7.1 第5次作业评讲

7.1.1 解答和证明部分

解¹ 设曲面上一点的坐标为$(x,y,-x^{2}y-{\mathrm{e}}^{2x})$，则原点到该点的距离平方为$$\begin{array}{}\text{(1)}& f(x,y)=x^2+y^2+{(x^{2}y+{\mathrm{e}}^{2x})}^2\end{array}$$ 显然$\underset{(x,y)\to \mathrm{\infty }}{lim}f(x,y)=+\mathrm{\infty }$，故$f(x,y)$在${\mathbb{R}}^2$上有最小值。驻点方程为$$\begin{array}{}\text{(2)}& \mathrm{\nabla }f=2\left(\begin{array}{c}x+(x^{2}y+{\mathrm{e}}^{2x})(2xy+2{\mathrm{e}}^{2x})\\ y+(x^{2}y+{\mathrm{e}}^{2x})\cdot x^2\end{array}\right)=0\end{array}$$ 通过第二个方程解得$$\begin{array}{}\text{(3)}& y=-\frac{x^2{\mathrm{e}}^{2x}}{1+x^4}\end{array}$$ 代入第一个方程中化简可得$$\begin{array}{}\text{(4)}& g(x)=x(1+x^4{)}^2+2{\mathrm{e}}^{4x}(1-x^3+x^4)=0\end{array}$$ 求导可得$$\begin{array}{}\text{(5)}& g^{\prime }(x)=1+10x^4+9x^8+{\mathrm{e}}^{4x}(8-6x^2+8x^4)\ge 1>0\end{array}$$ 且$\underset{x\to -\mathrm{\infty }}{lim}g(x)=-\mathrm{\infty }$、$\underset{x\to +\mathrm{\infty }}{lim}g(x)=+\mathrm{\infty }$，故$g(x)$在$\mathbb{R}$上有唯一零点$x^{\ast }$。因此$(x^{\ast },y^{\ast })$为$f$在${\mathbb{R}}^2$上的唯一驻点，亦即最小值点。

为了计算得到$x^{\ast }$，我们可以使用Newton迭代法。设迭代初值$x_0=0$，利用Taylor展开构造迭代公式：$$\begin{array}{}\text{(6)}& x_{n+1}=x_n-\frac{g(x_n)}{g^{\prime }(x_n)}\end{array}$$ 利用Excel进行迭代计算，迭代至$x_8$即收敛，结果为$$\begin{array}{}\text{(7)}& x^{\ast }\approx x_8=-0.407091518544949\end{array}$$ 代入原式可得最短距离的平方为$$\begin{array}{}\text{(8)}& f(x^{\ast },y^{\ast })\approx 0.356727682455897\end{array}$$ 故最短距离为$d_{\min }=\sqrt{f(x^{\ast },y^{\ast })}=0.597266843593295$。 $◻$

解² 构造Lagrange函数$$\begin{array}{}\text{(9)}& L(x,y,z,\lambda )=x^2+y^2+z^2+\lambda (x^{2}y+{\mathrm{e}}^{2x}+z)\end{array}$$ 驻点方程为$$\begin{array}{}\text{(10)}& \mathrm{\nabla }L=\left(\begin{array}{c}2x+\lambda (2xy+2{\mathrm{e}}^{2x})\\ 2y+\lambda x^2\\ 2z+\lambda \\ x^{2}y+{\mathrm{e}}^{2x}+z\end{array}\right)=0⟹\{\begin{array}{l}\lambda =0\\ x^{2}y+{\mathrm{e}}^{2x}+z=0\end{array}⟹z=-x^{2}y-{\mathrm{e}}^{2x}\end{array}$$ 消去$z,\lambda$后，得到的方程与解¹相同。 $◻$

解¹ 显然$x,y,z>1$，构造Lagrange函数$$\begin{array}{}\text{(11)}& L(x,y,z,\lambda )=x+y+z+\lambda (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1)\end{array}$$ 驻点方程为$$\begin{array}{}\text{(12)}& \mathrm{\nabla }L=\left(\begin{array}{c}1-\frac{\lambda }{x^2}\\ 1-\frac{\lambda }{y^2}\\ 1-\frac{\lambda }{z^2}\\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\end{array}\right)=0⟹\{\begin{array}{l}x=y=z\\ \frac{3}{x}-1=0\end{array}⟹x=y=z=3,\lambda =9\end{array}$$ $L$关于$(x,y,z)$的Hesse矩阵为$$\begin{array}{}\text{(13)}& H=\left(\begin{array}{ccc}\frac{2\lambda }{x^3}& 0& 0\\ 0& \frac{2\lambda }{y^3}& 0\\ 0& 0& \frac{2\lambda }{z^3}\end{array}\right)=\frac{2}{3}I\end{array}$$ 其显然正定，故$(x,y,z)=(3,3,3)$为极小值点。

记$D=\{(x,y,z)\in {\mathbb{R}}^3|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\}$，其为闭集。由于$f(x,y,z):=x+y+z$连续，且$\underset{D\ni (x,y,z)\to \mathrm{\infty }}{lim}=+\mathrm{\infty }$，故$f$在闭集$D$上有最小值，即$(3,3,3)$为$f$在$D$上的最小值点。

对于上确界，令$y=z=\frac{2x}{x-1}>0$，此时$$\begin{array}{}\text{(14)}& x+y+z=\frac{x(x+3)}{x-1}\end{array}$$ 当$x\to 1^{+}$或$x\to +\mathrm{\infty }$时，都有$x+y+z\to +\mathrm{\infty }$。故$x+y+z$的取值范围为$[9,+\mathrm{\infty })$。 $◻$

解² 从约束条件中解出$z$可得$$\begin{array}{}\text{(15)}& z=\frac{1}{1-x^{-1}-y^{-1}}=\frac{xy}{xy-x-y}>0\end{array}$$ 则$x,y$应当满足$$\begin{array}{}\text{(16)}& x>1,y>\frac{x}{x-1}\end{array}$$ 构造目标函数$$\begin{array}{}\text{(17)}& f(x,y)=x+y+z(x,y)=x+y+\frac{xy}{xy-x-y}\end{array}$$ 驻点方程为$$\begin{array}{}\text{(18)}& \mathrm{\nabla }f=\left(\begin{array}{c}\frac{x(y-1)(xy-x-2y)}{(xy-x-y{)}^2}\\ \frac{y(x-1)(xy-y-2x)}{(xy-x-y{)}^2}\end{array}\right)=0⟹\{\begin{array}{l}xy-x-2y=0\\ xy-y-2x=0\end{array}⟹x=y=3\end{array}$$ $f$的Hesse矩阵为$$\begin{array}{}\text{(19)}& H=\frac{2}{(xy-x-y{)}^3}\left(\begin{array}{cc}{y}^2(y-1)& xy\\ xy& x^2(x-1)\end{array}\right)=\frac{2}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\end{array}$$ 其显然正定，故$(x,y)=(3,3)$为极小值点，此时$z=3$。后续讨论同理。 $◻$

解³ 由权方和不等式可得$$\begin{array}{}\text{(20)}& x+y+z=\frac{1^2}{x^{-1}}+\frac{1^2}{y^{-1}}+\frac{1^2}{z^{-1}}\ge \frac{(1+1+1{)}^2}{x^{-1}+y^{-1}+z^{-1}}=9\end{array}$$ 后续讨论同理。 $◻$

解 曲面$f(x,y,z)=\sqrt{x}+\sqrt{y}+\sqrt{z}-1=0$上的一点$(x_0,y_0,z_0)$处的切平面为$$\begin{array}{}\text{(21)}& \frac{x-x_0}{2\sqrt{x_0}}+\frac{y-y_0}{2\sqrt{y_0}}+\frac{z-z_0}{2\sqrt{z_0}}=0⟹\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}+\frac{z}{\sqrt{z_0}}=\sqrt{x_0}+\sqrt{y_0}+\sqrt{z_0}=1\end{array}$$ 该平面与三个坐标平面围成的四面体体积为$$\begin{array}{}\text{(22)}& V=\frac{1}{6}\sqrt{x_0{y}_0{z}_0}\end{array}$$ 令$(a,b,c)=(\sqrt{x_0},\sqrt{y_0},\sqrt{z_0})$，则原题转化为：正数$a,b,c$满足$a+b+c=1$，求$V=\frac{1}{6}abc$的最大值。由AM-GM不等式可得$$\begin{array}{}\text{(23)}& V=\frac{1}{6}abc\le \frac{1}{6}{\left(\frac{a+b+c}{3}\right)}^3=\frac{1}{6}\cdot \frac{1}{27}=\frac{1}{162}\end{array}$$ 此时$a=b=c=\frac{1}{3}$，则切平面为$x+y+z=\frac{1}{3}$。 $◻$

解 参见例5.3.9，取值范围为$[\frac{3}{2},2)$，最小值在等边三角形时取得，上确界在等腰三角形的腰与底的比值趋于$+\mathrm{\infty }$时以极限形式取得。 $◻$

解 由于被积函数关于$(x,y)$在积分域二元连续，故可交换积分顺序，计算可得$$\begin{array}{}\text{(25)}& {\int }_{-\pi /2}^{\pi /2}{\mathrm{e}}^y\sin \frac{x}{y}\mathrm{d}x={-y{\mathrm{e}}^y\cos \frac{x}{y}|}_{x=-\pi /2}^{\pi /2}=-y{\mathrm{e}}^y\cos \frac{\pi }{2y}+y{\mathrm{e}}^y\cos \frac{-\pi }{2y}=0\end{array}$$ 故原积分为$0$。 $◻$

解¹(利用$I(0)$) 限制$x\in [\epsilon ,+\mathrm{\infty })$，其中$\epsilon >0$，被积函数对参数求偏导可得$$\begin{array}{}\text{(27)}& {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\partial }{\partial x}\frac{\ln (1+x^2{t}^2)}{1+t^2}\mathrm{d}t={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2x{t}^2}{(1+t^2)(1+x^2{t}^2)}\mathrm{d}t\end{array}$$ 注意到$$\begin{array}{}\text{(28)}& \frac{2x{t}^2}{(1+t^2)(1+x^2{t}^2)}\le \frac{2}{\epsilon (1+t^2)}\frac{x^2{t}^2}{1+x^2{t}^2}\le \frac{2}{\epsilon (1+t^2)}\end{array}$$ 由Weierstrass强函数判别法知原积分关于参数$x\in [\epsilon ,+\mathrm{\infty })$一致收敛。计算可得，当$x\ne 1$时，有$$\begin{array}{}\text{(29)}& \begin{array}{rl}{I}^{\prime }(x)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2x}{1-x^2}(\frac{1}{1+x^2{t}^2}-\frac{1}{1+t^2})\mathrm{d}t\\ & =\frac{2}{1-x^2}{(\arctan xt-x\arctan t)}_{t=-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{2\pi (1-x)}{1-x^2}=\frac{2\pi }{1+x}\end{array}\end{array}$$ 当$x=1$时，有$$\begin{array}{}\text{(30)}& \begin{array}{rl}{I}^{\prime }(1)& =4{\int }_0^{+\mathrm{\infty }}\frac{t^2}{(1+t^2{)}^2}\mathrm{d}t\stackrel{s=t^{-1}}{=}4{\int }_0^{+\mathrm{\infty }}\frac{s^{-2}}{(1+s^{-2}{)}^2}\frac{\mathrm{d}s}{s^2}=4{\int }_0^{+\mathrm{\infty }}\frac{1}{(s^2+1{)}^2}\mathrm{d}s\\ 2I^{\prime }(1)& =4{\int }_0^{+\mathrm{\infty }}\frac{t^2+1}{(1+t^2{)}^2}\mathrm{d}t=2\pi ⟹I^{\prime }(1)=\pi =\frac{2\pi }{1+1}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(31)}& I(x)=I(\epsilon )+{\int }_{\epsilon }^x\frac{2\pi }{1+t}\mathrm{d}t=I(\epsilon )+2\pi \ln \frac{1+x}{1+\epsilon }\end{array}$$ 令$\epsilon \to 0^{+}$，设$A>0$满足$x>A⟹\ln (1+x)<x^{1/4}$，注意到$$\begin{array}{}\text{(32)}& \begin{array}{rl}\underset{\epsilon \to 0^{+}}{lim}I(\epsilon )& =2\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{+\mathrm{\infty }}\frac{\ln (1+{\epsilon }^2{t}^2)}{1+t^2}\mathrm{d}t\le 2\underset{\epsilon \to 0^{+}}{lim}[{\int }_0^A\frac{{\epsilon }^2{t}^2}{1+t^2}\mathrm{d}t+{\int }_A^{+\mathrm{\infty }}\frac{\sqrt{\epsilon t}}{1+t^2}\mathrm{d}t]\\ & \le 2\underset{\epsilon \to 0^{+}}{lim}[{\epsilon }^{2}A+\sqrt{\epsilon }{\int }_A^{+\mathrm{\infty }}\frac{\mathrm{d}t}{t^{3/2}}]=0\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(33)}& I(x)=2\pi \ln (1+x)\end{array}$$ $◻$

解²(利用$I(1)$) 当$x\ne 0$时，限制$x\in [\epsilon ,+\mathrm{\infty })$，其中$\epsilon >0$，类似上例可得$$\begin{array}{}\text{(34)}& I(x)=I(x_0)+{\int }_{x_0}^x\frac{2\pi }{1+t}\mathrm{d}t=I(x_0)+2\pi \ln \frac{1+x}{1+x_0}\end{array}$$ 注意到$$\begin{array}{}\text{(35)}& \begin{array}{rl}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\ln (1+t^2)}{1+t^2}\mathrm{d}t& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\ln (1+t^2)\mathrm{d}\arctan t\stackrel{\theta =\arctan t}{=}{\int }_{-\pi /2}^{\pi /2}\ln (1+{\tan }^2\theta )\\ & =-4{\int }_0^{\pi /2}\ln \cos \theta \mathrm{d}\theta =-4\cdot (-\frac{\pi }{2}\ln 2)=2\pi \ln 2\end{array}\end{array}$$ 代入$x_0=1$可得$$\begin{array}{}\text{(36)}& I(x)=2\pi \ln 2+2\pi \ln \frac{1+x}{2}=2\pi \ln (1+x)\end{array}$$ $◻$

解³(升幂) 设$y=\sqrt{x}\in [0,+\mathrm{\infty })$，考虑计算$$\begin{array}{}\text{(37)}& J(y)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\ln (1+y^4{t}^2)}{1+t^2}\mathrm{d}t\end{array}$$ 被积函数对参数求偏导可得$$\begin{array}{}\text{(38)}& {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\partial }{\partial y}\frac{\ln (1+y^4{t}^2)}{1+t^2}\mathrm{d}t={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{4y^3{t}^2}{(1+t^2)(1+y^4{t}^2)}\mathrm{d}t=\frac{2\pi }{1+y^2}\cdot 2y\end{array}$$ 注意到$$\begin{array}{}\text{(39)}& \frac{4y^3{t}^2}{1+y^4{t}^2}=\frac{t^2}{\frac{1}{4}(\frac{1}{y^3}+\frac{y{t}^2}{3}+\frac{y{t}^2}{3}+\frac{y{t}^2}{3})}\le \frac{t^2}{\sqrt[4]{\frac{1}{y^3}\cdot \frac{y{t}^2}{3}\cdot \frac{y{t}^2}{3}\cdot \frac{y{t}^2}{3}}}=3^{3/4}\sqrt{|t|}\end{array}$$ 故$$\begin{array}{}\text{(40)}& {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{4y^3{t}^2}{(1+t^2)(1+y^4{t}^2)}\mathrm{d}t\le 2\cdot 3^{3/4}{\int }_0^{+\mathrm{\infty }}\frac{\sqrt{t}}{1+t^2}\mathrm{d}t\end{array}$$ 关于参数$y\in [0,+\mathrm{\infty })$一致收敛，因此$$\begin{array}{}\text{(41)}& J(y)=J(0)+{\int }_0^y\frac{2\pi }{1+t^2}\cdot 2t\mathrm{d}t=2\pi \ln (1+y^2)⟹I(x)=2\pi \ln (1+x)\end{array}$$ $◻$

注 本题最大的难点在于：若直接将被积函数对$x$求偏导，得到的积分对$x\in [0,+\mathrm{\infty })$并不一致收敛！正确的做法是：(1)考虑使用内闭一致收敛，利用极限延伸至$x=0$；(2)与(1)相似，但是费些力气计算出$I(1)$；(3)考虑将$x^2$升幂为$y^4$。