115年台北大學統計碩士班-基礎數學詳解

 國立臺北大學115學年度碩士班一般入學考試

系(所)組別:統計學系
科目:基礎數學

I. Calculus(50% total; 10% each)

解答:$$L=   (\csc \frac{x}{2})^{\tan^2 \frac{x}{2}} \Rightarrow \ln L=  \tan^2 \frac{x}{2} \ln(\csc \frac{x}{2}) ={\ln (\csc(x/2)) \over \cot^2(x/2)}\\ v={\pi\over 2}-{x\over 2} \Rightarrow \lim_{x\to \pi} \ln L=  \lim_{v\to 0} \frac{\ln(\sec v)}{\tan^2 v} = \lim_{v\to 0} \frac{-\ln(\cos v)}{\tan^2 v}  = \lim_{v\to 0} \frac{{d\over dv}(-\ln(\cos v))}{{d\over dv}\tan^2 v} \\= \lim_{v\to 0} \frac{-\frac{-\sin v}{\cos v}}{2\tan v \sec^2 v} = \lim_{v\to 0} \frac{\tan v}{2\tan v \sec^2 v} = \lim_{v\to 0} \frac{1}{2\sec^2 v} = \frac{1}{2}  \Rightarrow \lim_{x\to \pi} L=e^{1/2} =\bbox[red, 2pt]{\sqrt e}$$

解答:$$ \lim_{n\to\infty} (\frac{\pi}{\sqrt{n^2+1}} + \frac{\pi}{\sqrt{n^2+2^2}} + \cdots + \frac{\pi}{\sqrt{n^2+n^2}}) = \lim_{n\to\infty} \sum_{k=1}^n \frac{\pi}{\sqrt{n^2+k^2}} = \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n \frac{\pi}{\sqrt{1+(\frac{k}{n})^2}} \\= \int_0^1 \frac{\pi}{\sqrt{1+x^2}} dx  = \pi \left[ \ln(x + \sqrt{1+x^2}) \right]_0^1 =  \pi \left( \ln(1+\sqrt 2) -\ln 1\right) = \bbox[red, 2pt]{\pi \ln(1+\sqrt{2}) }$$

解答:$$ 2(x^2+y^2)^2 + (x^2-1) = 25(x^2-y^2) + 2^x \\\Rightarrow 4(x^2+y^2)(2x + 2y \frac{dy}{dx}) + 2x = 25(2x - 2y \frac{dy}{dx}) + 2^x \ln 2 \\ \cases{x=3\\ y=1} \Rightarrow 4\cdot 10\cdot (6+2{dy\over dx}) +6=25(6-2{dy\over dx}) +8\ln 2 \Rightarrow 246 + 80\frac{dy}{dx} = 150 - 50\frac{dy}{dx} + 8 \ln 2 \\ \Rightarrow  \frac{dy}{dx} = \frac{-96 + 8 \ln 2}{130} = \bbox[red, 2pt]{\frac{-48 + 4 \ln 2}{65} }$$

解答:$$u=\sqrt{4+e^x} \Rightarrow u^2=4+e^x \Rightarrow 2udu=e^x\,dx \Rightarrow dx ={2u\over u^2-4}du \\ \Rightarrow \int{1\over \sqrt{4+e^x}} \,dx = \int \frac{1}{u} \cdot\frac{2u}{u^2 - 4} du = \int \frac{2}{u^2 - 4} du =\int \left( \frac{1/2}{u-2} - \frac{1/2}{u+2} \right) du\\= \frac{1}{2} \ln\vert{}u-2\vert{} - \frac{1}{2} \ln\vert{}u+2\vert{} + C = \bbox[red, 2pt]{\frac{1}{2} \ln \left\vert{} \frac{u-2}{u+2} \right\vert{} + C }$$

解答:$$\cases{x= r\cos \theta \\ y=r\sin \theta} \Rightarrow  \int_0^1 \int_0^{\sqrt{1-x^2}} e^{x^2+y^2} dy dx = \int_0^{\frac{\pi}{2}} \int_0^1 e^{r^2} r dr d\theta = \int_0^{\frac{\pi}{2}} \left. \left[ {1\over 2}e^{r^2} \right] \right|_0^1\,d\theta\\= \int_0^{\frac{\pi}{2}} \frac{1}{2}(e - 1) d\theta = \frac{1}{2}(e - 1)  \cdot {\pi\over 2} = \bbox[red, 2pt]{\frac{\pi}{4}(e - 1)}$$

II. Linear Algebra

解答:$$\underline{w} \in U = \text{span}\{\underline{x}, \underline{y}\} \Rightarrow  \underline{w} = c_1\underline{x} + c_2\underline{y} ，\text{ where } c_1 \text{ and } c_2 \text{ are scalars.} \\   \underline{v} \in A=  \underline{x}\underline{y}^T + \underline{y}\underline{x}^T \Rightarrow  (\underline{x}\underline{y}^T + \underline{y}\underline{x}^T)\underline{v} = \underline{0} \Rightarrow  \underline{x}(\underline{y}^T\underline{v}) + \underline{y}(\underline{x}^T\underline{v}) = \underline{0} \\ \text{Since }\underline{x} \text{ and }\underline{y} \text{ are linearly indepent, }\underline{y}^T\underline{v}= \underline{x}^T\underline{v}=0 \Rightarrow (\underline{y}^T\underline{v})^T = (\underline{x}^T\underline{v})^T=0 \Rightarrow \underline{v}^T\underline{y} = \underline{v}^T \underline{x} =0\\ \Rightarrow \underline{v}^T\underline{w} = \underline{v}^T(c_1\underline{x} + c_2\underline{y}) = c_1(\underline{v}^T\underline{x}) + c_2(\underline{v}^T\underline{y}) =c_1\cdot 0+c_2\cdot 0 =0 \Rightarrow   \underline{v}^T\underline{w} = 0 \\ \Rightarrow  \text{ every }\underline{v} \in N(A)\text{ is orthogonal to every }\underline{w} \in U \quad \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) }P \text{ is the projection matrix } \Rightarrow P=  A(A^TA)^{-1}A^T  = (QR)((QR)^T(QR))^{-1}(QR)^T \\  = (QR)(R^TQ^TQR)^{-1}(R^TQ^T)= (QR)(R^TIR)^{-1}(R^TQ^T)   = QR(R^TR)^{-1}R^TQ^T \\  = Q R R^{-1} (R^T)^{-1} R^T Q^T =Q(I) (I)Q^T=QQ^T \Rightarrow P=QQ^T \quad \bbox[red, 2pt]{QED.} \\ \textbf{(b) }  \underline{x} = P\underline{b}= (QQ^T)b =Q(Q^Tb)  = [\underline{q_1} \quad \underline{q_2} \quad \cdots \quad \underline{q_n}] \begin{bmatrix} \underline{q_1}^T \underline{b} \\ \underline{q_2}^T \underline{b} \\ \vdots \\ \underline{q_n}^T \underline{b} \end{bmatrix}  \\ = (\underline{q_1}^T \underline{b}) \underline{q_1} + (\underline{q_2}^T \underline{b}) \underline{q_2} + \cdots + (\underline{q_n}^T \underline{b}) \underline{q_n} \quad \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) } L(\underline{e}_1) = (\underline{e}_1 \cdot \underline{u})\underline{u} = \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} \right) \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} = \cos\theta \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta \\ \sin\theta\cos\theta \end{bmatrix} \\ L(\underline{e}_2) = (\underline{e}_2 \cdot \underline{u})\underline{u} = \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} \right) \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} = \sin\theta \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} = \begin{bmatrix} \sin\theta\cos\theta \\ \sin^2\theta \end{bmatrix} \\ \Rightarrow  A = \bbox[red, 2pt]{\begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} } \\ \textbf{(b) } A_1 = \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} , A_2 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , A_3 = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\\ \Rightarrow A= A_3A_2 A_1 =\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} =  \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ 0 & 0 \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} }$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解