# Binomial Theorem/Multiindex

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## Theorem

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

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

Then:

- $\displaystyle \left({x + y}\right)^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}$

where $\displaystyle {n \choose k}$ is a binomial coefficient.

## Proof

First of all, by definition of multiindexed powers:

- $\displaystyle \left(x + y\right)^\alpha = \prod_{k \mathop = 1}^n\left(x_k + y_k\right)^{\alpha_k}$

Then:

\(\ds \left(x + y\right)^\alpha\) | \(=\) | \(\ds \prod_{k \mathop = 1}^n \sum_{\beta_k \mathop = 0}^{\alpha_k} {\alpha_k \choose \beta_k} x_k^{\alpha_k - \beta_k} y_k^{\beta_k}\) | Binomial Theorem for Integral Indices | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} {\alpha_1 \choose \beta_1} \cdots {\alpha_n \choose \beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}\) | expanding out the product |

On the other hand:

\(\ds \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}\) | \(=\) | \(\ds \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha_1 \choose \beta_1} \cdots {\alpha_n \choose \beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}\) | by definition of Power and Binomial Coefficient | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} {\alpha_1 \choose \ell_1} \cdots {\alpha_n \choose \ell_n} x_1^{\alpha_1 - \ell_1} \cdots x_n^{\alpha_n - \ell_n} y_1^{\ell_1} \cdots y_n^{\ell_n}\) | by definition of Ordering on Multiindices |

This shows that:

- $\displaystyle \left({x + y}\right)^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}$

as required.

$\blacksquare$