给定实数 $a$，设实多项式序列 $|f_n(x) |$ 满足 $\begin{cases}
f_0(x)=1\\
f_{n+1}(x)=xf_n(x)+f_n(ax),n=0,1,2,\cdots\\
\end{cases}$
（1）求证：$f_n(x)=x^nf_n(\dfrac{1}{x}),n=0,1,2,\cdots$
（2）求 $f_n(x)$ 的明显表达式．
f_0(x)=1\\
f_{n+1}(x)=xf_n(x)+f_n(ax),n=0,1,2,\cdots\\
\end{cases}$
（1）求证：$f_n(x)=x^nf_n(\dfrac{1}{x}),n=0,1,2,\cdots$
（2）求 $f_n(x)$ 的明显表达式．
【难度】
【出处】
1999第14届CMO试题
【标注】
【答案】
略
【解析】
（i）约定记 $F_k(x)=(x-1)f_k(x)+f_k(ax)-a^kxf_k(\dfrac{x}{a})$ 首先指出 $F_{k+1}(x)=x F_{k}(x)+F_{k}(a x)$ 事实上
$F_{k+1}(x)-x F_{k}(x)=(x-1) f_{k+1}(x)+f_{k+1}(a x)-a^{k+1} x f_{k+1}\left(\dfrac{x}{a}\right)-x(x-1) f_{k}(x)-x f_{k}(a x)+a^{k} x f_{k}(a)\\
=(x-1)\left(x f_{k}(x)+f_{k}(a x)\right)+\left(a x f_{k}(a x)+f_{k}\left(a^{2} x\right)\right)-a^{k+1} x\left(\dfrac{x}{a} f_{k}\left(\dfrac{x}{a}\right)+f_{k}(x)\right)-x(x-1) f_{k}(x)-$
$x f_{k}(a x)+a^{k} x^{2} f_{k}\left(\dfrac{x}{a}\right)=(a x-1) f_{k}(a x)+f_{k}\left(a^{2} x\right)-a^{k+1} x f_{k}(x)=F_{k}(a x)$ 因为 $F_{0}(x)=0$ 所以 $F_{n}(x)=0, n=0,1,2, \cdots$
（ii）利用（i）中的结论，我们归纳证明 $f_{k}(x)=x^{n} f_{n}\left(\dfrac{1}{x}\right), n=0,1,2, \cdots$
首先，显然 $f_{0}(x)=x^{0} f_{0}\left(\dfrac{1}{x}\right)$ 假定已证得 $f_{k}(x)=x^{k} f_{k}\left(\dfrac{1}{x}\right)$ 则有
$f_{k+1}(x)-x^{k+1} f_{k+1}\left(\dfrac{1}{x}\right)=f_{k+1}(x)-x^{k+1} f_{k+1}\left(\dfrac{1}{x}\right)-\left(f_{k}(x)-x^{k} f_{k}\left(\dfrac{1}{x}\right)\right)$
$=(x-1) f_{k}(x)+f_{k}(a x)-x^{k+1}\left(\dfrac{1}{x} f_{k}\left(\dfrac{1}{x}\right)+f_{k}\left(\dfrac{a}{x}\right)\right)+x^{k} f_{k}\left(\dfrac{1}{a}\right) )$
$=(x-1) f_{k}(x)+f_{k}(a x)-a^{k} x\left(\left(\dfrac{x}{a}\right) f_{k}\left(\dfrac{a}{x}\right)\right)=(x-1) f_{k}(x)+f_{k}(a x)-a^{k} x f_{k}\left(\dfrac{x}{a}\right)=0$
（iii）不妨设 $\displaystyle f_{k}(x)=\sum\limits_{j=1}^{k} b_{j}^{(k)} x^{j}, k=0,1, \cdots$
由题目的条件容易看出 $b_{k}^{(k)}=b_{k-1}^{(k-1)}=\cdots=b_{0}^{(0)}=1$
另据（ii）中证明得的结论可得 $b_{k-j}^{(k)}=b_{j}^{(k)}, k=0,1, \cdots ; j=0,1, \cdots, k$
特别地 $b_{0}^{(k)}=b_{k}^{(k)}=1, k=0,1,2, \cdots$
在式子 $f_{n}(x)=x f_{n-1}(x)+f_{n-1}(a x)$ 之中，比较两边 $x^{j}$ 和 $x^{n-j}$ 的系数，分别得到
$\begin{aligned} b_{j}^{(n)} &=b_{j-1}^{(n-1)}+a^{j} b_{j}^{(n-1)} , b_{n-j}^{(n)} =b_{n-j-1}^{(n-1)}+a^{n-j} b_{n-j}^{(n-1)} \end{aligned}$ 利用等式关系
$\begin{aligned} b_{n-j}^{(n)} =b_{j}^{(n)} , b_{n-j-1}^{(n-1)} =b_{j}^{(n-1)} , b_{n-j}^{(n-1)} =b_{j-1}^{(n-1)} \end{aligned}$ 可以从前列两式中消去 $b_{n-1-j}^{(n-1)}=b_{j}^{(n-1)}$，从而得到
$\begin{aligned} b_{j}^{\langle n)}=& \frac{a^{n}-1}{a^{j}-1} b_{j-1}^{(n-1)}= \frac{a^{n}-1}{a^{j}-1} \cdot \dfrac{a^{(n-1)}-1}{a^{j-1}-1} b_{j-2}^{(n-2)}=\cdots=\frac{a^{n}-1}{a^{j}-1} \cdot \dfrac{a^{(n-1)}-1}{a^{j-1}-1} \cdots \cdot \frac{a^{n-j+1}-1}{a-1} b_{0}^{(n-j)}= \dfrac{a^{n}-1}{a^{j}-1} \cdot \frac{a^{(n-1)}-1}{a^{j-1}-1} \dots \cdot \dfrac{a^{n-j+1}-1}{a-1} \end{aligned}$
我们得到 $\displaystyle f_{n}(x)=1+\sum\limits_{j=1}^{n}\left(\frac{a^{n}-1}{a^{j}-1} \cdots \frac{a^{n-j+1}}{a-1}-\frac{1}{1}\right) x^{j}$
$F_{k+1}(x)-x F_{k}(x)=(x-1) f_{k+1}(x)+f_{k+1}(a x)-a^{k+1} x f_{k+1}\left(\dfrac{x}{a}\right)-x(x-1) f_{k}(x)-x f_{k}(a x)+a^{k} x f_{k}(a)\\
=(x-1)\left(x f_{k}(x)+f_{k}(a x)\right)+\left(a x f_{k}(a x)+f_{k}\left(a^{2} x\right)\right)-a^{k+1} x\left(\dfrac{x}{a} f_{k}\left(\dfrac{x}{a}\right)+f_{k}(x)\right)-x(x-1) f_{k}(x)-$
$x f_{k}(a x)+a^{k} x^{2} f_{k}\left(\dfrac{x}{a}\right)=(a x-1) f_{k}(a x)+f_{k}\left(a^{2} x\right)-a^{k+1} x f_{k}(x)=F_{k}(a x)$ 因为 $F_{0}(x)=0$ 所以 $F_{n}(x)=0, n=0,1,2, \cdots$
（ii）利用（i）中的结论，我们归纳证明 $f_{k}(x)=x^{n} f_{n}\left(\dfrac{1}{x}\right), n=0,1,2, \cdots$
首先，显然 $f_{0}(x)=x^{0} f_{0}\left(\dfrac{1}{x}\right)$ 假定已证得 $f_{k}(x)=x^{k} f_{k}\left(\dfrac{1}{x}\right)$ 则有
$f_{k+1}(x)-x^{k+1} f_{k+1}\left(\dfrac{1}{x}\right)=f_{k+1}(x)-x^{k+1} f_{k+1}\left(\dfrac{1}{x}\right)-\left(f_{k}(x)-x^{k} f_{k}\left(\dfrac{1}{x}\right)\right)$
$=(x-1) f_{k}(x)+f_{k}(a x)-x^{k+1}\left(\dfrac{1}{x} f_{k}\left(\dfrac{1}{x}\right)+f_{k}\left(\dfrac{a}{x}\right)\right)+x^{k} f_{k}\left(\dfrac{1}{a}\right) )$
$=(x-1) f_{k}(x)+f_{k}(a x)-a^{k} x\left(\left(\dfrac{x}{a}\right) f_{k}\left(\dfrac{a}{x}\right)\right)=(x-1) f_{k}(x)+f_{k}(a x)-a^{k} x f_{k}\left(\dfrac{x}{a}\right)=0$
（iii）不妨设 $\displaystyle f_{k}(x)=\sum\limits_{j=1}^{k} b_{j}^{(k)} x^{j}, k=0,1, \cdots$
由题目的条件容易看出 $b_{k}^{(k)}=b_{k-1}^{(k-1)}=\cdots=b_{0}^{(0)}=1$
另据（ii）中证明得的结论可得 $b_{k-j}^{(k)}=b_{j}^{(k)}, k=0,1, \cdots ; j=0,1, \cdots, k$
特别地 $b_{0}^{(k)}=b_{k}^{(k)}=1, k=0,1,2, \cdots$
在式子 $f_{n}(x)=x f_{n-1}(x)+f_{n-1}(a x)$ 之中，比较两边 $x^{j}$ 和 $x^{n-j}$ 的系数，分别得到
$\begin{aligned} b_{j}^{(n)} &=b_{j-1}^{(n-1)}+a^{j} b_{j}^{(n-1)} , b_{n-j}^{(n)} =b_{n-j-1}^{(n-1)}+a^{n-j} b_{n-j}^{(n-1)} \end{aligned}$ 利用等式关系
$\begin{aligned} b_{n-j}^{(n)} =b_{j}^{(n)} , b_{n-j-1}^{(n-1)} =b_{j}^{(n-1)} , b_{n-j}^{(n-1)} =b_{j-1}^{(n-1)} \end{aligned}$ 可以从前列两式中消去 $b_{n-1-j}^{(n-1)}=b_{j}^{(n-1)}$，从而得到
$\begin{aligned} b_{j}^{\langle n)}=& \frac{a^{n}-1}{a^{j}-1} b_{j-1}^{(n-1)}= \frac{a^{n}-1}{a^{j}-1} \cdot \dfrac{a^{(n-1)}-1}{a^{j-1}-1} b_{j-2}^{(n-2)}=\cdots=\frac{a^{n}-1}{a^{j}-1} \cdot \dfrac{a^{(n-1)}-1}{a^{j-1}-1} \cdots \cdot \frac{a^{n-j+1}-1}{a-1} b_{0}^{(n-j)}= \dfrac{a^{n}-1}{a^{j}-1} \cdot \frac{a^{(n-1)}-1}{a^{j-1}-1} \dots \cdot \dfrac{a^{n-j+1}-1}{a-1} \end{aligned}$
我们得到 $\displaystyle f_{n}(x)=1+\sum\limits_{j=1}^{n}\left(\frac{a^{n}-1}{a^{j}-1} \cdots \frac{a^{n-j+1}}{a-1}-\frac{1}{1}\right) x^{j}$
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