114年台北大學統計碩士班-基礎數學詳解

 國立臺北大學114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

一、(50%，10% each) Calculus (清楚完整作答，始得完全分數)

解答:$$L=\left( {x+a\over x-a}\right)^{5x} \Rightarrow \ln L={ \ln({x+a\over x-a})\over 1/5x} \Rightarrow \lim_{x\to \infty} \ln L =\lim_{x\to \infty}{ (\ln({x+a\over x-a}))'\over (1/5x)'}  =\lim_{x\to \infty} { -2a/(x^2-a^2)\over -1/5x^2} \\= \lim_{x\to \infty}{10ax^2 \over x^2-a^2} =10a \Rightarrow \lim_{x\to \infty} L= \bbox[red, 2pt]{e^{10a}}$$

解答:$$f(x)= \cos{[2\sin{x\over 2}]} \Rightarrow f'(x)=-\cos {x\over 2} \cdot \sin{[2\sin{x\over 2}]} =0 \Rightarrow \cases{\cos(x/2)=0 \\ \sin(2\sin(x/2))=0} \\ \Rightarrow \cases{x= \pi\\ x=0,2\pi  } \Rightarrow \cases{f(\pi)= \cos 2 \\ f(0) =f(2\pi)=1} \Rightarrow \bbox[red, 2pt]{ \cases{\text{local minimum: }\cos 2\\ \text{absolute maximum: }1 }}$$

解答:$$u=3^x \Rightarrow du =\ln 3\cdot 3^x\,dx =\ln 3\cdot u\,dx \Rightarrow dx={1\over u\ln 3}\,dx \Rightarrow \int {1\over (1+3^x)^2}\,dx = \int {1\over \ln 3\cdot u(1+u)^2}\,du \\={1\over \ln3} \int \left( {1\over u}-{1\over 1+u} -{1\over (1+u)^2} \right)\,du ={1\over \ln 3}\left(\ln u-\ln(1+u)+{1\over 1+u} \right)+C \\= \bbox[red, 2pt]{{1\over \ln 3}\left(\ln 3^x-\ln(1+3^x)+{1\over 1+3^x} \right)+C}$$

解答:$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_R {1\over \sqrt{x^2+y^2}}\,dA =\int_0^{2\pi} \int_0^3 {1\over r}\cdot r\;\,drd\theta =\int_0^{2\pi} 3\,d\theta = \bbox[red, 2pt]{6\pi}$$

解答:$$f(x)=(x-90)^2+364 \Rightarrow f'(x)=2(x-90)=0 \Rightarrow x=90 \Rightarrow 速度為\bbox[red, 2pt]{90\text{ km/hour}} 最省油$$

二、(50%) Linear Algebra (請依序解答並須清楚完整，始得完全分數)

解答:$$\textbf{a. }\cases{v_1=(0,2,1,1)^T\\ v_2=(1,1,1,0)^T}, \text{ by Gram-Schmidt process, we have }\cases{e_1=(0, \sqrt 6/3, \sqrt 6/6, \sqrt 6/6)^T \\ e_2= (\sqrt 6/3, 0, \sqrt 6/6, -\sqrt 6/6)^T} \\ \Rightarrow Q=[e_1 e_2] = \begin{bmatrix}0,{\sqrt 6\over 3}\\{\sqrt 6\over 3},0\\ {\sqrt 6\over 6},{\sqrt 6\over 6} \\ {\sqrt 6\over 6},-{\sqrt 6\over 6}\end{bmatrix} \Rightarrow R= Q^TA = \begin{bmatrix} 0& {\sqrt 6\over 3}& {\sqrt 6\over 6} &{\sqrt 6\over 6} \\{\sqrt 6\over 3}& 0& {\sqrt 6\over 6} &-{\sqrt 6\over 6}\end{bmatrix} \begin{bmatrix}0& 1\\ 2& 1\\ 1& 1 \\1& 0\end{bmatrix} =\begin{bmatrix} \sqrt 6&{\sqrt 6\over 2}\\ 0& {\sqrt 6\over 2}\end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{Q= \begin{bmatrix}0,{\sqrt 6\over 3}\\{\sqrt 6\over 3},0\\ {\sqrt 6\over 6},{\sqrt 6\over 6} \\ {\sqrt 6\over 6},-{\sqrt 6\over 6}\end{bmatrix}, R=\begin{bmatrix} \sqrt 6&{\sqrt 6\over 2}\\ 0& {\sqrt 6\over 2}\end{bmatrix}} \\\textbf{b. }\bbox[red, 2pt]{C(A) =\left\{ a \begin{bmatrix} 0\\ 2\\ 1\\ 1\end{bmatrix} +b\begin{bmatrix} 1\\ 1\\ 1\\ 0 \end{bmatrix}  \middle|\, a,b\in \mathbb R\right\}  }\\ A^T\mathbf x =0 \Rightarrow \begin{bmatrix} 0& 2&1& 1\\ 1& 1& 1& 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\x_3\\x_4\end{bmatrix} =0 \Rightarrow \cases{2x_2 +x_3+ x_4= 0\\ x_1+ x_2+x_3=0} \Rightarrow \cases{x_1=-x_2-x_3\\ x_4=-2x_2-x_3} \\ \quad \Rightarrow \mathbf x= \begin{bmatrix} -x_2-x_3\\ x_2 \\x_3\\-2x_2-x_3\end{bmatrix} =x_2\begin{bmatrix} -1\\ 1 \\0 \\-2 \end{bmatrix} +x_3\begin{bmatrix} -1\\ 0 \\1\\-1\end{bmatrix} \Rightarrow \bbox[red, 2pt]{N(A^T) =\left\{s\begin{bmatrix} -1\\ 1 \\0\\-2\end{bmatrix}+t\begin{bmatrix} -1\\ 0 \\1\\-1 \end{bmatrix} \middle |\,s,t \in \mathbb R\right\}} \\\textbf{c. }\cases{\vec u\in C(A) \\ \vec v\in N(A^T)} \Rightarrow \cases{\vec u=(b,2a+b,a+b,a)^T, a,b\in \mathbb R\\ \vec v=(-s-t,s,t,-2s-t)^T, s,t \in \mathbb R} \Rightarrow \vec u\cdot \vec v=0 \Rightarrow C(A) \bot N(A^T),\bbox[red, 2pt]{QED.}$$

解答:$$\textbf{a. }L: y=mx \Rightarrow \text{ }\vec v=(1,m) \Rightarrow \text{proj}_L(\vec x) ={\vec x\cdot \vec v\over ||\vec v||^2} \vec v={(x_1, x_2) \cdot (1, m) \over 1+m^2} (1,m) ={x_1+mx_2\over 1+m^2} (1,m) \\= \bbox[red, 2pt]{ \left( {x_1+ mx_2\over 1+m^2} , {mx_1+m^2 x_2 \over 1+m^2}\right)} \\\textbf{b. } \vec v= \begin{bmatrix} 1\\m\end{bmatrix} \Rightarrow \vec v\vec v^T = \begin{bmatrix}1 \\m \end{bmatrix} [1\; m] = \begin{bmatrix}1& m\\ m& m^2 \end{bmatrix} \Rightarrow P={\vec v \vec v^T \over \vec v^T \vec v} = {\begin{bmatrix}1& m\\ m& m^2 \end{bmatrix}\over 1+ m^2} = \bbox[red, 2pt] {  \begin{bmatrix} \displaystyle {1\over 1+m^2} & \displaystyle {m\over 1+m^2} \\\displaystyle {m\over 1+m^2}& \displaystyle {m^2\over 1+m^2} \end{bmatrix}}$$

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