华中师范大学2010年数学分析试题解答

一、         计算题

1：解：由于$f(2x)=f(x)cos x$，则

$f(x)=f(frac{x}{2})cos frac{x}{2}$

$f(frac{x}{2})=f(frac{x}{4})cos frac{x}{4}$

$vdots $

$f(frac{x}{{{2}^{n-1}}})=f(frac{x}{{{2}^{n}}})cos frac{x}{{{2}^{n}}}$

$vdots $

于是

$f(x)=f(frac{x}{{{2}^{n}}})cos frac{x}{2}cos frac{x}{4}cos frac{x}{8}cdots cos frac{x}{{{2}^{n}}}$                             （1）

而

$cos frac{x}{2}cos frac{x}{4}cos frac{x}{8}cdots cos frac{x}{{{2}^{n}}}=frac{cos frac{x}{2}cos frac{x}{4}cos frac{x}{8}cdots cos frac{x}{{{2}^{n}}}sin frac{x}{{{2}^{n}}}}{sin frac{x}{{{2}^{n}}}}$

$=frac{1}{2}frac{cos frac{x}{2}cos frac{x}{4}cos frac{x}{8}cdots cos frac{x}{{{2}^{n-1}}}sin frac{x}{{{2}^{n-1}}}}{sin frac{x}{{{2}^{n}}}}=cdots =frac{frac{1}{{{2}^{n}}}sin x}{sin frac{x}{{{2}^{n}}}}$

于是设${{x}_{n}}=frac{x}{{{2}^{n}}},xin R$，则$underset{nto +infty }{mathop{lim }},{{x}_{n}}=0$，令两边$nto +infty $，有

$f(x)=underset{nto +infty }{mathop{lim }},f({{x}_{n}})underset{nto +infty }{mathop{lim }},frac{frac{1}{{{2}^{n}}}sin x}{sin frac{x}{{{2}^{n}}}}$

由归结原则知：

$f(x)=underset{xto 0}{mathop{lim }},f(x)underset{nto +infty }{mathop{lim }},frac{frac{1}{{{2}^{n}}}sin x}{sin frac{x}{{{2}^{n}}}}=underset{xto 0}{mathop{lim }},f(x)cdot frac{sin x}{x}=f(0)cdot frac{sin x}{x}=frac{sin x}{x}$

于是$f(x)=left{begin{array}{ll}
frac{{sin x}}{x}, & hbox{$x ne 0$;} \
0, & hbox{$x = 0$.}
end{array}
right.$

$

2：解：由于

${{a}_{n}}=int_{0}^{frac{pi }{4}}{{{tan }^{n}}}xdx=int_{0}^{frac{pi }{4}}{{{tan }^{n-2}}}xcdot frac{1-{{cos }^{2}}x}{{{cos }^{2}}x}dx=int_{0}^{frac{pi }{4}}{{{tan }^{n-2}}}xd(tan x)-int_{0}^{frac{pi }{4}}{{{tan }^{n-2}}}xdx$

$=frac{1}{n-1}{{(tan x)}^{n-1}}|_{0}^{frac{pi }{4}}-{{a}_{n-2}}=frac{1}{n-1}-{{a}_{n-2}}$

于是${{a}_{n}}+{{a}_{n+2}}=frac{1}{n+1}$

则[sumlimits_{n=1}^{+infty }{frac{{{a}_{n}}+{{a}_{n+2}}}{n}}=sumlimits_{n=1}^{+infty }{frac{1}{n(n+1)}}=sumlimits_{n=1}^{+infty }{(frac{1}{n}-frac{1}{n+1})}=1]

3：解：由于平面$x+y+z=0$的单位法向量为$(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$

由$Stokes$公式可知：

[oint_{L}{(y-z)dx+(z-x)dy+(x-y)dz=iintlimits_{D}{(-1-1)dydz+(-1-1)dzdx+(-1-1)dxdy}}]

其中$D$是由$L$所包围的图形

再由两类曲面积分之间的关系可知：

$oint_{L}{(y-z)dx+(z-x)dy+(x-y)dz=-2iintlimits_{S}{(frac{1}{sqrt{3}}}}+frac{1}{sqrt{3}}+frac{1}{sqrt{3}})dS=-2sqrt{3}pi $

二、

（1）证明：由于$f(x)={{x}^{alpha }}$在$(0,+infty )$上连续可导，且$(0,+infty )=(0,1]cup [1,+infty )$

于是由微分中值定理可知：当$0<alpha le 1$时，对任意的${{x}_{1}},{{x}_{2}}in [1,+infty )$，由于

$left| f({{x}_{1}})-f({{x}_{2}}) right|=left| f'(xi ) right|left| {{x}_{1}}-{{x}_{2}} right|$，其中$xi $在${{x}_{1}},{{x}_{2}}$之间

由于$f'(x)=alpha {{x}^{alpha -1}}le alpha $，于是

$left| f({{x}_{1}})-f({{x}_{2}}) right|le left| {{x}_{1}}-{{x}_{2}} right|$

于是对任意的$varepsilon >0$，令$delta =varepsilon $，当$left| {{x}_{1}}-{{x}_{2}} right|<delta $时，有

$left| f({{x}_{1}})-f({{x}_{2}}) right|le left| {{x}_{1}}-{{x}_{2}} right|<varepsilon $

即$f(x)$在$[1,+infty )$上一致收敛

而$f(x)$在$[0,1]$上连续，则$f(x)$在$[0,1]$上一致连续

于是$f(x)$在$(0,1]$上一致连续

则当$0<alpha le 1$时，$f(x)$在$(0,+infty )$上一致连续

（2）当$alpha >1$时，$exists {{varepsilon }_{0}}=frac{1}{2}$，对$x_{n}^{'}={{n}^{frac{1}{alpha }}},x_{n}^{''}={{(n+1)}^{frac{1}{alpha }}}$

令$m=frac{1}{alpha },t=frac{1}{n}$

由于

$underset{nto +infty }{mathop{lim }},left| x_{n}^{'}-x_{n}^{''} right|=underset{tto {{0}^{+}}}{mathop{lim }},frac{1-{{(t+1)}^{m}}}{{{t}^{m}}}=underset{tto {{0}^{+}}}{mathop{lim }},frac{-m{{(t+1)}^{m-1}}}{m{{t}^{m-1}}}=underset{tto {{0}^{+}}}{mathop{lim }},{{(frac{t}{t+1})}^{1-m}}=0,(0<m<1)$

而$left| f(x_{n}^{'})-f(x_{n}^{''}) right|=1>{{varepsilon }_{0}}$

于是当$alpha >1$时，$f(x)$在$(0,+infty )$上不一致收敛

三、

（1）证明：令$u=xy$，则

$int_{A}^{+infty }{frac{sin xy}{y}}dy=int_{Ax}^{+infty }{frac{sin u}{u}du}$，其中$A>0$

而$int_{0}^{+infty }{frac{sin u}{u}}du$收敛，故对任意的$varepsilon >0$，存在$M>0$，当$A'>M$时，有

$left| int_{A'}^{+infty }{frac{sin u}{u}du} right|<varepsilon $

令$Adelta >M$，则当$A>frac{M}{delta }$时，对一切$xin [delta ,+infty )$，有

$left| int_{A}^{+infty }{frac{sin xy}{y}dy} right|<varepsilon $

于是$int_{0}^{+infty }{frac{sin xy}{y}dy}$在$[delta ,+infty )$上一致收敛

（2）由于$int_{0}^{+infty }{frac{sin u}{u}}du$收敛，则对任意的正数${{varepsilon }_{0}}$及$M$，总存在$x>0$，使得

$left| int_{Mx}^{+infty }{frac{sin u}{u}du-int_{0}^{+infty }{frac{sin u}{u}du}} right|<{{varepsilon }_{0}}$

即

$int_{0}^{+infty }{frac{sin u}{u}du}-{{varepsilon }_{0}}<int_{Mx}^{+infty }{frac{sin u}{u}du}<int_{0}^{+infty }{frac{sin u}{u}du}+{{varepsilon }_{0}}$

令${{varepsilon }_{0}}=frac{1}{2}int_{0}^{+infty }{frac{sin u}{u}du}$，于是

$int_{M}^{+infty }{frac{sin xy}{y}dy=int_{Mx}^{+infty }{frac{sin u}{u}du}>2{{varepsilon }_{0}}}-{{varepsilon }_{0}}={{varepsilon }_{0}}$

所以$int_{0}^{+infty }{frac{sin xy}{y}dy}$在$(0,+infty )$上不一致收敛

四、证明：由题可知：

$frac{f(c)-f(a)}{c-a}=frac{f(b)-f(c)}{b-c}$

由于$f(x)$在$[a,b]$上连续，在$(a,b)$内可导，由拉格朗日中值定理可知：

存在${{xi }_{1}}in (a,c),{{xi }_{2}}in (c,b)$，使得

$f'({{xi }_{1}})=frac{f(c)-f(a)}{c-a}=f'({{xi }_{2}})=frac{f(b)-f(c)}{b-c}$

而$f'(x)$在$[a,b]$上连续，在$(a,b)$内可导，再由罗尔定理可知：

存在$xi in ({{xi }_{1}},{{xi }_{2}})subset (a,b)$，使得$f''(xi )=0$

五、

（1）由于$f_{n}^{'}(x)$在$U(x)cap [a,b]$上一致有界

于是存在${{M}_{i}}>0$，对任一切$xin U({{x}_{i}})cap [a,b]$，有$left| f_{n}^{'}(x) right|le {{M}_{i}}$

而$[a,b]$为有界闭区间，由有限覆盖定理可知

存在有限个开区间$U({{x}_{i}}),i=1,2,cdots ,m$，使得$[a,b]subset underset{i=1}{overset{m}{mathop bigcup }},(U({{x}_{i}}))$

令$M=max {{{M}_{1}},{{M}_{2}},cdots ,{{M}_{m}}}$，于是对一切$xin [a,b]$，有$left| f_{n}^{'}(x) right|le M$

即${f_{n}^{'}(x)}$在$[a,b]$上一致有界

（3）证明：对任意的$varepsilon >0$，作$[a,b]$的分划

$Delta :a={{x}_{0}}<{{x}_{1}}<cdots <{{x}_{k}}=b,left| Delta  right|<frac{varepsilon }{3M}$

由题可知，存在$N>0$，当$m,n>N$时，有$left| {{f}_{n}}({{x}_{i}})-{{f}_{m}}({{x}_{i}}) right|<frac{varepsilon }{3},i=0,1,2,cdots ,k$

于是对任意的$xin [{{x}_{{{i}_{0}}}},{{x}_{{{i}_{0}}+1}}]subset [a,b]$，当$n,m>N$时，由微分中值定理有

$left| {{f}_{n}}(x)-{{f}_{m}}(x) right|le left| {{f}_{n}}(x)-{{f}_{n}}({{x}_{{{i}_{0}}}}) right|+left| {{f}_{n}}({{x}_{{{i}_{0}}}})-{{f}_{m}}({{x}_{{{i}_{0}}}}) right|+left| {{f}_{m}}({{x}_{{{i}_{0}}}})-{{f}_{m}}(x) right|$

$le left| f_{n}^{'}({{xi }_{{{i}_{0}}}}) right|left| x-{{x}_{{{i}_{0}}}} right|+left| {{f}_{n}}({{x}_{{{i}_{0}}}})-{{f}_{m}}({{x}_{{{i}_{0}}}}) right|+left| f_{n}^{'}({{eta }_{{{i}_{0}}}}) right|left| x-{{x}_{{{i}_{0}}}} right|$

$<Mcdot frac{varepsilon }{3}+frac{varepsilon }{3}+Mcdot frac{varepsilon }{3}=varepsilon $

即${{{f}_{n}}(x)}$在$[a,b]$上一致收敛

六、

（1）解：由于$left| xy right|le frac{{{x}^{2}}+{{y}^{2}}}{2}$，则$left| xyln ({{x}^{2}}+{{y}^{2}}) right|le frac{1}{2}left| ({{x}^{2}}+{{y}^{2}})ln ({{x}^{2}}+{{y}^{2}}) right|$

设$t={{x}^{2}}+{{y}^{2}}ge 0$

由于$underset{tto {{0}^{+}}}{mathop{lim }},left| tln t right|=underset{tto {{0}^{+}}}{mathop{lim }},frac{ln t}{frac{1}{t}}=-underset{tto {{0}^{+}}}{mathop{lim }},t=0$

于是$underset{(x,y)to (0,0)}{mathop{lim }},f(x,y)=underset{(x,y)to (0,0)}{mathop{lim }},xyln ({{x}^{2}}+{{y}^{2}})=0=f(0,0)$

于是$f(x,y)$在$(0,0)$连续

（2）解：由于

${{f}_{x}}(0,0)=underset{Delta xto 0}{mathop{lim }},frac{f(Delta x,0)-f(0,0)}{Delta x}=0,{{f}_{y}}(0,0)=underset{Delta yto 0}{mathop{lim }},frac{f(0,Delta y)-f(0,0)}{Delta y}=0$

于是${{f}_{x}}(0,0),{{f}_{y}}(0,0)$都存在

（4）解：由于

$underset{rho to 0}{mathop{lim }},frac{f(Delta x,Delta y)-f(0,0)-{{f}_{x}}(0,0)Delta x-{{f}_{y}}(0,0)Delta y}{rho }=underset{rho to 0}{mathop{lim }},frac{Delta xDelta yln [{{(Delta x)}^{2}}+{{(Delta y)}^{2}}]}{sqrt{{{(Delta x)}^{2}}+{{(Delta y)}^{2}}}}$

由于[left| frac{Delta xDelta yln [{{(Delta x)}^{2}}+{{(Delta y)}^{2}}]}{sqrt{{{(Delta x)}^{2}}+{{(Delta y)}^{2}}}} right|le frac{1}{2}sqrt{{{(Delta x)}^{2}}+{{(Delta y)}^{2}}}left| ln [{{(Delta x)}^{2}}+{{(Delta y)}^{2}}] right|]

设$t=sqrt{{{(Delta x)}^{2}}+{{(Delta y)}^{2}}}$，则[left| frac{Delta xDelta yln [{{(Delta x)}^{2}}+{{(Delta y)}^{2}}]}{sqrt{{{(Delta x)}^{2}}+{{(Delta y)}^{2}}}} right|le left| tln t right|]

由（1）可知，$underset{rho to 0}{mathop{lim }},frac{f(Delta x,Delta y)-f(0,0)-{{f}_{x}}(0,0)Delta x-{{f}_{y}}(0,0)Delta y}{rho }=0$

于是$f(x,y)$在$(0,0)$可微

七、

（1）证明：由于$int_{0}^{+infty }{frac{1}{f(x)}}dx$收敛，于是$underset{nto +infty }{mathop{lim }},int_{n}^{n+1}{frac{1}{f(x)}}dx=0$

由积分中值定理可知，存在${{x}_{n}}in [n,n+1]$，使得

$underset{nto +infty }{mathop{lim }},int_{n}^{n+1}{frac{1}{f(x)}}dx=underset{nto +infty }{mathop{lim }},frac{1}{f({{x}_{n}})}=0$且$f(x)>0$

即$underset{nto +infty }{mathop{lim }},f({{x}_{n}})=+infty $，且${{{x}_{n}}}$严格单调递增，$underset{nto +infty }{mathop{lim }},{{x}_{n}}=+infty $

（2）先证施瓦兹（Schwarz）不等式

若$f(x),g(x)$在$[a,b]$上可积，则$[int_{a}^{b}{f(x)g(x)dx{{]}^{2}}}le int_{a}^{b}{{{f}^{2}}(x)}dxint_{a}^{b}{{{g}^{2}}(x)}dx$

证明：对任意的$tin R$，由于

${{[f(x)-tg(x)]}^{2}}ge 0Rightarrow int_{a}^{b}{{{[f(x)-tg(x)]}^{2}}dxge 0}$

于是

$int_{a}^{b}{{{g}^{2}}}(x)dxcdot {{t}^{2}}-2int_{a}^{b}{f(x)g(x)dxcdot t+int_{a}^{b}{{{f}^{2}}}(x)dx}ge 0$

上式对于任意的$tin R$恒成立，于是

$Delta =4{{[int_{a}^{b}{f(x)g(x)dx}]}^{2}}-4int_{a}^{b}{{{f}^{2}}(x)}dxint_{a}^{b}{{{g}^{2}}(x)}dxle 0$

即

$[int_{a}^{b}{f(x)g(x)dx{{]}^{2}}}le int_{a}^{b}{{{f}^{2}}(x)}dxint_{a}^{b}{{{g}^{2}}(x)}dx$

则对任意的$x>0$

$frac{{{x}^{2}}}{4}={{(int_{frac{x}{2}}^{x}{sqrt{f(t)cdot frac{1}{f(t)}}}dt)}^{2}}le (int_{frac{x}{2}}^{x}{f(t)dt)}(int_{frac{x}{2}}^{x}{frac{1}{f(t)}dt)}le (int_{0}^{x}{f(t)dt)}(int_{frac{x}{2}}^{x}{frac{1}{f(t)}dt)}$

即

$frac{1}{4}le (frac{1}{{{x}^{2}}}int_{0}^{x}{f(x)dx)(int_{frac{x}{2}}^{x}{frac{1}{f(x)}}}dx)$

令$xto +infty $，，由于$underset{xto +infty }{mathop{lim }},int_{frac{x}{2}}^{x}{frac{1}{f(x)}}dx=0$，于是

$underset{xto +infty }{mathop{lim }},frac{1}{{{x}^{2}}}int_{0}^{x}{f(x)dx}=+infty $

八、

（1）证明：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：

$iintlimits_{partial V}{gcdot frac{partial f}{partial n}dS}=iintlimits_{partial V}{g[frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)}dS$

$=iintlimits_{partial V}{g[frac{partial f}{partial x}dydz+frac{partial f}{partial y}dzdx+frac{partial f}{partial z}dxdy]}$

$=iiint_{V}{gcdot Delta fdxdydz+iiint_{V}{nabla gcdot nabla fdxdydz}}$

即

$iiint_{V}{(nabla fcdot nabla g)dxdydz=iintlimits_{partial V}{gcdot frac{partial f}{partial n}}}dS-iiint_{V}{(gcdot Delta f)dxdydz}$

（2）设$r=sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$，于是

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

$=iiint_{V}{nabla rcdot nabla fdxdydz}$

由（1）知：

$iiint_{V}{nabla rcdot nabla fdxdydz}=iintlimits_{partial V}{rcdot frac{partial f}{partial n}}dS-iiint_{V}{(rcdot Delta f)dxdydz}$

$=iintlimits_{partial V}{[r[frac{partial f}{partial x}cos (n,x)+frac{partial f}{partial y}cos (n,y)+frac{partial f}{partial z}cos (n,z)}]]dS-iiint_{V}{(rcdot Delta f)dxdydz}$

$=iintlimits_{partial V}{[frac{partial f}{partial x}dydz+frac{partial f}{partial y}dzdx+frac{partial f}{partial z}dxdy]}-iiint_{V}{(rcdot Delta f)dxdydz}$

$=iiint_{V}{Delta fdxdydz-iiint_{V}{(rcdot Delta f)dxdydz}}$

$=iiint_{V}{(1-r)cdot Delta fdxdydz}$

$=iiint_{V}{(1-r)cdot {{r}^{2}}dxdydz}$

设

$x=Rsin varphi cos theta ,y=Rsin varphi sin theta ,z=Rcos varphi ,0le theta le 2pi ,0le varphi le pi ,0le Rle 1$

于是

$iiint_{V}{(1-r)cdot {{r}^{2}}dxdydz}=int_{0}^{2pi }{dtheta int_{0}^{pi }{dvarphi int_{0}^{1}{(1-R){{R}^{2}}}}}cdot {{R}^{2}}sin varphi dR=frac{2}{15}pi $

即

$iiint_{V}{(frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}frac{partial f}{partial x}+frac{y}{sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}frac{partial f}{partial y}+frac{z}{sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}frac{partial f}{partial z})dxdydz}=frac{2}{15}pi $