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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}-1\le {{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,$\sqrt{2}-1\le {{x}_{2k-1}}\le \frac{1}{2}$,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:$\sqrt{2}-1\le {{x}_{2n+1}}\le \frac{1}{2}$
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,$\sqrt{2}-1\le {{x}_{2n+1}}\le \frac{1}{2}$,由单调有界原理可知:$\{{{x}_{2n+1}}\}$收敛
于是不妨设$\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{2n+1}}=l\in [\sqrt{2}-1,\frac{1}{2}]$,由${{x}_{2k+3}}=\frac{1}{2}[1-\frac{1}{4}{{(1-x_{2k+1}^{2})}^{2}}]$,两边对$n\to +\infty $
则$l=\frac{1}{2}[1-\frac{1}{4}{{(1-{{l}^{2}})}^{2}}]$,求得$l=\sqrt{2}-1$
即$\underset{x\to \infty }{\mathop{\lim }}\,{{x}_{2n+1}}=\sqrt{2}-1=A$
同理可证$\underset{x\to \infty }{\mathop{\lim }}\,{{x}_{2n}}=\sqrt{2}-1=A$
于是$\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{n}}=\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{2n}}=\underset{n\to +\infty }{\mathop{\lim }}\,{{x}_{2n+1}}=A$
(3)证明:不妨设${{a}_{n}}={{x}_{n}}-A={{x}_{n}}-(\sqrt{2}-1)$
于是
$\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{{{a}_{n+1}}}{{{a}_{n}}} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{{{x}_{n+1}}-A}{{{x}_{n}}-A} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\left| \frac{\frac{1}{2}(1-x_{n}^{2})-A}{{{x}_{n}}-A} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{2}\left| \frac{1-x_{n}^{2}-(2\sqrt{2}-2)}{{{x}_{n}}-(\sqrt{2}-1)} \right|=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{2}\left| \frac{x_{n}^{2}-{{(\sqrt{2}-1)}^{2}}}{{{x}_{n}}-(\sqrt{2}-1)} \right|=A1 $
于是$\sum\limits_{n=1}^{+\infty }{({{x}_{n}}}-A)$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 3\left| \sin \left| f({{x}_{1}}) \right|-\sin \left| f({{x}_{1}}) \right| \right|\le 3\left| \sin f({{x}_{1}})-\sin f({{x}_{2}}) \right|\le 3\left| 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$x\in (a,b)$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$x\in (a,b)$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$x\in [a,b]$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$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}=-\frac{f(b)-f(a)}{b-a},x\in [a,b]$
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$F\left( x,y \right)={{x}^{2}}+y-\cos \left( xy \right) $,
显然,有$F\left( 0,1 \right)=0 $,
${{F}_{y}}\left( x,y \right)=1+x\sin \left( xy \right) $,
${{F}_{y}}\left( 0,1 \right)=1\ne 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=y\left( x \right)$,$y\left( 0 \right)=1 $,
满足$F\left( x,y\left( x \right) \right)\equiv 0 $,$x\in U(0) $
(3)证明:由(1)可知:
${{F}_{x}}\left( x,y \right)=2x+y\sin \left( xy \right) $,
${y}\left( x \right)=-\frac{{{F}_{x}}\left( x,y \right)}{{{F}_{y}}\left( x,y \right)}=-\frac{2x+y\sin \left( xy \right)}{1+x\sin \left( xy \right)} $,
当$0x\delta $,($\delta 0 $充分小)时,有${y}\left( x \right)0 $,$y\left( x \right) $在$\left[ 0,\delta \right] $上严格单调递减;
当$-\delta x0 $时,有${y}\left( x \right)0 $,$y\left( 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 x\to 0}{\mathop{\lim }}\,\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\Delta x\cos \frac{1}{\left| \Delta x \right|}=0 $
${{f}_{y}}(0,0)=\underset{\Delta y\to 0}{\mathop{\lim }}\,\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=\underset{\Delta y\to 0}{\mathop{\lim }}\,\Delta y\cos \frac{1}{\left| \Delta y \right|}=0\ $
(2)当$(x,y)\ne (0,0)$时,
${{f}_{x}}(x,y)=2x\cos \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)=2y\cos \frac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\sin \frac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}} $
于是${f_x}(x,y)=\left\{\begin{array}{ll}2x\cos \frac{1}{{\sqrt {{x^2} + {y^2}} }} + \frac{x}{{\sqrt {{x^2} + {y^2}} }}\sin \frac{1}{{\sqrt {{x^2} + {y^2}} }}, \hbox{${x^2} + {y^2} \ne 0$} \\0, \hbox{${x^2} + {y^2} = 0$.}\end{array}\right.$
${f_y}(x,y)=\left\{\begin{array}{ll}2y\cos \frac{1}{{\sqrt {{x^2} + {y^2}} }} +
\frac{y}{{\sqrt {{x^2} + {y^2}} }}\sin \frac{1}{{\sqrt {{x^2} + {y^2}} }},,
\hbox{${x^2} + {y^2} \ne 0$} \\ 0, \hbox{${x^2} + {y^2} = 0$.}\end{array}\right.$
(2)设$y=kx$,于是$\underset{x\to 0}{\mathop{\lim
}}\,\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\frac{1}{\sqrt{1+{{k}^{2}}}}$与$k$有关
于是$\underset{x\to 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$\underset{\Delta x\to 0,\Delta y\to
0}{\mathop{\lim }}\,\frac{\Delta u-du}{\sqrt{\Delta {{x}^{2}}+\Delta
{{y}^{2}}}}=\underset{\Delta x\to 0,\Delta y\to 0}{\mathop{\lim
}}\,\sqrt{\Delta {{x}^{2}}+\Delta {{y}^{2}}}\cos \frac{1}{\sqrt{\Delta
{{x}^{2}}+\Delta {{y}^{2}}}}=0$
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{n\to +\infty }{\mathop{\lim
}}\,\sqrt[n]{{{a}_{n}}}={{e}^{\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\ln
{{a}_{n}}}{n}}}={{e}^{\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\ln
2({{e}^{\frac{1}{n}}}-1-\frac{1}{n})}{n}}}\overset{n=\frac{1}{x}}{\mathop{=}}\,{{e}^{\underset{x\to
+{{0}^{+}}}{\mathop{\lim }}\,\frac{\ln 2({{e}^{x}}-1-x)}{\frac{1}{x}}}}={{e}^{-\underset{x\to
+{{0}^{+}}}{\mathop{\lim
}}\,\frac{{{x}^{2}}({{e}^{x}}-1)}{{{e}^{x}}-1-x}}}={{e}^{-\underset{x\to
+{{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,$y(\frac{1}{n})-1-\frac{1}{n}=2({{e}^{\frac{1}{n}}}-1-\frac{1}{n})=\sum\limits_{n=1}^{+\infty
}{\frac{2}{{{n}^{2}}}+}\sum\limits_{n=1}^{+\infty }{o(\frac{2}{{{n}^{2}}}})$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$u={{t}^{2}}\Rightarrow
t=\sqrt{u},dt=\frac{1}{2\sqrt{u}}du$,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$x0$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,$M\in D$
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:
$[\frac{\partial
Q}{\partial x}-\frac{\partial P}{\partial y}]{{|}_{M}}\pi
{{r}^{2}}=-P({{x}_{1}},{{y}_{0}})\cdot 2r\Rightarrow [\frac{\partial
Q}{\partial x}-\frac{\partial P}{\partial y}]{{|}_{M}}\frac{\pi
r}{2}=-P({{x}_{1}},{{y}_{0}})$
令$r\to 0$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$r\to 0$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,$\frac{\partial Q}{\partial x}=0$