國立臺灣大學八十九學年度轉學生入學考試試題詳解

1. Power series $\displaystyle\sum a_nx^n=\sum_{n=1}^{\infty}\frac{n!}{n^n}x^n$. Find its radius of convergence $r$ so that this power series converges absolutely when $\lvert x\rvert< r$. Is it also convergent when $x=+r$, $x=-r$?　($20/100$)
訣竅
由比值審歛法(Ratio Test)求出收斂半徑後，利用不等式做出適當的估計以判斷收斂區間的端點是否能使級數收斂。
解法

設 $\displaystyle a_n=\frac{n!}{n^n}$，由比值審歛法可得收斂半徑

$\displaystyle r=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim_{n\to\infty}\frac{\displaystyle\frac{n!}{n^n}}{\displaystyle\frac{\left(n+1\right)!}{\left(n+1\right)^{n+1}}}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$.

對於 $x=e$ 的情形，我們考慮如下的事實

$\displaystyle e^n=\sum_{k=0}^{\infty}\frac{n^k}{k!}$

因此可以知道對所有正整數 $n$ 恆有 $\displaystyle e^n\geq\frac{n^n}{n!}$，即有 $\displaystyle\frac{n!}{\displaystyle\left(\frac{n}{e}\right)^n}\geq1$。如此易知 $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}e^n$ 發散。

而對 $x=-e$ 的情形可以援引上述的事實，可以注意到 $\displaystyle\lim_{n\to\infty}\frac{n!}{n^n}\left(-e\right)^n$ 不收斂至零，因此該級數亦發散。



2. Vector field $\vec{F}=\left(e^x-y^3\right)\vec{i}+\left(e^y+x^3\right)\vec{j}+e^z\vec{k}$. Closed curve $c=\left\{\vec{r}=\cos t\vec{i}+\sin t\vec{j}+\sin2t\vec{k},0\leq t\leq2\pi\right\}$. $c$ lies on the surface $z=2xy$. Find the line integral $\displaystyle\oint_c\vec{F}\cdot d\vec{r}=?$　($20/100$)
訣竅
根據參數化直接計算即可。
解法
參數化後代入後可以計算如下：
$\displaystyle\begin{aligned}\oint_c\vec{F}\cdot d\vec{r}=&\int_0^{2\pi}\left(e^{\cos t}-\sin^3t,e^{\sin t}+\cos^3t,e^{\sin2t}\right)\cdot\left(-\sin t,\cos t,2\cos2t\right)dt\\=&\int_0^{2\pi}\left(-\sin te^{\cos t}+\cos te^{\sin t}+\sin^4t+\cos^4t+2\cos2te^{\sin2t}\right)dt\\=&\left.e^{\cos t}+e^{\sin t}+e^{\sin2t}\right|_0^{2\pi}+\int_0^{2\pi}\left[\left(\frac{1-\cos2t}2\right)^2+\left(\frac{1+\cos2t}2\right)^2\right]dt\\=&\frac{1}2\int_0^{2\pi}\left(1+\cos^22t\right)dt\\=&\frac{1}2\int_0^{2\pi}\left(1+\frac{1+\cos4t}2\right)dt\\=&\left.\frac{1}2\left(\frac{3t}2+\frac{\sin4t}8\right)\right|_0^{2\pi}\\=&\frac{3\pi}2\end{aligned}$
事實上 $\vec{F}=\vec{F}_c+\vec{F}_{nc}$，其中 $\vec{F}_c=\left(e^x,e^y,e^z\right)$，而 $\vec{F}_{nc}=\left(-y^3,x^3,0\right)$。因此我們求線積分所得的量乃由非保守力作功而得。


3. $\displaystyle u=u\left(t,x,y\right)=\frac{1}te^{-\frac{x^2+y^2}{4t}}$, $\displaystyle\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=?$　($20/100$)
訣竅
直接計算即可。另外可以注意到本題為熱核(Heat Kernel)代入熱方程的驗算過程，為偏微分方程(Partial Differential Equation)的基本習題之一。
解法
藉由直接計算，我們可以得到一階偏導數如下：
$\begin{aligned}&\frac{\partial u}{\partial t}=-\frac1{t^2}e^{-\frac{x^2+y^2}{4t}}+\frac1te^{-\frac{x^2+y^2}{4t}}\times\frac{x^2+y^2}{4t^2}\\&\frac{\partial u}{\partial x}=-\frac1te^{-\frac{x^2+y^2}{4t}}\times\frac{x}{2t}\\&\frac{\partial u}{\partial y}=-\frac1te^{-\frac{x^2+y^2}{4t}}\times\frac{y}{2t}\end{aligned}$
接著我們再計算一些二階偏導數如下：
$\displaystyle\begin{aligned}&\frac{\partial^2u}{\partial x^2}=-\frac{1}{2t^2}e^{-\frac{x^2+y^2}{4t}}+\frac{x^2}{4t^3}e^{-\frac{x^2+y^2}{4t}}\\&\frac{\partial^2u}{\partial y^2}=-\frac1{2t^2}e^{-\frac{x^2+y^2}{4t}}+\frac{y^2}{4t^3}e^{-\frac{x^2+y^2}{4t}}\end{aligned}$
由此可知 $\displaystyle\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0$。


4. Cone $z^2=2x^2+y^2$. Find a vector $\vec{a}$ perpendicular to which the cone has circular cone sections.　($20/100$)
訣竅
考慮曲面函數後計算其其梯度即為法向量。
解法
設 $F\left(x,y,z\right)=2x^2+y^2-z^2$，考慮曲面 $F\left(x,y,z\right)=0$，則垂直於此曲面的方向為 $\nabla F\left(x,y,z\right)=\left(4x,2y,-2z\right)=\left(4x,2y,-2\sqrt{2x^2+y^2}\right)$。隨意地取 $\left(x,y\right)=\left(1,1\right)$，如此即得 $\vec{a}=\left(4,2,-2\sqrt{3}\right)$。


5. Implicitly solve $\left\{\begin{aligned}xy^2+zu+v^2=3\\x^3z+2y-uv=2\\xu+yv-xyz=1\end{aligned}\right.$　for $x=x\left(u,v\right)$, $y=y\left(u,v\right)$ and $z=z\left(u,v\right)$ at the point $\left(x,y,z,u,v\right)=\left(1,1,1,1,1\right)$.　$\displaystyle\frac{\partial y}{\partial u}=?$　($20/100$)
訣竅
應用隱函數定理即可。
解法

考慮函數 $F:\mathbb{R}^5\to\mathbb{R}^3$ 如下：

$F\left(x,y,z,u,v\right)=\left[\begin{aligned}F_1\left(x,y,z,u,v\right)\\F_2\left(x,y,z,u,v\right)\\F_3\left(x,y,z,u,v\right)\end{aligned}\right]=\left[\begin{aligned} &xy^2+zx+v^2-3\\&x^3z+2y-uv-2\\&xu+yv-xyz-1\end{aligned}\right]$

易知 $F\left(1,1,1,1,1\right)=\left[\begin{aligned}0\\0\\0\end{aligned}\right]$。現在我們對變數 $\left(x,y,z\right)$ 求 $F$ 的 Jacobian 行列式以檢查是否符合隱函數定理的條件：

$\det\begin{bmatrix}\displaystyle\frac{\partial F_1}{\partial x}&\displaystyle\frac{\partial F_1}{\partial y}&\displaystyle\frac{\partial F_1}{\partial z}\\\displaystyle\frac{\partial F_2}{\partial x}&\displaystyle\frac{\partial F_2}{\partial y}&\displaystyle\frac{\partial F_2}{\partial z}\\\displaystyle\frac{\partial F_3}{\partial x}&\displaystyle\frac{\partial F_3}{\partial y}&\displaystyle\frac{\partial F_3}{\partial z}\end{bmatrix}=\det\begin{bmatrix}y^2+z&2xy&x\\3x^2z&2&x^3\\u-yz&v-xz&-xy\end{bmatrix}$

可知在 $\left(1,1,1,1,1\right)$ 的行列式值為 $4$，從而知道在 $\left(1,1,1,1,1\right)$ 附近可用 $u,v$ 解出 $x,y,z$，亦即 $\left(u,v\right)$ 在 $\left(1,1\right)$ 附近時有$ x=x\left(u,v\right)$、$y=y\left(u,v\right)$、$z=z\left(u,v\right)$。

因此我們有 $F\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right),u,v\right)=\left[\begin{aligned}0\\0\\0\end{aligned}\right]$，由此計算對 $v$ 的偏微分可得

$\displaystyle\left\{\begin{aligned} &y^2\frac{\partial x}{\partial v}+2xy\frac{\partial y}{\partial v}+u\frac{\partial z}{\partial v}+2v=0\\&3x^2z\frac{\partial x}{\partial v}+x^3\frac{\partial z}{\partial v}+2\frac{\partial y}{\partial v}-u=0\\&u\frac{\partial x}{\partial v}+v\frac{\partial y}{\partial v}+y-yz\frac{\partial x}{\partial v}-xz\frac{\partial y}{\partial v}-xy\frac{\partial z}{\partial v}=0\end{aligned}\right.$

代入 $\left(x,y,z,u,v\right)=\left(1,1,1,1,1\right)$ 後可得

$\left\{\begin{aligned} &\left.\frac{\partial x}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+2\left.\frac{\partial y}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+\left.\frac{\partial z}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+2=0\\&3\left.\frac{\partial x}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+2\left.\frac{\partial y}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+\left.\frac{\partial z}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}-1=0\\&-\left.\frac{\partial z}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}+1=0\end{aligned}\right.$

最終可解得

$\displaystyle\left.\frac{\partial x}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}=\frac{3}{2}$　、　$\displaystyle\left.\frac{\partial y}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}=-\frac94$　、　$\displaystyle\left.\frac{\partial z}{\partial v}\right|_{\left(u,v\right)=\left(1,1\right)}=1$