问题及解答

设 $f(x)$ 是关于 $x$ 的 $k$ 次多项式,

\[
f(x)=a_k x^k+a_{k-1}x^{k-1}+\cdots+a_1 x+a_0,\qquad (a_k,\ldots,a_0)\in\mathbb{R}^{k+1},\ a_k\neq 0.
\]

写出 $f(\lambda x+\mu)-f(\mu)$, 这里 $\lambda,\mu\in\mathbb{R}$. 

\[
f(\lambda x+\mu)=a_k(\lambda x+\mu)^k+a_{k-1}(\lambda x+\mu)^{k-1}+\cdots+a_1(\lambda x+\mu)+a_0,
\]

\[
(\lambda x+\mu)^n=\sum_{i=0}^{n}C_n^i(\lambda x)^{n-i}\mu^i,
\]

\[
f(\mu)=a_k \mu^k+a_{k-1}\mu^{k-1}+\cdots+a_1 \mu+a_0,
\]

于是

\[
\begin{split}
f(\lambda x+\mu)-f(\mu)&=\sum_{j=1}^{k}\Bigl(a_j(\lambda x+\mu)^j-a_j \mu^j\Bigr)\\
&=\sum_{j=1}^{k}\Bigl(a_j\sum_{i=0}^{j}C_j^i(\lambda x)^{j-i}\mu^i-a_j\mu^j\Bigr)\\
&=\sum_{j=1}^{k}\Bigl(a_j\sum_{i=0}^{j-1}C_j^i(\lambda x)^{j-i}\mu^i\Bigr)\\
\end{split}
\]

例如

当 $k=1$ 时,

\[
a_1(\lambda x+\mu)-a_1 \mu=a_1 \lambda x.
\]

当 $k=2$ 时,

\[
\begin{split}
&a_2(\lambda x+\mu)^2+a_1(\lambda x+\mu)-a_2 \mu^2-a_1 \mu\\
=&a_2(\lambda^2 x^2+2\lambda\mu x)+a_1 \lambda x\\
=&a_2 \lambda^2 x^2+(2a_2 \lambda\mu+a_1 \lambda)x.
\end{split}
\]

当 $k=3$ 时,

\[
\begin{split}
&a_3(\lambda x+\mu)^3+a_2(\lambda x+\mu)^2+a_1(\lambda x+\mu)-a_3 \mu^3-a_2 \mu^2-a_1 \mu\\
=&a_3\bigl[(\lambda x+\mu)^3-\mu^3\bigr]+a_2 \lambda^2 x^2+(2a_2 \lambda\mu+a_1 \lambda)x\\
=&a_3\bigl[\lambda^3 x^3+3\lambda^2 \mu x^2+3\lambda\mu^2 x\bigr]+a_2 \lambda^2 x^2+(2a_2 \lambda\mu+a_1 \lambda)x\\
=&a_3 \lambda^3 x^3+(3a_3 \lambda^2 \mu+a_2 \lambda^2)x^2+(3a_3 \lambda\mu^2+2a_2 \lambda\mu+a_1 \lambda)x.
\end{split}
\]

当 $k=4$ 时,

\[
\begin{split}
&a_4(\lambda x+\mu)^4+a_3(\lambda x+\mu)^3+a_2(\lambda x+\mu)^2+a_1(\lambda x+\mu)-a_4 \mu^4-a_3 \mu^3-a_2 \mu^2-a_1 \mu\\
=&a_4\bigl[(\lambda x+\mu)^4-a_4 \mu^4\bigr]+a_3 \lambda^3 x^3+(3a_3 \lambda^2 \mu+a_2 \lambda^2)x^2+(3a_3 \lambda\mu^2+2a_2 \lambda\mu+a_1 \lambda)x\\
=&a_4\bigl[(\lambda x)^4+4(\lambda x)^3 \mu+6(\lambda x)^2 \mu^2+4(\lambda x)\mu^3\bigr]+a_3 \lambda^3 x^3+(3a_3 \lambda^2 \mu+a_2 \lambda^2)x^2+(3a_3 \lambda\mu^2+2a_2 \lambda\mu+a_1 \lambda)x\\
=&a_4 \lambda^4 x^4+(4a_4 \lambda^3 \mu+a_3 \lambda^3)x^3+(6a_4 \lambda^2\mu^2+3a_3 \lambda^2 \mu+a_2 \lambda^2)x^2+(4a_4 \lambda\mu^3+3a_3 \lambda\mu^2+2a_2 \lambda\mu+a_1 \lambda)x.
\end{split}
\]

可以用归纳法写出并证明一般形式.

\[
(a_k \lambda^k)x^k+(ka_k\lambda^{k-1}\mu+a_{k-1}\lambda^{k-1})x^{k-1}+\cdots+(\cdots)x
\]