2
$\begingroup$

The following is for an exam preparation exercise in induction

Problem: Let $f(x)=|x| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2}$ and $g(x)=|x|^{2k+1}$.
Let $N \leq n$ and let $k \in \mathbb{Z}$, show by the principle of complete induction that $\forall x \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$ $$ \large \frac{\partial^Ng(x) }{\partial x_{i_1}\partial x_{i_2}\cdot \dots \cdot \partial_{x_{i_N}}} = cx_{i_1} x_{i_2} \dots x_{i_N} |x|^{2k+1-2N} \tag{*} $$ You can assume that all indices $x_{i_j}$ are different

My approach: I assume that this exercise is fairly straightforward, however the notation of double indices ruins it for me.

I suppose that I can just fix an $i \in \lbrace 1, \dots , n \rbrace$ and compute the partial derivate with respect to said index. Large $N$ then just denotes the order of partial derivatives being computed and since all $x_{i_j}$ are supposed to be different I end up with all possible permutations.


Induction over $n \in \mathbb{N}$

Base case: I can "cheat" here and say let $n=0$ then the statement is trivial, namely $g(x)=g(x)$. Alternatively I could also assume that $N \leq n =1$ and for $N=1$ simply take the partial derivative with respect to $x_i$ of $g$ and see that the relation (*) holds.

Induction step: $n \leadsto n+1$. Suppose that * is true $\forall x \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$ and $N \leq n$. I want to show that * is still true for $n+1$ that is $N \leq n+1$. For $N \leq n$, I know that (*) is true by assumption, so let $N = n+1$

I want to compute the $N+1$-th partial derivative of $g$ with respect to $x_{i_{N+1}}$

I suppose I have to do that by using the definition of the partial derivative, maybe I can directly apply it to the right hand side of (*) however (but I wouldn't know why)

Let me denote the left hand side of (*) by $H_N(x)$ then $$H_{N+1}(x)= \lim_{h \to 0} \frac{H_N(x_1,x_2 , \dots , x_i + h, \dots x_n , x_{n+1})-H_N(x_1, x_2, \dots , x_i , \dots, x_{n+1})}{h} $$ So I think it is about time to make use of the induction hypothesis and plug it into the expression above $$\lim_{h \to 0} \frac{cx_{i_1}\cdot ...\cdot (x_{i_i}+h)\cdot ... \cdot x_{i_N} f(x_1, \dots , x_i + h, \dots ,x_n)^{2k+1-2N}-c x_{i_1}... x_{i_N}f(x)^{2k+1-2N}}{h} $$

I could multiply the above expression out, but I wouldn't know how to proceed, furthermore I believe that my process was entirely wrong, because I wont obtain the desired expression for $N+1$ in (*)


Update: I suppose for the induction step I should rather try to argument like this, if * is valid and $x \in \mathbb{R}^{n+1} \setminus \lbrace 0 \rbrace$ then I can already tell because of * that the $n$-th partial derivative of $g$ is continuous and therefore by the Lemma of Schwarz I can freely interchange any order of computation of the mixed partial derivatives.

If I then take $\frac{\partial}{\partial x_{i_{N+1}}}$ of the right hand side, the result would follow.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.