2022年138套数学分析高等代数考研真题参考解答勘误
第5页第1(2)题
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购书 / 答疑 / pdf1 / pdf2 / 容易算得 有有理根 , 进而可以进一步分解. 最后得到
f(x)=(x-1)(2x+1)(x^2-x+1).
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
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### 第11页第3(5)题(i)
(ii) 的上面, $\displaystyle f$ 化为了标准形 $\displaystyle y_1^2+y_2^2-2y_3^2$.
### 第12页第 1 行
$y_1^2+y_2^2-2y_3^2=1$.
### 第14页倒数第 3 行
$n$ 元二次型
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
f(x_1,\cdots,x_n)=\left|\begin{array}{cccccccccc}
a_{11}&a_{12}&\cdots&a_{1n}&x_1\\
a_{21}&a_{22}&\cdots&a_{2n}&x_2\\
\vdots&\vdots&&\vdots&{\color{red}\vdots}\\
a_{n1}&a_{n2}&\cdots&a_{nn}&x_n\\
{\color{red}x_1}&{\color{red}x_2}&{\color{red}\cdots}&{\color{red}x_n}&{\color{red}0}\end{array}\right|
=\left|\begin{array}{cccccccccc}A&X\\
X^\mathrm{T}&0\end{array}\right|
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第17页第4题(1)
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
|f(x)|\leq \int_a^x |f'(t){\color{red}|}\mathrm{ d} t, \forall\ x\in [a,b];
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第26页第1题往前第2行
这就证明了 $\displaystyle \dim V_\sigma={\color{red}r}=\mathrm{tr} \left(\frac{1}{m}\sum_{i=1}^m \sigma^i\right)$.
### 第29页第7题往前第2行
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\mathrm{Hess} z|_{(2,1)}=\left(\begin{array}{cccccccccc}\frac{8}{{\color{red}x^3}}&1\\
1&\frac{4}{y^3}\end{array}\right)_{(2,1)}=\left(\begin{array}{cccccccccc}1&1\\
1&4\end{array}\right)
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第30页第8题证明(1)第5行
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
&\int_\varepsilon^A \frac{f(ax)-f(bx)}{x}\mathrm{ d} x
=\int_\varepsilon^A \frac{f(ax)}{ax}\mathrm{ d} (ax)
-\int_\varepsilon^A \frac{{\color{red}f}(bx)}{bx}\mathrm{ d} (bx)\\
=&\int_{a\varepsilon}^{aA}\frac{f(t)}{t}\mathrm{ d} t
-\int_{b\varepsilon}^{bA} \frac{f(t)}{t}\mathrm{ d} t
=\int_{a\varepsilon}^{b\varepsilon}\frac{f(t)}{t}\mathrm{ d} t
-\int_{aA}^{bA} \frac{f(t)}{t}\mathrm{ d} t\\
\xlongequal[\tiny\mbox{分中值}]{\tiny\mbox{第一积}}&f(\xi)\int_{a\varepsilon}^{b\varepsilon}\frac{1}{t}\mathrm{ d} t
-f(\eta)\int_{aA}^{bA}\frac{1}{t}\mathrm{ d} t
=[f(\xi)-f(\eta)] \ln \frac{b}{a}.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第37页第6题第2行
将其扩充为 $\displaystyle {\color{red}V}$ 的一组基 $\displaystyle \varepsilon_1,\cdots,\varepsilon_n$
### 第38页倒数第2行
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
{\color{red}-}E=AB-A+B-E=(A+E)(B-E)
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第38页倒数第1行
而 $\displaystyle {\color{red}1}$ 不是 $\displaystyle B$ 的特征值.
### 第39页导数第3行
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
b_{n{\color{red}+1}}=a_{2(n+1)} > a_{2n+1} > a_{2n}=b_n,\quad b_n=a_{2n} > 0.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第288页(3)最后一行
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
{\color{red}\sqrt{\pi}}\left(\sqrt{b}-\sqrt{a}\right).
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第350页(4)
::: block-1
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设 $\displaystyle S$ 所围立体为 $\displaystyle \varOmega$, 则
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
(5-5)^2+2\cdot 0^2+(0+1)^2 < 3.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
故 $\displaystyle (5,0,0)$ 在 $\displaystyle \varOmega$ 的内部. 设
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
(P,Q,R)=\frac{(x-5,y,z)}{\left[(x-5)^2+y^2+z^2\right]^\frac{3}{2}}\Rightarrow P_x+Q_y+R_z=0.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
故
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\mbox{原式}\xlongequal{\tiny\mbox{Gauss}}& \iint_{(x-5)^2+y^2+z^2=\varepsilon^2}P\mathrm{ d} y\mathrm{ d} z+Q\mathrm{ d} z\mathrm{ d} x+R\mathrm{ d} x\mathrm{ d} y\\
=&\frac{1}{\varepsilon^3}\iint_{(x-5)^2+y^2+z^2=\varepsilon^2}(x-5)\mathrm{ d} y\mathrm{ d} z+y\mathrm{ d} z\mathrm{ d} x+z\mathrm{ d} x\mathrm{ d} y\\
\xlongequal{\tiny\mbox{Gauss}}&\frac{1}{\varepsilon^3}\iiint_{(x-5)^2+y^2+z^2\leq \varepsilon^2}
3\mathrm{ d} x\mathrm{ d} y\mathrm{ d} z=4\pi.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
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### 第383页第1题
::: block-1
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由
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\frac{\max\left\{a,b,c\right\}}{3^\frac{1}{x}}\leq \left(\frac{a^x+b^x+c^x}{3}\right)^\frac{1}{x}\leq \max\left\{a,b,c\right\}
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
及迫敛性知原式 $\displaystyle =\max\left\{a,b,c\right\}$.
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### 第518页第13题
::: block-1
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设 $\displaystyle x=\cos\theta, y=\sin\theta$, 则
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
I&=\int_{-\pi}^\pi \frac{-\sin \theta(-\sin\theta)+\cos\theta\cos \theta}{3\cos^2\theta+\sin^2\theta}\mathrm{ d} \theta
\xlongequal{\theta-\frac{\pi}{2}=u}2\int_{-\frac{\pi}{2}}^\frac{\pi}{2}
\frac{\mathrm{ d} u}{2-\cos 2u}\\
&\xlongequal{\tiny\mbox{对称性}} 4\int_0^\frac{\pi}{2}\frac{\mathrm{ d} u}{2-\cos 2u}
\xlongequal{\tan u=s}4\int_0^\infty \frac{1}{2-\frac{1-s^2}{1+s^2}}
\frac{\mathrm{ d} s}{1+s^2}\\
&=4\int_0^\infty \frac{\mathrm{ d} s}{3s^2+1}
\xlongequal{\sqrt{3}s=v}4\int_0^\infty \frac{\frac{1}{\sqrt{3}}\mathrm{ d} v}{1+v^2}
=\frac{4}{\sqrt{3}}\cdot \frac{\pi}{2}=\frac{2\pi}{\sqrt{3}}.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
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### 第761页第1(1)题
请加上如下参考解答:
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\mbox{原式}&=\mathrm{exp}\left[\lim_{x\to 0^+}x\ln x\right]
=\mathrm{exp}\left[\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}\right]\\
&\xlongequal{\tiny\mbox{L'Hospital}} \mathrm{exp}\left[\lim_{x\to 0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\right]
=\mathrm{e}^0=1.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
### 第771页第7题
设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 的 $\displaystyle (i,j)$ 元满足
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
a_{ij}=\left\{\begin{array}{llllllllllll}2,&i=j+1,\\
1,&\mbox{其它}.\end{array}\right.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
求 $\displaystyle A$ 的特征多项式和极小多项式.
::: block-1
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由
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
A=ee^\mathrm{T}+N, N=\left(\begin{array}{cccccccccc}0&&&\\
1&\ddots&&\\
&\ddots&\ddots&\\
&&1&0\end{array}\right)
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
知 $\displaystyle A$ 的特征多项式为
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
&|\lambda E-A|=|\lambda E-N-ee^\mathrm{T}|
=|\lambda E-N|\cdot \left|E-(\lambda E-N)^{-1}ee^\mathrm{T}\right|\\
=&\lambda^n \left[1-e^\mathrm{T}(\lambda E-N)^{-1}e\right]
=\lambda^n \left[1-\frac{1}{\lambda}e^\mathrm{T} \left(E-\frac{N}{\lambda}\right)^{-1}e\right]\\
=&\lambda^n\left[1-\frac{1}{\lambda}e^\mathrm{T}\left(E+\frac{N}{\lambda}+\cdots+\frac{N^{n-1}}{\lambda^{n-1}}\right)e\right]\\
=&\lambda^n-\lambda^{n-1}\left(n+\frac{n-1}{\lambda}+\cdots+\frac{1}{\lambda^{n-1}}\right)
=\lambda^n-\sum_{k=0}^{n-1}(k+1)\lambda^k.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
注: 上式对 $\displaystyle \forall\ \lambda\neq 0$ 都成立, 而由连续性知对 $\displaystyle \forall\ \lambda\in\mathbb{C}$ 都成立. 又将 $\displaystyle \lambda E-A$ 的第 $\displaystyle n$ 列 $\displaystyle \cdot (-1)$ 加到其它各列知
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\lambda E-A\to\left(\begin{array}{cccccccccc}\lambda&0&0&\cdots&0&-1\\
-1&\lambda&0&\cdots&0&-1\\
0&-1&\lambda&\cdots&0&-1\\
\vdots&\vdots&\vdots&&\vdots&\vdots\\
0&0&0&\cdots&\lambda&-1\\
-\lambda&-\lambda&-\lambda&\cdots&-\lambda-1&\lambda-1\end{array}\right).
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
注意到左上角有一个 $\displaystyle n-1$ 阶子式为 $\displaystyle \lambda^{n-1}$. 设 $\displaystyle \lambda E-A$ 的行列式因子为
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
D_1(\lambda),\cdots,D_n(\lambda)=|\lambda E-A|,
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
则 $\displaystyle D_{n-1}(\lambda)\mid \lambda^{n-1}, D_{n-1}(\lambda)\mid |\lambda E-A|$. 由
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\left(\lambda^n, |\lambda E-A|\right)=1
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
知 $\displaystyle D_{n-1}(\lambda)=1$, 而 $\displaystyle A$ 的行列式因子为
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
1,\cdots,1,|\lambda E-A|,
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
不变因子为
$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
1,\cdots,1,|\lambda E-A|.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
故 $\displaystyle A$ 的极小极小多项式也为 $\displaystyle |\lambda E-A|$.
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### 第781页第5行
$\varphi_n(x)\to \varphi_n(x{\color{red}+1})$.
### 第894页第1(1)题
解答第一个等号后面少了一个 $\displaystyle \ln$:
::: block-1
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$\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{equation*}\begin{aligned}
\mbox{原式}=&\mathrm{exp}\left[\lim_{x\to 0}\frac{1}{\mathrm{e}^x-1}{\color{red}\ln}\ln \frac{1+x}{x}\right]
\xlongequal[\tiny\mbox{代换}]{\tiny\mbox{等价}} \mathrm{exp}\left\{\lim_{x\to 0}\frac{1}{x} \left[\frac{\ln (1+x)}{x}-1\right]\right\}\\
=&\mathrm{exp}\left[\lim_{x\to 0}\frac{\ln (1+x)-x}{x^2}\right]
\xlongequal[\tiny\mbox{展开}]{\tiny\mbox{Taylor}} \mathrm{e}^{-\frac{1}{2}}.
\tiny\boxed{\begin{array}{c}\mbox{跟锦数学}\\\mbox{微信公众号}\end{array}}\end{aligned}\end{equation*}$$
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