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华中师范大学2012年数学分析试题解答
一、
(1)证明:由${{x}_{1}}=frac{1}{2},{{x}_{2}}=frac{3}{8},{{x}_{3}}=frac{55}{128},cdots $,猜测${{{x}_{2n+1}}}$单调递减,${{{x}_{2n}}}$单调递增
下用数归法先证$sqrt{2}-1le {{x}_{2n+1}}le frac{1}{2},frac{3}{8}le {{x}_{2n+1}}le sqrt{2}-1$
(1)当$n=1$时,$sqrt{2}-1<{{x}_{1}}=frac{1}{2}$恒成立
(2)设$n=2k-1$时,$sqrt{2}-1le {{x}_{2k-1}}le frac{1}{2}$,则
[{{x}_{2k+1}}=frac{1}{2}(1-x_{2k}^{2})=frac{1}{2}[1-frac{1}{4}{{(1-x_{2k-1}^{2})}^{2}}]in [sqrt{2}-1,frac{1}{2})]
即当$n=2k+1$时也恒成立
由(1)(2)可知:$sqrt{2}-1le {{x}_{2n+1}}le frac{1}{2}$
同理可证$frac{3}{8}le {{x}_{2n+1}}le sqrt{2}-1$
再证${{{x}_{2n+1}}}$单调递减,${{{x}_{2n}}}$单调递增,同样采用数学归纳法
(1)当$n=1$时,${{x}_{1}}=frac{1}{2}>frac{55}{128}={{x}_{3}}$恒成立
(2)设$n=2k-1$时,${{x}_{2k-1}}ge {{x}_{2k+1}}$,则
${{x}_{2k+3}}-{{x}_{2k+1}}=frac{1}{8}[{{(1-x_{2k-1}^{2})}^{2}}-{{(1-x_{2k+1}^{2})}^{2}}]=frac{1}{8}(x_{2k+1}^{2}-x_{2k-1}^{2})(2-x_{2k+1}^{2}-x_{2k-1}^{2})le 0$
即当$n=2k+1$时也恒成立
由(1)(2)可知:${{{x}_{2n+1}}}$单调递减
同理可证:${{{x}_{2n}}}$单调递增
于是${{{x}_{2n+1}}}$单调递减,$sqrt{2}-1le {{x}_{2n+1}}le frac{1}{2}$,由单调有界原理可知:${{{x}_{2n+1}}}$收敛
于是不妨设$underset{nto +infty }{mathop{lim }},{{x}_{2n+1}}=lin [sqrt{2}-1,frac{1}{2}]$,由${{x}_{2k+3}}=frac{1}{2}[1-frac{1}{4}{{(1-x_{2k+1}^{2})}^{2}}]$,两边对$nto +infty $
则$l=frac{1}{2}[1-frac{1}{4}{{(1-{{l}^{2}})}^{2}}]$,求得$l=sqrt{2}-1$
即$underset{xto infty }{mathop{lim }},{{x}_{2n+1}}=sqrt{2}-1=A$
同理可证$underset{xto infty }{mathop{lim }},{{x}_{2n}}=sqrt{2}-1=A$
于是$underset{nto +infty }{mathop{lim }},{{x}_{n}}=underset{nto +infty }{mathop{lim }},{{x}_{2n}}=underset{nto +infty }{mathop{lim }},{{x}_{2n+1}}=A$
(3)证明:不妨设${{a}_{n}}={{x}_{n}}-A={{x}_{n}}-(sqrt{2}-1)$
于是
$underset{nto +infty }{mathop{lim }},left| frac{{{a}_{n+1}}}{{{a}_{n}}} right|=underset{nto +infty }{mathop{lim }},left| frac{{{x}_{n+1}}-A}{{{x}_{n}}-A} right|=underset{nto +infty }{mathop{lim }},left| frac{frac{1}{2}(1-x_{n}^{2})-A}{{{x}_{n}}-A} right|=underset{nto +infty }{mathop{lim }},frac{1}{2}left| frac{1-x_{n}^{2}-(2sqrt{2}-2)}{{{x}_{n}}-(sqrt{2}-1)} right|=underset{nto +infty }{mathop{lim }},frac{1}{2}left| frac{x_{n}^{2}-{{(sqrt{2}-1)}^{2}}}{{{x}_{n}}-(sqrt{2}-1)} right|=A<1 $
于是$sumlimits_{n=1}^{+infty }{({{x}_{n}}}-A)$绝对收敛
二、
(1)成立,理由如下:
要证明${{sin }^{3}}left| f(x) right|$在$I$上一致连续,只需证明$left| f(x) right|$在$I$上一致连续即可
由于$f(x)$在$I$上一致连续,则对任意的$varepsilon >0$,任意的${{x}_{1}},{{x}_{2}}in I$,存在$delta >0$,当
$left| {{x}_{1}}-{{x}_{2}} right|<delta $时,有$left| f({{x}_{1}})-f({{x}_{2}}) right|<varepsilon $
于是对任意的$varepsilon >0$,任意的${{x}_{1}},{{x}_{2}}in I$,存在$delta >0$,当
$left| {{x}_{1}}-{{x}_{2}} right|<delta $时,有$left| left| f({{x}_{1}}) right|-left| f({{x}_{2}}) right| right|le left| f({{x}_{1}})-f({{x}_{2}}) right|<varepsilon $
即$left| f(x) right|$在$I$上一致连续即可
对任意的$varepsilon >0$,任意的${{x}_{1}},{{x}_{2}}in I$,存在$delta >0$,当
$left| {{x}_{1}}-{{x}_{2}} right|<delta $时,有
$left| {{sin }^{3}}left| f({{x}_{1}}) right|-{{sin }^{3}}left| f({{x}_{1}}) right| right|=left| (sin left| f({{x}_{1}}) right|-sin left| f({{x}_{1}}) right|)({{sin }^{2}}left| f({{x}_{1}}) right|+sin left| f({{x}_{1}}) right|sin left| f({{x}_{2}}) right|+{{sin }^{2}}left| f({{x}_{2}}) right| right|$$le 3left| sin left| f({{x}_{1}}) right|-sin left| f({{x}_{1}}) right| right|le 3left| sin f({{x}_{1}})-sin f({{x}_{2}}) right|le 3left| f({{x}_{1}})-f({{x}_{2}}) right|$
于是由复合函数的一致收敛性可知:${{sin }^{3}}left| f(x) right|$在$I$上一致连续
(4)成立,理由如下
要证明${{sin }^{3}}f(x)$在$I$上一致连续,只需证明$f(x)$在$I$上一致连续即可
由于$left| f(x) right|$在$I$上一致连续,则对任意的$varepsilon >0$,任意的${{x}_{1}},{{x}_{2}}in I$,存在$delta >0$,当
$left| {{x}_{1}}-{{x}_{2}} right|<delta $时,由$left| left| f({{x}_{1}}) right|-left| f({{x}_{2}}) right| right|<frac{varepsilon }{2}$
1:若$f({{x}_{1}}),f({{x}_{2}})$同号,则有$left| f({{x}_{1}})-f({{x}_{2}}) right|<frac{varepsilon }{2}<varepsilon $
2:若$f({{x}_{1}}),f({{x}_{2}})$同号,由$f(x)$连续,则存在$y$在${{x}_{1}},{{x}_{2}}$之间使得$f(y)=0$
于是$left| f({{x}_{1}})-f({{x}_{2}}) right|le left| f({{x}_{1}}) right|+left| f({{x}_{2}}) right|=left| left| f({{x}_{1}}) right|-left| f(y) right| right|+left| left| f({{x}_{2}}) right|-left| f(y) right| right|<varepsilon $
由1,2可知,$f(x)$在$I$上一致连续
再利用复合函数的一致连续性可知,${{sin }^{3}}f(x)$在$I$上一致连续
三、证明:
充分性:反证法:假设对任意的$xin (a,b)$都有$f'(x)le frac{f(b)-f(a)}{b-a}$
令$g(x)=f(x)-frac{f(b)-f(a)}{b-a}(x-a)$
则当$xin (a,b)$时有$g'(x)=f'(x)-frac{f(b)-f(a)}{b-a}le 0$
于是$g(x)$在$[a,b]$上单调递减
而$g(a)=g(b)=f(a)$
从而当$xin [a,b]$时,$g(x)=f(a)$
于是$f(x)=frac{f(b)-f(a)}{b-a}(x-a)+f(a)$矛盾
从而必存在$xi in (a,b)$,使得$f'(xi )>frac{f(b)-f(a)}{b-a}$
必要性:
令$g(x)=f(x)-frac{f(b)-f(a)}{b-a}(x-a)$
则$g(a)=g(b)=f(a)$
反证法:
1:若$f(x)$为常函数,则$g'(x)=f'(x)-frac{f(b)-f(a)}{b-a}=-frac{f(b)-f(a)}{b-a},xin [a,b]$
这与存在$xi in (a,b) $,使得$f'(xi )>frac{f(b)-f(a)}{b-a}$矛盾
2:若$f(x)$为线性函数,可知$g'(x)=f'(x)-frac{f(b)-f(a)}{b-a}$为常数
又由于存在$xi in (a,b) $,使得$f'(xi )>frac{f(b)-f(a)}{b-a}$可知,$g'(x)>0$
于是$g(x)$在$[a,b]$上严格单调递增
而$g(a)=g(b)=f(a)$矛盾
从而由1,2可知,$f(x)$不为常函数或线性函数
四、
(1)证明:设$Fleft( x,y right)={{x}^{2}}+y-cos left( xy right) $,
显然,有$Fleft( 0,1 right)=0 $,
${{F}_{y}}left( x,y right)=1+xsin left( xy right) $,
${{F}_{y}}left( 0,1 right)=1ne 0 $,由隐函数存在定理,
存在$delta >0 $,存在$left[ -delta ,delta right] $上的连续可微的函数$y=yleft( x right)$,$yleft( 0 right)=1 $,
满足$Fleft( x,yleft( x right) right)equiv 0 $,$xin U(0) $
(3)证明:由(1)可知:
${{F}_{x}}left( x,y right)=2x+ysin left( xy right) $,
${y}'left( x right)=-frac{{{F}_{x}}left( x,y right)}{{{F}_{y}}left( x,y right)}=-frac{2x+ysin left( xy right)}{1+xsin left( xy right)} $,
当$0<x<delta $,($delta >0 $充分小)时,有${y}'left( x right)<0 $,$yleft( x right) $在$left[ 0,delta right] $上严格单调递减;
当$-delta <x<0 $时,有${y}'left( x right)>0 $,$yleft( x right) $在$left[ -delta ,0 right] $上严格单调递增,
五、
(1)解:(1)由偏导数的定义:
${{f}_{x}}(0,0)=underset{Delta xto 0}{mathop{lim }},frac{f(Delta x,0)-f(0,0)}{Delta x}=underset{Delta xto 0}{mathop{lim }},Delta xcos frac{1}{left| Delta x right|}=0 $
${{f}_{y}}(0,0)=underset{Delta yto 0}{mathop{lim }},frac{f(0,Delta y)-f(0,0)}{Delta y}=underset{Delta yto 0}{mathop{lim }},Delta ycos frac{1}{left| Delta y right|}=0 $
(2)当$(x,y)ne (0,0)$时,
${{f}_{x}}(x,y)=2xcos frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}}+frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}}}sin frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}} $
${{f}_{y}}(x,y)=2ycos frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}}+frac{y}{sqrt{{{x}^{2}}+{{y}^{2}}}}sin frac{1}{sqrt{{{x}^{2}}+{{y}^{2}}}} $
于是${f_x}(x,y)=left{begin{array}{ll}
2xcos frac{1}{{sqrt {{x^2} + {y^2}} }} + frac{x}{{sqrt {{x^2} + {y^2}} }}sin frac{1}{{sqrt {{x^2} + {y^2}} }},& hbox{${x^2} + {y^2} ne 0$} \
0,& hbox{${x^2} + {y^2} = 0$.}
end{array}
right.$
${f_y}(x,y)=left{begin{array}{ll}
2ycos frac{1}{{sqrt {{x^2} + {y^2}} }} + frac{y}{{sqrt {{x^2} + {y^2}} }}sin frac{1}{{sqrt {{x^2} + {y^2}} }},,& hbox{${x^2} + {y^2} ne 0$} \
0,& hbox{${x^2} + {y^2} = 0$.}
end{array}
right.$
(2)设$y=kx$,于是$underset{xto 0}{mathop{lim }},frac{x}{sqrt{{{x}^{2}}+{{y}^{2}}}}=frac{1}{sqrt{1+{{k}^{2}}}}$与$k$有关
于是$underset{xto 0}{mathop{lim }},{{f}_{x}}(x,0) $不存在,故${{f}_{x}}(x,y) $在(0,0)不连续。
同理${{f}_{y}}(x,y) $ 在(0,0)也不连续。
(3)设$u=f(x,y) $,则在(0,0)点有
$Delta u-du=[f(Delta x,Delta y)-f(0,0)]-[{{f}_{x}}(0,0)Delta x+{{f}_{y}}(0,0)Delta y]=(Delta {{x}^{2}}+Delta {{y}^{2}})cos frac{1}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}} $
因$underset{Delta xto 0,Delta yto 0}{mathop{lim }},frac{Delta u-du}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}}=underset{Delta xto 0,Delta yto 0}{mathop{lim }},sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}cos frac{1}{sqrt{Delta {{x}^{2}}+Delta {{y}^{2}}}}=0$
故$f(x,y) $在(0,0)可微。
六、解:由于$y'=x+y,y(0)=1$
利用常数变易法求得$y=-1-x+2{{e}^{x}}$
记${{a}_{n}}=y(frac{1}{n})-1-frac{1}{n}=2({{e}^{frac{1}{n}}}-1-frac{1}{n})$
由$underset{nto +infty }{mathop{lim }},sqrt[n]{{{a}_{n}}}={{e}^{underset{nto +infty }{mathop{lim }},frac{ln {{a}_{n}}}{n}}}={{e}^{underset{nto +infty }{mathop{lim }},frac{ln 2({{e}^{frac{1}{n}}}-1-frac{1}{n})}{n}}}overset{n=frac{1}{x}}{mathop{=}},{{e}^{underset{xto +{{0}^{+}}}{mathop{lim }},frac{ln 2({{e}^{x}}-1-x)}{frac{1}{x}}}}={{e}^{-underset{xto +{{0}^{+}}}{mathop{lim }},frac{{{x}^{2}}({{e}^{x}}-1)}{{{e}^{x}}-1-x}}}={{e}^{-underset{xto +{{0}^{+}}}{mathop{lim }},frac{{{x}^{2}}[x+o(x)]}{frac{{{x}^{2}}}{2}+o({{x}^{2}})}}}=1 $
于是幂级数的收敛半径$R=1$
又由于$x=1$时,$y(frac{1}{n})-1-frac{1}{n}=2({{e}^{frac{1}{n}}}-1-frac{1}{n})=sumlimits_{n=1}^{+infty }{frac{2}{{{n}^{2}}}+}sumlimits_{n=1}^{+infty }{o(frac{2}{{{n}^{2}}}})$收敛
$x=-1$时,由莱布利兹判别法可知级数收敛
于是该幂级数的收敛域为$[-1,1]$
七、
(1)证明:令$u={{t}^{2}}Rightarrow t=sqrt{u},dt=frac{1}{2sqrt{u}}du$,于是$int_{x}^{x+c}{sin {{t}^{2}}}dt=frac{1}{2}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{sin u}{sqrt{u}}}du $
而函数$frac{1}{sqrt{u}}$在$[{{x}^{2}},{{(x+c)}^{2}}]$上递减,且$frac{1}{sqrt{u}}ge 0$,由积分第二中值定理可知:
存在$xi in [{{x}^{2}},{{(x+c)}^{2}}]$,使得
$int_{x}^{x+c}{sin {{t}^{2}}}dt=frac{1}{2}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{sin u}{sqrt{u}}}du=frac{1}{2x}int_{{{x}^{2}}}^{xi }{sin udu=frac{1}{2x}(cos {{x}^{2}}-cos xi )} $
故$left| int_{x}^{x+c}{sin {{t}^{2}}}dt right|le frac{1}{x}$
(2)能,理由如下
证明:
令$f(x)=int_{x}^{x+c}{sin {{t}^{2}}}dt=frac{1}{2}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{sin u}{sqrt{u}}}du=-frac{1}{2}frac{cos u}{sqrt{u}}|_{{{x}^{2}}}^{{{(x+c)}^{2}}}-frac{1}{4}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{sin u}{{{u}^{frac{3}{2}}}}}du$
$=frac{cos {{x}^{2}}}{2x}-frac{cos {{(x+c)}^{2}}}{2(x+c)}-frac{1}{4}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{sin u}{{{u}^{frac{3}{2}}}}}du$
存在${{u}_{0}}in [{{x}^{2}},{{(x+c)}^{2}}]$,使得$left| frac{cos {{u}_{0}}}{u_{0}^{frac{3}{2}}} right|<frac{1}{{{u}^{frac{3}{2}}}}$,因此当$x>0$时,有
$left| f(x) right|le left| frac{cos {{x}^{2}}}{2x} right|+left| frac{cos {{(x+c)}^{2}}}{2(x+c)} right|+frac{1}{4}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{left| frac{cos u}{{{u}^{frac{3}{2}}}} right|}du<frac{1}{2x}+frac{1}{2(x+c)}+frac{1}{4}int_{{{x}^{2}}}^{{{(x+c)}^{2}}}{frac{1}{{{u}^{frac{3}{2}}}}}du $
$=frac{1}{2x}+frac{1}{2(x+c)}+frac{1}{4}(-2{{u}^{-frac{1}{2}}})|_{{{x}^{2}}}^{{{(x+c)}^{2}}}=frac{1}{x}$
八、证明:对${{R}^{2}}$上任意一点$({{x}_{0}},{{y}_{0}})$,令${{L}_{1}}={x|{{(x-{{x}_{0}})}^{2}}={{r}^{2}}}$,方向取逆时针
由格林公式可知:
$int_{L}{Pdx+Qdy=}int_{L+{{L}_{1}}}{Pdx+Qdy=iint_{D}{[frac{partial Q}{partial x}}}-frac{partial P}{partial y}]dxdy=[frac{partial Q}{partial x}-frac{partial P}{partial y}]{{|}_{M}}pi {{r}^{2}}=0$
其中$D$是由$L+{{L}_{1}}$包围的图形,$Min D$
另一方面由积分中值定理可知:
$int_{L}{Pdx+Qdy=}-int_{{{L}_{1}}}{P(x,{{y}_{0}}})dx=-P({{x}_{1}},{{y}_{0}})cdot 2r$
其中$({{x}_{1}},{{y}_{0}})in {{L}_{1}}$
比较这两个式子知:
$[frac{partial Q}{partial x}-frac{partial P}{partial y}]{{|}_{M}}pi {{r}^{2}}=-P({{x}_{1}},{{y}_{0}})cdot 2rRightarrow [frac{partial Q}{partial x}-frac{partial P}{partial y}]{{|}_{M}}frac{pi r}{2}=-P({{x}_{1}},{{y}_{0}})$
令$rto 0$可知:$P({{x}_{0}},{{y}_{0}})=0$,由$({{x}_{0}},{{y}_{0}})$的任意性可知,$P(x,y)=0$
从而有 $[frac{partial Q}{partial x}-frac{partial P}{partial y}]{{|}_{M}}$,令$rto 0$可知$[frac{partial Q}{partial x}-frac{partial P}{partial y}]{{|}_{({{x}_{0}},{{y}_{0}})}}=0$,由$({{x}_{0}},{{y}_{0}})$的任意性可知,$frac{partial Q}{partial x}=0$
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