数学分析习题练习【谢绝回贴】

云南大学2024


解   可求得切平面的法向量为$(1,2,-1)$.
      于是切平面为:$x+2y-z=0.$

云南大学2024


解  首先\[\frac{\partial f}{\partial x}=ye^{x^3y^2}+\int_{0}^{xy}t^2e^{xt^2}dt,\]因而\[\frac{\partial^2 f}{\partial x\partial y}=e^{x^3y^2}+2x^3y^2e^{x^3y^2}+x^3y^2e^{x^3y^2},\]\[\frac{\partial^2 f}{\partial x\partial y}|_{(1,1)}=4e.\]

云南大学2024


解 \[y'=\tan x.\]
\begin{align*}s
&=\int_{0}^{\frac{\pi }{4}}\sqrt{\tan ^2x+1}dx \\
&=\int_{0}^{\frac{\pi }{4}}\frac{1}{\cos x}dx \\
&=\int_{0}^{\frac{\pi }{4}}\frac{\cos xdx}{1-\sin^2 x} \\
&=\frac{1}{2}\int_{0}^{\frac{\pi }{4}}[\frac{\cos xdx}{1-\sin x}+\frac{\cos xdx}{1+\sin x}] \\
&=\frac{1}{2}\ln |\frac{1+\sin x}{1-\sin x}||_0^{\frac{\pi }{4}}\\
&=\frac{1}{2}\ln\frac{2+\sqrt{2}}{2-\sqrt{2}}.
\end{align*}

云南大学2024


解
            分析：\[若要|\frac{x(x-1)}{x^2-1}-\frac{1}{2}|=|\frac{x-1}{x+1}|< |x-1|< \delta ,(\delta > 0)\]\[只要取\delta =\varepsilon ,即可.\]\[\therefore \forall \varepsilon > 0,\exists \delta > 0,\forall x,|x-1|< \delta ,s.t.\]\[|\frac{x(x-1)}{x^2-1}-\frac{1}{2}|<|x-1|<\varepsilon .\]

云南大学2024


解   \[\because n!=\sqrt{2n\pi }(\frac{n}{e})^n,\]\[\therefore \frac{a^nn!}{n^n}=\sqrt{2n\pi}(\frac{a}{e})^n.\]而\[\lim_{n\to\infty }\sqrt[n]{\sqrt{2n\pi}(\frac{a}{e})^n}=\frac{a}{e}.\]所以\[a\geqslant e时，原级数发散；a< e时，原级数收敛.\]

云南大学2024


证明     （1）、 \[由g(a)=g(b)=0,有\exists c\in (a,b),s.t.g(c)=0.连续函数介值定理.\]由Rolle定理，\[\exists x_1\in (a,c),\exists x_2\in (c,b),s.t.g'(x_1)=0,g'(x_2)=0.\]再次由Rolle定理，\[\exists x_0\in (x_1,x_2),s.t.g''(x_0)=0.与已知条件g''(x)\ne 0，矛盾.\]\[\therefore g''(x)\ne 0.\]
             （2）、令\[F(x)=f(x)g'(x)-f'(x)g(x),F(x)可导.\]而\[F(a)=F(b)=0,\]于是由Rolle定理，\[\exists \xi \in (a,b),s.t.F'(\xi)=f(\xi )g''(\xi )-f''(\xi )g(\xi )=0.\]即\[\frac{f(\xi )}{g(\xi )}=\frac{f''(\xi )}{g''(\xi )}.\]

云南大学2024


解   由已知\[z=ue^{ax+by},u=u(x,y).\]求微分\[\frac{\partial z}{\partial x}=\frac{\partial u}{\partial x}e^{ax+by}+aue^{ax+by},\]\[\frac{\partial z}{\partial y}=\frac{\partial u}{\partial y}e^{ax+by}+bue^{ax+by},\]\[\frac{\partial^2z}{\partial x\partial y}=\frac{\partial^2u}{\partial x\partial y}e^{ax+by}+b\frac{\partial u}{\partial x}e^{ax+by}+a\frac{\partial u}{\partial y}e^{ax+by}+abue^{ax+by}.\]再由已知\[ \frac{\partial^2z}{\partial x\partial y}+2\frac{\partial z}{\partial x}+3\frac{\partial z}{\partial y}+6z=0.\]代入得\[\frac{\partial^2u}{\partial x\partial y}e^{ax+by}+b\frac{\partial u}{\partial x}e^{ax+by}+a\frac{\partial u}{\partial y}e^{ax+by}+abue^{ax+by}\]\[+2\frac{\partial u}{\partial x}e^{ax+by}+2aue^{ax+by}+3\frac{\partial u}{\partial y}e^{ax+by}+3bue^{ax+by}+6ue^{ax+by}=0.\]比较系数，有\[ b+2=0,a+3=0,ab+2a+3b+6=0.\]\[b=-2,a=-3.\]

云南大学2024


解\[P=y(2-f(x^2+y^2)),Q=xf(x^2+y^2).\]\[\frac{\partial P}{\partial y}=2-f(x^2+y^2)-2y^2f'(x^2+y^2),\]\[\frac{\partial Q}{\partial x}=f(x^2+y^2)+2x^2f'(x^2+y^2).\]若曲线积分与路径L无关，则必有\[\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.\]于是\[2-f(x^2+y^2)-2y^2f'(x^2+y^2)=f(x^2+y^2)+2x^2f'(x^2+y^2).\]\[\therefore (x^2+y^2)f'(x^2+y^2)+f(x^2+y^2)=1.\]解微分方程.\[令t=x^2+y^2.\]则微分方程为\[f'(t)+\frac{1}{t}f(t)=\frac{1}{t}.\]通解\[f(t)=e^{-\int\frac{1}{t}dt}(\int\frac{1}{t}e^{\int\frac{1}{t}dt}dt+C)=1+\frac{C}{t}.\]\[由f(1)=0=1+C,得C=-1.\]因此，\[f(t)=1-\frac{1}{t}，\]即f\[f(x^2+y^2)=1-\frac{1}{x^2+y^2}.\]

云南大学2024


解   \begin{align*}S
&=\displaystyle\sum_{n=1}^{\infty }n(n+2) x^{n-1}\\
&=\displaystyle\sum_{n=1}^{\infty }n(n+1) x^{n-1}+\displaystyle\sum_{n=1}^{\infty }n x^{n-1}\\
&=\displaystyle\sum_{n=1}^{\infty }( x^{n+1})''+\displaystyle\sum_{n=1}^{\infty }(x^{n})' \\
&=(\displaystyle\sum_{n=1}^{\infty } x^{n+1})''+(\displaystyle\sum_{n=1}^{\infty }x^{n})' \\
&=(\frac{x^2}{1-x})''+(\frac{x}{1-x})' \\
&=\frac{2}{(1-x)^3}+\frac{1}{(1-x)^2} \\
&=\frac{3-x}{(1-x)^3} .
\end{align*}

云南大学2024


解 已知\[\int_{-\infty }^{+\infty }e^{-x^2}dx=\sqrt{\pi }.\]\[D:[-a,a]\times [-a,a]=D_1:[-a,y]\times [-a,a]\bigcup D_2:[-a,a]\times [-a,x].\]\begin{align*}\lim_{a\to+\infty }\iint\limits_{D} &=\lim_{a\to+\infty }[\iint\limits_{D_1}+\iint\limits_{D_2}] \\
&=\lim_{a\to+\infty }[\iint\limits_{D_1}\min \{x,y\}e^{-(x^2+y^2)}dxdy+\iint\limits_{D_2}\min \{x,y\}e^{-(x^2+y^2)}dxdy] \\
&=\lim_{a\to+\infty }[\int_{-a}^{a}e^{-y^2}dy\int_{-a}^{y}xe^{-x^2}dx+\int_{-a}^{a}e^{-x^2}dx\int_{-a}^{x}ye^{-y^2}dy] \\
&=2\lim_{a\to+\infty }\int_{-a}^{a}e^{-y^2}dy\int_{-a}^{y}xe^{-x^2}dx,(对称性) \\
&=\lim_{a\to+\infty }\int_{-a}^{a}e^{-y^2}(-e^{-y^2}+e^{-a^2})dy \\
&=\lim_{a\to+\infty }[e^{-a^2}\int_{-a}^{a}e^{-y^2}dy-\int_{-a}^{a}e^{-2y^2}dy] \\
&=\lim_{a\to+\infty }[\pi e^{-a^2}-\int_{-a}^{a}e^{-2y^2}dy]\\
&=-\frac{1}{\sqrt{2}}\lim_{a\to+\infty }\int_{-a}^{a}e^{-(\sqrt{2}y)^2}d(\sqrt{2}y)\\
&=-\sqrt{\frac{\pi }{2}}.
\end{align*}