武汉大学2007年数学分析试题解答

武汉大学数分答案2007

一、（此题共6小题，每题6分，共36分）

1：

解：由于$1\le {{(n!)}^{\frac{1}{{{n}^{2}}}}}\le \sqrt[n]{n}$

而$\underset{n\to +\infty }{\mathop{\lim }}\,\sqrt[n]{n}=1$由迫敛性知：$\underset{n\to +\infty }{\mathop{\lim }}\,{{(n!)}^{\frac{1}{{{n}^{2}}}}}=1$

2：

解：由于$\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}-2x=\frac{1}{\sqrt{{{x}^{2}}+1}+x}-\frac{1}{\sqrt{{{x}^{2}}-1}+x}=\frac{\sqrt{{{x}^{2}}-1}-\sqrt{{{x}^{2}}+1}}{(\sqrt{{{x}^{2}}+1}+x)(\sqrt{{{x}^{2}}-1}+x)}$

$=\frac{-2}{(\sqrt{{{x}^{2}}+1}+x)(\sqrt{{{x}^{2}}-1}+x)(\sqrt{{{x}^{2}}-1}+\sqrt{{{x}^{2}}+1})}$

则$\underset{x\to +\infty }{\mathop{\lim }}\,{{x}^{\alpha }}(\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}-2x)=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{-2{{x}^{\alpha -3}}}{(\sqrt{{{x}^{2}}+1}+x)(\sqrt{{{x}^{2}}-1}+x)(\sqrt{{{x}^{2}}-1}+\sqrt{{{x}^{2}}+1})}$

于是

$\beta =\left\{\begin{array}{ll}
0, & \hbox{$\alpha<3$;} \\
- \frac{1}{4}, & \hbox{$\alpha=3$;} \\
+ \infty, & \hbox{$\alpha>3$.}
\end{array}
\right.$

3：

解：利用高阶导数的公式可知：

${{y}^{(50)}}(x)=C_{50}^{0}\cdot {{x}^{2}}\cdot {{(\cos 3x)}^{(50)}}+C_{50}^{1}\cdot 2x\cdot {{(\cos 3x)}^{(49)}}+C_{50}^{2}\cdot 2\cdot {{(\cos 3x)}^{(48)}}$

$={{3}^{48}}[(2450-9{{x}^{2}})\cos 3x-300x\sin 3x]$

4：

解：由于$ds=\sqrt{1+{{({{y}^{'}})}^{2}}}dx=\sqrt{1+{{(\frac{-2x}{1-{{x}^{2}}})}^{2}}}dx=\frac{1+{{x}^{2}}}{1-{{x}^{2}}}dx=(\frac{2}{1-{{x}^{2}}}-1)dx$

$s=\int_{0}^{\frac{1}{2}}{(\frac{2}{1-{{x}^{2}}}}-1)dx=[\ln \frac{1+x}{1-x}-x]|_{0}^{\frac{1}{2}}=\ln 3-\frac{1}{2}$

5：

解：$\int\limits_{0}^{2}{dx\int\limits_{x}^{2}{{{e}^{-{{y}^{2}}}}}dy}=\int_{0}^{2}{dy\int_{0}^{y}{{{e}^{-{{y}^{2}}}}}}dx=\int_{0}^{2}{y{{e}^{-{{y}^{2}}}}}dy=\frac{1-{{e}^{-4}}}{2}$

6：

解：不妨设${{a}_{n}}=\frac{({{n}^{2}}-1)}{{{2}^{n}}}$

由于$R=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{{{a}_{n}}}{{{a}_{n+1}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{{{n}^{2}}-1}{{{(n+1)}^{2}}-1}\cdot 2=2$

且$\sum\limits_{n=0}^{+\infty }{({{n}^{2}}}-1)$和$\sum\limits_{n=0}^{+\infty }{{{(-1)}^{n}}({{n}^{2}}}-1)$均发散

从而该级数的收敛区间为$(-2,2)$

设\[S(x)=\sum\limits_{n=0}^{+\infty }{\frac{({{n}^{2}}-1)}{{{2}^{n}}}{{x}^{n}}},x\in (-2,2)\]

则$S(x)=\sum\limits_{n=0}^{+\infty }{(n-1)(n+1){{(\frac{x}{2})}^{n}}}$

先积分一次后两边同除$\frac{x}{2}$可得：

\[\sum\limits_{n=0}^{+\infty }{\frac{({{n}^{2}}-1)}{{{2}^{n}}}{{x}^{n}}}=\frac{4(3x-2)}{{{(2-x)}^{3}}},x\in (-2,2)\]

二、解：设切点为$({{x}_{0}},{{y}_{0}},{{z}_{0}})$，设$f(x,y,z)=\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}+\frac{{{z}^{2}}}{{{c}^{2}}}$

从而${{f}_{x}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{x}_{0}}}{{{a}^{2}}},{{f}_{y}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{y}_{0}}}{{{b}^{2}}},{{f}_{z}}({{x}_{0}},{{y}_{0}},{{z}_{0}})=\frac{2{{z}_{0}}}{{{c}^{2}}}$

从而$\pi $的表达式为$\frac{2{{x}_{0}}}{{{a}^{2}}}(x-{{x}_{0}})+\frac{2{{y}_{0}}}{{{b}^{2}}}(y-{{y}_{0}})+\frac{2{{z}_{0}}}{{{c}^{2}}}(z-{{z}_{0}})=0$ 

且$\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}=1$，代入化简得：$\frac{{{x}_{0}}}{{{a}^{2}}}x+\frac{{{y}_{0}}}{{{b}^{2}}}y+\frac{{{z}_{0}}}{{{c}^{2}}}z=1$

于是$\pi $在第一象限的部分与三个坐标的坐标分别为

$(\frac{{{a}^{2}}}{{{x}_{0}}},0,0),(0,\frac{{{b}^{2}}}{{{y}_{0}}},0),(0,0,\frac{{{c}^{2}}}{{{z}_{0}}})$，可知${{x}_{0}},{{y}_{0}},{{z}_{0}}>0$

于是$V=\frac{1}{6}\cdot \frac{{{a}^{2}}}{{{x}_{0}}}\cdot \frac{{{b}^{2}}}{{{y}_{0}}}\cdot \frac{{{c}^{2}}}{{{z}_{0}}}$，且$\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}=1$

由广义均值不等式知：\[\frac{x_{0}^{2}}{{{a}^{2}}}+\frac{y_{0}^{2}}{{{b}^{2}}}+\frac{z_{0}^{2}}{{{c}^{2}}}\ge 3\sqrt[3]{\frac{x_{0}^{2}}{{{a}^{2}}}\cdot \frac{y_{0}^{2}}{{{b}^{2}}}\cdot \frac{z_{0}^{2}}{{{c}^{2}}}}\]  

当且仅当${{x}_{0}}=\frac{\sqrt{3}}{3}a,{{y}_{0}}=\frac{\sqrt{3}}{3}b,{{z}_{0}}=\frac{\sqrt{3}}{3}c$等号成立

于是当$\pi $的方程为$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\sqrt{3}$时，${{V}_{\min }}=\frac{\sqrt{3}}{2}abc$

三、解：由于

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}$

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial y}=-\frac{1}{\sqrt{y}}\frac{\partial z}{\partial u}+\frac{1}{\sqrt{y}}\frac{\partial z}{\partial v}$

\[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}=[\frac{\partial (\frac{\partial z}{\partial u})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial u})}{\partial v}\cdot \frac{\partial v}{\partial x}]+[\frac{\partial (\frac{\partial z}{\partial v})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial v})}{\partial v}\cdot \frac{\partial v}{\partial x}]=\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}+2\frac{{{\partial }^{2}}z}{\partial u\partial v}+\frac{{{\partial }^{2}}z}{\partial {{v}^{2}}}\]

\[\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=\frac{1}{2y\sqrt{y}}\frac{\partial z}{\partial u}-\frac{1}{\sqrt{y}}[\frac{\partial (\frac{\partial z}{\partial u})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial u})}{\partial v}\cdot \frac{\partial v}{\partial x}]-\frac{1}{2y\sqrt{y}}\frac{\partial z}{\partial u}+\frac{1}{\sqrt{y}}[\frac{\partial (\frac{\partial z}{\partial v})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial v})}{\partial v}\cdot \frac{\partial v}{\partial x}]\]\[=\frac{1}{y}(\frac{{{\partial }^{2}}z}{\partial {{u}^{2}}}-2\frac{{{\partial }^{2}}z}{\partial u\partial v}+\frac{{{\partial }^{2}}z}{\partial {{v}^{2}}})+\frac{1}{2y\sqrt{y}}(\frac{\partial z}{\partial u}-\frac{\partial z}{\partial u})\]

代入方程，得到：$4\frac{{{\partial }^{2}}z}{\partial u\partial v}=0\Rightarrow \frac{{{\partial }^{2}}z}{\partial u\partial v}=0$

四、解：

（1）不妨设$G(x,y,z)=F(x-y,y-xz)$，则当$x\ne 0$时，${{G}_{z}}=-x{{F}_{\eta }}(x-y,y-xz)\ne 0$

于是由隐函数定理可知：方程$F(x-y,y-xz)=0$在$x\ne 0$附近唯一确定隐函数

$z=z(x,y)$

（2）对方程两边求导可得：

$\left\{\begin{array}{ll}
{F_\xi } - {F_\eta }(z + x{z_x}) = 0 \\
- {F_\xi } + {F_\eta }(1 - x{z_y}) = 0 \\
- {F_{\xi \xi }} + {F_{\xi \eta }}(1 - x{z_y} + z + x{z_x}) + {F_{\eta \eta }}({x^2}{z_x}{z_y} - z - x{z_x} + xz{z_y}) - {F_\eta }({z_y} + x{z_{xy}}) = 0
\end{array}
\right.$

解得

$\left\{\begin{array}{ll}
{z_x} = \frac{{{F_\xi } - z{F_\eta }}}{{x{F_\eta }}} \\
{z_y} = \frac{{{F_\eta } - {F_\xi }}}{{x{F_\eta }}} \\
{z_{xy}} = \frac{{ - xF_\eta ^2{F_{\xi \xi }} + 2x{F_\xi }{F_\eta }{F_{\xi \eta }} - xF_\xi ^2{F_{\eta \eta }} - F_\eta ^2 + {F_\xi }{F_\eta }}}{{{x^2}F_\eta ^3}}
\end{array}
\right.$

五、证明：由（3）可知：

$\exists M>0$，使得$\left| {{b}_{n}}(x) \right|\le M,x\in [a,b],n\in {{N}^{*}}$

由（2）可知：

对$\forall \varepsilon >0$，$\exists N>0$，对$\forall n.m>N,x\in [a,b]$，都有$\left| \sum\limits_{k=n+1}^{m}{{{a}_{k}}(x)} \right|<\frac{\varepsilon }{M}$

于是对$\forall \varepsilon >0,\exists N>0$，对$\forall n.m>N,x\in [a,b]$，有

\[\sum\limits_{k=n+1}^{m}{\left| {{a}_{k}}(x){{b}_{k}}(x) \right|\le M\sum\limits_{k=n+1}^{m}{{{a}_{k}}(x)}}<\varepsilon \]

由柯西收敛准则可知：$\sum\limits_{n=1}^{+\infty }{|{{a}_{n}}(x)}{{b}_{n}}(x)|$在$[a,b]$上一致收敛

六、证明：令$G(x)=\int_{1}^{x}{{{t}^{2}}f(t)dt}$，由微分中值定理可知：

$\exists \xi \in (1,2)\subset [1,3]$，使得\[G(2)-G(1)=G'(\xi )\Rightarrow \int_{1}^{2}{{{x}^{2}}f(x)dx=}{{\xi }^{2}}f(\xi )\]

令$F(x)={{x}^{2}}f(x),x\in [1,3]$

由题可知：$F(1)=F(\xi )=F(3)$

于是由罗尔中值定理可知：存在$1<{{\xi }_{1}}<\xi <{{\xi }_{2}}<3$，使得$2f(x)+xf'(x)=0$

于是存在${{\xi }_{\text{1}}},{{\xi }_{2}}\in (1,3)$，${{\xi }_{1}}\ne {{\xi }_{2}}$，满足恒等式$2f(x)+xf'(x)=0$。

（注意，此题也可以利用积分中值定理，但是需要讨论）

七、证明：

（1）先证明$f'(x)$无第一类间断点

证明：反证法，若$f'(x)$在$x=a$处是第一类间断点，则$f'(x)$在$x=a$处存在左右极限

由中值定理可知，$f'(a)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\frac{f(x)-f(a)}{x-a}=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f'(\xi ),x<\xi <a$

于是当$x\to {{a}^{-}}$时$\xi \to {{a}^{-}}$

则$f'(a)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f'(\xi )=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f'(x)$

即$f'(x)$在$x=a$处左连续

同理可证：$f'(x)$在$x=a$处右连续

于是$f'(x)$在$x=a$处连续矛盾

从而$f'(x)$无第一类间断点

（2）单调函数的间断点只可能为第一类跳跃的

证明：若$f(x)$在${{U}^{0}}({{x}_{0}})$上为单调增函数，取${{U}^{0}}({{x}_{0}})\subset I$，$\exists {{x}_{1}},{{x}_{2}}\in I$，使得${{x}_{1}}<x<{{x}_{2}}$

则$f(x)$在${{U}^{0}}({{x}_{0}})$上为有界函数，于是有

$f({{x}_{0}}+0)=\underset{x\in {{U}^{0}}({{x}_{0}})}{\mathop{\inf }}\,f(x),f({{x}_{0}}-0)=\underset{x\in {{U}^{0}}({{x}_{0}})}{\mathop{\sup }}\,f(x)$

即$f(x)$在${{x}_{0}}$的左、右极限均存在

因此若${{x}_{0}}$为$f(x)$的间断点，必为$f(x)$的第一类间断点

若$f(x)$为单调减函数，只需令$F(x)=-f(x)$，同理可证

于是单调函数的间断点只可能为第一类跳跃的

由（1）（2）可知，单调的导函数一定是连续的，于是$f'(x)$在$[a,b]$上连续

则$f'(x)$在$[a,b]$上有界，即存在$M>0$，对一切$x\in [a,b]$，都有$\left| f'(x) \right|\le M$

于是对任意$a\le x,y\le b$，由积分中值定理可知：

存在$\xi \in (a,b)$，使得$\left| f(y)-f(x) \right|=\left| f'(\xi ) \right|\left| y-x \right|\le M\left| y-x \right|$

于是令$L=M$，即证

八、解：记$\Sigma $所围的立体为$\Omega $，由散度定理和球坐标变换可知：

设$x=r\sin \varphi \cos \theta ,y=r\sin \varphi \sin \theta ,z=r\cos \varphi ,0\le \theta \le 2\pi ,0\le \varphi \le \frac{\pi }{4},a\le r\le 2a$

于是$I=\iint\limits_{\Sigma }{xydydz+yzdzdx+z\sqrt{{{x}^{2}}+{{y}^{2}}}}dxdy=\iiint_{\Omega }{(y+z+\sqrt{{{x}^{2}}+{{y}^{2}}})dxdydz}$

$\text{=}\int_{0}^{2\pi }{d\theta \int_{0}^{\frac{\pi }{4}}{d\varphi \int_{a}^{2a}{(r\sin \theta \sin \varphi +r\cos \theta +r\sin \theta }}}){{r}^{2}}\sin \theta dr$

$=\frac{3\pi {{a}^{4}}}{2}\int_{0}^{\frac{\pi }{4}}{(\cos \theta \sin \theta +{{\sin }^{2}}\theta )d\theta =\frac{3{{\pi }^{2}}{{a}^{4}}}{16}}$

九、证明：由题知，对任意的$n\in {{N}^{*}},\left| {{a}_{n}} \right|\le \frac{M}{{{n}^{\alpha }}},\left| {{b}_{n}} \right|\le \frac{M}{{{n}^{\alpha }}}$

且当$\alpha >1$时，$\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{+\infty }{(\left| {{a}_{n}} \right|+\left| {{b}_{n}} \right|)}<+\infty $

于是由$M$判别法可知，$\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{+\infty }{({{a}_{n}}\cos nx+{{b}_{n}}\sin nx)}$一致收敛，进一步可知，其和函数连续

当$\alpha >2$时，$\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{+\infty }{(n\left| {{a}_{n}} \right|+n\left| {{b}_{n}} \right|)}<+\infty $

于是由$M$判别法可知，$\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{+\infty }{(-n{{a}_{n}}\sin nx+n{{b}_{n}}\cos nx)}$一致收敛，进一步可知，其和函数有有连续的导函数

十、证明：证明：由$f(x)$的连续性知，存在$M>1$,使得对$\forall x\in \left[ 0,1 \right]$有$\left| f(x) \right|\le M$，又由$\frac{2}{\pi }\int_{0}^{+\infty }{\frac{1}{{{y}^{2}}+1}dy=1}$知，对$\forall \varepsilon >0$，

$\exists N$，使得当$n>N$.时，有

$\left| \frac{2}{\pi }\int_{n}^{+\infty }{\frac{dy}{{{y}^{2}}+1}} \right|<\frac{\varepsilon }{3(M+\left| f(0) \right|)}$

再由f(x)的连续性知，对上述$\varepsilon >0,\exists \delta >0$（不妨设$\delta <1$），使得对$\forall x\in {{U}_{+}}(0,\delta )$，

有$\left| f(x)-f(0) \right|<\frac{\varepsilon }{3}$ 令$N=\left[ \frac{{{N}_{1}}}{\delta } \right] $，则当$n>N$时，有

$\left| \frac{2}{\pi }\int_{0}^{1}{\frac{n}{{{n}^{2}}{{x}^{2}}+1}f(x)dx}-f(0) \right|=\left| \frac{2}{\pi }\int_{0}^{n}{\frac{f(\frac{y}{n})-f(0)}{{{y}^{2}}+1}dy-\frac{2}{\pi }\int_{n}^{+\infty }{\frac{f(0)dy}{{{y}^{2}}+1}}} \right|$

$\le \frac{2}{\pi }\int_{0}^{{{N}_{1}}}{\frac{\left| f(\frac{y}{n})-f(0) \right|}{{{y}^{2}}+1}dy+}\frac{2}{\pi }\int_{{{N}_{1}}}^{n}{\frac{\left| f(\frac{y}{n})-f(0) \right|}{{{y}^{2}}+1}dy+}\frac{2}{\pi }\left| f(0) \right|\int_{n}^{+\infty }{\frac{dy}{{{y}^{2}}+1}}$

$<\frac{\varepsilon }{3}\frac{2}{\pi }\int_{0}^{+\infty }{\frac{dy}{{{y}^{2}}+1}+(M+\left| f(0) \right|)\frac{2}{\pi }\int_{{{N}_{1}}}^{+\infty }{\frac{dy}{{{y}^{2}}+1}+\left| f(0) \right|\frac{2}{\pi }\int_{n}^{+\infty }{\frac{dy}{{{y}^{2}}+1}}}}$

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

由此知，$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{2}{\pi }\int_{0}^{1}{\frac{n}{{{n}^{2}}{{x}^{2}}+1}f(x)dx}=f(0)$

于是$\underset{n\to +\infty }{\mathop{\lim }}\,\int\limits_{0}^{1}{\frac{n}{1+{{n}^{2}}{{x}^{2}}}}f(x)dx=\frac{\pi }{2}f(0)$

 .