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华中师范大学2013年数学分析试题解答
一、
(1)解:不妨设函数$f(x)$定义在区间$[a,+infty )$上,则极限$underset{xto +infty }{mathop{lim }},f(x)=A$的充要条件是:对任意的数列${{{x}_{n}}}subset [a,+infty )$,且$underset{nto +infty }{mathop{lim }},{{x}_{n}}=+infty $,极限$underset{nto +infty }{mathop{lim }},f({{x}_{n}})$都存在且都为$A$
(2)证明:对任意的$xin (0,+infty )$,由$f(alpha x)=f(x)$得
$f(x)=f(alpha x)=f({{alpha }^{2}}x)=cdots =f({{alpha }^{n}}x)=cdots $
记${{x}_{n}}={{alpha }^{n}}x$,由$underset{nto infty }{mathop{lim }},{{alpha }^{n}}x=+infty $及归结原则可知:
$underset{nto infty }{mathop{lim }},f({{x}_{n}})=underset{xto +infty }{mathop{lim }},f(x)=1,xin (0,+infty )$
又由于$f(x)$在$x=0$处连续,则
$f(0)=underset{xto {{0}^{+}}}{mathop{lim }},f(x)=1$
于是$f(x)=1,xin [0,+infty )$
二、
(1)证明:由于函数$f(u,v)$在$G$上一致连续,于是对任意的$varepsilon >0$,存在${{delta }_{1}}>0$,对任意的$({{u}_{1}},{{v}_{1}}),({{u}_{2}},{{v}_{2}})in M$,当$left| {{u}_{1}}-{{v}_{1}} right|<{{delta }_{1}},left| {{u}_{2}}-{{v}_{2}} right|<{{delta }_{1}}$时,就有
$left| f({{u}_{1}},{{v}_{1}})-f({{u}_{2}},{{v}_{2}}) right|<varepsilon $
又由于函数$u=g(x),v=h(x)$在$I$上一致连续,于是存在$delta >0$,使得对任意的${{x}_{1}},{{x}_{2}}in I$,当$left| {{x}_{1}}-{{x}_{2}} right|<delta $时,且$g({{x}_{1}}),g({{x}_{2}}),h({{x}_{1}}),h({{x}_{2}})in G,$,有
$left| g({{x}_{1}})-g({{x}_{2}}) right|<{{delta }_{1}},left| h({{x}_{1}})-h({{x}_{2}}) right|<{{delta }_{1}}$
于是$left| f(g({{x}_{1}}),h({{x}_{1}})-f(g({{x}_{2}}),h({{x}_{2}}) right|<varepsilon $
即$f(g(x),h(x))$在$I$上一致收敛
(2)不妨设$f(x)=cos x,g(x)=sin x,h(x)=sqrt{x}$
易证$f(x),g(x),h(x)$在$[0,+infty )$上一致收敛
于是由(1)可知,$cos (sin x+sqrt{x})$在$[0,+infty )$上一致收敛
三、证明:先证明以下结论:必存在${{xi }_{1}},eta in (a,b)$,使得$f'({{xi }_{1}})<frac{f(b)-f(a)}{b-a}<f'(eta )$
反证法:假设对任意的$xin (a,b)$都有$f'(x)ge frac{f(b)-f(a)}{b-a}$
令$g(x)=f(x)-frac{f(b)-f(a)}{b-a}(x-a)$
则当$xin (a,b)$时有$g'(x)=f'(x)-frac{f(b)-f(a)}{b-a}ge 0$
于是$g(x)$在$[a,b]$上单调递增
而$g(a)=g(b)=f(a)$
从而当$xin [a,b]$时,$g(x)=f(a)$
于是$f(x)=frac{f(b)-f(a)}{b-a}(x-a)+f(a)$矛盾
从而必存在${{xi }_{1}}in (a,b)$,使得$f'({{xi }_{1}})<frac{f(b)-f(a)}{b-a}$
同理可证必存在$eta in (a,b)$,使得$frac{f(b)-f(a)}{b-a}<f'(eta )$
于是必存在${{xi }_{1}},eta in (a,b)$,使得$f'({{xi }_{1}})<frac{f(b)-f(a)}{b-a}<f'(eta )$
于是令$left| f'(xi ) right|=max {left| f'({{xi }_{1}}) right|,left| f'(eta ) right|}$
于是必存在$xi in (a,b)$,使得$left| f'(xi ) right|>left| frac{f(b)-f(a)}{b-a} right|$
四、
(1)证明:令$F(x,y)=2f(xy)-f(x)-f(y)$,则
1:$F(x,y)$在$U(1)$内连续
2:$F(1,1)=2f(1)-f(1)-f(1)=0$满足
3:$F(x,y)$在$U(1)$内有连续的导数,且${{F}_{y}}=2xf'(xy)-f'(y),xin U(1)$
4:${{F}_{y}}(1,1)=2f'(1)-f'(1)=f'(1)ne 0$
由上知:$2f(xy)=f(x)+f(y)$在$U(1)$内能确定唯一连续可微的隐函数$y=g(x)$
(2)由于${{F}_{x}}=2yf'(xy)-f'(x)$,则${{F}_{x}}(1,1)=f'(1)={{F}_{y}}(1,1)$
于是$g'(1)=-frac{{{F}_{x}}(1,1)}{{{F}_{y}}(1,1)}=-1$
五、
(1)当$(x,y)=(0,0)$时,由偏导数的定义:
${{f}_{x}}(0,0)=underset{Delta xto 0}{mathop{lim }},frac{f(Delta x,0)-f(0,0)}{Delta x}=underset{Delta xto 0}{mathop{lim }},Delta xcos frac{1}{left| Delta x right|}=0 $
${{f}_{y}}(0,0)=underset{Delta yto 0}{mathop{lim }},frac{f(0,Delta y)-f(0,0)}{Delta y}=underset{Delta yto 0}{mathop{lim }},Delta ycos frac{1}{left| Delta y right|}=0 $
当$(x,y)ne (0,0)$时,
${{f}_{x}}(x,y)=ycos frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}}+frac{2{{x}^{2}}y}{{{({{x}^{2}}+{{y}^{2}})}^{frac{3}{2}}}}sin frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}} $
${{f}_{y}}(x,y)=xcos frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}}+frac{2x{{y}^{2}}}{{{({{x}^{2}}+{{y}^{2}})}^{frac{3}{2}}}}sin frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}} $
于是${f_x}(x,y)=left{begin{array}{ll}
ycos frac{1}{{sqrt {{x^2} + {y^2}} }} + frac{{2{x^2}y}}{{{{({x^2} + {y^2})}^{frac{3}{2}}}}}sin frac{1}{{sqrt {{x^2} + {y^2}} }},& hbox{${x^2} + {y^2} ne 0$} \
0,& hbox{${x^2} + {y^2} = 0$.}
end{array}
right.$
${f_y}(x,y)=left{begin{array}{ll}
xcos frac{1}{{sqrt {{x^2} + {y^2}} }} + frac{{2x{y^2}}}{{{{({x^2} + {y^2})}^{frac{3}{2}}}}}sin frac{1}{{sqrt {{x^2} + {y^2}} }},& hbox{${x^2} + {y^2} ne 0$} \
0,& hbox{${x^2} + {y^2} = 0$.}
end{array}
right.$
(2)设$y=kx$,于是$underset{xto 0}{mathop{lim }},frac{2{{x}^{2}}y}{{{({{x}^{2}}+{{y}^{2}})}^{frac{3}{2}}}}=frac{2k}{{{(1+{{k}^{2}})}^{frac{3}{2}}}}$与$k$有关
于是$underset{xto 0}{mathop{lim }},{{f}_{x}}(x,0) $不存在,故${{f}_{x}}(x,y) $ 在(0,0)不连续。
同理${{f}_{y}}(x,y) $在(0,0)也不连续。
(3)设$u=f(x,y) $,则在(0,0)点有
$Delta u-du=[f(Delta x,Delta y)-f(0,0)]-[{{f}_{x}}(0,0)Delta x+{{f}_{y}}(0,0)Delta y]=(Delta {{x}^{2}}+Delta {{y}^{2}})cos frac{1}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}}$
因$underset{Delta xto 0Delta yto 0}{mathop{lim }},frac{Delta u-du}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}}=underset{Delta xto 0Delta yto 0}{mathop{lim }},sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}cos frac{1}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}}=0 $
故$f(x,y) $在(0,0)可微。
六、证法一:由于对$forall {{x}_{0}}in [a,b],underset{nto infty }{mathop{lim }},{{f}_{n}}({{x}_{0}})=f({{x}_{0}}),{{f}_{n}}({{x}_{0}})le {{f}_{n+1}}({{x}_{0}})$
于是对$forall varepsilon >0$,$exists {{n}_{k}}>0$,当$n>{{n}_{k}}$时有$0le left| f({{x}_{0}})-{{f}_{n}}({{x}_{0}}) right|le frac{varepsilon }{3}$
由递增性可知:$f({{x}_{0}})ge {{f}_{n}}({{x}_{0}})$
从而$0le f({{x}_{0}})-{{f}_{n}}({{x}_{0}})le frac{varepsilon }{3}$
又由于$f(x)$和${{f}_{n}}(x)$在$x={{x}_{0}}$处连续,由连续定义可知:
$forall varepsilon >0$,$exists {{delta }_{{{x}_{0}}}}>0$,当$left| x-{{x}_{0}} right|<{{delta }_{{{x}_{0}}}}$时有
$left| f(x)-f({{x}_{0}}) right|le frac{varepsilon }{3},left| {{f}_{{{n}_{k}}}}(x)-{{f}_{{{n}_{k}}}}({{x}_{0}}) right|le frac{varepsilon }{3} $同时成立
于是对$forall varepsilon >0$,$xin [a,b]$,当$n>{{n}_{k}}$且$left| x-{{x}_{0}} right|<{{delta }_{{{x}_{0}}}}$时,有
$left| f(x)-{{f}_{n}}(x) right|=f(x)-{{f}_{n}}(x)le f(x)-{{f}_{{{n}_{k}}}}(x)=left| f(x)-{{f}_{{{n}_{k}}}}(x) right|$
$le left| f(x)-f({{x}_{0}}) right|+left| f({{x}_{0}})-{{f}_{{{n}_{k}}}}({{x}_{0}}) right|+left| {{f}_{{{n}_{k}}}}({{x}_{0}})-{{f}_{{{n}_{k}}}}(x) right|<varepsilon $
于是,令${{Delta }_{{{x}_{0}}}}={x|left| x-{{x}_{0}} right|<{{delta }_{{{x}_{0}}}}}$可知,${{{f}_{n}}(x)}$在${{Delta }_{{{x}_{0}}}}$上一致收敛于$f(x)$
且有开区间族${{{Delta }_{{{x}_{0}}}}|{{x}_{0}}in [a,b]}$覆盖了$[a,b]$区间
而$f(x)$在$[a,b]$上连续,由海涅——博雷尔定理可知:
从开区间族${{{Delta }_{{{x}_{0}}}}|{{x}_{0}}in [a,b]}$可选出有限个${{Delta }_{{{x}_{1}}}},{{Delta }_{{{x}_{2}}}},cdots ,{{Delta }_{{{x}_{k}}}}$,使得$[a,b]subset bigcuplimits_{i=1}^{k}{{{Delta }_{{{x}_{i}}}}}$
当$xin {{Delta }_{{{x}_{i}}}}cap [a,b]$,且$n>{{n}_{{{k}_{i}}}}$时有$left| f(x)-{{f}_{n}}(x) right|<varepsilon $
令$N=max {{{n}_{{{k}_{i}}}}|1le ile k}$,当$n>N$且$xin [a,b]$时有$left| f(x)-{{f}_{n}}(x) right|<varepsilon $
即${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
同理可证当${{{f}_{n}}(x)}$是单调递减函数时,${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
综上所述:${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
证法二:反证法,同样不妨设${{{f}_{n}}(x)}$是单调递减函数,
设$exists {{varepsilon }_{0}}>0$,对$forall nin {{N}^{*}}$,$exists {{{x}_{n}}}in [a,b]$,使得$left| f({{x}_{n}})-{{f}_{n}}({{x}_{n}}) right|ge {{varepsilon }_{0}}$
由于${{{x}_{n}}}$有界,由维尔斯特拉斯聚点定理可知,必存在收敛子列${{{x}_{{{n}_{k}}}}}$收敛于$[a,b]$的某个值${{x}_{0}}$,且$underset{nto infty }{mathop{lim }},{{f}_{n}}({{x}_{0}})=f({{x}_{0}}),{{f}_{n}}({{x}_{0}})le {{f}_{n+1}}({{x}_{0}})$
于是对$exists {{n}_{{{k}_{p}}}}>0$,当$n>{{n}_{{{k}_{p}}}}$时有$0le f({{x}_{0}})-{{f}_{{{n}_{k}}}}({{x}_{0}})<frac{{{varepsilon }_{0}}}{3}$
不妨设${{n}_{{{k}_{{{p}_{1}}}}}}>{{n}_{{{k}_{p}}}}$
又由于$f(x)$和${{f}_{n}}(x)$在$x={{x}_{0}}$处连续,由连续定义可知:
$exists {{delta }_{0}}>0$,当$left| x-{{x}_{0}} right|<{{delta }_{0}}$,$xin [a,b]$时有
[left| f(x)-f({{x}_{0}}) right|<frac{{{varepsilon }_{0}}}{3},left| {{f}_{{{n}_{{{k}_{{{p}_{1}}}}}}}}(x)-{{f}_{{{n}_{{{k}_{{{p}_{1}}}}}}}}({{x}_{0}}) right|<frac{{{varepsilon }_{0}}}{3}]同时成立
显然,又因为${{{x}_{{{n}_{k}}}}}to {{x}_{0}}(nto +infty )$
于是$exists K>0$且$K>{{k}_{{{p}_{1}}}}$,当${{n}_{k}}>{{n}_{K}}$时$left| {{x}_{{{n}_{k}}}}-{{x}_{0}} right|<{{delta }_{0}}$
于是对当${{n}_{k}}>{{n}_{K}}$时
$left| f({{x}_{{{n}_{k}}}})-{{f}_{n}}({{x}_{{{n}_{k}}}}) right|<{{varepsilon }_{0}}$矛盾
即${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
同理可证当${{{f}_{n}}(x)}$是单调递减函数时,${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
综上所述:${{{f}_{n}}(x)}$在$[a,b]$上一致收敛于$f(x)$
七、
(1)证明:对$ [a,b] $上的任意划分$Delta $:$a={{x}_{0}}<{{x}_{1}}<cdots <{{x}_{m-1}}<{{x}_{m}}=b $,记${{m}_{i}},{{M}_{i}},{{omega }_{i}} $分别为函数$f(x) $在$ [{{x}_{i}},{{x}_{i+1}}] $上的最小值、最大值和振幅,且$left| {{m}_{i}} right|le M $,则
$int_{,a}^{,b}{f(x)sin nxdx}$$=sumlimits_{k=1}^{m}{int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{f(x)sin nxdx}} $
$=sumlimits_{k=1}^{m}{(int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{(f(x)-{{m}_{i}})sin nxdx}}+int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{{{m}_{i}}sin nxdx}) $,
由$f(x) $在$ [a,b] $上可积且绝对可积,则存在$delta >0 $,当$left| Delta right|<delta $时,有
$left| sumlimits_{k=1}^{m}{int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{(f(x)-{{m}_{i}})sin nxdx}} right|le sumlimits_{k=1}^{m}{int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{left| f(x)-{{m}_{i}} right|left| sin nx right|dx}} $
$le sumlimits_{k=1}^{m}{int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{{{omega }_{i}}Delta {{x}_{i}}}}<varepsilon $;
对于上述确定的划分
$left| sumlimits_{k=1}^{m}{int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{{{m}_{k}}sin nxdx}} right|=left| sumlimits_{k=1}^{m}{{{m}_{k}}int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{sin nxdx}} right|le sumlimits_{k=1}^{m}{left| {{m}_{k}} right|left| int_{,{{x}_{k-1}}}^{,{{x}_{k}}}{sin nxdx} right|}$
$=sumlimits_{k=1}^{m}{left| {{m}_{k}} right|}frac{1}{n}left| cos (nb)-cos (na) right|le frac{2}{n}sumlimits_{k=1}^{m}{left| {{m}_{k}} right|}$
存在${{N}_{1}}>0 $,当$n>{{N}_{1}} $时$frac{2}{n}sumlimits_{k=1}^{m}{left| {{m}_{k}} right|}<varepsilon $,故当$n>max {N,{{N}_{1}}} $时,有
$left| int_{,a}^{,b}{f(x)sin nxdx} right|<2varepsilon $,
即$underset{nto +infty }{mathop{lim }},$$int_{a}^{,b}{f(x)sin nxdx}=0 $;
(2)证明:由于$int_{a}^{+infty }{left| f(x) right|}dx$收敛,于是对任意的$varepsilon >0$,存在$G>a$,使得
$int_{G}^{+infty }{left| f(x)sin nx right|}dx<int_{G}^{+infty }{left| f(x) right|}dx<frac{varepsilon }{2}$
而$underset{nto +infty }{mathop{lim }}, $ $int_{a}^{,G}{f(x)sin nxdx}=0 $
于是对上述$varepsilon >0$,存在$N>0$,当$n>N$时$left| int_{a}^{G}{f(x)sin nxdx} right|<frac{varepsilon }{2}$
于是对任意的$varepsilon >0$,存在存在$N>0$,当$n>N$时
$left| int_{a}^{+infty }{f(x)sin nxdx} right|le left| int_{a}^{G}{f(x)sin nxdx} right|+left| int_{G}^{+infty }{f(x)sin nxdx} right|<varepsilon $
即$underset{nto +infty }{mathop{lim }},$$int_{a}^{,+infty }{f(x)sin nxdx}=0 $
八、
(1)证明:由于$
int_Lleft|begin{array}{cccc}
{frac{{partial f}}{{partial overrightarrow n }}} & {frac{{partial g}}{{partial overrightarrow n }}}\
f & g
end{array}right|ds
$$int_L {[g cdot } frac{{partial f}}{{partial overrightarrow n }} - f cdot frac{{partial g}}{{partial overrightarrow n }}]ds$
而$frac{partial f}{partial overrightarrow{n}}=frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)=frac{partial f}{partial x}cdot frac{dy}{ds}-frac{partial f}{partial y}cdot frac{dx}{ds} $
于是
$int_{L}{gcdot frac{partial f}{partial overrightarrow{n}}}ds=oint_{L}{[gcdot frac{partial f}{partial x}}dy-gcdot frac{partial f}{partial y}dx]overset{}{mathop{text{=}}},iintlimits_{D}{gcdot Delta fdxdy}+iintlimits_{D}{[frac{partial g}{partial x}cdot frac{partial f}{partial x}+frac{partial g}{partial y}cdot frac{partial f}{partial y}]dxdy} $
同理可得:
$int_{L}{fcdot frac{partial g}{partial overrightarrow{n}}}ds=iintlimits_{D}{fcdot Delta gdxdy}+iintlimits_{D}{[frac{partial g}{partial x}cdot frac{partial f}{partial x}+frac{partial g}{partial y}cdot frac{partial f}{partial y}]dxdy} $
于是$
int_Lleft|begin{array}{cccc}
{frac{{partial f}}{{partial overrightarrow n }}} & {frac{{partial g}}{{partial overrightarrow n }}}\
f & g
end{array}right|ds
$$
=iint_Dleft|begin{array}{cccc}
{Delta f} & {Delta g}\
f & g
end{array}right|dxdy
$
(2)证明:由于对任意的$(x,y)in {{R}^{2}}backslash {({{x}_{0}},{{y}_{0}})}$,$Delta g=0$
于是记以$({{x}_{0}},{{y}_{0}})$为中心,半径为$varepsilon >0$(充分小)的圆周为${{C}_{varepsilon }}$,所围成的区域为${{D}_{1}}$,由(1)可知:
$int_{L + {C_varepsilon }} {left[ {f cdot frac{{partial ln r}}{{partial mathop nlimits^ to }} - ln r cdot frac{{partial f}}{{partial mathop nlimits^ to }}} right]} ds = oint_{L + {C_varepsilon }} {[f cdot frac{{partial ln r}}{{partial mathop nlimits^ to }} - ln r cdot frac{{partial f}}{{partial mathop nlimits^ to }}]ds} $$
=iint_{{D_1}}left|begin{array}{cccc}
{Delta g} & {Delta f}\
g & f
end{array}right|dxdy
$
所以$oint_{L}{[fcdot frac{partial ln r}{partial overset{to }{mathop{n}},}-ln rcdot frac{partial f}{partial overset{to }{mathop{n}},}]ds}=oint_{{{C}_{varepsilon }}}{[fcdot frac{partial ln r}{partial overset{to }{mathop{n}},}-ln rcdot frac{partial f}{partial overset{to }{mathop{n}},}]ds}overset{Delta }{mathop{=}},{{I}_{1}}-{{I}_{2}}$
其中${{I}_{1}}=oint_{{{C}_{varepsilon }}}{ufrac{partial g}{partial overrightarrow{n}}}ds$
${{I}_{2}}=oint_{{{C}_{varepsilon }}}{gfrac{partial f}{partial overrightarrow{n}}}ds=ln varepsilon oint_{{{C}_{varepsilon }}}{[frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)]ds=ln varepsilon iint_{{{D}_{2}}}{Delta udxdy=0}}$
其中${{D}_{2}}$是由${{C}_{varepsilon }}$所包围的闭区域
注意到在${{C}_{varepsilon }}$上$frac{partial ln r}{partial overrightarrow{n}}{{|}_{r=varepsilon }}=frac{partial ln r}{partial r}{{|}_{r=varepsilon }}=frac{1}{r}{{|}_{r=varepsilon }}=frac{1}{varepsilon }$
于是${{I}_{1}}=oint_{{{C}_{varepsilon }}}{ffrac{partial ln r}{partial overrightarrow{n}}}ds=frac{1}{varepsilon }oint_{{{C}_{varepsilon }}}{fds}=frac{1}{varepsilon }f(Q)oint_{{{C}_{varepsilon }}}{ds=2pi }f(Q)$(积分第一中值定理)
其中点$Q$是${{D}_{2}}$内任意一点
于是令$varepsilon to 0$可知,$f(Q)to f({{x}_{0}},{{y}_{0}})$
即$f({{x}_{0}},{{y}_{0}})=frac{1}{2pi }int_{L}{left[ fcdot frac{partial ln r}{partial overset{to }{mathop{n}},}-ln rcdot frac{partial f}{partial overset{to }{mathop{n}},} right]}ds$
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