國立臺灣大學一百一拾學年度研究所碩士班入學考試試題：微積分(C)

※注意：請於試卷內之「非選擇題作答區」依序作答，並應註明作答之大題及小題題號。
※Please show the detailed calculation process for the questions whenever necessary.
※If the answers are with decimal numbers, please round to the second decimal place, e.g., $99.37$ or $2.43\%$.

1. ($10\%$) Find $k$ such that $f\left(x\right)=kx^2$ is a probability density function over the interval $\left[2,5\right]$. Then write the probability density function.
訣竅
按照機率密度的定義求解即可。
解法
按照定義可知
$\displaystyle1=\int_2^5f\left(x\right)\,dx=k\int_2^5x^2\,dx=\left.\frac{k}3x^3\right|_2^5=39k.$
因此 $\displaystyle k=\frac1{39}$，如此機率密度函數為 $\displaystyle f\left(x\right)=\begin{cases}\displaystyle\frac{x^2}{39},&\text{if}~x\in\left[2,5\right],\\0,&\text{otherwise}.\end{cases}$。


2. ($10\%$) Find the area of the region $E$ bounded by the ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
訣竅
利用對稱性可計算第一象限的面積，其中需運用三角代換法處理之。
解法

整理方程式考慮在第一象限的曲線為 $\displaystyle y=b\sqrt{1-\frac{x^2}{a^2}}$，如此所求的面積可表達為

$\displaystyle A=4\int_0^ab\sqrt{1-\frac{x^2}{a^2}}dx$.

令 $x=a\sin\theta$，那麼

• 當 $x=0$ 時有 $\theta=0$；
• 當 $x=a$ 時有 $\displaystyle\theta=\frac\pi2$；
• 求導有 $dx=a\cos\theta\,d\theta$。

據此所求的面積可改寫並計算如下

$\displaystyle A=4b\int_0^{\frac\pi2}\sqrt{1-\sin^2\theta}\cdot a\cos\theta\,d\theta=2ab\int_0^{\frac\pi2}\left(1+\cos2\theta\right)d\theta=2ab\cdot\frac\pi2=\pi ab$.



3. ($10\%$) Solve $\displaystyle\frac{dy}{dx}+\frac{y}x=e^{2x}$.
訣竅
運用積分因子法求解微分方程即可。
解法

對給定的方程同乘以 $x$ 後可得

$\displaystyle\frac{d}{dx}\left(xy\right)=x\frac{dy}{dx}+y=xe^{2x}$

如此取不定積分可得

$\displaystyle xy=\frac12xe^{2x}-\frac14e^{2x}+\frac{C}4$

同除以 $x$ 後得所求函數為

$\displaystyle y=\frac{2xe^{2x}-e^{2x}+C}{4x}$



4. ($10\%$) Suppose that the real-valued function $f:I\to R$ is integrable, where $I=\left[a,b\right]\times\left[c,d\right]$ is a rectangle in the plane $R^2$. To use Fubini's Theorem to calculate the integral $\displaystyle\int_If$, what conditions are required? State and explain Fubini's Theorem in details.
訣竅
回憶關於 Fubini 定理的敘述。
解法

設 $f$ 在 $I$ 上可積分且對每個 $x\in\left[a,b\right]$ 而言，函數 $f\left(x,\cdot\right)$ 在 $\left[c,d\right]$ 上可積、對每個 $y\in\left[c,d\right]$ 而言，函數 $f\left(\cdot,y\right)$ 在 $\left[a,b\right]$ 上可積，則有

$\displaystyle\int_If=\int_a^b\int_c^df\left(x,y\right)\,dy\,dx=\int_c^d\int_a^bf\left(x,y\right)\,dx\,dy.$



5. ($10\%$) What are metric spaces? When is a mapping between two metric spaces called continuous?
訣竅
根據賦距空間以及連續函數的定義回答即可。
解法
我們稱 $\left(X,d\right)$ 為賦距空間，意指雙變數函數 $d:X\times X\to\mathbb R$ 滿足下列條件
• 對任何 $x\in X$ 恆有 $d\left(x,x\right)=0$；
• 若 $d\left(x,y\right)=0$，則 $x=y$；
• 對任何 $x,y\in X$ 恆有 $d\left(x,y\right)=d\left(y,x\right)$；
• 對任何 $x,y,z\in X$ 恆有 $d\left(x,y\right)\leq d\left(x,z\right)+d\left(z,x\right)$。
設 $\left(X,d\right)$ 與 $\left(Y,\tau\right)$ 為兩個賦距空間而 $f:X\to Y$ 為連續函數，其定義為：「對任何 $\varepsilon>0$，存在 $\delta>0$ 使得當 $d\left(x,y\right)<\delta$ 時有 $\tau\left(f\left(x\right),f\left(y\right)\right)<\varepsilon$」。


6. ($10\%$) In February 2021, an automobile manufacturer sells in America, Europe, and Asia, charging a different price in each of the three markets. The price function for cars sold in America is $p=20-0.2x$ (for $0\leq x\leq100$), the price function for cars sold in Europe is $q=16-0.1y$ (for $0\leq y\leq160$), and the price function for cars sold in Asia is $r=12-0.1z$ (for $0\leq z\leq120$), all in thousands of dollars, where $x$, $y$, and $z$ are the numbers of cars sold in America, Europe, and Asia, respectively. The company's production cost function is $C=22+4\left(x+y+z\right)$ thousand dollars.
   1. ($2\%$) Find the company's profit function $P\left(x,y,z\right)$, which should be the sum of the revenues from different markets minus production costs, and each revenue is price times quantity.
   2. ($8\%$) Suppose the automobile manufacturer's productivity capacity is $100$ cars in this month. How many cars should be sold in each market to maximize profit and what is the maximized profit?
訣竅
依據題意設定出獲利函數。隨後根據限制條件使用 Lagrange 乘子法求解。
解法
1. 公司的獲利函數 $P$ 可表達如下

   $P\left(x,y,z\right)=xp+yq+zr-C=16x-0.2x^2+12y-0.1y^2+8z-0.1z^2-22$.

2. 根據題意知生產限制為 $x+y+z\leq100$，如此加入鬆弛變量 $t$ 使得 $x+y+z+t^2=100$。因此我們考慮 Lagrange 乘子函數為

   $F\left(x,y,z,t,\lambda\right)=16x-0.2x^2+12y-0.1y^2+8z-0.1z^2-22+\lambda\left(x+y+z+t^2-100\right)$.

   據此解下列的聯立方程組

   $\left\{\begin{aligned}&F_x\left(x,y,z,t,\lambda\right)=16-0.4x+\lambda=0,\\&F_y\left(x,y,z,t,\lambda\right)=12-0.2y+\lambda=0,\\&F_z\left(x,y,z,t,\lambda\right)=8-0.2z+\lambda=0,\\&F_t\left(x,y,z,t,\lambda\right)=2\lambda t=0,\\&F_\lambda\left(x,y,z,t,\lambda\right)=x+y+z+t^2-100=0.\end{aligned}\right.$

   由第四式，可知 $\lambda=0$ 或 $t=0$。
   • 當 $\lambda=0$ 時則有 $x=40$、$y=60$、$z=40$，但這就表明 $x+y+z=140>100$，矛盾。
   • 當 $t=0$ 時便有 $x+y+z=100$。那麼將第二式乘以 $2$ 與第三式乘以 $2$ 後將前三式相加便得 $\displaystyle\lambda=-\frac{16}5$，如此解得 $\displaystyle\left(x,y,z\right)=\left(32,44,24\right)$。
   因此當美國賣 $32$ 輛、歐洲賣 $44$ 輛而亞洲賣 $24$ 輛時會有最大利潤。此時利潤為 $P\left(32,44,24\right)=754$ 千元。


7. ($10\%$) A company's marginal cost function is $MC\left(x\right)=xe^{-x/2}$ and fixed costs are $200$, where $x$ is a nonnegative real number that denotes the amount of units produced. Find its cost as a function of $x$. (Hint: Marginal cost is the cost added by producing one additional unit of a product or service.)
訣竅
根據邊際成本的數學定義使用分部積分法計算即可。
解法

按照微積分基本定理可知

$\displaystyle C\left(x\right)=C\left(0\right)+\int_0^xMC\left(t\right)\,dt=200+\int_0^xte^{-t/2}\,dt=200+\left(-2te^{-t/2}-4e^{-t/2}\right)\Big|_0^x=204-2xe^{-x/2}-4e^{-x/2}$.



8. ($30\%$) Consider a power call or put option with the payoff at the maturity $T$ as

   $C_T=\max\left(S_T^i-X^i,0\right)$ and $P_T=\max\left(X^i-S_T^i,0\right)$,

   respectively, where $S_t$ is the stock price at time $t$, $i$ is a positive-integer exponent, and $X^i$ denotes the strike price. Under the Black-Scholes framework, their respective value functions toady ($t=0$) are

   $C_0=\Pi S_0^iN\left(d_i\right)-\exp\left(-rT\right)X^iN\left(d_0\right),$

   $P_0=\exp\left(-rT\right)X^iN\left(-d_0\right)-\Pi S_0^iN\left(-d_i\right),$

   where $\Pi=\Pi\left(i,r,\sigma,T\right)=\exp\left(-rT+irT+\left(i^2-i\right)\sigma^2T/2\right)$, $\displaystyle d_i=\frac{\ln\left(\frac{S_0}X\right)+\left(r-\frac{\sigma^2}2\right)T+i\sigma^2T}{\sigma\sqrt{T}}$, $\displaystyle d_0=\frac{\ln\left(\frac{S_0}{X}\right)+\left(r-\frac{\sigma^2}2\right)T}{\sigma\sqrt{T}}$, $r$ is the risk-free interest rate, $\sigma$ is the stock price volatility, and $N\left(\cdot\right)$ is the cumulative distribution function of the standard normal distribution defined as

   $\displaystyle N\left(d\right)=\int_{-\infty}^dn\left(x\right)\,dx=\int_{-\infty}^d\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}\,dx,$

   where $n\left(\cdot\right)$ is the probability density function of the standard normal distribution.
   1. ($10\%$) Prove that $\Pi S_0^in\left(d_i\right)=\exp\left(-rT\right)X^in\left(d_0\right)$.
   2. ($8\%$) Derive and express $\displaystyle\frac{\partial C_0}{\partial S_0}$ and $\displaystyle\frac{\partial P_0}{\partial S_0}$ as $AN\left(B\right)$ and $CN\left(D\right)$, respectively. What are $A$, $B$, $C$, and $D$?
   3. ($8\%$) Derive and express $\displaystyle\frac{\partial^2C_0}{\partial S_0^2}$ and $\displaystyle\frac{\partial^2P_0}{\partial S_0^2}$ as $EN\left(F\right)+Gn\left(H\right)$ and $IN\left(J\right)+Kn\left(L\right)$, respectively. What are $E$, $F$, $G$, $H$, $I$, $J$, $K$, and $L$?
   4. ($4\%$) In what condition does $\displaystyle\frac{\partial^2C_0}{\partial S_0^2}$ equal $\displaystyle\frac{\partial^2P_0}{\partial S_0^2}$?
訣竅
運用多變數函數的連鎖律求解即可。
解法

1. 首先可以發現

   $\displaystyle\frac{d_0+d_i}2=\frac{\ln\left(S_0/X\right)+\left(r-\sigma^2/2\right)T+i\sigma^2T/2}{\sigma\sqrt{T}}$

   接著同乘以 $d_i-d_0$ 便有

   $\displaystyle\frac{d_i^2-d_0^2}2=\frac{\ln\left(S_0/X\right)+\left(r-\sigma^2/2\right)T+i\sigma^2T/2}{\sigma\sqrt T}\cdot\frac{i\sigma^2T}{\sigma\sqrt T}=\ln\left(\frac{S_0^i}{X^i}\right)+\left(ir-\frac{i\sigma^2}2\right)T+\frac{i^2\sigma^2T}2.$

   隨後取自然指數便有

   $\displaystyle e^{\frac{d_i^2-d_0^2}2}=\frac{S_0^i}{X^i}\cdot\exp\left(\left(ir-\frac{i\sigma^2}2\right)T+\frac{i^2\sigma^2T}2\right)$

   移項並同乘以 $\displaystyle\frac{\exp\left(-rT\right)}{\sqrt{2\pi}}$ 便有

   $\displaystyle\begin{aligned}\exp\left(-rT\right)X^in\left(d_0\right)&=\exp\left(-rT\right)X^i\frac1{\sqrt{2\pi}}e^{-\frac{d_0^2}2}\\&=\exp\left(-rT\right)S_0^i\exp\left(\left(ir-\frac{i\sigma^2}2\right)T+\frac{i^2\sigma^2T}2\right)\cdot\frac1{\sqrt{2\pi}}e^{-\frac{d_i^2}2}\\&=\exp\left(-rT+irT+\frac{\left(i^2-i\right)\sigma^2T}2\right)S_0^in\left(d_i\right)=\Pi S_0^in\left(d_i\right).\end{aligned}$

2. 首先將 $C_0$ 對 $S_0$ 求導可得

   $\displaystyle\begin{aligned}\frac{\partial C_0}{\partial S_0}&=i\Pi S_0^{i-1}N\left(d_0\right)+\Pi S_0^in\left(d_i\right)\cdot\frac{\partial d_i}{\partial S_0}-\exp\left(-rT\right)X^in\left(d_0\right)\cdot\frac{\partial d_0}{\partial S_0}\\&=i\Pi S_0^{i-1}N\left(d_0\right)+\frac1{S_0\sigma\sqrt T}\left[\Pi S_0^in\left(d_i\right)-\exp\left(-rT\right)X^in\left(d_0\right)\right]=i\Pi S_0^{i-1}N\left(d_0\right).\end{aligned}$

   因此 $A=i\Pi S_0^{i-1}$、$B=d_0$。另一方面，將 $P_0$ 對 $S_0$ 求偏導可得

   $\displaystyle\begin{aligned}\frac{\partial P_0}{\partial S_0}&=\exp\left(-rT\right)X^in\left(-d_0\right)\cdot-\frac{\partial d_0}{\partial S_0}-i\Pi S_0^{i-1}N\left(-d_i\right)-\Pi S_0^in\left(-d_i\right)\cdot-\frac{\partial d_i}{\partial S_0}\\&=-\frac1{S_0\sigma\sqrt{T}}\left[\exp\left(-rT\right)X^in\left(-d_0\right)-\Pi S_0^in\left(-d_i\right)\right]-i\Pi S_0^{i-1}N\left(-d_i\right)=-i\Pi S_0^{i-1}N\left(-d_i\right),\end{aligned}$

   其中我們用了 $n\left(-x\right)=n\left(x\right)$。因此 $C=-i\Pi S_0^{i-1}$、$D=-d_i$。
3. 進一步由前一小題的結果計算二階偏導函數。首先有

   $\displaystyle\begin{aligned}\frac{\partial^2C_0}{\partial S_0^2}&=i\Pi\frac{\partial}{\partial S_0}\left(S_0^{i-1}N\left(d_0\right)\right)=i\left(i-1\right)\Pi S_0^{i-2}N\left(d_0\right)+i\Pi S_0^{i-1}n\left(d_0\right)\cdot\frac1{S_0\sigma\sqrt T}\\&=i\left(i-1\right)\Pi S_0^{i-2}N\left(d_0\right)+\frac{i\Pi S_0^{i-2}}{\sigma\sqrt T}n\left(d_0\right).\end{aligned}$

   因此 $E=i\left(i-1\right)\Pi S_0^{i-2}$、$F=d_0$、$\displaystyle G=\frac{i\Pi S_0^{i-2}}{\sigma\sqrt T}$、$H=d_0$。另一方面則有

   $\displaystyle\begin{aligned}\frac{\partial^2P_0}{\partial S_0^2}&=-i\Pi\frac{\partial}{\partial S_0}\left(S_0^{i-1}N\left(-d_i\right)\right)=-i\left(i-1\right)\Pi S_0^{i-2}N\left(-d_i\right)-i\Pi S_0^{i-1}n\left(-d_i\right)\cdot-\frac1{S_0\sigma\sqrt T}\\&=-i\left(i-1\right)\Pi S_0^{i-2}N\left(-d_i\right)+\frac{i\Pi S_0^{i-2}}{\sigma\sqrt T}n\left(d_i\right).\end{aligned}$

   因此 $I=-i\left(i-1\right)\Pi S_0^{i-2}$、$J=-d_i$、$\displaystyle K=\frac{i\Pi S_0^{i-2}}{\sigma\sqrt T}$、$L=d_i$。
4. 要使 $\displaystyle\frac{\partial^2C_0}{\partial S_0^2}=\frac{\partial^2P_0}{\partial S_0^2}$，這等價於

   $\displaystyle i\left(i-1\right)\Pi S_0^{i-2}\left[N\left(d_0\right)+N\left(-d_i\right)\right]+\frac{i\Pi S_0^{i-2}}{\sigma\sqrt T}\left[n\left(d_0\right)-n\left(d_i\right)\right]=0.$

   由於 $i$、$\Pi$、$S_0^{i-2}$ 恆不為零，故等價於

   $\displaystyle\left(i-1\right)\left[N\left(d_0\right)+N\left(-d_i\right)\right]+\frac{n\left(d_0\right)-n\left(d_i\right)}{\sigma\sqrt T}=0.$