许以超习题

$$\begin{example} 设域$F$上多项式$f(x)$被$x-1,x-2,x-3$除后,余式分别为$4,8,16$.试求$f(x)$被$(x-1)(x-2)(x-3)$除后的余式. \end{example}$$
$$\begin{solution} $12$. \end{solution}$$

$$\begin{example} 给定正整数$n$,试证:存在正整数$m$,使得域$\mathbb{F}$上多项式 \[ \left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right) =1+x+x^2+\cdots +x^m. \] \end{example}$$
$$\begin{solution} 注意到 \begin{align*} \left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right) &=\frac{\left( 1-x \right) \left( 1+x \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right)}{1-x} \\ &=\frac{\left( 1-x^2 \right) \left( 1+x^2 \right) \cdots \left( 1+x^{2^n} \right)}{1-x}=\cdots =\frac{1-x^{2^{n+1}}}{1-x} \\ &=1+x+x^2+\cdots +x^{2^{n+1}-1}, \end{align*} 即$m=2^{n+1}-1$. \end{solution}$$

$$\begin{example} 试计算复多项式 \[ x^n+\left( a+b \right) x^{n-1}+\left( a^2+ab+b^2 \right) x^{n-2}+\cdots +\left( a^n+a^{n-1}b+\cdots +ab^{n-1}+b^n \right) \] 的根的方幂和$S_1,S_2,\cdots,S_n$. \end{example}$$
$$\begin{solution} 等价于$x^n+\frac{a^2-b^2}{a-b}x^{n-1}+\frac{a^3-b^3}{a-b}x^{n-2}+\cdots +\frac{a^n-b^n}{a-b}$. \end{solution}$$

$$\begin{example} 对任意$x_j\in (0,1/2],j=1,2,\cdots,n$和正整数$n$,证明不等式 \[ \frac{\prod_{j=1}^n{x_j}}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\left( \sum_{j=1}^n{\left( 1-x_j \right)} \right) ^n}.\] \end{example}$$
\begin{solution}
等价于证明
$$\frac{\prod_{j=1}^n{x_j}}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\left( n-\sum_{j=1}^n{x_j} \right) ^n}\Leftrightarrow \frac{\left( n-\sum_{j=1}^n{x_j} \right) ^n}{\left( \sum_{j=1}^n{x_j} \right) ^n}\le \frac{\prod_{j=1}^n{\left( 1-x_j \right)}}{\prod_{j=1}^n{x_j}},$$
即
$$\left( \frac{n}{\sum_{j=1}^n{x_j}}-1 \right) ^n\le \prod_{j=1}^n{\left( \frac{1}{x_j}-1 \right)}\Leftrightarrow \ln \left( \frac{n}{\sum_{j=1}^n{x_j}}-1 \right) \le \frac{\sum_{j=1}^n{\ln \left( \frac{1}{x_j}-1 \right)}}{n}.$$

考虑辅助函数$f\left( x \right) =\ln \left( \frac{1}{x}-1 \right)$,由Jensen不等式即可.
\end{solution}

$$\begin{example} 计算 \[ \sum_{j=1}^{n-1}{\frac{1}{1-\exp \left\{ \frac{2\pi ij}{n} \right\}}}. \] \end{example}$$
\begin{solution}
(2011年清华金秋营)设$\varepsilon_n=e^{\frac{2\pi i}{n}}$,试求: $\sum_{k=0}^{n-1}\frac{1}{1-\varepsilon_n^kt}, \sum_{k=1}^{n-1}\frac{1}{1-\varepsilon_n^k}, \sum_{k=1}^{n-1}\frac{1}{(1-\varepsilon_n^k)(1-\varepsilon_n^{-k})}$.

记$\varepsilon_n=e^{\frac{2\pi i}{n}}$,则$1,\varepsilon_n,\varepsilon_n^2,\cdots,\varepsilon_n^{n-1}$为$x^n=1$的$n$个根.

由韦达定理可知
$$\sum_{i=1}^{n}\prod \varepsilon_n^{k_1}\cdots \varepsilon_n^{k_i}=0,\quad 1\leq i<n.$$
由$x^n-1=\left( x-1 \right) \left( x-\varepsilon _n \right) \left( x-\varepsilon _{n}^{2} \right) \cdots \left( x-\varepsilon _{n}^{n-1} \right)$可得
$$\frac{1}{x^n}-1=\left( \frac{1}{x}-1 \right) \left( \frac{1}{x}-\varepsilon _n \right) \left( \frac{1}{x}-\varepsilon _{n}^{2} \right) \cdots \left( \frac{1}{x}-\varepsilon _{n}^{n-1} \right),$$
于是
$$1-x^n=\left( 1-x \right) \left( 1-\varepsilon _n x \right) \left( 1-\varepsilon _{n}^{2}x^2 \right) \cdots \left( 1-\varepsilon _{n}^{n-1}x^{n-1} \right).$$
因此
$$\begin{align*} \sum_{k=1}^{n-1}{\frac{1}{1-\varepsilon _{n}^{k}t}} &=\frac{n}{1-t^n}-\frac{1}{1-t}=\frac{n-\left( 1+t+t^2+\cdots +t^{n-1} \right)}{1-t^n} \\ &=\frac{\left( 1-t \right) +\left( 1-t^2 \right) +\cdots +\left( 1-t^{n-1} \right)}{1-t^n} \\ &=\frac{\left( n-1 \right) +\left( n-2 \right) t+\left( n-3 \right) t^2+\cdots +t^{n-2}}{1+t+t^2+\cdots +t^{n-1}}. \end{align*}$$

令$t=1$,有
$$\sum_{k=1}^{n-1}{\frac{1}{1-\varepsilon _{n}^{k}}}=\frac{\left( n-1 \right) +\left( n-2 \right) +\cdots +1}{n}=\frac{n-1}{2}.$$

而
$$\begin{align*} \sum_{k=1}^{n-1}{\frac{1}{\left( 1-\varepsilon _{n}^{k} \right) \left( 1-\varepsilon _{n}^{-k} \right)}} &=\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\cos \frac{2k\pi}{n}-i\sin \frac{2k\pi}{n} \right) \left( 1-\cos \frac{2k\pi}{n}+i\sin \frac{2k\pi}{n} \right)}} \\ &=\sum_{k=1}^{n-1}{\frac{1}{\left( 1-\cos \frac{2k\pi}{n} \right) ^2+\sin ^2\frac{2k\pi}{n}}}=\sum_{k=1}^{n-1}{\frac{1}{2-2\cos \frac{2k\pi}{n}}} \\ &=\sum_{k=1}^{n-1}{\frac{1}{4\sin ^2\frac{k\pi}{n}}}=\frac{n-1}{4}+\frac{1}{4}\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}, \end{align*}$$

又因为
$$\left( \cos \frac{k\pi}{n}+i\sin \frac{k\pi}{n} \right) ^n=\left( -1 \right) ^k=\sum_{j=0}^n{C_{n}^{j}\left( \cos \frac{k\pi}{n} \right) ^{n-j}\left( i\sin \frac{k\pi}{n} \right) ^j},$$
当$n=2m$时,
$$\begin{align*} &\sum_{j=1}^m{C_{2m}^{2j-1}\left( \cos \frac{k\pi}{n} \right) ^{2m-2j+1}\left( i\sin \frac{k\pi}{n} \right) ^{2j-1}}=0, \\ &\sum_{j=1}^m{C_{2m}^{2j-1}\left( \cot \frac{k\pi}{n} \right) ^{2m-2j}\left( -1 \right) ^j}=0, \end{align*}$$
所以$\cot ^2\frac{\pi}{n},\cot ^2\frac{2\pi}{n},\cdots,\cot ^2\frac{\left( m-1 \right) \pi}{n}$为多项式$\sum_{j=1}^m{C_{2m}^{2j-1}x^{m-j}\left( -1 \right) ^j}=0$的根,因此
$$\sum_{k=1}^{m-1}{\cot ^2\frac{k\pi}{n}}=\frac{C_{2m}^{3}}{C_{2m}^{1}}=\frac{\left( m-1 \right) \left( 2m-1 \right)}{3},$$
故
$$\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}=\frac{\left( 2m-2 \right) \left( 2m-1 \right)}{3}=\frac{\left( n-2 \right) \left( n-1 \right)}{3},$$
当$n=2m+1$时,类似可得
$$\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}}=2\frac{C_{2m+1}^{3}}{C_{2m+1}^{1}}=\frac{\left( 2m-1 \right) 2m}{3}=\frac{\left( n-2 \right) \left( n-1 \right)}{3},$$
因此
$$\begin{align*} \sum_{k=1}^{n-1}{\frac{1}{\left( 1-\varepsilon _{n}^{k} \right) \left( 1-\varepsilon _{n}^{-k} \right)}} &=\frac{n-1}{4}+\frac{1}{4}\sum_{k=1}^{n-1}{\cot ^2\frac{k\pi}{n}} \\ &=\frac{n-1}{4}+\frac{1}{4}\cdot \frac{\left( n-2 \right) \left( n-1 \right)}{3}=\frac{n^2-1}{12}. \end{align*}$$
\end{solution}