國立中央大學113學年度碩士班考試入學試題
所別:統計研究所碩士班
科目:基礎數學
解答:$$\textbf{(a) }u=x-1 \Rightarrow I=\int_{-\infty}^\infty (u+1)^2 e^{-u^2/2}\,du =\int_{-\infty}^\infty u^2 e^{-u^2/2}\,du +2\int_{-\infty}^\infty u e^{-u^2/2}\,du +\int_{-\infty}^\infty e^{-u^2/2}\,du\\ f(u)=ue^{-u^2/2} \text{ is an odd function, then }\int_{-\infty}^\infty u e^{-u^2 /2}\, du=0\\ I_2= \int_{-\infty}^\infty e^{-u^2/2}\,du \Rightarrow I_2^2 =\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)/2} \,dxdy = \int_0^{2\pi} \int_0^\infty re^{-r^2/2 }\,drd \theta =\int_0^{2\pi} \left. \left[ - e^{-r^2/2}\right] \right|_0^\infty \,d\theta\\\qquad =\int_0^{2\pi} 1\,d\theta= 2\pi \Rightarrow I_2= \sqrt{2\pi} \\\cases{v=u\\ dw = ue^{-u^2/2}\,du} \Rightarrow \cases{dv =du\\ w=-e^{-u^2/2}} \Rightarrow I_1=\int_{-\infty}^\infty u^2e^{-u^2/2}\, du = \left. \left[ -ue^{-u^2/2} \right] \right|_{-\infty}^\infty +\int_{-\infty}^\infty e^{-u^2/2}\,du \\ =0+I_2=\sqrt{2\pi} \Rightarrow I=I_1+ I_2 = \sqrt{2\pi}+ \sqrt{2\pi} =\bbox[red, 2pt]{2\sqrt{2\pi}} \\\textbf{(b) } \int_0^1 x^5(1-x)^6\,dx = \int_0^1 x^5(1-6x+ 15x^2-20x^3+ 15x^4-6x^5+x^6)\,dx \\\quad =\int_0^1 (x^5-6x^6+ 15x^7-20x^8+ 15x^9-6x^{10}+x^{11})\,dx \\\quad = \left. \left[{1\over 6}x^6-{6\over 7}x^7 +{15\over 8}x^8-{20\over 9}x^9 +{15\over 10}x^{10}-{6\over 11}x^{11}+ {1\over 12}x^{12}\right] \right|_0^1 \\={1\over 6} -{6\over 7} +{15\over 8} -{20\over 9} +{15\over 10} -{6\over 11} + {1\over 12} =\bbox[red, 2pt]{1\over 5544}$$
解答:$$\textbf{(a) } {1\over n^2+1} \lt {1\over n^2}, \forall n \in \mathbb N \text{ and } \sum_{n=1}^\infty {1\over n^2} \text{ is convergent} \Rightarrow \sum_{n=1}^\infty {1\over n^2+1} \text{ is }\bbox[red, 2pt] {\text{convergent}}\\ \textbf{(b) } a_n={3^n\over n\cdot 5^n} \Rightarrow \lim_{n\to \infty} {a_{n+1} \over a_n} = \lim_{n\to \infty} \left( {3^{n+1} \over (n+1)5^{n+1}} \cdot {n5^n\over 3^n} \right) =\lim_{n\to \infty} \left({3\over 5}\cdot {n\over n+1} \right) ={3\over 5}\lt 1 \\\qquad \Rightarrow \text{It is }\bbox[red, 2pt]{\text{ convergent}}$$
解答:$$z=x+y \Rightarrow \cases{f(x,y)=x^2+y^2+ (x+y)^2\\ g(x,y) ={x^2\over 4}+{y^2\over 5}+{(x+y)^2\over 25}-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ g=0}\\ \Rightarrow \cases{2x+2(x+y) =\lambda({x\over 2}+ {2\over 25}(x+y)) \cdots(1)\\ 2y+2(x+y) =\lambda({2\over 5}y+{2\over 25}(x+y)) \cdots(2)\\ {x^2\over 4}+{y^2\over 5}+{(x+y)^2\over 25}=1} \Rightarrow {(1) \over (2)} ={2x+y\over x+2y} ={{29\over 50}x+{2\over 25}y \over {2\over 25}x+ {12\over 25}y} \\ \Rightarrow 21x^2+ 10xy-6y^2= 0 \Rightarrow (3x-2y)(7x+8y)=0 \Rightarrow \cases{y=3x/2\\ y=-7x/8} \Rightarrow \cases{z=5x/2\\ z=x/8} \\ \Rightarrow \cases{g(x,3x/2) ={19\over 20}x^2-1=0 \Rightarrow x^2=20/19 \Rightarrow x^2+y^z+z^2= 10\\ g(x,-7x/8)= {323\over 800}x^2-1=0 \Rightarrow x^2=800/323 \Rightarrow x^2+y^2+z^2=75/17}\\ \Rightarrow \bbox[red, 2pt]{\cases{\text{maximum: }10 \\ \text{minimum: }75/17}}$$
解答:$$A=\begin{bmatrix}4 & 6 & 0 \\-3 & -5 & 0 \\-3 & -6 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I) = -(\lambda-1)^2(\lambda+2)=0 \Rightarrow \lambda=1,-2\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 3 & 6 & 0 \\-3 & -6 & 0 \\-3 & -6 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow x_1+2x_2=0 \\\qquad \Rightarrow v=x_2 \begin{pmatrix}-2\\ 1\\ 0 \end{pmatrix}+x_3 \begin{pmatrix}0\\0\\ 1 \end{pmatrix}, \text{ choose }v_1=\begin{pmatrix}-2\\ 1\\ 0 \end{pmatrix}, v_2= \begin{pmatrix}0\\0\\ 1 \end{pmatrix}\\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} 6 & 6 & 0 \\-3 & -3 & 0 \\-3 & -6 & 3 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+ x_3=0 \\x_2=x_3} \\\qquad \Rightarrow v=x_3 \begin{pmatrix}-1\\ 1\\ 1 \end{pmatrix}, \text{ choose }v_3= \begin{pmatrix}-1\\ 1\\ 1 \end{pmatrix} \\ P=[v_1 v_2 v_3] ,D=\begin{bmatrix} \lambda_1 & 0 & 0 \\0 & \lambda_1 & 0 \\0 & 0 & \lambda_2 \end{bmatrix} \Rightarrow A=PDP^{-1} \Rightarrow A^{10}=PD^{10}P^{-1} \\ \Rightarrow A^{10} = \begin{pmatrix} -2 & -1 & 0 \\1 & 1 & 0 \\0 & 1 & 1 \end{pmatrix} \begin{pmatrix}1^{10} & 0 & 0 \\0 & (-2)^{10} & 0 \\0 & 0 & 1^{10} \end{pmatrix} \begin{pmatrix} -1 & -1 & 0 \\1 & 2 & 0 \\-1 & -2 & 1\end{pmatrix} = \bbox[red, 2pt]{\begin{bmatrix} -1022 & -2046 & 0 \\1023 & 2047 & 0 \\1023 & 2046 & 1 \end{bmatrix}}$$
解答:$$\textbf{(a) } \begin{pmatrix} A_{11} &C \\ 0& A_{22}\end{pmatrix} \begin{pmatrix} A_{11}^{-1} &-A_{11}^{-1} C A_{22}^{-1} \\ 0& A_{22}^{-1}\end{pmatrix} = \begin{pmatrix} A_{11}A_{11}^{-1} &-A_{11} A_{11}^{-1} C A_{22}^{-1}+CA_{22}^{-1} \\ 0& A_{22} A_{22}^{-1} \end{pmatrix} =\begin{pmatrix} I &0\\ 0& I\end{pmatrix} \\\qquad \Rightarrow A^{-1} =\begin{pmatrix} A_{11}^{-1} &-A_{11}^{-1} C A_{22}^{-1} \\ 0& A_{22}^{-1}\end{pmatrix} \\\qquad \begin{pmatrix} B_{11} &0 \\ C& B_{22}\end{pmatrix} \begin{pmatrix} B_{11}^{-1} & 0 \\ -B_{11}^{-1} C B_{22}^{-1}& B_{22}^{-1}\end{pmatrix} \\\begin{pmatrix} B_{11} & 0\\ C & B_{22} \end{pmatrix}\begin{pmatrix} B_{11}^{-1} & 0\\ -B_{22}^{-1}CB_{11}^{-1} & B_{22}^{-1}\end{pmatrix} =\begin{pmatrix} B_{11}B_{11}^{-1} & 0\\ CB_{11}^{-1}-B_{22}B_{22}^{-1}CB_{11}^{-1} & B_{22 }B_{22}^{-1} \end{pmatrix} =\begin{pmatrix} I & 0\\ 0 & I\end{pmatrix} \\ \qquad \Rightarrow B^{-1}= \begin{pmatrix} B_{11}^{-1} & 0\\ -B_{22}^{-1}CB_{11}^{-1} & B_{22}^{-1}\end{pmatrix} \quad \bbox[red, 2pt]{QED} \\\bbox[cyan,2pt]{題目有誤},B^{-1}應該是 \begin{pmatrix} B_{11}^{-1} & 0\\ -B_{22}^{-1}CB_{11}^{-1} & B_{22}^{-1} \end{pmatrix} \\\textbf{(b) } \cases{B_{11}=\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix} \\C= \begin{pmatrix} 3 & 7 \\ 2 & 5 \end{pmatrix} \\B_{22} =\begin{pmatrix} 2 & 3\\ 1 & 2 \end{pmatrix}} \Rightarrow \cases{B_{11}^{-1} =\begin{pmatrix} 2 & -1\\ -1 & 1 \end{pmatrix} \\B_{22}^{-1} =\begin{pmatrix} 2 & -3\\ -1 & 2 \end{pmatrix}} \Rightarrow \begin{pmatrix}1& 1& 0 & 0\\ 1& 2 & 0 & 0\\ 3& 7 & 2& 3\\ 2 & 5& 1& 2 \end{pmatrix} =\begin{pmatrix} B_{11} & 0\\ C & B_{22} \end{pmatrix} \\ \Rightarrow \begin{pmatrix}1& 1& 0 & 0\\ 1& 2 & 0 & 0\\ 3& 7 & 2& 3\\ 2 & 5& 1& 2 \end{pmatrix} ^{-1}=\begin{pmatrix} B_{11}^{-1} & 0\\ -B_{22}^{-1}CB_{11}^{-1} & B_{22}^{-1} \end{pmatrix} =\bbox[red, 2pt] {\begin{pmatrix}2& -1& 0 & 0\\ -1& 1& 0 & 0\\ -1& 1& 2& -3\\ 1& -2 & -1& 2 \end{pmatrix}}$$
解答:$$\textbf{(a) } Ax=\lambda x \Rightarrow A^2x=A(Ax) =A(\lambda x)= \lambda A(x) =\lambda^2 x \Rightarrow \lambda^2 \text{ is the eigenvalue of }A^2\\ \quad A^3x=A(A^2x)=A(\lambda^2x) =\lambda^2 Ax= \lambda^3x \Rightarrow \lambda^3 \text{ is the eigenvalue of }A^3\\ \Rightarrow A^kx=A(A(\cdots (Ax))) =\lambda^k x \Rightarrow \lambda^k \text{ is the eigenvalue of }A^k \quad \bbox[red, 2pt]{QED} \\\textbf{(b) }AX= \lambda X \Rightarrow A^kX=\lambda^k X \Rightarrow A \text{ and }A^k \text{ have the same eigenvectors}\\\qquad \Rightarrow \bbox[red, 2pt]X \text{ is the eigenvector of }A^k \text{ associated with }\lambda^k$$
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解題僅供參考，碩士班歷年試題及詳解
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