115年中山大學資工碩士班-工程數學詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:工程數學【資工系碩士班乙組】

A. Ordinary differential equations

解答:$${dz\over dx}= z^2x+x=(z^2+1)x \Rightarrow \int {1\over z^2+1} \,dz=\int x\,dx \Rightarrow \tan^{-1}z ={1\over 2}x^2+c_1 \\ \Rightarrow z=\tan \left( {1\over 2}x^2+c_1 \right) \Rightarrow z(0)= \tan c_1=0 \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt]{z=\tan {x^2\over 2}}$$

解答:$$\cases{v_1'=v_2\\ v_2'=2v_1+v_2} \Rightarrow   \begin{bmatrix}v_1'\\ v_2'\end{bmatrix}= \begin{bmatrix}0&1\\2& 1 \end{bmatrix} \begin{bmatrix}v_1\\ v_2 \end{bmatrix} \Rightarrow \mathbf v'= A\mathbf v \Rightarrow A= \begin{bmatrix}0&1\\2& 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \\ \Rightarrow (\lambda+1)(\lambda-2) =0 \Rightarrow \cases{\lambda_1=-1\\ \lambda_2=2} \Rightarrow \text{ eigenvectors: }\cases{v_1= \begin{bmatrix}-1\\1 \end{bmatrix} \\v_2= \begin{bmatrix}1\\2 \end{bmatrix}} \\ \Rightarrow \mathbf v=c_1e^{\lambda_1 t} v_1+c_2 e^{\lambda_2 t}v_2 =c_1 e^{-t} \begin{bmatrix}-1\\1 \end{bmatrix}+c_2 e^{2t} \begin{bmatrix}1\\ 2 \end{bmatrix} \Rightarrow \mathbf v(0) = \begin{bmatrix}1\\ 2 \end{bmatrix} = \begin{bmatrix}-c_1\\ c_1 \end{bmatrix}+ \begin{bmatrix}c_2\\2c_2 \end{bmatrix} \\ \Rightarrow \cases{-c_1+c_2=1\\ c_1+2c_2=2 } \cases{c_1=0\\ c_2=1} \Rightarrow \mathbf v=e^{2t} \begin{bmatrix}1\\ 2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{v_1(t) =e^{2t} \\v_2(t)=2e^{2t}}}$$

B. Linear algebra

解答:$$M= \begin{bmatrix}1&-1\\ 1& 1 \end{bmatrix} = \begin{bmatrix}1& -1\\0&0 \end{bmatrix}+ \begin{bmatrix}0& 0\\1& 1 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{B=\begin{bmatrix}1& -1\\0&0 \end{bmatrix}, C=\begin{bmatrix}0& 0\\1& 1 \end{bmatrix}}$$

解答:$$P= [s_1\;s_2\; s_3] = \begin{bmatrix}1& 0& 0\\1& 1& 1\\0& 0& 1 \end{bmatrix}, D= \begin{bmatrix}\lambda_1& 0& 0\\ 0& \lambda_2& 0\\ 0& 0& \lambda_3 \end{bmatrix} =\begin{bmatrix}-1& 0& 0\\ 0& 0& 0\\ 0& 0& 1 \end{bmatrix}\\ \Rightarrow A=PDP^{-1} = \begin{bmatrix}1& 0& 0\\1& 1& 1\\0& 0& 1 \end{bmatrix} \begin{bmatrix}-1& 0& 0\\ 0& 0& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1& 0& 0\\ -1&1& -1\\0&0&1 \end{bmatrix} = \bbox[red, 2pt]{ \begin{bmatrix}-1& 0& 0\\-1& 0& 1\\0&0& 1 \end{bmatrix}}$$

C. Fourier series and Fourier integral

解答:$$f(t)= \begin{cases} -1,& -\pi\lt t\lt 0\\ 1,& 0\lt t\lt \pi\end{cases} \Rightarrow f(t+2\pi)=f(t) \Rightarrow f(-t)=-f(t) \Rightarrow f(t) \text{ is odd} \Rightarrow a_n=0\\ \Rightarrow b_n={1\over \pi} \int_{-\pi}^\pi f(t) \sin(nt)\,dt = {2\over \pi} \int_0^\pi \sin(nt)\,dt ={2\over n\pi}(1-(-1)^n) \\ \Rightarrow f(t)= \sum_{k=1}^\infty{4\over (2k-1)\pi} \sin((2k-1)t) \Rightarrow f(\pi/2) ={4\over \pi} \alpha \Rightarrow {4\over \pi} \alpha=1 \Rightarrow \alpha=\bbox[red, 2pt]{\pi\over 4}$$

解答:$$g(t) =\begin{cases} 1,& -1\lt t\lt 1\\ 0, &\text{otherwise}\end{cases} \Rightarrow g(t) \text{ is even} \Rightarrow B(\omega)=0 \Rightarrow A(\omega)={1\over \pi}\int_{-1}^1 \cos(\omega t)\,dt={2\over \pi}\cdot {\sin(\omega) \over \omega} \\ \Rightarrow g(t) =\int_0^\infty \left( {2\over \pi}\cdot {\sin(\omega) \over \omega}  \right) \cos(\omega t)\,d\omega \Rightarrow g(0)= {2\over \pi}\beta =1 \Rightarrow \beta= \bbox[red, 2pt]{\pi\over 2}$$

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