Product of Generating Functions

Theorem

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Let $\map H z$ be the generating function for the sequence $\sequence {b_n}$.

Then $\map G z \map H z$ is the generating function for the sequence $\sequence {c_n}$, where:

$\forall n \in \Z_{\ge 0}: c_n = \ds \sum_{k \mathop = 0}^n a_k b_{n - k}$

General Rule

Let $\map {G_0} z, \map {G_1} z, \map {G_2} z, \ldots$ be any number of generating functions (up to countably infinite) for the sequences $\sequence {a_0 n}, \sequence {a_1 n}, \sequence {a_2 n}, \ldots$

Then:

\(\ds \prod_{j \mathop \ge 0} \map {G_j} z\)

\(=\)

\(\ds \prod_{j \mathop \ge 0} \sum_{k \mathop \ge 0} a_{j k} z^k\)

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop \ge 0} z^n \sum_{\substack {k_0, k_1, k_2, \ldots \mathop \ge 0 \\ k_0 \mathop + k_1 \mathop + \mathop \cdots \mathop = n} } \paren {\prod_{j \mathop \ge 0} a_{j k} }\)

Proof

By definition of generating function:

$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$

$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$

Then:

$\begin {array}{c|ccccc}

\times & b_0 & b_1 z & b_2 z^2 & b_3 z^3 & \cdots \\ \hline a_0 & a_0 b_0 & a_0 b_1 z & a_0 b_2 z^2 & a_0 b_3 z^3 & \cdots \\ a_1 z & a_1 b_0 z & a_1 b_1 z^2 & a_1 b_2 z^3 & a_1 b_3 z^4 & \cdots \\ a_2 z^2 & a_2 b_0 z^2 & a_2 b_1 z^3 & a_2 b_2 z^4 & a_2 b_3 z^5 & \cdots \\ a_3 z^3 & a_3 b_0 z^3 & a_3 b_1 z^4 & a_3 b_2 z^5 & a_3 b_3 z^6 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end {array}$

The result follows by definition of generating function.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: $(6)$