Does the equality $\partial^\alpha(x^\alpha)(0)=\alpha!$ hold?

Question

Do we have $\partial^\alpha(x^\beta)(0)=\alpha!=\beta!$ if $\alpha=\beta$ and $0$ else?

I tried to proof it on induction, can include my attempts if needed, but they seem to have failed anyway...

Answer 1

This is correct. The easiest way to do it is to separate the partial derivatives accordingly. Write $x = (x_1,\ldots,x_n)$ and $\alpha = (\alpha_1,\ldots, \alpha_n)$, then $\partial^{\alpha} = \partial_{x_1}^{\alpha_1}\cdots\partial_{x_n}^{\alpha_n}$. From this it isn't hard to see that you need only to consider the one-dimensional case since $x^{\alpha} = x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ and the partial derivatives commute. If you are unsure, note that

$$\partial^{\alpha} x^{\beta} = (\partial_{x_1}^{\alpha_1}\cdots \partial_{x_n}^{\alpha_n})(x_1^{\beta_1}\cdots x_n^{\beta_n}) = (\partial_{x_1}^{\alpha_1}(x_1^{\beta_1}))\cdots (\partial_{x_n}^{\alpha_n}(x_n^{\beta_n})).$$

Reducing ourselves to computing $\partial_{x_i}^{\alpha_i}(x_i^{\beta_i})$ greatly simplifies the problem at hand. Computing this, we have

$$\partial_{x_i}^{\alpha_i}(x_i^{\beta_i}) = \begin{cases} \dfrac{\beta_i!}{(\beta_i-\alpha_i)!}x^{\beta_i-\alpha_i} & \beta_i \ge \alpha_i \\ 0 & \beta_i < \alpha_i \end{cases}. $$

Taking $\beta_i = \alpha_i$, we get that $\partial_{x_i}^{\alpha_i}(x_i^{\beta_i})(0) = \alpha_i! = \beta_i!$ if $\alpha_i=\beta_i$ and $0$ otherwise. By definition, $\alpha! = \alpha_1!\cdots\alpha_n!$ so the result follows.