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 $a_{i, k} \in R$:
- $p_i = \ds \sum_{k \mathop \in Z} a_{i, k} X^k$
where $Z$ comprises the multiindices of natural numbers over $J$.
Then:
- $\ds \prod_{i \mathop = 1}^n p_i = \sum_{k \mathop \in Z} b_k X^k$
where:
- $\ds b_k := \sum_{k_1 + \cdots + k_n \mathop = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }$
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:
- $\ds \prod_{i \mathop = 1}^{n - 1} p_i = \sum_{k \mathop \in Z} c_d X^k$
where:
- $\ds c_d := \sum_{k_1 + \cdots + k_{n - 1} = d} \paren {\prod_{i \mathop = 1}^{n - 1} a_{i, k_i} }$
Now we need to show that the result is true for the product $\ds \prod_{i \mathop = 1}^n p_i$.
Induction Step
This is our induction step:
Let $b_k$ be the coefficient of $X^k$ in $\ds \prod_{i \mathop = 1}^n p_i$.
Then:
\(\ds b_k\) | \(=\) | \(\ds \sum_{d + k_n \mathop = k} c_d a_{n, k_n}\) | Definition of Multiplication of Polynomial Forms | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{d + k_n \mathop = k} \ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^{n - 1} a_{i, k_i} } a_{n, k_n}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{d + k_n \mathop = k} \ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\substack {k_1 + \dotsb + k_{n - 1} \mathop = d \\ d + k_n \mathop = k} } \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k_1 + \cdots + k_n = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }\) |
The result follows by induction.
$\blacksquare$