112年北大統計碩士班-基礎數學

國立臺北大學112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%)CALCULUS

解答：$$L=\left({1\over x} \right)^x \Rightarrow \ln L=x\ln {1\over x}=\cfrac{\ln {1\over x}}{1\over x} \\ \lim_{x \rightarrow 0^+}\cfrac{\ln {1\over x}}{1\over x}  = \lim_{x \rightarrow 0^+}\cfrac{(\ln {1\over x})'}{({1\over x})'}  = \lim_{x \rightarrow 0^+}\cfrac{-{1\over x}}{-1\over x^2}  =0 \Rightarrow \lim_{x \rightarrow 0^+} L=e^0=\bbox[red, 2pt]1  $$

解答：$$f(x)=\int_x^{x^2} \sqrt{1+t^3}\,dt  \Rightarrow f'(x)= \sqrt{1+x^6}\cdot (2x)-\sqrt{1+x^3}\cdot 1 \\= \bbox[red, 2pt] {2x\sqrt{1+x^6}-\sqrt{1+x^3}}$$

解答：$$\textbf{(a)}\;x=\sin u \Rightarrow \cases{\sqrt{1-x^2}=\cos x \\dx= \cos u\,du} \Rightarrow \int_{-1}^0 {1+x\over \sqrt{1-x^2}} \,dx =\int_{-\pi/2}^0 {1+\sin u\over \cos u}\cdot \cos u\,du \\ \quad =\int_{-\pi/2}^0 (1+\sin u)\,du =\left. \left[ u-\cos u \right] \right|_{-\pi/2}^0 =\bbox[red, 2pt]{{\pi\over 2}-1}\\ \textbf{(b)}\;f(x)={1+x\over \sqrt{1-x^2}} \Rightarrow f'(x)={1\over \sqrt{1-x^2}} +{x(1+x) \over (1-x^2)^{3/2}} \Rightarrow \cases{f(0)=1\\ f'(0)=1} \\\quad \Rightarrow \int_{-1}^0 (f(0)+f'(0)x)\, dx =\int_{-1}^0 (1+x)\,dx =\left. \left[ x+{1\over 2}x^2 \right] \right|_{-1}^0 =\bbox[red, 2pt]{1\over 2}$$

解答：$$L(\sigma)=\prod_{i=1}^n{1\over \sqrt{2\pi}\sigma} e^{-{1\over 2}\left({x_i-\mu\over \sigma} \right)^2} =\left(2\pi \sigma^2 \right)^{-n/2} e^{-{1\over 2}\sum\left({x_i-\mu\over \sigma} \right)^2} \\ \Rightarrow \ln L(\sigma) =-{n\over 2}\ln(2 \pi)-{n\over 2}\ln \sigma^2-{1\over 2\sigma^2} \sum_{i=1}^n\left(x_i-\mu  \right)^2 \\ \Rightarrow \frac{\partial  }{\partial \sigma^2} \ln L(\sigma) =-{n\over 2\sigma^2} +{1\over 2\sigma^4} \sum_{i=1}^n\left(x_i-\mu  \right)^2 ={1\over 2\sigma^2} \left[ {1\over \sigma^2} \sum_{i=1}^n\left(x_i-\mu  \right)^2-n\right] =0 \\ \Rightarrow \sigma^2=\cfrac{\sum_{i=1}^n\left(x_i-\mu  \right)^2}{n} \Rightarrow \bbox[red, 2pt]{\sigma= \sqrt{ \cfrac{ \sum_{i=1}^n \left(x_i-\mu  \right)^2}{n}}}$$

二、（50％）

解答：$$\textbf{(b)}\; \cases{C^{-1}B^{-1}BC = C^{-1}C=I \\ BCC^{-1}B^{-1}= BB^{-1}=I}\Rightarrow C^{-1}B^{-1}=(BC)^{-1},\bbox[red, 2pt]{Q.E.D.}$$

解答：$$\textbf{(a)}\; \det(A-\lambda I)= \left|\begin{matrix}0-\lambda & 2 & -1 \\2 & 3-\lambda & -2 \\-1 & -2 & 0-\lambda \end{matrix} \right| =-\lambda^3 + 3\lambda^2+9 \lambda+5 \\\Rightarrow \text{the characteristic polynomial of }A: \bbox[red, 2pt]{-\lambda^3 + 3\lambda^2+9 \lambda+5} \\ \det(A-\lambda I)=0 \Rightarrow -(\lambda+1)^2 (\lambda-5)=0 \Rightarrow \text{eighenvalues: }\lambda =\bbox[red, 2pt]{-1,5}\\ \textbf{(b)}\;\lambda_1= -1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & 2 & -1 \\2 & 4 & -2 \\-1 & -2 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1+2x_2=x_3\\ \qquad \Rightarrow v=x_2 \begin{pmatrix} -2\\ 1\\ 0\end{pmatrix}+ x_3\begin{pmatrix} 1\\ 0 \\ 1\end{pmatrix} \Rightarrow E_{-1}(A)= \bbox[red, 2pt]{\left\{ s \begin{pmatrix} -2\\ 1\\ 0\end{pmatrix}+ t \begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}\mid s,t\in \mathbb R\right\}} \\ \lambda_2=5 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-5 & 2 & -1 \\2 & -2 & -2 \\-1 & -2 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1+x_3=0\\ x_2+2x_3=0} \\\qquad \Rightarrow v=x_3\begin{pmatrix}-1\\ -2\\ 1 \end{pmatrix} \Rightarrow E_{5}(A)= \bbox[red, 2pt]{\left\{\begin{pmatrix}-1\\ -2\\ 1 \end{pmatrix} \mid k\in \mathbb R \right\}} \\ \textbf{(c)}\; A=\left[ \begin{matrix} 0 & 2 & -1 \\2 & 3 & -2 \\-1 & -2 & 0 \end{matrix} \right] =\left[ \begin{matrix} -2 & 1 & -1 \\1 & 0 & -2 \\
0 & 1 & 1\end{matrix} \right] \left[ \begin{matrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{matrix}\right] \left[ \begin{matrix} -2 & 1 & -1 \\1 & 0 & -2 \\ 0 & 1 & 1\end{matrix} \right]^{-1} \\ \left[ \begin{matrix} -2 & 1 & 1 \\1 & 0 & -2 \\ 0 & 1 & 1 \end{matrix} \right] \xrightarrow{\text{By Gram-Schmidt process} } \left[ \begin{matrix} \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{30}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} & \frac{-2}{\sqrt{6}} \\0 & \frac{5}{\sqrt{30}} & \frac{1}{\sqrt{6}} \end{matrix} \right] \\ \Rightarrow A=\left[ \begin{matrix} \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{30}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} & \frac{-2}{\sqrt{6}} \\0 & \frac{5}{\sqrt{30}} & \frac{1}{\sqrt{6}} \end{matrix} \right] \left[ \begin{matrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{matrix}\right] \left[ \begin{matrix} \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{30}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} & \frac{-2}{\sqrt{6}} \\0 & \frac{5}{\sqrt{30}} & \frac{1}{\sqrt{6}} \end{matrix} \right]^T \\ \Rightarrow \bbox[red, 2pt]{P=\left[ \begin{matrix} \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{30}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} & \frac{-2}{\sqrt{6}} \\0 & \frac{5}{\sqrt{30}} & \frac{1}{\sqrt{6}} \end{matrix} \right]}, \bbox[red, 2pt]{D= \left[ \begin{matrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{matrix}\right]}$$

解答：$$B \text{ is a basis of }V \Rightarrow \vec w=\sum_{i=1}^n c_i \vec v_i, \forall \vec w \in V \Rightarrow T(\vec w)=T\left( \sum_{i=1}^n c_i \vec v_i\right) =\sum_{i=1}^n c_i T(\vec v_i) \\\qquad \Rightarrow T(\vec w) \in Span \left\{T(\vec v_i)\right\}_{i=1}^n \Rightarrow Range(T)\subseteq Span \left\{T(\vec v_i)\right\}_{i=1}^n\\ \vec w' \in Span \left\{T(\vec v_i)\right\}_{i=1}^n \Rightarrow \vec w'=\sum_{i=1}^n c_i T(\vec v_i) =T\left( \sum_{i=1}^n c_i \vec v_i\right) \Rightarrow \vec w' \in Range(T) \\\qquad \Rightarrow Span \left\{T(\vec v_i)\right\}_{i=1}^n \subseteq Range(T) \\ \Rightarrow T(B) \text{ spans the range of }T, \bbox[red, 2pt]{Q.E.D.}$$

解答：$$\textbf{(a)}\;\cases{x^TAx = \langle x,Ax \rangle =\langle Ax,x \rangle \\ x^TAx= x^T(-A^T)x =-(x^TA^T)x=-(Ax)^Tx=-\langle Ax,x\rangle} \\ \quad \Rightarrow \langle Ax,x \rangle=-\langle Ax,x \rangle \Rightarrow \langle Ax,x \rangle=0 \Rightarrow x^TAx=0,\bbox[red, 2pt]{Q.E.D.} \\\textbf{(b)}\; \text{If }I+A \text{ is not invertible, we can find nonzero vector }x, \text{ such that } (I+A)x=0 \\\quad \Rightarrow x+Ax=0 \Rightarrow Ax=-x \Rightarrow x^TAx =\langle Ax,x\rangle =\langle -x, x\rangle=0\quad (by \textbf{(a)})\\ \quad \Rightarrow x=0, \text{ a contradiction.} \Rightarrow I+A\text{ is invertible}, \bbox[red, 2pt]{Q.E.D.}$$

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