} a_{n+1}z^n = z^{-1}\left( \sum_{n\geq 1} a_n z^n \right) = z^{-1}(A(z) - 1). \end{equation}