# Coefficients of Polynomial Product

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## Theorem

Let $J$ be a set.

Let $p_1, \ldots p_n$ be polynomial forms in the indeterminates $\set {X_j : j \in J}$ over a commutative ring $R$.

Suppose that for each $i$ with $1 \le i \le n$, we have, for appropriate $a_{i, k} \in R$:

$p_i = \displaystyle \sum_{k \mathop \in Z} a_{i, k} X^k$

where $Z$ comprises the multiindices of natural numbers over $J$.

Then:

$\displaystyle \prod_{i \mathop = 1}^n p_i = \displaystyle \sum_{k \mathop \in Z} b_k X^k$

where:

$\displaystyle b_k := \sum_{k_1 + \cdots + k_n \mathop = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$

## Proof

We proceed by induction over $n \ge 1$.

### Basis for the Induction

If $n = 1$ the result is trivially true.

This establishes the basis for the induction.

### Induction Hypothesis

This is our induction hypothesis:

$\displaystyle \prod_{i \mathop = 1}^{n - 1} p_i = \displaystyle \sum_{k \mathop \in Z} c_d X^k$

where:

$\displaystyle c_d := \sum_{k_1 + \cdots + k_{n - 1} = d} \paren {\prod_{i \mathop = 1}^{n - 1} a_{i, k_i} }$

Now we need to show that the result is true for the product $\displaystyle \prod_{i \mathop = 1}^n p_i$.

### Induction Step

This is our induction step:

Let $b_k$ be the coefficient of $X^k$ in $\displaystyle \prod_{i \mathop = 1}^n p_i$.

Then:

 $\displaystyle b_k$ $=$ $\displaystyle \sum_{d + k_n \mathop = k} c_d a_{n, k_n}$ Definition of Multiplication of Polynomial Forms $\displaystyle$ $=$ $\displaystyle \sum_{d + k_n \mathop = k}\ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^{n-1} a_{i, k_i} } a_{n, k_n}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{d + k_n \mathop = k}\ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$ $\displaystyle$ $=$ $\displaystyle \sum_{\substack {k_1 + \dotsb + k_{n - 1} \mathop = d \\ d + k_n \mathop = k} } \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$ $\displaystyle$ $=$ $\displaystyle \sum_{k_1 + \cdots + k_n = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$

The result follows by induction.

$\blacksquare$