115年政治大學統計碩士班-基礎數學詳解

 國立政治大學115學年度碩士班暨碩士在職專班招生考試試題

考試科目：基礎數學
系所別：統計學系 精算科學組

Part I: Calculus

解答:$$\textbf{(a) }f(x)={1\over 1-x} = \sum_{n=0}^\infty x^n \Rightarrow f'(x)= {1\over (1-x)^2} =\sum_{n=1}^\infty nx^{n-1} =\sum_{n=0}^\infty (n+1)x^n \\ \Rightarrow f'(-{1\over 3}) ={9\over 16}= \sum_{n=0}^\infty (n+1)\cdot (-{1\over 3})^n \Rightarrow \sum_{n=0}^\infty (-1)^n {n+1\over 3^n} =\bbox[red, 2pt] {9\over 16} \\\textbf{(b)} \sum_{n=0}^\infty \left( {(-1)^n\over n+1}+{(-1)^n\over n+2} \right) = \left( 1+{1\over 2} \right)- \left( {1\over 2}+{1\over 3} \right)+ \left( {1\over 3}+{1\over 4} \right)- \left( {1\over 4}+{1\over 5} \right)+\cdots \\= 1+ \left( {1\over 2}-{1\over 2} \right) + \left( -{1\over 3}+{1\over 3} \right)+ \left( {1\over 4}-{1\over 4} \right) + \cdots =\bbox[red, 2pt]1$$

解答:$$\int_0^{2\pi} \cos^2(x)\,dx = \int_0^{2\pi} {1\over 2}(\cos(2x)+1)\,dx = {1\over 2} \left. \left[ {1\over 2}\sin (2x)+x \right] \right|_0^{2\pi} =\bbox[red, 2pt]{\pi}$$

解答:$$u=-\ln x \Rightarrow x=e^{-u} \Rightarrow dx=-e^{-u} \,du \Rightarrow \Gamma(n,k)= \int_0^1 (-\ln x)^{k-1} x^{n-1}\,dx \\= \int_{\infty}^0 u^{k-1}(e^{-u})^{n-1}(-e^{-u})\,du =\int_0^\infty u^{k-1}e^{-un}\,du \\ t=un \Rightarrow u={t\over n} \Rightarrow du={dt\over n} \Rightarrow \Gamma(n,k)= \int_0^\infty  \left( {t\over n} \right)^{k-1} e^{-t} \left( {1\over n} \right)\,dt ={1\over n^{k}} \int_0^\infty t^{k-1}e^{-t}\,dt = {\Gamma(k) \over n^k} \\ \Rightarrow \bbox[red, 2pt]{\Gamma(n,k) ={\Gamma(k) \over n^k}}$$

解答:$$\sin x=x-{x^3\over 6}+O(x^5) \Rightarrow x-\sin x={x^3\over 6}-O(x^5) \\ e^x=1+x+{1\over 2}x^2+ {1\over 6}x^3+O(x^4) \Rightarrow e^{\sin x}=1+\sin x+{1\over 2} \sin^2 x+{1\over 6}\sin^3 x+O(x^4) \\=1+ \left( x-{x^3\over 6}+O(x^5) \right)+{1\over 2} \left( x-{x^3\over 6}+O(x^5) \right)^2+{1\over 6} \left( x-{x^3\over 6}+O(x^5) \right)^3+O(x^4) \\=1+x+{1\over 2}x^2+O(x^4) \Rightarrow e^x-e^{\sin x} ={x^3\over 6}+O(x^4) \\ \Rightarrow \lim_{x\to 0}{e^x-e^{\sin x} \over x-\sin x} = \lim_{x\to 0}{{x^3\over 6}+O(x^4) \over {x^3\over 6}-O(x^5)} =\bbox[red, 2pt]1$$

Part II: Linear Algebra

解答:$$\textbf{(a) }\bbox[red, 2pt]{True}: R(A)=R(B) \Rightarrow \dim(R(A)) =\dim(R(B)) \Rightarrow \text{rank}(A)= \text{rank}(B) \\\textbf{(b) }\bbox[red, 2pt]{False}:\text{Let} A= \begin{bmatrix}1&0\\0& 1\\0& 0 \end{bmatrix}, \text{ then }Ax=0 \Rightarrow x= \begin{bmatrix}0\\0 \end{bmatrix}, \text{ but }A \text{ is not square and not invertible.} \\\textbf{(c) }\bbox[red, 2pt]{False}: S=\{(1,0),(0,1),(1,1)\} \text{ spans }\mathbb R^2, \text{ but }(1,1)=(1,0)+(0,1) \Rightarrow \text{linearly dependent} \\\textbf{(d) }\bbox[red, 2pt]{True}: \text{Let }\lambda \text{ be an eigenvalue of }A \text{ with eigenvector }v\ne 0 \Rightarrow Av=\lambda v \Rightarrow v^*Av=v^*\lambda v =\lambda v^*v \\\qquad \Rightarrow \lambda={v^*Av \in \mathbb R\over v^*v \in \mathbb R} \in  \mathbb R$$

解答:$$\textbf{(a) }\mathbf x\in \mathbb R^n \Rightarrow \mathbf x= A\mathbf x+( \mathbf x- A \mathbf x) \Rightarrow \cases{A \mathbf x\in \text{Range}(A) \\ A( \mathbf x-A \mathbf x) =A \mathbf x- A^2 \mathbf x =A \mathbf x-A \mathbf x=0 \Rightarrow  \mathbf x-A \mathbf x\in Null(A)} \\ \Rightarrow \text{ we have shown the sum spans the space} \\  \mathbf y\in \text{Range}(A)\cap  \text{Null}(A) \Rightarrow \cases{\mathbf y\in \text{Range}(A) \Rightarrow \exists \mathbf x\in \mathbb R^n \text{ such that }\ \mathbf y=A\mathbf x\\ \mathbf y\in \text{Null}(A) \Rightarrow A\mathbf y=0} \Rightarrow A(A\mathbf x)=0 \\\qquad \Rightarrow A^2 \mathbf x=0 \Rightarrow A\mathbf x= \mathbf y=\mathbf 0 \Rightarrow \text{Range}(A)\cap  \text{Null}(A) =\{\mathbf 0\} \\ \Rightarrow \mathbb R^n= \text{Range}(A)\oplus   \text{Null}(A)  \quad \bbox[red, 2pt]{QED.} \\ \textbf{(b) }\text{Let }\cases{\mathbf u\in \text{Range}(A) \\\mathbf v \in \text{Null}(A)}, \text{ then }\cases{\mathbf u= A\mathbf x, \text{ for some }\mathbf x\in \mathbb R^n\\ A\mathbf v=0} \Rightarrow \mathbf u\cdot \mathbf v =\mathbf u^T\cdot \mathbf v= (A\mathbf x)^T\mathbf v =\mathbf x^TA^T \mathbf v\\ \qquad  =\mathbf x^TA \mathbf v =\mathbf x^T 0=0 \Rightarrow \mathbb R^n= \text{Range}(A) \oplus^\bot \text{Null}(A) \quad \bbox[red, 2pt]{QED.}$$

解答:$$\det(A-\lambda I) =-(\lambda-2)(\lambda^2-4\lambda+2)=0 \Rightarrow \cases{\lambda_1=2\gt 0\\ \lambda_2=2+\sqrt 2 \gt 0\\ \lambda_3=2-\sqrt 2\gt 0} \Rightarrow \bbox[red, 2pt]{A \text{ is positive definite}}$$

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解題僅供參考，碩士班歷年試題及詳解