华中师范大学2007年数学分析试题解答

一、         计算题

1：解：由于

$underset{xto {{0}^{+}}}{mathop{lim }},frac{ln (1-x)}{-x}=underset{xto {{0}^{+}}}{mathop{lim }},frac{1}{1-x}=1,underset{xto {{0}^{+}}}{mathop{lim }},frac{{{e}^{{{x}^{2}}}}-1}{{{x}^{2}}}=underset{xto {{0}^{+}}}{mathop{lim }},frac{2x{{e}^{{{x}^{2}}}}}{2x}=1$

于是

$underset{xto {{0}^{+}}}{mathop{lim }},frac{{{ln }^{3}}(1-x)sin [sin (ln frac{1}{x})]}{{{e}^{{{x}^{2}}}}-1}=underset{xto {{0}^{+}}}{mathop{lim }},frac{{{(-x)}^{3}}}{{{x}^{2}}}sin [sin (ln frac{1}{x})]=0$

2：不妨设${{t}_{1}}={{x}^{x}},{{t}_{2}}={{x}^{ln x}}$，于是

$ln {{t}_{1}}=xln x,ln {{t}_{2}}={{(ln x)}^{2}}$

则$frac{t_{1}^{'}}{{{t}_{1}}}=ln x+1,frac{t_{2}^{'}}{{{t}_{2}}}=frac{2ln x}{x}$

即$t_{1}^{'}=(ln x+1){{x}^{x}},t_{2}^{'}=frac{2ln x}{2}{{x}^{ln x}}$

于是$y'=frac{2ln x}{x}{{x}^{ln x}}+(ln x+1){{x}^{x}}$

3：解：令$t={{x}^{4}}$，于是

$int_{0}^{+infty }{{{e}^{-{{x}^{4}}}}}dxcdot int_{0}^{+infty }{{{x}^{2}}{{e}^{-{{x}^{4}}}}dx=frac{1}{16}}int_{0}^{+infty }{{{t}^{-frac{3}{4}}}}{{e}^{-t}}cdot int_{0}^{+infty }{{{t}^{-frac{1}{4}}}}{{e}^{-t}}dt$

$=frac{1}{16}Gamma (frac{1}{4})Gamma (frac{3}{4})overset{}{mathop{=}},frac{1}{16}cdot frac{pi }{sin frac{pi }{4}}=frac{pi }{8sqrt{2}}$

此处介绍余元公式：

$forall pin (0,1)$，有$Gamma (p)Gamma (1-p)=frac{pi }{sin ppi }$

4：解：

$F'(x)={{f}_{x}}[f(x,x),f(x,x)]cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]+{{f}_{y}}[f(x,x),f(x,x)]cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]$

于是$F'(1)={{(a+b)}^{2}}$

5：解：不妨设$x=rcos theta ,y=rsin theta ,0le rle 1,0le theta le 2pi $

于是

$iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=int_{0}^{2pi }{dtheta int_{0}^{1}{{{r}^{4}}}({{sin }^{3}}theta +{{cos }^{3}}theta ){{e}^{{{r}^{2}}}}dr=}int_{0}^{2pi }{({{sin }^{3}}theta +{{cos }^{3}}theta )}dtheta cdot int_{0}^{1}{{{r}^{4}}{{e}^{r}}dr}$

而

$int_{0}^{2pi }{({{sin }^{3}}theta +{{cos }^{3}}theta )}dtheta =int_{0}^{2pi }{(sin theta +cos theta )(1+sin theta cos theta )dtheta }$

$=int_{0}^{2pi }{(sin theta +cos theta )dtheta +int_{0}^{2pi }{{{sin }^{2}}}}theta cos theta dtheta +int_{0}^{2pi }{sin theta {{cos }^{2}}}theta dtheta $

$=int_{0}^{2pi }{{{sin }^{2}}}theta d(sin theta )-int_{0}^{2pi }{{{cos }^{2}}}theta d(cos theta )$

$=frac{1}{3}{{sin }^{3}}theta |_{0}^{2pi }-frac{1}{3}{{cos }^{3}}theta |_{0}^{2pi }$

$=0$

则$iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=0$

6：解：记${{L}_{1}}:0le xle 2,y=0$，$D$是由$L$和${{L}_{1}}$所围成的封闭曲线

而

$int_{{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=int_{0}^{2}{{{e}^{x}}}}dx={{e}^{x}}|_{0}^{2}={{e}^{2}}-1$

另一方面，由格林公式可知：

$int_{L+{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=iint_{D}{({{e}^{x}}sin y-{{e}^{x}}sin y)dxdy=0}}$

于是

$int_{L}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=-}int_{{{L}_{1}}}{{{e}^{x}}sin ydy-{{e}^{x}}cos ydx=}1-{{e}^{2}}$

二、

（1）证明：由于$sqrt{x}f'(x)$在$(0,+infty )$上有界，则存在$M>0$，使得

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

对任意的$varepsilon >0$，对任意的${{x}_{1}},{{x}_{2}}in (0,+infty )$，不妨设${{x}_{1}}<{{x}_{2}}$

于是

$left| f({{x}_{1}})-f({{x}_{2}}) right|=left| int_{{{x}_{1}}}^{{{x}_{2}}}{f'(t)dt} right|le int_{{{x}_{1}}}^{{{x}_{2}}}{left| f'(t) right|}dt=int_{{{x}_{1}}}^{{{x}_{2}}}{left| sqrt{t}f'(t) right|}cdot frac{1}{sqrt{t}}dt$

$le 2Mint_{{{x}_{1}}}^{{{x}_{2}}}{frac{1}{sqrt{t}}}dt=2M(sqrt{{{x}_{2}}}-sqrt{{{x}_{1}}}le 2Msqrt{{{x}_{2}}-{{x}_{1}}}$

令$delta =frac{{{varepsilon }^{2}}}{4{{M}^{2}}}$，当$left| {{x}_{1}}-{{x}_{2}} right|<delta $时，$left| f({{x}_{1}})-f({{x}_{2}}) right|<varepsilon $

即$f(x)$在$(0,+infty )$上一致收敛

（或利用柯西中值定理，令$g(x)=sqrt{x}$）

（2）由（1）可知，对任意的$varepsilon >0$，令${{x}_{1}}to {{0}^{+}}$，当$0<x<delta $时，有

$left| f(x)-f(0) right|<varepsilon $

由极限定义可知：

$underset{xto {{0}^{+}}}{mathop{lim }},f(x)=f({{0}^{+}})$存在

（3）能，理由如下：

证明：由于$underset{xto {{0}^{+}}}{mathop{lim }},sqrt{x}f'(x)$存在，则存在$M>0$，存在${{X}_{1}}>0$，当$x<{{X}_{1}}$时，$left| sqrt{x}f'(x) right|le M$

由（1）可知：$f(x)$在$(0,{{X}_{1}})$上一致收敛

而$underset{xto +infty }{mathop{lim }},sqrt{x}f'(x)$存在，同理可证$f(x)$在$({{X}_{2}},+infty )$上一致收敛

而$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上连续，则$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上一致连续

于是$f(x)$在$(0,+infty )$上一致连续

三、

（1）证明：对任意的$xin (a,+infty )$，则$f(x)$在$[x,x+1]$上可导

记$underset{xto +infty }{mathop{lim }},f(x)=A$

$underset{xto +infty }{mathop{lim }},f(x+1)=A$，则

$underset{xto +infty }{mathop{lim }},[f(x+1)-f(x)]=0$

由拉格朗日中值定理可知：存在$x<xi <x+1$，使得

$f(x+1)-f(x)=f'(xi )[(x+1)-x]=f'(xi )$

且当$xto +infty $时，$xi to +infty $，于是

$underset{xto +infty }{mathop{lim }},f'(xi )=underset{xto +infty }{mathop{lim }},[f(x+1)-f(x)]=1$

由归结原则可知：

$underset{xto +infty }{mathop{lim }},f'(xi )=underset{xto +infty }{mathop{lim }},f'(x)=0$

（2）可以推出，理由如下：

证明：应用$Taylor$公式可知：对任意的$m=1,2,3,4$

$f(x+m)=f(x)+mf'(x)+frac{{{m}^{2}}}{2!}f''(x)+frac{{{m}^{3}}}{3!}f'''(x)+frac{{{m}^{4}}}{4!}f''''({{xi }_{m}}(x)),x<{{xi }_{m}}(x)<x+m$

由于$f'(x),f''(x),f'''(x)$均可用$f(x+m)-f(x)$及$f''''({{xi }_{m}}(x))$表达

而$underset{xto +infty }{mathop{lim }},[f(x+m)-f(x)]$与$underset{xto +infty }{mathop{lim }},f''''({{xi }_{m}}(x))=underset{xto +infty }{mathop{lim }},f''''(x)$均存在有限

故$underset{xto +infty }{mathop{lim }},{{f}^{(k)}}(x)$均存在有限，不妨记

$underset{xto +infty }{mathop{lim }},{{f}^{(k)}}(x)={{A}_{k}},k=0,1,2,3,4$

在上面的$Taylor$公式中，令$xto +infty $，则${{xi }_{m}}(x)to +infty $，于是有

$0=underset{xto +infty }{mathop{lim }},[f(x+m)-f(x)]=munderset{xto +infty }{mathop{lim }},f'(x)+frac{{{m}^{2}}}{2!}underset{xto +infty }{mathop{lim }},f''(x)+frac{{{m}^{3}}}{3!}underset{xto +infty }{mathop{lim }},f'''(x)+frac{{{m}^{4}}}{4!}underset{xto +infty }{mathop{lim }},f''''({{xi }_{m}}(x))$

即$0={{A}_{0}}-{{A}_{0}}=m{{A}_{1}}+frac{{{m}^{2}}}{2!}{{A}_{2}}+frac{{{m}^{3}}}{3!}{{A}_{3}}+frac{{{m}^{4}}}{4!}{{A}^{4}},m=1,2,3,4$

列举出来为

$left{begin{array}{ll}
{A_1} + frac{1}{{2!}}{A_2} + frac{1}{{3!}}{A_3} + frac{1}{{4!}}{A_4} = 0, \
2{A_1} + frac{{{2^2}}}{{2!}}{A_2} + frac{{{2^3}}}{{3!}}{A_3} + frac{{{2^4}}}{{4!}}{A_4} = 0, \
3{A_1} + frac{{{3^2}}}{{2!}}{A_2} + frac{{{3^3}}}{{3!}}{A_3} + frac{{{3^4}}}{{4!}}{A_4} = 0, \
4{A_1} + frac{{{4^2}}}{{2!}}{A_2} + frac{{{4^3}}}{{3!}}{A_3} + frac{{{4^4}}}{{4!}}{A_4} = 0,
end{array}
right.$
$Rightarrowleft{begin{array}{ll}
{A_1} + frac{1}{{2!}}{A_2} + frac{1}{{3!}}{A_3} + frac{1}{{4!}}{A_4} = 0, \
{A_1} + frac{2}{{2!}}{A_2} + frac{{{2^2}}}{{3!}}{A_3} + frac{{{2^3}}}{{4!}}{A_4} = 0, \
{A_1} + frac{3}{{2!}}{A_2} + frac{{{3^2}}}{{3!}}{A_3} + frac{{{3^3}}}{{4!}}{A_4} = 0, \
{A_1} + frac{4}{{2!}}{A_2} + frac{{{4^2}}}{{3!}}{A_3} + frac{{{4^3}}}{{4!}}{A_4} = 0,
end{array}
right.$

方程组的系数行列式为

$
left|begin{array}{cccc}
1 & {frac{1}{{2!}}} & {frac{1}{{3!}}} & {frac{1}{{4!}}} \
1 & {frac{2}{{2!}}} & {frac{{{2^2}}}{{3!}}} & {frac{{{2^3}}}{{4!}}} \
1 & {frac{3}{{2!}}} & {frac{{{3^2}}}{{3!}}} & {frac{{{3^3}}}{{4!}}} \
1 & {frac{4}{{2!}}} & {frac{{{4^2}}}{{3!}}} & {frac{{{4^3}}}{{4!}}}
end{array}right|
$$
=frac{1}{{1!2!3!4!}}left|begin{array}{cccc}
1 & 1 & 1 & 1 \
1 & 2 & {{2^2}} & {{2^3}} \
1 & 3 & {{3^2}} & {{3^3}} \
1 & 4 & {{4^2}} & {{4^3}}
end{array}right|
$

=$frac{1}{{1!2!3!4!}}(2 - 1)(3 - 1)(3 - 2)(4 - 1)(4 - 2)(4 - 3) = frac{1}{{4!}} ne 0$

即$underset{xto +infty }{mathop{lim }},{{f}^{(k)}}(x)=0,k=1,2,3,4$

（或利用第二数学归纳法证明）

四、证明：证明：由题设知$f(x)$在$ [0,+infty ) $上必有界,设$left| f(x) right|le M$

对$forall varepsilon >0$,有

$left| frac{1}{x}int_{0}^{x}{f(t)dt}-A right|$

$=left| int_{0}^{1}{(f(yx)-A)dy} right|$

$le int_{0}^{frac{varepsilon }{2(M+A)}}{left| f(yx)-A right|}dy+int_{frac{varepsilon }{2(M+A)}}^{1}{left| f(yx)-A right|}dy$

由$underset{xto +infty }{mathop{lim }},f(x)=A$知

对上述$varepsilon >0,exists {{X}_{1}}>0, $使得当$x>{{X}_{1}}$时有$left| f(x)-A right|<frac{varepsilon }{2}$

令$text{X}=frac{2(M+A)}{varepsilon }{{X}_{1}}$,则当$x>X$时,有$int_{frac{varepsilon }{2(M+A)}}^{1}{left| f(yx)-A right|}dy<frac{varepsilon }{2}$

于是

$left| frac{1}{x}int_{0}^{x}{f(t)dt}-A right|<frac{varepsilon }{2}+frac{varepsilon }{2}=varepsilon $

因此

$underset{xto +infty }{mathop{lim }},frac{1}{x}int_{0}^{x}{f(t)dt}=A$.

（或直接利用定义）

五、证明：

充分性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛，令$0<S=underset{nto +infty }{mathop{lim }},{{S}_{n}}=sumlimits_{n=1}^{+infty }{{{a}_{n}}}<+infty $

由于

$underset{nto +infty }{mathop{lim }},frac{frac{{{a}_{n}}}{{{S}_{n}}}}{{{a}_{n}}}=underset{nto +infty }{mathop{lim }},frac{1}{{{S}_{n}}}=frac{1}{S}<+infty $

由比较判别法可知：$sumlimits_{n=1}^{+infty }{frac{{{a}_{n}}}{{{S}_{n}}}}$收敛

必要性：反证法：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$发散

考虑

$sumlimits_{k=n+1}^{n+p}{frac{{{a}_{k}}}{{{S}_{k}}}}=frac{{{a}_{n+1}}}{{{S}_{n+1}}}+frac{{{a}_{n+2}}}{{{S}_{n+2}}}+cdots +frac{{{a}_{n+p}}}{{{S}_{n+p}}}>frac{sumlimits_{k=n+1}^{n+p}{{{a}_{k}}}}{{{S}_{n+p}}}=frac{{{S}_{n+p}}-{{S}_{n}}}{{{S}_{n+p}}}=1-frac{{{S}_{n}}}{{{S}_{n+p}}}$

由于$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$发散，则对任意的$nin {{N}^{*}}$，存在$pin {{N}^{*}}$，使得

$frac{{{S}_{n}}}{{{S}_{n+p}}}<frac{1}{2}$

从而$sumlimits_{k=n+1}^{n+p}{frac{{{a}_{k}}}{{{S}_{k}}}}>frac{1}{2}$

由柯西收敛准则知：$sumlimits_{n=1}^{+infty }{frac{{{a}_{n}}}{{{S}_{n}}}}$发散，矛盾

从而$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛

六、

（1）证明：由于

$2sin frac{x}{2}(frac{1}{2}+sumlimits_{k=1}^{n}{cos kx)=sin }frac{x}{2}+(sin frac{3x}{2}-sin frac{x}{2})+cdots +[sin (n+frac{1}{2})x-sin (n-frac{1}{2})x]=sin (n+frac{1}{2})x$

当$xin [alpha ,2pi -alpha ]$时，$sin frac{x}{2}ne 0$

于是

$frac{1}{2}+sumlimits_{k=1}^{n}{cos kx}=frac{sin (n+frac{1}{2})x}{2sin frac{x}{2}}$

于是$sumlimits_{n=1}^{+infty }{cos nx}$部分和在$xin [alpha ,2pi -alpha ]$上一致有界

令${{u}_{n}}(x)=cos nx,{{v}_{n}}(x)={{a}_{n}}$

由狄利克雷判别法知：$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$xin [alpha ,2pi -alpha ]$上一致收敛

（2）证明：

充分性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛，由于$left| {{a}_{n}}cos nx right|le left| {{a}_{n}} right|={{a}_{n}}$

由$M$判别法可知：$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$[0,2pi ]$上一致收敛

必要性：若$sumlimits_{n=1}^{+infty }{{{a}_{n}}}cos nx$在$[0,2pi ]$上一致收敛，令$x=0$

即$sumlimits_{n=1}^{+infty }{{{a}_{n}}}$收敛收敛

七、证明：由于

${{u}_{x}}=z+x{{z}_{x}}+yf'(z){{z}_{x}}+g'(z){{z}_{x}}=z$

${{u}_{xx}}={{z}_{x}}$

${{u}_{xy}}={{z}_{y}}$

${{u}_{y}}=x{{z}_{y}}+f(z)+yf'(z){{z}_{y}}+g'(z){{z}_{y}}=f(z)$

${{u}_{yy}}=f'(z){{z}_{y}}$

${{u}_{yx}}=f'(z){{z}_{x}}$

于是

$frac{{{partial }^{2}}u}{partial {{x}^{2}}}cdot frac{{{partial }^{2}}u}{partial {{y}^{2}}}-{{left( frac{{{partial }^{2}}u}{partial xpartial y} right)}^{2}}=0$

八、证明：

（1）由于$frac{partial f}{partial n}=frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)$

由第一型和第二型曲面积分的关系和$Guass$公式可知：

$iintlimits_{S}{frac{partial f}{partial n}}dS=iintlimits_{S}{frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)}dS$

$=iintlimits_{S}{frac{partial f}{partial x}dydz+frac{partial f}{partial y}dzdx+frac{partial f}{partial z}dxdy}$

$=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}$

即$iintlimits_{S}{frac{partial f}{partial n}}dS=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}$

（2）充分性：若对任意的$S$，$iintlimits_{S}{frac{partial f}{partial n}}dS$，总有$iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}=0$

不妨设$g(x,y,z)=frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}$，由积分中值定理可知：存在$(xi ,eta ,zeta )in V$

[iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}})dxdydz=g(xi ,eta ,zeta )V=0]

于是由$V$的任意性可知：$g(xi ,eta ,zeta )=0$，再由$(xi ,eta ,zeta )$的任意性知：

$frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}=0$

必要性：若$frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}}=0$，则$iintlimits_{S}{frac{partial f}{partial n}}dS=iiint_{V}{(frac{{{partial }^{2}}f}{partial {{x}^{2}}}+frac{{{partial }^{2}}f}{partial {{y}^{2}}}+frac{{{partial }^{2}}f}{partial {{z}^{2}}})dxdydz}=0$