数学分析习题练习【谢绝回贴】

西南大学2024 数学分析


解  用夹逼法\[\because \frac{1}{n\sqrt{2+\frac{1}{n^2}}}< \frac{1}{\sqrt{2n^2+k}}< \frac{1}{n\sqrt{2+\frac{1}{n}}},\]\[ \frac{n}{n\sqrt{2+\frac{1}{n^2}}}<\frac{1}{\sqrt{2n^2+1}}+\frac{1}{\sqrt{2n^2+2}}+\cdots +\frac{1}{\sqrt{2n^2+n}}< \frac{n}{n\sqrt{2+\frac{1}{n}}}.\]\[\therefore \lim_{n\to\infty }(\frac{1}{\sqrt{2n^2+1}}+\frac{1}{\sqrt{2n^2+2}}+\cdots +\frac{1}{\sqrt{2n^2+n}})=\frac{\sqrt{2}}{2}.\]

西南大学2022 数学分析


解  令\[J(s)=\int_{0}^{1}\frac{\ln(1+sx)}{1+x^2}dx.\]
              显然$J(s)$在$[0,1]$上一致收敛.如此\[\begin{align*}J'(s)
&=\int_{0}^{1}\frac{x}{(1+x^2)(1+sx)}dx \\
&=\frac{1}{1+s^2}\int_{0}^{1}[\frac{x+s}{1+x^2}-\frac{s}{1+sx}]dx \\
&=\frac{1}{1+s^2}[\frac{1}{2}\ln(1+x^2)|_0^1+s\arctan x|_0^1-\ln(1+sx)|_0^1] \\
&=\frac{1}{1+s^2}[\frac{1}{2}\ln2+\frac{s\pi}{2}-\ln(1+s)].
\end{align*}\]于是\[J(s)=\frac{1}{2}\ln2\arctan s+\frac{\pi}{4}\ln(1+s^2)-J.\]\[J=J(1)=\frac{1}{2}(\frac{1}{2}\ln2\arctan 1+\frac{\pi}{4}\ln2)=\frac{\pi}{4}\ln2.\]

西南大学2022 数学分析


解 送分题\[\frac{\partial z}{\partial x}=yf'_1+f'_2.\]\[\frac{\partial z}{\partial y}=xf'_1-f'_2.\]

西南大学2022 数学分析


解\[D=D_1+D_2.\]\[D_1:y^2=x,x=1.\]\[D_2:y^2=x,x=1,y=x-2.\]\[I_1=\iint\limits_{D_1}xydxdy=\int_{-1}^{1}ydy\int_{y^2}^{1}xdx=\frac{1}{2}\int_{-1}^{1}y(1-y^4)dy=0.\]\[\begin{align*}I_2
&=\iint\limits_{D_2}xydxdy=\int_{1}^{2}xdx\int_{x-2}^{\sqrt{x}}ydy \\
&=\frac{1}{2}\int_{1}^{2}(-x^3+3x^2-4x)dx=-\frac{7}{8}.
\end{align*}\]\[I=I_1+I_2=-\frac{7}{8}.\]

西南大学2022 数学分析


解\[\Omega :0\le x,y,z\le a.\]\[\begin{align*}I=\unicode{8751}_{S}
&=\iiint\limits_{\Omega }(y+x)dxdydz,高斯公式\\
&=\int_{0}^{a}dx\int_{0}^{a}dy\int_{0}^{a}(y+x)dz \\
&=a^4.
\end{align*}\]

中科大2023数分2


解  \[\because \partial _xz=1+y\partial _z\varphi \partial _xz,\partial _xz=\frac{1}{1-y\partial _z\varphi}.\]\[\partial _xu=\partial _zf\partial _xz=\frac{\partial _zf}{1-y\partial _z\varphi}.\]\[\therefore \partial _x(\varphi ^2(z)\partial _xu)=\partial _x(\frac{\varphi ^2(z)\partial _zf}{1-y\partial _z\varphi})\]

中科大2023数分2


解   曲面的最大方向导数即为曲面梯度.\[|\grad f(x,y)|^2=f_x^2+f_y^2=(1+y)^2+(1+x)^2.\]求条件极值.令\[F(x,y,\lambda )=(1+y)^2+(1+x)^2-\lambda (x^2+xy+y^2-3).\]\[F_x=2(1+x)-\lambda (2x+y)=0.\]\[F_y=2(1+y)-\lambda (x+2y)=0.\]\[F_\lambda =x^2+xy+y^2-3=0.\]
             $\Rightarrow (1,1)$时，曲面的最大方向导数\[\grad f(1,1)=\sqrt{(1+y)^2+(1+x)^2}|_{(1,1)}=2\sqrt{2}.\]

中科大2023数分2


解    坐标变换\[x=a\rho \cos \alpha ,y=b\rho \sin \alpha ,0\le r\le \rho \le R\le +\infty .\]则\[I=\iint_{D}\frac{e^{-(\frac{x^2}{a^2}+\frac{y^2}{b^2})}}{\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}}}dxdy=ab\int_{0}^{2\pi}d\alpha \int_{r}^{R }\frac{e^{-\rho ^2}}{\rho }\rho d\rho =2ab\pi \int_{r}^{R }e^{-\rho ^2} d\rho .\]有\[\lim_{R \to+\infty,r\to0 }I=2ab\pi \int_{0}^{+\infty  }e^{-\rho ^2} d\rho =ab\pi\sqrt{\pi}.\]

中科大2023数分2


解  \[用高斯公式。先添加一个平面Z=0^-,方向向下，与\Sigma_+组成一个闭合体\Sigma_0.则\]\[\begin{align*}I=\iint\limits_{\Sigma _+}
&=\iint\limits_{\Sigma _0}-\iint\limits_{Z=0^+}\\
&=\iiint_{V}(\frac{z}{a^2}+\frac{z}{b^2}+2\frac{z}{c^2})dV+0 \\
&=(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{c^2})\iiint_{V}zdV \\
&=(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{c^2})\iint_{S}dxdy\int_{0}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}zdz \\
&=(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{c^2})\iint_{S}\frac{c^2}{2}(1-\frac{x^2}{a^2}-\frac{y^2}{b^2})dxdy \\
&=\frac{abc^2}{2}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{c^2})\int_{0}^{2\pi }d\theta \int_{0}^{1}rdr,(x=ar\cos \theta ,y=br\sin \theta ,J=abr)\\
&=\frac{abc^2\pi }{2}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{c^2}).
\end{align*}\]

中科大2023数分2


证明   由已知条件，将$f(x,y)$在$(1,1)$处泰勒展开.\[f(x,y)=f(1,1)+f_x(1,1)(x-1)+f_y(1,1)(y-1)+\frac{1}{2}[f_{xx}(1,1)(x-1)^2+2f_{xy}(1,1)(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]+R_3.\]而\[f(1,1)=0，f_x(1,1)=0，f_y(1,1)=0.\]所以有\[\begin{align*}I
&=\iint_{[0,1]^2}f(x,y)dxdy \\
&\le \iint_{[0,1]^2}\frac{1}{2}[f_{xx}(1,1)(x-1)^2+2f_{xy}(1,1)(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]dxdy \\
&\le \frac{M}{2}[\int_{0}^{1} dy\int_{0}^{1} (x-1)^2dx+2\int_{0}^{1}(y-1) dy\int_{0}^{1} (x-1)dx+\int_{0}^{1} dx\int_{0}^{1} (y-1)^2dy]\\
&=\frac{7M}{12} .
\end{align*}\]