How to formulate the product of two generating functions without their final terms?

Question

I know that if we have two generating functions like so:

$A(z) = \sum_{n=0}^\infty a_nz^n$ and $B(z) = \sum_{n=0}^\infty b_nz^n$

Then we can write

$A(z)B(z) = \sum_{n=0}^\infty(a_0b_n + a_1b_{n-1} + ... + a_nb_0)z^n$

How can we formulate two generating functions $A(z)$ and $B(z)$ such that their product is:

$\sum_{n=0}^\infty(a_0b_{n-1} + a_1b_{n-2} + ... + a_{n-1}b_0)z^n$

I think this may be an issue in my understanding of generating functions. From what I understand, generating functions are infinite. But the equation above calls for the product of two generating functions without their final terms. Something like:

$ A(z) = (\sum_{n=0}^\infty a_nz^n) - a_nz^n $

But I know this notation doesn't make sense. Can anyone give me some guidance here?

Answer 1

We know that \begin{align*} a_0b_n+a_1b_{n-1}+\cdots+a_nb_0=\sum_{k=0}^na_kb_{n-k}\tag{1} \end{align*} is the coefficient of $z^n$ of $A(z)B(z)$ $\quad (n\geq 0)$

Now, if we replace $n$ with $n-1$ in (1) we get the expression \begin{align*} a_0b_{n-1}+a_1b_{n-2}+\cdots+a_{n-1}b_0=\sum_{k=0}^{n-1}a_kb_{(n-1)-k}\tag{2} \end{align*} which is the coefficient of $z^{n-1}$ of $A(z)B(z)$ when $n\geq 1$ (!).

But (2) is also the expression OP is searching for. Since the expression (2) should be the coefficient of $z^n$ of the product of appropriate generating functions we could simply take

\begin{align*} zA(z)\quad \text{ and }\quad B(z) \end{align*}

Indeed, we observe

\begin{align*} \sum_{n= 1}^{\infty}&(a_0b_{n-1}+a_1b_{n-2}+\cdots+a_{n-1}b_0)z^n\\ &=\sum_{n=1}^{\infty}\left(\sum_{k=0}^{n-1}a_kb_{n-1-k}\right)z^n\\ &=z\sum_{n=1}^{\infty}\left(\sum_{k=0}^{n-1}a_kb_{n-1-k}\right)z^{n-1}\\ &=z\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_{n-k}\right)z^{n}\tag{3}\\ &=zA(z)B(z) \end{align*}

In (3) the index $n$ was shifted by one.