114年成大統計碩士班-數學詳解

 國立成功大學114學年度碩士班招生考試試題

系 所:統計學系
科目:數學

解答:$$\textbf{(1) }u= \sin x \Rightarrow du =\cos x \,dx \Rightarrow \int_0^{\pi/2} \sin^7 x\cos^5x \,dx =\int_0^{\pi/2} \sin^7 x(1-\sin^2 x)^2\cos x \,dx \\ \qquad = \int_0^1 u^7(1-u^2)^2\,du =\int_0^1(u^{11}-2u^9+u^7)\,du =\left. \left[ {1\over 12}u^{12}-{1\over 5}u^{10}+{1\over 8}u^8 \right] \right|_0^1\\ \qquad = {1\over 12}-{1\over 5}+{1\over 8} = \bbox[red, 2pt]{1\over 120} \\\textbf{(2) }\text{Changing the order of integration, }\int_0^3 \int_{\sqrt{1+x}}^2 \cos {x\over y+1}\,dydx = \int_1^2 \int_{0}^{y^2-1} \cos {x\over y+1} \,dx\,dy \\\qquad = \int_1^2 \left. \left[ (y+1) \sin{x\over y+1} \right] \right|_0^{y^2-1} \,dy =\int_1^2 (y+1)\sin {y^2-1\over y+1}\,dy =\int_1^2 (y+1)\sin (y-1)\,dy \\\qquad = \int_{0}^1 (u+2)\sin u\,d u =\left. \left[ \sin u-u\cos u -2\cos u\right] \right|_{0}^1 = \bbox[red, 2pt]{2+\sin 1-3\cos 1} \\\textbf{(3) }\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_0^{1-x-y} (x^2+y^2)\,dzdydx = \int_0^{2\pi} \int_0^1 \int_0^{1-r\cos\theta-r\sin \theta} r^3\,dzdrd\theta\\ \qquad  = \int_0^{2\pi} \int_0^1   r^3(1-r\cos\theta-r\sin \theta)\,drd\theta =\int_0^{2\pi}\left({1\over 4}-{1\over 5}(\cos \theta+ \sin \theta) \right)\,d\theta ={1\over 4}\cdot 2\pi = \bbox[red, 2pt]{\pi \over 2}$$

解答:$$e^x=\sum_{n=0}^\infty {1\over n!}x^n =1+\sum_{n=1}^\infty {1\over n!}x^n  \Rightarrow  x e^x =x + \sum_{n=1}^\infty {1\over n!}x^{n+1}  \\ \Rightarrow \int x e^x\,dx =\int\left( x + \sum_{n=1}^\infty {1\over n!}x^{n+1}\right)\,dx \Rightarrow f(x)=xe^x-e^x={1\over 2}x^2+\sum_{n=1}^\infty {1\over n!(n+2)}x^{n+2} +C\\ \Rightarrow f(0)=-1=C \Rightarrow f(x)=xe^x-e^x={1\over 2}x^2+\sum_{n=1}^\infty {1\over n!(n+2)}x^{n+2} -1 \\ \Rightarrow f(1)=0={1\over 2}+\sum_{n=1}^\infty {1\over n!(n+2)}  -1 \Rightarrow \sum_{n=1}^\infty {1\over n!(n+2)} ={1\over 2} \qquad \bbox[red, 2pt]{QED}$$

解答:$$y={\sin x\over 3+\cos^2 x} \Rightarrow V= 2\pi \int_0^\pi x\cdot {\sin x\over 3+\cos^2 x}\,dx \\ \cases{u=x \\dv =\displaystyle {\sin x\over 3+\cos^2 x}\,dx} \Rightarrow \cases{du=dx \\\ v=-\displaystyle {1\over \sqrt 3} \tan^{-1} {\cos x\over \sqrt 3}}\\ \Rightarrow V= 2\pi \left(  \left. \left[ -{x\over \sqrt 3} \tan^{-1}{\cos x\over \sqrt 3} \right] \right|_0^\pi \right) +{1\over \sqrt 3}\int_0^\pi \tan^{-1} {\cos x\over \sqrt 3}\,dx =2\pi \left( {\pi\over \sqrt 3}\tan^{-1}{1\over \sqrt 3} \right) \\= 2\pi \cdot {\pi\over \sqrt 3}\cdot {\pi\over 6} ={\pi^3\over 3\sqrt 3} = {\pi^b\over a} \Rightarrow (a,b) = \bbox[red, 2pt]{\left( 3\sqrt 3, 3 \right)}$$

解答:$$f(x)=e^x \Rightarrow f'(x) =e^x \Rightarrow S= 2\pi \int f(x) \sqrt{1+f'(x)^2}=2\pi \int_0^1 e^x \sqrt{1+e^{2x}} \,dx \\ u=e^x \Rightarrow du=e^x dx \Rightarrow S=2\pi\int_1^e \sqrt{1+u^2}\,du =2\pi \left. \left[ {1\over 2} u\sqrt{u^2+1}+{1\over 2}\ln \left( u+\sqrt{1+u^2} \right) \right] \right|_1^e \\= \bbox[red, 2pt]{\pi \left( e\sqrt{e^2+1} -\sqrt 2 +\ln \left( {e+\sqrt{e^2+1} \over 1+\sqrt 2} \right) \right)}$$

解答:$$\textbf{(1) } Q^TQ = I_n \Rightarrow rank(Q)=n \Rightarrow rank(P)= rank(QQ^T) =rank(Q)=n \\ \quad n\lt m \Rightarrow P \text{ is singular } \Rightarrow \det(P)= \bbox[red, 2pt]0 \\\textbf{(2) }P=QQ^T \Rightarrow P^2 =QQ^TQQ^T =QI_nQ^T =QQ^T=P \Rightarrow P^2=P \Rightarrow \lambda^2 =\lambda \\\quad \Rightarrow \lambda=0,1 \;(\lambda \in \text{Eigs}(P)) \\\quad  \lambda=1 \Rightarrow tr(P)=tr(QQ^T) =tr(Q^TQ) =tr(I_n) =n \Rightarrow \text{multiplicity of }(\lambda=1)=n \\ \lambda=0 \Rightarrow \text{multiplicity of }(\lambda=0)=m-n \;(P_{m\times m} \Rightarrow \deg(\det(P-\lambda I))=m) \\ \Rightarrow \bbox[red, 2pt]{\text{multiplicity of }(\lambda=1)=n, \text{multiplicity of }(\lambda=0)=m-n}$$

解答:$$A \text{ is diagonalizable }\Rightarrow A=PDP^{-1}, D_{ii} = \lambda_i (\text{eigenvalues of }A)  \Rightarrow A^k=PD^kP^{-1}, D_{ii}^k =\lambda_i^k \\k=2 \Rightarrow \cases{\lambda_i=0 \Rightarrow \lambda_i^2 =0^2 = \lambda_i\\ \lambda_i= 1\Rightarrow \lambda_i^2=1^2= \lambda_i} \Rightarrow D^2=D \Rightarrow A^2=A \Rightarrow \text{ the smallest integer }\bbox[red, 2pt]{k=2}$$

解答:$$\textbf{(1) }P_3 = \begin{pmatrix} \binom{0}{0} & \binom{1}{0} & \binom{2}{0} \\ \binom{1}{1} & \binom{2}{1} & \binom{3}{1} \\ \binom{2}{2} & \binom{3}{2} & \binom{4}{2} \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{pmatrix} \Rightarrow \det(P_3-\lambda I) =-(\lambda-1)(\lambda^2-8\lambda+1) =0 \\ \quad \Rightarrow \text{the eigenvalues of }P_3:\bbox[red, 2pt]{1, 4\pm \sqrt{15}} \\\textbf{(2) }A= L_nU_n \Rightarrow A_{ij} = \sum_{k=1}^n (L_{n})_{ik}(U_n)_{kj} = \sum_{k=1}^n {i-1\choose k-1} {j-1\choose k-1} = \sum_{k=1}^n {i-1\choose k-1} {j-1\choose j-k} \\\quad =\sum_{k=1}^n {i-1+j-1\choose j-1} =\sum_{k=1}^n {i+j-2\choose j-1} =(P_n)_{ij} \Rightarrow P_n= L_nU_n \quad \bbox[red, 2pt]{QED.} \\\textbf{(3) }P_n =L_n L_n^T \Rightarrow P_n^{-1} = \left( L_n^T \right)^{-1} L_n^{-1} = \left( L_n^{-1} \right)^T L_n^{-1} = \left( DL_nD \right)^{T} (DL_nD) = \left( D^TL_n^TD^T \right) (DL_nD) \\\quad = DL_n^T (DD)L_nD= DL_n^T (I)L_nD =DL_n^TL_nD =D(L_n^TL_n)D^{-1} \Rightarrow P_n^{-1} \text{ is similar to }L_n^TL_n \\ \quad \Rightarrow \text{Eigs}(P_n^{-1}) =\text{Eigs}(L_n^TL_n) =\text{Eigs}(L_nL_n^T) = \text{Eigs}(P_n ) \\\quad \Rightarrow \text{ the eigenvalues occur in  reciprocal pairs} \quad \bbox[red, 2pt]{QED.}\\ \quad \text{Where }D_{ij} = \begin{cases} (-1)^{i-1} & i=j\\ 0 & i\ne j\end{cases}.$$

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