PlanetMath: multinomial theorem (proof)

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multinomial theorem (proof)

(Proof)

Proof. The below proof of the multinomial theorem uses the binomial theorem and induction on . In addition, we shall use multi-index notation.

First, for , both sides equal . For the induction step, suppose the multinomial theorem holds for . Then the binomial theorem and the induction assumption yield

$\displaystyle (x_1+\cdots + x_k\,+\,x_{k+1})^n$	$\displaystyle =$	$\displaystyle \sum_{l=0}^n {n \choose l} (x_1+\cdots + x_k)^l x_{k+1}^{n-l}$
	$\displaystyle =$	$\displaystyle \sum_{l=0}^n {n \choose l} l! \sum_{\vert i\vert=l} \frac{x^i}{i!} x_{k+1}^{n-l}$
	$\displaystyle =$	$\displaystyle n! \sum_{l=0}^n \sum_{\vert i\vert=l} \frac{x^i x_{k+1}^{n-l}}{i! (n-l)!}$

where $x=(x_1,\ldots, x_k)$ and

is a multi-index in

. To complete the proof, we need to show that the sets

$\displaystyle A$	$\displaystyle =$	$\displaystyle \{ (i_1,\ldots,i_k, n-l)\in I^{k+1}_+ \mid l=0,\ldots, n,\, \vert(i_1,\ldots, i_k)\vert=l \},$
$\displaystyle B$	$\displaystyle =$	$\displaystyle \{j \in I^{k+1}_+ \mid \vert j\vert=n \}$

are equal. The inclusion $A \subset B$ is clear since

$\displaystyle \vert(i_1,\ldots,i_k, n-l)\vert = l + n-l = n.$

For $B \subset A$ , suppose $j=(j_1,\ldots, j_{k+1}) \in I^{k+1}_+$ , and $\vert j\vert=n$ . Let $l=\vert(j_1,\ldots, j_k)\vert$ . Then $l=n-j_{k+1}$ , so $j_{k+1} = n-l$ for some $l=0,\ldots, n$ . It follows that that

Let us define $y=(x_1,\cdots, x_{k+1})$ and let $j=(j_1,\ldots, j_{k+1})$ be a multi-index in $I_+^{k+1}$ . Then

$\displaystyle (x_1+\cdots + x_{k+1})^n$	$\displaystyle =$	$\displaystyle n! \sum_{\vert j\vert=n} \frac{x^{(j_1,\ldots, j_k)} x_{k+1}^{j_{k+1}}}{(j_1,\ldots, j_k)! j_{k+1}!}$
	$\displaystyle =$	$\displaystyle n! \sum_{\vert j\vert=n} \frac{y^j}{j!}.$