Problem 1845

[Solution] (1) Agreement
$$F_{k}(x)=(x-1) f_{k}(x)+f_{k}(a x)-a^{k} x f_{k}\left(\frac{x}{a}\right)$$

Firstly, we point out: $F_{k+1}(x)=x F_{k}(x)+F_{k}(a x)$. In fact,
$$\begin{aligned} & F_{k+1}(x)-x F_{k}(x) \\ = & (x-1) f_{k+1}(x)+f_{k+1}(a x)-a^{k+1} x f_{k+1}\left(\frac{x}{a}\right)-x(x- \\ & 1) f_{k}(x)-x f_{k}(a x)-a^{k} x^{2} f_{k}\left(\frac{x}{a}\right) \\ = & (x-1)\left(x f_{k}(x)+f_{k}(a x)\right)+\left(a x f_{k}(a x)+f_{k}\left(a^{2} x\right)\right)- \\ & \quad a^{k+1} x\left(\left(\frac{x}{a}\right) f_{k}\left(\frac{x}{a}\right)+f_{k}(x)\right)-x(x-1) f_{k}(x)- \\ & x f_{k}(a x)-a^{k} x^{2} f_{k}\left(\frac{x}{a}\right) \\ = & (a x-1) f_{k}(a x)+f_{k}\left(a^{2} x\right)-a^{k+1} x f_{k}(x) \\ = & F_{k}(a x) . \end{aligned}$$

Since $F_{0}(x)=0$, we have $F_{n}(x)=0, n=0,1,2, \cdots$.
Next, we prove by mathematical induction that
$$f_{n}(x)=x^{n} f_{n}\left(\frac{1}{x}\right), n=0,1,2, \cdots$$

First, it is clear that $f_{0}(x)=x^{0} f_{0}\left(\frac{1}{x}\right)$.
Assume that $f_{k}(x)=x^{k} f_{k}\left(\frac{1}{x}\right)$ has been proven. Then we have
$$\begin{aligned} & f_{k+1}(x)-x^{k+1} f_{k+1}\left(\frac{1}{x}\right) \\ = & f_{k+1}(x)-x^{k+1} f_{k+1}\left(\frac{1}{x}\right)-\left(f_{k}(x)-x^{k} f_{k}\left(\frac{1}{x}\right)\right) \\ = & (x-1) f_{k}(x)+f_{k}(a x)-x^{k+1}\left[\left(\frac{1}{x}\right) f_{k}\left(\frac{1}{x}\right)+f_{k}\left(\frac{a}{x}\right)\right]+ \\ & x^{k} f_{k}\left(\frac{1}{x}\right) \\ = & (x-1) f_{k}(x)+f_{k}(a x)-a^{k} \cdot x \cdot\left(\left(\frac{x}{a}\right)^{k} f_{k}\left(\frac{a}{x}\right)\right) \\ = & (x-1) f_{k}(x)+f_{k}(a x)-a^{k} x f_{k}\left(\frac{x}{a}\right) \\ = & F_{k}(x) \\ = & 0 \end{aligned}$$

That is,
$$f_{k+1}(x)=x^{k+1} f_{k+1}\left(\frac{1}{x}\right)$$

By the principle of mathematical induction, for any non-negative integer $n$, we have
$$f_{n}(x)=x^{n} f_{n}\left(\frac{1}{x}\right)$$
(2) From the given conditions, we know that the degree of $f_{k}(x)$ is no greater than $k$. Let us assume
$$f_{k}(x)=\sum_{j=1}^{k} b_{j}^{(k)} x^{j}, k=0,1,2, \cdots$$

From the given conditions, it is easy to see that
$$b_{0}^{(k)}=1, k=0,1,2, \cdots$$

From the conclusion proven in (1), we have
$$b_{k-j}^{(k)}=b_{j}^{(k)} \quad(k=0,1, \cdots ; j=0,1, \cdots k)$$

In particular,
$$b_{0}^{(k)}=b_{k}^{(k)}=1, k=0,1,2, \cdots$$

By comparing the coefficients of $x^{j}$ and $x^{n-j}$ on both sides of the equation
$$f_{n}(x)=x f_{n-1}(x)+f_{n-1}(a x)$$

we get
$$\left\{\begin{array}{l} b_{j}^{(n)}=b_{j-1}^{(n-1)}+a^{j} \cdot b_{j}^{(n-1)} \\ b_{n-j}^{(n)}=b_{n-j-1}^{(n-1)}+a^{n-j} b_{n-j}^{(n-1)} \end{array}\right.$$

That is, $\left\{\begin{array}{l}b_{j}^{(n)}=b_{j-1}^{(n-1)}+a^{j} \cdot b_{j}^{(n-1)}, \\ b_{j}^{(n)}=b_{j}^{(n-1)}+a^{n-j} b_{n-j}^{(n-1)} .\end{array}\right.$
By eliminating $b_{j}^{(n-1)}$ from the above system of equations, we get $\left(a^{j}-1\right) b_{j}^{(n)}=\left(a^{n}-1\right) b_{j-1}^{(n-1)}$,
Thus, we have
$$\begin{aligned} b_{j}^{(n)} & =\frac{a^{n}-1}{a^{j}-1} \cdot b_{j-1}^{(n-1)} \\ & =\frac{a^{n}-1}{a^{j}-1} \cdot \frac{a^{n-1}-1}{a^{j-1}-1} \cdot b_{j-2}^{(n-2)} \\ & \cdots \\ & =\frac{\left(a^{n}-1\right)\left(a^{n-1}-1\right) \cdots\left(a^{n-j+1}-1\right)}{\left(a^{j}-1\right)\left(a^{j-1}-1\right) \cdots(a-1)} b_{0}^{n-j} \\ & =\frac{\left(a^{n}-1\right)\left(a^{n-1}-1\right) \cdots\left(a^{n-j+1}-1\right)}{\left(a^{j}-1\right)\left(a^{j-1}-1\right) \cdots(a-1)} . \end{aligned}$$

Therefore, we obtain
$$f_{n}(x)=1+\sum_{j=1}^{n} \frac{\left(a^{n}-1\right)\left(a^{n-1}-1\right) \cdots\left(a^{n-j+1}-1\right)}{\left(a^{j}-1\right)\left(a^{j-1}-1\right) \cdots(a-1)} x^{j}$$