dz1281638775 发表于 2014-2-14 21:35:05

2014年北师大数分高代(回忆版)考研试题及参考解答



\title{2014年北京师范大学数科院数分高代考研真题(回忆版)}
\author{} %% 邮箱地址
\date{}
\maketitle
\renewcommand{\abstractname}{声\quad 明}
\renewcommand{\proofname}{证明}

\section{高等代数部分65分}
\quad\\
{\color{blue}1.(15').数域$F$上的多项式$f(x)$不可约,证明:$f(x)$在复数域内没有重根。}
\begin{proof}
假设$f(x)$在复数域内有重根$a$,则$f(a)=f'(a)=0$,所以,$f(x)$与$f'(x)$不互素,若不然,则$\exists u(x),v(x)$,使得
\begin{displaymath}
f(x)u(x)+f'(x)v(x)=1,
\end{displaymath}
令$\displaystyle x=a$,则得$\displaystyle 0=1$的矛盾.\\
因此,存在次数不小于1的公因式$\displaystyle d(x)$,满足$\displaystyle d(x)\mid f(x), d(x)\mid f(x)$\\
$\displaystyle \Rightarrow \partial^{\circ}(d(x)) \leqslant \partial ^{\circ}(f'(x)) < \partial^{\circ}(f(x))$\\
故$d(x)$是$f(x)$的非平凡因式,这与$f(x)$不可约矛盾.
\end{proof}
\textrm{\\}
\\
{\color{blue}2.(20').子空间$V_1$由$\alpha_1=(\qquad\qquad)'$,$\alpha_2=(\qquad\qquad)'$生成,

\qquad\qquad\quad$V_2$由$\beta_1=(\qquad\qquad)'$,$\beta_2=(\qquad\qquad)'$生成。\\
求$V_1+V_2$,$V_1 \cap V_2$的基和维数。} \\
\textrm{\\}
\quad\\
{\color{blue}3.(15').(1)(5').验证双线性非退化函数;}\\
{\color{blue}(2)(5').$Hom(V,V)\to Hom(V,V)^{*}$同构;}\\
{\color{blue}(3)(5').$l_V$在同构下的像。}\\
\textrm{\\}
\quad\
{\color{blue}4.(15').$S=(a_{ij})_{n \times n}$是一个随机矩阵,且满足下列条件:

\qquad (\textrm{i})$0 \leqslant a_{ij} , \forall i,j$;(\textrm{ii})$a_{i1}+a_{i2}+\cdots+a_{in}=1,(i=1,2,\cdots ,n)$}\\
\quad\\
{\color{blue}证明:(1)(5')$S$有特征值$1$。(2)(10')$S$的特征值$\mid \lambda\mid \leqslant 1$。 }
\begin{proof}
(1)因为
\begin{displaymath}
(1,1,\cdots,1)\cdot S= (1,1,\cdots) \Rightarrow
(1\cdot I- S)^{T}\cdot
\left( \begin{array}{c}
1 \\
1 \\
\vdots \\
1
\end{array} \right)=0
\end{displaymath}

\qquad 所以,$S$有特征值$1$.

\qquad (2)用反证法,若$S$有一个特征值$\lambda_0$满足$\mid\lambda_0\mid >1$,则$\lambda_0I-S$为严格对角占优矩阵,

从而,$\lambda_0I-S$可逆,这与$\lambda_0$为$S$的特征值矛盾.
\qquad\\

\textcolor{1.00,0.00,0.00}{(注:与1999年浙大高代第六题完全一致)}
\end{proof}
\textrm{\\}


\section{数学分析部分85分}
\quad\\
{\color{blue}5.(20').求$\displaystyle\int\!\!\!\int\!\!\!\int_{\Omega} \mid \sin(x+y^2)\mid \mbox{d} x\, \mbox{d} y\,\mbox{d} z$,$\Omega$由$x=0,y=0,x=\pi,y=\frac{\pi}{2},y=z$围成。 }\\
\begin{eqnarray}
\mbox{解:原式} & = & \int_0^{\frac{\pi}{2}} dy\int_{0}^{-y^2+\pi}dx\int_{0}^{y}\sin(x+y^2)dz+\int_0^{\frac{\pi}{2}} dy\int_{-y^2+\pi}^{\pi}dx\int_{0}^{y}-\sin(x+y^2) dz \nonumber\\
\quad & = & \int_{0}^{\frac{\pi}{2}}dy\int_{0}^{-y^2+\pi}y\sin(x+y^2)dx+\int_{0}^{\frac{\pi}{2}}dy\int_{-y^2+\pi}^{\pi}-y\sin(x+y^2)dx\nonumber\\
\quad &
= &\int_{0}^{\frac{\pi}{2}}(-y\cos \pi+y\cos y^2)dy+\int_{0}^{\frac{\pi}{2}}(-y\cos y^2-y\cos \pi)dy =\frac{\pi^2}{4}\nonumber
\end{eqnarray}

\textrm{\\}
\\
{\color{blue}6.(20').$\displaystyle F(y)=\int_{0}^{\infty} e^{-xy}\frac{\sin x}{x}\mbox{d} x,0\leqslant y <\infty$.}\\

\quad\\
{\color{blue}(1)证明$F(y)$在$[0,\infty)$内连续,在$(0,+\infty)$内可微.(2)求 $F(y)$在$[0,+\infty)$上的值及$F(0)$.}\\
解:(1)令$\displaystyle u(x,y)=\frac{e^{-xy}}{x}, \qquad v(x,y)=\sin x$,\\则$u(x,y)$关于$x$在$(0,+\infty)$上单调且$\displaystyle\mathop{\lim}_{x\to +\infty} \sup\{\mid u(x,y)\mid:y\in[0,+\infty) \}=0$,又存在$M=2>0$,使得$\forall y\in [0,+\infty),\forall b\in(0,+\infty),c>0,\left |\int_{b}^{b+c}v(x,y)dx\right| \leqslant M$,根据Dirchlet判别法,\\
$F(y)$关于$y$在$[0,+\infty)$上一致收敛.\\
取任意的$y_0\in $,由于$F(y)$在$[\alpha,\beta]$上一致收敛,且$\displaystyle e^{-xy}\frac{\sin x}{x}$在$\displaystyle $上连续,从而$\displaystyle F(y)$在$\displaystyle [\alpha,\beta]$上连续,由$\displaystyle y_0$的任意性可得,$\displaystyle F(y)$在$\displaystyle $上连续. (可微性需证 $\displaystyle \int_0^{\infty} (e^{-xy}\frac{\sin x}{x})_y dx=-\int_0^{+\infty} e^{-xy}\sin x dx$内闭一致收敛)
\\(2)$F(y)$在$(0,+\infty)$内可微,且
\begin{displaymath}
F'(y)=-\int_0^{+\infty} e^{-xy}\sin x dx ,\qquad \forall y\in (0,+\infty)
\end{displaymath}
对上式进行分部积分,得:
\begin{displaymath}
F'(y)=-\frac{1}{y^2}-\frac{F'(y)}{y^2} ,\qquad \forall y\in (0,+\infty)
\end{displaymath}
所以,
\begin{displaymath}
F'(y)=-\frac{1}{1+y^2},\qquad \forall y\in (0,+\infty)
\end{displaymath}
进一步,有
\begin{displaymath}
F(y)=F(0)-\arctan y, \qquad \forall y\in (0,+\infty)
\end{displaymath}
于是,
\begin{displaymath}
F(+\infty)=\lim_{y \to \infty}(F(0)-\arctan y)=0 \Rightarrow F(0)=\frac{\pi}{2}
\end{displaymath}
\textrm{\\}
\quad\\
{\color{blue}7.(20')对于函数$\displaystyle f(x,y)=\left\{ \begin{array}{ll}
\frac{x^2y}{x^2+y^2} &, x^2+y^2 \neq 0\\
0&, x^2+y^2 = 0
\end{array}\right.$.}\\
{\color{blue}(1)求$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\mbox{在}(0,0)$点的值;}\\
{\color{blue}(2) 证明$f(x,y)$在$(0,0)$处连续 ;}\\
{\color{blue}(3)证明$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\mbox{在}(0,0)$处不连续;}\\
{\color{blue}(4) 证明$f(x,y)$在$(0,0)$不可微 .}\\
解:(1)由于
\begin{displaymath}
\mathop{\lim}_{\Delta x \to 0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\mathop{\lim}_{\Delta x \to 0}\frac{0-0}{\Delta x}=0
\end{displaymath}
\begin{displaymath}
\mathop{\lim}_{\Delta y \to 0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=\mathop{\lim}_{\Delta y \to 0}\frac{0-0}{\Delta y}=0
\end{displaymath}
所以,$\displaystyle \frac{\partial f(0,0)}{\partial x}=0,\frac{\partial f(0,0)}{\partial y}=0$.\\
(2)由于
\begin{displaymath}
\frac{\mid x^2y\mid}{x^2+y^2} \leqslant \left|\frac{x^2y}{2xy}\right|=\frac{1}{2}\mid x\mid
\end{displaymath}
从而得到
\begin{displaymath}
\lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}=0=f(0,0)
\end{displaymath}
故$\displaystyle f(x,y)$在$(0,0)$处连续.\\
(3)当$\displaystyle x^2+y^2=0$时,由(1)知$\displaystyle f_x(0,0)=f_y(0,0)=0$;\\
当$\displaystyle x^2+y^2\neq 0$时,可得
\begin{displaymath}
f_x(x,y)=\frac{2xy^3}{(x^2+y^2)^2}, \quad f_y(x,y)=\frac{x^4-x^2y^2}{(x^2+y^2)^2}
\end{displaymath}
令$\displaystyle y=kx$,则
\begin{displaymath}
\mathop{\lim}_{(x,y) \to (0,0)} f_x(x,y)=\frac{2k^3}{(1+k^2)^2},\quad \mathop{\lim}_{(x,y) \to (0,0)} f_y(x,y)=\frac{1-k^2}{(1+k^2)^2}
\end{displaymath}
所以,$\displaystyle f_x(x,y),f_y(x,y)$在$\displaystyle (0,0)$点不连续.\\
(4)$\displaystyle f(0+\Delta x,0+\Delta)-f(0,0)-=f(\Delta x,\Delta y)=\frac{\Delta x^2 \Delta y}{\Delta x^2+\Delta y^2},$\\
令$\Delta y = k \Delta x,$则
\begin{displaymath}
\frac{f(\Delta x,\Delta y)}{\sqrt{\Delta x^2 +\Delta y^2}}=\frac{\Delta x^2 \Delta y}{(\Delta x^2+\Delta y^2)^{\frac{3}{2}}}=\frac{k}{(1+k^2)^{\frac{3}{2}}}
\end{displaymath}
即$\displaystyle f(0+\Delta x,0+\Delta y)-f(0,0)-\neq o(\sqrt{\Delta x^2+\Delta y^2})$\\
因此,$\displaystyle f$在$\displaystyle (0,0)$点不可微.
\textrm{\\}
\quad\\
{\color{blue}8.(25').$f \in C \mbox{且}f$在$(0,1)$上二阶可导,$f''(x)\neq 0$
\begin{displaymath}
\int_{0}^{1} f(t)dt=0, f(0)=f(1)>0
\end{displaymath}.}\\
{\color{blue}(1)$f''(x)>0$ . }\\
{\color{blue}(2)$f(x)=0$在$(0,1)$上恰有两根.}\\
{\color{blue}(3)$\exists \xi \in (0,1),s.t.\quad f'(\xi)=\int_{\xi}^{1} f(t) \mbox{d} t$.} \\

\begin{proof}
(1).由积分中值定理$\exists \xi \in (0,1) , s.t.\quad \int_{0}^{1} f(t) \mbox{d} t=f(\xi)=0$

\qquad 由Lagrange 中值定理:$f(\xi)-f(0)=f'(\xi_1)\cdot \xi <0,\quad \xi_1 \in (0,\xi)$

\qquad $\Rightarrow$$f'(\xi_1)<0,$同理 $f(1)-f(\xi)=f'(\xi_2)\cdot(1-\xi),\quad \xi_2 \in(\xi,1) $

\qquad $\Rightarrow$ $f'(\xi_2) >0,$再利用Lagrange 中值定理有:
\begin{displaymath}
   f'(\xi_2)-f'(\xi_1)=f''(\xi_3)\cdot(\xi_2-\xi_1) >0 \Rightarrow f''(\xi_3) >0,
\end{displaymath}

\qquad 又由导数的介值性及$f''(x)\neq 0(\forall x\in (0,1))$可知$f''(x)>0$(否则若$\exists x_0 \in (0,1),s.t,\quad $

\qquad $ f''(x_0)<0$由导数的介值性得到矛盾,即$\exists x_1,s.t. f''(x_1)=0.$)

\qquad (2)$(\textrm{i})$若$f(x)$在$(0,1)$上有三个根或以上,显然利用三次Rolle定理,即知$\exists x',s.t.f''(x')=0 $,矛盾

\qquad 由(1)$f(x)$有一个根$\xi$.

\qquad (\textrm{ii})若$f(x)$只有一个根,由$f(x)$连续及$f(0)=f(1)>0$,必有$f(x)\geqslant 0 (\forall x \in (0,1))$

\qquad 若$\exists x_1 \in (0,1),s.t,f(x_1)<0.$ 显然由介值定理知,$f(x)$在$(0,x_1)$和$(x_1,1)$上各至少有一根,矛盾.

\qquad 但是,若$f(x) \geqslant 0(\forall x \in (0,1))$由$f(x)\in C$且$f(0)=f(1) >0$,必有$\int_0^1 f(x) \mbox{d} x >0$矛盾.

\qquad 综上所述,$f(x)$在$(0,1)$上恰有两根.

\qquad (3)令$F(x)=f'(x)-\int_{x}^{1} f(t) \mbox{d} t$,由(2)显然$f'(0+) <0,f'(1-)>0$

\qquad 故$F(0)=f'(0)-\int_0^1f(t)\mbox{d} t <0,F(1-)=f'(1)>0$, 又$F(x)\in C$

\qquad 由介值定理,$\exists \xi \in (0,1),s.t. f'(\xi)=\int_{\xi}^{1} f(t)\mbox{d} t$


\textcolor{1.00,0.00,0.00}{(注:与2005首师大数分第八题完全一致)}
\end{proof}

当$k$为何值时,直线$y=kx+\sqrt{2}$与曲线$y=e^x$所围成的图形的面积最小?并求此最小面积。\\
证明$<x^2+1>$是$\mathbb{Z}_3$的素理想和极大理想.
\renewcommand{\proofname}{解}
\begin{proof}
(1)设这$1000$个产品的寿命为$X_1,X_2,\cdots,X_{1000}$,若保险公司亏本,则有
\begin{displaymath}
\sum_{i=1}^{1000} (100-2000\cdot \mathrm{1}_{\{x_i <1095\}})<0 \Rightarrow \sum_{i=1}^{1000}\mathrm{1}_{\{x_i <1095\}} >50
\end{displaymath}
故令$\displaystyle\xi_i=\mathrm{1}_{\{x_i <1095\}}$,则所求概率即为$\displaystyle \mathbb{P}\left(\sum_{i=1}^{1000}\xi >50\right)$\\
由于\\
$\displaystyle \mathbb{E}(\xi_i)=0\cdot \mathbb{P}(x_i \geqslant 1095)+1\cdot \mathbb{P}(x_i\leqslant 1095)=\int_{365}^{1095}\frac{1}{20000}e^{-\frac{1}{20000}(x-365)}=1-e^{-\frac{730}{20000}}$\\
$\displaystyle\mathbb{D}^2(\xi_i)=\mathbb{E}{\xi_i ^2}-(\mathbb{E}\xi_i)^2=\mathbb{E}\xi_i-(\mathbb{E}\xi_i)^2$,\\
$\displaystyle \mathbb{P}\left(\sum_{i=0}^{1000}>50\right)=\mathbb{P}\left(\sum_{i=1}^{1000}(\xi_i-\mathbb{E}\xi_i)>50-1000\mathbb{E}\xi_i\right)=\mathbb{P}\left(\frac{\mathop{\sum}_{i=1}^{1000}(\xi_i-\mathbb{E}\xi_i)}{1000}>\frac{50}{1000}-\mathbb{E}\xi_i\right)$\\
所以,根据中心极限定理,$\displaystyle\frac{\mathop{\sum}_{i=1}^{1000}(\xi_i-\mathbb{E}\xi_i)}{1000} \sim N\left(0,\frac{\mathbb{D}(\xi_i)}{1000^2}\right)$\\
(就是这样做的,剩下的计算问题略)



dz1281638775 发表于 2014-2-14 21:36:13

错误在所难免,请指正!:)

hylpy 发表于 2018-11-7 11:30:17

感谢分享
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