113年中正大學機械碩士班-工程數學詳解

 國立中正大學113學年度碩士班招生考試

科目名稱:工程數學
系所組別:機械工程學系乙組

解答: $$rref\left( \left[\begin{matrix}1 & 2 & 3\\2 & -1 & 3\\0 & 1 & -1\end{matrix}\right]\right)= I_3 \Rightarrow \text{linear independent} \quad\bbox[red, 2pt]{QED}$$

解答: $$\textbf{(1) }A= \begin{bmatrix} -1 & 1 & 0 \\1 & -1 & 0 \\ 0 & 0 & -2\end{bmatrix} \Rightarrow \det(A-\lambda I) =-\lambda(\lambda+2)^2 =0 \Rightarrow \lambda=0,-2 \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow \begin{bmatrix} -1 & 1 & 0 \\1 & -1 & 0 \\ 0 & 0 & -2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_2\\ x_3=0} \\\qquad \Rightarrow v= x_2\begin{pmatrix}1\\1\\0 \end{pmatrix} ,\text{ choose }v_1= \begin{pmatrix}1\\1\\0 \end{pmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow \begin{bmatrix} 1 & 1 & 0 \\1 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1+x_2=0 \\ \qquad \Rightarrow v=x_2\begin{pmatrix}-1\\1\\0 \end{pmatrix} +x_3 \begin{pmatrix}0 \\0 \\1 \end{pmatrix}, \text{ choose }v_2=\begin{pmatrix}-1\\1\\0 \end{pmatrix}, v_3= \begin{pmatrix}0 \\0 \\1 \end{pmatrix}\\ \qquad \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }0,-2}, \bbox[red, 2pt]{\text{ eigenvectors: }\begin{pmatrix}1 \\1\\0 \end{pmatrix},  \begin{pmatrix}-1\\1\\0 \end{pmatrix}, \begin{pmatrix}0 \\0 \\1 \end{pmatrix}} \\\textbf{(2) }A= \begin{bmatrix} -1 & 1 & 0 \\1 & -1 & 0 \\ 0 & 0 & -2\end{bmatrix} =\begin{bmatrix} 1 & -1 & 0 \\1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 0 & 0 & 0 \\0 & -2 & 0 \\ 0 & 0 & -2\end{bmatrix} \begin{bmatrix} 1/2 & 1/2 & 0 \\-1/2 & 1/2 & 0 \\ 0 & 0 & 1\end{bmatrix} \\\quad \Rightarrow e^{At} =\begin{bmatrix} 1 & -1 & 0 \\1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & e^{-2t} & 0 \\ 0 & 0 & e^{-2t}\end{bmatrix}\begin{bmatrix} 1/2 & 1/2 & 0 \\-1/2 & 1/2 & 0 \\ 0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 1/2(1+e^{-2t}) & 1/2(1-e^{-2t}) & 0 \\-1/2(1 +e^{-2t}) & 1/2(1+e^{-2t}) & 0 \\ 0 & 0 & e^{-2t}\end{bmatrix}\\ \quad X'=AX \Rightarrow X=e^{At} C =\begin{bmatrix} 1/2(1+e^{-2t}) & 1/2(1-e^{-2t}) & 0 \\1/2(1-e^{-2t}) & 1/2(1+e^{-2t}) & 0 \\ 0 & 0 & e^{-2t}\end{bmatrix} \begin{bmatrix} c_1 \\c_2 \\ c_3\end{bmatrix} \\\qquad = \begin{bmatrix} {1\over 2}c_1(1+e^{-2t}) + {1\over 2}c_2(1-e^{-2t})   \\{1\over 2}c_1(1-e^{-2t}) +{1\over 2} c_2(1+e^{-2t})  \\ c_3 e^{-2t}\end{bmatrix}\Rightarrow \bbox[red, 2pt] {\mathbf {X =\begin{bmatrix} c_4 + c_5e^{-2t}   \\c_4  -c_5e^{-2t}   \\ c_3 e^{-2t}\end{bmatrix}}}$$

解答: $$\textbf{(1) }f(x)=x^2 \Rightarrow f(-x)=f(x) \Rightarrow f\text{ is even} \Rightarrow b_n=0\\\quad a_0={1\over \pi} \int_{-\pi}^\pi x^2\,dx ={2\over \pi} \int_0^{\pi}x^2\,dx ={2\pi^2\over 3} \\ \quad a_n={1\over \pi} \int_{-\pi}^\pi x^2 \cos(nx)\,dx ={2\over \pi} \int_0^\pi x^2 \cos(nx)\,dx ={2\over \pi} \cdot {2\pi\over n^2}(-1)^n ={4\over n^2}(-1)^n \\ \Rightarrow f(x)\sim \bbox[red, 2pt]{{\pi^2\over 3}+\sum_{n=1}^\infty{4\over n^2}(-1)^n \cos(nx)} \\\textbf{(2)} \text{Parseval's equality: }{1\over \pi} \int_{-\pi}^\pi f(x)^2\,dx = {a_0^2\over 2}+ \sum_{n=1}^\infty (a_n^2+ b_n^2) \\ \quad \Rightarrow{1\over \pi}\int_{-\pi}^\pi x^4\,dx ={2\over 5}\pi^4={2\over 9}\pi^4+ \sum_{n=1}^\infty {16\over n^4} \Rightarrow {8\over 45}\pi^4 =16\sum_{n=1}^\infty {1\over n^4} \Rightarrow \sum_{n=1}^\infty {1\over n^4}={1\over 90}\pi^4 \quad \bbox[red, 2pt]{QED}$$

解答: $$\textbf{(1)  }{dx\over dt}+3x=0 \Rightarrow e^{3t}{dx\over dt} +3xe^{3t}=0 \Rightarrow (xe^{3t})'=0 \Rightarrow xe^{3t}=c_1 \Rightarrow \bbox[red, 2pt]{x_h =c_1e^{-3t}} \\\textbf{(2) }x_p=Ae^{-2t} \Rightarrow x_p'=-2Ae^{-2t} \Rightarrow x_p'+3x_p=Ae^{-2t}=e^{-2t} \Rightarrow A=1 \Rightarrow x_p=e^{-2t} \\\quad x=x_h+x_p =c_1e^{-3t}+e^{-2t} \Rightarrow x(0)=c_1+1=1 \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt]{x=e^{-2t}}$$

解答: $$\textbf{(1)  } L\{{dx_1\over dt}\} =sX_1(s)-x_1(0)=\bbox[red, 2pt]{sX_1(s)-1}\\ \textbf{(2)  } \cases{2x_1'-x_2'+3x_1=1\\ x_1'+x_2'-4x_2=e^{-t}} \Rightarrow \cases{L\{2x_1'-x_2'+3x_1 \}= L\{1\}\\ L\{x_1'+x_2'-4x_2\} = L\{e^{-t} \}}\\ \Rightarrow \cases{2(sX_1(s)-x_1(0))-(sX_2(s)-x_2(0)) +3X_1(s) =1/s \\sX_1(s) -x_1(0)+ sX_2(s)-x_2(0) -4X_2(s) =1/(s+1)} \\ \quad \Rightarrow \cases{(2s+3)X_1(s)-sX_2(s)={1\over s}+2\\ sX_1(s)+(s-4)X_2(s)={1\over s+1}+1} \Rightarrow \cases{(2s+3)(s-4)X_1(s)-s(s-4)X_2(s)={s-4\over s}+2(s-4)\\ s^2X_1(s)+ s(s-4)X_2(s)={s\over s+1}+s} \\ \Rightarrow (3s^2-5s-12)X_1(s) =3s-8+{s-4\over s}+{s\over s+1} \Rightarrow (s-3)(3s+4)X_1(s)= 3s-6-{4\over s}-{1\over s+1} \\ \Rightarrow X_1(s)={3s-6\over (s-3)(3s+4)}-{4\over s(s-3)(3s+4)}-{1\over (s+1) (s-3)(3s+4)} \\ \Rightarrow \bbox[red, 2pt]{X_1(s)={3s^3-3s^2-11s-4\over s(s+1)(s-3)(3s+4)}} \\ \Rightarrow X_2(s)={1\over (s+1)(s-4)}+ {1\over s-4}-{s\over s-4}X_1(s)\\={1\over (s+1)(s-4)}+ {1\over s-4}-{s\over s-4}\cdot {3s^3-3s^2-11s-4\over s(s+1)(s-3)(3s+4)} \\ \Rightarrow \bbox[red, 2pt]{X_2(s)={3s^3+s^2-22s-24\over (s-3)(s-4)(s+1)(3s+4)}}$$

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解題僅供參考，碩士班歷年試題及詳解