来源 [尊重原有作者劳动成果]
一、 计算题
1:解:由于
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\ln (1-x)}{-x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{1-x}=1,\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{e}^{{{x}^{2}}}}-1}{{{x}^{2}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2x{{e}^{{{x}^{2}}}}}{2x}=1$
于是
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{\ln }^{3}}(1-x)\sin [\sin (\ln \frac{1}{x})]}{{{e}^{{{x}^{2}}}}-1}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{(-x)}^{3}}}{{{x}^{2}}}\sin [\sin (\ln \frac{1}{x})]=0$
2:不妨设${{t}_{1}}={{x}^{x}},{{t}_{2}}={{x}^{\ln x}}$,于是
$\ln {{t}_{1}}=x\ln x,\ln {{t}_{2}}={{(\ln x)}^{2}}$
则$\frac{t_{1}^{}}{{{t}_{1}}}=\ln x+1,\frac{t_{2}^{}}{{{t}_{2}}}=\frac{2\ln x}{x}$
即$t_{1}^{}=(\ln x+1){{x}^{x}},t_{2}^{}=\frac{2\ln x}{2}{{x}^{\ln x}}$
于是$y=\frac{2\ln x}{x}{{x}^{\ln x}}+(\ln x+1){{x}^{x}}$
3:解:令$t={{x}^{4}}$,于是
$\int_{0}^{+\infty }{{{e}^{-{{x}^{4}}}}}dx\cdot \int_{0}^{+\infty }{{{x}^{2}}{{e}^{-{{x}^{4}}}}dx=\frac{1}{16}}\int_{0}^{+\infty }{{{t}^{-\frac{3}{4}}}}{{e}^{-t}}\cdot \int_{0}^{+\infty }{{{t}^{-\frac{1}{4}}}}{{e}^{-t}}dt$
$=\frac{1}{16}\Gamma (\frac{1}{4})\Gamma (\frac{3}{4})\overset{}{\mathop{=}}\,\frac{1}{16}\cdot \frac{\pi }{\sin \frac{\pi }{4}}=\frac{\pi }{8\sqrt{2}}$
此处介绍余元公式:
$\forall p\in (0,1)$,有$\Gamma (p)\Gamma (1-p)=\frac{\pi }{\sin p\pi }$
4:解:
$F(x)={{f}_{x}}[f(x,x),f(x,x)]\cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]+{{f}_{y}}[f(x,x),f(x,x)]\cdot [{{f}_{x}}(x,x)+{{f}_{y}}(x,x)]$
于是$F(1)={{(a+b)}^{2}}$
5:解:不妨设$x=r\cos \theta ,y=r\sin \theta ,0\le r\le 1,0\le \theta \le 2\pi $
于是
$\iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=\int_{0}^{2\pi }{d\theta \int_{0}^{1}{{{r}^{4}}}({{\sin }^{3}}\theta +{{\cos }^{3}}\theta ){{e}^{{{r}^{2}}}}dr=}\int_{0}^{2\pi }{({{\sin }^{3}}\theta +{{\cos }^{3}}\theta )}d\theta \cdot \int_{0}^{1}{{{r}^{4}}{{e}^{r}}dr}$
而
$\int_{0}^{2\pi }{({{\sin }^{3}}\theta +{{\cos }^{3}}\theta )}d\theta =\int_{0}^{2\pi }{(\sin \theta +\cos \theta )(1+\sin \theta \cos \theta )d\theta }$
$=\int_{0}^{2\pi }{(\sin \theta +\cos \theta )d\theta +\int_{0}^{2\pi }{{{\sin }^{2}}}}\theta \cos \theta d\theta +\int_{0}^{2\pi }{\sin \theta {{\cos }^{2}}}\theta d\theta $
$=\int_{0}^{2\pi }{{{\sin }^{2}}}\theta d(\sin \theta )-\int_{0}^{2\pi }{{{\cos }^{2}}}\theta d(\cos \theta )$
$=\frac{1}{3}{{\sin }^{3}}\theta |_{0}^{2\pi }-\frac{1}{3}{{\cos }^{3}}\theta |_{0}^{2\pi }$
$=0$
则$\iint_{D}{({{x}^{3}}+{{y}^{3}}){{e}^{{{x}^{2}}+{{y}^{2}}}}}dxdy=0$
6:解:记${{L}_{1}}:0\le x\le 2,y=0$,$D$是由$L$和${{L}_{1}}$所围成的封闭曲线
而
$\int_{{{L}_{1}}}{{{e}^{x}}\sin ydy-{{e}^{x}}\cos ydx=\int_{0}^{2}{{{e}^{x}}}}dx={{e}^{x}}|_{0}^{2}={{e}^{2}}-1$
另一方面,由格林公式可知:
$\int_{L+{{L}_{1}}}{{{e}^{x}}\sin ydy-{{e}^{x}}\cos ydx=\iint_{D}{({{e}^{x}}\sin y-{{e}^{x}}\sin y)dxdy=0}}$
于是
$\int_{L}{{{e}^{x}}\sin ydy-{{e}^{x}}\cos ydx=-}\int_{{{L}_{1}}}{{{e}^{x}}\sin ydy-{{e}^{x}}\cos ydx=}1-{{e}^{2}}$
二、
(1)证明:由于$\sqrt{x}f(x)$在$(0,+\infty )$上有界,则存在$M0$,使得
$\left| \sqrt{x}f(x) \right|\le M$
对任意的$\varepsilon 0$,对任意的${{x}_{1}},{{x}_{2}}\in (0,+\infty )$,不妨设${{x}_{1}}{{x}_{2}}$
于是
$\left| f({{x}_{1}})-f({{x}_{2}}) \right|=\left| \int_{{{x}_{1}}}^{{{x}_{2}}}{f(t)dt} \right|\le \int_{{{x}_{1}}}^{{{x}_{2}}}{\left| f(t) \right|}dt=\int_{{{x}_{1}}}^{{{x}_{2}}}{\left| \sqrt{t}f(t) \right|}\cdot \frac{1}{\sqrt{t}}dt$
$\le 2M\int_{{{x}_{1}}}^{{{x}_{2}}}{\frac{1}{\sqrt{t}}}dt=2M(\sqrt{{{x}_{2}}}-\sqrt{{{x}_{1}}}\le 2M\sqrt{{{x}_{2}}-{{x}_{1}}}$
令$\delta =\frac{{{\varepsilon }^{2}}}{4{{M}^{2}}}$,当$\left| {{x}_{1}}-{{x}_{2}} \right|\delta $时,$\left| f({{x}_{1}})-f({{x}_{2}}) \right|\varepsilon $
即$f(x)$在$(0,+\infty )$上一致收敛
(或利用柯西中值定理,令$g(x)=\sqrt{x}$)
(2)由(1)可知,对任意的$\varepsilon 0$,令${{x}_{1}}\to {{0}^{+}}$,当$0x\delta $时,有
$\left| f(x)-f(0) \right|\varepsilon $
由极限定义可知:
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f({{0}^{+}})$存在
(3)能,理由如下:
证明:由于$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt{x}f(x)$存在,则存在$M0$,存在${{X}_{1}}0$,当$x{{X}_{1}}$时,$\left| \sqrt{x}f(x) \right|\le M$
由(1)可知:$f(x)$在$(0,{{X}_{1}})$上一致收敛
而$\underset{x\to +\infty }{\mathop{\lim }}\,\sqrt{x}f(x)$存在,同理可证$f(x)$在$({{X}_{2}},+\infty )$上一致收敛
而$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上连续,则$f(x)$在$[{{X}_{1}},{{X}_{2}}]$上一致连续
于是$f(x)$在$(0,+\infty )$上一致连续
三、
(1)证明:对任意的$x\in (a,+\infty )$,则$f(x)$在$[x,x+1]$上可导
记$\underset{x\to +\infty }{\mathop{\lim }}\,f(x)=A$
$\underset{x\to +\infty }{\mathop{\lim }}\,f(x+1)=A$,则
$\underset{x\to +\infty }{\mathop{\lim }}\,[f(x+1)-f(x)]=0$
由拉格朗日中值定理可知:存在$x\xi x+1$,使得
$f(x+1)-f(x)=f(\xi )[(x+1)-x]=f(\xi )$
且当$x\to +\infty $时,$\xi \to +\infty $,于是
$\underset{x\to +\infty }{\mathop{\lim }}\,f(\xi )=\underset{x\to +\infty }{\mathop{\lim }}\,[f(x+1)-f(x)]=1$
由归结原则可知:
$\underset{x\to +\infty }{\mathop{\lim }}\,f(\xi )=\underset{x\to +\infty }{\mathop{\lim }}\,f(x)=0$
(2)可以推出,理由如下:
证明:应用$Taylor$公式可知:对任意的$m=1,2,3,4$
$f(x+m)=f(x)+mf(x)+\frac{{{m}^{2}}}{2!}f(x)+\frac{{{m}^{3}}}{3!}f(x)+\frac{{{m}^{4}}}{4!}f({{\xi }_{m}}(x)),x{{\xi }_{m}}(x)x+m$
由于$f(x),f(x),f(x)$均可用$f(x+m)-f(x)$及$f({{\xi }_{m}}(x))$表达
而$\underset{x\to +\infty }{\mathop{\lim }}\,[f(x+m)-f(x)]$与$\underset{x\to +\infty }{\mathop{\lim }}\,f({{\xi }_{m}}(x))=\underset{x\to +\infty }{\mathop{\lim }}\,f(x)$均存在有限
故$\underset{x\to +\infty }{\mathop{\lim }}\,{{f}^{(k)}}(x)$均存在有限,不妨记
$\underset{x\to +\infty }{\mathop{\lim }}\,{{f}^{(k)}}(x)={{A}_{k}},k=0,1,2,3,4$
在上面的$Taylor$公式中,令$x\to +\infty $,则${{\xi }_{m}}(x)\to +\infty $,于是有
$0=\underset{x\to +\infty }{\mathop{\lim }}\,[f(x+m)-f(x)]=m\underset{x\to +\infty }{\mathop{\lim }}\,f(x)+\frac{{{m}^{2}}}{2!}\underset{x\to +\infty }{\mathop{\lim }}\,f(x)+\frac{{{m}^{3}}}{3!}\underset{x\to +\infty }{\mathop{\lim }}\,f(x)+\frac{{{m}^{4}}}{4!}\underset{x\to +\infty }{\mathop{\lim }}\,f({{\xi }_{m}}(x))$
即$0={{A}_{0}}-{{A}_{0}}=m{{A}_{1}}+\frac{{{m}^{2}}}{2!}{{A}_{2}}+\frac{{{m}^{3}}}{3!}{{A}_{3}}+\frac{{{m}^{4}}}{4!}{{A}^{4}},m=1,2,3,4$
列举出来为
$\left\{\begin{array}{ll} {A_1} + \frac{1}{{2!}}{A_2} + \frac{1}{{3!}}{A_3} + \frac{1}{{4!}}{A_4} = 0, \\ 2{A_1} + \frac{{{2^2}}}{{2!}}{A_2} + \frac{{{2^3}}}{{3!}}{A_3} + \frac{{{2^4}}}{{4!}}{A_4} = 0, \\ 3{A_1} + \frac{{{3^2}}}{{2!}}{A_2} + \frac{{{3^3}}}{{3!}}{A_3} + \frac{{{3^4}}}{{4!}}{A_4} = 0, \\ 4{A_1} + \frac{{{4^2}}}{{2!}}{A_2} + \frac{{{4^3}}}{{3!}}{A_3} + \frac{{{4^4}}}{{4!}}{A_4} = 0, \end{array} \right.$ $\Rightarrow\left\{\begin{array}{ll} {A_1} + \frac{1}{{2!}}{A_2} + \frac{1}{{3!}}{A_3} + \frac{1}{{4!}}{A_4} = 0, \\ {A_1} + \frac{2}{{2!}}{A_2} + \frac{{{2^2}}}{{3!}}{A_3} + \frac{{{2^3}}}{{4!}}{A_4} = 0, \\ {A_1} + \frac{3}{{2!}}{A_2} + \frac{{{3^2}}}{{3!}}{A_3} + \frac{{{3^3}}}{{4!}}{A_4} = 0, \\ {A_1} + \frac{4}{{2!}}{A_2} + \frac{{{4^2}}}{{3!}}{A_3} + \frac{{{4^3}}}{{4!}}{A_4} = 0, \end{array} \right.$
方程组的系数行列式为
$ \left|\begin{array}{cccc} 1 {\frac{1}{{2!}}} {\frac{1}{{3!}}} {\frac{1}{{4!}}} \\ 1 {\frac{2}{{2!}}} {\frac{{{2^2}}}{{3!}}} {\frac{{{2^3}}}{{4!}}} \\ 1 {\frac{3}{{2!}}} {\frac{{{3^2}}}{{3!}}} {\frac{{{3^3}}}{{4!}}} \\ 1 {\frac{4}{{2!}}} {\frac{{{4^2}}}{{3!}}} {\frac{{{4^3}}}{{4!}}} \end{array}\right| $$ =\frac{1}{{1!2!3!4!}}\left|\begin{array}{cccc} 1 1 1 1 \\ 1 2 {{2^2}} {{2^3}} \\ 1 3 {{3^2}} {{3^3}} \\ 1 4 {{4^2}} {{4^3}} \end{array}\right| $
=$\frac{1}{{1!2!3!4!}}(2 - 1)(3 - 1)(3 - 2)(4 - 1)(4 - 2)(4 - 3) = \frac{1}{{4!}} \ne 0$
即$\underset{x\to +\infty }{\mathop{\lim }}\,{{f}^{(k)}}(x)=0,k=1,2,3,4$
(或利用第二数学归纳法证明)
四、证明:证明:由题设知$f(x)$在$ [0,+\infty ) $上必有界,设$\left| f(x) \right|\le M$
对$\forall \varepsilon 0$,有
$\left| \frac{1}{x}\int_{0}^{x}{f(t)dt}-A \right|$
$=\left| \int_{0}^{1}{(f(yx)-A)dy} \right|$
$\le \int_{0}^{\frac{\varepsilon }{2(M+A)}}{\left| f(yx)-A \right|}dy+\int_{\frac{\varepsilon }{2(M+A)}}^{1}{\left| f(yx)-A \right|}dy$
由$\underset{x\to +\infty }{\mathop{\lim }}\,f(x)=A$知
对上述$\varepsilon 0,\exists {{X}_{1}}0, $使得当$x{{X}_{1}}$时有$\left| f(x)-A \right|\frac{\varepsilon }{2}$
令$\text{X}=\frac{2(M+A)}{\varepsilon }{{X}_{1}}$,则当$xX$时,有$\int_{\frac{\varepsilon }{2(M+A)}}^{1}{\left| f(yx)-A \right|}dy\frac{\varepsilon }{2}$
于是
$\left| \frac{1}{x}\int_{0}^{x}{f(t)dt}-A \right|\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon $
因此
$\underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{x}\int_{0}^{x}{f(t)dt}=A$.
(或直接利用定义)
五、证明:
充分性:若$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$收敛,令$0S=\underset{n\to +\infty }{\mathop{\lim }}\,{{S}_{n}}=\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}+\infty $
由于
$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\frac{{{a}_{n}}}{{{S}_{n}}}}{{{a}_{n}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{1}{{{S}_{n}}}=\frac{1}{S}+\infty $
由比较判别法可知:$\sum\limits_{n=1}^{+\infty }{\frac{{{a}_{n}}}{{{S}_{n}}}}$收敛
必要性:反证法:若$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$发散
考虑
$\sum\limits_{k=n+1}^{n+p}{\frac{{{a}_{k}}}{{{S}_{k}}}}=\frac{{{a}_{n+1}}}{{{S}_{n+1}}}+\frac{{{a}_{n+2}}}{{{S}_{n+2}}}+\cdots +\frac{{{a}_{n+p}}}{{{S}_{n+p}}}\frac{\sum\limits_{k=n+1}^{n+p}{{{a}_{k}}}}{{{S}_{n+p}}}=\frac{{{S}_{n+p}}-{{S}_{n}}}{{{S}_{n+p}}}=1-\frac{{{S}_{n}}}{{{S}_{n+p}}}$
由于$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$发散,则对任意的$n\in {{N}^{*}}$,存在$p\in {{N}^{*}}$,使得
$\frac{{{S}_{n}}}{{{S}_{n+p}}}\frac{1}{2}$
从而$\sum\limits_{k=n+1}^{n+p}{\frac{{{a}_{k}}}{{{S}_{k}}}}\frac{1}{2}$
由柯西收敛准则知:$\sum\limits_{n=1}^{+\infty }{\frac{{{a}_{n}}}{{{S}_{n}}}}$发散,矛盾
从而$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$收敛
六、
(1)证明:由于
$2\sin \frac{x}{2}(\frac{1}{2}+\sum\limits_{k=1}^{n}{\cos kx)=\sin }\frac{x}{2}+(\sin \frac{3x}{2}-\sin \frac{x}{2})+\cdots +[\sin (n+\frac{1}{2})x-\sin (n-\frac{1}{2})x]=\sin (n+\frac{1}{2})x$
当$x\in [\alpha ,2\pi -\alpha ]$时,$\sin \frac{x}{2}\ne 0$
于是
$\frac{1}{2}+\sum\limits_{k=1}^{n}{\cos kx}=\frac{\sin (n+\frac{1}{2})x}{2\sin \frac{x}{2}}$
于是$\sum\limits_{n=1}^{+\infty }{\cos nx}$部分和在$x\in [\alpha ,2\pi -\alpha ]$上一致有界
令${{u}_{n}}(x)=\cos nx,{{v}_{n}}(x)={{a}_{n}}$
由狄利克雷判别法知:$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}\cos nx$在$x\in [\alpha ,2\pi -\alpha ]$上一致收敛
(2)证明:
充分性:若$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$收敛,由于$\left| {{a}_{n}}\cos nx \right|\le \left| {{a}_{n}} \right|={{a}_{n}}$
由$M$判别法可知:$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}\cos nx$在$[0,2\pi ]$上一致收敛
必要性:若$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}\cos nx$在$[0,2\pi ]$上一致收敛,令$x=0$
即$\sum\limits_{n=1}^{+\infty }{{{a}_{n}}}$收敛收敛
七、证明:由于
${{u}_{x}}=z+x{{z}_{x}}+yf(z){{z}_{x}}+g(z){{z}_{x}}=z$
${{u}_{xx}}={{z}_{x}}$
${{u}_{xy}}={{z}_{y}}$
${{u}_{y}}=x{{z}_{y}}+f(z)+yf(z){{z}_{y}}+g(z){{z}_{y}}=f(z)$
${{u}_{yy}}=f(z){{z}_{y}}$
${{u}_{yx}}=f(z){{z}_{x}}$
于是
$\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}\cdot \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}-{{\left( \frac{{{\partial }^{2}}u}{\partial x\partial y} \right)}^{2}}=0$
八、证明:
(1)由于$\frac{\partial f}{\partial n}=\frac{\partial f}{\partial x}\cos (n,x)+\frac{\partial f}{\partial y}\cos (n,y)+\frac{\partial f}{\partial z}\cos (n,z)$
由第一型和第二型曲面积分的关系和$Guass$公式可知:
$\iint\limits_{S}{\frac{\partial f}{\partial n}}dS=\iint\limits_{S}{\frac{\partial f}{\partial x}\cos (n,x)+\frac{\partial f}{\partial y}\cos (n,y)+\frac{\partial f}{\partial z}\cos (n,z)}dS$
$=\iint\limits_{S}{\frac{\partial f}{\partial x}dydz+\frac{\partial f}{\partial y}dzdx+\frac{\partial f}{\partial z}dxdy}$
$=\iiint_{V}{(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}})dxdydz}$
即$\iint\limits_{S}{\frac{\partial f}{\partial n}}dS=\iiint_{V}{(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}})dxdydz}$
(2)充分性:若对任意的$S$,$\iint\limits_{S}{\frac{\partial f}{\partial n}}dS$,总有$\iiint_{V}{(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}})dxdydz}=0$
不妨设$g(x,y,z)=\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}}$,由积分中值定理可知:存在$(\xi ,\eta ,\zeta )\in V$
\[\iiint_{V}{(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}}})dxdydz=g(\xi ,\eta ,\zeta )V=0\]
于是由$V$的任意性可知:$g(\xi ,\eta ,\zeta )=0$,再由$(\xi ,\eta ,\zeta )$的任意性知:
$\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}}=0$
必要性:若$\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}}=0$,则$\iint\limits_{S}{\frac{\partial f}{\partial n}}dS=\iiint_{V}{(\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}f}{\partial {{z}^{2}}})dxdydz}=0$