國立臺灣大學一百零六學年度研究所碩士班入學考試試題：微積分(C)

For Questions 1 to 7, show your calculations in detail on the answer sheet.

1. ($10\%$) Evaluate the limit $\displaystyle\lim_{x\to0}\frac{e^{3x}-1}x$.
訣竅
使用羅必達法則求解即可；亦可辨識其為導數的定義。
解法一

由羅必達法則立即有

$\displaystyle\lim_{x\to0}\frac{e^{3x}-1}x=\lim_{x\to0}\frac{3e^{3x}}1=3$

解法二

設 $f\left(x\right)=e^{3x}$，那麼所求的極限式可辨識為如下的導數值

$\displaystyle\lim_{x\to0}\frac{e^{3x}-1}{x}=\lim_{x\to0}\frac{f\left(x\right)-f\left(0\right)}{x-0}=f'\left(0\right)=3e^{3x}\Big|_{x=0}=3$



2. ($10\%$) Find the area of the region bounded by the graphs of $f\left(x\right)=-x^2+2x$ and $g\left(x\right)=3x^3-x^2-10x$.
訣竅
先找出兩曲線的交點並明確出兩者有界區域的範圍，隨後透過積分求出面積。
解法

為了找出兩圖形的交點，我們由代入消去法去解方程式 $-x^2+2x=3x^3-x^2-10x$，即 $x^3-4x=0$，能得 $x=0$ 與 $x=\pm2$。容易注意到當 $x\in\left(0,2\right)$ 上有 $f\left(x\right)>g\left(x\right)$，而當 $x\in\left(-2,0\right)$ 時有 $f\left(x\right)<g\left(x\right)$。故所求的面積為

$\displaystyle\begin{aligned}A&=\int_{-2}^0\left[g\left(x\right)-f\left(x\right)\right]dx+\int_0^2\left[f\left(x\right)-g\left(x\right)\right]dx=3\int_{-2}^0\left(x^3-4x\right)dx+3\int_0^2\left(4x-x^3\right)dx\\&=\left.3\left(\frac{x^4}4-2x^2\right)\right|_{-2}^0+\left.3\left(\frac{x^4}4-2x^2\right)\right|_0^2=12+12=24\end{aligned}$



3. ($10\%$) Evaluate $\displaystyle\int\frac{\sin\sqrt{x}}{\sqrt{x}}dx$.
訣竅
運用變數變換的概念求不定積分即可。
解法

直接由變數變換的概念演算可知

$\displaystyle\int\frac{\sin\sqrt{x}}{\sqrt{x}}dx=2\int\sin\sqrt{x}d\sqrt{x}=-2\cos\sqrt{x}+C$



4. ($10\%$) Find the values of $x$ for which the power series is convergent: $\displaystyle\sum_{n=1}^{+\infty}\left(-1\right)^{n+1}\frac{2^nx^n}{n3^n}$.
訣竅
運用比值審歛法的概念求出收斂半徑，隨後檢驗端點的收斂性。
解法
由比值審歛法的概念有
$\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\left|\frac{2^n}{n\cdot3^n}\cdot\frac{\left(n+1\right)3^{n+1}}{2^{n+1}}\right|=\frac32$
而對於端點的斂散性則檢驗如下
• 當 $\displaystyle x=-\frac32$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}\frac{-1}n$，此為調和級數從而發散；
• 當 $\displaystyle x=\frac32$ 時級數寫為 $\displaystyle\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}n$，此由交錯級數審歛法可知收斂。
因此給定的冪級數在區間 $\displaystyle\left(-\frac32,\frac32\right]$ 上收斂。


5. ($10\%$) Find $\displaystyle\frac{dy}{dx}$ if $y=\tanh^{-1}\left(\cos2x\right)$.
訣竅
利用隱函數微分法求解；亦可將反雙曲函數的形式明確獲得後應用連鎖律即可。
解法一

由條件可知

$\displaystyle1-2\left(1+e^{2y}\right)^{-1}=\frac{e^{2y}-1}{e^{2y}+1}=\tanh y=\cos2x$

如此使用隱函數微分有

$\displaystyle2\left(1+e^{2y}\right)^{-2}\cdot2e^{2y}\cdot\frac{dy}{dx}=-2\sin2x$

故解得 $\displaystyle\frac{dy}{dx}=-\frac1{e^{2y}}\cdot\frac{\sin2x}{2}\cdot\left(1+e^{2y}\right)^2$。又利用條件可注意到 $\displaystyle e^{2y}=\frac{1+\cos2x}{1-\cos2x}$，故

$\displaystyle\frac{dy}{dx}=-\frac{1-\cos2x}{1+\cos2x}\cdot\frac{\sin2x}2\cdot\frac4{\left(1-\cos2x\right)^2}=-2\csc2x$

解法二

承解法一的過程可知

$\displaystyle y=\frac12\ln\left|\frac{1+\cos2x}{1-\cos2x}\right|$

因此求導便有

$\displaystyle\frac{dy}{dx}=\frac12\cdot\frac{1-\cos2x}{1+\cos2x}\cdot\frac{-2\sin2x\left(1-\cos2x\right)-\left(1+\cos2x\right)\cdot2\sin2x}{\left(1-\cos2x\right)^2}=-\frac{2\sin2x}{\left(1+\cos2x\right)\left(1-\cos2x\right)}=-2\csc2x$



6. ($35\%$) Let $w_T$ denote the terminal wealth of a uni-dollar investment after $T$ periods. Consider two possible cumulative distribution functions (CDFs) for $w_T$ as follows.

   $\begin{cases}\ln w_T\sim N\left(T\mu_F,T\sigma_F^2\right)&\text{if the CDF}~F\left(w_T\right)~\text{is considered}\\\ln w_T\sim N\left(T\mu_G,T\sigma_G^2\right)&\text{if the CDF}~G\left(w_T\right)~\text{is considered}\end{cases}$

   where $N\left(\mu,\sigma^2\right)$ represents the normal distribution with the mean to be $\mu$ and the variance to be $\sigma^2$. For any utility function $U'\left(\cdot\right)\geq0$, the expected utility under $F$ or $G$ is

   $\displaystyle E_PU\left(w_T\right)=\int_0^{\infty}U\left(w_T\right)dP\left(w_T\right)$ for $P=F$ or $G$.

   Last, $\displaystyle\Phi\left(c\right)\equiv\int_{-\infty}^c\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}dx$ is the CDF for the standard normal distribution.
   1. ($5\%$) Find $w_T^*$ such that $F\left(w_T^*\right)=G\left(w_T^*\right)$.
   2. ($10\%$) Suppose that $\sigma_F>\sigma_G$. Prove $\displaystyle\int_0^{w_T^*}\left[G\left(w_T\right)-F\left(w_T\right)\right]dw_T<0$.
   3. ($20\%$) Consider a utility function

      $U\left(w_T\right)=\begin{cases}w_T&\text{if}~w_T\leq M\\M\left[\ln\left(w_t/M\right)+1\right]&\text{if}~w_T>M\end{cases}$

      where $M$ is a positive constant, and the expected utility under $F$ is

      $\displaystyle E_FU\left(w_T\right)=\int_0^MU\left(w_T\right)dF\left(w_T\right)+\int_M^{\infty}U\left(w_T\right)dF\left(w_T\right)=Ⅰ+Ⅱ$.

      1. ($5\%$) Prove that $\displaystyle0<Ⅰ<M\Phi\left(\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}\right)$.
      2. ($10\%$) If $Ⅱ$ can be expressed as $M\left[\alpha\Phi\left(\beta\right)+\gamma e^{-\delta^2/2}\right]$, what are $\alpha$, $\beta$, $\gamma$, and $\delta$?
      3. ($5\%$) What is $\displaystyle\lim_{T\to\infty}\frac1TE_FU\left(w_T\right)$?
訣竅
運用累積分配函數的定義表達出給定的函數形式從而解出交點，隨後證明兩者的大小關係並獲得所證的不等式。最後一小題則沿著題幹所給出的資訊並進行高斯積分的配方處理即能獲得證明。
解法

1. 按照對數常態分配及其累積分配函數的定義可知

   $\displaystyle\begin{aligned}&F\left(x\right)=\frac1{\sqrt{2\pi T}\sigma_F}\int_{-\infty}^x\frac{\exp\left[-\left(\ln s-T\mu_F\right)^2/\left(2T\sigma_F^2\right)\right]}sds,\\&G\left(x\right)=\frac1{\sqrt{2\pi T}\sigma_G}\int_{-\infty}^x\frac{\exp\left[-\left(\ln s-T\mu_G\right)^2/\left(2T\sigma_G^2\right)\right]}sds.\end{aligned}$

   運用變數代換法，函數 $F$ 與 $G$ 可分別改寫為

   $\displaystyle F\left(x\right)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\left(\ln x-T\mu_F\right)/\left(\sqrt{T}\sigma_F\right)}e^{-\frac{t^2}2}dt$,　$\displaystyle G\left(x\right)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\left(\ln x-T\mu_G\right)/\left(\sqrt{T}\sigma_G\right)}e^{-\frac{t^2}2}dt$.

   運用標準常態分佈的累積函數表達可知

   $\displaystyle F\left(x\right)=\Phi\left(\frac{\ln x-T\mu_F}{\sqrt{T}\sigma_F}\right)$,　$\displaystyle G\left(x\right)=\Phi\left(\frac{\ln x-T\mu_G}{\sqrt{T}\sigma_G}\right)$

   為了解方程式 $F\left(w_T^*\right)=G\left(w_T^*\right)$，我們應解方程式

   $\displaystyle\frac{\ln w_T^*-T\mu_F}{\sqrt{T}\sigma_F}=\frac{\ln w_T^*-T\mu_G}{\sqrt{T}\sigma_G}$

   假若 $\sigma_F\neq\sigma_G$ 則可解得

   $\displaystyle w_T^*=\exp\left(\frac{T\left(\mu_G\sigma_F-\mu_F\sigma_G\right)}{\sigma_G-\sigma_F}\right)$

   假若 $\sigma_F=\sigma_G$ 且 $\mu_F=\mu_G$，則無窮多組解；但 $\sigma_F=\sigma_G$ 且 $\mu_F\neq\mu_G$，則無解。
2. 假若 $\sigma_F>\sigma_G$，那麼容易發現當 $0< w_T<w_T^*$ 時有不等式

   $\displaystyle \ln w_T<\ln w_T^*=\frac{T\left(\mu_G\sigma_F-\mu_F\sigma_G\right)}{\sigma_G-\sigma_F}$

   此表明

   $\displaystyle\frac{\ln w_T-T\mu_F}{\sqrt{T}\sigma_F}<\frac{\ln w_T-T\mu_G}{\sqrt{T}\sigma_G}$

   從而 $F\left(w_T\right)-G\left(w_T\right)<0$，因此這個函數在 $\left[0,w_T^*\right]$ 上的積分恆負，證明完畢。
3. 1. 對於 $Ⅰ$ 而言，容易看出當 $w_T\in\left[0,M\right]$ 時 $U\left(w_T\right)=w_T$。又 $F'\left(w_T\right)>0$，這就說明了

      $\displaystyle Ⅰ=\int_0^MU\left(w_T\right)dF\left(w_T\right)=\int_0^Mw_TF'\left(w_T\right)dw_T>0$

      另一方面，可以由分部積分法看出

      $\displaystyle Ⅰ=w_TF\left(w_T\right)\Big|_{w_T=0}^{w_T=M}-\int_0^MF\left(w_T\right)dw_T<MF\left(M\right)=M\Phi\left(\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}\right)$

      證明完畢。
   2. 直接將已知的資訊代入後有

      $\displaystyle Ⅱ=M\int_M^{\infty}\left(1+\ln\frac{w_T}{M}\right)\cdot\frac1{w_T\sqrt{2\pi T}\sigma_F}\exp\left[-\frac{\left(\left(\ln w_T-T\mu_F\right)/\left(\sqrt{T}\sigma_F\right)\right)^2}2\right]dw_T$

      使用變數變換，令 $\displaystyle s=\frac{\ln w_T-T\mu_F}{\sqrt{T}\sigma_F}$，那麼
      • 當 $w_T\to\infty$ 時有 $s\to\infty$；
      • 當 $w_T=M$ 時有 $\displaystyle s=\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}$；
      • 求導有 $\displaystyle ds=\frac{dw_T}{w_T\sqrt{T}\sigma_F}$。
      那麼 $Ⅱ$ 可改寫並計算如下

      $\displaystyle\begin{aligned}Ⅱ&=M\int_{\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}}^{\infty}\left(s\sqrt{T}\sigma_F+1+T\mu_F-\ln M\right)\cdot\frac1{\sqrt{2\pi}}e^{-\frac{s^2}2}ds\\&=M\int_{\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}}^{\infty}\frac{\sqrt{T}\sigma_F}{\sqrt{2\pi}}\cdot se^{-\frac{s^2}2}ds+M\left(1+T\mu_F-\ln M\right)\int_{\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}}^{\infty}\frac1{\sqrt{2\pi}}e^{-\frac{s^2}2}ds\\&=\left.M\cdot-\frac{\sqrt{T}\sigma_F}{\sqrt{2\pi}}e^{-\frac{s^2}2}\right|_{\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}}^{\infty}+M\left(1+T\mu_F-\ln M\right)\Phi\left(-\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}\right)\\&=M\left[\left(1+T\mu_F-\ln M\right)\Phi\left(\frac{T\mu_F-\ln M}{\sqrt{T}\sigma_F}\right)+\frac{\sqrt{T}\sigma_F}{\sqrt{2\pi}}\exp\left(-\frac{\left(\left(\ln M-T\mu_F\right)^2/\left(\sqrt{T}\sigma_F\right)\right)^2}2\right)\right]\end{aligned}$

      故取

      $\displaystyle \alpha=1+T\mu_F-\ln M$,　$\displaystyle\beta=\frac{T\mu_F-\ln M}{\sqrt{T}\sigma_F}$,　$\displaystyle\gamma=\frac{\sqrt{T}\sigma_F}{\sqrt{2\pi}}$

      以及 $\displaystyle\delta=\frac{\ln M-T\mu_F}{\sqrt{T}\sigma_F}$。
   3. 由於根據6.3.1以及累積分配函數的特性可知 $Ⅰ<M$，因此 $Ⅰ/T$ 隨 $T$ 趨於無窮而趨於零。又又利用6.3.2的結果可以看出 $Ⅱ/T$ 隨 $T$ 趨於無窮而趨於 $M\mu_F$，因此所求

      $\displaystyle\lim_{T\to\infty}\frac1TE_FU\left(w_T\right)=\lim_{T\to\infty}\left(\frac{Ⅰ}T+\frac{Ⅱ}T\right)=M\mu_F$



7. ($15\%$) Define $\displaystyle\Phi\left(c\right)\equiv\int_{-\infty}^c\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}dx$ to be the cumulative distribution function for the standard normal distribution and $a>0$ and $b>0$.
   1. ($5\%$) What is $\displaystyle\lim_{y\to\infty}e^{ay}\Phi\left(-b\sqrt{y}\right)$ if $a>b^2/2$?
   2. ($5\%$) What is $\displaystyle\lim_{y\to\infty}e^{ay}\Phi\left(-b\sqrt{y}\right)$ if $a\leq b^2/2$?
   3. ($5\%$) What is $\displaystyle\lim_{y\to\infty}e^{\frac{b^2y}2}\sqrt{y}\Phi\left(-b\sqrt{y}\right)$?
訣竅
運用羅必達法則求極限。
解法

1. 運用羅必達法則可知

   $\displaystyle\lim_{y\to\infty}e^{ay}\Phi\left(-b\sqrt{y}\right)=\lim_{y\to\infty}\frac{\displaystyle\int_{-\infty}^{-b\sqrt{y}}\exp\left(-x^2/2\right)dx}{\sqrt{2\pi}e^{-ay}}=\lim_{y\to\infty}\frac{\displaystyle e^{-\frac{b^2y}2}\cdot-\frac{b}{2\sqrt{y}}}{-a\sqrt{2\pi}e^{-ay}}=\frac{b}{2a\sqrt{2\pi}}\lim_{y\to\infty}\frac{e^{(a-b^2/2)y}}{\sqrt{y}}=\infty$

   其中最後一個等號是因為 $a-b^2/2>0$ 所致。
2. 運用羅必達法則可知

   $\displaystyle\lim_{y\to\infty}e^{ay}\Phi\left(-b\sqrt{y}\right)=\lim_{y\to\infty}\frac{\displaystyle\int_{-\infty}^{-b\sqrt{y}}\exp\left(-x^2/2\right)dx}{\sqrt{2\pi}e^{-ay}}=\lim_{y\to\infty}\frac{\displaystyle e^{-\frac{b^2y}2}\cdot-\frac{b}{2\sqrt{y}}}{-a\sqrt{2\pi}e^{-ay}}=\frac{b}{2a\sqrt{2\pi}}\lim_{y\to\infty}\frac{e^{(a-b^2/2)y}}{\sqrt{y}}=0$

   其中最後一個等號是因為 $a-b^2/2\leq0$ 所致。
3. 運用羅必達可知

   $\displaystyle\begin{aligned}\lim_{y\to\infty}e^{\frac{b^2y}2}\sqrt{y}\Phi\left(-b\sqrt{y}\right)&=\lim_{y\to\infty}\frac{\displaystyle\int_{-\infty}^{-b\sqrt{y}}\exp\left(-x^2/2\right)dx}{\sqrt{2\pi}\exp\left(-b^2y/2\right)/\sqrt{y}}\\&=\lim_{y\to\infty}\frac{\displaystyle e^{-\frac{b^2y}2}\cdot-\frac{b}{2\sqrt{y}}}{\displaystyle-\frac{b^2\sqrt{2\pi}e^{-\frac{b^2y}2}}{2\sqrt{y}}-\frac{\sqrt{2\pi}e^{-\frac{b^2y}2}}{2y\sqrt{y}}}=\lim_{y\to\infty}\frac{by}{b^2y\sqrt{2\pi}+\sqrt{2\pi}}=\frac1{b\sqrt{2\pi}}\end{aligned}$