國立臺灣大學一百零六學年度轉學生入學考試試題詳解

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1. ($20\%$) Determine the convergence or divergence of the following series:

   $\displaystyle\sum_{n=1}^{\infty}\left(\frac{\cos\left(\pi n\right)}{\tanh\left(\pi n\right)}\frac{1-\cosh\left(\frac1n\right)}{\sin\left(\frac1n\right)}\right)$.

訣竅
運用交錯級數審歛法即可。
解法

由於 $\displaystyle\cos\left(\pi n\right)=\left(-1\right)^n$，因此本題之級數可寫為 $\displaystyle\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{\cosh\left(\frac1n\right)-1}{\sin\left(\frac{1}{n}\right)\tanh\left(\pi n\right)}$，此為交錯級數。

設 $\displaystyle f\left(x\right)=\frac{\cosh\left(\frac{1}{x}\right)-1}{\sin\left(\frac1x\right)\tanh\left(\pi x\right)}=\left[\cosh\left(\frac{1}{x}\right)-1\right]\csc\left(\frac1x\right)\coth\left(\pi x\right)$，並且記 $\displaystyle g\left(x\right)=\cosh\left(\frac1x\right)-1$、$\displaystyle h\left(x\right)=\csc\left(\frac{1}{x}\right)$、$k\left(x\right)=\coth\left(\pi x\right)$，如此對 $x\geq1$ 有

$\begin{aligned} &g'\left(x\right)=-\frac{\sinh\left(1/x\right)}{x^2}<0,\\&h'\left(x\right)=-\frac{\csc\left(1/x\right)\cot\left(1/x\right)}{x^2}<0,\\&k'\left(x\right)=-\pi^2\mbox{csch}^2\left(\pi x\right)<0.\end{aligned}$

且由 $g$、$h$、$k$ 在 $x\geq1$ 時非負，因此可知 $f$ 在 $\geq1$ 時遞減。另一方面由 $\displaystyle\lim_{x\to\infty}k\left(x\right)=1$、$\displaystyle\lim_{x\to\infty}g\left(x\right)h\left(x\right)=0$（由泰勒展開觀察之），因此有 $\displaystyle\lim_{x\to\infty}f\left(x\right)=0$。

因此由交錯級數審歛法，交錯之正項遞減且收斂至零，因此此級數收斂。



2. ($20\%$) For a second continuously differentiable function, $f\left(x,y,z\right)$, defined on the space $\mathbb{R}^3$, its Laplacian is defined to be

   $\displaystyle\bigtriangleup f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}$.

   Derive the formula of $\bigtriangleup f$ in the spherical coordinate system.
訣竅
熟悉球極座標變換並按照多變數連鎖律仔細計算之即可。
解法

由球極座標變換，設 $\left\{\begin{aligned} &x=\rho\sin\phi\cos\theta\\&y=\rho\sin\phi\sin\theta\\&z=\rho\cos\phi\end{aligned}\right.$，即有 $\left\{\begin{aligned} &\rho=\sqrt{x^2+y^2+z^2}\\&\phi=\arccos\left(\frac{z}{\rho}\right)\\&\theta=\arctan\left(\frac{y}x\right)\end{aligned}\right.$。
首先分別表達 $f$ 對 $x$、$y$ 與 $z$ 的一階偏導數如下

$\displaystyle\begin{aligned}&\frac{\partial f}{\partial x}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial x}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial x}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial x}\\&\frac{\partial f}{\partial y}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial y}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial y}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial y}\\&\frac{\partial f}{\partial z}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial z}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial z}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial z}\end{aligned}$

因此我們計算下列各項偏導有

$\begin{array}{lll}\displaystyle\frac{\partial\rho}{\partial x}=\sin\phi\cos\theta,\quad&\displaystyle\frac{\partial\phi}{\partial x}=\frac{\cos\theta\cos\phi}{\rho},\quad&\displaystyle\frac{\partial\theta}{\partial x}=-\frac{\sin\theta}{\rho\sin\phi},\\\displaystyle\frac{\partial\rho}{\partial y}=\sin\phi\sin\theta,\quad&\displaystyle\frac{\partial\phi}{\partial y}=\frac{\sin\theta\cos\phi}{\rho},\quad&\displaystyle\frac{\partial\theta}{\partial y}=\frac{\cos\theta}{\rho\sin\phi},\\\displaystyle\frac{\partial\rho}{\partial z}=\cos\phi,\quad&\displaystyle\frac{\partial\phi}{\partial z}=-\frac{\sin\phi}{\rho},\quad&\displaystyle\frac{\partial\theta}{\partial z}=0.\end{array}$

由此可得

$\begin{array}{l}\displaystyle\frac{\partial f}{\partial x}=\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\\\displaystyle\frac{\partial f}{\partial y}=\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\\\displaystyle\frac{\partial f}{\partial z}=\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}{\rho}\frac{\partial f}{\partial\phi}\end{array}$

據此計算二階偏導數。首先是 $\displaystyle\frac{\partial^2f}{\partial x^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial x^2}=&\sin\phi\cos\theta\frac{\partial}{\partial\rho}\frac{\partial f}{\partial x}+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial}{\partial\phi}\frac{\partial f}{\partial x}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial}{\partial\theta}\frac{\partial f}{\partial x}\\=&\sin\phi\cos\theta\frac{\partial}{\partial\rho}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial}{\partial\phi}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial}{\partial\theta}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\=&\sin^2\phi\cos^2\theta\frac{\partial^2f}{\partial\rho^2}+\frac{\cos^2\theta\cos^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{\sin^2\theta}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}\\&+\frac{2\cos^2\theta\sin\phi\cos\phi}{\rho}\frac{\partial^2f}{\partial\rho\partial\phi}-\frac{2\sin\theta\cos\theta}{\rho}\frac{\partial^2f}{\partial\rho\partial\theta}-\frac{2\sin\theta\cos\theta\cos\phi}{\rho^2\sin\phi}\frac{\partial^2f}{\partial\phi\partial\theta}\\&+\frac{\cos^2\theta\cos^2\phi+\sin^2\theta}{\rho}\frac{\partial f}{\partial\rho}+\frac{\sin^2\theta\cos\phi-2\sin\phi^2\cos\phi\cos^2\theta}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}\\&+\frac{2\sin\theta\cos\theta}{\rho^2\sin^2\phi}\frac{\partial f}{\partial\theta}\end{aligned}$

接著是 $\displaystyle\frac{\partial^2f}{\partial y^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial y^2}=&\sin\phi\sin\theta\frac{\partial}{\partial\rho}\frac{\partial f}{\partial y}+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial}{\partial\phi}\frac{\partial f}{\partial y}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial}{\partial\theta}\frac{\partial f}{\partial y}\\=&\sin\phi\sin\theta\frac{\partial}{\partial\rho}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial}{\partial\phi}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial}{\partial\theta}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}{\rho}\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\=&\sin^2\phi\sin^2\theta\frac{\partial^2f}{\partial\rho^2}+\frac{\sin^2\theta\cos^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{\cos^2\theta}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}\\&+\frac{2\sin^2\theta\sin\phi\cos\phi}{\rho}\frac{\partial^2f}{\partial\rho\partial\phi}+\frac{2\sin\theta\cos\theta}{\rho}\frac{\partial^2f}{\partial\rho\partial\theta}+\frac{2\sin\theta\cos\theta\cos\phi}{\rho^2\sin\phi}\frac{\partial^2f}{\partial\phi\partial\theta}\\&+\frac{\sin^2\theta\cos^2\phi+\cos^2\theta}{\rho^2}\frac{\partial f}{\partial\rho}+\frac{\cos^2\theta\cos\phi-2\sin^2\theta\sin^2\phi\cos\theta}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}\\&-\frac{2\sin\theta\cos\theta}{\rho^2\sin^2\phi}\frac{\partial f}{\partial\theta}\end{aligned}$

最後是 $\displaystyle\frac{\partial^2f}{\partial z^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial z^2}=&\cos\phi\frac{\partial}{\partial\rho}\frac{\partial f}{\partial z}-\frac{\sin\phi}{\rho}\frac{\partial}{\partial\phi}\frac{\partial f}{\partial z}\\=&\cos\phi\frac{\partial}{\partial\rho}\left(\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}{\rho}\frac{\partial f}{\partial\phi}\right)-\frac{\sin\phi}{\rho}\frac{\partial}{\partial\phi}\left(\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}{\rho}\frac{\partial f}{\partial\phi}\right)\\=&\cos^2\phi\frac{\partial^2f}{\partial\rho^2}+\frac{\sin^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}-\frac{2\sin\phi\cos\phi}{\rho}\frac{\partial^2f}{\partial\rho\partial\phi}+\frac{\sin^2\phi}{\rho}\frac{\partial f}{\partial\rho}+\frac{2\sin\phi\cos\phi}{\rho^2}\frac{\partial f}{\partial\phi}\end{aligned}$

藉由以上的計算將三者加總可得

$\displaystyle\bigtriangleup f=\frac{\partial^2f}{\partial\rho^2}+\frac{1}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{1}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}+\frac2\rho\frac{\partial f}{\partial\rho}+\frac{\cos\phi}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}$



3. ($20\%$) Suppose that $f\left(x\right)>0$ is a continuously differentiable function defined on $x>1$. Let $S$ be the surface of revolution of the graph $y=f\left(x\right)$ about the $x$-axis. Let $E\subset\mathbb{R}^3$ be the solid enclosed by $S$ and $x=1$. More precisely,

   $E=\left\{\left(x,y,z\right)\in\mathbb{R}^3|x\geq1\mbox{ and }\sqrt{y^2+z^2}\leq f\left(x\right)\mbox{ for each }x\geq1\right\}$.

   If the surface area of $S$ is finite, prove that the volume of $E$ must also be finite.
訣竅
此定理之逆敘述為經典之悖論：存在一旋轉體其體積有限但表面積無限。在估算體積時應充分留意表面積有限可先得到函數值有界，並由此估量體積。
解法
由於表面積 $A$ 有限，因此我們可以注意到
$\begin{aligned}\displaystyle\lim_{t\to\infty}\sup_{x\geq t}\left[f^2\left(x\right)-f^2\left(1\right)\right]&=\limsup_{x\to\infty}\int_1^x\left(f^2\left(s\right)\right)'ds\leq\int_1^{\infty}\left|\left(f^2\left(s\right)\right)'\right|ds\\&=\int_1^{\infty}2f\left(s\right)\left|f'\left(s\right)\right|ds\leq\int_1^{\infty}2f\left(s\right)\sqrt{1+\left(\frac{dy}{dx}\left(s\right)\right)^2}ds=\frac{A}{\pi}<\infty\end{aligned}$
因此存在 $t_0$ 使得 $\displaystyle\sup_{x\geq t_0}f\left(x\right)$ 有限，如此由 $f$ 的連續性可知 $f$ 在 $\left[1,\infty\right)$ 上有界（記為 $M$），因此 $E$ 的體積可被估算如下：
$\displaystyle\int_1^{\infty}\pi f^2\left(x\right)dx\leq\int_1^{\infty}\frac{M}{2}\cdot2\pi f\left(x\right)dx\leq\frac{M}2\int_1^{\infty}2\pi f\left(x\right)\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx=\frac{MA}2$.
如此由 $A$ 的有限性即可推得 $V$ 的有限性，證明完畢。


4. Let $C$ be the curve defined by $\left\{\left(x,y\right)\in\mathbb{R}^2|x^4+2x^2y^2+y^4-2x\left(x^2-y^2\right)=0\right\}$.
   1. ($10\%$) Sketch the curve $C$.
   2. Calculate the area of the region enclosed by $C$.
訣竅
運用極座標變換後描點作圖並且運用極座標下的面積公式計算其面積。
解法

1. 運用極座標變換，令 $\left\{\begin{aligned}x=&r\cos\theta\\y=&r\sin\theta\end{aligned}\right.$，如此方程可改寫為

   $r^4\cos^4\theta+2r^4\cos^2\theta\sin^2\theta+r^4\sin^4\theta-2r^3\cos\theta\cos2\theta=0$

   即有 $r=2\cos\theta\cos2\theta$。描點繪圖如下：
   
2. 運用面積公式可得

   $\displaystyle\begin{aligned}A=&\frac{1}{2}\int_0^{\pi}r^2d\theta\\=&2\int_0^{\pi}\cos^2\theta\cos^22\theta d\theta\\=&\frac{1}{8}\int_0^{\pi}\left(1+\cos2\theta\right)\left(1+\cos4\theta\right)d\theta\\=&\frac{1}{2}\int_0^{\pi}\left(1+\cos2\theta+\cos4\theta+\cos2\theta\cos4\theta\right)d\theta\\=&\frac{\pi}{2}+\frac{1}{2}\int_0^{\pi}\frac{\cos6\theta+\cos2\theta}{2}d\theta\\=&\frac{\pi}{2}\end{aligned}$



5. ($20\%$) Let $a,b,c$ be three positive numbers with $a>b>c$. Let $\Sigma$ be the surface defined by

   $\displaystyle\frac{\left(x-b\right)^2}{a^2}+\frac{\left(y-c\right)^2}{b^2}+\frac{z^2}{c^2}=4$.

   Note that $\Sigma$ encloses a bounded solid. Set $\vec{n}$ to be the unit outer normal with respect to that solid. Consider the vector field

   $\displaystyle\vec{F}=\left(\frac{x}{\left(x^2+y^2+z^2\right)^{\frac32}}+\cos\left(yz\right),\frac{y}{\left(x^2+y^2+z^2\right)^{\frac32}}+y,\frac{z}{\left(x^2+y^2+z^2\right)^{\frac32}}+e^{z^2}\right)$.

   Calculate the flux integral $\displaystyle\int_{\Sigma}\vec{F}\cdot\vec{n}\,dS$.
訣竅
運用高斯散度定理並且對奇異點圈出另外處理。
解法

首先，記 $\vec{G}$ 與 $\vec{H}$ 如下

$\displaystyle\begin{aligned} &\vec{G}=\left(\frac{x}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}},\frac{y}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}},\frac{z}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\right),\\&\vec{H}=\left(\cos\left(yz\right),y,e^{z^2}\right).\end{aligned}$

並且設 $\Omega$ 為 $\Sigma$ 所包圍之區域，那麼可以發現 $\vec{H}$ 在 $\Omega$ 上有一階地連續偏導函數，因此直接使用高斯散度定理可得

$\displaystyle\begin{aligned}\int_{\Sigma}\vec{H}\cdot\vec{n}\,dS&=\iiint_{\Omega}\nabla\cdot\vec{H}\,dV\\&=\iiint_{\Omega}\left(1+2ze^{z^2}\right)dV\\&=\iiint_{\Omega}1\,dV+\iiint_{\Omega}2ze^{z^2}dV\\&=\frac43\pi\cdot2a\cdot2b\cdot2c+\iint_D\int_{-c\sqrt{4-\frac{\left(x-b\right)^2}{a^2}-\frac{\left(y-c\right)^2}{b^2}}}^{c\sqrt{4-\frac{\left(x-b\right)^2}{a^2}-\frac{\left(y-c\right)^2}{b^2}}}2ze^{z^2}dzdA=\frac{32}{3}\pi abc,\end{aligned}$

其中 $\displaystyle D=\left\{\left(x,y\right)\in\mathbb{R}^2:\frac{\left(x-b\right)^2}{a^2}+\frac{\left(y-c\right)^2}{b^2}\leq4\right\}$。另一方面，因為 $\left(0,0,0\right)\in\Omega$，因此 $\vec{G}$ 在 $\Omega$ 上沒有連續地一階偏導數。可以取充分小的正數 $\varepsilon$ 使得 $B_\varepsilon(\vec{0})\subset\Omega$，從而 $\vec{G}$ 在 $\Omega-B_\varepsilon\left(\vec{0}\right)$ 上有連續地一階偏導函數，那麼在其上使用高斯散度定理可以直接得到

$\displaystyle\int_{\Sigma\cup\partial B_\varepsilon(\vec{0})}\vec{G}\cdot\vec{n}\,dS=\iiint_{\Omega-B_\varepsilon(\vec{0})}\nabla\cdot\vec{G}\,dV=0$

因此

$\displaystyle\int_{\Sigma}\vec{G}\cdot\vec{n}\,dS=\int_{\partial B_\varepsilon(\vec{0})}\vec{G}\cdot\vec{n}\,dS,$

其中 $\vec{n}$ 均表示所積分之區域指向外的單位法向量（在不同的積分中可能所指之方向會隨積分範圍而有所不同）。藉由直接計算可知

$\displaystyle\int_{\partial B_\varepsilon(\vec{0})}\vec{G}\cdot\vec{n}\,dS=\iint_{\partial B_\varepsilon(\vec{0})}\frac{\left(x,y,z\right)}{\varepsilon^3}\cdot\frac{\left(x,y,z\right)}{\varepsilon}dS=4\pi\varepsilon^2\cdot\frac{1}{\varepsilon^2}=4\pi$.

因此本題之所求為

$\displaystyle\int_{\Sigma}\vec{F}\cdot\vec{n}\,dS=\int_{\Sigma}\vec{G}\cdot\vec{n}\,dS+\int_{\Sigma}\vec{H}\cdot\vec{n}\,dS=4\pi+\frac{32}3\pi abc$.