武汉大学2009年数学分析试题解答

一．1.解$\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+\cdots +n} \right) $

$=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=2}^{n}{\frac{2}{k\left( k+1 \right)}} $$=2\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1}{2}-\frac{1}{n+1} \right)=1$；

2.解 $\underset{x\to 0}{\mathop{\lim }}\,\frac{\int_{0}^{x}{\left( x-t \right)\sin {{t}^{2}}dt}}{x\int_{0}^{x}{\sin {{t}^{2}}dt}}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\int_{0}^{x}{\sin {{t}^{2}}dt}-\int_{0}^{x}{t\sin {{t}^{2}}dt}}{x\int_{0}^{x}{\sin {{t}^{2}}dt}}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{\int_{0}^{x}{\sin {{t}^{2}}dt}}{\int_{0}^{x}{\sin {{t}^{2}}dt}+x\sin {{x}^{2}}}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{\sin {{x}^{2}}+\sin {{x}^{2}}+2{{x}^{2}}\cos {{x}^{2}}}$

$=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{\sin {{x}^{2}}}{{{x}^{2}}}}{\frac{2\sin {{x}^{2}}}{{{x}^{2}}}+2\cos {{x}^{2}}}=\frac{1}{4}$；

3.解 因为$\sin t=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{t}^{2n+1}}}$，

$F\left( x \right)=\frac{1}{x}\int_{0}^{x}{\frac{\sin t}{t}dt}$$=\frac{1}{x}\int_{0}^{x}{\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{t}^{2n}}}dt}$

   $=\frac{1}{x}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}\frac{1}{\left( 2n+1 \right)}{{x}^{2n+1}}}$

   $=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\left( 2n+1 \right)}^{2}}\left( 2n \right)!}{{x}^{2n}}}$，

于是${{F}^{\left( 2n \right)}}\left( 0 \right)=\frac{{{\left( -1 \right)}^{n}}}{{{\left( 2n+1 \right)}^{2}}}$，${{F}^{\left( 2n-1 \right)}}\left( 0 \right)=0$，

所以${{F}^{\left( 4 \right)}}\left( 0 \right)=\frac{1}{25}$，${{F}^{\left( 9 \right)}}\left( 0 \right)=0$.

4. $\frac{\partial z}{\partial x}=-\frac{\left( yz+xyz \right){{e}^{x+y+z}}}{\left( xy+xyz \right){{e}^{x+y+z}}}=-\frac{z\left( 1+x \right)}{x\left( 1+z \right)}=-\frac{1+\frac{1}{x}}{1+\frac{1}{z}}$，$\frac{\partial z}{\partial y}=-\frac{1+\frac{1}{y}}{1+\frac{1}{z}}$；

$\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}=\frac{\frac{1}{{{x}^{2}}}\left( 1+\frac{1}{z} \right)-\frac{{{z}_{x}}}{{{z}^{2}}}\left( 1+\frac{1}{x} \right)}{{{\left( 1+\frac{1}{z} \right)}^{2}}}$$=\frac{z\left[ {{\left( 1+z \right)}^{2}}+{{\left( 1+x \right)}^{2}} \right]}{{{x}^{2}}{{\left( 1+z \right)}^{3}}}$；

$\frac{{{\partial }^{2}}z}{\partial x\partial y}=\frac{\frac{{{z}_{y}}}{{{z}^{2}}}\left( 1+\frac{1}{x} \right)}{{{\left( 1+\frac{1}{z} \right)}^{2}}}=\frac{z\left( 1+x \right)\left( 1+y \right)}{xy{{\left( 1+z \right)}^{3}}}$.

5.解 $\iint\limits_{D}{\ln \frac{{{x}^{3}}}{y}dxdy}$$=\int_{1}^{2}{dx\int_{1}^{x}{3\ln xdy}}-\int_{1}^{2}{dy\int_{y}^{2}{\ln ydx}}$

$=3\int_{1}^{2}{\left( x-1 \right)\ln xdx}-\int_{1}^{2}{\left( 2-y \right)\ln ydy}$

$=\int_{1}^{2}{\left( 4x-5 \right)\ln xdx}$

$=\int_{1}^{2}{{{\left( 2{{x}^{2}}-5x \right)}^{\prime }}\ln xdx}$

$=\left. \left[ \left( 2{{x}^{2}}-5x \right)\ln x-\left( {{x}^{2}}-5x \right) \right] \right|_{1}^{2}$

$=-2\ln 2+2$.

二．证明：对于每一个$x\in \left[ 0,1 \right] $，由于$\left\{ {{I}_{u}} \right\}$覆盖了$\left[ 0,1 \right] $，

存在${{I}_{{{u}_{x}}}}\in \left\{ {{I}_{u}} \right\}$，使得$x\in {{I}_{{{u}_{x}}}}$，

又由${{I}_{{{u}_{x}}}}$是开区间，存在${{\delta }_{x}}>0$，

使得$x\in U\left( x,{{\delta }_{x}} \right)\subset U\left( x,2{{\delta }_{x}} \right)\subset {{I}_{{{u}_{x}}}}$，

显然开集族$\left\{ U\left( x,{{\delta }_{x}} \right):x\in \left[ 0,1 \right] \right\}$亦覆盖了区间$\left[ 0,1 \right] $，

根据有限覆盖定理，

存在有限个${{x}_{1}},{{x}_{2}},\cdots ,{{x}_{N}}$，使得

$U\left( {{x}_{1}},{{\delta }_{{{x}_{1}}}} \right),U\left( {{x}_{2}},{{\delta }_{{{x}_{2}}}} \right),\cdots ,U\left( {{x}_{N}},{{\delta }_{{{x}_{N}}}} \right) $就能覆盖了$\left[ 0,1 \right] $，

取$\delta =\min \left\{ {{\delta }_{{{x}_{1}}}},{{\delta }_{{{x}_{2}}}},\cdots ,{{\delta }_{{{x}_{N}}}} \right\}$，

对任意${{y}_{1}},{{y}_{2}}\in \left[ 0,1 \right] $，

当$\left| {{y}_{1}}-{{y}_{2}} \right|{x}_{k}},{{\delta }_{{{x}_{k}}}} \right) $，有${{y}_{1}}\in U\left( {{x}_{k}},{{\delta }_{{{x}_{k}}}} \right) $，{y}_{2}}-{{x}_{k}} \right|\le \left| {{y}_{1}}-{{y}_{2}} \right|+\left| {{y}_{1}}-{{x}_{k}} \right|{\delta }_{{{x}_{k}}}}\le 2{{\delta }_{{{x}_{k}}}}$，{y}_{1}},{{y}_{2}}\in U\left( {{x}_{k}},2{{\delta }_{{{x}_{k}}}} \right)\subset {{I}_{{{u}_{{{x}_{k}}}}}}$，{a}^{-}} \right)=+\infty $矛盾，{x}^{2}}+{{y}^{2}}\le y,x\ge 0 \right\}$，{x}^{2}}+{{\left( y-\frac{1}{2} \right)}^{2}}\le {{\left( \frac{1}{2} \right)}^{2}},x\ge 0 \right\}$，{\left( \frac{1}{2} \right)}^{2}}=\frac{\pi }{8}$，{x}^{2}}-{{y}^{2}}}dxdy}$$=\int_{0}^{\frac{\pi }{2}}{d\theta \int_{0}^{\sin \theta }{\sqrt{1-{{r}^{2}}}rdr}}${\left( 1-{{r}^{2}} \right)}^{\frac{3}{2}}} \right] \right|_{0}^{\sin \theta }d\theta }$$=\frac{1}{3}\int_{0}^{\frac{\pi }{2}}{\left( 1-{{\cos }^{3}}\theta  \right)d\theta }${\sin }^{2}}\theta  \right)d\sin \theta }=\frac{\pi }{6}-\frac{2}{9}$,{x}^{2}}-{{y}^{2}}}dxdy}-\frac{8}{\pi }\iint\limits_{D}{f\left( x,y \right)dxdy}\cdot \iint\limits_{D}{dxdy}${x}^{2}}-{{y}^{2}}}-\frac{8}{\pi }\left( \frac{\pi }{12}-\frac{1}{9} \right) $.{{x}^{2}}+{{y}^{2}}}}\frac{\partial f}{\partial x}+\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\frac{\partial f}{\partial y} \right)dxdy}${f}_{x}}+\sin \theta {{f}_{y}} \right)rd\theta }}$$=\int_{0}^{1}{dr\int\limits_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{{{f}_{x}}dy-{{f}_{y}}dx}}${{x}^{2}}+{{y}^{2}}={{r}^{2}}}{\left( {{f}_{xx}}+{{f}_{yy}} \right)dxdy}}$$=\int_{0}^{1}{dr\iint\limits_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}dxdy}}${{\rho }^{4}}\rho d\rho }}}$$=2\pi \cdot \frac{1}{6}\int_{0}^{1}{{{r}^{6}}dr}$$=2\pi \cdot \frac{1}{6}\cdot \frac{1}{7}=\frac{\pi }{21}$.{a}_{n}}=+\infty $，对任给的$A0$，存在${{N}_{1}}\in {{N}^{*}}$，使得凡是$n>{{N}_{1}}$时有${{a}_{n}}>3A$；设$n>{{N}_{1}}$，这时

$\frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}{n}$$=\frac{({{a}_{1}}+\cdots +{{a}_{{{N}_{1}}}})+{{a}_{{{N}_{1}}+1}}+\cdots +{{a}_{n}}}{n}$

$>\frac{{{a}_{1}}+\cdots +{{a}_{{{N}_{1}}}}}{n}+3A\frac{n-{{N}_{1}}}{n}$，又因$\frac{{{a}_{1}}+\cdots +{{a}_{{{N}_{1}}}}}{n}\to 0$，$\frac{n-{{N}_{1}}}{n}=1-\frac{{{N}_{1}}}{n}\to 1$，（$n\to \infty $），

故可取正整数$N>{{N}_{1}}$，使当$n>N$时，恒有$|\frac{{{a}_{1}}+\cdots +{{a}_{{{N}_{1}}}}}{n}|{N}_{1}}}{n}=1-\frac{{{N}_{1}}}{n}\frac{1}{2}$

于是，当$n>N$时，恒有$\frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}{n}>A$，由此可知$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}{n}=+\infty $ 。

显然，$\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=-\infty $等价于$\underset{n\to \infty }{\mathop{\lim }}\,(-{{a}_{n}})=+\infty $ 。

定理  如果$\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=-\infty $，则$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}{n}=-\infty $。

七、解 （1）$\sum\limits_{n=1}^{\infty }{{{u}_{n}}\left( x \right)} $，$\left[ 0,b \right] $，

所以${{\beta }_{n}}=\underset{x\in \left[ 0,+\infty  \right)}{\mathop{\sup }}\,\left| {{u}_{n}}\left( x \right) \right|$，$\ge {{u}_{n}}\left( {{e}^{{{n}^{3}}}} \right)=\frac{1}{{{n}^{3}}}\ln \left( 1+{{n}^{3}}{{e}^{{{n}^{3}}}} \right) $，

（2）当$\ge \frac{1}{{{n}^{3}}}\ln {{e}^{{{n}^{3}}}}=1$时，$\left\{ {{u}_{n}}\left( x \right) \right\}$，

$\left[ 0,+\infty  \right) $，

当$0$不存在，$\sum\limits_{n=1}^{\infty }{{{u}_{n}}\left( x \right)} $在$\left[ 0,+\infty  \right) $处不连续，

当$S\left( x \right) $不存在，$\left[ 0,+\infty  \right) $在${{{u}'}_{n}}\left( x \right)=\frac{{{n}^{3}}}{{{n}^{3}}\left( 1+{{n}^{3}}x \right)} $处不连续，

（3）当$\sum\limits_{n=1}^{\infty }{{{{{u}'}}_{n}}\left( 0 \right)} $时，

$\frac{f\left( \Delta x,\Delta y \right)-\left[ f(0,0)+{{f}_{x}}\left( 0,0 \right)\Delta x+{{f}_{y}}\left( 0,0 \right)\Delta y \right]}{\sqrt{{{\left( \Delta x \right)}^{2}}+{{\left( \Delta y \right)}^{2}}}}$

$0{{u}'}_{n}}\left( x \right)\le \frac{1}{{{n}^{3}}x}=\frac{1}{x}\frac{1}{{{n}^{3}}}$，{{{{u}'}}_{n}}\left( x \right)} $在点$\left( 0,+\infty  \right) $处可微，且$df\left( 0,0 \right)=0$ 。{\left( \frac{\partial }{\partial x}+\frac{\partial }{\partial y} \right)}^{2}}z=\left( \frac{\partial }{\partial x}+\frac{\partial }{\partial y} \right)\left( \left( x+y \right)\frac{\partial z}{\partial u} \right) ${\left( x+y \right)}^{2}}\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}$，{{v}^{2}}+4u}}{2}$，$y=\frac{-v\pm \sqrt{{{v}^{2}}+4u}}{2}$，{{v}^{2}}+4u}$，$\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+\frac{1}{{{v}^{2}}+4u}\frac{\partial z}{\partial u}=0$.0$，当$x\in \left[ 0,b \right] $时，

$0\le {{u}_{n}}\left( x \right)=\frac{1}{{{n}^{3}}}\ln \left( 1+{{n}^{3}}x \right) $

       $\le \frac{1}{{{n}^{3}}}\ln \left( 1+{{n}^{3}}b \right) $

        $\le \frac{1}{{{n}^{2}}}\frac{1}{n}\ln \left( 2{{n}^{3}}b \right)=\frac{1}{{{n}^{2}}}\frac{\ln \left( 2b \right)+3\ln n}{n}$      $\left( {{n}^{3}}\ge \frac{1}{6} \right) $

        $\le \frac{1}{{{n}^{2}}}$ ， （$n$充分大）

而$\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{2}}}}$收敛，所以$\sum\limits_{n=1}^{\infty }{{{u}_{n}}\left( x \right)} $在$\left[ 0,b \right] $上是一致收敛的.

因为${{\beta }_{n}}=\underset{x\in \left[ 0,+\infty  \right)}{\mathop{\sup }}\,\left| {{u}_{n}}\left( x \right) \right|\ge {{u}_{n}}\left( {{e}^{{{n}^{3}}}} \right)=\frac{1}{{{n}^{3}}}\ln \left( 1+{{n}^{3}}{{e}^{{{n}^{3}}}} \right) $

                  $\ge \frac{1}{{{n}^{3}}}\ln {{e}^{{{n}^{3}}}}=1$，

所以$\left\{ {{u}_{n}}\left( x \right) \right\}$在$\left[ 0,+\infty  \right) $上不一致收敛于$0$，

从而$\sum\limits_{n=1}^{\infty }{{{u}_{n}}\left( x \right)} $在$\left[ 0,+\infty  \right) $上不一致收敛.

（2）显然$S\left( x \right) $在$\left[ 0,+\infty  \right) $上连续，${{{u}'}_{n}}\left( x \right)=\frac{{{n}^{3}}}{{{n}^{3}}\left( 1+{{n}^{3}}x \right)} $，

显然$\sum\limits_{n=1}^{\infty }{{{{{u}'}}_{n}}\left( 0 \right)} $发散，对于$x>0$，$0{{u}'}_{n}}\left( x \right)\le \frac{1}{{{n}^{3}}x}=\frac{1}{x}\frac{1}{{{n}^{3}}}$，{{{{u}'}}_{n}}\left( x \right)} $在$\left( 0,+\infty  \right) $上收敛，0$，当$x\in \left[ \delta ,+\infty  \right) $时，$0{{u}'}_{n}}\left( x \right)\le \frac{1}{{{n}^{3}}x}\le \frac{1}{\delta }\frac{1}{{{n}^{3}}}$，{{{{u}'}}_{n}}\left( x \right)} $在$\left[ \delta ,+\infty  \right) $上一致收敛，且有${S}'\left( x \right)=\sum\limits_{n=1}^{\infty }{{{{{u}'}}_{n}}\left( x \right)} $，0$的任意性，可知，$S\left( x \right) $在$\left( 0,+\infty  \right) $上可微