武汉大学2009年数学分析试题解答

武汉大学2009年数学分析试题解答

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

$=underset{n o infty }{mathop{lim }}\,sumlimits_{k=2}^{n}{frac{2}{kleft( k+1 ight)}} $$=2underset{n o infty }{mathop{lim }}\,left( frac{1}{2}-frac{1}{n+1} ight)=1$；

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

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

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

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

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

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

$Fleft( x ight)=frac{1}{x}int_{0}^{x}{frac{sin t}{t}dt}$$=frac{1}{x}int_{0}^{x}{sumlimits_{n=0}^{infty }{frac{{{left( -1 ight)}^{n}}}{left( 2n+1 ight)!}{{t}^{2n}}}dt}$

   $=frac{1}{x}sumlimits_{n=0}^{infty }{frac{{{left( -1 ight)}^{n}}}{left( 2n+1 ight)!}frac{1}{left( 2n+1 ight)}{{x}^{2n+1}}}$

   $=sumlimits_{n=0}^{infty }{frac{{{left( -1 ight)}^{n}}}{{{left( 2n+1 ight)}^{2}}left( 2n ight)!}{{x}^{2n}}}$，

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

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

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

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

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

5.解 $iintlimits_{D}{ln frac{{{x}^{3}}}{y}dxdy}$$=int_{1}^{2}{dxint_{1}^{x}{3ln xdy}}-int_{1}^{2}{dyint_{y}^{2}{ln ydx}}$

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

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

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

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

$=-2ln 2+2$.

二．证明：对于每一个$xin left[ 0,1 ight] $，由于$left{ {{I}_{u}} ight}$覆盖了$left[ 0,1 ight] $，

存在${{I}_{{{u}_{x}}}}in left{ {{I}_{u}} ight}$，使得$xin {{I}_{{{u}_{x}}}}$，

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

使得$xin Uleft( x,{{delta }_{x}} ight)subset Uleft( x,2{{delta }_{x}} ight)subset {{I}_{{{u}_{x}}}}$，

显然开集族$left{ Uleft( x,{{delta }_{x}} ight):xin left[ 0,1 ight] ight}$亦覆盖了区间$left[ 0,1 ight] $，

根据有限覆盖定理，

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

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

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

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

当$left| {{y}_{1}}-{{y}_{2}} ight|<delta $时，

存在$Uleft( {{x}_{k}},{{delta }_{{{x}_{k}}}} ight) $，有${{y}_{1}}in Uleft( {{x}_{k}},{{delta }_{{{x}_{k}}}} ight) $，

$left| {{y}_{2}}-{{x}_{k}} ight|le left| {{y}_{1}}-{{y}_{2}} ight|+left| {{y}_{1}}-{{x}_{k}} ight|<delta +{{delta }_{{{x}_{k}}}}le 2{{delta }_{{{x}_{k}}}}$，

于是${{y}_{1}},{{y}_{2}}in Uleft( {{x}_{k}},2{{delta }_{{{x}_{k}}}} ight)subset {{I}_{{{u}_{{{x}_{k}}}}}}$，

结论得证.

（2）例如$left{ left( frac{1}{n+1},frac{1}{n} ight) ight}$是$left( 0,1 ight) $的一个开覆盖，就没有这个性质.

三.证明 用反证法，假设结果不真，

存在$M>0$，使得${f}'left( x ight)le M$，$xin left( a-delta ,a ight) $.

对任意$a-delta <x<y<a$，

由Lagrange中值定理，

$fleft( y ight)=fleft( x ight)+{f}'left( xi  ight)left( y-x ight) $

    $le fleft( x ight)+Mleft( y-x ight)le fleft( x ight)+Ma$，

从而$fleft( x ight) $在$a$的某个左邻域内有上界，这与$fleft( {{a}^{-}} ight)=+infty $矛盾，

所以命题成立.

四.解 $D=left{ left( x,y ight):{{x}^{2}}+{{y}^{2}}le y,xge 0 ight}$，

$D=left{ left( x,y ight):{{x}^{2}}+{{left( y-frac{1}{2} ight)}^{2}}le {{left( frac{1}{2} ight)}^{2}},xge 0 ight}$，

$sigma left( D ight)=frac{1}{2}pi {{left( frac{1}{2} ight)}^{2}}=frac{pi }{8}$，

$z=iintlimits_{D}{sqrt{1-{{x}^{2}}-{{y}^{2}}}dxdy}$$=int_{0}^{frac{pi }{2}}{d heta int_{0}^{sin heta }{sqrt{1-{{r}^{2}}}rdr}}$

 $=int_{0}^{frac{pi }{2}}{left. left[ -frac{1}{3}{{left( 1-{{r}^{2}} ight)}^{frac{3}{2}}} ight] ight|_{0}^{sin heta }d heta }$$=frac{1}{3}int_{0}^{frac{pi }{2}}{left( 1-{{cos }^{3}} heta  ight)d heta }$

 $=frac{pi }{6}-frac{1}{3}int_{0}^{frac{pi }{2}}{left( 1-{{sin }^{2}} heta  ight)dsin heta }=frac{pi }{6}-frac{2}{9}$,

由$iintlimits_{D}{fleft( x,y ight)dxdy}=iintlimits_{D}{sqrt{1-{{x}^{2}}-{{y}^{2}}}dxdy}-frac{8}{pi }iintlimits_{D}{fleft( x,y ight)dxdy}cdot iintlimits_{D}{dxdy}$

             $=frac{pi }{6}-frac{2}{9}-iintlimits_{D}{fleft( x,y ight)dxdy}$，

所以$iintlimits_{D}{fleft( x,y ight)dxdy}=frac{pi }{12}-frac{1}{9}$，

所以$fleft( x,y ight)=sqrt{1-{{x}^{2}}-{{y}^{2}}}-frac{8}{pi }left( frac{pi }{12}-frac{1}{9} ight) $.

五．解 利用极坐标和$Gauss$公式，$I=iintlimits_{D}{left( frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}}}frac{partial f}{partial x}+frac{y}{sqrt{{{x}^{2}}+{{y}^{2}}}}frac{partial f}{partial y} ight)dxdy}$

$=int_{0}^{1}{drint_{0}^{2pi }{left( cos heta {{f}_{x}}+sin heta {{f}_{y}} ight)rd heta }}$$=int_{0}^{1}{drintlimits_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{{{f}_{x}}dy-{{f}_{y}}dx}}$

$=int_{0}^{1}{driintlimits_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{left( {{f}_{xx}}+{{f}_{yy}} ight)dxdy}}$$=int_{0}^{1}{driintlimits_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{{{left( {{x}^{2}}+{{y}^{2}} ight)}^{2}}dxdy}}$

$=int_{0}^{1}{drint_{0}^{2pi }{d heta int_{0}^{r}{{{ ho }^{4}} ho d ho }}}$$=2pi cdot frac{1}{6}int_{0}^{1}{{{r}^{6}}dr}$$=2pi cdot frac{1}{6}cdot frac{1}{7}=frac{pi }{21}$.

六．  证明  因为$underset{n o infty }{mathop{lim }}\,{{a}_{n}}=+infty $，对任给的$A>0$，存在${{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}$

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

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

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

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

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

七、解 （1）$sumlimits_{n=1}^{infty }{{{u}_{n}}left( x ight)} $，$left[ 0,b ight] $，

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

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

$left[ 0,+infty  ight) $，

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

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

（3）当$sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( 0 ight)} $时，

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

$0<{{{u}'}_{n}}left( x ight)le frac{1}{{{n}^{3}}x}=frac{1}{x}frac{1}{{{n}^{3}}}$，

所以$sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( x ight)} $在点$left( 0,+infty  ight) $处可微，且$dfleft( 0,0 ight)=0$ 。

八．（1）解 $left( frac{partial }{partial x}+frac{partial }{partial y} ight)z=left( x+y ight)frac{partial z}{partial u}$，

${{left( frac{partial }{partial x}+frac{partial }{partial y} ight)}^{2}}z=left( frac{partial }{partial x}+frac{partial }{partial y} ight)left( left( x+y ight)frac{partial z}{partial u} ight) $

          $=2frac{partial z}{partial u}+{{left( x+y ight)}^{2}}frac{{{partial }^{2}}z}{partial {{u}^{2}}}$，

由$x=frac{vpm sqrt{{{v}^{2}}+4u}}{2}$，$y=frac{-vpm sqrt{{{v}^{2}}+4u}}{2}$，

可知$x+y=sqrt{{{v}^{2}}+4u}$，$frac{{{partial }^{2}}z}{partial {{u}^{2}}}+frac{1}{{{v}^{2}}+4u}frac{partial z}{partial u}=0$.

（2）由$frac{partial left( u,v ight)}{partial left( x,y ight)}=-left( x+y ight) $，变量在$x+y=0$上失效，因为在此雅克比行列式为$0$.

九.（1）证明 对任意$b>0$，当$xin left[ 0,b ight] $时，

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

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

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

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

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

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

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

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

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

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

显然$sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( 0 ight)} $发散，对于$x>0$，$0<{{{u}'}_{n}}left( x ight)le frac{1}{{{n}^{3}}x}=frac{1}{x}frac{1}{{{n}^{3}}}$，

可见$sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( x ight)} $在$left( 0,+infty  ight) $上收敛，

对任意$delta >0$，当$xin left[ delta ,+infty  ight) $时，$0<{{{u}'}_{n}}left( x ight)le frac{1}{{{n}^{3}}x}le frac{1}{delta }frac{1}{{{n}^{3}}}$，

由此知，$sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( x ight)} $在$left[ delta ,+infty  ight) $上一致收敛，且有${S}'left( x ight)=sumlimits_{n=1}^{infty }{{{{{u}'}}_{n}}left( x ight)} $，

所以$Sleft( x ight) $在$left[ delta ,+infty  ight) $上连续可微，

由$delta >0$的任意性，可知，$Sleft( x ight) $在$left( 0,+infty  ight) $上可微