- ($10\%$) Solve the function $A\left(t\right)$ if
$A'\left(t\right)=\alpha A^2\left(t\right)+\beta A\left(t\right)-1$, and $A\left(0\right)=0$,
where $\alpha$ and $\beta$ are real numbers. - 若 $\beta^2+4\alpha>0$，那麼設方程的兩實根為 $\displaystyle\frac{-\beta-\sqrt{\beta^2+4\alpha}}{2\alpha}=x_1<x_2:=\frac{-\beta+\sqrt{\beta^2+4\alpha}}{2\alpha}$，如此給定的微分方程可寫為
$A'\left(t\right)=\alpha\left(A\left(t\right)-x_1\right)\left(A\left(t\right)-x_2\right)$
那麼便有$\displaystyle\left(\frac1{A-x_2}-\frac1{A-x_1}\right)dA=\frac{dA}{\left(A-x_1\right)\left(A-x_2\right)}=\alpha\left(x_2-x_1\right)dt$
取積分可得$\displaystyle\ln\left|\frac{A-x_2}{A-x_1}\right|=\alpha\left(x_2-x_1\right)t+\ln\left|\frac{x_2}{x_1}\right|$
取自然指數有$\displaystyle\frac{A-x_2}{A-x_1}=\frac{x_2}{x_1}e^{\alpha\left(x_2-x_1\right)t}$
移項整理可得$\displaystyle A\left(t\right)=\frac{x_1x_2\left(e^{\alpha\left(x_2-x_1\right)t}-1\right)}{x_2e^{\alpha\left(x_2-x_1\right)t}-x_1}=\frac{2\left(1-e^{t\sqrt{\beta^2+4\alpha}}\right)}{\left(-\beta+\sqrt{\beta^2+4\alpha}\right)e^{t\sqrt{\beta^2+4\alpha}}+\beta+\sqrt{\beta^2+4\alpha}}$
- 若 $\beta^2+4\alpha=0$，那麼方程式有相同的實根 $\displaystyle-\frac\beta{2\alpha}$，如此給定的微分方程可寫為
$\displaystyle A'\left(t\right)=\alpha\left(A\left(t\right)+\frac\beta{2\alpha}\right)^2$
那麼移項有$\displaystyle\frac{dA}{\displaystyle\left(A+\frac\beta{2\alpha}\right)^2}=\alpha dt$
取積分可得$\displaystyle-\frac1{\displaystyle A+\frac\beta{2\alpha}}=\alpha t-\frac{2\alpha}\beta$
可解得$\displaystyle A\left(t\right)=\frac{\beta^2t}{2\alpha\left(2-\beta t\right)}$
- 若 $\beta^2+4\alpha<0$，那麼方程式有兩共軛複根 $\displaystyle\frac{-\beta+i\sqrt{-\beta^2-4\alpha}}{2\alpha}$ 與 $\displaystyle\frac{-\beta-i\sqrt{-\beta^2-4\alpha}}{2\alpha}$，如此給定的微分方程可寫為
$\displaystyle\frac{dA}{\displaystyle\left(A+\frac\beta{2\alpha}\right)^2+\left(\frac{\sqrt{-\beta^2-4\alpha}}{2\alpha}\right)^2}=\alpha dt$
取積分有$\displaystyle\frac{2\alpha}{\sqrt{-\beta^2-4\alpha}}\tan^{-1}\frac{2\alpha A+\beta}{\sqrt{-\beta^2-4\alpha}}=\alpha t+\frac{2\alpha}{\sqrt{-\beta^2-4\alpha}}\tan^{-1}\frac{\beta}{\sqrt{-\beta^2-4\alpha}}$
$\displaystyle A\left(t\right)=\frac1{2\alpha}\left[-\beta+\sqrt{-\beta^2-4\alpha}\tan\left(\frac{t\sqrt{-\beta^2-4\alpha}}2+\tan^{-1}\frac{\beta}{\sqrt{-\beta^2-4\alpha}}\right)\right]$
- ($10\%$) Given
$\displaystyle y_m=\int\frac{dx}{\left(x^2+4\right)^m}$,
express $y_m$ as $A+By_{m-1}$, where $A$ and $B$ are functions of $x$ and $m$. - ($10\%$) Find the Taylor series about $x=0$ for the following integral:
$\displaystyle\int x^2e^{-x^2}dx$.
- ($20\%$) The Black-Scholes formula for a call option with six input parameters $\left(S,X,r,q,\sigma,T\right)$ is as follows.
$c\left(S,X,r,q,\sigma,T\right)=Se^{-qT}N\left(d_1\right)-Xe^{-rT}N\left(d_2\right)$,
where$\displaystyle d_1=\frac{\ln\left(S/X\right)+\left(r-q+\sigma^2/2\right)T}{\sigma\sqrt{T}}$ and $\displaystyle d_2=\frac{\ln\left(S/X\right)+\left(r-q-\sigma^2/2\right)T}{\sigma\sqrt{T}}=d_1-\sigma\sqrt{T}$,
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.- ($5\%$) Derive and express $\displaystyle\frac{\partial c}{\partial S}$ as the form of $e^AN\left(B\right)$. What are $A$ and $B$?
- ($5\%$) Derive and express $\displaystyle\frac{\partial^2c}{\partial S^2}$ as the form of $\displaystyle\frac{n\left(C\right)e^D}{E}$. What are $C$, $D$, and $E$?
- ($5\%$) Derive and express $\displaystyle\frac{\partial c}{\partial\sigma}$ as the form of $Fe^Gn\left(H\right)$. What are $F$, $G$, and $H$?
- ($5\%$) Derive and express $\displaystyle\frac{\partial c}{\partial r}$ as the form of $Ie^JN\left(K\right)$. What are $I$, $J$, and $K$?
- 由給定的方程求偏導有
$\displaystyle\begin{aligned}\frac{\partial c}{\partial S}&=e^{-qT}N\left(d_1\right)+Se^{-qT}N'\left(d_1\right)\cdot\frac{\partial d_1}{\partial S}-Xe^{-rT}N'\left(d_2\right)\cdot\frac{\partial d_2}{\partial S}\\&=e^{-qT}N\left(d_1\right)+\frac{Se^{-qT}n\left(d_1\right)-Xe^{-rT}n\left(d_2\right)}{S\sigma\sqrt{T}}\end{aligned}$
其中我們注意到$\displaystyle d_1+d_2=\frac{2\left[\ln\left(S/X\right)+\left(r-q\right)T\right]}{\sigma\sqrt{T}}=\frac{2\left[\ln\left(S/X\right)+\left(r-q\right)T\right]}{d_1-d_2}$
這表明$\displaystyle e^{\frac{d_1^2-d_2^2}2}=\frac{S}Xe^{\left(r-q\right)T}$
即有 $Se^{-qT}e^{-d_1^2/2}=Xe^{-rT}e^{-d_2^2/2}$，因此 $Se^{-qT}n\left(d_1\right)-Xe^{-rT}n\left(d_2\right)=0$，故得 $\displaystyle\frac{\partial c}{\partial S}=e^{-qT}N\left(d_1\right)$，因此 $A=-qT$ 且 $B=d_1$。 - 繼續使用連鎖律求二階偏導函數有
$\displaystyle\frac{\partial^2c}{\partial S^2}=e^{-qT}n\left(d_1\right)\cdot\frac1{S\sigma\sqrt{T}}=\frac{n\left(d_1\right)e^{-qT}}{S\sigma\sqrt{T}}$
因此 $C=d_1$、$D=-qT$、$E=S\sigma\sqrt{T}$。 - 使用連鎖律求導可知
$\displaystyle\begin{aligned}\frac{\partial c}{\partial\sigma}&=Se^{-qT}n\left(d_1\right)\frac{\partial d_1}{\partial\sigma}-Xe^{-rT}n\left(d_2\right)\frac{\partial\left(d_1-\sigma\sqrt{T}\right)}{\partial\sigma}\\&=Se^{-qT}n\left(d_1\right)\left(-\frac{d_1}{\sigma}+\sqrt{T}\right)-Xe^{-rT}n\left(d_2\right)\left(-\frac{d_1}{\sigma}\right)=S\sqrt{T}e^{-qT}n\left(d_1\right)\end{aligned}$
因此 $F=S\sqrt{T}$、$G=-qT$、$H=d_1$。 - 同樣使用連鎖律計算可得
$\displaystyle\frac{\partial c}{\partial r}=Se^{-qT}n\left(d_1\right)\cdot\frac{\sqrt{T}}\sigma+Xe^{-rT}N\left(d_2\right)-Xe^{-rT}n\left(d_2\right)\cdot\frac{\sqrt{T}}{\sigma}=Xe^{-rT}N\left(d_2\right)$
故取 $I=X$、$J=-rT$、$K=d_2$。 - ($10\%$) Represent $\left(1-x\right)^{-2}$ in a Maclaurin series for $-1<x<1$.
- ($10\%$) Find the equation of the line tangent to the curve $x=2t^3-15t^2+24t+7$, $y=t^2+t+1$ at $t=2$.
- ($10\%$) Find the maximum and minimum values of $f\left(x,y\right)=xy^2$ subject to the condition $x^2+y^2=1$.
- 若 $y=0$，則第三式給出 $x=\pm1$；
- 若 $x=-\lambda$，則第一式有 $y^2=2\lambda^2$，從而於第三式有 $3\lambda^2=1$，即 $\displaystyle\lambda=\pm\frac{\sqrt3}3$，進而得四座標 $\displaystyle\left(x,y\right)=\left(\frac{\sqrt3}3,\pm\frac{\sqrt6}3\right)$ 與 $\displaystyle\left(-\frac{\sqrt3}3,\pm\frac{\sqrt6}3\right)$。
- ($10\%$) Evaluate $\displaystyle\int\left(\tan^5x\right)\left(\sec^4x\right)dx$.
- ($10\%$) Determine the interval of convergence of $\displaystyle\sum_{n=1}^{+\infty}\frac{x^n}{2+n^2}$.
- 當 $x=1$ 時級數可寫為 $\displaystyle\sum_{n=1}^{\infty}\frac1{2+n^2}<\sum_{n=1}^{\infty}\frac1{n^2}$ 收斂。
- 當 $x=-1$ 時可透過前一點知道其絕對收斂，故此點也收斂。
訣竅
利用分離變量法以及一元二次方程式的概念按係數討論來處理。解法
假若 $\alpha=\beta=0$，那麼微分方程為 $A'\left(t\right)=-1$，取積分並使用 $A\left(0\right)=0$ 得 $A\left(t\right)=-t$。假若 $\alpha=0$ 但 $\beta\neq0$ 時同乘以 $e^{-\beta t}$ 有$\left(e^{-\beta t}A\left(t\right)\right)'=e^{-\beta t}A'\left(t\right)-\beta e^{-\beta t}A\left(t\right)=-e^{-\beta t}$
同取積分有$\displaystyle e^{-\beta t}A\left(t\right)=\frac{e^{-\beta t}-1}{\beta}$
故得 $\displaystyle A\left(t\right)=\frac{1-e^{\beta t}}\beta$。現設 $\alpha\neq0$，並考慮下列三種情形：
訣竅
運用分部積分法改寫以求得遞迴關係式。解法
直接使用分部積分法可知$\displaystyle\begin{aligned}y_{m-1}&=\frac{x}{\left(x^2+4\right)^{m-1}}-\int x\left[\left(x^2+4\right)^{-m+1}\right]'dx=\frac{x}{\left(x^2+4\right)^{m-1}}+2\left(m-1\right)\int\frac{x^2}{\left(x^2+4\right)^m}dx\\&=\frac{x}{\left(x^2+4\right)^{m-1}}+2\left(m-1\right)\int\frac{x^2+4}{\left(x^2+4\right)^m}dx-8\left(m-1\right)\int\frac{dx}{\left(x^2+4\right)^m}\\&=\frac{x}{\left(x^2+4\right)^{m-1}}+2\left(m-1\right)y_{m-1}-8\left(m-1\right)y_m\end{aligned}$
由此整理可得$\displaystyle y_m=\frac{2m-3}{8\left(m-1\right)}y_{m-1}+\frac{x}{8\left(m-1\right)\left(x^2+4\right)^{m-1}}$
因此 $\displaystyle A=\frac{x}{8\left(m-1\right)\left(x^2+4\right)^{m-1}}$、$\displaystyle B=\frac{2m-3}{8\left(m-1\right)}$。訣竅
經由自然指數的泰勒展開式進行適當的變形改寫即可。解法
由自然指數的泰勒展開式可知$\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$
用 $-x^2$ 取代 $x$ 可得$\displaystyle e^{-x^2}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^nx^{2n}}{n!}$
兩邊同乘以 $x^2$ 有$\displaystyle x^2e^{-x^2}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^nx^{2n+2}}{n!}$
最後同取不定積分有$\displaystyle\int x^2e^{-x^2}dx=C+\sum_{n=0}^{\infty}\frac{\left(-1\right)^nx^{2n+3}}{n!\left(2n+3\right)}$
訣竅
按照題意直接使用多變數函數的連鎖律演算即可。解法
訣竅
由經典的無窮等比級數導出給定函數的級數表達式。解法
首先注意到$\displaystyle\left(1-x\right)^{-1}=\sum_{n=0}^{\infty}x^n$
兩邊同時微分可得$\displaystyle\left(1-x\right)^{-2}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}\left(n+1\right)x^n$
訣竅
運用連鎖律求參數曲線的斜率，隨後使用點斜式寫出切線方程式。解法
使用連鎖律可知$\displaystyle\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{6t^2-30t+24}$
那麼在 $t=2$ 時的斜率為 $\displaystyle\left.\frac{dy}{dx}\right|_{t=2}=-\frac5{12}$。故切線方程式為$\displaystyle y-y\left(2\right)=-\frac5{12}\left(x-x\left(2\right)\right)$
即 $5x+12y=167$。訣竅
由初等不等式估算極值即可；亦可使用拉格朗日乘子法求解。解法一
使用算術幾何不等式可知$\displaystyle\frac13=\frac{\displaystyle x^2+\frac{y^2}2+\frac{y^2}2}3\geq\sqrt[3]{x^2\cdot\frac{y^2}2\cdot\frac{y^2}2}=2^{-2/3}\left(xy^2\right)^{2/3}$
如此可知$\displaystyle-\frac{2\sqrt3}9=-\frac2{3\sqrt3}=-\left(\frac{2^{2/3}}3\right)^{3/2}\leq xy^2\leq\left(\frac{2^{2/3}}3\right)^{3/2}=\frac2{3\sqrt3}=\frac{2\sqrt3}9$
此時等號成立條件為 $\displaystyle x^2=y^2/2$，故分別解得 $\displaystyle\left(x,y\right)=\left(\frac{\sqrt3}3,\pm\frac{\sqrt6}3\right)$ 與 $\displaystyle\left(x,y\right)=\left(-\frac{\sqrt3}3,\pm\frac{\sqrt6}3\right)$。解法二
設定拉格朗日乘子函數如下$F\left(x,y,\lambda\right)=xy^2+\lambda\left(x^2+y^2-1\right)$
據此解聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,\lambda\right)=y^2+2\lambda x=0\\&F_y\left(x,y,\lambda\right)=2xy+2\lambda y=0\\&F_{\lambda}\left(x,y,\lambda\right)=x^2+y^2-1=0\end{aligned}\right.$
由第二式可知 $2y\left(x+\lambda\right)=0$。訣竅
使用三角恆等式以及基本三角函數的反導函數即可求解。解法
改寫後可直接計算如下$\displaystyle\int\left(\tan^5x\right)\left(\sec^4x\right)dx=\int\tan^5x\left(1+\tan^2x\right)\sec^2xdx=\frac{\tan^6x}6+\frac{\tan^8x}8+C$
訣竅
由比值審歛法的概念先確定收斂半徑，隨後檢查端點。解法
首先由比值審歛法能知$\displaystyle\lim_{n\to\infty}\left|\frac{x^{n+1}}{2+\left(n+1\right)^2}\div\frac{x^n}{2+n^2}\right|<1$
即得 $\left|x\right|<1$，因此收斂半徑為 $1$。現檢查端點如下
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