Help with function $f_r(x^q)=q^rx^{q-1}$

Question

Let $r,q$ be a positive integers. I am looking for a function $f_r(x^q)$ such that it is satisfied $$ f_r (x^q)=q^r x^{q-1}$$ (without explicit dependence on $q$ of course, and for $r>1$).

I started by considering $f_r(x^q)=x^{r-1}(x^q)^{(r)}$, and then tried to subtract the terms that appear in the derivative. But I cannot find a closed/brief expression.

Any help is appreciated.

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A bit of what I've tried: I know $(x^q)^{(r)}=q(q-1)\cdots (q-r+1)x^{q-r}=\frac{q!}{(q-r)!}x^{q-r}$, where I can see the factor $q^r$, next if I multiply by $x^{r-1}$ I naturally get $$x^{r-1}(x^q)^{(r)}=q(q-1)\cdots (q-r+1)x^{q-1},$$ but so far I can't get rid off the unwanted terms. It is important that I do not make explicit use of $q$ so I suspect some iteration must be involved.

Answer 1

Probably not what you want, but consider this:

Let $h : \mathbb R[X] \to \mathbb R[X]$ be defined by $$h(P)= X P'(X) \,.$$

Define now $h_1=h$ and recursively $$h_r=h_{r-1} \circ h \,.$$

Then, $h, h_r$ are independent of $q$ and satisfy $$h_r(X^q)=q^rX^q$$

Take $f_r(P(X)) = \frac{h_r(P(X))}{X}$.

Note that $h$ is a linear function and $h(\mathbb P_n) \subset \mathbb P_n$, where $\mathbb P_n$ is the subspace of polynomials of degree at most $n$. Then, you can represent $h$ as a matrix, and composition becomes power of the corresponding matrix.

And if needed, to make sense of division by $X$ you can either go to rational functions, or restrict $h$ to the subspace of polynomials for which $P(0)=0$.

Answer 2

Reading this problem again, here is a simple construction to map the monomial $x^q$ to $q^{r-1}x^{q-1}$. Consider the formal derivative operator $D(x^q) =q x^{q-1}$; then $(Dx)(x^q)=(q+1)x^q$. Consequently $$(Dx)^r D(x^q) = (Dx)^r(qx^{q-1})=(Dx)^{r-1}q^2 x^{q-1}=\cdots = q^r x^{r-1}.$$ (Equivalently we may write this as $D(xD)^{r-1}$.)