# Binomial Theorem/Multiindex

## Theorem

Let $\alpha$ be a multiindex, indexed by $\set {1, \ldots, n}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.

Let $x = \tuple {x_1, \ldots, x_n}$ and $y = \tuple {y_1, \ldots, y_n}$ be ordered tuples of real numbers.

Then:

$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$

where $\dbinom n k$ is a binomial coefficient.

## Proof

First of all, by definition of multiindexed powers:

$\ds \paren {x + y}^\alpha = \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{\alpha_k}$

Then:

 $\ds \paren {x + y}^\alpha$ $=$ $\ds \prod_{k \mathop = 1}^n \sum_{\beta_k \mathop = 0}^{\alpha_k} \dbinom {\alpha_k} {\beta_k} x_k^{\alpha_k - \beta_k} y_k^{\beta_k}$ Binomial Theorem for Integral Indices $\ds$ $=$ $\ds \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \dbinom {\alpha_1} {\beta_1} \cdots \dbinom {\alpha_n} {\beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}$ expanding out the product

On the other hand:

 $\ds \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$ $=$ $\ds \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom {\alpha_1} {\beta_1} \cdots \dbinom {\alpha_n} {\beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}$ Definition of Power and Definition of Binomial Coefficient $\ds$ $=$ $\ds \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \dbinom {\alpha_1} {\ell_1} \cdots \dbinom {\alpha_n} {\ell_n} x_1^{\alpha_1 - \ell_1} \cdots x_n^{\alpha_n - \ell_n} y_1^{\ell_1} \cdots y_n^{\ell_n}$ Definition of Ordering on Multiindices

This shows that:

$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$

as required.

$\blacksquare$