國立臺灣大學一百零一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. To answer:
   1. ($5\%$) Evaluate $\displaystyle\lim_{x\to1}\frac{1+\cos\pi x}{x^2-2x+1}$
   2. ($5\%$) If $\displaystyle\phi\left(\alpha\right)=\int_{\alpha}^{\alpha^2}\frac{\sin\alpha x}xdx$, find $\phi'\left(\alpha\right)$ where $\alpha\neq0$.
   3. ($5\%$) Test for convergence: $\displaystyle\sum_{n=1}^{\infty}\frac1{n^p}$, $p=$ constant.
   4. ($5\%$) Determine whether $\displaystyle\int_{-1}^5\frac{dx}{\left(x-1\right)^3}$ converges. (i) in the usual sense, (ii) in the Cauchy principal value sense.
訣竅
第一小題可運用羅必達法則求解；第二小題先使用變數代換後透過微積分基本定理與連鎖律計算即可；第三題為標準透過積分審歛法檢驗 $p$ 級數的問題；第四題回憶瑕積分與科西主值的定義計算即可。
解法

1. 使用羅必達法則可知道

   $\displaystyle\lim_{x\to1}\frac{1+\cos\pi x}{x^2-2x+1}=\lim_{x\to1}\frac{-\pi\sin\pi x}{2x-2}=\lim_{x\to1}\frac{-\pi^2\cos\pi x}2=\frac{\pi^2}2$

2. 令 $u=\alpha x$，那麼
   • 當 $x=\alpha$ 時有 $u=\alpha^2$；
   • 當 $x=\alpha^2$ 時有 $u=\alpha^3$；
   • 求導有 $du=\alpha dx$。
   如此函數 $\phi$ 可改寫並表達如下

   $\displaystyle\phi\left(\alpha\right)=\int_{\alpha}^{\alpha^2}\frac{\sin\alpha x}xdx=\int_{\alpha^2}^{\alpha^3}\frac{\sin u}udu$

   如此使用微積分基本定理與連鎖律可知

   $\displaystyle\phi'\left(\alpha\right)=\frac{\sin\left(\alpha^3\right)}{\alpha^3}\cdot3\alpha^2-\frac{\sin\left(\alpha^2\right)}{\alpha^2}\cdot2\alpha=\frac{3\sin\left(\alpha^3\right)-2\sin\left(\alpha^2\right)}\alpha$

3. 當 $p\leq0$ 時，數列極限 $\displaystyle\lim_{n\to\infty}\frac1{n^p}$ 不趨於零，故級數發散。而當 $p>0$ 時有函數 $\displaystyle\frac1{x^p}$ 遞減且 $\displaystyle\lim_{x\to\infty}\frac1{x^p}=0$。針對 $p$ 我們可區分三種情形如下
   • 當 $0<p<1$ 時，可以計算發現

     $\displaystyle\int_1^{\infty}\frac1{x^p}dx=\left.\frac{x^{-p+1}}{-p+1}\right|_1^{\infty}=\infty$

     由積分審歛法知給定的級數發散。
   • 當 $p=1$ 時，可計算發現

     $\displaystyle\int_1^{\infty}\frac1xdx=\ln x\Big|_1^{\infty}=\infty$

     由積分審歛法知給定的級數發散。
   • 當 $p>1$ 時，可以計算發現

     $\displaystyle\int_1^{\infty}\frac1{x^p}dx=\left.\frac{x^{-p+1}}{-p+1}\right|_1^{\infty}=\frac1{p-1}<\infty$

     由積分審歛法知給定的級數收斂。
   綜上可知當 $p>1$ 時給定級數收斂，而當 $p\leq1$ 時給定級數發散。
4. 1. 按通常意義可知給定的瑕積分應表達如下

      $\displaystyle\int_{-1}^5\frac{dx}{\left(x-1\right)^3}=\lim_{a\to1^-}\int_{-1}^a\frac{dx}{\left(x-1\right)^3}+\lim_{a\to1^+}\int_a^5\frac{dx}{\left(x-1\right)^3}$

      然而直接計算將發現

      $\displaystyle\begin{aligned}&\lim_{a\to1^-}\int_{-1}^a\frac{dx}{\left(x-1\right)^3}=\left.-\lim_{a\to1^-}\frac1{2\left(x-1\right)^2}\right|_{-1}^a=\lim_{a\to1^-}\left(\frac18-\frac1{2\left(a-1\right)^2}\right)=-\infty\\&\lim_{a\to1^+}\int_a^5\frac{dx}{\left(x-1\right)^3}=-\left.\lim_{a\to1^+}\frac1{2\left(x-1\right)^2}\right|_a^5=\lim_{a\to1^+}\left(\frac1{2\left(a-1\right)^2}-\frac1{32}\right)=\infty\end{aligned}$

      故給定的瑕積分按照通常的意義發散。
   2. 按照柯西主值意義下的瑕積分應透過下式計算

      $\displaystyle\begin{aligned}\text{PV}\int_{-1}^5\frac{dx}{\left(x-1\right)^3}&=\lim_{\varepsilon\to0^+}\left(\int_{-1}^{1-\varepsilon}\frac{dx}{\left(x-1\right)^3}+\int_{1+\varepsilon}^5\frac{dx}{\left(x-1\right)^3}\right)\\&=\lim_{\varepsilon\to0^+}\left[\left.-\frac1{2\left(x-1\right)^2}\right|_{-1}^{1-\varepsilon}+\left.-\frac1{2\left(x-1\right)^2}\right|_{1+\varepsilon}^5\right]\\&=\lim_{\varepsilon\to0^+}\left[-\frac1{2\varepsilon^2}+\frac18-\frac1{32}+\frac1{2\varepsilon^2}\right]=\frac3{32}\end{aligned}$



2. ($10\%$) Solve $y''+2y'+2y=-10xe^x+5\sin x$. [hint: general solution using undetermined coefficients]
訣竅
先透過輔助方程解出齊次微分方程的解，隨後透過待定係數法求出非齊次方程的特解，最終獲得通解。
解法

先考慮齊次微分方程 $y_h''+2y_h'+2y_h=0$，其所對應的輔助方程為 $\lambda^2+2\lambda+2=0$，可解得 $\lambda=-1\pm i$，從而有

$y_h\left(x\right)=c_1e^{-x}\sin x+c_2e^{-x}\cos x$

而對於原方程的非齊次項，我們考慮特解 $y_p$ 具有如下的形式

$y_p\left(x\right)=Axe^x+Be^x+C\sin x+D\cos x$

據此代入原方程中有

$5Axe^x+\left(4B+4A\right)e^x+\left(C-2D\right)\sin x+\left(D+2C\right)\cos x=-10xe^x+5\sin x$

經由比較係數有下列的線性方程組

$\left\{\begin{aligned}&5A=-10\\&4B+4A=0\\&C-2D=5\\&D+2C=0\end{aligned}\right.$

可解得 $A=-2$、$B=2$、$C=1$、$D=-2$，故取特解 $y_p\left(x\right)=-2xe^x+2e^x+\sin x-2\cos x$。因此給定的微分方程的通解為

$y\left(x\right)=y_h\left(x\right)+y_p\left(x\right)=c_1e^{-x}\sin x+c_2e^{-x}\cos x-2xe^x+2e^x+\sin x-2\cos x$



3. ($10\%$) Let $x$ be a random variable with a finite expected value, $\mu=E\left(x\right)$. If $\varphi\left(x\right)$ is a twice differentiable convex function, then to show: $E\left[\varphi\left(x\right)\right]\geq\varphi\left[E\left(x\right)\right]$.
訣竅
此為經典的 Jensen 不等式。
解法

設 $a$ 與 $b$ 為擴充實數，即 $a,b\in\overline{\mathbb{R}}=\mathbb{R}\cup\left\{\pm\infty\right\}$，且設 $x$ 為連續的隨機變數並定義於 $\left(a,b\right)$ 上。按題設有 $\displaystyle\mu=\int_a^bxf_X\left(x\right)dx$，其中 $f_X$ 為對應的機率密度函數。由此所需證明的不等式為 $\displaystyle\int_a^b\varphi\left(x\right)f_X\left(x\right)dx\geq\varphi\left(\mu\right)$。

回憶關於凸函數的定義以及 $\phi$ 可二次求導，故知

$\phi\left(x\right)\geq\phi\left(\mu\right)+\phi'\left(\mu\right)\left(x-\mu\right)$

兩邊同乘以 $f_X\left(x\right)$ 後在 $\left[a,b\right]$ 上取積分可知

$\displaystyle\int_a^b\varphi\left(x\right)f_X\left(x\right)dx\geq\phi\left(\mu\right)\int_a^bf_X\left(x\right)dx+\varphi'\left(\mu\right)\left[\int_a^bxf_X\left(x\right)dx-\mu\int_a^bf_X\left(x\right)dx\right]=\varphi\left(\mu\right)$

其中我們使用了 $\displaystyle\int_a^bf_X\left(x\right)dx=1$ 與 $\displaystyle\int_a^bxf_X\left(x\right)dx=\mu$。證明完畢。



4. ($10\%$) If $f\left(x\right)$ and $g\left(x\right)$ are continuous on the closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$. To show: there exists a point $c$, $a<c<b$, such that $\left[f\left(b\right)-f\left(a\right)\right]g'\left(c\right)=\left[g\left(b\right)-g\left(a\right)\right]f'\left(c\right)$.
訣竅
此為柯西均值定理，可運用 Rolle 定理證明之。
解法
考慮函數 $h:\left[a,b\right]\to\mathbb{R}$ 定義如下
$h\left(x\right)=\left[f\left(b\right)-f\left(a\right)\right]g\left(x\right)-\left[g\left(b\right)-g\left(a\right)\right]f\left(x\right)$
那麼容易發現
$\begin{aligned}h\left(b\right)&=\left[f\left(b\right)-f\left(a\right)\right]g\left(b\right)-\left[g\left(b\right)-g\left(a\right)\right]f\left(b\right)=g\left(a\right)f\left(b\right)-g\left(b\right)f\left(a\right)\\&=\left[f\left(b\right)-f\left(a\right)\right]g\left(a\right)-\left[g\left(b\right)-g\left(a\right)\right]f\left(a\right)=h\left(a\right)\end{aligned}$
且因 $f$ 與 $g$ 在 $\left[a,b\right]$ 上連續而在 $\left(a,b\right)$ 上可導，故 $h$ 也如此，從而由 Rolle 定理可知存在 $c\in\left(a,b\right)$ 滿足 $h'\left(c\right)=0$，此即所欲證。


5. ($12\%$) A telephone company is planning to introduce two new types of executive communication systems that it hopes to sell to its commercial customers. It is estimated that if the first type of system is priced at $x$ hundred dollars per system and the second type at $y$ hundred dollars per system, approximately $40-8x+5y$ customers will buy the first type and $50+9x-7y$ customers will buy the second type. If the cost of manufacturing the first type is $\$1,000$ per system and the cost of manufacturing the second type is $\$2,000$ per system, how should the telephone company price the systems to generate the largest possible profit?
   1. ($4\%$) Derive the objective function of the total profit and calculate the first-order partial derivatives with respect to $x$ and $y$, respectively.
   2. ($4\%$) Solve the optimal values of $x$ and $y$.
   3. ($4\%$) Examine the second-order partial derivatives to verify the optimal values of $x$ and $y$ can generate the largest total profit.
訣竅
運用線性規劃的概念解此題。
解法
1. 以總利潤為目標函數為

   $P\left(x,y\right)=\left(x-10\right)\left(40-8x+5y\right)+\left(y-20\right)\left(50+9x-7y\right)=-8x^2+14xy-7y^2-60x+140y-1400$

   單位為百元。再者分別計算一階偏導函數有

   $\left\{\begin{aligned}&P_x\left(x,y\right)=-16x+14y-60\\&P_y\left(x,y\right)=14x-14y+140\end{aligned}\right.$

2. 解一階偏導函數為零之處，即解聯立方程組

   $\left\{\begin{aligned}&-16x+14y-60=0\\&14x-14y+140=0\end{aligned}\right.$

   可知 $\left(x,y\right)=\left(40,50\right)$。
3. 現計算二階判別式如下

   $D\left(x,y\right)=\begin{vmatrix}P_{xx}\left(x,y\right)&P_{xy}\left(x,y\right)\\P_{yx}\left(x,y\right)&P_{yy}\left(x,y\right)\end{vmatrix}=\begin{vmatrix}-16&14\\14&-14\end{vmatrix}=28>0$

   且 $P_{xx}<0$，故 $P$ 在 $\left(x,y\right)=\left(40,50\right)$ 處達到絕對極大值。


6. ($12\%$) Consider a variable $x$ which follows a Poisson distribution with the intensity of events occurring to be $\lambda$. The probability that there are exactly $k$ events occurring is

   $\displaystyle\text{prob}\left(x=k\right)=\frac{\lambda^ke^{-\lambda}}{k!}$.

   Moreover, the intensity parameter, $\lambda$, is a nonnegative variable and follows a gamma distribution, i.e., $\lambda\sim\text{Gamma}\left(\alpha,\beta\right)$, where $\alpha$ and $\beta$ are positive real numbers. Consequently, the probability density function for $\lambda$ is as follows.

   $\displaystyle f\left(\lambda\right)=\beta^{\alpha}\frac1{\Gamma\left(\alpha\right)}\lambda^{\alpha-1}e^{-\beta\lambda}$,

   where $\Gamma\left(Z\right)$ is the gamma function and defined as

   $\displaystyle\Gamma\left(Z\right)=\int_0^{\infty}e^{-t}t^{Z-1}dt$,

   if $Z$ is a complex number with a positive real part.

   For the above Gamma-Poisson mixture distribution, the probability that there are exactly $k$ events occurring for the variable $x$ can be expressed in the following form.

   $\displaystyle\text{prob}\left(x=k\right)=\frac{\Gamma\left(a\right)}{\Gamma\left(b\right)k!}\left(c\right)^{\alpha}\left(d\right)^k$.

   Solve $a$, $b$, $c$, and $d$ as functions of $\alpha$, $\beta$, $\lambda$, and $k$.

訣竅
運用連續-離散聯合機率密度的概念求解即可。
解法
按聯合機率密度的概念可知
$\displaystyle\text{prob}\left(x=k\right)=\int_0^{\infty}\frac{\lambda^ke^{-\lambda}}{k!}f\left(\lambda\right)d\lambda=\frac{\beta^{\alpha}}{\Gamma\left(\alpha\right)k!}\int_0^{\infty}\lambda^{k+\alpha-1}e^{-\left(\beta+1\right)\lambda}d\lambda$
令 $t=\left(\beta+1\right)\lambda$，那麼上式可改寫為
$\displaystyle\text{prob}\left(x=k\right)=\frac{\beta^{\alpha}}{\Gamma\left(\alpha\right)k!\left(\beta+1\right)^{k+\alpha}}\int_0^{\infty}t^{k+\alpha-1}e^{-t}dt=\frac{\Gamma\left(k+\alpha\right)}{\Gamma\left(\alpha\right)k!}\left(\frac{\beta}{1+\beta}\right)^{\alpha}\left(\frac1{1+\beta}\right)^k$
故 $a=k+\alpha$、$b=\alpha$、$\displaystyle c=\frac{\beta}{1+\beta}$、$\displaystyle d=\frac1{1+\beta}$。


7. ($26\%$) The Gaussian Quadrature (GQ) method provides a numerical way to generate the approximation of the definite integral of any function $f\left(x\right)$ as follows.

   $\displaystyle\int_{-1}^1f\left(x\right)dx\approx\sum_{i=1}^nw_if\left(x_i\right)$.

   Different from the traditional numerical integration methods, such as the rectangular rule, trapezoid rule, or Simpson's rule, the GQ method gives users the freedom to choose not only the location of the abscissas, $x_i$, but also the weighting coefficients, $w_i$, such that GQ can generate the exact integral results of the polynomial functions with degree $2n-1$ or less in $\left[-1,1\right]$. Therefore, $x_i$ and $w_i$ can be solved based on the following $2n$ equations.

   $\displaystyle\int_{-1}^1x^ldx=\sum_{i=1}^nw_ix_i^l$ for $l=0,1,\dots,2n-1$.

   For the following problems, consider the case of $n=2$.
   1. ($12\%$) Solve $w_1$, $w_2$, $x_1$, and $x_2$.
   2. ($6\%$) Verify that for any cubic polynomial function, $f\left(x\right)=ax^3+bx^2+cx+d$, then GQ method can generate the exact integral results.
   3. ($8\%$) Suppose $f\left(x\right)=e^{-\frac19x^2+\log_3\left|x\right|}$ is not integral. Apply the GQ method to estimate

      $\displaystyle\int_{-3+2\sqrt3}^{3+2\sqrt3}f\left(x\right)dx$.

訣竅
第一小題僅需針對題意表達出聯立方程組即可計算求解；第二小題則運運前述結果進行驗算；最後一小題則運用先前獲得的數據進行求解即可，其中需留意 GQ 法所適用的範圍，故取適當的變數代換處理之。
解法

1. 由於 $n=2$，我們取 $l=0,1,2,3$ 則有下列四式

   $\displaystyle\left\{\begin{aligned}&2=\int_{-1}^1x^0dx=w_1+w_2\\&0=\int_{-1}^1x^1dx=w_1x_1+w_2x_2\\&\frac23=\int_{-1}^1x^2dx=w_1x_1^2+w_2x_2^2\\&0=\int_{-1}^1x^3dx=w_1x_1^3+w_2x_2^3\end{aligned}\right.$

   由第一式可知 $w_2=2-w_1$，如此第二式至第四式可寫為

   $\left\{\begin{aligned}&w_1\left(x_1-x_2\right)+2x_2=w_1x_1+\left(2-w_1\right)x_2=0\\&w_1\left(x_1^2-x_2^2\right)+2x_2^2=w_1x_1^2+\left(2-w_1\right)x_2^2=\frac23\\&w_1\left(x_1^3-x_2^3\right)+2x_2^3=w_1x_1^3+\left(2-w_1\right)x_2^3=0\end{aligned}\right.$

   那麼有 $w_1\left(x_1-x_2\right)=-2x_2$。分別同乘以 $x_1+x_2$ 與 $x_1^2+x_1x_2+x_2^2$ 可得

   $\displaystyle -2\left(x_1+x_2\right)x_2+2x_2^2=\frac23$,　$-2\left(x_1^2+x_1x_2+x_2^2\right)x_2+2x_2^3=0$.

   即 $\displaystyle x_1x_2=-\frac13$ 與 $\left(x_1+x_2\right)x_1x_2=0$。容易知道 $x_1+x_2=0$ 且由 $\displaystyle x_1x_2=-\frac13$ 得 $\displaystyle\left(x_1,x_2\right)=\left(\frac1{\sqrt3},-\frac1{\sqrt3}\right)$ 或 $\displaystyle\left(x_1,x_2\right)=\left(-\frac1{\sqrt3},\frac1{\sqrt3}\right)$。若 $\displaystyle\left(x_1,x_2\right)=\left(\frac1{\sqrt3},-\frac1{\sqrt3}\right)$，那麼可知

   $\displaystyle w_1=-\frac{2x_2}{x_1-x_2}=1$

   從而 $w_2=2-w_1=1$。類似地，對於 $\displaystyle\left(x_1,x_2\right)=\left(-\frac1{\sqrt3},\frac1{\sqrt3}\right)$ 也可得到 $w_1=w_2=1$。
2. 現利用前一小題的結果進行檢驗如下

   $\displaystyle\int_{-1}^1f\left(x\right)dx=2\int_0^1\left(bx^2+d\right)dx=\left.2\left(\frac{bx^3}3+dx\right)\right|_0^1=\frac{2b}3+2d$

   另一方面有

   $\displaystyle\sum_{i=1}^2w_if\left(x_i\right)=f\left(\frac1{\sqrt3}\right)+f\left(-\frac1{\sqrt3}\right)=\left[\frac{a}{3\sqrt3}+\frac{b}3+\frac{c}{\sqrt3}+d\right]+\left[-\frac{a}{3\sqrt3}+\frac{b}3-\frac{c}{\sqrt3}+d\right]=\frac{2b}3+2d$

   兩者相符，驗證完畢。
3. 由於給定的辦法適用於區間 $\left[-1,1\right]$，故我們先運用變數變換處理區域。令 $\displaystyle u=\frac{x-2\sqrt3}3$，那麼所求的定積分可改寫為

   $\displaystyle\int_{-3+2\sqrt3}^{3+2\sqrt3}f\left(x\right)dx=3\int_{-1}^1f\left(3u+2\sqrt3\right)du=3\int_{-1}^1g\left(u\right)du$

   其中 $g\left(u\right)=f\left(3u+2\sqrt3\right)$，即 $\displaystyle g\left(u\right)=\exp\left[-\frac19\left(3u+2\sqrt3\right)^2+\log_3\left|3u+2\sqrt3\right|\right]$。如此運用 GQ 法可估計定積分如下

   $\displaystyle\int_{-3+2\sqrt3}^{3+2\sqrt3}f\left(x\right)dx\approx3g\left(\frac1{\sqrt3}\right)+3g\left(-\frac1{\sqrt3}\right)=3f\left(3\sqrt3\right)+3f\left(\sqrt3\right)=3+3e^{-3/2}$

【註】　第三小題之敘述有誤。事實上函數 $e^{-\frac19x^2+\log_3\left|x\right|}$ 仍然可積分，但其反導函數無法以初等函數的四則運算與合成進行表達。