Answer to Question #95780 in Calculus for Rachel

Question

Let a and β be multi-indices such that β ≤ a. Prove that ∂^(β) x^(a) = (a,β) β!x^(a - β) if β≤a

Answer 1

First we prove statement for one-dimensional case, that is $(x^a)^{(b)}=\binom{a}bb!x^{a-b}$, where $b\le a$.

We apply mathematical induction on $b\le a$.

Indeed, $(x^a)^{(0)}=\binom{a}00!x^{a-0}=x^a$. Suppose that $(x^a)^{(b)}=\binom{a}bb!x^{a-b}$ for all $b\le k$ , where $0\le k\le a$.

If $k=a$ , then our statement is proved. If $k<a$ , then $(x^a)^{(k+1)}=\bigl((x^a)^{(k)}\bigr)'$ .

By the induction hypothesis we have $(x^a)^{(k)}=\binom{a}kk!x^{a-k}$ , so $(x^a)^{(k+1)}=\bigl((x^a)^{(k)}\bigr)'=\bigl(\binom{a}kk!x^{a-k}\bigr)'=$

$=\binom{a}kk!\cdot(a-k)x^{a-k-1}$

Consider $\binom{a}kk!\cdot(a-k)$ .

$\binom{a}kk!\cdot(a-k)=\frac{a!}{k!(a-k)!}k!(a-k)=\frac{a!}{(a-k-1)!}=$

$=\frac{a!}{(a-k-1)!(k+1)!}(k+1)!=\binom{a}{k+1}(k+1)!$

So $(x^a)^{(k+1)}=\binom{a}kk!\cdot(a-k)x^{a-k-1}=\binom{a}{k+1}(k+1)!x^{a-k-1}$ .

By the mathematical induction we obtain $(x^a)^{(b)}=\binom{a}bb!x^{a-b}$ for all $b\le a$ .

Consider multi-indices. We prove by induction on $n$ that $\frac{\partial^{|\beta|} x^\alpha}{\partial x^\beta}=\binom{\alpha}{\beta}\beta!x^{\alpha-\beta}$ , where $x=(x_1,\ldots,x_n)$, $\alpha=(\alpha_1,\ldots,\alpha_n)$, $\beta=(\beta_1,\ldots,\beta_n)$ and $\beta\le\alpha$.

We proved above this statement for $n=1$. Suppose that the statement is true for $n=k$, where $k\ge 1$ .

Let $x=(x_1,\ldots,x_k,x_{k+1})$, $\alpha=(\alpha_1,\ldots,\alpha_k,\alpha_{k+1})$, $\beta=(\beta_1,\ldots,\beta_k,\beta_{k+1})$ and $\beta\le\alpha$.

Introduce new denotations: $x_0=(x_1,\ldots,x_k)$, $\alpha_0=(\alpha_1,\ldots,\alpha_k)$, $\beta_0=(\beta_1,\ldots,\beta_k)$.

We have $\frac{\partial^{|\beta|} x^\alpha}{\partial x^\beta}=\frac{\partial^{|\beta_0|+\beta_{k+1}} x_0^{\alpha_0}x_{k+1}^{\alpha_{k+1}}}{\partial x_0^{\beta_0}\partial x_{k+1}^{\beta_{k+1}}}=\frac{\partial^{\beta_{k+1}}}{\partial x_{k+1}^{\beta_{k+1}}}\bigl(\frac{\partial^{|\beta_0|} x_0^{\alpha_0}x_{k+1}^{\alpha_{k+1}}}{\partial x_0^{\beta_0}}\bigr)=$ $=\frac{\partial^{\beta_{k+1}}}{\partial x_{k+1}^{\beta_{k+1}}}\bigl(x_{k+1}^{\alpha_{k+1}}\frac{\partial^{|\beta_0|} x_0^{\alpha_0}}{\partial x_0^{\beta_0}}\bigr)=\frac{\partial^{|\beta_0|} x_0^{\alpha_0}}{\partial x_0^{\beta_0}}\frac{\partial^{\beta_{k+1}} x_{k+1}^{\alpha_{k+1}}}{\partial x_{k+1}^{\beta_{k+1}}}$Since $\beta\le\alpha$, we obtain $\beta_0\le\alpha_0$ and $\beta_{k+1}\le\alpha_{k+1}$, so $\frac{\partial^{\beta_{k+1}} x_{k+1}^{\alpha_{k+1}}}{\partial x_{k+1}^{\beta_{k+1}}}=\binom{\alpha_{k+1}}{\beta_{k+1}}\beta_{k+1}!x_{k+1}^{\alpha_{k+1}-\beta_{k+1}}$ and by induction hypothesis we have $\frac{\partial^{|\beta_0|} x_0^{\alpha_0}}{\partial x_0^{\beta_0}}=\binom{\alpha_0}{\beta_0}\beta_0!x_0^{\alpha_0-\beta_0}$ . We obtain $\frac{\partial^{|\beta|} x^\alpha}{\partial x^\beta}=\frac{\partial^{|\beta_0|} x_0^{\alpha_0}}{\partial x_0^{\beta_0}}\frac{\partial^{\beta_{k+1}} x_{k+1}^{\alpha_{k+1}}}{\partial x_{k+1}^{\beta_{k+1}}}=$ $=\binom{\alpha_0}{\beta_0}\beta_0!x_0^{\alpha_0-\beta_0}\cdot\binom{\alpha_{k+1}}{\beta_{k+1}}\beta_{k+1}!x_{k+1}^{\alpha_{k+1}-\beta_{k+1}}=$ $=\binom{\alpha_0}{\beta_0}\binom{\alpha_{k+1}}{\beta_{k+1}}\beta_0!\beta_{k+1}!x_0^{\alpha_0-\beta_0}x_{k+1}^{\alpha_{k+1}-\beta_{k+1}}=\binom{\alpha}{\beta}\beta!x^{\alpha-\beta}$ By the mathematical induction we obtain $\frac{\partial^{|\beta_0|} x_0^{\alpha_0}}{\partial x_0^{\beta_0}}=\binom{\alpha}{\beta}\beta!x^{\alpha-\beta}$ for all $n$, $x=(x_1,\ldots,x_n)$, $\alpha=(\alpha_1,\ldots,\alpha_n)$, $\beta=(\beta_1,\ldots,\beta_n)$, where $\beta\le\alpha$ .