陈洪葛的博客

1.设$A = \left( {\begin{array}{*{20}{c}}
1&2&0&0\\
1&3&0&0\\
0&0&0&2\\
0&0&{ - 1}&0
\end{array}} \right)$且$\left[\left(\frac{1}{2}A\right)^{*}\right]^{-1}BA=6AB+12E$,求$B$.

解：我们有$A^{*}A=AA^{*}=|A|E$,而
\[A = \left( {\begin{array}{*{20}{c}}
X&{}\\
{}&Y
\end{array}} \right)\]
其中
\[ X = \left( {\begin{array}{*{20}{c}}
1&2\\
1&3
\end{array}} \right)\qquad Y = \left( {\begin{array}{*{20}{c}}
0&2\\
{-1}&0
\end{array}} \right)\]
则
\[ |A|=|X||Y|=1\cdot 2=2\]
而
\[ BA=3\cdot |A|B+6A^{*}=6B+6A^{*}\]
\[\Rightarrow B(A-6E)=6A^{*}\]
\[\Rightarrow B(A-6E)A=12E \]
\[ B=12[(A-6E)A]^{-1} \]
而
\[\left( {A - 6E} \right)A = \left( {\begin{array}{*{20}{c}}
{ - 5}&2&{}&{}\\
1&{ - 3}&{}&{}\\
{}&{}&{ - 6}&2\\
{}&{}&{ - 1}&{ - 6}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
1&2&{}&{}\\
1&3&{}&{}\\
{}&{}&0&2\\
{}&{}&{ - 1}&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 3}&{ - 4}&{}&{}\\
{ - 2}&{ - 7}&{}&{}\\
{}&{}&{ - 2}&{ - 12}\\
{}&{}&6&{ - 2}
\end{array}} \right)\]
\[{\left( {\begin{array}{*{20}{c}}
{ - 3}&{ - 4}&{}&{}\\
{ - 2}&{ - 7}&{}&{}\\
{}&{}&{ - 2}&{ - 12}\\
{}&{}&6&{ - 2}
\end{array}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{ - \frac{7}{{13}}}&{\frac{4}{{13}}}&{}&{}\\
{\frac{2}{{13}}}&{ - \frac{3}{{13}}}&{}&{}\\
{}&{}&{ - \frac{1}{{38}}}&{\frac{6}{{38}}}\\
{}&{}&{ - \frac{3}{{38}}}&{ - \frac{1}{{38}}}
\end{array}} \right)\]
所以
\[B = 12\left( {\begin{array}{*{20}{c}}
{ - \frac{7}{{13}}}&{\frac{4}{{13}}}&{}&{}\\
{\frac{2}{{13}}}&{ - \frac{3}{{13}}}&{}&{}\\
{}&{}&{ - \frac{1}{{38}}}&{\frac{6}{{38}}}\\
{}&{}&{ - \frac{3}{{38}}}&{ - \frac{1}{{38}}}
\end{array}} \right)\]
二、计算$D = \left| {\begin{array}{*{20}{c}}
{{s_0}}&{{s_1}}& \cdots &{{s_{n - 1}}}\\
{{s_1}}&{{s_2}}& \cdots &{{s_n}}\\
 \vdots & \vdots & \cdots & \vdots \\
{{s_n}}&{{s_{n + 1}}}& \cdots &{{s_{2n - 1}}}
\end{array}\begin{array}{*{20}{c}}
1\\
x\\
 \vdots \\
{{x^n}}
\end{array}} \right|$,其中${s_k} = x_1^k + x_2^k +  \cdots x_n^k$

解：注意到
\begin{align*}
D &= \left| {\begin{array}{*{20}{c}}
{1 + 1 +  \cdots  + 1}&{{x_1} + {x_2} +  \cdots  + {x_n}}& \cdots &{x_1^{n - 1} + x_2^{n - 1} +  \cdots  + x_n^{n - 1}}&1\\
{{x_1} + {x_2} +  \cdots  + {x_n}}&{x_1^2 + x_2^2 +  \cdots  + x_n^2}& \cdots &{x_1^n + x_2^n +  \cdots  + x_n^n}&x\\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
{x_1^n + x_2^n +  \cdots  + x_n^n}&{x_1^{n + 1} + x_2^{n + 1} +  \cdots  + x_n^{n + 1}}& \cdots &{x_1^{2n - 1} + x_2^{2n - 1} +  \cdots  + x_n^{2n - 1}}&{{x^n}}
\end{array}} \right|\\
 &= \left| {\begin{array}{*{20}{c}}
1&1& \cdots &1&1\\
{{x_1}}&{{x_2}}& \cdots &{{x_n}}&x\\
{x_1^2}&{x_2^2}& \cdots &{x_n^2}&{{x^2}}\\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
{x_1^n}&{x_2^n}& \cdots &{x_n^n}&{{x^n}}
\end{array}} \right| \cdot \left| {\begin{array}{*{20}{c}}
1&{{x_1}}&{x_1^2}& \cdots &{x_1^{n - 1}}&0\\
1&{{x_2}}&{x_2^2}& \cdots &{x_2^{n - 1}}&0\\
1&{{x_3}}&{x_3^2}& \cdots &{x_3^{n - 1}}&0\\
 \vdots & \vdots & \vdots & \cdots & \vdots &0\\
1&{{x_n}}&{x_n^2}& \cdots &{x_n^{n - 1}}&0\\
0&0&0& \cdots &0&1
\end{array}} \right|\\
&=V(x_{1},x_{2},\cdots,x_{n},x)\cdot V(x_{1},x_{2},\cdots,x_{n})
\end{align*}
三、有$\alpha_{1},\alpha_{2},\cdots,\alpha_{s},\alpha_{s+1}$,且$\beta_{i}=\alpha_{i}+t_{i}\alpha_{s+1},i=1,2,\cdots,s$, 证明如果$\beta_{1},\beta_{2},\cdots,\beta_{s}$,线性无关，则$\alpha_{1},\alpha_{2},\cdots,\alpha_{s+1}$必定线性无关。

五、设$sl_{n}(F)$是$M(F)$上$A,B$矩阵满足$AB-BA$生成的子空间，证明$\dim(sl_{n}(F))=n^2-1$.

证明：
\[ sl_{n}(F)=\{AB-BA|A,B\in M(F)\} \]
对任意的$X\in sl_{n}(F)$,存在$A,B\in M(F)$,使得
\[ X=AB-BA \]
我们注意到
\[ \text{tr}(X)=\text{tr}(AB-BA)=0 \]
假设
\[X = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}& \cdots & \cdots &{{a_{1n}}}\\
{{a_{21}}}&{{a_{22}}}& \cdots & \cdots &{{a_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots \\
{{a_{n1}}}&{{a_{n2}}}& \cdots & \cdots &{{a_{nn}}}
\end{array}} \right)\]
则有$a_{11}+a_{22}+\cdots+a_{nn}=0$,于是，当$a_{22},a_{33},\cdots,a_{nn}$确定了，$a_{11}$自然就确定了。下证明
\[{E_{ij}} = \left( {\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}\\
{}&{}&{}&{}&{}\\
{}&{}&{}&1&{}\\
{}&{}&{}&{}&{}\\
{}&{}&{}&{}&{}
\end{array}} \right)(i\neq j)\]
\[{E_{ii}} = \left( {\begin{array}{*{20}{c}}
{ - 1}&{}&{}&{}&{}&{}\\
{}& \ddots &{}&{}&{}&{}\\
{}&{}&1&{}&{}&{}\\
{}&{}&{}& \ddots &{}&{}\\
{}&{}&{}&{}& \ddots &{}\\
{}&{}&{}&{}&{}&0
\end{array}} \right)\qquad (i\geq 2)\]
这里$E_{ij}$表示第$i$行第$j$列为$1$，其他为$0$的矩阵，$E_{ii}$表示第一行第一列为$-1$,第$i$行第$i$列为$1$，其余元素为$0$,的矩阵。不难计算发现所有的这2类矩阵数目为$n^2-1$,下证明它们可以构成$sl_{n}(F)$的一组基。
\[ \sum_{ij\neq 1}k_{ij}E_{ij}=0\]
\[\left( {\begin{array}{*{20}{c}}
{ - \sum\limits_{i = 2}^n {{k_{ii}}} }&{{k_{12}}}&{{k_{13}}}& \cdots & \cdots &{{k_{1n}}}\\
{{k_{21}}}&{{k_{22}}}&{{k_{23}}}& \cdots & \cdots &{{k_{2n}}}\\
 \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{k_{n1}}}&{{k_{n2}}}& \cdots & \cdots & \cdots &{{k_{nn}}}
\end{array}} \right) = 0\]
显然有$k_{ij}=0$,因此，这$n^2-1$个矩阵线性无关，而对任意的$X\in sl_{n}(F)$,有
\[ X=\sum_{ij\neq 1}a_{ij}E_{ij}\]
故$E_{12},\cdots,E_{1n},\cdots E_{nn}$为$sl_{n}(F)$的一组基。马上得到$\dim(sl_{n}(F))=n^2-1$。

六、设数域$K$上的$n$维线性空间$V$到$m$维线性空间上的所有线性映射组成线性空间$Hom_{k}(V,V')$.证明
(1)$Hom_{k}(V,V')$是线性空间；
(2)$Hom_{k}(V,V')$的维数为$mn$.

证明：（1）先验证线性空间的8条运算法则。对任意的$\alpha\in V$,及$\forall f,g,h\in Hom_{k}(V,V')$
1
$(f+g)(\alpha)=f(\alpha)+g(\alpha)=g(\alpha)+f(\alpha)=(g+f)(\alpha)\qquad \forall f,g\in Hom_{k}(V,V')$
2
$((f+g)+h)(\alpha)=(f+g)(\alpha)+h(\alpha)=f(\alpha)+g(\alpha)+h(\alpha)=f(\alpha)+(g+h)(\alpha)=(f+(g+h))(\alpha) $
3在$Hom_{k}(V,V')$中存在一个变换$0$,使得$f+0=f,\forall f\in Hom_{k}(V,V')$.
4.对于$Hom_{k}(V,V')$中的每个元素$f$,都存在$-f\in Hom_{k}(V,V')$,使得$f+(-f)=0$.
5.对于$Hom_{k}(V,V')$中的每个元素$f$，有$1f=f$
6.$k(lf)(\alpha)=klf(\alpha)=(kl)f(\alpha),\forall k,l\in K,\forall \alpha\in V$.
7.$(k+l)f(\alpha)=kf(\alpha)+lf(\alpha),\forall k,l \in K,\forall \alpha \in V$.
8.$(k(f+g))(\alpha)=kf(\alpha)+kg(\alpha),\forall k\in K,\forall \alpha \in V$.
所以$Hom_{k}(V,V')$是线性空间；
（2）我们在$V$中找一组基，记为$\eta_{1},\eta_{2},\cdots,\eta_{n}$,$V'$中找一组基，记为$\nu_{1},\nu_{2},\cdots,\nu_{m}$,则对任意的$\alpha\in V$,
\[ \alpha=\sum_{k=1}x_{k}\eta_{k} \]
\[ f(\alpha)=\sum_{k=1}^{n}x_{k}f(\eta_{k})\]
而对于$f(\eta_{k})\in V'$,
\[ f(\eta_{k})=\sum_{j=1}^{m}y_{kj}\nu_{j}\]
所以
\[ f(\alpha)=\sum_{k=1}^{n}\sum_{j=1}^{m}x_{k}y_{kj}\nu_{j}\]
\[ f(\eta_{1},\eta_{2},\cdots,\eta_{n})=(\nu_{1},\nu_{2},\cdots,\nu_{m})\left( {\begin{array}{*{20}{c}}
{{y_{11}}}&{{y_{12}}}& \cdots & \cdots & \cdots &{{y_{1n}}}\\
{{y_{21}}}&{{y_{22}}}& \cdots & \cdots & \cdots &{{y_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{y_{m1}}}&{{y_{m2}}}& \cdots & \cdots & \cdots &{{y_{mn}}}
\end{array}} \right)\]
记
\[ A=\left( {\begin{array}{*{20}{c}}
{{y_{11}}}&{{y_{12}}}& \cdots & \cdots & \cdots &{{y_{1n}}}\\
{{y_{21}}}&{{y_{22}}}& \cdots & \cdots & \cdots &{{y_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{y_{m1}}}&{{y_{m2}}}& \cdots & \cdots & \cdots &{{y_{mn}}}
\end{array}} \right)\]
构造映射$\phi:f\to A$,则$\varphi$是从$Hom_{n}(V,V')$到$M_{m\times n}$上的映射。则有
\[ \phi(f+g)=\phi(f)+\phi(g) \]
\[ \phi(kf)=k\phi(f) \]
对于任意的$f,g$满足
\[ \phi(f)=\phi(g)\]
则有
\[ f=g \]
对任意的$A\in M_{m\times n}$,存在$f\in Hom_{n}(V,V')$,使得$\phi(f)=A$,所以
\[ Hom_{n}(V,V')\cong M_{m\times n}\]
而我们知道$\dim(M_{m\times n})=mn$,故$Hom_{k}(V,V')$的维数为$mn$.

七、已知
\[F = \left( {\begin{array}{*{20}{c}}
0&{}&{}&{}&{}&{ - {c_0}}\\
1&0&{}&{}&{}&{ - {c_1}}\\
{}&1&0&{}&{}& \vdots \\
{}&{}&1& \ddots &{}&{ - {c_{n - 3}}}\\
{}&{}&{}& \ddots &0&{ - {c_{n - 2}}}\\
{}&{}&{}&{}&1&{ - {c_{n - 1}}}
\end{array}} \right)\]
(1)求$F$的特征多项式$f(x)$与最小多项式$m(x)$;
(2)求所有与$F$可交换的矩阵.

八、设$\varphi$是复数域上的线性变换，$\epsilon$为恒等变换，$\lambda_{0}$为$\varphi$的一个特征值，$\lambda_{0}$在$\varphi$的最小多项式中的重数$$m_{0}=\min_{k}\{k\in\mathbf{N}^{+}|\text{ker}(\lambda_{0}\epsilon-\varphi)^{k}=\text{ker}(\lambda_{0}\epsilon-\varphi)^{k+1}\}$$

 

问题：$x_{1},x_{2}>0$,$\{x_{n}\}$满足
\[ x_{n+1}=x_{1}^{\frac{1}{n}}+x_{2}^{\frac{1}{n}}+\cdots+x_{n}^{\frac{1}{n}}\]
求
\[ \lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}\]
解：设$a=\max\{x_{1},x_{2},4\}$,由Holder不等式。可以得到
\[ x_{n+1}^{n}\leq \left(\sum_{k=1}^{n}x_{k}\right)\cdot n^{n-1}\]
\[ x_{n+1}\leq  \left(\sum_{k=1}^{n}x_{k}\right)^{\frac{1}{n}}\cdot n^{\frac{n-1}{n}}\]
这样，用数学归纳法可以证明
\[ 0<a_{n}<an \]
而Taylor公式告诉我们
\[ x_{k}^{\frac{1}{n}}=1+\frac{1}{n}\ln x_{k}+o\left(\frac{1}{n}\right)\]
于是
\[ x_{n}=n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)\]
\begin{align*}
\lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}&=\lim_{n\to\infty}\frac{\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}-1 }{\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln \prod_{k=1}^{n-1}x_{k}-(n-1)}{(n-1)\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{n\ln(n+1)-(n-1)\ln{n}}\qquad (O.Stolz)\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{\ln(n+1)}\\
&=\lim_{n\to\infty}\frac{\ln \frac{x_{n+1}}{x_{n}}}{\frac{1}{n+1}}\qquad (O.Stolz)\\
&=1
\end{align*}
最后一步只要注意到
\[ \frac{x_{n+1}}{x_{n}}=\frac{n+\frac{1}{n}\ln\prod_{k=1}^{n}x_{k}+o\left(\frac{1}{n}\right)}{n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)}\]
利用上面的放缩，可以看到
\[ \ln\left(\frac{x_{n+1}}{x_{n}}\right)\sim \ln\left(1+\frac{1}{n-1}\right)\quad (n\to+\infty) \]

设$k$是一个大于$1$的整数，且$x_0>0$,满足$x_{n+1}=x_{n}+\frac{1}{\sqrt[k]{x_{n}}}$,求
\[ \lim_{x\to\infty}\frac{x_{n}^{k+1}}{n^{k}}\]
求证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}\]
证明：
我们将要证明
\[  \lim_{n\to\infty}\frac{x^{k+1}_{n}}{n^{k}}=\left(\frac{k+1}{k}\right)^{k}\]
为此，只要证明
\[  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
而归纳得到
\[ x_{n+1}>x_{n},x_{n}>0\]
因此，$x_{n}$的单调递增序列。设$A=\lim_{n\to+\infty}x_{n}$,则有$A=+\infty$.说明$x_{n}\to+\infty$,$n\to+\infty$.
\[ (x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}})=x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}} \right)^{1+\frac{1}{k}}-1\right)=x_{n}^{1+\frac{1}{k}}\left(\exp\left[\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}} \right) \right]-1 \right)\sim \frac{k+1}{k}(n\to+\infty)\]
所以，由O.Stolz定理
\[  \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
对于加强版本，我们先看一个事实，对任意一个收敛的序列，比如说
\[ \lim_{n\to\infty}a_{n}=A \]
那么，
\[ \lim_{n\to\infty}\frac{(a_{n}^{k}-A^{k})}{a_{n}-A}=kA^{k-1}\]
这样，设$a_{n}=\frac{x_{n}^{\frac{k+1}{k}}}{n}$,就有$\displaystyle \lim_{n\to\infty}a_{n}=\frac{k+1}{k}=A$.
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}^{k}-A^{k})= \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)\cdot k\cdot \left(\frac{k+1}{k}\right)^{k-1}\]
于是，只要证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)=\frac{1}{2k}\]
就是
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{1}{2k}\]
\[ \frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\]
这时，可以用O.Stolz定理了。就有
\begin{align*}
&\lim_{n\to\infty}\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\\
&\overbrace{=}^{O.Stolz}\lim_{n\to\infty}\frac{x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}}-A}{\ln\left(1+\frac{1}{n} \right)}\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}}\right)^{1+\frac{1}{k}}-1\right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}}\right) } -1\right)-A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}+o\left( \left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right) \right) }-1 \right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{k+1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)+\frac{1}{2}\left(\frac{k+1}{k}\right)^{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)^{2} \right)-A\right]\\
&=\lim_{n\to\infty}\frac{k+1}{2k^{2}}\frac{n}{x_{n}^{1+\frac{1}{k}}}\\
&=\frac{1}{2k}
\end{align*}
因此，命题得证。

Problems is the heart of Math!

_______________________________________________________

西西每日一题

（2006PTN）
1设$k$是一个大于$1$的整数，且$x_0>0$,满足$x_{n+1}=x_{n}+\frac{1}{\sqrt[k]{x_{n}}}$,求
\[ \lim_{x\to\infty}\frac{x_{n}^{k+1}}{n^{k}}\]
求证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}\]

2 正数序列$\{a_{n}\}$满足
\[ a_{n}\sum_{n=1}^{\infty}a_{n}^{p}=1\qquad n\geq 1 \]
求证：
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(1-n(p+1)a_{n}^{p+1})=\frac{p+2}{2(p+1)}\qquad p>-1 \]

3.熟悉的IMO预选题
\[\dfrac{1}{1+x_{1}}+\dfrac{1}{1+x_{1}+x_{2}}+\cdots+\dfrac{1}{1+x_{1}+x_{2}+\cdots+x_{n}}<\sqrt{\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}}+\cdots+\dfrac{1}{x_{n}}}\]
加强到
\[\dfrac{1}{1+x_{1}}+\dfrac{1}{1+x_{1}+x_{2}}+\cdots+\dfrac{1}{1+x_{1}+x_{2}+\cdots+x_{n}}<\left(1-\dfrac{\ln{n}}{2n}\right)\sqrt{\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}}+\cdots+\dfrac{1}{x_{n}}}
\]

4.
设$a_{i}>0$,$i=1,2,\cdots,n$,$n\geq 3$,求证
\[\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le\left(2-\dfrac{7\ln{2}}{8\ln{n}}\right)\sum_{k=1}^{n}\dfrac{1}{a_{k}}
\]

5.已知实系数多项式$f(x)=\displaystyle\sum_{i=0}^{n}a_{i}x^i$,且满足对于任意$|x|\leq 1$,都有$|f(x)|\leq 1$.
证明：
（1）
\[|a_{n}|+|a_{n-1}|\le 2^{n-1}\]
（2）
\[ |a_{n}|+|a_{n-1}|+\cdots+|a_{1}|+|a_{0}| \le \frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}\]

6.(代数，求行列式)
\[\left| {\begin{array}{*{20}{c}}
9&{26}&{24}&0& \cdots &0&0\\
1&9&{26}&{24}& \cdots &0&0\\
0&1&9&{26}& \cdots &0&0\\
0& \ddots &1&9& \ddots & \ddots & \vdots \\
 \vdots & \ddots & \ddots & \ddots & \ddots & \ddots &{24}\\
0&0& \ddots & \ddots & \ddots &9&{26}\\
0&0& \cdots &0&0&1&9
\end{array}} \right|\]
\[\left| {\begin{array}{*{20}{c}}
x&1&{}&{}&{}&{}\\
{ - n}&{x - 2}&2&{}&{}&{}\\
{}&{ - \left( {n - 1} \right)}&{x - 4}& \ddots &{}&{}\\
{}&{}& \ddots & \ddots &{n - 1}&{}\\
{}&{}&{}&{ - 2}&{x - 2\left( {n - 1} \right)}&n\\
{}&{}&{}&{}&{ - 1}&{x - 2n}
\end{array}} \right|\]
其中未填的为0。(李炯生，线性代数上习题)

7 设矩阵\[A = \left| {\begin{array}{*{20}{c}}
{2012}&1&1& \cdots &1\\
1&{2012}&1& \cdots &1\\
1&1&{2012}& \cdots &1\\
 \vdots & \vdots & \ddots & \cdots & \vdots \\
1&1&1& \cdots &{2012}
\end{array}} \right|\]
设$f(x)=\frac{1}{2}(x+1)^{2013}$,求$|f(A)|$.

8.设$f(x)$在$[0,1]$上连续，且$\int_{0}^{1}f(x)dx=1$,若$\lambda\in(0,1)$,证明存在不同的$\xi,\eta$,使得
\[ \lambda f(\xi)+(1-\lambda)f(\eta)=1 \]

9.设函数$f(x)=\frac{1}{5}(x^3+4x^2+6x-6)$,若数列$\{x_n\}$满足$x_{n+1}=f(x_n)$,$n=0,1,2,\cdots$.且$\lim_{n\to\infty}x_{n}$存在，求$x_0$范围，并求该极限.

10.对$x\neq 0$,且$|x|$充分小，证明：存在唯一的$\theta(x)\in[0,1]$使得
\[ \sin{x}=x-\frac{x^3}{6}\cos(x\theta(x)) \]
并求$\lim_{x\to 0}\theta(x)$

11.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。

12.设$f(x)$满足$f''(x)>0$,$\int_{0}^{1}f(x)dx=0$,证明$\forall x\in[0,1]$,必有
\[ |f(x)|\leq \max\{f(0),f(1)\} \]

13.设数列$\{x_n\}$满足$0<x_0<\pi$,且$x_n=\sin{x_{n-1}}$,$n=1,2,\cdots$,求证
\[ x_{n}=\sqrt{\frac{3}{n}}\left[1-\frac{3}{10}\cdot\frac{\ln{n}}{n}-\frac{C}{2n}+\frac{271\ln^2{n}}{200n^2}+\frac{\beta\ln{n}+\gamma}{n^2}+O\left(\frac{\ln^3{n}}{n^3}\right) \right]\]
$C=C(x_0)$,且$\lim_{x\to0^{+}}C(x_0)=+\infty$,$\beta=\frac{9}{20}C-\frac{9}{50}.\gamma=\frac{3}{8}C^2-\frac{3}{10}C+\frac{79}{700}$
$x,y,z>0$,求证：
\[ (x^2+y^2+z^2)\left[\frac{x^2}{(x^2+yz)^2}+\frac{y^2}{(y^2+xz)^2}+\frac{z^2}{(z^2+xy)^2}\right]\geq \frac{9}{4}\]

14计算积分(西西博客)
\[\int_0^{+\infty}  {\frac{{x\cos x}}{{\left( {1 + {x^2}} \right)\left( {2 + {x^2}} \right)}}dx} \]

15
求
\[ \sum_{n=1}^{\infty}\frac{\ln{2}-\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)}{n}\]

16
\[ \int_{0}^{1}\left\{\frac{1}{x}\right\}^{4}dx \]

17
设数列$\{x_{n}\}$满足$x_0>0$,且$x_1+x_2+\cdots+x_n=\frac{1}{\sqrt{x_{n+1}}}$,$n\geq 0$,求
\[ \lim_{n\to\infty}\left[\frac{n\left(n^2x_{n}^{3}-\frac{1}{9}\right)}{\ln{n}}\right] \]

18
若数列$\{a_{n}\}$满足$a_{n+3}=a_{n+1}+a_{n}$.
\[ A=\lim_{n\to\infty}\frac{\sqrt{a_{n+5}^2+a_{n+4}^2+a_{n+3}^2-a_{n+2}^2+a_{n+1}^2-a_{n}^2}}{a_{n+6}}\]
\[ B=\lim_{n\to\infty}\frac{3^{n}}{\prod_{k=1}^{n}\frac{\arcsin\left(\frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}} \right)}{\arctan\left(\frac{1}{\sqrt{3k+1}}\right)}}\]
求$A+B$.

19证明
\[ \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2+1}\right)=\arctan\left[\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}} \right)\cdot\frac{\exp\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)+\exp\left(-\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}{\exp\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)-\exp\left(-\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)} \right]-\frac{\pi}{8} \]

20
设$a,b,c$是三个正数，且满足$abc=\frac{1}{16}$,求证
\[ \int_{0}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}dx\leq \pi \]

21
设$a_n=\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}$,求证
\[ \lim_{n\to\infty}\sqrt{n}\sqrt[n]{\left(\lim_{n\to\infty}a_{n}\right)-a_{n}}\]
存在，并计算该极限。

22
设$a_n=L_{n}-\frac{4}{\pi^2}\ln{n}$,且$\displaystyle L_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|\frac{\sin\left(n+\frac{1}{2}\right)}{\sin{\frac{x}{2}}}\right|$,证明
$\{a_{n}\}$是有界数列。

23
已知$a_1+a_2+a_3+\cdots+a_n=0$,并满足$a_i\neq \frac{\pi}{2}+k\pi,k\in \mathbb{Z}$,且$f(x)=\frac{\cos{2x}+\sin{2x}+(-1)^{n-1}\sin{2x}+(-1)^{n}\cos{2x}}{2}$
求\[D = \left| {\begin{array}{*{20}{c}}
1&1& \cdots & \cdots & \cdots &1\\
{\tan {a_1}}&{\tan {a_2}}& \cdots & \cdots & \cdots &{\tan {a_n}}\\
{{{\tan }^2}{a_1}}&{{{\tan }^2}{a_2}}& \cdots & \cdots & \cdots &{{{\tan }^2}{a_n}}\\
 \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
{{{\tan }^{n - 2}}{a_1}}&{{{\tan }^{n - 2}}{a_2}}& \cdots & \cdots & \cdots &{{{\tan }^{n - 2}}{a_n}}\\
{f\left( {{a_1}} \right)}&{f\left( {{a_2}} \right)}& \cdots & \cdots & \cdots &{f\left( {{a_n}} \right)}
\end{array}} \right|\]
的取值范围。

24
设$f(x)$是多项式函数，若对于任意的$x\in \mathbb{R}$,只要有$f(x)+f'(x)-Kf''(x)\geq 0$ 恒成立，就一定有$f(x)\geq 0$,求$K$的最小值。

24
证明
\[ \int_{0}^{\frac{\pi}{4}}(\tan{x})^{a}\geq \frac{\pi}{4+2a\pi},a>0 \]
\[ \frac{\pi^2}{12-\sqrt{3}}\left(3^{2-\frac{\sqrt{3}}{6}\pi}-\frac{1}{3}\right)\leq \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{x}{\sin{x}}dx\leq \frac{7\pi^2\ln3}{72\ln3-36\ln 2}\]

25
设$a$是正常数，证明方程
\[ \frac{1}{x}+\frac{2x}{x^2-1^2}+\frac{2x}{x^2-2^2}+\cdots+\frac{2x}{x^2-n^2}=a \]
在$[n,+\infty)$有唯一的根，并求极限$\lim_{n\to\infty}\frac{x_n}{n}$.

26
设$\displaystyle f(n)=\int_{1}^{n}\frac{(x-1)^n}{x^{4}-2x^3+3x^2-2x+1}dx$,求当$f(n)$达到最小时。计算
\[ \int_{1}^{\sqrt{3}}x^{nx^2+1}+\ln\left(x^{nx^{nx^2+1}}\right)dx\]
的值.

27
设$x_1>0$,$x_{n+1}=\frac{1}{n^{x_{n}}}+\sqrt[n]{x_{n}}$,求
\[ \lim_{n\to\infty}\frac{\ln{n}}{x_{1}+x_{2}+x_{3}+\cdots+x_{n}} \]

28
设$\displaystyle a_{n}=\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^4 dx$,证明
\[ \lim_{n\to\infty}\frac{a_{n}}{n^2}=\ln 2 \]

29.$\displaystyle \lim_{h\to 0^{+}}h^{n}\int_{c}^{\infty}\cdots\int_{c}^{\infty}\left(\int_{c}^{\infty}\cdots\int_{c}^{\infty}f(x_1,\cdots,x_n,y_1,\cdots,y_n)dy_1dy_2\cdots dy_n \right)^{p}dx_1\cdots dx_n $,其中
\[ f(x_1,\cdots,x_n,y_1,\cdots,y_n)=\left(\prod_{k=1}^{n}x_{k}y_{k}\right)^{-\frac{1}{4}}\exp\left(\sum_{k=1}^{n}\sqrt{x_{k}y_{k}}-\frac{\sum_{k=1}^{n}(x_{k}+y_{k})}{2}-\frac{h\sum_{k=1}^{n}y_{k}}{p} \right)\]
$c>0,p\geq 1$,$n$,$p$是常数。

30.设$a_{n+3}=a_{n+2}+\dfrac{a_{n}}{2n+6}$,且$a_0=a_1=a_2=1 $,求极限
\[ \lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}} \]
同时证明
\[ \lim_{n\to\infty}n\left(\frac{a_{n}}{\sqrt{n}}-\frac{2}{\sqrt{\pi}\cdot\sqrt[4]{e^{3}}}\right)=\frac{7}{4\sqrt{\pi}\cdot\sqrt[4]{e^{3}}}\]

31.若$a,b>0$,求证
\[\left| {\begin{array}{*{20}{c}}
{\Gamma \left( a \right)}&{\Gamma \left( {a + b} \right)}&{\Gamma \left( {a + 2b} \right)}\\
{\Gamma \left( {a + b} \right)}&{\Gamma \left( {a + 2b} \right)}&{\Gamma \left( {a + 3b} \right)}\\
{\Gamma \left( {a + 2b} \right)}&{\Gamma \left( {a + 3b} \right)}&{\Gamma \left( {a + 4b} \right)}
\end{array}} \right| > 0\]
32.证明：$SU(l+1)$($A_{l}$代数),有任意的$\alpha$,满足
\[ g_{\alpha}\equiv\frac{\alpha(g,\alpha)}{(\alpha,\alpha)}=1 \]

设级数
\[ \sum_{n=1}^{\infty}2^{-n}a_{n+1}(a_{1}a_{2}\cdots a_{n}a_{n+1})^{-\frac{1}{2}}\]
收敛，证明数列
\[ x_{n}=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}\]
也收敛，这里$a_{1},a_{2},\cdots,a_{n},\cdots $为正实数。

证明：注意到$\{x_{n}\}$是单调递增的。而又有
\begin{align*}
&x_{n+1}-x_{n}\\
&=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}\\
&=\frac{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}\\
&<\frac{\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}}\\
&<\frac{\sqrt{a_{3}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{3}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}\cdot 2\sqrt{a_{2}}}\\
&<\cdots\\
&<\frac{\sqrt{a_{n+1}}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n}}}\\
&=\frac{a_{n+1}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n+1}}}
\end{align*}
而由于级数收敛，所以数列$\{x_{n}\}$收敛。