武汉大学2012年数学分析试题解答

武汉大学2012年数学分析试题解答

1：计算

（1）  解：由[1-frac{1}{sumlimits_{i=0}^{k}{(2i+1)}}=1-frac{1}{k(k+1)+k}=frac{k(k+2)}{{{(k+1)}^{2}}}=frac{k}{k+1}cdot frac{k+2}{k+1}]

从而$(1-frac{1}{1+3})(1-frac{1}{1+3+5})cdots (1-frac{1}{1+3+cdots +(2n+1)})=frac{1}{n+1}cdot frac{n+2}{2}$

于是$underset{n o +infty }{mathop{lim }}\,(1-frac{1}{1+3})(1-frac{1}{1+3+5})cdots (1-frac{1}{1+3+cdots +(2n+1)})=frac{1}{2}$

（2）  解：原极限$=underset{x o 0}{mathop{lim }}\,frac{intlimits_{0}^{x}{sin ({{t}^{2}})dt}}{4{{x}^{3}}}=underset{x o 0}{mathop{lim }}\,frac{sin ({{x}^{2}})}{12{{x}^{2}}}=frac{1}{12}$

（3）  解：由于$F(x)=frac{1}{x}intlimits_{0}^{x}{cos {{t}^{2}}}dt=frac{1}{x}intlimits_{0}^{x}{sumlimits_{n=0}^{infty }{frac{{{(-1)}^{n}}}{(2n)!}{{t}^{4n}}}}dt=sumlimits_{n=0}^{infty }{frac{{{(-1)}^{n}}}{(2n)!}cdot frac{{{x}^{4n}}}{4n+1}}$

从而${{F}^{(8)}}(0)=frac{1}{4!}cdot frac{1}{9}cdot 8!=frac{8!}{216}=frac{560}{3},{{F}^{(10)}}(0)=0$  

（4）  解：[{{z}_{x}}=yf_{1}^{'}+f_{2}^{'}]

[{{z}_{xx}}=y(yf_{11}^{''}+f_{12}^{''})+(yf_{21}^{''}+f_{22}^{''})={{y}^{2}}f_{11}^{''}+2yf_{12}^{''}+f_{22}^{''}]

[{{z}_{xy}}=f_{1}^{'}+xyf_{11}^{''}+(x+y)f_{12}^{''}+f_{22}^{''}]

（5）  解：原式[=iint_{D}{ln ydxdy-2iint_{D}{operatorname{lnx}dxdy}}]

[=intlimits_{0}^{1}{ln ydyintlimits_{y}^{2}{dx}}-2intlimits_{0}^{1}{ln xdyintlimits_{1}^{x}{dx}}]

[=intlimits_{1}^{2}{(2-y)ln ydy-2intlimits_{1}^{2}{(x-1)ln xdx}}]

[=intlimits_{1}^{2}{(4-3x)ln xdx}]

[=2ln 2-frac{7}{4}]

2：证明：设${{S}_{n}}=sumlimits_{i=1}^{n}{a_{i}^{n}}$

由于$underset{n o infty }{mathop{lim }}\,{{a}_{n}}>0$,则存在${{N}_{1}}>0$,当$n>{{N}_{1}}$时,使得${{a}_{n}}>0$

而${{a}_{n+1}}ge {{a}_{n}}$,则存在${{N}_{2}}>0$,当$n>{{N}_{2}}$时,使得${{S}_{n}}ge 0$

且${{N}_{2}}ge {{N}_{1}}$，于是

当$n>{{N}_{2}}$时,必有$a_{n}^{n}le sumlimits_{i=1}^{n}{a_{i}^{n}}le na_{n}^{n}$

即${{a}_{n}}le sqrt[n]{{{S}_{n}}}le sqrt[n]{n}{{a}_{n}}$

两边取极限，由迫敛性知：$underset{n o +infty }{mathop{lim }}\,sqrt[n]{sumlimits_{k=1}^{n}{a_{k}^{n}}}=a$

3：证明：反证法。若$underset{xin (a,b)}{mathop{sup }}\,left| f'(x) ight|=M<infty $

则  $underset{xin (a,b)}{mathop{sup }}\,left| f'(x) ight|=underset{xin (a,b)}{mathop{sup }}\,left| f(frac{a+b}{2})+f'(xi )(x-frac{a+b}{2}) ight|$

     $le left| f(frac{a+b}{2}) ight|+Mcdot frac{b-a}{2}<infty $

矛盾，即假设不成立，原命题成立！

4：证明：由

$intlimits_{a}^{b}{f(x)intlimits_{a}^{b}{left| g(x)-g(t) ight|}}dtdx=intlimits_{a}^{b}{f(x)intlimits_{a}^{x}{left| g(x)-g(t) ight|}}dtdx+intlimits_{a}^{b}{f(x)intlimits_{x}^{b}{left| g(x)-g(t) ight|}}dtdx$

[=intlimits_{a}^{b}{intlimits_{t}^{b}{f(x)left| g(x)-g(t) ight|}}dtdx+intlimits_{a}^{b}{intlimits_{t}^{b}{f(t)left| g(x)-g(t) ight|}}dtdx]

[=intlimits_{a}^{b}{intlimits_{t}^{b}{left| f(x)-f(t) ight|left| g(x)-g(t) ight|}}dtdxge 0]

即知：$(b-a)intlimits_{a}^{b}{f(x)g(x)dxge }intlimits_{a}^{b}{f(x)dx}intlimits_{a}^{b}{g(x)dx}$

5：解：（1）${{f}_{x}}(0,0)=underset{x o 0}{mathop{lim }}\,frac{f(x,0)-f(0,0)}{x}=0,{{f}_{x}}(0,0)=underset{x o 0}{mathop{lim }}\,frac{f(0,y)-f(0,0)}{y}=0$

即存在，且都为0

（2）当${{x}^{2}}+{{y}^{2}} e 0$时，[{{f}_{x}}(x,y)=frac{2x{{y}^{3}}}{{{({{x}^{2}}+{{y}^{2}})}^{2}}},{{f}_{y}}(x,y)=frac{{{x}^{2}}({{x}^{2}}-{{y}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}]

而$underset{y=kx o 0}{mathop{lim }}\,{{f}_{x}}(x,y)=frac{2{{k}^{3}}}{{{(1+{{k}^{2}})}^{2}}},underset{y=kx o 0}{mathop{lim }}\,{{f}_{y}}(x,y)=frac{1-{{k}^{2}}}{{{(1+{{k}^{2}})}^{2}}}$与$k$有关

从而 ${{f}_{x}}(x,y),{{f}_{y}}(x,y)$在$(0,0)$不连续。

（1）  由 $underset{y=kx o 0}{mathop{lim }}\,frac{f(x,y)-f(0,0)-{{f}_{x}}(0,0)x-{{f}_{y}}(0,0)y}{sqrt{{{x}^{2}}+{{y}^{2}}}}=frac{k}{{{(1+{{k}^{2}})}^{frac{3}{2}}}}$

即知$f(x,y)$在$(0,0)$上不可微

6：证明：设三个单参数曲面族分别为

   $F(x,y,z)=xz-uy=0$

   $G(x,y,z)=sqrt{{{x}^{2}}+{{y}^{2}}}+sqrt{{{y}^{2}}+{{z}^{2}}}-v=0$

   $H(x,y,z)=sqrt{{{x}^{2}}+{{y}^{2}}}-sqrt{{{y}^{2}}+{{z}^{2}}}-w=0$

则在它们的公共点$(x,y,z)$ 处的法向量分别为

  ${{n}_{F}}(x,y,z,u)=(z,-u,x)=(z,-frac{xz}{y},x)$

  ${{n}_{G}}(x,y,z,v)=(frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}}},frac{y}{sqrt{{{x}^{2}}+{{y}^{2}}}}+frac{y}{sqrt{{{y}^{2}}+{{z}^{2}}}},frac{z}{sqrt{{{y}^{2}}+{{z}^{2}}}})$

  ${{n}_{H}}(x,y,z,v)=(frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}}},frac{y}{sqrt{{{x}^{2}}+{{y}^{2}}}}-frac{y}{sqrt{{{y}^{2}}+{{z}^{2}}}},frac{-z}{sqrt{{{y}^{2}}+{{z}^{2}}}})$

易验证$leftlangle {{n}_{F}},{{n}_{G}} ight angle =leftlangle {{n}_{G}},{{n}_{H}} ight angle =leftlangle {{n}_{H}},{{n}_{F}} ight angle =0$，从而结论成立。

7：证明：由

$frac{partial (u,v)}{partial (x,y)}$
$
=left|egin{array}{cccc}
{{u_x}} & {{u_y}} \
{{v_x}} & {{v_y}}
end{array} ight|
$
$
=left|egin{array}{cccc}
{f'(x)} & 0 \
{f(x) + xf'(x)} & -1
end{array} ight|
$
$=-f'(x)$

而$f'({{x}_{0}}) e 0$且在$({{x}_{0}},{{y}_{0}})$附近是局部可逆的，从而

$u=f(x)Rightarrow x=g(u)$

$v=xf(x)-v=ug(u)-v$

8：解：（1）当$x=0$时，$S(x)=0;$当$x>0$ 时，由

$underset{y o infty }{mathop{lim }}\,frac{u(x,y)}{{}^{1}/{}_{{{y}^{2}}}}=underset{y o infty }{mathop{lim }}\,frac{ln (1+{{y}^{3}}x)}{y}=underset{y o infty }{mathop{lim }}\,frac{3{{y}^{2}}x}{1+{{y}^{3}}x}=0$

知$S(x)$有定义；当$x<0$时，$1+{{y}^{3}}x>0Rightarrow y<frac{-1}{sqrt[3]{x}}$

知$S(x)$不存在，综上所述，$S(x)$的定义域为$[0,+infty )$

（2）  $frac{ln (1+{{y}^{3}}x)}{{{y}^{3}}}le frac{ln (1+{{y}^{3}}b)}{{{y}^{3}}}(0le xle b)$，即知$S(x)$在有界区间$[0,b]$上是一致收敛

（2）   又由[intlimits_{n}^{2n}{frac{ln (1+{{y}^{3}}{{e}^{{{n}^{2}}}})}{{{y}^{3}}}}dyge intlimits_{n}^{2n}{frac{{{operatorname{lne}}^{{{n}^{2}}}}}{{{(2n)}^{3}}}}dy=frac{1}{4}] ，即知在$(0,+infty )$上不是一致收敛的

（3）  由${{u}_{x}}(x,y)=frac{1}{1+{{y}^{3}}x}le frac{1}{{{y}^{3}}a}(0<ale x<infty )$

即知$S(x)$在$(0,+infty )$上可微，又由

[underset{x o {{0}^{+}}}{mathop{lim }}\,frac{S(x)-S(0)}{x}=underset{x o {{0}^{+}}}{mathop{lim }}\,intlimits_{1}^{+infty }{frac{ln (1+{{y}^{3}}x)}{{{y}^{3}}x}}dy]

[=underset{x o {{0}^{+}}}{mathop{lim }}\,intlimits_{x}^{+infty }{frac{ln (1+u)}{u}cdot frac{1}{3{{x}^{frac{1}{3}}}{{u}^{frac{2}{3}}}}}du=underset{x o {{0}^{+}}}{mathop{lim }}\,frac{1}{3{{x}^{frac{1}{3}}}}intlimits_{0}^{+infty }{frac{ln (1+u)}{u}}du=infty ]

知$S(x)$在$x=0$处不可导