multi-index derivative of a power

Theorem If $i, k$ are multi-indices in $\mathbb{N}^{n}$ , and $x=(x_{1},\ldots,x_{n})$ , then

\displaystyle\partial^{i}x^{k}=\left\{\begin{array}[]{ll}\frac{k!}{(k-i)!}x^{k% -i}&\mbox{if}\,\,i\leq k,\\ 0&\mbox{otherwise}.\end{array}\right.

Proof. The proof follows from the corresponding rule for the ordinary derivative; if $i, k$ are in $0,1,2,\ldots$ , then

\displaystyle\frac{d^{i}}{dx^{i}}x^{k}=\left\{\begin{array}[]{ll}\frac{k!}{(k-% i)!}x^{k-i}&\mbox{if}\,\,i\leq k,\\ 0&\mbox{otherwise.}\end{array}\right.

(1)

Suppose $i=(i_{1},\ldots,i_{n})$ , $k=(k_{1},\ldots,k_{n})$ , and $x=(x_{1},\ldots,x_{n})$ . Then we have that

	$\displaystyle\partial^{i}x^{k}$	$\displaystyle=$	$\displaystyle\frac{\partial^{\|i\|}}{\partial x_{1}^{i_{1}}\cdots\partial x_{n}^% {i_{n}}}x_{1}^{k_{1}}\cdots x_{n}^{k_{n}}$
		$\displaystyle=$	$\displaystyle\frac{\partial^{i_{1}}}{\partial x_{1}^{i_{1}}}x_{1}^{k_{1}}\cdot% \cdots\cdot\frac{\partial^{i_{n}}}{\partial x_{n}^{i_{n}}}x_{n}^{k_{n}}.$

For each $r=1,\ldots,n$ , the function $x_{r}^{k_{r}}$ only depends on $x_{r}$ . In the above, each partial differentiation $\partial/\partial x_{r}$ therefore reduces to the corresponding ordinary differentiation $d/dx_{r}$ . Hence, from equation 1, it follows that $\partial^{i}x^{k}$ vanishes if $i_{r}>k_{r}$ for any $r=1,\ldots,n$ . If this is not the case, i.e., if $i\leq k$ as multi-indices, then for each $r$ ,

\frac{d^{i_{r}}}{dx_{r}^{i_{r}}}x_{r}^{k_{r}}=\frac{k_{r}!}{(k_{r}-i_{r})!}x_{% r}^{k_{r}-i_{r}},

and the theorem follows. $\Box$

Title	multi-index derivative of a power
Canonical name	MultiindexDerivativeOfAPower
Date of creation	2013-03-22 13:42:01
Last modified on	2013-03-22 13:42:01
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	9
Author	matte (1858)
Entry type	Theorem
Classification	msc 05-00