一、选择题

(本题共 10 小题, 每小题 5 分, 共 50 分. 在每小题给出的四个选项中, 只有一项符合题目 要求,把所选项前的字母填在题后的括号内. )

(1)

数一2021 函数 $\displaystyle f(x)=\left\{\begin{array}{ll}\frac{\mathrm{e}^{x}-1}{x}, & x \neq 0, \\ 1, & x=0\end{array}\right.$ 在 $x=0$ 处 ( ) (A) 连续且取极大值.(B) 连续且取极小值.(C) 可导且导数等于 0 .(D) 可导且导数不为 0 .

(1)

• 数一2021 答 应选 D.
• 思路
  • 这道题先使用导数的定义，
  • 在连续使用两次洛必达求得求得极限
• 因为 $\displaystyle \begin{aligned} f'(0) \xlongequal[]{导数定义}\lim_{x \to 0} \frac{f(x) - f(0)}{x} \xlongequal[f(0)=1]{f(x)=\frac{e^{x}-1}{x}}\lim_{x \to 0} \frac{\frac{e^x - 1}{x} - x}{x} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{通分}\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \xlongequal[或洛必达]{泰勒}\lim_{x \to 0} \frac{1}{2}x^2 = \frac{1}{2} \end{aligned}$

(2)

数一2021 设函数 $f(x, y)$ 可微, 且 $f\left(x+1, \mathrm{e}^{x}\right)=x(x+1)^{2}, f\left(x, x^{2}\right)=2 x^{2} \ln x$, 则 $\mathrm{d} f(1,1)=(\quad)$ (A) $\mathrm{d} x+\mathrm{d} y$.(B) $\mathrm{d} x-\mathrm{d} y$.(C) $\mathrm{d} y$.(D) $-\mathrm{d} y$.

(2)

• 解

• 根据全微分的定义：$\mathrm{d} f(1,1)=f_1^{\prime}(1,1) \mathrm{d} x+f_2^{\prime}(1,1) \mathrm{d} y$

• 需要对 $f(x+1, \mathrm{e}^x) = x(x+1)^2$ 和 $f(x, x^2) = 2x^2 \ln x$ 分别对 $x$ 求导

  • 对 $f(x+1, \mathrm{e}^x) = x(x+1)^2\xrightarrow[]{等式左右都对x求导}$
    • $\displaystyle \begin{aligned}f_1'(x+1, \mathrm{e}^x) + f_2'(x+1, \mathrm{e}^x) \cdot \mathrm{e}^x = (x+1)^2 + 2x(x+1)\end{aligned}$
    • 代入 $x=0$ 得到: $f_1'(1,1) + f_2'(1,1) = 1$
  • 对 $f(x, x^2) = 2x^2 \ln x\xrightarrow[]{等式左右都对x求导}$
    • $\displaystyle \begin{aligned}f_1'(x, x^2) + 2x \cdot f_2'(x, x^2) = 4x \ln x + 2x\end{aligned}$
    • 代入 $x=1$ ，得到 $f_1'(1,1) + 2f_2'(1,1) = 2$
  • 联立方程组得到 $f_1'(1,1) = 0$, $f_2'(1,1) = 1$
    • 写出全微分方程$z=\frac{\partial z}{\partial u} d u+\frac{\partial z}{\partial v} d v$
    • $\mathrm{d}f(1,1) = f_1'(1,1)\mathrm{d}x + f_2'(1,1)\mathrm{d}y=0dx+1\mathrm{d}y$

高昆仑版

(3)

数一2021 设函数 $f(x)=\frac{\sin x}{1+x^{2}}$ 在 $x=0$ 处的 3 次泰勒多项式为 $a x+b x^{2}+c x^{3}$, 则 ( ) (A) $a=1, b=0, c=-\frac{7}{6}$.(B) $a=1, b=0, c=\frac{7}{6}$.(C) $a=-1, b=-1, c=-\frac{7}{6}$.(D) $a=-1, b=-1, c=\frac{7}{6}$.

(3)

• 数一2021

• 问题是求函数 $f(x)=\frac{\sin x}{1+x^2}$ 在 $x=0$ 处的3次泰勒多项式 $a x + b x^2 + c x^3$。

  • 只需要展开到3次

• $\displaystyle \begin{aligned} f(x) = \sin x \cdot \frac{1}{1 + x^2} = \left( x - \frac{1}{6}x^3 + \ldots \right) \cdot \left( 1 - x^2 + \ldots \right) \end{aligned}$

  • $\displaystyle \begin{aligned} = x - x^3 - \frac{1}{6}x^3 + \ldots \end{aligned}$
  • $\displaystyle \begin{aligned}= x - \frac{7}{6}x^3 + \ldots\end{aligned}$

• 对比3次泰勒多项式: $a x + b x^2 + c x^3$。

  • 正确答案是 (B) $a=1, b=0, c=-\frac{7}{6}$

• $\displaystyle \begin{aligned} \sin x=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2 n+1}}{(2 n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots+(-1)^n \frac{x^{2 n+1}}{(2 n+1)!}+\cdots,-\infty<x<+\infty . \end{aligned}$

• $\displaystyle \begin{aligned} \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n=1-x+x^2-x^3+\cdots+(-1)^n x^n+\cdots,-1<x<1 . \end{aligned}$

(4)

数一2021 设函数 $f(x)$ 在区间 $[0,1]$ 上连续, 则 $\displaystyle \int_{0}^{1} f(x) \mathrm{d} x=(\quad)$ (A) $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(\frac{2 k-1}{2 n}\right) \frac{1}{2 n}$. (B) $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(\frac{2 k-1}{2 n}\right) \frac{1}{n}$. (C) $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{2 n} f\left(\frac{k-1}{2 n}\right) \frac{1}{n}$. (D) $\displaystyle \lim _{n \rightarrow \infty} \sum_{k=1}^{2 n} f\left(\frac{k}{2 n}\right) \frac{2}{n}$.

(4)

• • 只需要$\frac{1}{n} 与\frac{i}{n}$对应，或$\frac{1}{2n} 与 \frac{i}{2n}$对应，秒杀B
• 分析选项 (B)
  • 将 $[0,1]$ 分为 $n$ 等份
    • 每个小区间为 $\left[\frac{k-1}{n}, \frac{k}{n}\right]$
    • 区间长度为 $\frac{1}{n}$
    • 在 $\left[\frac{k-1}{n}, \frac{k}{n}\right]$ 中取中点 $f\left(\frac{\frac{k}{n}+\frac{k-1}{n}}{2}\right)$=$\frac{2k-1}{2n}$
      • $\displaystyle \begin{aligned} = \lim_{n \to \infty} \left( \frac{1}{n} \cdot f\left( \frac{0 + \frac{1}{n}}{2} \right) + \frac{1}{n} \cdot f\left( \frac{\frac{1}{n} + \frac{2}{n}}{2} \right) + \ldots + \frac{1}{n} \cdot f\left( \frac{\frac{n-1}{n} + 1}{2} \right) \right) \end{aligned}$
      • $\displaystyle \begin{aligned} = \lim_{n \to \infty} \left( \frac{1}{n} \cdot f\left( \frac{1}{2n} \right) + \frac{1}{n} \cdot f\left( \frac{3}{2n} \right) + \ldots + \frac{1}{n} \cdot f\left( \frac{2n-1}{2n} \right) \right) \end{aligned}$
    • 根据定积分定义
      • 得 $\displaystyle \int_0^1 f(x) \mathrm{d} x = \lim_{n \rightarrow \infty} \sum_{k=1}^n f\left(\frac{2k-1}{2n}\right) \cdot \frac{1}{n}$
• 分析其他选项 (A), (C), (D)
  • 选项 (A)，得到 $\displaystyle \frac{1}{2} \int_0^1 f(x) \mathrm{d} x$
  • 选项 (C)，得到 $\displaystyle 2 \int_0^1 f(x) \mathrm{d} x$
  • 选项 (D)，得到 $\displaystyle 4 \int_0^1 f(x) \mathrm{d} x$
• 结论:答案是 (B)

(5)

数一2021 二次型 $f\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}-\left(x_{3}-x_{1}\right)^{2}$ 的正惯性指数与负惯性指数依次 为 $(\quad)$(A) 2,0 .(B) 1,1 .(C) 2,1 .(D) 1,2 .

(5)

• • 这个问题是关于计算二次型的正惯性指数和负惯性指数的。解题步骤如下：

• 展开并化简二次型 $f(x_1, x_2, x_3)$

• 将 $f(x_1, x_2, x_3)$ 中的完全平方拆开
  $\displaystyle \begin{aligned}f\left(x_1, x_2, x_3\right)= & \left(x_1+x_2\right)^2+\left(x_2+x_3\right)^2-\left(x_3-x_1\right)^2 \\& =x_1^2+2 x_1 x_2+x_2^2+x_2^2+2 x_2 x_3+x_3^2-x_3^2+2 x_1 x_3-x_1^2 \\& =2 x_2^2+2 x_1 x_2+2 x_2 x_3+2 x_1 x_3 .\end{aligned}$

  • 得到 $2x_2^2 + 2x_1x_2 + 2x_2x_3 + 2x_1x_3$
    • 写成矩阵 $\displaystyle A = \left[\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 0\end{array}\right]$
  • 计算特征多项式 $|\lambda E - A|$
    • $\displaystyle \begin{aligned}|\lambda \boldsymbol{E}-\boldsymbol{A}|=\left|\begin{array}{ccc}\lambda & -1 & -1 \\-1 & \lambda-2 & -1 \\-1 & -1 & \lambda\end{array}\right|\xlongequal[]{\begin {array}{l}r_1-r_{3}两行相减为0\\并且形成公因式\end{array}}\left|\begin{array}{ccc}\lambda+1 & 0 & -\lambda-1 \\-1 & \lambda-2 & -1 \\-1 & -1 & \lambda\end{array}\right|\end{aligned}$
      • $\displaystyle \begin{aligned}\xlongequal[]{c_{3}+c_{1}}\mid \begin{array}{ccc}\lambda+1 & 0 & 0 \\-1 & \lambda-2 & -2 \\-1 & -1 & \lambda-1\end{array}\end{aligned}$
    • $\displaystyle \begin{aligned}\xlongequal[]{按照第一行展开}(\lambda+1)\left|\begin{array}{ccc}1 & 0 & 0 \\-1 & \lambda-2 & -2 \\-1 & -1 & \lambda-1\end{array}\right|\end{aligned}$
    • $\displaystyle \begin{array}{l} =(\lambda+1)[(\lambda-2)(\lambda-1)-2] \xrightarrow[]{先展开\left[\lambda^2-3 \lambda+2-2\right]}\end{array}$
    • $\displaystyle \begin{aligned}\xlongequal[]{因式分解}\lambda(\lambda+1)(\lambda-3)\end{aligned}$

• 计算得到 $\lambda(\lambda + 1)(\lambda - 3)$确定正惯性指数和负惯性指数

  • 特征值为 $0, -1, 3$
    • 正惯性指数 $p = 1$ （一个正特征值）
    • 负惯性指数 $q = 1$ （一个负特征值）

(6)

已知 $\displaystyle \boldsymbol{\alpha}_{1}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right), \boldsymbol{\alpha}_{2}=\left(\begin{array}{l}1 \\ 2 \\ 1\end{array}\right), \boldsymbol{\alpha}_{3}=\left(\begin{array}{l}3 \\ 1 \\ 2\end{array}\right)$, 记 $\underbrace{\boldsymbol{\beta}_{1}=\boldsymbol{\alpha}_{1}, \boldsymbol{\beta}_{2}=\boldsymbol{\alpha}_{2}-k \boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{3}=\boldsymbol{\alpha}_{3}-l_{1} \boldsymbol{\beta}_{1}-l_{2} \boldsymbol{\beta}_{2}}_{正好对应施密特正交化}$, 若 $\boldsymbol{\beta}_{1}$, $\boldsymbol{\beta}_{2}, \boldsymbol{\beta}_{3}$ 两两正交, 则 $l_{1}, l_{2}$ 依次为 $(\quad)$ (A) $\frac{5}{2}, \frac{1}{2}$. (B) $-\frac{5}{2}, \frac{1}{2}$. (C) $\frac{5}{2},-\frac{1}{2}$. (D) $-\frac{5}{2},-\frac{1}{2}$.

(6)

• • $\boldsymbol{\beta}_3 = \boldsymbol{\alpha}_3 - \underbrace{\frac{(\boldsymbol{\beta}_1, \boldsymbol{\alpha}_3)}{(\boldsymbol{\beta}_1, \boldsymbol{\beta}_1)}}_{l_1}\boldsymbol{\beta}_1 - \underbrace{\frac{(\boldsymbol{\beta}_2, \boldsymbol{\alpha}_3)}{(\boldsymbol{\beta}_2, \boldsymbol{\beta}_2)}}_{l_2}\boldsymbol{\beta}_2$
• 进行施密特正交化:对 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 进行正交化处理
  • 令 $\displaystyle \boldsymbol{\beta}_1 = \boldsymbol{\alpha}_1=\left(\begin{array}{l}1\\ 0 \\ 1\end{array}\right)$
  • 计算 $\boldsymbol{\beta}_2 = \boldsymbol{\alpha}_2 - \frac{(\boldsymbol{\beta}_1, \boldsymbol{\alpha}_2)}{(\boldsymbol{\beta}_1, \boldsymbol{\beta}_1)} \boldsymbol{\beta}_1$
    • 计算内积 $\displaystyle (\boldsymbol{\beta}_1, \boldsymbol{\alpha}_2)=\boldsymbol{\beta}_1^{\mathrm{T}} \boldsymbol{\alpha}_2=(1,0,1)\left(\begin{array}{l}1 \\ 2 \\ 1\end{array}\right)=2$
    • 计算内积 $\displaystyle (\boldsymbol{\beta}_1, \boldsymbol{\beta}_1) = \boldsymbol{\beta}_1^{\mathrm{T}} \boldsymbol{\beta}_1=(1,0,1)\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=2$
    • 得到 $\displaystyle \boldsymbol{\beta}_2 =\left(\begin{array}{l}1 \\2 \\1\end{array}\right)-\frac{2}{2}\left(\begin{array}{l}1 \\0 \\1\end{array}\right)=\left(\begin{array}{l}0 \\2 \\0\end{array}\right)$
  • 计算 $\boldsymbol{\beta}_3 = \boldsymbol{\alpha}_3 - \frac{(\boldsymbol{\beta}_1, \boldsymbol{\alpha}_3)}{(\boldsymbol{\beta}_1, \boldsymbol{\beta}_1)} \boldsymbol{\beta}_1 - \frac{(\boldsymbol{\beta}_2, \boldsymbol{\alpha}_3)}{(\boldsymbol{\beta}_2, \boldsymbol{\beta}_2)} \boldsymbol{\beta}_2$
    • 计算内积$\displaystyle (\boldsymbol{\beta}_1, \boldsymbol{\alpha}_3) =\boldsymbol{\beta}_1^{\mathrm{T}} \boldsymbol{\alpha}_3=(1,0,1)\left(\begin{array}{l}3 \\1 \\2\end{array}\right)=5$
    • 计算内积 $\displaystyle \left(\boldsymbol{\beta}_2, \boldsymbol{\alpha}_3\right)=\boldsymbol{\beta}_2^{\mathrm{T}} \boldsymbol{\alpha}_3=(0,2,0)\left(\begin{array}{l}3 \\1 \\2\end{array}\right)=2 \text {, }$
    • 计算内积 $\displaystyle \left(\boldsymbol{\beta}_2, \boldsymbol{\beta}_2\right)=\boldsymbol{\beta}_2^{\mathrm{T}} \boldsymbol{\beta}_2=(0,2,0)\left(\begin{array}{l}0 \\ 2 \\ 0\end{array}\right)=4$
  • 得到 $\boldsymbol{\beta}_3 = \boldsymbol{\alpha}_3 - \frac{5}{2} \boldsymbol{\beta}_1 - \frac{1}{2} \boldsymbol{\beta}_2$
    • 确定 $l_1$ 和 $l_2$
      • 由上述计算可知 $l_1 = \frac{5}{2}$, $l_2 = \frac{1}{2}$

(7)

设 $\boldsymbol{A}, \boldsymbol{B}$ 为 $n$ 阶实矩阵, 下列不成立的是 ( ) ( A) $\displaystyle r\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A}^{\mathrm{T}} \boldsymbol{A}\end{array}\right)=2 r(\boldsymbol{A})$. (B) $\displaystyle r\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{A} \boldsymbol{B} \\ \boldsymbol{O} & \boldsymbol{A}^{\mathrm{T}}\end{array}\right)=2 r(\boldsymbol{A})$. ( C) $\displaystyle r\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{B} \boldsymbol{A} \\ \boldsymbol{O} & \boldsymbol{A} \boldsymbol{A}^{\mathrm{T}}\end{array}\right)=2 r(\boldsymbol{A})$ (D) $\displaystyle r\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{B} \boldsymbol{A} & \boldsymbol{A}^{\mathrm{T}}\end{array}\right)=2 r(\boldsymbol{A})$.

• (7)

• 答 应选 C.
• 解 由矩阵的秩的性质知, $r(\boldsymbol{A})=r\left(\boldsymbol{A}^{\mathrm{T}}\right)=r\left(\boldsymbol{A A ^ { \mathrm { T } }}\right)=r\left(\boldsymbol{A}^{\mathrm{T}} \boldsymbol{A}\right)$.
• 故 $\displaystyle r\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A}^{\mathrm{T}} \boldsymbol{A}\end{array}\right)=r(A)+r\left(A^T A\right)=2 r(A)$, 选项 B 成立;
• 对于(B)，因为$\displaystyle \begin{aligned} AB \end{aligned}$ 的列变换可由 $\displaystyle \begin{aligned} A \end{aligned}$ 的列变换消去，于是
  • $\displaystyle \begin{aligned} r\left[\begin{array}{cc} A & AB \\ O & A^T \end{array}\right] = r\left[\begin{array}{cc} A & O \\ O & A^T \end{array}\right] = r(A) + r(A^T) = 2r(A). \end{aligned}$
• 对于(C)，$BA$为行变换，无法用$A$的列变换消去
  • 而$\displaystyle \begin{aligned} r\left[\begin{array}{cc} A & BA \\ O & AA^T \end{array}\right] \text{未必等于} r\left[\begin{array}{cc} A & O \\ O & AA^T \end{array}\right]. \end{aligned}$
• 对于(D)，因为 $\displaystyle \begin{aligned} BA \end{aligned}$ 的行变换可由 $\displaystyle \begin{aligned} A \end{aligned}$ 的行变换消去，于是
  • $\displaystyle \begin{aligned} r\left[\begin{array}{ll} A & O \\ BA & A^T \end{array}\right] = r\left[\begin{array}{ll} A & O \\ O & A^T \end{array}\right] = r(A) + r(A^T) = 2r(A). \end{aligned}$

(8)

设 $A, B$ 为随机事件, 且 $0<P(B)<1$, 下列命题中为假命题的是 $(\quad)$ (A) 若 $P(A \mid B)=P(A)$, 则 $P(A \mid \bar{B})=P(A)$. (B) 若 $P(A \mid B)>P(A)$, 则 $P(\bar{A} \mid \bar{B})>P(\bar{A})$. (C) 若 $P(A \mid B)>P(A \mid \bar{B})$, 则 $P(A \mid B)>P(A)$. (D) 若 $P(A \mid A \cup B)>P(\bar{A} \mid A \cup B)$, 则 $P(A)>P(B)$.

小崔版

• 思路:
  • 左侧的式子恒等变形，右侧的式子也恒等变形
  • 最终能变成同一个式子
• A
  • 左侧:$\displaystyle \begin{aligned} \frac{P(A B)}{P(B)}=P(A) \xrightarrow[]{移项} & P(A B)=P(A) P(B) \\\end{aligned}$
  • 右侧:$\displaystyle \begin{aligned}\frac{P(A \bar{B})}{P(\bar{B})}=P(A) \xrightarrow[]{移项} & P(A \bar{B})=P(A) P(\bar{B})\xrightarrow[P(\bar{B})=1-P(B)]{取对立事件}\end{aligned}$
    • $\displaystyle \begin{aligned} P(A)-P(A B)=P(A)(1-P(B) )=P(A)-P(A)P(B)\end{aligned}$
    • $\displaystyle \begin{aligned}\Leftrightarrow & P(A B)=P(A) P(B)\end{aligned}$
• B
  • 左侧:若 $\displaystyle \begin{aligned} P(A \mid B) > P(A) \end{aligned}$，则 $\displaystyle \begin{aligned} \xrightarrow[\frac{P(AB)}{P(B)} > P(A)]{条件概率}P(AB) > P(A)P(B) \end{aligned}$
  • 右侧:$\displaystyle \begin{aligned} \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} & >1-P(A) \Leftrightarrow \frac{1-P(A+B)}{1-P(B)}>1-P(A) \end{aligned}$
    • $\displaystyle \begin{aligned}\frac{1-P(A)-P(B)+P(A B)}{1-P(B)}>1-P(A)\end{aligned}$
    • $\displaystyle \begin{aligned} \underline{1-P(A)-P(B)}+P(A B)>\underline{1-P(A)-P(B)} +P(A)P(B) \end{aligned}$
• C
  • 左侧:$\displaystyle \begin{aligned}\frac{P(A B)}{P(B)}>\frac{P(A)-P(A B)}{1-P(B)}\end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[]{交叉相乘}P(A B)-P(B) P(A B)>P(A) P(B)-P(B) P(A B)\end{aligned}$
  • 右侧:同B选项左侧
• D不成立

(9)

设 $\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \cdots,\left(X_{n}, Y_{n}\right)$ 为来自总体 $\displaystyle N\left(\mu_{1}, \mu_{2} ; \sigma_{1}^{2}, \sigma_{2}^{2} ; \rho\right)$ 的简单随机样本, 令 $\displaystyle \theta=\mu_{1}-\mu_{2}, \bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i}, \bar{Y}=\frac{1}{n} \sum_{i=1}^{n} Y_{i}, \hat{\theta}=\bar{X}-\bar{Y}$, 则 $(\quad)$ (A) $\hat{\theta}$ 是 $\theta$ 的无偏估计, $\displaystyle D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}}{n}$. (B) $\hat{\theta}$ 不是 $\theta$ 的无偏估计, $\displaystyle D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}}{n}$. (C) $\hat{\theta}$ 是 $\theta$ 的无偏估计, $\displaystyle D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}}{n}$. (D) $\hat{\theta}$ 不是 $\theta$ 的无偏估计, $\displaystyle D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}}{n}$.

(9)

• • $\rho=\frac{\operatorname{Cov}(X, Y)}{\sqrt{D(X)} \sqrt{D(Y)}}=\frac{E(X Y)-E(X) E(Y)}{\sqrt{D(X)} \sqrt{D(Y)}} .$
• 求样本期望
  • $\displaystyle \begin{aligned} \text{由} E(\hat{\theta}) = E(\bar{X}-\bar{Y}) = E\bar{X} - E\bar{Y} \angle EX - EY = \mu_{1} - \mu_{2} = \theta. \end{aligned}$
• 求样本方差
  • $\displaystyle \begin{aligned} D(\hat{\theta}) \xlongequal[]{\theta=\mu_{1}-\mu_{2}} D(\bar{X}-\bar{Y}) \xlongequal[]{性质} D\bar{X} + D\bar{Y} - 2\operatorname{Cov}(\bar{X},\bar{Y}) \end{aligned}$
    • $\displaystyle \operatorname{Cov}(\bar{X}, \bar{Y}) = \frac{\rho \sigma_1 \sigma_2}{n}$
  • $\displaystyle \begin{aligned} \xlongequal[]{D(\bar{X}) = \frac{\sigma_1^2}{n}，D(\bar{Y}) = \frac{\sigma_2^2}{n}}\left(\frac{1}{n} DX + \frac{1}{n} DY - 2 \operatorname{Cov}\left(\frac{1}{n}\sum_{i=1}^{n} X_{i}, \frac{1}{n}\sum_{i=1}^{n} Y_{i}\right)\right. \end{aligned}$
  • $\displaystyle \begin{aligned} = \frac{1}{n}\sigma_{1}^{2} + \frac{1}{n}\sigma_{2}^{2} - 2 \cdot \frac{1}{n} \cdot \frac{1}{n} \operatorname{Cov}\left(\sum_{i=1}^{n} X_{i}, \sum_{i=1}^{n} Y_{i}\right) \end{aligned}$
    • $\displaystyle \left\{\begin{array}{l}X_1, Y_1 \\ \cdots \ldots \\ X_1, Y_n\end{array}, \cdots,\left\{\begin{array}{l}X_n, Y_1 \\ \cdots \ldots \\ X_n, Y_n\end{array}\right.\right.$
  • $\displaystyle \begin{aligned} \xlongequal[]{处理协方差}\frac{1}{n}\sigma_{1}^{2} + \frac{1}{n}\sigma_{2}^{2} - 2 \cdot \frac{1}{n} \cdot \frac{1}{n} \cdot n \operatorname{Cov}(X_{1}, Y_{1}) \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\rho=\frac{\operatorname{Cov}(X, Y)}{\sqrt{D(X)} \sqrt{D(Y)}}}\frac{1}{n}\sigma_{1}^{2} + \frac{1}{n}\sigma_{2}^{2} - \frac{2}{n} \cdot \rho \cdot \sigma_{1} \cdot \sigma_{2}. \end{aligned}$
• 综上，$\hat{\theta}$ 是 $\theta$ 的无偏估计，其方差为 $\displaystyle \frac{\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2}{n}$。因此，正确答案是选项 (C)。

(10)

设 $X_{1}, X_{2}, \cdots, X_{16}$ 是来自总体 $N(\mu, 4)$ 的简单随机样本, 考虑假设检验问题 $H_{0}: \mu \leqslant 10$, $H_{1}: \mu>10, \Phi(x)$ 表示标准正态分布函数, 若该检验问题的拒绝域为 $W=\{\bar{X}>11\}$, 其中 $\displaystyle \bar{X}=\frac{1}{16} \sum_{i=1}^{16} X_{i}$, 则 $\mu=11.5$ 时, 该检验犯第二类错误的概率为 $(\quad)$ ( A) $1-\Phi(0.5)$. ( B ) $1-\Phi(1)$. ( C) $1-\Phi(1.5)$. (D) $1-\Phi(2)$.

(10)

• [01:07:54](file:///C:/Users/wangpanfeng/Videos/01.%E6%95%B0%E5%AD%A6%E4%B8%80/08.2021%E5%B9%B4%E6%95%B0%E4%B8%80%E7%9C%9F%E9%A2%98/01.2021%E5%B9%B4%E8%80%83%E7%A0%94%E6%95%B0%E5%AD%A6%E7%9C%9F%E9%A2%98%E9%80%89%E6%8B%A9%E9%A2%98%EF%BC%88%E6%95%B0%E4%B8%80%EF%BC%89.mp4#t=1:07:54) 答 应选 A.

• (1)两类错误

  • (1)第一类错误 (弃真错误): 当原假设 $H_0$ 为真时, 但检验结果为拒绝 $H_0$;
    • 拒绝了真的假设
  • (2)第二类错误 (存伪错误): 当原假设 $H_0$ 不正确时, 但检验结果为接受 $H_0$.
    • 接受了错误的假设

• 根据已知条件, $\mu=11.5$, 原假设 $H_0: \mu \leqslant 10$ 不为真.

  • 由于该检验问题的拒绝域为 $W=\{\bar{X}>11\}$,
    • 得到接受域：$\bar{X} \leqslant 11$
  • 故当 $\bar{X} \leqslant 11$ 时, 不拒绝 $H_0$. 此时, 该检验犯了第二类错误, 其概率为 $P\{\bar{X} \leqslant 11\}$.
    • 接受了错误的假设

• 下面我们计算 $P\{\bar{X} \leqslant 11\}$ 。

  • 由样本均值 $\displaystyle \bar{X} = \frac{1}{16} \sum_{i=1}^{16} X_i$,得$n = 16$
    • 故 $\bar{X} \sim N\left(\mu, \frac{4}{16}\right)$,
    • $\xrightarrow[]{\mu=11.5}\bar{X} \sim N\left(11.5, \frac{1}{4}\right)$，标准化可得, $\frac{\bar{X}-11.5}{\frac{1}{2}} \sim N(0,1)$.
  • $P\{\bar{X} \leqslant 11\}=P\left\{\frac{\bar{X}-11.5}{\frac{1}{2}} \leqslant \frac{11-11.5}{\frac{1}{2}}\right\}=\Phi(-1)=1-\Phi(1)$
    • 选项中没有$\Phi(-1)$ ,只有$\Phi(1)$
    • 犯第二类错误落在接受域的概率：$1 - \Phi(1)$

• (2018)设总体 $X$ 服从正态分布 $\displaystyle N\left(\mu, \sigma^{2}\right) . X_{1}, X_{2}, \cdots, X_{n}$ 是来自总体 $X$ 的简单随机样本,据此样本检验 假设 $H_{0}: \mu=\mu_{0}, H_{1}: \mu \neq \mu_{0}$, 则 ( )

  • (A) 如果在检验水平 $\alpha=0.05$ 下拒绝 $H_{0}$, 那么 $\alpha=0.01$ 下必拒绝 $H_{0}$.
  • (B) 如果在检验水平 $\alpha=0.05$ 下拒绝 $H_{0}$, 那么 $\alpha=0.01$ 下必接受 $H_{0}$.
  • (C) 如果在检验水平 $\alpha=0.05$ 下接受 $H_{0}$, 那么 $\alpha=0.01$ 下必拒绝 $H_{0}$.
  • (D) 如果在检验水平 $\alpha=0.05$ 下接受 $H_{0}$, 那么 $\alpha=0.01$ 下必接受 $H_{0}$.

二、填空题

(本题共 6 小题,每小题 5 分,共 30 分, 把答案填在题中横线.上.)}

(11)

$\displaystyle \int_{0}^{+\infty} \frac{1}{x^{2}+2 x+2} \mathrm{~d} x=$

(11)

答 应填 $\frac{\pi}{4}$.

高昆仑版

• $\displaystyle \begin{aligned} \int_{0}^{+\infty} \frac{1}{x^{2}+2x+2} \, dx = \int_{0}^{+\infty} \frac{1}{(x+1)^{2}+1} \, dx \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\int \frac1{1+u^2}du=\arctan(u)} \left. \arctan(x+1) \right|_{0}^{+\infty} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[\arctan(1) = \frac{\pi}{4}]{\arctan(+\infty) = \frac{\pi}{2}}\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \end{aligned}$

gpt版

• 计算积分 $\displaystyle \int_{0}^{+\infty} \frac{1}{x^2 + 2x + 2} \, dx$
  • 式子化简
    • 分母化简：$x^2 + 2x + 2$ 拆成$x^2 + 2x + 1 + 1$
      • 完全平方：$(x + 1)^2 + 1$
    • 新积分形式：$\displaystyle \int_0^{+\infty} \frac{1}{(x + 1)^2 + 1} \, d(x+1)$
  • 计算积分
    • 使用反正切函数：$\displaystyle \int \frac1{1+u^2}du=\arctan(u)$
    • 积分结果：$=\left. \arctan(x + 1) \right|_0^{+\infty}\xlongequal[\arctan(1) = \frac{\pi}{4}]{\arctan(+\infty) = \frac{\pi}{2}}\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

(12)

设函数 $y=y(x)$ 由参数方程 $\displaystyle \left\{\begin{array}{l}x=2 \mathrm{e}^{t}+t+1, \\ y=4(t-1) \mathrm{e}^{t}+t^{2}\end{array}\right.$ 确定, 则 $\left.\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right|_{t=0}=$

• $\displaystyle \begin{aligned} \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{4e^t + 4(t-1)e^t + 2t}{2e^t + 1} \approx 2t, \end{aligned}$
• $\displaystyle \begin{aligned} \left. \frac{d^2y}{dx^2} \right|_{t=0} = \left. \frac{d(2t)}{dt} \cdot \frac{1}{x'(t)} \right|_{t=0} = \left. \frac{2}{2e^t + 1} \right|_{t=0} = \frac{2}{3}. \end{aligned}$

(13)

欧拉方程 $x^{2} y^{\prime \prime}+x y^{\prime}-4 y=0$ 满足条件 $y(1)=1, y^{\prime}(1)=2$ 的解为 $y=$

(13)

• • $\displaystyle \begin{aligned} x = e^t, \quad x^2 y'' = D(D-1) y, \quad x y' = D y, \quad D = \frac{d}{dt} \end{aligned}$
    • $\displaystyle \begin{aligned} D(D-1) y + D y - 4y = 0 \end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[]{整理} D^2 y - 4y = 0 \end{aligned}$
    • $\displaystyle \begin{aligned} \xrightarrow[]{D = \frac{d}{dt} }\frac{d^2 y}{dt^2} - 4y = 0 \end{aligned}$，从而$\displaystyle \begin{aligned} \xrightarrow[]{求特征方程}\lambda^2 - 4 = 0 \end{aligned}$，得$\displaystyle \begin{aligned} \lambda = \pm 2 \end{aligned}$

• $\displaystyle \begin{aligned}\xrightarrow[]{通解公式} y = C_1 e^{2t} + C_2 e^{-2t} \end{aligned}$

  • $\displaystyle \begin{aligned} \begin{cases} y = C_1 x^2 + C_2 x^{-2} & \leftarrow y(1) = 1 \\y' = 2C_1 x - 2C_2 x^{-3} & \leftarrow y'(1) = 2\end{cases} \end{aligned}$
  • $\displaystyle \begin{aligned} \begin{cases}1 = C_1 + C_2 \\1 = C_1 - C_2 \\\end{cases} \end{aligned}$，得$\displaystyle \begin{aligned} & C_1=1 \\ & C_2=0\end{aligned}$
  • $\xrightarrow[]{代入原式}y = x^2$

• 2004 年数一试题欧拉方程
  $x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+4 x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=0(x>0)$ 的通解为 $\qquad$

(14)

设 $\displaystyle \Sigma$ 为空间区域 $\left\{(x, y, z) \mid x^{2}+4 y^{2} \leqslant 4,0 \leqslant z \leqslant 2\right\}$ 表面的外侧, 则曲面积分 $\displaystyle \iint_{\Sigma} x^{2} \mathrm{~d} y \mathrm{~d} z+$ $y^{2} \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y=$

(14)

• • 
    
• 解 利用高斯公式，得
  • $\displaystyle \begin{aligned} \text{原式} = \oint_\Sigma x^2 \, dy \, dz + y^2 \, dz \, dx + z \, dx \, dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{高斯} \iiint_\Omega (2x + 2y + 1) \, dv \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[x，y]{奇函数为0}0 + \iiint_\Omega 1 \, dv = V_\Omega \xlongequal[\pi ab×h]{求圆柱体体积} \pi \cdot 2 \cdot 1 \cdot 2 \cdot \frac{4}{4} \pi. \end{aligned}$

(15)

设 $\boldsymbol{A}=\left(a_{i j}\right)$ 为 3 阶矩阵, $A_{i j}$ 为元素 $a_{i j}$ 的代数余子式, 若 $\boldsymbol{A}$ 的每行元素之和均为 2 , 且 $|\boldsymbol{A}|=$ 3 , 则 $A_{11}+A_{21}+A_{31}=$

(15)

• 答案

• 方法2：构造一个行列式

  • 由每行元素之和等于2→往这里恒等变形，和构造新的行列式
  • $\displaystyle \begin{aligned} & (2)|A|=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|\xlongequal[]{c_1+c_2+c_3}\left|\begin{array}{lll}a_{11}+a_{12}+a_{13} & a_{12} & a_{13} \\ a_{21}+a_{22}+a_{23} & a_{22} & a_{23} \\ a_{31}+a_{32}+a_{33} & a_{32} & a_{33}\end{array}\right| \end{aligned}$
    • 将每行元素之和等于2代入，得$\displaystyle =\left|\begin{array}{lll}2 & a_{12} & a_{13} \\ 2 & a_{22} & a_{23} \\ 2 & a_{32} & a_{33}\end{array}\right|$
    • 按照第一列展开$=2 A_{11}+2 A_{21}+2 A_{31}=3$
      • 则答案:$A_{11}+A_{21}+A_{31}=\frac{3}{2}$

• 用元素和代数余子式的乘积这条路走不通，计算量太大
  $\displaystyle \begin{aligned} & {[\text { 分析 }] A_{i j} \rightarrow \text { 概念 。 }} \end{aligned}$

• $\displaystyle \begin{aligned} \rightarrow \text { 定理 }\left\{\begin{array}{l}|A|=a_{11} A_{11}+a_{12} A_{12}+\cdots+a_{1 n} A_{1 n} \\a_{11} A_{21}+a_{12} A_{22}+\cdots+a_{1 n} A_{2 n}=0\end{array}\right. \end{aligned}$

• $\displaystyle \begin{aligned} \end{aligned}$

• 方法3:

  • A的每行元素之和为2，说明2是一个特征值，(1,1,1)ᵀ是一个特征向量
    • $\displaystyle A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]=2\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
  • $|A|=3$说明A可逆
  • 代数余子式的两个角度
    • 从伴随矩阵分析(对应方法一)
    • 从行列式的展开分析(对应方法二)
  • 接下来用伴随矩阵分析

• $\nabla$ 方法一: 用伴随矩阵∶题中所求, 是伴随矩阵第一行元素的和

  • 当 $\displaystyle A^*=\left[\begin{array}{}\left(A_{11}\right. & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$
    • 要求的是 则伴随矩阵第1，2，3行的和可以写为$\displaystyle A^*\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}(1=A_{11}+A_{21}+A_{31}=\frac{3}{2}) \\ (2) \\ (3)\end{array}\right]$
  • $\nabla$ 题中给大家的是, A矩阵, 每行元素的和
    • $A$ 每行元素之和均为 2，即$\displaystyle \left\{\begin{array}{l}a_{11}+a_{12}+a_{13}=2 \\a_{21}+a_{22}+a_{23}=2 \\a_{31}+a_{32}+a_{33}=2\end{array}\right.$
    • 这其实是一个乘法$\displaystyle \left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array}\right]\left[\begin{array}{l}1 \\1 \\1\end{array}\right]=\left[\begin{array}{l}2 \\2 \\2\end{array}\right]\xrightarrow[]{将矩阵简写}A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]$
    • 另一种写法，$\displaystyle A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]\xrightarrow[]{用A^*左乘}\begin{aligned}& \text { } \\& A^*\left[\begin{array}{l}2 \\2 \\2\end{array}\right]=\underline{A^* A}\left[\begin{array}{l}1 \\1 \\1\end{array}\right]\xrightarrow[]{A^{*}\cdot A=|A|}|A| E\left[\begin{array}{l}1 \\1 \\1\end{array}\right]=3\left[\begin{array}{l}1 \\1 \\1\end{array}\right] \end{aligned}$
      • 式子两边倍除2，得到$\displaystyle \therefore A^*\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}\frac{3}{2} \\ \frac{3}{2} \\ \frac{2}{2}\end{array}\right]$
      • 从而伴随矩阵的第一行元素:$A_{11}+A_{21}+A_{31}=\frac{3}{2}$

(16)

甲、乙两个盒子中各装有 2 个红球和 2 个白球, 先从甲盒中任取一球, 观察颜色后放入乙盒, 再从乙盒中任取一球, 令 $X, Y$ 分别表示从甲盒和乙盒中取到的红球的个数, 则 $X$ 与 $Y$ 的相关系数为

(16)

• • 分析 本题主要考查相关系数的计算.
• (解) 根据相关系数的计算公式， $X$ 与 $Y$ 的相关系数为
  $\rho=\frac{\operatorname{Cov}(X, Y)}{\sqrt{D(X)} \sqrt{D(Y)}}=\frac{E(X Y)-E(X) E(Y)}{\sqrt{D(X)} \sqrt{D(Y)}} .$
  • 
    
    • $\displaystyle \begin{aligned} & P\{X=0 \cdot Y=0\} =\frac{2}{4} \cdot \frac{3}{5}=\frac{3}{10}\end{aligned}$
    • $\displaystyle \begin{aligned} P\{X=1, Y=0\}=\frac{2}{4} \cdot \frac{2}{5}=\frac{2}{10} \end{aligned}$
    • $\displaystyle \begin{aligned} P\{X=0 \cdot Y=1\}=\frac{2}{4} \cdot \frac{2}{5}=\frac{2}{10} \end{aligned}$
    • $\displaystyle \begin{aligned} P\{X=1, Y=1\}=\frac{2}{4} \cdot \frac{3}{5}=\frac{3}{10} \end{aligned}$
• 下面分别计算 $X, Y$ 的分布律与数字特征.取球模型为等可能模型。
  • 从甲盒中取到红球的概率
    $X$ 的可能取值为 0,1 . 取到白球, 则 $X=0$; 取到红球, 则 $X=1$.
    • $P\{X=0\}=\frac{1}{2}, \quad P\{X=1\}=\frac{1}{2} .$
      • $E(X)=0 \times \frac{1}{2}+1 \times \frac{1}{2}=\frac{1}{2}$
      • $E(X^2)=0 \times \frac{1}{2}+1^2 \times \frac{1}{2}=\frac{1}{2}$
      • $\displaystyle \begin{aligned}D(X)=E(X^2)-[E(X)]^2=\frac{1}{2}-\left(\frac{1}{2}\right)^2=\frac{1}{4}\end{aligned}$
  • 从乙盒中取到红球的概率
    $Y$ 的可能取值为 0,1 .
    • 分析Y的概率
      • 若从甲盒中取出的是白球, 则后来乙盒中共有 2 个红球和 3 个白球, 取到红球的概率为 $\frac{2}{5}$,
        • 即在 $X=0$ 发生的条件下 $Y=1$ 发生的概率为 $\frac{2}{5}$;
      • 若从甲盒中取出的是红球, 则后来乙盒中共有 3 个红球和 2 个白球, 取到红球的概率为 $\frac{3}{5}$,
        • 即在 $X=1$ 发生的条件下 $Y=1$ 发生的概率为 $\frac{3}{5}$.
    • $\displaystyle \begin{aligned}P\{Y=0\}=\frac{1}{2}, \quad P\{Y=1\}=\frac{1}{2} .\end{aligned}$
      • $\displaystyle \begin{aligned} P\{Y=1\} & =P\{Y=1 \mid X=0\} P\{X=0\}+P\{Y=1 \mid X=1\} P\{X=1\} \end{aligned}$
        • $\displaystyle \begin{aligned} =\frac{2}{5} \times \frac{1}{2}+\frac{3}{5} \times \frac{1}{2}=\frac{1}{2} \end{aligned}$
      • $P\{Y=0\}=1-P\{Y=1\}=1-\frac{1}{2}=\frac{1}{2} .$
    • 同理, $E(Y)=\frac{1}{2}, E\left(Y^2\right)=\frac{1}{2}, D(Y)=\frac{1}{2}-\left(\frac{1}{2}\right)^2=\frac{1}{4}$.
  • $X Y$ 的可能取值为 0,1
    • $\displaystyle \begin{aligned} & P\{X Y=1\}=P\{X=1, Y=1\}=P\{Y=1 \mid X=1\} P\{X=1\}=\frac{3}{5} \times \frac{1}{2}=\frac{3}{10} . \end{aligned}$
    • $\displaystyle \begin{aligned} P\{X Y=0\}=1-\frac{3}{10}=\frac{7}{10} . \end{aligned}$
    • 于是, $E(X Y)=P\{X Y=1\} \times 1+P\{X Y=0\} \times 0$
      • $\displaystyle \begin{aligned}=\frac{3}{10} \times 1+\frac{7}{16} \times 0=\frac{3}{10}\end{aligned}$
• 因此,$\displaystyle \begin{aligned}\rho=\frac{E(X Y)-E(X) E(Y)}{\sqrt{D(X)} \sqrt{D(Y)}}=\frac{\frac{3}{10}-\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}}=\frac{\frac{1}{20}}{\frac{1}{4}}=\frac{1}{5} .\end{aligned}$

三、解答题

(本题共 6 小题, 共 70 分, 解答应写出文字说明、证明过程或演算步骤.)

(17)

(本题满分 10 分)

求极限 $\displaystyle \lim _{x \rightarrow 0}\left(\frac{1+\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{\mathrm{e}^{x}-1}-\frac{1}{\sin x}\right)$.

(17)

• 原式
  • $\displaystyle \begin{aligned} = \lim_{x \to 0} \frac{\sin x + \sin x \int_0^x e^{t^2} dt - e^x + 1}{ (e^x - 1) \sin x} \xrightarrow[]{\left.\begin{array}{l}e^x-1 \sim x \\\sin x \sim x\end{array}\right\} x^2}\end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[乘法求导]{洛必达}\lim_{x \to 0} \frac{\cos x + \cos x \int_0^x e^{t^2} dt + \sin x \cdot e^{x^2} - e^x}{2x} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[就洛必达几次]{分母有几次} \lim_{x \to 0} \frac{-\sin x - \sin x \int_0^x e^{t^2} dt + \cos x \cdot e^{x^2} + \cos x \cdot e^{x^2} + \sin x \cdot e^{x^2} \cdot 2x - e^x}{2} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\sin 0=0, \cos 0=1} \frac{1}{2}. \end{aligned}$

(18)

(本题满分 12 分)

设 $u_{n}(x)=\mathrm{e}^{-n x}+\frac{x^{n+1}}{n(n+1)}(n=1,2, \cdots)$, 求级数 $\displaystyle \sum_{n=1}^{\infty} u_{n}(x)$ 的收敛域及和函数.

(18)

• 设 $\displaystyle \begin{aligned} u_n(x) = e^{-nx} + \frac{x^{n+1}}{n(n+1)} \end{aligned}$（$\displaystyle \begin{aligned} n = 1, 2, \cdots \end{aligned}$），求级数 $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} u_n(x) \end{aligned}$ 的收敛域及和函数。

• $\displaystyle \begin{aligned} S(x) = \sum_{n=1}^{\infty} \left[ e^{-nx} + \frac{1}{n(n+1)} x^{n+1} \right] \end{aligned}$

  • $\displaystyle \begin{aligned} = \sum_{n=1}^{\infty} e^{-nx} + \sum_{n=1}^{\infty} \frac{1}{n(n+1)} x^{n+1} \end{aligned}$

• 求收敛域

  • $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} e^{-nx} = e^{-x} + e^{-2x} + \cdots + e^{-nx} + \cdots = \frac{e^{-x}}{1 - e^{-x}}, \quad -1 < e^{-x} < 1, \text{即 } x > 0. \end{aligned}$
  • 而$\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}, \text{因为} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n(n+1)}{(n+1)(n+2)} = 1, \text{于是 } R = 1. \end{aligned}$
    • 又 $\displaystyle \begin{aligned} x = 1 \end{aligned}$ 和 $\displaystyle \begin{aligned} x = -1 \end{aligned}$ 时，$\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \end{aligned}$ 和 $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)} \end{aligned}$ 都收敛，
    • 所以 $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)} \end{aligned}$ 的收敛域为 $\displaystyle \begin{aligned}\ [-1, 1 ] \end{aligned}$
  • 所以级数 $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} u_n(x) \end{aligned}$ 的收敛域为 $\displaystyle \begin{aligned} (0,1] \end{aligned}$。令 $\displaystyle \begin{aligned} S(x) = \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)} \end{aligned}$，$\displaystyle \begin{aligned} 0 < x \leq 1 \end{aligned}$。

• 求和函数$\displaystyle \begin{aligned} S(x)= \sum_{n=1}^{\infty} e^{-nx} + \sum_{n=1}^{\infty} \frac{1}{n(n+1)} x^{n+1} \end{aligned}$

  • 第一部分
    • $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} e^{-nx} = \sum_{n=1}^{\infty} t^n, \quad -1 < t < 1 \xrightarrow[]{\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}}\frac{e^{-x}}{1 - e^{-x}}\end{aligned}$
  • 第二部分
    • 逐项求导
      • 当 $\displaystyle \begin{aligned} 0 < x < 1 \end{aligned}$ 时，$\displaystyle \begin{aligned} S'(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) \end{aligned}$，
        因 $\displaystyle \begin{aligned} \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} \end{aligned}$，$\displaystyle \begin{aligned} -1 < x < 1 \end{aligned}$。
    • 逐项积分
      • $\displaystyle \begin{aligned} \text{于是 } S(x) = \int_0^x -\ln(1-t) \, dt + S(0) = -t\ln(1-t) \bigg|_0^x + \int_0^x t \cdot \frac{-1}{1-t} \, dt + 0 \end{aligned}$
      • $\displaystyle \begin{aligned} = -x\ln(1-x) + \int_0^x \frac{1-t-1}{1-t} \, dt = -x\ln(1-x) + x + \ln(1-x).\end{aligned}$

• 补点:而 $\displaystyle \begin{aligned} S(x) \end{aligned}$ 在收敛域上是连续的，于是 $\displaystyle \begin{aligned} S(1) = \lim_{x \to 1} S(x) \end{aligned}$

  • $\displaystyle \begin{aligned} = \lim_{x \to 1^-} \left[ -x\ln(1-x) + x + \ln(1-x) \right] \end{aligned}$
  • $\displaystyle \begin{aligned} = \lim_{x \to 1^-} \underbrace{(1-x)}_{=0}\ln(1-x) + 1 \end{aligned}$
  • (令 $\displaystyle \begin{aligned} 1-x = t \end{aligned}$) $\displaystyle \begin{aligned} = \lim_{t \to 0^+} t\ln t + 1 = 0 + 1 = 1 \end{aligned}$。

• 综上可知，级数 $\displaystyle \begin{aligned} \sum_{n=1}^{\infty} u_n(x) \end{aligned}$ 的和函数为

  • $\displaystyle \begin{aligned} \begin{cases}\frac{e^{-x}}{1-e^{-x}} - x\ln(1-x) + x + \ln(1-x), & 0 < x < 1, \\\frac{e^{-1}}{1-e^{-1}} + 1, & x = 1.\end{cases} \end{aligned}$

小猪佩奇

• $\displaystyle \begin{aligned} \frac{1}{n \cdot (n+1)} = \frac{1}{n} - \frac{1}{n+1} \end{aligned}$
• $\displaystyle \begin{aligned}\sum_{n=1}^{\infty} \frac{1}{n(n+1)} x^{n+1}\end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{裂项}\sum_{n=1}^{\infty} \left[ \frac{x^{n+1}}{n} - \frac{1}{n+1} x^{n+1} \right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{向分母看齐}x \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{套公式} x \cdot [-\ln(1-x)] + \ln(1-x) + x \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{合并同类项} (1-x) \cdot \ln(1-x) + x \end{aligned}$

(19)

(本题满分 12 分)

已知曲线 $\displaystyle C:\left\{\begin{array}{l}x^{2}+2 y^{2}-z=6, \\ 4 x+2 y+z=30,\end{array}\right.$ 求 $C$ 上的点到 $x O y$ 坐标面距离的最大值.

(19)

• • 分析 本题可以利用拉格朗日乘数法求解.点 $(x, y, z)$ 到 $x O y$ 面的距离为 $|z|$ ，故目标函数可设为 $z^2$ 。本题中的约束条件为点 $(x, y, z)$ 落在曲线 $C$ 上，故约束条件有两个等式.
• $\xrightarrow[]{建立拉格朗日函数}L(x, y, z, \lambda, \mu)=|z|+\lambda\left(x^2+2 y^2-z-6\right)+\mu(4 x+2 y+z-30)$
• $\displaystyle \begin{aligned} \xrightarrow[]{绝对值变成平方} L(x, y, z, \lambda, \mu)=z^2+\lambda\left(x^2+2 y^2-z-6\right)+\mu(4 x+2 y+z-30) \end{aligned}$
  • $\displaystyle \begin{aligned} \left\{\begin{array}{l}①L_x=2 \lambda x+4 \mu=0 \\②L_y=4 \lambda y+2 \mu=0 \\③L_z=2 z-\lambda+\mu=0 \\④L_\lambda=x^2+2 y^2-z-6=0 \\⑤L_\mu=4 x+2 y+z-30=0\end{array}\right. \end{aligned}$

gpt版

• 转化问题：求 $C$ 上点到 $xOy$ 面的距离，实际上是求 $z$ 坐标的最大值。

  • 曲线 $C$ 的方程组为 $\{x^2 + 2y^2 - z = 6, 4x + 2y + z = 30\}$。

• 构造拉格朗日函数 $L(x, y, z, \lambda, \mu)$。

  • $L(x, y, z, \lambda, \mu) = z + \lambda(4x + 2y + z - 30) + \mu(x^2 + 2y^2 - z - 6)$。

• 消元法求解驻点：对 $L$ 求偏导并令其等于零。

  • $\frac{\partial L}{\partial x} = 4\lambda + 2\mu x = 0$。
  • $\frac{\partial L}{\partial y} = 2\lambda + 4\mu y = 0$。
  • $\frac{\partial L}{\partial z} = 1 + \lambda - \mu = 0$。
  • $\frac{\partial L}{\partial \lambda} = 4x + 2y + z - 30 = 0$。
  • $\frac{\partial L}{\partial \mu} = x^2 + 2y^2 - z - 6 = 0$。

• 消元法求解驻点的坐标。

  • 解得两组解：$(x, y, z) = (4, 1, 12)$ 和 $(-8, -2, 66)$。

• 确定最大值。

  • 对比两组解，$z$ 的最大值为 66。

(20)

设 $D \subset \mathbf{R}^{2}$ 是有界单连通闭区域, $I(D)=\iint_{D}\left(4-x^{2}-y^{2}\right) \mathrm{d} x \mathrm{~d} y$ 取得最大值的积 分区域记为 $D_{1}$. (I) 求 $I\left(D_{1}\right)$ 的值;

(II) 计算 $\displaystyle \int_{\partial D_{1}} \frac{\left(x \mathrm{e}^{x^{2}+4 y^{2}}+y\right) \mathrm{d} x+\left(4 y \mathrm{e}^{x^{2}+4 y^{2}}-x\right) \mathrm{d} y}{x^{2}+4 y^{2}}$, 其中 $\partial D_{1}$ 是 $D_{1}$ 的正向边界.

(20)

• (1)
  

  • 区域 $\displaystyle \begin{aligned} D_1 \end{aligned}$ 应当可以使被积函数 $\displaystyle \begin{aligned} f(x, y) = 4 - x^2 - y^2 \geq 0 \end{aligned}$，
  • 且 $\displaystyle \begin{aligned} D_1 \end{aligned}$ 之外还必须可以使被积函数 $\displaystyle \begin{aligned} f(x, y) = 4 - x^2 - y^2 < 0 \end{aligned}$。
  • 所以 $\displaystyle \begin{aligned} D_1 = \{(x, y) | x^2 + y^2 \leq 4\} \end{aligned}$。

• $\displaystyle \begin{aligned} I(D_1) = \iint_{D_1} (4 - x^2 - y^2) \, dx \, dy = \int_0^{2\pi} d\theta \int_0^2 (4 - r^2) r \, dr \xlongequal[]{\left.\left(2 r^2-\frac{r^4}{4}\right)\right|_0 ^2=8-4=4}8\pi. \end{aligned}$

• (2)

• 记所求积分为 $I, P(x, y)=\frac{x \mathrm{e}^{x^2+4 y^2}+y}{x^2+4 y^2}, Q(x, y)=\frac{4 y \mathrm{e}^{x^2+4 y^2}-x}{x^2+4 y^2}$,
  则当 $(x, y) \neq(0,0)$ 时,

  • $\displaystyle \begin{aligned} \frac{\partial Q}{\partial x} & =\frac{\left(8 x y \mathrm{e}^{x^2+4 y^2}-1\right)\left(x^2+4 y^2\right)-\left(4 y \mathrm{e}^{x^2+4 y^2}-x\right) \cdot 2 x}{\left(x^2+4 y^2\right)^2} =\frac{\partial P}{\partial y}\end{aligned}$
  • 于是, $\displaystyle \begin{aligned}\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\end{aligned}$

• 挖点用格林
  

  • 由格林公式，$\displaystyle \begin{aligned}\oint_L P d x+Q d y=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d \sigma\end{aligned}$
    • 挖点前的全部区域:
      $\displaystyle \begin{aligned} \oint_{L+L^{\prime}} \frac{\left(x \mathrm{e}^{x^2+4 y^2}+y\right) \mathrm{d} x+\left(4 y \mathrm{e}^{x^2+4 y^2}-x\right) \mathrm{d} y}{x^2+4 y^2} \xlongequal{\text { 格林公式 }} \iint_{D_1 \backslash D_0} 0 \mathrm{~d} x \mathrm{~d} y=0, \end{aligned}$
  • 挖点后的区域
    $\displaystyle \begin{aligned} \oint_{L} & =\underline{\oint_{L+L'}}-\oint_{L^{'内}} \end{aligned}$
    • $\displaystyle \begin{aligned} =0+\oint_{L^{'外}} \end{aligned}$
    • $\displaystyle \begin{aligned}=\oint_{L^{\prime－}} \frac{\left(x \mathrm{e}^{x^2+4 y^2}+y\right) \mathrm{d} x+\left(4 y \mathrm{e}^{x^2+4 y^2}-x\right) \mathrm{d} y}{x^2+4 y^2}\end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[x^2+4 y^2=\varepsilon^2]{\varepsilon^2} \oint_{L^{\prime}}\left(x \mathrm{e}^{\varepsilon^2}+y\right) \mathrm{d} x+\left(4 y \mathrm{e}^{\varepsilon^2}-x\right) \mathrm{d} y \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal{\text { 格林公式 }} \frac{1}{\varepsilon^2} \iint_{D_0}\left[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right] \mathrm{d} x \mathrm{~d} y=\frac{1}{\varepsilon^2} \iint_{D_0}(-1-1) \mathrm{d} x \mathrm{~d} y \end{aligned}$
    • $\displaystyle \begin{aligned} =-\frac{2}{\varepsilon^2} \cdot D_0 \text { 的面积 }\xlongequal[]{\pi ab}-\frac{2}{\varepsilon^2} \times \pi \times \varepsilon \times \frac{\varepsilon}{2}=-\pi . \end{aligned}$

• 2000 年数一试题

  • 计算曲线积分 $I=\oint_L \frac{x \mathrm{~d} y-y \mathrm{~d} x}{4 x^2+y^2}$ ，其中 $L$ 是以点 $(1,0)$ 为中心， $R$ 为半径的圆周 $(R>1)$, 取逆时针方向.

• 2020 年数一试题

  • 计算曲线积分 $I=\oint_L \frac{4 x-y}{4 x^2+y^2} \mathrm{~d} x+\frac{x+y}{4 x^2+y^2} \mathrm{~d} y$, 其中 $L$ 为 $x^2+y^2=2$, 方向为逆时针方向.

(21)

(本题满分 12 分)

已知 $\displaystyle \boldsymbol{A}=\left(\begin{array}{ccc}a & 1 & -1 \\ 1 & a & -1 \\ -1 & -1 & a\end{array}\right)$. ( I ) 求正交矩阵 $\boldsymbol{P}$,使得 $\boldsymbol{P}^{\mathrm{T}} \boldsymbol{A P}$ 为对角矩阵;

(II) 求正定矩阵 $\boldsymbol{C}$,使得 $\boldsymbol{C}^{2}=(a+3) \boldsymbol{E}-\boldsymbol{A}$, 其中 $\boldsymbol{E}$ 为 3 阶单位矩阵.

(21)

• 解 (1)

• $\displaystyle |A-\lambda E|=\left|\begin{array}{ccc}a-\lambda & 1 & -1 \\ 1 & a-\lambda & -1 \\ -1 & -1 & a-\lambda\end{array}\right|\xlongequal[]{c1+c3}\left|\begin{array}{ccc}a-\lambda-1 & 1 & -1 \\ 0 & a-\lambda & -1 \\ a-\lambda-1 & -1 & a-\lambda\end{array}\right|$

  • $\displaystyle \begin{aligned} \xlongequal[]{r3-r1}\left|\begin{array}{ccc}a-\lambda-1 & 1 & -1 \\0 & a-\lambda & -1 \\0 & -2 & a-\lambda+1\end{array}\right| \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{按第一列展开}(a-\lambda-1)[(a-\lambda)(a-\lambda+1)-2]\end{aligned}$
  • $\displaystyle \begin{aligned}=(a-\lambda-1)\left[(a-\lambda)^2+(a-\lambda)-2\right]\end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\begin{aligned}& t+2 \\& t-1\end{aligned}}(a-\lambda-1)[a-\lambda+2][a-\lambda-1] \end{aligned}$
  • $\lambda_1=\lambda_2=a-1 . \quad \lambda_3=a+2$

• 当二重根$\displaystyle \begin{aligned} \begin{gathered}\lambda=a-1 \end{gathered} \end{aligned}$，$\displaystyle \begin{aligned}(A-(a-1) E) x=0 \end{aligned}$

  • $\displaystyle \begin{aligned}\left(\begin{array}{ccc}1 & 1 & -1 \\1 & 1 & -1 \\-1 & -1 & 1\end{array}\right) \xrightarrow[]{2，3两行成比例}\left(\begin{array}{ccc}1 & 1 & -1 \\0 & 0 & 0 \\0 & 0 & 0\end{array}\right)\end{aligned}$
  • $\displaystyle \begin{aligned}x=k_1\left(\begin{array}{c}-1 \\1 \\0\end{array}\right)+k_2\left(\begin{array}{l}1 \\0 \\1\end{array}\right) \quad\left(k_1 ， k_2 \text { 不全为 } 0\right)\end{aligned}$

• $\displaystyle \begin{aligned} \text { 当 } \lambda=a+2 \text {， }(A-(a+2) E) x=0 \end{aligned}$

  • $\displaystyle \begin{aligned} \xrightarrow[]{} \left(\begin{array}{ccc}-2 & 1 & -1 \\1 & -2 & -1 \\-1 & -1 & -2\end{array}\right) \xrightarrow[]{行互换化简}\left(\begin{array}{ccc}1 & 1 & 2 \\1 & -2 & -1 \\-2 & 1 & -1\end{array}\right) \end{aligned}$
  • $\displaystyle \begin{aligned}\xrightarrow[]{\begin{aligned}& r_2-r_1 \\& r_3+2 r_1\end{aligned}}\left(\begin{array}{ccc}1 & 1 & 2 \\0 & -3 & -3 \\0 & 3 & 3\end{array}\right) \xrightarrow[]{2，3两行成比例}\left(\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 1 \\0 & 0 & 0\end{array}\right)\end{aligned}$
  • $\displaystyle \begin{aligned} \therefore x=k_3\left(\begin{array}{c}-1 \\-1 \\1\end{array}\right)\end{aligned}$

• 先施密特正交化

  • $\displaystyle \begin{aligned} \boldsymbol{\beta}_1=\boldsymbol{\xi}_1=\left(\begin{array}{c}-1 \\1 \\0\end{array}\right), \end{aligned}$
  • $\displaystyle \begin{aligned} \boldsymbol{\beta}_2=\boldsymbol{\xi}_2-\frac{\left(\boldsymbol{\beta}_1, \boldsymbol{\xi}_2\right)}{\left\|\boldsymbol{\beta}_1\right\|^2} \boldsymbol{\beta}_1=\left(\begin{array}{l}1 \\0 \\1\end{array}\right)-\frac{(-1)}{2}\left(\begin{array}{c}-1 \\1 \\0\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}1 \\1 \\2\end{array}\right), \end{aligned}$
  • $\displaystyle \begin{aligned} \boldsymbol{\beta}_3=\boldsymbol{\xi}_3=\left(\begin{array}{c}-1 \\-1 \\1\end{array}\right)\end{aligned}$
    • 实对称矩阵不同特征值的特征向量相互正交，$\boldsymbol{\beta}_3$不用算了

• $\displaystyle \begin{aligned} \text { 将 } \boldsymbol{\beta}_1, \boldsymbol{\beta}_2, \boldsymbol{\beta}_3 \text { 单位化. } \end{aligned}$

  • $\displaystyle \begin{aligned} \boldsymbol{\varepsilon}_1=\frac{\boldsymbol{\beta}_1}{\left\|\boldsymbol{\beta}_1\right\|}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}-1 \\1 \\0\end{array}\right), \end{aligned}$
  • $\displaystyle \begin{aligned}\boldsymbol{\varepsilon}_2=\frac{\boldsymbol{\beta}_2}{\left\|\boldsymbol{\beta}_2\right\|}=\frac{1}{\sqrt{6}}\left(\begin{array}{c}1 \\1 \\2\end{array}\right)\end{aligned}$
  • $\displaystyle \begin{aligned}\boldsymbol{\varepsilon}_3=\frac{\boldsymbol{\beta}_3}{\left\|\boldsymbol{\beta}_3\right\|}=\frac{1}{\sqrt{3}}\left(\begin{array}{c}-1 \\-1 \\1\end{array}\right) .\end{aligned}$

• 存在正交矩阵 P ，使得\displaystyle \begin{aligned}P^T A P = \Lambda = \left(\begin{array} a-1 & & \\ & a-1 \\ & & a+2 \end{array}\right)\end{aligned}

(2)

• $\boldsymbol{C}^{2}=(a+3) \boldsymbol{E}-\boldsymbol{A}$

  • $\displaystyle (a+3) E-A=\left(\begin{array}{ccc}3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 3\end{array}\right)$
  • $\displaystyle \begin{aligned} C = Q \Lambda^{\frac{1}{k}} Q^{\top}， A = C^k \end{aligned}$

• 下面就是要把施密特正交公式背下来∶施密特正交化的步骤

  • β1就是α1∶$\beta_1=\alpha_1$
  • β2就是α2减β1∶$\beta_2=\alpha_2-\frac{\left(\alpha_2, \beta_1\right)}{\left(\beta_1, \beta_1\right)} \beta_1$
    • 求β2的系数
    • 分子是α2和β1做内积
    • 分母就是β1向量的坐标平方和
  • β3就是α3减β1再减β2∶$\beta_3=\alpha_3-\frac{\left(\alpha_3, \beta_1\right)}{\left(\beta_1, \beta_1\right)} \beta_1-\frac{\left(\alpha_3, \beta_2\right)}{\left(\beta_2, \beta_2\right)} \beta_2$
    • 求β3的系数
    • β1的系数∶分子是α3和β1的内积，分母仍然是β1的坐标平方和
    • 这样得到的三个向量肯定两两垂直，
  • 这三个系数起什么作用？要保证这三个向量一定是两两垂直的
    • 这个系数非常有规律，容易记

• 下一步单位化

  • $\displaystyle \begin{aligned} & \gamma_1=\frac{\beta_1}{\left\|\beta_1\right\|} \quad \gamma_2=\frac{\beta_2}{\left\|\beta_2\right\|} \quad \gamma_3=\frac{\beta_3}{\left\|\beta_3\right\|} \\ & 写成通式，即\gamma_i=\frac{\beta_i}{\left\|\beta_i\right\|}(i=1,2,3) \end{aligned}$
  • 单位化后，得到的就是将两两垂直的向量，变成长度是1的单位向量
    • 拼起来之后，就是正交矩阵

• 若 $\displaystyle \begin{aligned} n \end{aligned}$ 阶矩阵 $\displaystyle \begin{aligned} A \end{aligned}$ 可相似对角化/可正交对角化，且其特征值分别为 $\displaystyle \begin{aligned} \lambda_1, \lambda_2, \cdots, \lambda_n \end{aligned}$，要计算 $\displaystyle \begin{aligned} C \end{aligned}$ 使得 $\displaystyle \begin{aligned} A = C^k \end{aligned}$？

• $\displaystyle \begin{aligned} \exists P，使得 P^{-1} A P = \Lambda \rightarrow A = P \Lambda P^{-1} = P \left( \Lambda^{\frac{1}{k}} \cdot \Lambda^{\frac{1}{k}} \cdots \Lambda^{\frac{1}{k}} \right) P^{-1} \end{aligned}$

  • $\displaystyle \begin{aligned} = P \Lambda^{\frac{1}{k}} P^{-1} \cdot P \Lambda^{\frac{1}{k}} P^{-1} \cdots \cdot P \Lambda^{\frac{1}{k}} P^{-1} \end{aligned}$
  • $\displaystyle \begin{aligned} \therefore C = P \Lambda^{\frac{1}{k}} P^{-1} \cdot A = C^k \end{aligned}$

• $\displaystyle \begin{aligned} \exists Q，使得 Q^{\top} A Q = \Lambda. \rightarrow A = Q \Lambda Q^{\top} = Q \left( \Lambda^{\frac{1}{k}} \right) Q^{\top} \end{aligned}$

  • $\displaystyle \begin{aligned} = Q \Lambda^{\frac{1}{k}} Q^{\top} \cdots \cdot Q \Lambda^{\frac{1}{k}} Q^{\top} \end{aligned}$
  • $\displaystyle \begin{aligned} C = Q \Lambda^{\frac{1}{k}} Q^{\top}， A = C^k \end{aligned}$

(22)

(本题满分 12 分)

在区间 $(0,2)$ 上随机取一点, 将该区间分成两段,较短一段的长度记为 $X$, 较长一段的长度记 为 $Y$, 令 $Z=\frac{Y}{X}$. (I) 求 $X$ 的概率密度; (II) 求 $Z$ 的概率密度; (III) 求 $E\left(\frac{X}{Y}\right)$.

(22)

• 解 为了解释这个答案，我们将分步骤来展示解题过程及其逻辑关系。以下是解析： (I) 求 $X$ 的概率密度
• 定义 $X$
  • $X = \min\{V, 2-V\}$，表示较短一段的长度,则x的取值范围为(0,1)
    • 如果超过1，就成了较长的一段，而不是较短的一段
    • x为长度，而不是(0，2)上的区间
  • 若随机变量 $X \sim U(a, b)$ 则 $X$ 概率密度函数
    • $\displaystyle f(x)=\left\{\begin{array}{ll}\frac{1}{b-a}, & a<x<b \\0, & \text { 其他. }\end{array}\right.$
  • $x \sim \cup(0,1)$:x服从(0,1)的均匀分布,套公式
    • 概率密度=区间长度分之一$\displaystyle \therefore f_x(x)=\left\{\begin{array}{lc}1 , & 0<x<1 \\ 0 & \text { else }\end{array}\right.$

(II) 求 $Z$ 的概率密度

• 定义 $Z$
  • $Z = \frac{Y}{X} \xlongequal[]{X+Y=2} \frac{2-X}{X}=\frac{2}{X}-1$。
• 如何讨论
  • 如果x是离散型,则不用画进图中,直接用全集分解来分类讨论
  • 只有连续型的范围,才画入图中
    • 均匀分布,如何画图
    • 正态分布,如何画图
    • 指数分布
  • 如果X和Y都是离散型,如何画图
  • 如果X和Y都是连续型,如何画图
    • x,y,z都存在,则画出(x,y)的区间,然后让z=f(x,y)在这个区间从下往上刷
  • 如果X和Y一个离散，一个连续,如何画图
    • 只剩(x和z)或(y和z)
    • 将x=g(z)反函数之后，就是从左往右刷
    • 没有用反函数，还是z=f(x)的时候，就从下往上刷
• 由x的范围,y的范围和函数Z=g(X,Y)三者围成的区间画图
  
  • 然后直接三件套
    • 当 $z<1$ 时, $\xrightarrow[]{写出分布函数}F_Z(z)\xlongequal[]{定义}P\{Z \leqslant z\}\xlongequal[]{转化}0$;
    • 当 $z \geqslant 1$ 时,$\xrightarrow[]{写出分布函数}F_Z(z)\xlongequal[]{定义}P\{Z \leqslant z\}\xlongequal[Z=\frac{2}{X}-1]{转化}P\left\{\frac{2}{X}-1 \leqslant z\right\}\xlongequal[]{反函数}P\left\{X \geqslant \frac{2}{z+1}\right\}$
      • $\displaystyle \xlongequal[]{根据x的积分上下限}\int_{\frac{2}{z+1}}^1 1 \mathrm{~d} x=1-\frac{2}{z+1} .$
        
• $Z$ 的概率密度=分布函数求导
  • 利用 $X$ 的概率密度和 $Z$ 与 $X$ 的关系来计算：
  • $\displaystyle f_Z(z) = \begin{cases}\frac{2}{(1+z)^2}, & z > 1, \\ 0, & \text{其他}.\end{cases}$

(III) 求 $E\left(\frac{X}{Y}\right)$

• 计算期望值:求谁的期望，在谁前面乘概率密度进行积分
  • $\displaystyle E\left(\frac{X}{Y}\right) =E\left(\frac{X}{2-X}\right) =\int_{-\infty}^{+\infty} \frac{x}{2-x} \cdot f_X(x) \mathrm{d} x\xlongequal[何处求积分]{f_X(x) =1}\int_0^1 \frac{x}{2-x} dx=2 \ln 2 - 1$

小崔版

• $\displaystyle \begin{aligned} &\text { (1) 令点为 } M . M \sim U(0.2) \end{aligned}$
  • $\displaystyle \begin{aligned} f_M(m)=\left\{\begin{array}{cc}\frac{1}{2} & 0<m<2 \\0 & \text { 其 } 0\end{array} . \quad F_M(m)=\left\{\begin{array}{cc}0 & m<0 \\\frac{m}{2} & 0 \leqslant m<2 \\1 & 2 \leqslant m\end{array}\right.\right. \end{aligned}$
  • $\displaystyle \begin{aligned} X=\min \{M, 2-M\} \end{aligned}$
  • $\displaystyle \begin{aligned} F_x(x) & =P\{X \leqslant x\}=P\{\min \{M, 2-M\} \leqslant x\} \end{aligned}$
    • $\displaystyle \begin{aligned} =1-P\{\min \{M, 2-M\}>x\}=1-P\{M>x .2-M>x\} \end{aligned}$
    • $\displaystyle \begin{aligned} =1-P\{M>x ，M<2-x\} \end{aligned}$
  • $\displaystyle \begin{aligned} & \left\{\begin{array}{l}1^{\circ} \text { 若 } x<0 \text { 时. } F_x(x)=0 . \\2^{\circ} \text { 若 } 0 \leqslant x<1 \text { 时. } F_x(x)=1-P\{x<M<2-x\}=1-\int_x^{2-x} \frac{1}{2} d m=x \\3^{\circ} \text { 若 } 1 \leqslant x \text { 时. } F_x(x)=1 \end{array} . \$ .\right. \end{aligned}$
    • $\displaystyle \begin{aligned}\Rightarrow f_x(x)= \begin{cases}1 & 0<x<1 \\0 & \text { 其他 }\end{cases}\end{aligned}$

(2)

• $\displaystyle \begin{aligned} F_2(z) & =P\{Z\leqslant z\}=P\left\{\frac{2-X}{X} \leqslant z\right\}=P\left\{\frac{2}{X}-1 \leqslant z\right\} \end{aligned}$
  • $\displaystyle \begin{aligned} =P\left\{x \geqslant \frac{2}{z+1}\right\} \end{aligned}$
    • $\displaystyle \begin{aligned} z<1 \text { 时. } F_2(z)=0 \end{aligned}$
    • $\displaystyle \begin{aligned} 1 \leqslant z \text { 时. } F_2(z)=P\left\{x \geqslant \frac{2}{z+1}\right\}\xlongequal[x密度为1]{x属于(0，1)} \int_{\frac{2}{z+1}}^1 1 d x=1-\frac{2}{3+1}\end{aligned}$

(3)

• $\displaystyle \begin{aligned} E\left(\frac{X}{Y}\right) & =E\left(\frac{X}{2-X}\right)=\int_{-\infty}^{+\infty} \frac{x}{2-x} \cdot f_x(x) d x \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{f_X(x) =1}\int_0^1 \frac{x}{2-x} d x\xlongequal[]{-2+2}\int_0^1-1+\frac{2}{2-x} d x \end{aligned}$
  • $\displaystyle \begin{aligned} =-x-\left.2 \ln |2-x|\right|_0 ^1=2 \ln 2-1 \end{aligned}$