武汉大学2008年数学分析试题解答

武汉大学2008年数学分析试题解答

一：计算题

1.$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\ln x\ln \left( 1-x \right)=-\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x\ln x=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\ln \frac{1}{x}}{\frac{1}{x}}=\underset{y\to +\infty }{\mathop{\lim }}\,\frac{\ln y}{y}=0$

2.原极限$=\underset{x\to 0}{\mathop{\lim }}\,{{x}^{1-n}}\left( 1-\sqrt{1+x} \right)\cdots \left( 1-\sqrt[n]{1+x} \right)$

       $=\underset{x\to 0}{\mathop{\lim }}\,{{x}^{1-n}}{{\left( -1 \right)}^{n-1}}\frac{1}{n!}{{x}^{n-1}}=\frac{{{\left( -1 \right)}^{n-1}}}{n!}$

3.$\frac{dy}{dx}=\frac{\cos t}{3{{t}^{2}}+1}$

$\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\frac{-\sin t\left( 3{{t}^{2}}+1 \right)-6t\cos t}{{{\left( 3{{t}^{2}}+1 \right)}^{2}}}}{3{{t}^{2}}+1}=-\frac{\sin t\left( 3{{t}^{2}}+1 \right)+6t\cos t}{{{\left( 3{{t}^{2}}+1 \right)}^{3}}}$

4.${{f}^{n}}\left( x \right)={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\frac{n}{2}}}{{e}^{ax}}\sin \left( bx+n\arctan \frac{b}{a} \right)$

5.$\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{{{n}^{2}}}{\frac{n}{{{n}^{2}}+{{k}^{2}}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{{{n}^{2}}}{\frac{1}{1+{{\left( \frac{k}{n} \right)}^{2}}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\int_{0}^{n}{\frac{dx}{1+{{x}^{2}}}}=\int_{0}^{+\infty }{\frac{dx}{1+{{x}^{2}}}}=\frac{\pi }{2}$

二：证明：

由于$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt{x}{f}'\left( x \right)=a$，可知：$\exists M>0,\delta >0\left( <<1 \right),\forall x\in \left( 0,\delta  \right],$均有

$\left| \sqrt{x}{f}'\left( x \right) \right|\le M$

故对$\forall {{x}_{1}},{{x}_{2}}\in \left( 0,\delta  \right],$则存在${{x}_{3}}\in \left( 0,\delta  \right]$，有

$\frac{f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right)}{\sqrt{{{x}_{1}}}-\sqrt{{{x}_{2}}}}=2\sqrt{{{x}_{3}}}{f}'\left( {{x}_{3}} \right)$

即

$\left| f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right) \right|\le 2M\left| \sqrt{{{x}_{1}}}-\sqrt{{{x}_{2}}} \right|\le 2M\sqrt{\left| {{x}_{1}}-{{x}_{2}} \right|}$

从而可知，对$\forall \varepsilon >0\left( {{\left( \frac{\varepsilon }{2M} \right)}^{2}}<<\delta  \right),\forall {{x}_{1}},{{x}_{2}}\in \left( 0,\delta  \right]$，且$\left| {{x}_{1}}-{{x}_{2}} \right|<\left( \frac{\varepsilon }{2M} \right)$有

$\left| f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right) \right|<\varepsilon$

故$f\left( x \right)$在$\left( 0,\delta  \right]$上一致收敛

而$f(x)$在$[\delta ,1]$上连续，则$f\left( x \right)$在$[\delta ,1]$上一致收敛

于是$f\left( x \right)$在$(0,1]$上一致收敛

三：证明：由分析可知$\exists {{x}_{0}}\in \left[ a,b \right],M>0,\forall x\in \left[ a,b \right],0<f\left( x \right)\le f\left( {{x}_{0}} \right)=M$，

从而知${{\left( \int_{a}^{b}{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}\le M{{\left( b-a \right)}^{\frac{1}{n}}}$不妨设${{x}_{0}}\in \left( a,b \right)$，则

$\forall \varepsilon >0,\exists \delta >0,\left( {{x}_{0}}-\delta ,{{x}_{0}}+\delta  \right)\subseteq \left[ a,b \right]$，且$\forall x\in \left( {{x}_{0}}-\delta ,{{x}_{0}}+\delta  \right),f\left( x \right)\ge M-\varepsilon$ ，从而

${{\left( \int_{a}^{b}{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}\ge {{\left( \int_{{{x}_{0}}-\delta }^{{{x}_{0}}+\delta }{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}\ge {{\left( 2\delta  \right)}^{\frac{1}{n}}}\left( M-\varepsilon  \right)$，从而

$M-\varepsilon \le \underset{n\to +\infty }{\mathop{\lim }}\,{{\left( \int_{a}^{b}{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}\le M$，由于$\varepsilon$任意性，即$\underset{n\to +\infty }{\mathop{\lim }}\,{{\left( \int_{a}^{b}{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}=M$，即$\underset{n\to +\infty }{\mathop{\lim }}\,{{\left( \int_{a}^{b}{{{\left( f\left( x \right) \right)}^{n}}dx} \right)}^{\frac{1}{n}}}=\underset{x\in \left[ a,b \right]}{\mathop{\max }}\,f\left( x \right)$

四：证明：

1.由于$\left| {{e}^{-n}}\cos {{n}^{2}}x \right|\le \left| {{e}^{-n}} \right|$，而级数$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}$收敛，从而可知函数项级数$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}\cos \left( {{n}^{2}}x \right)$在$\left( -\infty ,+\infty  \right)$上一致收敛

2.$\forall k\in {{\mathbb{N}}^{*}}$，$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}\frac{{{d}^{k}}\left( \cos \left( {{n}^{2}}x \right) \right)}{d{{x}^{k}}}=\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}{{n}^{2k}}\cos \left( {{n}^{2}}x+\frac{n\pi }{2} \right)$，

而$\left| {{e}^{-n}}{{n}^{2k}}\cos \left( {{n}^{2}}x+\frac{n\pi }{2} \right) \right|\le {{e}^{-n}}{{n}^{2k}}$，而级数$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}{{n}^{2k}}$收敛，从而可知函数项级数$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}\frac{{{d}^{k}}\left( \cos \left( {{n}^{2}}x \right) \right)}{d{{x}^{k}}}$在$\left( -\infty ,+\infty  \right)$上一致收敛，由于$k$的任意性

$\sum\limits_{n=0}^{\infty }{{{e}^{-n}}}\frac{{{d}^{k}}\left( \cos \left( {{n}^{2}}x \right) \right)}{d{{x}^{k}}}\left( k=1,2,\cdots  \right)$在$\left( -\infty ,+\infty  \right)$上一致收敛

3.由分析可知${{f}^{\left( 2k \right)}}\left( 0 \right)={{\left( -1 \right)}^{k}}\sum\limits_{n=0}^{\infty }{{{e}^{-n}}{{n}^{4k}}},{{f}^{\left( 2k+1 \right)}}\left( 0 \right)=0$，从而可知$f\left( x \right)$在$x=0$的

$Taylor$级数为$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\sum\limits_{n=0}^{\infty }{{{e}^{-n}}{{n}^{4k}}}}{\left( 2k \right)!}}{{x}^{2k}}$

4.由于

 

$\sqrt[2k]{\left| \frac{{{\left( -1 \right)}^{k}}\sum\limits_{n=0}^{\infty }{{{e}^{-n}}{{n}^{4k}}}}{\left( 2k \right)!} \right|}\ge \sqrt[2k]{\left| \frac{{{e}^{-2k}}{{\left( 2k \right)}^{4k}}}{\left( 2k \right)!} \right|}\ge \frac{2k}{e}\to +\infty$从而可知$f\left( x \right)$在

$x=0$的$Taylor$级数为$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\sum\limits_{n=0}^{\infty }{{{e}^{-n}}{{n}^{4k}}}}{\left( 2k \right)!}}{{x}^{2k}}$

五、证明：由于$f(x)$在$[a,b]$上连续，则$f(x)$在$[a,b]$上一致连续

于是对$\forall \varepsilon >0,\exists \delta >0,$对任意的$x',x''\in [a,b]$，当$\left| x'-x'' \right|<\delta$时，有

$$\left| f(x')-f(x'') \right|<\frac{\varepsilon }{2}$$

于是将$[a,b]$区间$k$等分，$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{k}}=b$，使得$\frac{b-a}{k}<\delta$

且$\Delta {{x}_{i}}={{x}_{i}}-{{x}_{i-1}}=\frac{b-a}{k}$，$i=1,2,\cdots ,k$于是有$\left| f({{x}_{i}})-f({{x}_{i-1}}) \right|<\frac{\varepsilon }{2}$

同时对任意的$x\in [{{x}_{i-1}},{{x}_{i}}]$，有$\left| f(x)-f({{x}_{i}}) \right|<\frac{\varepsilon }{2},\left| f(x)-f({{x}_{i\text{-}1}}) \right|<\frac{\varepsilon }{2}$

又由于$\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}({{x}_{i}})=f({{x}_{i}})\Rightarrow \exists {{N}_{i}}>0$，当$n>{{N}_{i}}$时有$\left| {{f}_{n}}({{x}_{i}})-f({{x}_{i}}) \right|<\frac{\varepsilon }{2}$，其中$i=1,2,\cdots ,k$

于是令$N=\max \{{{N}_{1}},{{N}_{2}},\cdots ,{{N}_{k}}\}$，对任意的$x\in [{{x}_{i-1}},{{x}_{i}}]$，当$n>N$时有

$\left| {{f}_{n}}({{x}_{i-1}})-f(x) \right|\le \left| {{f}_{n}}({{x}_{i-1}})-f({{x}_{i-1}}) \right|+\left| f({{x}_{i-1}})-f(x) \right|<\varepsilon$

$\left| {{f}_{n}}({{x}_{i}})-f(x) \right|\le \left| {{f}_{n}}({{x}_{i}})-f({{x}_{i}}) \right|+\left| f({{x}_{i}})-f(x) \right|<\varepsilon$

对任意的$x\in [a,b]$，必有$x\in [{{x}_{i-1}},{{x}_{i}}]$，$i=1,2,\cdots ,k$，当$n>N$时，有$f(x)$的单调性知： $$\left| {{f}_{n}}(x)-f(x) \right|\le \max \{\left| {{f}_{n}}({{x}_{i-1}})-f(x) \right|\left| {{f}_{n}}({{x}_{i}})-f(x) \right|\text{ }\!\!\}\!\!\text{ }\varepsilon$$

于是$\{{{f}_{n}}(x)\}$在$[a,b]$上一致收敛于$f(x)$

六：解：1.由分析可知

$\frac{\partial w}{\partial x}=y\frac{\partial F}{\partial u},\frac{\partial w}{\partial y}=x\frac{\partial F}{\partial u}+z\frac{\partial F}{\partial v},\frac{\partial w}{\partial z}=y\frac{\partial F}{\partial v}$

显然有

$x\frac{\partial w}{\partial x}+z\frac{\partial w}{\partial z}=y\frac{\partial w}{\partial y}$

2.如果$F\left( {{u}_{1}},\cdots ,{{u}_{m}} \right)$是具有连续偏导数的多元函数，则原偏微分方程的完全积分为

$w=F\left( {{x}_{1}}y,\cdots ,{{x}_{m}}y \right)$

七：证明：

${F}'\left( t \right)=2t\int_{{{t}^{2}}-t}^{{{t}^{2}}+t}{\sin \left( {{t}^{4}}-{{t}^{2}}+{{y}^{2}} \right)dy}+\int_{0}^{{{t}^{2}}}{\left( \sin \left[ {{\left( x+t \right)}^{2}}+{{x}^{2}}-{{t}^{2}} \right]-\sin \left[ {{\left( x-t \right)}^{2}}+{{x}^{2}}-{{t}^{2}} \right] \right)dx-}$

$\int_{0}^{{{t}^{2}}}{dx}\int_{x-t}^{x+t}{2t\cos \left( {{x}^{2}}+{{y}^{2}}-{{t}^{2}} \right)}dy$

八：解：由分析可知

$I=\iiint\limits_{V}{\left( 2z+\sqrt{{{x}^{2}}+{{y}^{2}}} \right)}dxdydz$

$=\int_{0}^{\frac{\pi }{4}}{d\theta }\int_{a}^{2a}{d\rho }\int_{0}^{2\pi }{\left( \rho \sin \theta +\rho \cos \theta  \right)}{{\rho }^{2}}\sin \theta d\varphi$

$=2\pi \frac{{{\left( 2a \right)}^{4}}-{{a}^{4}}}{4}\left( \frac{\pi }{8}-\frac{1}{4}+\frac{1}{4} \right)=\frac{15{{\pi }^{2}}{{a}^{4}}}{16}$