多项式问题

\begin{example}
给定一个正整数$n$,设$n$个实数$a_1,a_2,\cdots,a_n$满足下列$n$个方程:
$$\sum_{i=1}^{n}\frac{a_i}{i+j}=\frac{4}{2j+1}\quad (j=1,2,\cdots,n)$$
化简和式$S=\sum_{i=1}^{n}\frac{a_i}{2i+1}$.
\end{example}
\begin{proof}
由
$$
\sum_{i=1}^n{\frac{a_i}{2i+1}}=\frac{1}{2}\sum_{i=1}^n{\frac{a_i}{i+\frac{1}{2}}}=\left. \frac{1}{2}\sum_{i=1}^n{\frac{a_i}{i+x}} \right|_{x=\frac{1}{2}}
$$
及已知条件,我们计算
$$
\sum_{i=1}^n{\frac{a_i}{x+i}}-\frac{4}{2x+1}=\frac{P\left( x \right)}{\left( 2x+1 \right) \prod_{i=1}^n{\left( x+i \right)}},
$$
其中$P(x)$是次数不超过$n$的整系数多项式,并且$P(1)=P(2)=\cdots=P(n)=0$,由唯一性,得$P(x)=C\prod_{i=1}^n{\left( x-i \right)}$,其中$C$是待定常数.从而,有
\[
\sum_{i=1}^n{\frac{a_i}{x+i}}=\frac{4}{2x+1}+\frac{C\prod_{i=1}^n{\left( x-i \right)}}{\left( 2x+1 \right) \prod_{i=1}^n{\left( x+i \right)}} \tag{1}
\]
化成整式,得
\[
\left( 2x+1 \right) \prod_{i=1}^n{\left( x+i \right)}\sum_{i=1}^n{\frac{a_i}{x+i}}=4\prod_{i=1}^n{\left( x+i \right)}+C\prod_{i=1}^n{\left( x-i \right)}\tag{2}
\]
因为由(1)式恒等(除$x=-1,-2,\cdots,-n,-\frac{1}{2}$外)与唯一性,可知(2)式也恒等(对一切复数$x$),所以在(2)式中取$x=-\frac{1}{2}$,得
$$
C\left( -\frac{1}{2}-1 \right) \left( -\frac{1}{2}-2 \right) \cdots \left( -\frac{1}{2}-n \right) +4\left( -\frac{1}{2}+1 \right) \left( -\frac{1}{2}+2 \right) \cdots \left( -\frac{1}{2}+n \right) =0.
$$
则
$$
C=-\frac{4\left( -\frac{1}{2}+1 \right) \left( -\frac{1}{2}+2 \right) \cdots \left( -\frac{1}{2}+n \right)}{\left( -1 \right) ^n\left( \frac{1}{2}+1 \right) \left( \frac{1}{2}+2 \right) \cdots \left( \frac{1}{2}+n \right)}=\left( -1 \right) ^{n-1}\frac{4\cdot \left( 2n-1 \right) !!}{\left( 2n+1 \right) !!}=\left( -1 \right) ^{n-1}\frac{4}{2n+1}.
$$
故
$$
\sum_{i=1}^n{\frac{a_i}{x+i}}=\frac{4}{2x+1}+\frac{\left( -1 \right) ^{n-1}}{2n+1}\cdot \frac{4\prod_{i=1}^n{\left( x-i \right)}}{\left( 2x+1 \right) \prod_{i=1}^n{\left( x+i \right)}}.
$$
取$x=\frac{1}{2}$,得
\begin{align*}
2S &=\sum_{i=1}^n{\frac{a_i}{\frac{1}{2}+i}}=2+\frac{\left( -1 \right) ^{n-1}}{2n+1}\cdot \frac{4\prod_{i=1}^n{\left( \frac{1}{2}-i \right)}}{2\prod_{i=1}^n{\left( \frac{1}{2}+i \right)}}
\\
&=2+\frac{\left( -1 \right) ^{n-1}}{2n+1}\cdot \frac{2\left( -1 \right) ^n\prod_{i=1}^n{\left( 2i-1 \right)}}{\prod_{i=1}^n{\left( 2i+1 \right)}}=2-\frac{2}{\left( 2n+1 \right) ^2}.
\end{align*}
故
$$
S=1-\frac{1}{\left( 2n+1 \right) ^2}.
$$
\end{proof}

\begin{example}
(2016年首届卓越大学联盟高校大学生数学竞赛非数学类)考虑级数$\sum_{n=1}^{\infty}\sin(n!\pi x)$.证明:

(1)在任何区间$(a,b)$内都存在使得级数收敛的点.

(2)在任何区间$(a,b)$内都存在使得级数发散的无理数点.
\end{example}
\begin{proof}
(1)由有理数的稠密性可知,在任何区间$(a,b)$内都有有理数.任取一有理数$x=\frac{q}{p}\in (a,b)$,则所讨论的级数在$x$处都收敛,这是因为当$n$充分大时, $n!\frac{q}{p}$均为整数,从而$\sin\left(n!\pi\frac{q}{p}\right)=0$.此时级数$\sum_{n=1}^{\infty}\sin(n!\pi x)$收敛.

(2)首先有级数在$x=e$处发散.这是因为:由
$$e=1+1+\frac{1}{2!}+\cdots+\frac{1}{n!}
+\frac{1}{(n+1)!}+\cdots$$
知
$$
\sin \left( n!\pi e \right) =\sin \left( k\pi +\frac{\pi}{n+1}+O\left( \frac{1}{n^2} \right) \right) \sim \frac{\pi}{n+1},
$$
其中$k$是个整数.故级数在该点发散.

由于$e$是无理数,对任意区间$(a,b)$,在其内任取一有理点$r$,然后取一充分大的整数$m$,使得$r+\frac{e}{m}\in (a,b) $.则$r+\frac{e}{m}$即为所求的无理数.级数$\sum_{n=1}^{\infty}\sin(n!\pi x)$在这些点均发散.
\end{proof}

\begin{Example}
设$f(x)=a_1\sin x+a_2\sin 2x+\cdots+a_n\sin nx$,其中$a_1,a_2,\cdots,a_n$是实数, $n$是正整数.如果对所有实数$x$都有$|f(x)|\leqslant |\sin x|$,求证:
$$|a_1+2a_2+\cdots+na_n|\leqslant 1.$$
\end{Example}
\begin{proof}
令$M=|a_1|+|a_2|+\cdots+|a_n|$.对于正整数$k$ ($1\leqslant k\leqslant n$),由于$\lim_{n\to 0}\frac{\sin kx}{\sin x}=k$,所以任给$\varepsilon>0$,存在实数$x$,使$\sin x\neq 0$,且$\left| \frac{\sin kx}{\sin x}-k \right|<\frac{\varepsilon}{M}$, $k=1,2,\cdots,n$.由此可得
\begin{align*}
1 &\geqslant \left| \frac{f\left( x \right)}{\sin x} \right|=\left| \sum_{k=1}^n{\frac{\sin kx}{\sin x}} \right|
\\
&=\left| \sum_{k=1}^n{ka_k}-\sum_{k=1}^n{\left( \frac{\sin kx}{\sin x}-k \right) a_k} \right|
\\
&\geqslant \left| \sum_{k=1}^n{ka_k} \right|-\sum_{k=1}^n{\left| \frac{\sin kx}{\sin x}-k \right|\left| a_k \right|}\geqslant \left| \sum_{k=1}^n{ka_k} \right|-\varepsilon,
\end{align*}
由$\varepsilon$的任意性可知所求证的不等式成立.
\end{proof}

\begin{Example}
已知$a_1,a_2,\cdots,a_n\in\mathbb{R}$,设$f(x)=\sum_{k=1}^{n}a_k\sin kx$,若对于任意$x\in\mathbb{R}$都有$|f(x)|\leqslant 1$,证明对任意$x\in\mathbb{R}$有$|f(x)|\leqslant n|\sin x|$.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
(1996中国数学奥林匹克,常庚哲提供,难度: \score{4}{5})设$n\in \mathbb{N}$, $x_0=0$, $x_i>0,i=1,2,\cdots,n$且$\sum_{i=1}^{n}x_i=1$.求证:
$$1\leqslant\sum_{i=1}^{n}
\frac{x_i}{\sqrt{1+x_0+x_1+\cdots+x_{i-1}}
\sqrt{x_i+\cdots+x_n}}<\frac{\pi}{2}.$$
\end{Example}
\begin{proof}
(谈谈第11届数学冬令营的一道题目,常庚哲)

因为$\sum_{i=1}^{n}x_i=1$,故由均值不等式有
$$
\sqrt{\left( 1+x_0+x_1+\cdots +x_{i-1} \right) \left( x_i+\cdots +x_n \right)}\leqslant \frac{1+x_0+x_1+\cdots +x_n}{2}=1,
$$
由此即得所求证的第一个不等式.

因为$0\leqslant x_0+x_1+\cdots +x_i\leqslant 1$, $i=0,1,\cdots,n$,故可令$\theta_i=\arcsin(x_0+x_1+\cdots +x_i)$, $i=0,1,\cdots,n$,于是$\theta_i\in\left[0,\frac{\pi}{2}\right]$且有
$0=\theta_0<\theta_1<\cdots<\theta_n=\frac{\pi}{2}$.因而有
\begin{align*}
x_i&=\sin \theta _i-\sin \theta _{i-1}=2\cos \frac{\theta _i+\theta _{i-1}}{2}\sin \frac{\theta _i-\theta _{i-1}}{2}
\\
&<2\cos \theta _{i-1}\sin \frac{\theta _i-\theta _{i-1}}{2}.
\end{align*}

因为对$x\in\left[0,\frac{\pi}{2}\right]$,有$\sin x<x$,所以有
$$
x_i<2\cos \theta _{i-1}\cdot \frac{\theta _i-\theta _{i-1}}{2}=\left( \theta _i-\theta _{i-1} \right) \cos \theta _{i-1}.
$$
故得
$$\frac{x_i}{\cos \theta _{i-1}}<\theta _i-\theta _{i-1},\quad i=0,1,\cdots,n.$$
在上式两端对$i$从$1$到$n$求和,得到
\[\sum_{i=1}^{n}\frac{x_i}{\cos \theta _{i-1}}<\theta _n-\theta _0=\frac{\pi}{2}.\tag{1}\]
由$\theta _i$定义知$\sin \theta _i=x_0+x_1+\cdots +x_i$,所以有
\begin{align*}
\cos \theta _{i-1} &=\sqrt{1-\sin ^2\theta _{i-1}}
\\
&=\sqrt{1-\left( x_0+x_1+\cdots +x_{i-1} \right) ^2}
\\
&=\sqrt{\left( 1+x_0+x_1+\cdots +x_{i-1} \right) \left( x_i+\cdots +x_n \right)} \tag{2}
\end{align*}
将(2)代入(1)即得所欲证.

\textbf{证法二.}事实上,令$y_i=x_0+x_1+\cdots +x_{i-1}$,则
\begin{align*}
&\sum_{i=1}^n{\frac{x_i}{\sqrt{1+x_0+x_1+\cdots +x_{i-1}}\sqrt{x_i+\cdots +x_n}}}
\\
&=\sum_{i=1}^n{\frac{x_i}{\sqrt{1+x_0+x_1+\cdots +x_{i-1}}\sqrt{1-\left( x_0+x_1+\cdots +x_{i-1} \right)}}}
\\
&=\sum_{i=1}^n{\frac{x_i}{\sqrt{1-\left( x_0+x_1+\cdots +x_{i-1} \right) ^2}}}=\sum_{i=1}^n{\frac{y_{i+1}-y_i}{\sqrt{1-\left( y_i \right) ^2}}}
\\
&<\int_0^1{\frac{dx}{\sqrt{1-x^2}}}=\frac{\pi}{2}.
\end{align*}
\end{proof}

\begin{Example}
对任意$x\in \mathbb{R}$及$n\in \mathbb{N}_+$.求证:

(1) $\left| \sum_{k=1}^n{\frac{\sin kx}{k}} \right|\le 2\sqrt{\pi}$;

(2) $\sum_{k=1}^n{\frac{\left| \sin kx \right|}{k}}\ge \left| \sin nx \right|$.
\end{Example}
\begin{Proof}
(1)定理4[一致有界性] 设$x\in\mathbb{R}$,则存在与 $x$无关的常数$M>0$使得
$$\left|\sum\limits_{k=1}^n\dfrac{\sin kx}{k}\right|<M,\forall n\in\mathbb{N}^\ast,\forall x.$$
Motivation: 回顾不等式$\sin t\leq t\, (t\geq 0)$以及
\begin{align*}
\sum\limits_{k=m+1}^n\sin kx&=\dfrac{1}{2\sin\frac{x}{2}}\sum\limits_{k=m+1}^n \left[\cos\left(k+\dfrac{1}{2}\right)x-\cos\left(k-\dfrac{1}{2}\right)x\right] \\ &=\dfrac{\cos\left(m+\dfrac{1}{2}\right)x-\cos\left(n+\dfrac{1}{2}\right)x}{2\sin\frac{x}{2}},\\ \Rightarrow \left|\sum\limits_{k=m+1}^n\sin kx\right|&\leq\dfrac{1}{\left|\sin\frac{x}{2}\right|} \end{align*}
注意$\sin t\leq t$在$t$比较小的范围内是比较“好”的, 而当$t$比较大的时候直接用三角函数有界性即可放缩. 所以当$t$比较大的时候, 可以考虑采用Abel变换以及三角级数的求和.

证明：由三角函数的周期性, 只需要考虑$(-\pi,\pi)$区间. 又由$\sin kx$是奇函数, 故只需要考虑$(0,\pi)$区间(另一半区间相当于取相反数).

(1)若$n\leq\dfrac{\sqrt{2\pi}}{x}$,根据不等式 $\sin t<t\, (t>0)$得
$$\sum\limits_{k=1}^n\dfrac{\sin kx}{k} \leq\sum\limits_{k=1}^n\dfrac{kx}{k}=nx\leq\sqrt{2\pi}<2\sqrt{2\pi}.$$

(2)若$n>\dfrac{\sqrt{2\pi}}{x}$,取正整数$m$满足 $mx<\sqrt{2\pi}<(m+1)x$,则$m<n$.把级数拆成两个部分:
$$\sum\limits_{k=1}^n\dfrac{\sin kx}{k}=\sum\limits_{k=1}^m\dfrac{\sin kx}{k} +\sum\limits_{k=m+1}^n\dfrac{\sin kx}{k}.$$
根据前一定理, 我们在这里不需要加绝对值.

由不等式$\sin t<t\, (t>0)$,得
$$\sum\limits_{k=1}^m\dfrac{\sin kx}{k} \leq\sum\limits_{k=1}^m\dfrac{kx}{k}=mx.$$

回顾Abel求和, 设$S_n=\sum\limits_{k=1}^n\sin kx$,则
\begin{align*} \sum\limits_{k=m+1}^n\dfrac{\sin kx}{k} &= \sum\limits_{k=m+1}^n\dfrac{S_k}{k}-\sum\limits_{k=m+1}^n\dfrac{S_{k-1}}{k} \\ &=\sum\limits_{k=m+1}^n\dfrac{S_k}{k}-\sum\limits_{k=m}^{n-1}\dfrac{S_{k}}{k+1} \\ &=\left[\sum\limits_{k=m+1}^{n-1}S_k\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\right] +\dfrac{S_n}{n}-\dfrac{S_m}{m+1}. \end{align*}
根据$x\in(0,\pi)$以及前一定理可得
\begin{align*} \sum\limits_{k=m+1}^n\dfrac{\sin kx}{k} &\leq\sum\limits_{k=m+1}^{n-1}S_k\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right) +\dfrac{S_n}{n}-\dfrac{S_m}{m+1} \\ &\leq \dfrac{1}{\sin\frac{x}{2}}\left(\dfrac{1}{m+1}-\dfrac{1}{n}\right) +\dfrac{1}{\sin\frac{x}{2}}\cdot\left(\dfrac{1}{n}+\dfrac{1}{m+1}\right) \\ &=\dfrac{2}{(m+1)\sin\frac{x}{2}}. \end{align*}
注意当$x\in\left(0,\frac{\pi}{2}\right)$时有 $\sin x>\dfrac{2x}{\pi}$,所以
$$\sum\limits_{k=m+1}^n\dfrac{\sin kx}{k}\leq \dfrac{2}{(m+1)\sin\frac{x}{2}}<\dfrac{2\pi}{(m+1)x}.$$
综上,
$$\sum\limits_{k=1}^n\dfrac{\sin kx}{k}<mx+\dfrac{2\pi}{(m+1)x} <\sqrt{2\pi}+\dfrac{2\pi}{\sqrt{2\pi}}=2\sqrt{2\pi}.$$

这里正整数$m$是待定的(与$x$有关), 取$mx<\sqrt{2\pi}<(m+1)x$,则上界$M$可以取为$2\sqrt{2\pi}$.

注：稍微修改某一步, 可以把上界$M$变小为$2\sqrt{\pi}$.

(2)利用第二数学归纳法,假设$n=1,2,\cdots,m$时,不等式都成立,则
\[
\sum_{k=1}^n{\frac{\left| \sin kx \right|}{k}}\ge \left| \sin nx \right|,\quad n=1,2,\cdots,m.
\]
由$|\sin(x+y)|\leq |\sin x\cos y|+|\cos x\sin y|\leq |\sin x|+|\sin y|$可知
\begin{align*}
&\left( m+1 \right) \left( \left| \sin x \right|+\frac{\left| \sin 2x \right|}{2}+\cdots +\frac{\left| \sin mx \right|}{m} \right) \ge 2\sum_{k=1}^m{\left| \sin kx \right|}
\\
&=\sum_{k=1}^m{\left( \left| \sin kx \right|+\left| \sin \left( m+1-k \right) x \right| \right)}\ge \sum_{k=1}^m{\left| \sin \left( m+1 \right) x \right|}=m\left| \sin \left( m+1 \right) x \right|.
\end{align*}
因此
\[
\sum_{k=1}^m{\frac{\left| \sin kx \right|}{k}}\ge \frac{m}{m+1}\left| \sin \left( m+1 \right) x \right|,
\]
故
\[\sum_{k=1}^{m+1}{\frac{\left| \sin kx \right|}{k}}\ge \frac{m}{m+1}\left| \sin \left( m+1 \right) x \right|+\frac{\left| \sin \left( m+1 \right) x \right|}{m+1}=\left| \sin \left( m+1 \right) x \right|.\]
\end{Proof}

\begin{Example}
设$\{a_n\}$为非增的正数列,求证:若对$n\geq 2001,na_n<1$,则对任意自然数$m\geq 2001,x\in \mathbb{R}$,有$\left| \sum_{k=2001}^m{a_k\sin kx} \right|\le 1+\pi$.
\end{Example}
\begin{Proof}
1
\end{Proof}

\begin{Example}
(1981年第10届美国数学奥林匹克) Show that for any positive real $x$,\[[nx]\ge \sum_{k=1}^{n}\left(\frac{[kx]}{k}\right),\]where $[x]$ denotes the greatest integer not exceeding $x$.
\end{Example}
\begin{Proof}
由Abel恒等式可知
\[
\sum_{k=1}^n{\left[ kx \right]}=\sum_{k=1}^n{k\cdot \frac{\left[ kx \right]}{k}}=n\sum_{k=1}^n{\frac{\left[ kx \right]}{k}}+\sum_{k=1}^{n-1}{\left( -1 \right) \left( \sum_{i=1}^k{\frac{\left[ ix \right]}{i}} \right)},
\]
于是
\[
n\sum_{k=1}^n{\frac{\left[ kx \right]}{k}}=\sum_{k=1}^n{\left[ kx \right]}+\sum_{k=1}^{n-1}{\left( \sum_{i=1}^k{\frac{\left[ ix \right]}{i}} \right)}.
\]
事实上,利用第二数学归纳法,假设$1,2,\cdots,n-1$时,不等式都成立,由$[x]+[y]\leq [x+y]$可知
\begin{align*}
n\sum_{k=1}^n{\frac{\left[ kx \right]}{k}} &=\sum_{k=1}^n{\left[ kx \right]}+\sum_{k=1}^{n-1}{\left( \sum_{i=1}^k{\frac{\left[ ix \right]}{i}} \right)}\\
&\le \sum_{k=1}^n{\left[ kx \right]}+\sum_{k=1}^{n-1}{\left[ kx \right]}=\left[ nx \right] +2\sum_{k=1}^{n-1}{\left[ kx \right]}\\
&=\left[ nx \right] +\sum_{k=1}^{n-1}{\left( \left[ kx \right] +\left[ \left( n-k \right) x \right] \right)}\le \left[ nx \right] +\sum_{k=1}^{n-1}{\left[ nx \right]}=n\left[ nx \right].
\end{align*}
不等式左右两边同除$n$,可得
\[
\sum_{k=1}^n{\frac{\left[ kx \right]}{k}}\le \left[ nx \right].
\]
\end{Proof}

\begin{Example}
(2008年中国数学奥林匹克,难度: \score{4}{5})给定正整数$n$,及实数$x_1\leqslant x_2\leqslant\cdots\leqslant x_n$, $y_1\geqslant y_2\geqslant\cdots\geqslant y_n$,满足
$$\sum_{i=1}^{n}ix_i=\sum_{i=1}^{n}iy_i.$$
证明:对任意实数$\alpha$,有
$$\sum_{i=1}^{n}x_i[i\alpha]=\sum_{i=1}^{n}y_i[i\alpha].$$
这里, $[\beta]$表示不超过实数$\beta$的最大整数. (朱华伟提供)
\end{Example}
\begin{Proof}

\end{Proof}

由$\cos \left( \alpha \pm \beta \right) =\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$可推导积化和差公式
\begin{align*}
\cos \alpha \cos \beta &=\frac{1}{2}\left[ \cos \left( \alpha +\beta \right) +\cos \left( \alpha -\beta \right) \right],
\\
\sin \alpha \sin \beta &=\frac{1}{2}\left[ \cos \left( \alpha -\beta \right) -\cos \left( \alpha +\beta \right) \right].
\end{align*}
由$\sin \left( \alpha \pm \beta \right) =\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$可推导积化和差公式
\begin{align*}
\sin \alpha \cos \beta &=\frac{1}{2}\left[ \sin \left( \alpha +\beta \right) +\sin \left( \alpha -\beta \right) \right],
\\
\cos \alpha \sin \beta &=\frac{1}{2}\left[ \sin \left( \alpha +\beta \right) -\sin \left( \alpha -\beta \right) \right].
\end{align*}

利用
\begin{align*}
\cos \alpha +\cos \beta &=\cos \left( \frac{\alpha +\beta}{2}+\frac{\alpha -\beta}{2} \right) +\cos \left( \frac{\alpha +\beta}{2}-\frac{\alpha -\beta}{2} \right)
\\
&=2\cos \frac{\alpha +\beta}{2}\cos \frac{\alpha -\beta}{2}.
\end{align*}
可得

和差化积公式:
\begin{align*}
\cos \alpha +\cos \beta &=2\cos \frac{\alpha +\beta}{2}\cos \frac{\alpha -\beta}{2},
\\
\cos \alpha -\cos \beta &=-2\sin \frac{\alpha +\beta}{2}\sin \frac{\alpha -\beta}{2},
\\
\sin \alpha +\sin \beta &=2\sin \frac{\alpha +\beta}{2}\cos \frac{\alpha -\beta}{2},
\\
\sin \alpha -\sin \beta &=2\cos \frac{\alpha +\beta}{2}\sin \frac{\alpha -\beta}{2},
\end{align*}
其他的一些公式:
\begin{align*}
\sin 3\alpha &=3\sin \alpha -4\sin ^3\alpha =4\sin \left( 60^{\circ}-\alpha \right) \cdot \sin \alpha \sin \left( 60^{\circ}+\alpha \right) ,
\\
\cos 3\alpha &=4\cos ^3\alpha -3\cos \alpha =4\cos \left( 60^{\circ}-\alpha \right) \cdot \cos \alpha \cos \left( 60^{\circ}+\alpha \right) ,
\\
\tan 3\alpha &=\tan \left( 60^{\circ}-\alpha \right) \cdot \tan \alpha \tan \left( 60^{\circ}+\alpha \right) ,
\\
\sin ^2\alpha -\sin ^2\beta &=\sin \left( \alpha +\beta \right) \sin \left( \alpha -\beta \right) ,
\\
\cos ^2\alpha -\cos ^2\beta &=-\sin \left( \alpha +\beta \right) \sin \left( \alpha -\beta \right) ,
\\
a\sin x+b\cos x &=\sqrt{a^2+b^2}\sin \left( x+\theta \right),\quad \text{其中}\tan\theta=\frac{b}{a}.
\end{align*}

\begin{Example}%辞典三角P319
(三角等差数列求和)求证:
\begin{align*}
S_n &=\sin a+\sin \left( a+d \right) +\sin \left( a+2d \right) +\cdots +\sin \left( a+\left( n-1 \right) d \right)
\\
&=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\sin \left( a+\frac{\left( n-1 \right) d}{2} \right),
\\
C_n &=\cos a+\cos \left( a+d \right) +\cos \left( a+2d \right) +\cdots +\cos \left( a+\left( n-1 \right) d \right)
\\
&=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\cos \left( a+\frac{\left( n-1 \right) d}{2} \right).
\end{align*}
\end{Example}
\begin{proof}
\textbf{证法一.}
(1)利用
$$
2\sin \frac{d}{2}\sin \left( a+kd \right) =\cos \left( a+\frac{2k-1}{2}d \right) -\cos \left( a+\frac{2k+1}{2}d \right)
$$
可得
\begin{align*}
2\sin \frac{d}{2}S_n &=\sum_{k=0}^{n-1}{2\sin \frac{d}{2}\sin \left( a+kd \right)}=\sum_{k=0}^{n-1}{\left[ \cos \left( a+\frac{2k-1}{2}d \right) -\cos \left( a+\frac{2k+1}{2}d \right) \right]}
\\
&=\cos \left( a-\frac{d}{2} \right) -\cos \left( a+\frac{2n-1}{2}d \right) =2\sin \frac{nd}{2}\sin \left( a+\frac{n-1}{2}d \right).
\end{align*}
于是
$$
S_n=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\sin \left( a+\frac{n-1}{2}d \right).
$$

另一方面,
$$
C_n=\sum_{k=0}^{n-1}{\sin \left( \frac{\pi}{2}+a+kd \right)}=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\sin \left( \frac{\pi}{2}+a+\frac{n-1}{2}d \right) =\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\cos \left( a+\frac{n-1}{2}d \right).
$$

或者利用
$$
2\sin \frac{d}{2}\cos \left( a+kd \right) =\sin \left( a+\frac{2k+1}{2}d \right) -\sin \left( a+\frac{2k-1}{2}d \right)
$$
也可以.

\textbf{证法二.}注意到
\begin{align*}
C_n+iS_n &=\sum_{k=0}^{n-1}{\left[ \cos \left( a+kd \right) +i\sin \left( a+kd \right) \right]}=\sum_{k=0}^{n-1}{e^{i\left( a+kd \right)}}=e^{ia}\frac{1-e^{ind}}{1-e^{id}}
\\
&=\left( \cos a+i\sin a \right) \frac{1-\cos nd-i\sin nd}{1-\cos d-i\sin d}
\\
&=\left( \cos a+i\sin a \right) \frac{2\sin \frac{nd}{2}\left( \sin \frac{nd}{2}-i\cos \frac{nd}{2} \right)}{2\sin \frac{d}{2}\left( \sin \frac{d}{2}-i\cos \frac{d}{2} \right)}
\\
&=e^{ia}\frac{-ie^{i\frac{nd}{2}}}{-ie^{i\frac{d}{2}}}\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}=e^{i\left( a+\frac{n-1}{2}d \right)}\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}
\\
&=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\left[ \cos \left( a+\frac{n-1}{2}d \right) +i\sin \left( a+\frac{n-1}{2}d \right) \right].
\end{align*}
对比等式两边的实部和虚部可得
$$
S_n=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\sin \left( a+\frac{n-1}{2}d \right) ,\quad C_n=\frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}}\cos \left( a+\frac{n-1}{2}d \right).
$$
\end{proof}

令$d=a$,可得
\begin{align*}
\sin a+\sin \left( 2a \right) +\sin \left( 3a \right) +\cdots +\sin \left( na \right) &=\frac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\sin \left( \frac{n+1}{2}a \right) ,
\\
\cos a+\cos \left( 2a \right) +\cos \left( 3a \right) +\cdots +\cos \left( na \right) &=\frac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\cos \left( \frac{n+1}{2}a \right) .
\end{align*}

若令$d=2a$,可得
\begin{align*}
\sin a+\sin \left( 3a \right) +\sin \left( 5a \right) +\cdots +\sin \left( 2n-1 \right) a &=\frac{\sin ^2na}{\sin a},
\\
\cos a+\cos \left( 3a \right) +\cos \left( 5a \right) +\cdots +\cos \left( 2n-1 \right) a &=\frac{\sin \left( na \right) \cos \left( na \right)}{\sin a}=\frac{\sin \left( 2na \right)}{2\sin a}.
\end{align*}

利用
$$
\frac{\sin \left( 2n-1 \right) a}{2\sin a}=\frac{\sin \left( 2na \right) \cos a-\cos \left( 2na \right) \sin a}{2\sin a}
$$
可得

若令$a=0,d=2d$,可得
$$
1+\cos \left( 2d \right) +\cos \left( 4d \right) +\cdots +\cos \left( 2n-2 \right) d=\frac{\sin nd}{\sin d}\cos \left( n-1 \right) d=\frac{\sin \left( 2n-1 \right) d+\sin d}{2\sin d},
$$
故
$$
\frac{\sin \left( 2n-1 \right) d}{\sin d}=1+2\sum_{k=1}^{n-1}{\cos 2kd},\quad \frac{\sin \left( 2nd \right)}{\sin d}=2\sum_{k=1}^n{\cos \left( 2k-1 \right) d}.
$$

\begin{Example}
(北京市第四届大学生数学竞赛(1992年))设$f(x)$在$[0,\pi]$上连续,在$(0,\pi)$内可导,且
$$\int_{0}^{\pi}f(x)\cos xdx=\int_{0}^{\pi}f(x)\sin xdx=0,$$
求证:存在$\xi\in (0,\pi)$,使得$f'(\xi)=0$.
\end{Example}
\begin{Solution}
因为在$(0,\pi)$内$\sin x>0$,又$\int_{0}^{\pi}f(x)\sin xdx=0$,由积分的保号性可知, $f(x)$在$(0,\pi)$内必有零点.

设$a\in (0,\pi)$是$f(x)$在$(0,\pi)$内的唯一零点,则当$x\neq a$, $x\in (0,\pi)$时, $\sin (x-a)f(x)$必恒正或恒负(否则$f(x)$必另有零点),于是可推出$\int_{0}^{\pi}f(x)\sin (x-a)dx\neq 0$,但由已知条件可得
\begin{align*}
\int_{0}^{\pi}f(x)\sin (x-a)dx
&=\int_{0}^{\pi}f(x)(\sin x\cos a-\cos x\sin a)dx\\
&=\cos a\int_{0}^{\pi}f(x)\sin xdx-\sin a\int_{0}^{\pi}f(x)\cos xdx
=0
\end{align*}
矛盾,由此可知,函数$f(x)$在$(0,\pi)$内零点个数不止一个.于是由罗尔定理知:在函数$f(x)$得两个零点之间必然存在导函数等于$0$的点,即存在$\xi\in(0,\pi)$,使得$f'(\xi)=0$.
\end{Solution}

\begin{Example}
设$f(x)$是一个不恒为零实系数多项式,如果存在$n\ (\geqslant 2)$个点$x_1,x_2,\cdots,x_n\in (0,\pi)$.使
$$\sum_{k=1}^{n} f(x_k)\sin x_k=\sum_{k=1}^{n} f(x_k)\cos x_k=0.$$
求证: $f(x)$在$(0,\pi)$内至少有两个不同的根.
\end{Example}
\begin{Proof}
首先证明$f(x)$在$(0,\pi)$内必有零点.若不然,则$f(x)$在$(0,\pi)$内同号.因$\sin x_k>0\ (k=1,2,\cdots,n)$,所以
$$\sum_{k=1}^{n} f(x_k)\sin x_k\neq 0,$$
与所设矛盾.

其次证明$f(x)$至少两个不同的根.反设$f(x)$只有一个根$\alpha$,则不妨设$f(x)$在$(0,\alpha)$中为负,而在$(\alpha,\pi)$中$f(x)$为正,又设
$0<x_1<x_2<\cdots<x_m\leqslant \alpha, \alpha\leqslant x_{m+1}<x_{m+2}<\cdots<x_n<\pi$.
则$\sin(x_i-\alpha)\leqslant 0,\ (i=1,2,\cdots,m)$,
$\sin(x_j-\alpha)\geqslant 0,\ (j=m+1,m+2,\cdots,n)$.
且上不等式至少有一个不等号.

故$f(x_i)\sin(x_i-\alpha)\geqslant 0$,
$f(x_j)\sin(x_j-\alpha)\geqslant 0$,至少有一个成立不等号.

所以
$$\sum_{k=1}^{n} f(x_k)\sin (x_k-\alpha)
=\sum_{i=1}^{m} f(x_i)\sin (x_i-\alpha)
+\sum_{j=m+1}^{n} f(x_j)\sin (x_j-\alpha)>0.$$

另一方面,由已知又有
$$\sum_{k=1}^{n} f(x_k)\sin (x_k-\alpha)
=\cos \alpha\sum_{k=1}^{n} f(x_k)\sin x_k
-\sin \alpha\sum_{k=1}^{n} f(x_k)\cos x_k=0.$$
这与上面得到的事实相矛盾,即原结论得证.
\end{Proof}

\begin{Example}
若$\int_{a}^{b}f(x)dx=\int_{a}^{b}xf(x)dx=\int_{a}^{b}x^2f(x)dx=0$,证明:存在相异的三个点使得
\[f(x_1)=f(x_2)=f(x_3)=0.\]
\end{Example}
\begin{Proof}
显然$f(x)$在$(a,b)$上存在零点,否则$\int_ {a}^{b}f(x)dx\neq 0$.利用反证法.设$f(x)$在$(a,b)$上只有一个零点,记为$l$,不妨设在$x\in (a,l)$时有$f(x)<0$,在$x\in (l,b)$时有$f(x)>0$,则
\[
\int_a^b{\left( x-l \right) f\left( x \right) dx}=\int_a^l{\left( x-l \right) f\left( x \right) dx}+\int_a^l{\left( x-l \right) f\left( x \right) dx}>0.\]
这与
\[
\int_a^b{\left( x-l \right) f\left( x \right) dx}=\int_a^b{xf\left( x \right) dx}-l\int_a^b{f\left( x \right) dx}=0
\]
矛盾.

设$f(x)$在$(a,b)$上有两个零点,记为$m,n$.不妨设在$x\in (a,m)$时有$f(x)>0$,在$x\in (m,n)$时有$f(x)<0$,在$x\in (n,b)$时有$f(x)>0$,则
\begin{align*}
\int_a^b{\left( x-m \right) \left( x-n \right) f\left( x \right) dx}=&\int_a^m{\left( x-m \right) \left( x-n \right) f\left( x \right) dx}
\\
&+\int_m^n{\left( x-m \right) \left( x-n \right) f\left( x \right) dx}+\int_n^b{\left( x-m \right) \left( x-n \right) f\left( x \right) dx}>0,
\end{align*}
这与
\begin{align*}
\int_a^b{\left( x-m \right) \left( x-n \right) f\left( x \right) dx}= &\int_a^b{x^2f\left( x \right) dx}
\\
&-\left( m+n \right) \int_a^b{xf\left( x \right) dx}+mn\int_a^b{f\left( x \right) dx}=0
\end{align*}
矛盾.因此$f(x)$在$(a,b)$上至少有三个零点.
\end{Proof}

\begin{Example}
设$f\in C[a,b]$,且满足条件
$$\int_{a}^{b} x^kf(x)dx=0\quad (k=0,1,\cdots,n)$$
证明:函数$f$在$(a,b)$内至少有$n+1$个不同的零点.
\end{Example}
\begin{Proof}
(谢惠民上册P316)
\end{Proof}

\begin{Example}
设$f(x)=\sum_{k=0}^{n}a_{k}\cos{kx}$.其中系数$a_i(i=0,1,\cdots,n)$都是常数,且$a_{n}>|a_{0}|+|a_{1}|+\cdots+|a_{n-1}|$,证明:$f^{n}(x)$在$(0,2\pi)$内至少有$n$个零点.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
设函数$f$是以$2\pi$为周期的连续函数,不恒等于$0$,且
$$\int_ {-\pi}^{\pi}f(x)\sin kxdx=\int_ {-\pi}^{\pi}f(x)\cos kxdx=0,\quad k=0,1,\cdots,n.$$
证明: $f$在任何长度大于$2\pi$的区间上至少改变符号$2n+2$次.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}

\end{Example}
\begin{Proof}

\end{Proof}

---

\begin{Example}
给定正整数$n>1$,记$\xi_j=\cos\frac{2j\pi}{n}+i\sin\frac{2j\pi} {n}$,
$j=1,2,\cdots,n$.
试计算: $$\prod_{1\leqslant j<k\leqslant n}(\xi_j-\xi_k)^2.$$
\end{Example}
\begin{Proof}
由等式
\[
x^n-1=\left( x-\xi _1 \right) \left( x-\xi _2 \right) \cdots \left( x-\xi _n \right) \tag{1}
\]
可得
$$\xi _1\xi _2\cdots \xi _n=(-1)^{n-1}.$$
对(1)求导可得
$$
nx^{n-1}=\sum_{i=1}^n{\prod_{k\ne i}{\left( x-\xi _k \right)}}.
$$
分别令$x=\xi _1,\xi _2,\cdots, \xi _n$可得
$$
\begin{cases}
nx_{1}^{n-1}=\left( \xi _1-\xi _2 \right) \left( \xi _1-\xi _3 \right) \cdots \left( \xi _1-\xi _n \right),\\
nx_{2}^{n-1}=\left( \xi _2-\xi _1 \right) \left( \xi _2-\xi _3 \right) \cdots \left( \xi _2-\xi _n \right),\\
\cdots \cdots \cdots \cdots \cdots \cdots\\
nx_{n}^{n-1}=\left( \xi _n-\xi _1 \right) \left( \xi _n-\xi _2 \right) \cdots \left( \xi _n-\xi _{n-1} \right).
\end{cases}
$$
将上述式子相乘可得
\begin{align*}
n^n\left( \xi _1\xi _2\cdots \xi _n \right) ^{n-1} &=\left( -1 \right) ^{1+2+\cdots +n-1}\prod_{1\leqslant j<k\leqslant n-1}{\left( \xi _j-\xi _k \right) ^2}
\\
&=\left( -1 \right) ^{\frac{n\left( n-1 \right)}{2}}\prod_{1\leqslant j<k\leqslant n}{\left( \xi _j-\xi _k \right) ^2},
\end{align*}
因此
$$
\prod_{1\leqslant j<k\leqslant n}{\left( \xi _j-\xi _k \right) ^2}=\left( -1 \right) ^{\frac{\left( n-1 \right) \left( n-2 \right)}{2}}n^n.
$$

\textbf{证法二.}记$z=e^{\frac{2\pi i}{n}}$,构造Fourier矩阵
$$
V=\left( \begin{matrix}
1& 1& 1& \cdots& 1\\
1& z& z^2& \cdots& z^{n-1}\\
1& z^2& z^4& \cdots& z^{2\left( n-1 \right)}\\
\vdots& \vdots& \vdots& & \vdots\\
1& z^{n-1}& z^{2\left( n-1 \right)}& \cdots& z^{\left( n-1 \right) \times \left( n-1 \right)}\\
\end{matrix} \right),
$$
则Fourier行列式为
$$
\det \left( V \right) =\prod_{0\leqslant j<k\leqslant n-1}{\left( z^k-z^j \right)},
$$
则原式$=[\det \left( V \right)]^2$.用$V^\ast$表示$V$的共轭转置,则
\[|\det \left( V \right)|^2=\det(VV^\ast).\]
考察$VV^\ast$的$(i,j)$元
$$V(i,j)=1+z^{(i-j)}+z^{2(i-j)}+\cdots++z^{(n-1)(i-j)}.$$
当$\alpha\in \mathbb{Z},n\in \mathbb{N}^\ast$时,
$$\sum_{k=1}^{n}e^{\frac{2\pi ik\alpha}{n}}=\frac{1-e^{2\pi i\alpha}}{1-e^{\frac{2\pi i\alpha}{n}}}=\begin{cases}
0, & \mbox{若}\ n\nmid \alpha, \\
n, & \mbox{若}\ n\mid \alpha.
\end{cases}$$
可知在$VV^\ast$中仅对角线非零,为$n$,从而
$$|\det(V)|=n^{\frac{n}{2}}.$$
令$\alpha=e^{\frac{\pi i}{n}},z=\alpha^2$,则
$$
z^k-z^j=\alpha ^{2k}-\alpha ^{2j}=\alpha ^{k+j}\left( \alpha ^{k-j}-\alpha ^{-\left( k-j \right)} \right) =\alpha ^{k+j}\left( 2i\sin \frac{k-j}{n}\pi \right),
$$
则
$$
\det \left( V \right) =\prod_{0\leqslant j<k\leqslant n-1}{\alpha ^{k+j}\left( 2i\sin \frac{k-j}{n}\pi \right)}=M\prod_{0\leqslant j<k\leqslant n-1}{\left( 2\sin \frac{k-j}{n}\pi \right)},
$$
其中
$$
M=i^{\frac{n\left( n+1 \right)}{2}}\prod_{0\leqslant j<k\leqslant n-1}{\alpha ^{k+j}}.
$$
显然$|M|=1$,且$\sin \frac{k-j}{n}\pi>0$,则
$$
\prod_{0\leqslant j<k\leqslant n-1}{\left( 2\sin \frac{k-j}{n}\pi \right)}=n^{\frac{n}{2}}.
$$
仅需计算$M$的具体表达式,事实上只需计算$\sum_{0\leqslant j<k\leqslant n-1}{\left( k+j \right)}$.而
\begin{align*}
\sum_{0\leqslant j<k\leqslant n-1}{\left( k+j \right)} &=\sum_{j=0}^{n-2}{\sum_{k=j+1}^{n-1}{\left( k+j \right)}}=\sum_{j=0}^{n-2}{\sum_{k=j+1}^{n-1}{k}}+\sum_{j=0}^{n-2}{\sum_{k=j+1}^{n-1}{j}}
\\
&=\sum_{j=0}^{n-2}{\frac{\left( n-j-1 \right) \left( n-j \right)}{2}}+\sum_{j=0}^{n-2}{j\left( n-j-1 \right)}
\\
&=\frac{1}{2}\sum_{j=0}^{n-2}{\left( n^2+2nj-3j^2-n-2j \right)}=\frac{n\left( n-1 \right) ^2}{2},
\end{align*}
则
\begin{align*}
M &=i^{\frac{n\left( n+1 \right)}{2}}\prod_{0\leqslant j<k\leqslant n-1}{\alpha ^{k+j}}=i^{\frac{n\left( n+1 \right)}{2}}\alpha ^{\frac{n\left( n-1 \right) ^2}{2}}
\\
&=i^{\frac{n\left( n+1 \right)}{2}}e^{\frac{\pi i}{n}\cdot \frac{n\left( n-1 \right) ^2}{2}}=i^{\frac{n\left( n+1 \right)}{2}+\left( n-1 \right) ^2}=i^{\frac{\left( n-1 \right) \left( 3n-2 \right)}{2}}.
\end{align*}
最终得到
$$
\det \left( V \right) =i^{\frac{\left( n-1 \right) \left( 3n-2 \right)}{2}}n^{\frac{n}{2}},\quad \prod_{1\leqslant j<k\leqslant n}{\left( \xi _j-\xi _k \right) ^2}=\left( -1 \right) ^{\frac{\left( n-1 \right) \left( n-2 \right)}{2}}n^n.
$$
\end{Proof}

\begin{Example}
(1997, 2000年国家集训队测试)给定正整数$n>1$,记$\xi_j=\cos\frac{2j\pi}{n}+i\sin\frac{2j\pi} {n}$,
$j=1,2,\cdots,n-1$.
试计算: $$\prod_{1\leqslant j<k\leqslant n-1}(\xi_j-\xi_k)^2.$$
\end{Example}

\begin{Proof}
注意到$\xi_n=1$,且
$$
\prod_{1\leqslant j\leqslant n-1}{\left( \xi _j-\xi _n \right) ^2}=\left[ \left( \xi _n-\xi _1 \right) \left( \xi _n-\xi _2 \right) \cdots \left( \xi _n-\xi _{n-1} \right) \right] ^2=n^2,
$$
因此
$$
\prod_{1\leqslant j\leqslant n-1}{\left( \xi _j-\xi _n \right) ^2}=\frac{1}{n^2}\prod_{1\leqslant j<k\leqslant n}{\left( \xi _j-\xi _k \right) ^2}=\left( -1 \right) ^{\frac{\left( n-1 \right) \left( n-2 \right)}{2}}n^{n-2}.
$$
\end{Proof}

\begin{Example}
证明:
$$I=\iint_{0<x<y<\pi}\ln |\sin (x-y)|dxdy=-\frac{1}{2}\pi^2\ln 2.$$
\end{Example}
\begin{proof}
由上述证明可知
$$
\prod_{0\leqslant j<k\leqslant n-1}{\left( 2\sin \frac{k-j}{n}\pi \right) ^2}=n^n.
$$
取对数可得
$$
2\sum_{0\leqslant j<k\leqslant n-1}{\ln 2}+2\sum_{0\leqslant j<k\leqslant n-1}{\ln \left| \sin \frac{k-j}{n}\pi \right|}=n\ln n.
$$
以$\frac{\pi^2}{n^2}$乘上式可得
$$
\frac{\pi ^2}{n^2}\sum_{0\leqslant j<k\leqslant n-1}{\ln \left| \sin \left( \frac{j\pi}{n}-\frac{k\pi}{n} \right) \right|}=\frac{\pi ^2}{2n}\ln n-\frac{\left( n-1 \right) \pi ^2}{2n}\ln 2.
$$
令$n\to\infty$可得
$$
\iint_{0<x<y<\pi}{\ln \left| \sin \left( x-y \right) \right|dxdy}=-\frac{1}{2}\pi ^2\ln 2.
$$
\end{proof}

\begin{Example}
$$\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}
=\frac{\pi^2}{\sin^2(\pi x)}.$$
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设正整数$n>2$, $\alpha_1,\alpha_2,\cdots,\alpha_n
\in\mathbb{R}$,求$\sum_{1\leqslant i<j\leqslant n}\cos^2(\alpha_i-\alpha_j)$的最小值.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
若$m,n>1$为整数,求证$S=\sum_{i=1}^{n}\frac{1}{m+i}$不是整数.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设正整数$a,b$不能表为同一个正整数的正整数次幂($a>1$),证明$\log_ab$是无理数.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
对正整数$n$,记$P_n=\sin^n\theta+\cos^n\theta$.已知$P_1$是有理数,求证$P_n$恒为有理数.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
证明$f_n(x)=\cos (n(\arccos x))$是关于$x$的$n$次多项式.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
证明正方形可剖分成任意个数多于$5$个的小正方形.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设数列$a_0,a_1,\cdots,a_n$满足$a_0=\frac{1}{2},a_{k+1}=a_k+\frac{1}{n}a_k^2$ ($k=1,2,\cdots,n$),其中$n$是一个给定的正整数.试证: $1-\frac{1}{n}<a_n<1$.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
对于正整数$n$证明$a_n=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}<\frac{1}{\sqrt{3n}}$.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设数列$a_1,a_2,\cdots$的每一项都是正整数,且$a_1=1,a_k<1+a_1+a_2+\cdots+a_{k-1}$ ($k>1$).
试证:所有的正整数都可以表成这个数列的若干项之和.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设正整数$a_0<a_1<\cdots<a_n$,证明
$$\frac{1}{[a_0,a_1]}+\frac{1}{[a_1,a_2]}+\cdots
+\frac{1}{[a_{n-1},a_n]}\leqslant 1-\frac{1}{2^n},$$
其中$[a,b]$表示$a$与$b$的最小公倍数.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设$a_1,a_2,\cdots$是严格递增且无界的正数数列,求证对足够大的$k$有:

(1) $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots
+\frac{a_k}{a_{k+1}}<k-1$;

(2) $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots
+\frac{a_k}{a_{k+1}}<k-1987$.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
设$p>1,a_i\geqslant 0$ ($i=1,2,\cdots,n$).求证:
$$
\sum_{k=1}^n{\left( \frac{a_1+a_2+\cdots +a_k}{k} \right) ^p}\leqslant \left( \frac{p}{p-1} \right) ^p\sum_{k=1}^n{a_{k}^{p}},
$$
等号仅当所有的$a_k=0$时成立.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
给定$13$个不同的实数$a_1,a_2,\cdots,a_{13}$.求证:其中至少有两个$a_i,a_j$ ($i\neq j$)满足
$$0<\frac{a_i-a_j}{1+a_ia_j}
<\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}.$$
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
$x_0,a,b\in \mathbb{N}^\ast$, $x_{n+1}=ax_n+b$ ($n=0,1,2,\cdots$).证明: $x_n$不可能全是素数.
\end{Example}
\begin{proof}

\end{proof}

\begin{Example}
是否存在函数$f:\mathbb{R}\to \mathbb{R}$,使得$f(f(x))=x^2-2,\forall x\in \mathbb{R}$ ?
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
证明:存在无穷有理数集$M$,使对任意$x\in M,x^2,x^2-5,x^2+5$都是有理数的平方.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
求证: $\left|\sum_{k=1}^{n}\frac{1}{k}\sin kx\right|<\pi+1$, $\forall x\in [0,\pi]$.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
(1997, 2000年国家集训队测试)给定正整数$n>1$,记$\xi_j=\cos\frac{2j\pi}{n}+i\sin\frac{2j\pi} {n}$,
$j=1,2,\cdots,n-1$.
试计算: $$\prod_{1\leqslant j<k\leqslant n-1}(\xi_j-\xi_k)^2.$$
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
设$p=4k+3$是质数($k>1$),约定记$\omega=\cos\frac{2\pi}{p} +i\sin\frac{2\pi}{p}$.考察复数$\beta=\sum_{j=1}^{\frac{p-1}{2}}\omega^{j^2}$,试求这复数的模$|\beta|$.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
设函数$f(x)$对$(0,1)$中任意有理数$p,q$有$f\left(\frac{p+q}{2}\right)\leqslant \frac{1}{2}[f(p)+f(q)]$.求证:对所有有理数$\lambda,x_1,x_2\in (0,1)$,有
$f(\lambda x_1+(1-\lambda)x_2)\leqslant \lambda f(x_1)+(1-\lambda)f(x_2)$.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
(1996年国家集训队测试)设$P(x)=\sum_{i=0}^na_ix^i(1-x)^{n-i}$,且
$a_0+\sum_{a_i<0}\left(1-\frac{i}{n}\right)a_i>0$
与$a_n+\sum_{a_i<0}\frac{i}{n}a_i>0$同时成,这里$\sum_{a_i<0}$表示对一切负系数$a_i$求和,求证: $P(x)>0$对$x\in [0,1]$成立.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
(1993年国家集训队测试)已知$a_1=a_2=\frac{1}{3}$,
$$a_n=\frac{(1-2a_{n-2})a_{n-1}^2}{2a_{n-1}^2-4a_{n-2}a_{n-1}^2+a_{n-2}},\quad n=3,4,5,\cdots$$

(1)求数列$\{a_n\}$的通项公式.

(2)证明$\frac{1}{a_n}-2$是整数的平方.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
(2002年国家集训队测试)已知$a_1=1,a_2=5,a_{n+1}=\frac{a_na_{n-1}}{\sqrt{a_n^2+a_{n-1}^2+1}}$,求$\{a_n\}$的通项公式.
\end{Example}
\begin{Proof}

\end{Proof}

\begin{Example}
设$\displaystyle p(x)=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}$,若$\displaystyle a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}>0$, 且$\displaystyle a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}>0 $,求证：$\forall x\in[0,1]$,有$p(x)>0$.
\end{Example}
\begin{Proof}
由Weight-AM-GM,有 \[ \left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n}\geq (1-x)^{n-i}x^{i} \] 这时对$p(x)$,有 \begin{align*} p(x)&=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\ &=a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{i=1}^{n-1}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\ &\geq a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\ &\geq a_{0}(1-x)^{n}+ a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}\left[\left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n} \right]\\ &=(1-x)^{n}\left[a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}\right]+x^{n}\left[a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}\right]\\ &>0 \end{align*}
\end{Proof}

Vo Quoc Ba Can的一个漂亮的引理:

\begin{Example}
For all postive real numbers $a_{1},a_{2},\cdots,a_{n}$,the following inequality holds
\[ \sum_{k=1}^{n}{a^{\frac{k}{k+1}}_{k}}\leq \sum_{k=1}^{n}{a_{k}}+\sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}} \]
\end{Example}
\begin{Proof}
We begin with a preliminary result.\\
\textbf{Lemma}.If
\[ m_{k}=\left\{
\begin{array}{ll}
1, & \hbox{if \ $k=1$;} \\
2\sqrt{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}, & \hbox{if \ $k>1$ .}
\end{array}
\right.
\]
then
\[ a_{k}^{\frac{k}{k+1}}\leq a_{k}+m_{k}\sqrt{a_{k}} \]
Proof,If $k=1$,this is trivially true.Otherwise,if $k>1$,the inequality can be written as
\[ a_{k}^{\frac{k}{k+1}}\left(1-a^{\frac{1}{k+1}}_{k} \right)\leq m_{k}\sqrt{a_{k}} \]
For the nontrivial case $0<a_{k}<1$,this inequality is equivalent to
\[ a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k} \right)^{2}\leq m^{2}_{k} \]
To prove it,we use the AM-GM Inequality,as follows\\
\begin{align*}
a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k}\right)^{2}
&=4(k-1)^{k-1}\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)^{k-1}\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)^{2}\\
&\leq4(k-1)^{k-1}\left[\frac{(k-1)\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)+2\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)}{k+1}\right]^{k+1}\\
&=\frac{4(k-1)^{k-1}}{(k+1)^{k+1}}\\
&=m^{2}_{k}
\end{align*}
and so the lemma is proved.\\
Returning to our problem,we see that it suffices to show that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}}\]
But the \emph{Cauchy-Schwarz} Inequality gives us that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\left(\sum_{k=1}^{n}{m^{2}_{k}}\right)\left(\sum_{k=1}^{n}a_{k}\right)}\]
So,it remains to prove that
\[ \sum_{k=1}^{n}{m^{2}_{k}}\leq \frac{2(\pi^2-3)}{9}\]
or
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
Now,because
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}=(k+1)^{2}\left(\frac{k+1}{k-1}\right)^{k-1}=(k+1)^2\left(1+\frac{2}{k-1}\right)^{k-1}\]
\emph{Bernoulli's} Inequality yields
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}\geq 3(k+1)^{2} \]
and thus
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{1}{4}+\frac{1}{3}\sum_{k=2}^{n}{\frac{1}{(k+1)^{2}}}=\frac{1}{3}\sum_{k=1}^{n+1}{\frac{1}{k^2}}-\frac{1}{6}\]
On the other hand
\[ \lim_{n\rightarrow \infty}{\left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)}=\zeta{(2)}=\frac{\pi^2}{6}\]
and therefore it follows that
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
which finishes our proof.Notice also that equality does not occur.
\end{Proof}