國立臺灣大學九十八學年度研究所碩士班入學考試試題：微積分乙（不含線性代數）

1. ($10\%$) Show that ther series $\displaystyle\sum_{n=0}^{\infty}\frac1{2n+1}$ diverges.
訣竅
利用極限比較審斂法即可判斷。
解法
設 $\displaystyle a_n=\frac1{2n+1}$、$\displaystyle b_n=\frac1n$，容易知道 $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\frac12$，故兩級數 $\displaystyle\sum_{n=1}^{\infty}a_n$ 與 $\displaystyle\sum_{n=1}^{\infty}b_n$ 有相同的歛散性。而後者為調和級數（$p$ 級數在 $p=1$ 的情形）故發散，因此題目給定的級數也為發散級數。


2. ($10\%$) The population of a town is increasing at the rate of $400te^{0.02t}$ people per year, where $t$ is the number of years from now. Find the total gain in population during the next $5$ years.
訣竅
依照題意表達出給定的量，由此可看出使用微積分基本定理表達淨人口成長。
解法

設 $P\left(t\right)$ 表在時刻 $t$ 的人口數，那麼按照題意有 $P'\left(t\right)=400te^{0.02t}$，故由微積分基本定理以及分部積分法計算五年後的人口淨成長為

$\displaystyle\begin{aligned}P\left(5\right)-P\left(0\right)&=\int_0^5P'\left(t\right)dt=400\int_0^5te^{0.02t}dt=20000\int_0^5tde^{0.02t}\\&=20000\left(te^{0.02t}\Big|_0^5-\int_0^5e^{0.02t}dt\right)=20000\left(5e^{0.1}-50e^{0.02t}\Big|_0^5\right)\\&=10^5\left(e^{0.1}-10e^{0.1}+10\right)=10^5\left(10-9e^{0.1}\right)\end{aligned}$



3. ($10\%$) Find the solution of $y'+\left(\cos x\right)y=\cos x$, $y\left(0\right)=2$.
訣竅
運用積分因子法求解微分方程。
解法
給定方程的積分因子為 $e^{\int\cos xdx}=e^{\sin x}$，故兩邊同乘以 $e^{\sin x}$ 可得
$\displaystyle\frac{d}{dx}\left(y\left(x\right)e^{\sin x}\right)=e^{\sin x}\frac{dy}{dx}\left(x\right)+e^{\sin x}\left(\cos x\right)y\left(x\right)=e^{\sin x}\cos x$
故在區間 $\left[0,x\right]$ 上取定積分可得
$\displaystyle e^{\sin x}y\left(x\right)-y\left(0\right)=e^{\sin x}-1$
由 $y\left(0\right)=2$ 可解得 $y\left(x\right)=1+e^{-\sin x}$。


4. ($20\%$) Suppose $x_1$ and $x_2$ are bivariate normally distributed and the definitions of corresponding probability density functions are as follows.

   $\displaystyle\begin{aligned}&f\left(x_1\right)=\frac1{\sqrt{2\pi\sigma_1^2}}e^{\textstyle-\frac{\left(x_1-\mu_1\right)^2}{2\sigma_1^2}}\\&f\left(x_2\right)=\frac1{\sqrt{2\pi\sigma_2^2}}e^{\textstyle-\frac{\left(x_2-\mu_2\right)^2}{2\sigma_2^2}}\\&f\left(x_1,x_2\right)=\frac1{2\pi\sqrt{\sigma_1^2\sigma_2^2\left(1-\rho^2\right)}}e^{\textstyle-\frac1{2\left(1-\rho^2\right)}\left(\frac{\left(x_1-\mu_1\right)^2}{\sigma_1^2}-2\rho\frac{\left(x_1-\mu_1\right)\left(x_2-\mu_2\right)}{\sigma_1\sigma_2}+\frac{\left(x_2-\mu_2\right)^2}{\sigma_2^2}\right)}\end{aligned}$

   1. ($5\%$) Calculate $\displaystyle\int_{-\infty}^{\infty}e^{cx_1}f\left(x_1\right)dx_1$.
   2. ($5\%$) According to the definition of the conditional probability density function,

      $\displaystyle f\left(x_2\mid x_1\right)=\frac{f\left(x_1,x_2\right)}{f\left(x_1\right)}$,

      show that

      $\displaystyle f\left(x_2\mid x_1\right)=\frac1{\sqrt{2\pi\sigma_2^2\left(1-\rho^2\right)}}e^{\textstyle-\frac{\left(x_2-\left(\mu_2+\rho\frac{\sigma_2}{\sigma_1}\left(x_1-\mu_1\right)\right)\right)^2}{2\sigma_2^2\left(1-\rho^2\right)}}$

   3. ($10\%$) Calculate $\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{x_1}e^{cx_2}f\left(x_1,x_2\right)dx_1dx_2$.
訣竅
第一小題將被積分函數重新配方後運用變數代換法求解；第二小題可直接利用定義計算；第三小題則將前兩小題的結果組合使用即可。
解法

1. 按照題目給定的函數整理並配方計算如下

   $\displaystyle\begin{aligned}\int_{-\infty}^{\infty}e^{cx_1}f\left(x_1\right)dx_1&=\frac1{\sqrt{2\pi\sigma_1^2}}\int_{-\infty}^{\infty}\exp\left[-\frac{\left(x_1-\mu_1\right)^2}{2\sigma_1^2}+cx_1\right]dx_1\\&=\frac1{\sqrt{2\pi\sigma_1^2}}\int_{-\infty}^{\infty}\exp\left[-\frac1{2\sigma_1^2}\left(x_1-\mu_1-\sigma_1^2c\right)^2+\frac{2\mu_1c+\sigma_1^2c^2}2\right]dx_1\\&=\sqrt{\frac{\exp\left(2\mu_1c+\sigma_1^2c^2\right)}{2\pi\sigma_1^2}}\int_{-\infty}^{\infty}\exp\left[-\frac1{2\sigma_1^2}\left(x_1-\mu_1-\sigma_1^2c\right)^2\right]dx_1\\&=\sqrt{\frac{\exp\left(2\mu_1c+\sigma_1^2c^2\right)}{2\pi\sigma_1^2}}\cdot\sqrt2\sigma_1\sqrt\pi=\sqrt{\exp\left(2\mu_1c+\sigma_1^2c^2\right)}\end{aligned}$

2. 按照定義可知

   $\displaystyle\begin{aligned}&\,f\left(x_2\mid x_1\right)=\frac{f\left(x_1,x_2\right)}{f\left(x_1\right)}\\=&\,\frac{\sqrt{2\pi\sigma_1^2}}{2\pi\sqrt{\sigma_1^2\sigma_2^2\left(1-\rho^2\right)}}\exp\left\{-\frac1{2\left(1-\rho^2\right)}\left[\frac{\left(x_1-\mu_1\right)^2}{\sigma_1^2}-2\rho\frac{\left(x_1-\mu_1\right)\left(x_2-\mu_2\right)}{\sigma_1\sigma_2}+\frac{\left(x_2-\mu_2\right)^2}{\sigma_2^2}\right]+\frac{\left(x_1-\mu_1\right)^2}{2\sigma_1^2}\right\}\\=&\,\frac1{\sqrt{2\pi\sigma_2^2\left(1-\rho^2\right)}}\exp\left\{-\frac1{2\sigma_2^2\left(1-\rho^2\right)}\left[\left(x_2-\mu_2\right)^2-\frac{2\rho\sigma_2}{\sigma_1}\left(x_1-\mu_1\right)\left(x_2-\mu_2\right)\right]-\frac{\rho^2\left(x_1-\mu_1\right)^2}{2\sigma_1^2\left(1-\rho^2\right)}\right\}\\=&\,\frac1{\sqrt{2\pi\sigma_2^2\left(1-\rho^2\right)}}\exp\left[-\frac1{2\sigma_2^2\left(1-\rho^2\right)}\left[\left(x_2-\mu_2\right)-\frac{\rho\sigma_2}{\sigma_1}\left(x_1-\mu_1\right)\right]^2\right]\\=&\,\frac1{\sqrt{2\pi\sigma_2^2\left(1-\rho^2\right)}}\exp\left[-\frac{\left[x_2-\left(\mu_2+\rho\frac{\sigma_2}{\sigma_1}\left(x_1-\mu_1\right)\right)\right]^2}{2\sigma_2^2\left(1-\rho^2\right)}\right]\end{aligned}$

   計算完畢。
3. 利用第二題的結果並交換積分次序，所求可以改寫如下

   $\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{x_1}e^{cx_2}f\left(x_1,x_2\right)dx_1dx_2=\int_{-\infty}^{\infty}\left[e^{x_1}f\left(x_1\right)\int_{-\infty}^{\infty}e^{cx_2}f\left(x_2\mid x_1\right)dx_2\right]dx_1$

   接著運用第一小題的結論可知

   $\displaystyle\int_{-\infty}^{\infty}e^{cx_2}f\left(x_2\mid x_1\right)dx_2=\sqrt{\exp\left[2c\left(\mu_2+\rho\frac{\sigma_2}{\sigma_1}\left(x_1-\mu_1\right)\right)+\sigma_2^2\left(1-\rho^2\right)c^2\right]}$

   如此所求為

   $\displaystyle\begin{aligned}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{x_1}e^{cx_2}f\left(x_1,x_2\right)dx_1dx_2&=\sqrt{\exp\left(\sigma_2^2\left(1-\rho^2\right)c^2+2c\mu_2-2c\rho\frac{\sigma_2}{\sigma_1}\mu_1\right)}\int_{-\infty}^{\infty}e^{\left(c\rho\frac{\sigma_2}{\sigma_1}+1\right)x_1}f\left(x_1\right)dx_1\\&=\sqrt{\exp\left(\sigma_2^2\left(1-\rho^2\right)c^2+2c\mu_2-2c\rho\frac{\sigma_2}{\sigma_1}\mu_1+2\mu_1\left(c\rho\frac{\sigma_2}{\sigma_1}+1\right)+\sigma_1^2\left(c\rho\frac{\sigma_2}{\sigma_1}+1\right)^2\right)}\\&=\sqrt{\exp\left(c^2\sigma_2^2+2c\mu_2+2\mu_1+2c\rho\sigma_1\sigma_2+\sigma_1^2\right)}\end{aligned}$



5. ($10\%$) If $f$ is differentiable at some number $z$, show that, for any positive numbers $a$ and $b$ with $a<b$,

   $\displaystyle f'\left(z\right)=\lim_{h\to0}\frac{a^2f\left(z+bh\right)-b^2f\left(z+ah\right)+\left(b^2-a^2\right)f\left(z\right)}{\left(a^2b-b^2a\right)h}$.

訣竅
運用導數的定義去改寫即可。
解法
首先由導數的定義注意到
$\displaystyle\lim_{h\to0}\frac{a^2f\left(z+bh\right)-a^2f\left(z\right)}{bh}=a^2f'\left(z\right),\qquad\lim_{h\to0}\frac{b^2f\left(z+ah\right)-b^2f\left(z\right)}{ah}=b^2f'\left(z\right)$
將兩式分別乘以 $b$ 與 $a$ 後相減有
$\displaystyle\lim_{h\to0}\frac{a^2f\left(z+bh\right)-b^2f\left(z+ah\right)+\left(b^2-a^2\right)f\left(z\right)}h=\left(a^2b-b^2a\right)f'\left(z\right)$
同除以 $a^2b-b^2a$ 便完成證明。


6. ($10\%$) Compute the line integral:

   $\displaystyle\int_C\frac{\left(x^3+x^2+x+1\right)}{x^4}dx$

   where $C$ is the lower quarter-circle centered at $0$ joining $\displaystyle\frac{-1-i}{\sqrt2}$ and $\displaystyle\frac{1-i}{\sqrt2}$ in the positive (counterclockwise) sense.
訣竅
將其參數化後代入求解。
解法

令 $x=\cos\theta$ 且 $y=\sin\theta$，其參數範圍為 $\displaystyle\theta\in\left[\frac{5\pi}4,\frac{7\pi}4\right]$，故所求的線積分可表達並計算如下

$\displaystyle\begin{aligned}\int_C\frac{x^3+x^2+x+1}{x^4}dx&=\int_{\frac{5\pi}4}^{\frac{7\pi}4}\left(\frac{\sin^3\theta+\sin^2\theta+\sin\theta+1}{\sin^4\theta}\right)\cos\theta d\theta\\&=\left.\left(\ln\left|\sin\theta\right|-\frac1{\sin\theta}-\frac1{2\sin^2\theta}-\frac1{3\sin^3\theta}\right)\right|_{\frac{5\pi}4}^{\frac{7\pi}4}=0\end{aligned}$



7. ($10\%$)
   1. ($5\%$) Evaluate the following integral:

      $\displaystyle\int_{-1}^2\int_{-z}^{z+2}\left(x+2z^2\right)dxdz$

   2. ($5\%$) Switch the order of $x$ and $z$ in the above integrals, i.e., rewrite the above integral into a summation of terms of the form (You don't have to evaluate the integral).

      $\displaystyle\int_?^?\int_?^?\left(x+2z^2\right)dzdx$

訣竅
第一小題可直接計算迭代積分即可；第二小題則按其題意改寫積分範圍後交換積分次序。
解法

1. 直接計算定積分可知

   $\displaystyle\begin{aligned}\int_{-1}^2\int_{-z}^{z+2}\left(x+2z^2\right)dxdz&=\int_{-1}^2\left.\left(\frac{x^2}2+2xz^2\right)\right|_{x=-z}^{x=z+2}dz=\int_{-1}^2\left(\frac{\left(z+2\right)^2}2+2\left(z+2\right)z^2-\frac{\left(-z\right)^2}2-2\left(-z\right)z^2\right)dz\\&=\int_{-1}^2\left(4z^3+4z^2+2z+2\right)dz=\left.\left(z^4+\frac{4z^3}3+z^2+2z\right)\right|_{-1}^2=36\end{aligned}$

2. 原積分範圍為 $\left\{\begin{aligned}&-z\leq x\leq z+2\\&-1\leq z\leq2\end{aligned}\right.$，由第一式可知 $x-2\leq z$ 且 $-x\leq z$。為了比較 $x-2$ 與 $-x$ 的大小，我們可區分範圍 $x\in\left[-2,1\right]$ 時有 $-x\geq x-2$，從而 $-x\leq z\leq2$；而當 $x\in\left[1,4\right]$ 時有 $x-2\geq-x$，從而 $x-2\leq z\leq2$，故所求為

   $\displaystyle\int_{-1}^2\int_{-z}^{z+2}\left(x+2z^2\right)dxdz=\int_{-2}^1\int_{-x}^2\left(x+2z^2\right)dzdx+\int_1^4\int_{x-2}^2\left(x+2z^2\right)dzdx$



8. ($20\%$) Find the limit:
   1. ($10\%$)

      $\displaystyle\lim_{h\to0}\frac{-3h^3+8h^2-7h+11}{4h^3-h^2+5}$

   2. ($10\%$)

      $\displaystyle\lim_{z\to0}\frac{z^2+2\cos z-2}{z^4}$

訣竅
第一小題可以注意到在 $h=0$ 處連續故直接代入即可；第二小題可應用羅必達法則或餘弦函數的泰勒展開式求解。
解法

1. 運用極限的四則運算定理可知

   $\displaystyle\lim_{h\to0}\frac{-3h^3+8h^2-7h+11}{4h^3-h^2+5}=\frac{\displaystyle\lim_{h\to0}\left(-3h^3+8h^2-7h+11\right)}{\displaystyle\lim_{h\to0}\left(4h^3-h^2+5\right)}=\frac{11}5$

2. 【方法一】　運用羅必達法則可知

   $\displaystyle\lim_{z\to0}\frac{z^2+2\cos z-2}{z^4}=\lim_{z\to0}\frac{2z-2\sin z}{4z^3}=\lim_{z\to0}\frac{1-\cos z}{6z^2}=\lim_{z\to0}\frac{\sin z}{12z}=\frac1{12}$

   【方法二】　運用餘弦函數的泰勒展開式可知

   $\displaystyle\lim_{z\to0}\frac{z^2+2\cos z-2}{z^4}=\lim_{z\to0}\frac{\displaystyle z^2+2\left(1-\frac{z^2}2+\frac{z^4}{24}-\cdots\right)-2}{z^4}=\lim_{z\to0}\frac{\displaystyle\frac{z^4}{12}-\cdots}{z^4}=\frac1{12}$