武汉大学2011年数学分析试题解答
1：计算题
（1） 解：原极限$ ext{=}underset{n o infty }{mathop{lim }}\,frac{sqrt[n]{n}}{n}cdot {{n}^{1-alpha }}={{e}^{-1}}underset{n o infty }{mathop{lim }}\,{{n}^{1-alpha }}=$$left{egin{array}
+infty, & hbox{$0<alpha<1$;} \
e^{-1}, & hbox{$alpha=1$;} \
0, & hbox{$alpha>1$.}
end{array}
ight.$
(解释一下：$underset{n o infty }{mathop{lim
}}\,frac{sqrt[n]{n}}{n}={{e}^{underset{n o infty }{mathop{lim
}}\,frac{1}{n}sumlimits_{i=1}^{n}{Infrac{i}{n}}}}={{e}^{intlimits_{0}^{1}{Inxdx}}}={{e}^{-1}}$
（来源于数学分析上册第二章课后习题）
（2） 解：考虑等价无穷小：$underset{x o 0}{mathop{lim }}\,frac{1-cos x}{frac{1}{2}{{x}^{2}}}=1$ ，则
$1-cos sqrt{ an x-sin x}=2{{sin }^{2}}frac{sqrt{ an x-sin x}}{2}sim frac{1}{2}( an x-sin x)=frac{sin x}{2cos x}(1-cos x)sim frac{1}{4}{{x}^{3}}$
另一方面：[sqrt[3]{1+{{x}^{3}}}-sqrt[3]{1-{{x}^{3}}}=frac{2{{x}^{3}}}{{{(sqrt[3]{1+{{x}^{3}}})}^{2}}+sqrt[3]{(1+{{x}^{3}})(1-{{x}^{3}})}+{{(sqrt[3]{1-{{x}^{3}}})}^{2}}}sim frac{2{{x}^{3}}}{3}]
从而原式 =$frac{3}{8}$
（3） 法一：
解：原式[=int{frac{1+cos x}{sqrt{1+cos x}}dx=}int{frac{2{{cos }^{2}}frac{x}{2}}{sqrt{1+cos x}}dx=}int{frac{2{{cos }^{2}}frac{x}{2}(1-{{sin }^{2}}frac{x}{2})}{{{cos }^{2}}frac{x}{2}sqrt{1+cos x}}dx}]
[=int{frac{1+cos x-sin xsin frac{x}{2}cos frac{x}{2}}{{{cos }^{2}}frac{x}{2}sqrt{1+cos x}}dx=}int{sqrt{1+cos x}{{sec }^{2}}frac{x}{2}+frac{-sin x}{sqrt{1+cos x}} an frac{x}{2}dx}]
$=2int{d(sqrt{1+cos x} an frac{x}{2})=2}sqrt{1+cos x} an frac{x}{2}+C$
法二：由于$int{sqrt{ ext{1+}cos x}dx}=sqrt{2}int{left| cos frac{x}{2} ight|}dx$
考虑到$int{left| x ight|}dx=frac{{{x}^{2}}}{2}sgn x+C$
于是$int{sqrt{ ext{1+}cos x}dx}=2sqrt{2}sin frac{x}{2}sgn (cos frac{x}{2})+C$（$C$为常数）
法三：由于$int{sqrt{ ext{1+}cos x}dx}=sqrt{2}int{left| cos frac{x}{2} ight|}dx$
$overset{t=frac{x}{2}}{mathop{=}}\,2sqrt{2}int{left| cos t ight|}dt$
而
$int {left| {cos t}
ight|} dt=left{egin{array}{ll}
sin t + {c_k}, & hbox{$- frac{pi }{2} + 2kpi le t le frac{pi }{2} + 2kpi$;} \
- sin t + {d_k}, & hbox{$frac{pi }{2} + 2kpi le t le frac{{3pi }}{2} + 2kpi$.}
end{array}
ight.$+C为连续函数，其中C为常数
于是
$left{egin{array}{ll}
{c_k} + 1 = - 1 + {d_k} \
{c_{k + 1}} - 1 = 1 + {d_k}
end{array}
ight.$
，令${{c}_{0}}=0$
则${{c}_{k}}=4k,{{d}_{k}}=4k+2$
于是
$int {left| {cos t}
ight|} dt=left{egin{array}{ll}
sin t + 4k, & hbox{$ - frac{pi }{2} + 2kpi le t le frac{pi }{2} + 2kpi$;} \
- sin t + 4k + 2, & hbox{$frac{pi }{2} + 2kpi le t le frac{{3pi }}{2} + 2kpi$.}
end{array}
ight.$+C
即
$int {sqrt {{
m{1 + }}cos x} dx}=left{egin{array}{ll}
2sqrt 2 (sin frac{x}{2} + 4k), & hbox{$ - pi + 4kpi le x le pi + 4kpi$;} \
2sqrt 2 ( - sin frac{x}{2} + 4k + 2), & hbox{$pi + 4kpi le x le 3pi + 4kpi$.}
end{array}
ight.$+C，其中C为常数
（4） 解：[F(x,y)=xint_{frac{y}{x}}^{xy}{zf(z)dz-yint_{frac{y}{x}}^{xy}{f(z)dz}}]
则$F_{x}^{'}=int_{frac{y}{x}}^{xy}{zf(z)dz+x[xyf(xy)cdot y-frac{y}{x}f(frac{y}{x})cdot frac{-y}{{{x}^{2}}}]-y[f(xy)cdot y-f(frac{y}{x})cdot frac{-y}{{{x}^{2}}}]}$
$=int_{frac{y}{x}}^{xy}{zf(z)dz+({{x}^{2}}-1){{y}^{2}}f(xy)}$
$F_{xx}^{''}=xyf(xy)cdot y-frac{y}{x}f(frac{y}{x})cdot frac{-y}{{{x}^{2}}}+2x{{y}^{2}}f(xy)+({{x}^{2}}-1){{y}^{2}}f'(xy)cdot y$
$=3x{{y}^{2}}f(xy)+frac{{{y}^{2}}}{{{x}^{3}}}f(frac{y}{x})+({{x}^{2}}-1){{y}^{3}}f'(xy)$
（5） 解：原式$=intlimits_{0}^{1}{dyintlimits_{-1}^{{{y}^{2}}}{({{y}^{2}}-x)dx+}}intlimits_{0}^{1}{dyintlimits_{{{y}^{2}}}^{1}{(-{{y}^{2}}+x)dx}}=frac{6}{5}$
2：说明，原版的试卷中的题目可能有点问题，原版试题如下：
已知$f(x),g(x)$在$[a,b]$上连续，在$(a,b)$上可微，且$g'(x)$在$(a,b)$上无零点，证明：$exists xi in (a,b),st$ $frac{f'(xi )}{g'(xi )}=frac{f(b)-g(xi )}{g(xi )-g(a)}$
如果有思路的话，欢迎补充！
证明:作辅助函数$Fleft( x ight)=fleft( x ight)gleft( x ight)-gleft( b ight)fleft( x ight)-fleft( a ight)gleft( x ight) $
虽然$Fleft( x ight) $在$[a,b]$上连续，在$(a,b)$上可微, $Fleft( a ight)=Fleft( b ight)=-fleft( a ight)gleft( b ight) $
由罗尔中值定理,存在$xi in left( a,b ight)$使得${F}'left( xi ight)=0$
即${f}'left( xi ight)gleft( xi ight)+fleft( xi ight){g}'left( xi ight)-gleft( b ight){f}'left( xi ight)-fleft( a ight){g}'left( xi ight)=0$
整理$left[ fleft( a ight)-fleft( xi ight) ight]{g}'left( xi ight)-{f}'left( xi ight)left[ gleft( xi ight)-gleft( b ight) ight]=0$
即$frac{fleft( a ight)-fleft( xi ight)}{gleft( xi ight)-gleft( b ight)}=frac{{f}'left( xi ight)}{{g}'left( xi ight)} $,证得
3：（方法一）
证明：$left| frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+cdots +{{a}_{n}}}-b ight|=left| frac{{{a}_{1}}({{b}_{n}}-b)+{{a}_{2}}({{b}_{n-1}}-b)+cdots +{{a}_{n}}({{b}_{1}}-b)}{{{a}_{1}}+{{a}_{2}}+cdots +{{a}_{n}}} ight|$
$le frac{{{a}_{1}}left| {{b}_{n}}-b ight|+cdots +{{a}_{n-N}}left| {{b}_{N+1}}-b ight|}{{{a}_{1}}+cdots +{{a}_{n}}}+frac{{{a}_{n-N+1}}left| {{b}_{N}}-b ight|+cdots +{{a}_{n}}left| {{b}_{1}}-b ight|}{{{a}_{1}}+cdots +{{a}_{n}}}$
$le underset{N+1le kle n}{mathop max }\,left| {{b}_{k}}-b ight|+underset{N+1le kle n}{mathop max }\,left| {{b}_{k}}-b ight|cdot frac{N}{n-N+1}=I_{1}^{n}+I_{2}^{n}$
从而对$forall varepsilon >0,$先取定$N$使得$I_{1}^{n}<frac{varepsilon }{2}$，后让$n$充分大即有$I_{2}^{n}<frac{varepsilon }{2}$，于是有结论成立。
（方法二）
证明：设${{t}_{nk}}=frac{{{a}_{n-k+1}}}{{{a}_{1}}+{{a}_{2}}+cdots +{{a}_{n}}},k=1,2,cdots ,n,n=1,2,cdots $
则${{t}_{nk}}>0$且$sumlimits_{k=1}^{n}{{{t}_{nk}}}=1$
再由${{a}_{n}}ge 0,{{a}_{n}}le {{a}_{n-1}}Rightarrow {{a}_{1}}+{{a}_{2}}+cdots ext{+}{{a}_{n}}ge {{a}_{1}}+{{a}_{2}}+cdots ext{+}{{a}_{n-k+1}}ge (n-k+1){{a}_{n-k+1}}$
于是$0le {{t}_{nk}}le frac{{{a}_{n-k+1}}}{(n-k+1){{a}_{n-k+1}}}=frac{1}{(n-k+1)} o 0(n o +infty )$
由迫敛性知：$underset{n o +infty }{mathop{lim }}\,{{t}_{nk}}=0$
再由$underset{n o +infty }{mathop{lim }}\,{{b}_{n}}=b$，可知
$exists M>0,s.t$对任意的$nin {{N}^{*}},left| {{a}_{n}}-a ight|<M$
同时，对任意的$varepsilon >0,exists {{N}_{1}}in {{N}^{*}}$，当$n>{{N}_{1}}$时，有$left| {{a}_{n}}-a ight|<frac{varepsilon }{2}$
固定${{N}_{1}}$，由$underset{n o +infty }{mathop{lim }}\,{{t}_{nk}}=0$可知：
$exists {{N}_{2}}in {{N}^{*}}$，当$n>{{N}_{2}}$时，有$left| {{a}_{n}}-a ight|<frac{varepsilon }{2{{N}_{1}}M},k=1,2,cdots ,{{N}_{1}}$
令$N=max {{{N}_{1}},{{N}_{2}}}$，当$n>N$时，有
$left| frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+cdots +{{a}_{n}}}-b ight|=left| sumlimits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-b} ight|=left| sumlimits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-sumlimits_{k=1}^{n}{{{t}_{nk}}}b} ight|$
$le sumlimits_{k=1}^{n}{{{t}_{nk}}}left| {{b}_{k}}-b ight|<M({{t}_{n1}}+{{t}_{n2}}+cdots +{{t}_{n{{N}_{1}}}})+frac{varepsilon }{2}({{t}_{n,{{N}_{1}}+1}}+cdots +{{t}_{nn}})<varepsilon $
即$underset{n o +infty }{mathop{lim }}\,frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+cdots +{{a}_{n}}}=b$
（方法三，书上并无介绍此公式，不建议使用）：
直接利用$Stolz$进行计算即可
4：证明：${{a}_{k}}=frac{1}{2pi }intlimits_{-pi }^{pi }{f(x)cos kxdx=}frac{1}{2kpi }intlimits_{-pi }^{pi }{f(x)d(sinkx)=}frac{-1}{2kpi }intlimits_{-pi }^{pi }{f'(x)sinkxdx}$
$=frac{-1}{2{{k}^{2}}pi }intlimits_{-kpi }^{kpi }{f'(frac{t}{k})sintdt}$
则
${{a}_{2n}}=frac{-1}{8{{n}^{2}}pi }intlimits_{-2npi }^{2npi }{f'(frac{t}{2n})operatorname{sint}dt=}frac{-1}{8{{n}^{2}}pi }sumlimits_{m=-n}^{n-1}{[intlimits_{-2mpi }^{(2m+1)pi }{f'(frac{t}{2n})operatorname{sint}dt+intlimits_{(2m+1)pi }^{2(m+1)pi }{f'(frac{t}{2n})operatorname{sint}dt]=}}}$
[ge frac{-1}{8{{n}^{2}}pi }sumlimits_{m=-n}^{n-1}{[intlimits_{2mpi }^{(2m+1)pi }{f'(frac{t}{2n})sin tdt+}}intlimits_{(2m+1)pi }^{2(m+1)pi }{operatorname{sintdt}]}]
而
$sin tleft{egin{array}{ll}
> 0, & hbox{$ 2mpi < x < (2m + 1)pi$;} \
< 0, & hbox{$(2m + 1)pi < x < 2(m + 1)pi$.}
end{array}
ight.$
且$f$为凸函数
从而原式$ge frac{-1}{8{{n}^{2}}pi }sumlimits_{m=-n}^{n-1}{f'(frac{2m+1}{2n}pi )intlimits_{2mpi }^{2(m+1)pi }{sin tdt=0}}$
另一方面：
[{{a}_{2n+1}}=frac{-1}{2{{(2n+1)}^{2}}pi }intlimits_{-(2n+1)pi }^{(2n+1)pi }{f'(frac{t}{2n+1})sin tdt}]
[=frac{-1}{2{{(2n+1)}^{2}}pi }sumlimits_{m=-n}^{n}{[intlimits_{(2m-1)pi }^{2mpi }{f'(frac{t}{2n+1})sin tdt+intlimits_{2mpi }^{(2m+1)pi }{f'(frac{t}{2n+1})sin tdt}}}]
$le frac{-1}{2{{(2n+1)}^{2}}pi }sumlimits_{m=-n}^{n}{[intlimits_{(2m+1)pi }^{2mpi }{f'(frac{2m}{2n+1})sin tdt}}+intlimits_{2mpi }^{(2m+1)pi }{f'(frac{2m}{2n+1})sin tdt}]$
$==frac{-1}{2{{(2n+1)}^{2}}pi }sumlimits_{m=-n}^{n}{f'(frac{2m}{2n+1}pi )}intlimits_{(2m-1)pi }^{(2m+1)pi }{sin tdt=0}$
5：证明：
（1） 设${{f}_{k}}(x)=sumlimits_{n=1}^{k}{{{u}_{n}}(x),}$
因为
$left| f(x)-f(y) ight|le left| f(x)-{{f}_{K}}(x) ight|+left| {{f}_{K}}(x)-{{f}_{K}}(y) ight|+left| {{f}_{K}}(y)-f(y) ight|$
$le 2cdot sup left| left| {{f}_{K}}(z)-f(z) ight| ight|+left| {{f}_{K}}(x)-{{f}_{K}}(y) ight|equiv I_{1}^{K}+I_{2}^{K}$
从而对$forall varepsilon >0,$可先选定$K$使$I_{1}^{K}<frac{varepsilon }{2}$，后取$delta >0$ 使[left| x-y ight|<delta Rightarrow I_{2}^{K}<frac{varepsilon }{2}]
（2） 若条件改为逐点连续，则结论不成立。为此，仅需构造出${{f}_{k}}(x) o f(x)$
其中每个${{f}_{k}}(x)$一致收敛，但$f(x)$不一致收敛，甚至不连续的例子很多，例如：
${f_k}(x)left{egin{array}{ll}
- 1, & hbox{$x le frac{{ - 1}}{k}$;} \
kx, & hbox{$frac{{ - 1}}{k} < x < frac{1}{k}$;} \
1, & hbox{$x ge frac{1}{k}$.}
end{array}
ight.$
，恒有${{f}_{k}}(x) o sgn (x)$
6：证明：
（1）$intlimits_{c}^{d}{f(x,y)dy}$一致收敛
$Leftrightarrow exists F(x),forall varepsilon >0,exists delta >0,s.t.$
$left{egin{array}{ll}
x in [a,b] \
delta ' in (0,delta )
end{array}
ight.$
其有$Cauchy$准则：
$forall varepsilon >0,exists delta >0,s.t.$
$left{egin{array}{ll}
x in [a,b] \
delta ',delta '' in (0,delta )
end{array}
ight.$
$Rightarrow left| intlimits_{c+delta '}^{c+delta ''}{f(x,y)dy}
ight|<varepsilon $
（3） 仅需注意到$left| intlimits_{c+delta '}^{c+delta }{f(x,y)g(x,y)dy} ight|le underset{[a,b] imes [c,d]}{mathop{max }}\,left| g(x,y) ight|cdot intlimits_{c+delta '}^{c+delta ''}{left| f(x,y) ight|}dy$
后利用$Cauchy$即可得证
7：（1）解：$g=f(frac{x}{z})mp zf(frac{z}{x})+f(frac{y}{z})mp zf(frac{z}{y})$
${{g}_{x}}=frac{1}{z}f''(frac{x}{z})mp frac{{{z}^{2}}}{{{x}^{2}}}f''(frac{z}{x})$
${{g}_{xx}}=frac{1}{{{z}^{2}}}f''(frac{x}{z})mp frac{2{{z}^{2}}}{{{x}^{3}}}f'(frac{z}{x})mp frac{{{z}^{3}}}{{{x}^{4}}}f''(frac{z}{x})$
${{g}_{yy}}=frac{1}{{{z}^{2}}}f''(frac{y}{z})mp frac{{{z}^{2}}}{{{y}^{3}}}f'(frac{z}{y})mp frac{{{z}^{3}}}{{{y}^{4}}}f''(frac{z}{y})$
${{g}_{z}}=frac{-1}{{{z}^{2}}}f'(frac{x}{z})mp f(frac{z}{x})mp frac{z}{x}f'(frac{z}{x})$
$=frac{-y}{{{z}^{2}}}f'(frac{y}{z})mp f(frac{z}{y})mp frac{z}{y}f'(frac{z}{y})$
${{g}_{zz}}=frac{2x}{{{z}^{3}}}f'(frac{x}{z})+frac{{{x}^{2}}}{{{z}^{4}}}f''(frac{x}{z})mp frac{1}{x}f'(frac{z}{x})mp frac{1}{x}f'(frac{z}{x})mp frac{z}{{{x}^{2}}}f''(frac{z}{x})$
$+frac{2y}{{{z}^{3}}}f'(frac{y}{z})+frac{{{y}^{2}}}{{{z}^{4}}}f''(frac{y}{z})mp frac{1}{y}f'(frac{z}{y})mp frac{1}{y}f'(frac{z}{y})mp frac{z}{{{y}^{2}}}f''(frac{z}{y})$
于是
${{x}^{2}}{{g}_{xx}}+{{y}^{2}}{{g}_{yy}}-{{z}^{2}}{{g}_{zz}}=-frac{2x}{z}f(frac{x}{z})-frac{2y}{z}f(frac{y}{z})$
（2）设$0<{{a}_{1}}<{{a}_{2}},0<{{b}_{1}}<{{b}_{2}},1<{{c}_{1}}<{{c}_{2}}.$则
原式[=-2iiint_{Omega }{[frac{x}{z}f'(frac{x}{z})+frac{y}{z}f'(frac{y}{z})dxdydz}]
[=-2intlimits_{{{a}_{1}}}^{{{a}_{2}}}{duintlimits_{{{b}_{1}}}^{{{b}_{2}}}{dvintlimits_{{{c}_{1}}}^{{{c}_{2}}}{[frac{1}{u}f'(frac{1}{u})+frac{1}{v}f'(frac{1}{v})]cdot frac{2}{3}{{u}^{-2}}{{v}^{-2}}dw}}}]
（其中令$u=frac{z}{x},v=frac{z}{y},w={{z}^{3}}$ ）
$frac{-2({{c}_{2}}-{{c}_{1}})}{3}intlimits_{{{a}_{1}}}^{{{a}_{2}}}{duintlimits_{{{b}_{1}}}^{{{b}_{2}}}{[frac{1}{{{u}^{2}}v}f'(frac{1}{u})+frac{1}{u{{v}^{2}}}f'(frac{1}{v})}}]dv$
[=frac{-2({{c}_{2}}-{{c}_{1}})}{3}intlimits_{{{a}_{1}}}^{{{a}_{2}}}{{frac{1}{{{u}^{2}}}f'(frac{1}{u})Infrac{{{b}_{2}}}{{{b}_{1}}}-frac{1}{u}[f(frac{1}{{{b}_{2}}})-f(frac{1}{{{b}_{1}}})]Infrac{{{a}_{2}}}{{{a}_{1}}}}du}]
[=frac{-2({{c}_{2}}-{{c}_{1}})}{3}left{ -Infrac{{{b}_{2}}}{{{b}_{1}}}[ ight.f(frac{1}{{{a}_{2}}})-f(frac{1}{{{a}_{1}}})]-[f(frac{1}{{{b}_{2}}})-f(frac{1}{{{b}_{1}}})]Infrac{{{a}_{2}}}{{{a}_{1}}}}]
[=frac{2({{c}_{2}}-{{c}_{1}})}{3}left{ Infrac{{{b}_{2}}}{{{b}_{1}}}[ ight.f(frac{1}{{{a}_{2}}})+f(frac{1}{{{a}_{1}}})]-[f(frac{1}{{{b}_{2}}})-f(frac{1}{{{b}_{1}}})]Infrac{{{a}_{2}}}{{{a}_{1}}}}]
8：证明：[iint_{D}{(xfrac{partial u}{partial x}}+yfrac{partial u}{partial y})dxdy=intlimits_{0}^{1}{dr}int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{(}x{{u}_{x}}+y{{u}_{y}})dS]
[=intlimits_{0}^{1}{r}int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{frac{partial u}{partial n}}dS=intlimits_{0}^{1}{r}iint_{{{x}^{2}}+{{y}^{2}}le {{r}^{2}}}{ abla udxdy}]
[=intlimits_{0}^{1}{r}iint_{{{x}^{2}}+{{y}^{2}}le {{r}^{2}}}{cos (pi ({{x}^{2}}+{{y}^{2}}))dxdy}]
[=intlimits_{0}^{1}{r}iint_{{{x}^{2}}+{{y}^{2}}le {{r}^{2}}}{cos (pi {{s}^{2}})cdot 2pi sdS}]
[=intlimits_{0}^{1}{rsin (pi r)dr=frac{1}{pi }}]