Eufisky - The lost book

## 问题征解1

1.（高等代数）证明:实对称矩阵${A_n} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{1}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\\vdots & \vdots & \vdots &{}& \vdots \\{\frac{1}{n}}&{\frac{1}{n}}&{\frac{1}{n}}& \cdots &{\frac{1}{n}}\end{array}} \right]$的特征值都大于$0$,且小于等于$3+2\sqrt{2}$.

$\left\{ \begin{array}{l}b_1^2 + b_2^2 + b_3^2 + \cdots + b_n^2 = 1,\\b_2^2 + b_3^2 + \cdots + b_n^2 = \frac{1}{2},\\\vdots \\b_n^2 = \frac{1}{n}.\end{array} \right.$

$0\leq\frac{x^TB^TBx}{x^Tx}=\frac{(Bx)^TBx}{x^Tx}\leq 3+2\sqrt{2},$

$0\leq\sum_{k=1}^{n-1}{\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}}+\frac{\left(x_1+x_2+\cdots +x_n\right)^2}{n}\leq (3+2\sqrt{2})\sum_{k=1}^n{x_{k}^{2}}.$

$\left(\frac{x_{1}^{2}}{a_1}+\frac{x_{2}^{2}}{a_2}+\cdots +\frac{x_{k}^{2}}{a_k}\right)\left(a_1+a_2+\cdots +a_k\right)\geq\left(x_1+x_2+\cdots +x_k\right)^2.$

$\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}\leq\sum_{i=1}^k{\frac{a_1+a_2+\cdots +a_k}{k\left(k+1\right)a_i}x_{i}^{2}}.$

$\sum_{k=1}^{n-1}{\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}}+\frac{\left(x_1+x_2+\cdots+x_n\right)^2}{n}\leq\sum_{k=1}^n{y_kx_{k}^{2}},$

$y_k=\sum_{i=k}^{n-1}{\frac{a_1+a_2+\cdots +a_i}{i\left(i+1\right)a_k}}+\frac{a_1+a_2+\cdots +a_n}{na_k}.$

$y_k=\frac{1}{a_k}\left(\sum_{i=k}^{n-1}{\frac{1}{\left(i+1\right)\sqrt{i}}}+\frac{1}{\sqrt{n}}\right).$

\begin{align*}2\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)&=2\frac{\sqrt{i+1}-\sqrt{i}}{\sqrt{i}\cdot\sqrt{i+1}}=2\frac{1}{\sqrt{i}\cdot\sqrt{i+1}\left(\sqrt{i+1}+\sqrt{i}\right)}\\&\geq 2\frac{1}{\sqrt{i}\cdot\sqrt{i+1}\left(\sqrt{i+1}+\sqrt{i+1}\right)}=\frac{1}{\left(i+1\right)\sqrt{i}}.\end{align*}

\begin{align*}y_k&\leq\frac{1}{a_k}\left[\sum_{i=k}^{n-1}{2\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)}+\frac{1}{\sqrt{n}}\right]=\frac{1}{a_k}\left[2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{n}}\right)+\frac{1}{\sqrt{n}}\right]\\&\leq\frac{2}{a_k\sqrt{k}}=\frac{2}{\left(\sqrt{k}-\sqrt{k-1}\right)\sqrt{k}}=\frac{2\left(\sqrt{k}+\sqrt{k-1}\right)}{\sqrt{k}}\\&=2\left(1+\sqrt{1-\frac{1}{k}}\right)<4<3+2\sqrt{2}.\end{align*}

2.（高等概率论）Let the stochastic processes $\{X_k,1\leq k\leq n\}$ and $\{X'_k,1\leq k\leq n\}$ be independent of one another and have the same joint distributions. If $m_k$ is a median of $X_k,1\leq k\leq n$. Prove: for $\lambda>0$,

$$P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k} - {m_k}} \right| \ge \lambda } \right) \le 2P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k} - {X'_k}} \right| \ge \lambda } \right).$$

$A_k^+=\left\{S_{k-1}^{\ast}<\lambda , X_k-m_k\geq\lambda\right\},\quad A_k=\left\{S_{k-1}^{\ast}<\lambda , X_k-m_k\leq -\lambda\right\},$则有$A_k^+\cap A_k^-=\emptyset,A_k^+\cup A_k^-=A_k$,且$\forall 1\leq i<j\leq n, A_i^+\cap A_j^+=A_i^-\cap A_j^-=\emptyset$.又因为$A_i\cap A_j=\emptyset$,则$\forall 1\leq i<j\leq n,A_i^+\cap A_j^-=\emptyset$.再令$M_{k}^{+}=\left\{m_k-X'_k\geq 0\right\},M_{k}^{-}=\left\{m_k-X'_k\leq 0\right\}$且$B_k=\left\{\underset{1\leq j\leq k-1}{\max}\left| X_j-X'_j\right|<\lambda ,\left| X_k-X'_k\right|\geq\lambda\right\},$

$P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)=P\left(\bigcap_{k=1}^n{B_k}\right).$

$P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)=P\left(\bigcap_{k=1}^n{B_k^+}\right)+P\left(\bigcap_{k=1}^n{B_k^-}\right).$

\begin{align*}P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)&\geq P\left(\bigcap_{k=1}^n{\left(A_{k}^{+}\cap M_{k}^{+}\right)}\right)+P\left(\bigcap_{k=1}^n{\left(A_{k}^{-}\cap M_{k}^{-}\right)}\right)\\&=\sum_{k=1}^n{P\left(A_{k}^{+}\right)\cdot P\left(M_{k}^{+}\right)}+\sum_{k=1}^n{P\left(A_{k}^{-}\right)\cdot P\left(M_{k}^{-}\right)}\\&=\frac{1}{2}\sum_{k=1}^n{\left[P\left(A_{k}^{+}\right)+P\left(A_{k}^{-}\right)\right]}=\frac{1}{2}P\left(\bigcap_{k=1}^n{\left(A_{k}^{+}\cap A_{k}^{-}\right)}\right)\\&=\frac{1}{2}P\left(\bigcap_{k=1}^n{A_k}\right)=\frac{1}{2}P\left\{\underset{1\leq k\leq n}{\max}\left| X_k-m_k\right|\geq\lambda\right\}.\end{align*}

3.（高等代数）$F$为数域, $A,B,P\in M_n(F)$, $P$幂零且$(A-B)P=P(A-B),\qquad BP-PB=2(A-B).$

\begin{align*}B{G^n} - {G^n}B&= \left( {B{G^n} - GB{G^{n - 1}}} \right) + \left( {GB{G^{n - 1}} - {G^2}B{G^{n - 1}}} \right) + \cdots + \left( {{G^{n - 1}}BG - {G^n}B} \right)\\&= H{G^{n - 1}} + GH{G^{n - 2}} + \cdots + {G^{n - 1}}H = nH{G^{n - 1}},\end{align*}

$B{G^n} - {G^n}B=nH{G^{n - 1}}.$

\begin{align*}\left( {B + H} \right)Q - QB &= \left( {B + H} \right)\left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right) - \left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right)B\\&= \sum\limits_{n \ge 0} {\left[ {\left( {B + H} \right)\frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}} - \frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}}B} \right]} = \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {B{G^n} - {G^n}B + H{G^n}} \right)} \\&= \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {nH{G^{n - 1}} + H{G^n}} \right)} = \sum\limits_{n \ge 0} {{{\left( { - 1} \right)}^n}\left( {\frac{{H{G^{n - 1}}}}{{\left( {n - 1} \right)!}} + \frac{{H{G^n}}}{{n!}}} \right)} = 0.\end{align*}

## 北大本科06数分期中试题(李伟固命题)

1.给定实数$\lambda_i(1\leq i\leq n)$,满足$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$.令$f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}}$.证明: $f^{(k)}(0)>0,k=1,2,3,\cdots$.

$g\left( x \right) = \left( {\sum\limits_{i = 1}^n {{\lambda _i}} } \right)x + \left( {\frac{1}{2}\sum\limits_{i = 1}^n {\lambda _i^2} } \right){x^2} + \cdots + \left( {\frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} } \right){x^k} + \cdots$由函数幂级数展开的唯一性可知${g^{\left( k \right)}}\left( 0 \right) = \frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {x \in U\left( {0;\delta } \right)} \right)$.

${f^{\left( k \right)}}\left( x \right) = {\left( {{e^{g\left( x \right)}}} \right)^{\left( k \right)}} = \left( {\sum\limits_{j \in {N_ + },{k_i} \in {N_ + }} {{g^{\left( {{k_1}} \right)}}\left( x \right){g^{\left( {{k_2}} \right)}}\left( x \right) \cdots {g^{\left( {{k_j}} \right)}}\left( x \right)} } \right){e^{g\left( x \right)}}.$由${g^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$且$g(0)=0$,所以${f^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$.

${f^{\left( k \right)}}\left( 0 \right) = \sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}} \left( \text{其中}{{k_i} \in N} \right).$若能通过$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$得出$\sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}}>0$即可得到证明.

2.令$D = \left\{ {u = \left( {x,y} \right) \in {\mathbb{R}^2}\left| {\left\| u \right\| = \sqrt {{x^2} + {y^2}} \le \frac{1}{2}} \right.} \right\}$. $f(u)=f(x,y)$是全平面上的连续可微函数满足$\left\| {\nabla f\left( {0,0} \right)} \right\| = 1,\left\| {\nabla f\left( u \right) - \nabla f\left( v \right)} \right\| \le \left\| {u - v} \right\|$.那么对于任意的$u,v\in D$,证明函数$f|_D$在$D$中唯一点处达到其最大值.

$1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,$亦即$1 - \left\| u \right\| \le \left\| {\nabla f\left( u \right)} \right\|$,则$\nabla f\left( u \right) \ne \left( {0,0} \right)$,所以$f$的最大值只可能在边界上取得.

$\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {{u_1}} \right\| \le \left\| {\nabla f\left( {{u_1}} \right)} \right\| = \left| {{a_1}} \right|,$

\begin{align*}&{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2}\\= &{\left\| {{a_1}\overrightarrow {{r_1}} - {a_2}\overrightarrow {{r_2}} } \right\|^2} - {\left\| {\frac{1}{2}\overrightarrow {{r_1}} - \frac{1}{2}\overrightarrow {{r_2}} } \right\|^2} = a_1^2 + a_2^2 - 2{a_1}{a_2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} - \left( {\frac{1}{4} - \frac{1}{2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} + \frac{1}{4}} \right)\\= &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right)\overrightarrow {{r_1}} \overrightarrow {{r_2}} \left( \text{由于}{\overrightarrow {{r_1}}\text{与} \overrightarrow {{r_2}} \text{不同向},\text{所以}\overrightarrow {{r_1}} \overrightarrow {{r_2}} < 1,\text{且}2{a_1}{a_2} - \frac{1}{2} > 0} \right)\\> &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right) = {\left( {{a_1} - {a_2}} \right)^2} \ge 0,\end{align*}

3.讨论级数$\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}},\alpha\in \mathbb{R}$的收敛性.

(2)当$\alpha>0$时,由于$\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}} = 0$,我们讨论其前$2n$项和数列的收敛性即可,也就是级数$\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$的收敛性.

\begin{align*}&\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\= &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)\left( {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right) + \frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right) - \sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)} \right|\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\frac{{2\cos \left[ {\frac{{\ln \left( {2k - 1} \right) + \ln \left( {2k} \right)}}{2}} \right]\sin \left[ {\frac{{\ln \left( {2k - 1} \right) - \ln \left( {2k} \right)}}{2}} \right]}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha}}}.\end{align*}

3.求$\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\}.$

${f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{{k + 1}}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2kj\pi }}{{k + 1}}}}{k} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k}.$

${f_k}\left( 0 \right) = \sum\limits_{i = 1}^k {\frac{1}{i}} > - \frac{1}{2},{f_k}\left( {\frac{\pi }{2}} \right) = \sum\limits_{i = 1}^k {\frac{{\cos i\frac{\pi }{2}}}{i}} = \left\{ \begin{array}{l}\frac{1}{2}\left[ { - 1 + \left({\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s + 1\\\frac{1}{2}\left[ { - 1 + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s\end{array} \right.$,从而${f_k}\left( {\frac{\pi }{2}} \right) \ge - \frac{1}{2}$.

$\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\} = - \frac{1}{2}.$

4.函数$f(x)$在$[0,1]$上二次可导, $f(0)=2,f'(0)=-2,f(1)=1$.证明存在$c\in (0,1)$,使得$f(c)f'(c)+f''(c)=0$.

$F'\left( c \right) = f\left( c \right)f'\left( c \right) + f''\left( c \right) = 0.$

5.$A$和$B$是自然数$\mathbb{N}$的两个无穷子集,满足$A\cap B=\text{空集},A\cup B=\mathbb{N}$,对于任意的自然数$c>0$,是否存在两个递增的数列$\{a_n\},\{b_n\},\{a_n\}\in A,\{b_n\}\in B$,使得$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.