114年政大統計碩士班-基礎數學詳解

國立政治大學114學年度碩士班招生考試

考試科目:基礎數學
系所別:統計學系

Part I: Calculus

解答:$$x-c= (\sqrt[3]x)^3 - (\sqrt[3]c)^3 = \left( \sqrt[3]x  - \sqrt[3]c  \right) \left( (\sqrt[3]x)^2 + \sqrt[3]{xc}+ (\sqrt[3]c)^2 \right)\\ \Rightarrow |\sqrt[3]x  - \sqrt[3]c|= {|x-c|\over |\sqrt[3]{x^2} +\sqrt[3]{xc}+ \sqrt[3]{c^2}|} ={|x-c| \over \sqrt[3]{x^2} +\sqrt[3]{xc}+ \sqrt[3]{c^2}}\\ |x| \gt {|c|\over 2} \Rightarrow \sqrt[3]{x^2} +\sqrt[3]{xc}+ \sqrt[3]{c^2} \gt \sqrt[3]{ \left( {|c|\over 2} \right)^2} +0+\sqrt[3]{c^2} =\sqrt[3]{c^2} \left( 1+{1\over \sqrt[3]4} \right) =M \\ \Rightarrow |\sqrt[3]x  - \sqrt[3]c| \lt {|x-c|\over M} \lt \epsilon \Rightarrow |x-c| \lt M \epsilon\\ \forall \epsilon \gt 0, \text{ choose }\delta= \min \left\{{|c|\over 2}, \sqrt[3]{c^2} \left( 1+{1\over \sqrt[3]4} \right)\right \} , \text{ we have }\\ \text{if }|x-c| \lt \delta  \Rightarrow |f(x)-f(c)| = |\sqrt[3]x  - \sqrt[3]c|  \lt \epsilon \Rightarrow f(x)\text{ is continuous at }c\ne 0 \quad \bbox[red, 2pt]{QED.}$$
 

解答:$$\textbf{(a) }L = \lim_{n\to \infty} {1\over n^3} \left( \sum_{i=1}^{3n} \sqrt i \right)^2  = \lim_{n\to \infty} {1\over n^3} \left( \sum_{i=1}^{3n} \sqrt n\cdot \sqrt {i\over n} \right)^2 = \lim_{n\to \infty} {1\over n^2} \left( \sum_{i=1}^{3n}  \sqrt {i\over n} \right)^2\\ \quad = \lim_{n\to \infty}  \left( {1\over n}\sum_{i=1}^{3n}  \sqrt {i\over n} \right)^2 = \lim_{n\to \infty} \left( S_n \right)^2, \quad S_n= {1\over n}\sum_{i=1}^{3n}  \sqrt {i\over n} \\ \quad \Rightarrow \lim_{n\to \infty} S_n =\lim_{n\to \infty} {3\over 3n} \sum_{i=1}^{3n} \sqrt{3\cdot {i\over 3n}} = 3\int_0^1 \sqrt{3x} \,dx =2\sqrt 3 \\ \quad \Rightarrow L= \lim_{n\to \infty} \left( S_n \right)^2 = \left( \lim_{n\to \infty} S_n \right)^2 =(2\sqrt 3)^2= \bbox[red, 2pt]{12} \\\textbf{(b) }\cases{\log x= \ln x/\ln 10\\ x^x =e^{x\ln x}} \Rightarrow \lim_{x\to 1}{\log x-x^x+1 \over \log x-x+1} =\lim_{x\to 1}{(\log x-x^x+1)' \over (\log x-x+1)'}\\ \quad=\lim_{x\to 1}{{1\over x\ln 10}-(1+\ln x)x^x  \over {1\over x \ln 10}- 1} ={{1\over \ln 10}-1\over {1\over \ln 10}-1} = \bbox[red, 2pt]1$$

解答:$$a_{n+1} ={1+\sqrt{1+a_n^2} \over a_n} \Rightarrow \tan (b_{n+1}) ={1+ \sqrt{1+ \tan^2 (b_n)} \over \tan(b_n)} ={1+ \sec(b_n) \over \tan(b_n)} ={1+\displaystyle {1\over \cos(b_n)} \over \displaystyle {\sin(b_n) \over \cos(b_n)}} \\={1+\cos(b_n) \over \sin(b_n)} ={2\cos^2(b_n/2) \over 2\sin(b_n/2)  \cos(b_n/2)} = \cot \left( {b_n\over 2} \right) = \tan \left( {\pi\over 2}-{b_n\over 2} \right) \\ \Rightarrow b_{n+1} = {\pi\over 2}-{b_n\over 2} \Rightarrow b_{n+1}-{\pi\over 3} ={\pi\over 6}-{b_n\over 2} =-{1\over 2} \left( b_n-{\pi\over 3} \right) \\ \Rightarrow b_n-{\pi\over 3} =-{1\over 2}(b_{n-1}-{\pi\over 3}) = \left( -{1\over 2} \right)^2 \left( b_{n-2}-{\pi\over 3} \right) = \cdots =\left( -{1\over 2} \right)^n \left( b_{0}-{\pi\over 3} \right) \\ \Rightarrow \bbox[red, 2pt]{b_n=\left( -{1\over 2} \right)^n \left( b_{0}-{\pi\over 3} \right)+{\pi\over 3}}$$

解答:$$I= \int_{-\infty}^\infty e^{-z^2/2} \,dz \Rightarrow I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)/2} \,dxdy \\ \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow I^2 =\int_0^{2\pi} \int_0^\infty re^{-r^2}\,drd\theta = \int_0^{2\pi} {1\over 2}\,d\theta =\pi \Rightarrow I=\sqrt \pi \\\Rightarrow \int_{0}^\infty e^{-z^2/2} \,dz ={I\over 2} = \bbox[red, 2pt]{\sqrt \pi\over 2}$$

Part II: Calculus (應該是Linear Algebra)

解答:$$(a) \bbox[red, 2pt]{false}:\cases{\text{system }A:\cases{x+y=0\\ x-y=0} \\ \text{system }B:1\cdot(x+y)+1\cdot(x-y) =2x=0} \Rightarrow \cases{\text{solution of }A:\{(0,0)\} \\ \text{solution of }B:\{(0, y) \mid y\in \mathbb R\}} \\ \qquad \Rightarrow Sol(A) \ne Sol(B) \\(b)\bbox[red, 2pt]{true}:A \text{ and }B \text{ are invertible }\Rightarrow \cases{AB (B^{-1}A^{-1}) =I \\(B^{-1}A^{-1})AB=I} \Rightarrow AB \text{ is invertible} \\(c)\bbox[red, 2pt]{false}: \text{A basis of vector space  is a linearly independent set. But a set contains zero vector, }\\\qquad \text{then the set is automatically linearly dependent.} \\(d)\bbox[red, 2pt]{true}: \text{If a matrix is invertible and in row-reduced echelon form (RREF), then the matrix }\\\qquad \text{must be the identity matrix }I_n.$$

解答:$$"\Longrightarrow": A=0 \Rightarrow A^TA=0 \Rightarrow tr(A^TA)=0\\ "\Longleftarrow": A=[a_{i,j}]_{n\times n} \Rightarrow tr(A^TA) =(a_{1,1}^2+ a_{1,2}^2+\cdots+a_{1,n}^2) +(a_{2,1}^2+ a_{2,2}^2+ \cdots+ a_{2,n}^2) \\\qquad +\cdots +(a_{n,1}^2+a_{n,2}^2+ \cdots+ a_{n,2}^2) =0 \Rightarrow a_{i,j}=0, 1\le i,j \le n \Rightarrow A=0\\ \bbox[red, 2pt]{QED}$$

解答:$$\text{Null}(A^T) =\text{Range}(A-I) \Rightarrow A^T[(A-I)x] =0, \forall x \in \mathbb R^n \\ \Rightarrow (A^TA-A^T)x=0 \Rightarrow A^TA=A^T \Rightarrow (A^TA)^T=(A^T)^T \Rightarrow A^TA=A \\ \Rightarrow \cases{A^TA=A^T\\ A^TA=A} \Rightarrow A^T=A \quad \bbox[red, 2pt]{QED}$$

解答:$$\textbf{(a) }x\in \text{Null}(A^T) \Rightarrow A^T(x)=0 \Rightarrow A(x)=0 \Rightarrow (A-I)(-x) =-A(x)+x=x \\\qquad \Rightarrow x \in \text{Range}(A-I) \quad \bbox[red, 2pt]{QED.} \\\textbf{(b) }x\in \text{Range}(A-I) \Rightarrow \exists y \text{ such that }(A-I)(y)=x \Rightarrow A(y)-y=x \\ \quad \Rightarrow A^TA(y)-A^T(y)=A^T(x) \Rightarrow A^2(y)-A(y)= A(y)-A(y)=0=A^T(x) \\\quad \Rightarrow x\in  \text{Null}(A^T) \quad \bbox[red, 2pt]{QED.}$$

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解題僅供參考，碩士班歷年試題及詳解