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}