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

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

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

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

$frac{{{d}^{2}}y}{d{{x}^{2}}}=frac{frac{-sin tleft( 3{{t}^{2}}+1 ight)-6tcos t}{{{left( 3{{t}^{2}}+1 ight)}^{2}}}}{3{{t}^{2}}+1}=-frac{sin tleft( 3{{t}^{2}}+1 ight)+6tcos t}{{{left( 3{{t}^{2}}+1 ight)}^{3}}}$

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

5.$underset{n o +infty }{mathop{lim }}\,sumlimits_{k=1}^{{{n}^{2}}}{frac{n}{{{n}^{2}}+{{k}^{2}}}}=underset{n o +infty }{mathop{lim }}\,frac{1}{n}sumlimits_{k=1}^{{{n}^{2}}}{frac{1}{1+{{left( frac{k}{n} ight)}^{2}}}}=underset{n o +infty }{mathop{lim }}\,int_{0}^{n}{frac{dx}{1+{{x}^{2}}}}=int_{0}^{+infty }{frac{dx}{1+{{x}^{2}}}}=frac{pi }{2}$

二：证明：

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

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

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

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

即

$left| fleft( {{x}_{1}} ight)-fleft( {{x}_{2}} ight) ight|le 2Mleft| sqrt{{{x}_{1}}}-sqrt{{{x}_{2}}} ight|le 2Msqrt{left| {{x}_{1}}-{{x}_{2}} ight|}$

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

$left| fleft( {{x}_{1}} ight)-fleft( {{x}_{2}} ight) ight|<varepsilon $

故$fleft( x ight) $在$left( 0,delta  ight] $上一致收敛

而$f(x)$在$[delta ,1]$上连续，则$fleft( x ight) $在$[delta ,1]$上一致收敛

于是$fleft( x ight) $在$(0,1]$上一致收敛

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

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

$forall varepsilon >0,exists delta >0,left( {{x}_{0}}-delta ,{{x}_{0}}+delta  ight)subseteq left[ a,b ight]$，且$forall xin left( {{x}_{0}}-delta ,{{x}_{0}}+delta  ight),fleft( x ight)ge M-varepsilon $ ，从而

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

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

四：证明：

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

2.$forall kin {{mathbb{N}}^{*}}$，$sumlimits_{n=0}^{infty }{{{e}^{-n}}}frac{{{d}^{k}}left( cos left( {{n}^{2}}x ight) ight)}{d{{x}^{k}}}=sumlimits_{n=0}^{infty }{{{e}^{-n}}}{{n}^{2k}}cos left( {{n}^{2}}x+frac{npi }{2} ight)$，

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

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

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

$Taylor$级数为$sumlimits_{k=0}^{infty }{frac{{{left( -1 ight)}^{k}}sumlimits_{n=0}^{infty }{{{e}^{-n}}{{n}^{4k}}}}{left( 2k ight)!}}{{x}^{2k}}$

4.由于

$sqrt[2k]{left| frac{{{left( -1 ight)}^{k}}sumlimits_{n=0}^{infty }{{{e}^{-n}}{{n}^{4k}}}}{left( 2k ight)!} ight|}ge sqrt[2k]{left| frac{{{e}^{-2k}}{{left( 2k ight)}^{4k}}}{left( 2k ight)!} ight|}ge frac{2k}{e} o +infty $从而可知$fleft( x ight)$在

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

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'' ight|<delta $时，有

[left| f(x')-f(x'') ight|<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$left| f({{x}_{i}})-f({{x}_{i-1}}) ight|<frac{varepsilon }{2}$

同时对任意的$xin [{{x}_{i-1}},{{x}_{i}}]$，有$left| f(x)-f({{x}_{i}}) ight|<frac{varepsilon }{2},left| f(x)-f({{x}_{i ext{-}1}}) ight|<frac{varepsilon }{2}$

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

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

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

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

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

$frac{partial w}{partial x}=yfrac{partial F}{partial u},frac{partial w}{partial y}=xfrac{partial F}{partial u}+zfrac{partial F}{partial v},frac{partial w}{partial z}=yfrac{partial F}{partial v}$

显然有

$xfrac{partial w}{partial x}+zfrac{partial w}{partial z}=yfrac{partial w}{partial y}$

2.如果$Fleft( {{u}_{1}},cdots ,{{u}_{m}} ight)$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

$w=Fleft( {{x}_{1}}y,cdots ,{{x}_{m}}y ight)$

七：证明：

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

$int_{0}^{{{t}^{2}}}{dx}int_{x-t}^{x+t}{2tcos left( {{x}^{2}}+{{y}^{2}}-{{t}^{2}} ight)}dy$

八：解：由分析可知

$I=iiintlimits_{V}{left( 2z+sqrt{{{x}^{2}}+{{y}^{2}}} ight)}dxdydz$

$=int_{0}^{frac{pi }{4}}{d heta }int_{a}^{2a}{d ho }int_{0}^{2pi }{left( ho sin heta + ho cos heta  ight)}{{ ho }^{2}}sin heta dvarphi $

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