I'm just trying to understand how ordinary generating functions works.. For example, how can I get the sequence of this OGF:
$F(z) = \frac{z^{2}}{(1-z)^3}$
I organized it as follows:
$F(z)=\frac{z^{2}}{(1-z)^3}=z^2\left ( \frac{1}{(1-z)^3} \right )=z^2\left ( \frac{1}{(1-z)}\frac{1}{(1-z)}\frac{1}{(1-z)} \right)$
because I know the sequence of $\frac{1}{1-z}$, witch is $1, 1, 1, ..., 1, ...$
However, I'm not seeing a way to develop this any further and find a sequence.
For example, I know if $\frac{1}{(1-z)}\frac{1}{(1-z)}\frac{1}{(1-z)}$ generates the sequence $a_0, a_1, ..., a_n, ...$, when I multiply it by $z^2$ I will get the sequence $0, 0, a_0, a_1, ..., a_n, ...$, but I have no idea how to obtain the values for $a_0, a_1, ..., a_n$
Since $$F(z)=\sum_{n\ge0}a_nz^n=\frac{z^2}{(1-z)^3}$$ the idea is then to extract $a_n$ from the power series expansion of $\frac{z^2}{(1-z)^3}$.
One way to do it is to compute the first few derivatives and compare their values at $z=0$ to the coefficients of the corresponding Taylor series centered at $z=0$:
$$F(z)=\sum_{n\ge0}\frac{F^{(n)}(0)}{n!}z^n$$
We have
$$\begin{align*} F(z)=\frac{z^2}{(1-z)^3}&\implies F(0)=0\implies a_0=0\\[1ex] F'(z)=\frac{z^2+2z}{(1-z)^4}&\implies F'(0)=0\implies a_1=0\\[1ex] F''(z)=\frac{2z^2+8z+2}{(1-z)^5}&\implies F''(0)=2\implies a_2=1 \end{align*}$$
and so on.
Another method would be to manipulate (mainly via differentiation) the well-known geometric series,
$$\sum_{n\ge0}z^n=\frac1{1-z}$$
until it matches $F(z)$. This is easy to do if you first decompose $F(z)$ into partial fractions:
$$\frac{z^2}{(1-z)^3}=\frac1{1-z}-\frac2{(1-z)^2}+\frac1{(1-z)^3}$$
Then noticing that
$$\frac{\mathrm d}{\mathrm dz}\left[\frac1{1-z}\right]=\frac1{(1-z)^2}=\sum_{n\ge0}(n+1)z^n$$
$$\frac{\mathrm d^2}{\mathrm dz^2}\left[\frac1{1-z}\right]=\frac2{(1-z)^3}=\sum_{n\ge0}(n+1)(n+2)z^n$$
we find
$$F(z)=\sum_{n\ge0}\left(1-2(n+1)+\frac{(n+1)(n+2)}2\right)z^n=\sum_{n\ge0}\frac{n(n-1)}2z^n$$
$$\implies\{a_n\}_{n\ge0}=\{0,0,1,3,6,10,15,\ldots\}$$