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$1le {{(n!)}^{frac{1}{{{n}^{2}}}}}le sqrt[n]{n}$
而$underset{n o +infty }{mathop{lim }}\,sqrt[n]{n}=1$由迫敛性知：$underset{n o +infty }{mathop{lim }}\,{{(n!)}^{frac{1}{{{n}^{2}}}}}=1$
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$underset{x o +infty }{mathop{lim }}\,{{x}^{alpha }}(sqrt{{{x}^{2}}+1}+sqrt{{{x}^{2}}-1}-2x)=underset{x o +infty }{mathop{lim }}\,frac{-2{{x}^{alpha -3}}}{(sqrt{{{x}^{2}}+1}+x)(sqrt{{{x}^{2}}-1}+x)(sqrt{{{x}^{2}}-1}+sqrt{{{x}^{2}}+1})}$
于是
$eta =left{egin{array}{ll}
0, & hbox{$alpha<3$;} \
- frac{1}{4}, & hbox{$alpha=3$;} \
+ infty, & hbox{$alpha>3$.}
end{array}
ight.$
3：
解：利用高阶导数的公式可知：
${{y}^{(50)}}(x)=C_{50}^{0}cdot {{x}^{2}}cdot {{(cos 3x)}^{(50)}}+C_{50}^{1}cdot 2xcdot {{(cos 3x)}^{(49)}}+C_{50}^{2}cdot 2cdot {{(cos 3x)}^{(48)}}$
$={{3}^{48}}[(2450-9{{x}^{2}})cos 3x-300xsin 3x]$
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：$intlimits_{0}^{2}{dxintlimits_{x}^{2}{{{e}^{-{{y}^{2}}}}}dy}=int_{0}^{2}{dyint_{0}^{y}{{{e}^{-{{y}^{2}}}}}}dx=int_{0}^{2}{y{{e}^{-{{y}^{2}}}}}dy=frac{1-{{e}^{-4}}}{2}$
6：
解：不妨设${{a}_{n}}=frac{({{n}^{2}}-1)}{{{2}^{n}}}$
由于$R=underset{n o +infty }{mathop{lim }}\,frac{{{a}_{n}}}{{{a}_{n+1}}}=underset{n o +infty }{mathop{lim }}\,frac{{{n}^{2}}-1}{{{(n+1)}^{2}}-1}cdot 2=2$
且$sumlimits_{n=0}^{+infty }{({{n}^{2}}}-1)$和$sumlimits_{n=0}^{+infty }{{{(-1)}^{n}}({{n}^{2}}}-1)$均发散
从而该级数的收敛区间为$(-2,2)$
设[S(x)=sumlimits_{n=0}^{+infty }{frac{({{n}^{2}}-1)}{{{2}^{n}}}{{x}^{n}}},xin (-2,2)]
则$S(x)=sumlimits_{n=0}^{+infty }{(n-1)(n+1){{(frac{x}{2})}^{n}}}$
先积分一次后两边同除$frac{x}{2}$可得：
[sumlimits_{n=0}^{+infty }{frac{({{n}^{2}}-1)}{{{2}^{n}}}{{x}^{n}}}=frac{4(3x-2)}{{{(2-x)}^{3}}},xin (-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 3sqrt[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}}}+2frac{{{partial }^{2}}z}{partial upartial v}+frac{{{partial }^{2}}z}{partial {{v}^{2}}}]
[frac{{{partial }^{2}}z}{partial {{y}^{2}}}=frac{1}{2ysqrt{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}{2ysqrt{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}}}-2frac{{{partial }^{2}}z}{partial upartial v}+frac{{{partial }^{2}}z}{partial {{v}^{2}}})+frac{1}{2ysqrt{y}}(frac{partial z}{partial u}-frac{partial z}{partial u})]
代入方程，得到：$4frac{{{partial }^{2}}z}{partial upartial v}=0Rightarrow frac{{{partial }^{2}}z}{partial upartial v}=0$
四、解：
（1）不妨设$G(x,y,z)=F(x-y,y-xz)$，则当$x e 0$时，${{G}_{z}}=-x{{F}_{eta }}(x-y,y-xz) e 0$
于是由隐函数定理可知：方程$F(x-y,y-xz)=0$在$x e 0$附近唯一确定隐函数
$z=z(x,y)$
（2）对方程两边求导可得：
$left{egin{array}{ll}
{F_xi } - {F_eta }(z + x{z_x}) = 0 \
- {F_xi } + {F_eta }(1 - x{z_y}) = 0 \
- {F_{xi xi }} + {F_{xi eta }}(1 - x{z_y} + z + x{z_x}) + {F_{eta eta }}({x^2}{z_x}{z_y} - z - x{z_x} + xz{z_y}) - {F_eta }({z_y} + x{z_{xy}}) = 0
end{array}
ight.$
解得
$left{egin{array}{ll}
{z_x} = frac{{{F_xi } - z{F_eta }}}{{x{F_eta }}} \
{z_y} = frac{{{F_eta } - {F_xi }}}{{x{F_eta }}} \
{z_{xy}} = frac{{ - xF_eta ^2{F_{xi xi }} + 2x{F_xi }{F_eta }{F_{xi eta }} - xF_xi ^2{F_{eta eta }} - F_eta ^2 + {F_xi }{F_eta }}}{{{x^2}F_eta ^3}}
end{array}
ight.$
五、证明：由（3）可知：
$exists M>0$，使得$left| {{b}_{n}}(x) ight|le M,xin [a,b],nin {{N}^{*}}$
由（2）可知：
对$forall varepsilon >0$，$exists N>0$，对$forall n.m>N,xin [a,b]$，都有$left| sumlimits_{k=n+1}^{m}{{{a}_{k}}(x)} ight|<frac{varepsilon }{M}$
于是对$forall varepsilon >0,exists N>0$，对$forall n.m>N,xin [a,b]$，有
[sumlimits_{k=n+1}^{m}{left| {{a}_{k}}(x){{b}_{k}}(x) ight|le Msumlimits_{k=n+1}^{m}{{{a}_{k}}(x)}}<varepsilon ]
由柯西收敛准则可知：$sumlimits_{n=1}^{+infty }{|{{a}_{n}}(x)}{{b}_{n}}(x)|$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$F(x)={{x}^{2}}f(x),xin [1,3]$
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${{xi }_{ ext{1}}},{{xi }_{2}}in (1,3)$，${{xi }_{1}} e {{xi }_{2}}$，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，$f'(a)=underset{x o {{a}^{-}}}{mathop{lim }}\,frac{f(x)-f(a)}{x-a}=underset{x o {{a}^{-}}}{mathop{lim }}\,f'(xi ),x<xi <a$
于是当$x o {{a}^{-}}$时$xi o {{a}^{-}}$
则$f'(a)=underset{x o {{a}^{-}}}{mathop{lim }}\,f'(xi )=underset{x o {{a}^{-}}}{mathop{lim }}\,f'(x)$
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
$f({{x}_{0}}+0)=underset{xin {{U}^{0}}({{x}_{0}})}{mathop{inf }}\,f(x),f({{x}_{0}}-0)=underset{xin {{U}^{0}}({{x}_{0}})}{mathop{sup }}\,f(x)$
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$xin [a,b]$，都有$left| f'(x) ight|le M$
于是对任意$ale x,yle b$，由积分中值定理可知：
存在$xi in (a,b)$，使得$left| f(y)-f(x) ight|=left| f'(xi ) ight|left| y-x ight|le Mleft| y-x ight|$
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=rsin varphi cos heta ,y=rsin varphi sin heta ,z=rcos varphi ,0le heta le 2pi ,0le varphi le frac{pi }{4},ale rle 2a$
于是$I=iintlimits_{Sigma }{xydydz+yzdzdx+zsqrt{{{x}^{2}}+{{y}^{2}}}}dxdy=iiint_{Omega }{(y+z+sqrt{{{x}^{2}}+{{y}^{2}}})dxdydz}$
$ ext{=}int_{0}^{2pi }{d heta int_{0}^{frac{pi }{4}}{dvarphi int_{a}^{2a}{(rsin heta sin varphi +rcos heta +rsin heta }}}){{r}^{2}}sin heta dr$
$=frac{3pi {{a}^{4}}}{2}int_{0}^{frac{pi }{4}}{(cos heta sin heta +{{sin }^{2}} heta )d heta =frac{3{{pi }^{2}}{{a}^{4}}}{16}}$
九、证明：由题知，对任意的$nin {{N}^{*}},left| {{a}_{n}} ight|le frac{M}{{{n}^{alpha }}},left| {{b}_{n}} ight|le frac{M}{{{n}^{alpha }}}$
且当$alpha >1$时，$frac{{{a}_{0}}}{2}+sumlimits_{n=1}^{+infty }{(left| {{a}_{n}} ight|+left| {{b}_{n}} ight|)}<+infty $
于是由$M$判别法可知，$frac{{{a}_{0}}}{2}+sumlimits_{n=1}^{+infty }{({{a}_{n}}cos nx+{{b}_{n}}sin nx)}$一致收敛，进一步可知，其和函数连续
当$alpha >2$时，$frac{{{a}_{0}}}{2}+sumlimits_{n=1}^{+infty }{(nleft| {{a}_{n}} ight|+nleft| {{b}_{n}} ight|)}<+infty $
于是由$M$判别法可知，$frac{{{a}_{0}}}{2}+sumlimits_{n=1}^{+infty }{(-n{{a}_{n}}sin nx+n{{b}_{n}}cos nx)}$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$forall xin left[ 0,1 ight]$有$left| f(x) ight|le M$，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
$left| frac{2}{pi }int_{n}^{+infty }{frac{dy}{{{y}^{2}}+1}} ight|<frac{varepsilon }{3(M+left| f(0) ight|)}$
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$forall xin {{U}_{+}}(0,delta )$，
有$left| f(x)-f(0) ight|<frac{varepsilon }{3}$ 令$N=left[ frac{{{N}_{1}}}{delta } ight] $，则当$n>N$时，有
$left| frac{2}{pi }int_{0}^{1}{frac{n}{{{n}^{2}}{{x}^{2}}+1}f(x)dx}-f(0) ight|=left| frac{2}{pi }int_{0}^{n}{frac{f(frac{y}{n})-f(0)}{{{y}^{2}}+1}dy-frac{2}{pi }int_{n}^{+infty }{frac{f(0)dy}{{{y}^{2}}+1}}} ight|$
$le frac{2}{pi }int_{0}^{{{N}_{1}}}{frac{left| f(frac{y}{n})-f(0) ight|}{{{y}^{2}}+1}dy+}frac{2}{pi }int_{{{N}_{1}}}^{n}{frac{left| f(frac{y}{n})-f(0) ight|}{{{y}^{2}}+1}dy+}frac{2}{pi }left| f(0) ight|int_{n}^{+infty }{frac{dy}{{{y}^{2}}+1}}$
$<frac{varepsilon }{3}frac{2}{pi }int_{0}^{+infty }{frac{dy}{{{y}^{2}}+1}+(M+left| f(0) ight|)frac{2}{pi }int_{{{N}_{1}}}^{+infty }{frac{dy}{{{y}^{2}}+1}+left| f(0) ight|frac{2}{pi }int_{n}^{+infty }{frac{dy}{{{y}^{2}}+1}}}}$
$<frac{varepsilon }{3}+frac{varepsilon }{3}+frac{varepsilon }{3}=varepsilon $
由此知，$underset{n o +infty }{mathop{lim }}\,frac{2}{pi }int_{0}^{1}{frac{n}{{{n}^{2}}{{x}^{2}}+1}f(x)dx}=f(0)$
于是$underset{n o +infty }{mathop{lim }}\,intlimits_{0}^{1}{frac{n}{1+{{n}^{2}}{{x}^{2}}}}f(x)dx=frac{pi }{2}f(0)$
.