110年中山大學海下碩士班-工程數學詳解

 國立中山大學110學年度碩士班招生考試

科目名稱:工程數學【海下所碩士班】

解答:$$\lim_{x\to \infty} {\ln x\over x^{a+1}} =\lim_{x\to \infty} {\frac{d }{dx}\ln x\over \frac{d }{dx}x^{a+1}} =\lim_{x\to \infty} {1/ x\over (a+1)x^{a}} =\lim_{x\to \infty} {1\over (a+1)x^{a+1}} = \bbox[red, 2pt] 0$$

解答:$$\textbf{(1) }v_1=u_1=<3,1> \Rightarrow e_1={v_1\over |v_1|} =<{3\over \sqrt{10}},{1\over \sqrt{10}}> =<{3\sqrt{10} \over 10} , {\sqrt{10} \over 10}> \\ v_2=u_2-(u_2\cdot e_1)e_1 =<1,1>-{2\sqrt{10}\over 5}<{3\sqrt{10} \over 10} , {\sqrt{10} \over 10}> =<-{1\over 5},{3\over 5}>\\\qquad  \Rightarrow e_2= {v_2\over |v_2|} =<-{\sqrt{10}\over 10},{3\sqrt{10}\over 10}> \\ \Rightarrow \text{ an orthonormal basis: }\{e_1,e_2\} = \bbox[red, 2pt]{\left\{<{3\sqrt{10} \over 10} , {\sqrt{10} \over 10}> ,<-{\sqrt{10}\over 10},{3\sqrt{10}\over 10}> \right\}} \\\textbf{(2) }v_1=u_1=<1,1,1> \Rightarrow e_1={v_1\over |v_1|}  =<{\sqrt 3\over 3}, {\sqrt 3\over 3}, {\sqrt 3\over 3}>\\\quad v_2=u_2-(u_2\cdot e_1)e_1 =<1,2,2>-{5\sqrt 3\over 3} <{\sqrt 3\over 3}, {\sqrt 3\over 3}, {\sqrt 3\over 3}>= <-{2\over 3},{1\over 3}, {1\over 3}>\\ \qquad \Rightarrow e_2={v_2\over |v_2|} =<-{\sqrt 6\over 3},{\sqrt 6\over 6}, {\sqrt 6\over 6}>\\ v_3=u_3-(u_3\cdot e_1)e_1-(u_3\cdot e_2)e_2 =<0,{1\over 2},-{1\over 2}> \Rightarrow e_3={v_3\over |v_3|} =<0,{\sqrt 2\over 2} ,-{\sqrt 2\over 2}> \\ \Rightarrow \text{ an orthonormal basis: }\{e_1,e_2, e_3\} = \bbox[red, 2pt]{\left\{<{\sqrt 3\over 3}, {\sqrt 3\over 3}, {\sqrt 3\over 3}>, <-{\sqrt 6\over 3},{\sqrt 6\over 6}, {\sqrt 6\over 6}> <0,{\sqrt 2\over 2} ,-{\sqrt 2\over 2}>\right\}}$$

解答:$$A=\begin{bmatrix} 2 & 2& -1\\ 1& -1& 0\\ 0& 1& 0\end{bmatrix} \Rightarrow A^T= \begin{bmatrix} 2 & 1& 0\\ 2& -1& 1\\ -1& 0& 0\end{bmatrix} \Rightarrow \cases{B={1\over 2}(A+A^T) \\C={1\over 2}(A-A^T)} \\ \Rightarrow \cases{ \bbox[red, 2pt]{B= \begin{bmatrix} 2 & 3/2& -1/2\\ 3/2& -1& 1/2\\ -1/2& 1/2& 0\end{bmatrix}} \\[1ex] \bbox[red, 2pt]{C= \begin{bmatrix} 0 & 1/2& -1/2\\ -1/2& 0& -1/2\\ 1/2& 1/2& 0\end{bmatrix}}}$$

解答:$${dy\over dx}+y=x \Rightarrow e^x {dy\over dx}+ ye^x=xe^x \Rightarrow \left(e^x y \right)'=xe^x \Rightarrow e^x y= \int xe^x\,dx = xe^x -e^x+ c_1\\ \Rightarrow y=x-1+{c_1\over e^x} \Rightarrow y(0)=-1+c_1=3 \Rightarrow c_1=4 \Rightarrow \bbox[red, 2pt]{y=x-1+4e^{-x}}$$

解答:$$\textbf{(1) }L\{ \sin(2t)\} =\int_0^\infty e^{-st} \sin(2t)\,dt ={1\over 2i}\int_0^\infty e^{-st} \left(e^{2ti}  -e^{-2ti}\right)\,dt ={1\over 2i}\int_0^\infty   \left(e^{(2i-s)t}  -e^{-(2i+s)t}\right)\,dt \\\quad = {1\over 2i} \left. \left[ {1\over 2i-s} e^{(2i-s)t}+{1 \over 2i+s}e^{-(2i+s) t} \right] \right|_0^\infty =-{1\over 2i} \left( {1\over 2i-s}+{1\over 2i+s}\right) \\ \quad =-{1\over 2i}\left( {2i+s\over -4-s^2}+ {2i-s\over -4-s^2}\right) ={1\over 2i}\cdot {4i\over s^2+4} =\bbox[red, 2pt]{2\over s^2+4} \\\textbf{(2) }L\{ {dy\over dt}\} +3L\{y\} =13L\{ \sin(2t)\} \Rightarrow sY(s)-6+3Y(s)=13\cdot {2\over s^2+4} \\ \quad \Rightarrow (s+3)Y(s)={26\over s^2+4}+6 \Rightarrow Y(s)={26\over (s^2+4)(s+3)} +{6\over s+3} \\\quad \Rightarrow y(t) =L^{-1}\{Y(s)\} =L^{-1} \left\{ {26\over (s^2+4)(s+3)} +{6\over s+3}\right\}   =L^{-1}\left\{ {-2s+6\over s^2+4 } +{8\over s+3}\right\} \\ \quad \Rightarrow \bbox[red, 2pt]{y(t)=-2\cos(2t)+ 3\sin(2t)+ 8e^{-3t}}$$

解答:$$\text{Bessel’s equation of order  }ν\text{  is given by} \bbox[red, 2pt]{x^2y''+xy'+(x^2-v^2)y=0}$$

解答:$$y''x^2+(x^2-81)y=-xy' \Rightarrow y''x^2+xy'+(x^2-9^2)y=0 \text{ is a Bessel's equation of order }v=9\\ \Rightarrow \bbox[red, 2pt]{y=c_1J_9(x)+ c_2Y_9(x)}, \text{ where }J_9(x)= \sum_{n=0}^\infty {(-1)^n \over n! \Gamma(n+10)} \left({x\over 2} \right)^{2n+9},\\\qquad  J_{-9}(x) =\sum_{n=0}^\infty {(-1)^n \over n! \Gamma(n-8)} \left({x\over 2} \right)^{2n-9} \text{ and }Y_9(x) =\lim_{v\to 9} {\cos v\pi J_v(x)-J_{-v}(x) \over \sin v\pi }$$

解答:$$\textbf{Case I }k=0: y''=0 \Rightarrow y=c_1x+ c_2 \Rightarrow \cases{y(0)=c_2=0 \\ y(L)=c_1L+ c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow y=0\\ \textbf{Cases II }k\gt 0: y''+ky=0 \Rightarrow y=c_1 \cos(\sqrt kx) +c_2\sin(\sqrt kx) \Rightarrow y(0)=c_1=0 \\\qquad \Rightarrow y(L) =c_2 \sin(\sqrt kL) =0 \Rightarrow \sqrt kL= n\pi \Rightarrow k={n^2\pi^2 \over L^2} ,n\in \mathbb N \\ \textbf{Case III }k\lt 0: y''+ky=0 \Rightarrow y=c_1e^{\sqrt kx} +c_2e^{-\sqrt k x} \Rightarrow y(0)=c_1+c_2=0 \Rightarrow c_2=-c_1\\ \qquad \Rightarrow y=c_1e^{\sqrt k x}-c_1e^{-\sqrt k x} \Rightarrow y(L) = c_1e^{\sqrt k L}-c_1e^{-\sqrt k L} =0 \Rightarrow c_1(e^{2\sqrt kL}-1) =0 \\\qquad \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow y=0\\ \text{Summary, eigenfunctions: } \bbox[red, 2pt]{y_n(x) =c_n \sin({n\pi x\over L}), n\in \mathbb N}$$

解答:$$f(t)\text{ with period }2 \Rightarrow f(t)=\begin{cases}1& 0\le t\le 1\\ -1& 1\le t\le 2 \end{cases} \;\equiv  f(t) =\begin{cases}1& 0\le t\le 1\\ -1& -1\le t\le 0 \end{cases} \\ \Rightarrow c_n= {1\over 2} \int_{-1}^1 f(t)e^{-i n\pi t }\,dt ={1 \over 2}\left( \int_{-1}^0 -e^{-in\pi t} \,dx+\int_0^1 e^{-i n\pi t} \,dt\right) \\= {1\over 2} \left( \left. \left[ {1\over in\pi} e^{-in \pi t} \right] \right|_{-1}^0 + \left. \left[ -{1\over in \pi }e^{-in\pi t}\right] \right|_0^1 \right) ={1\over 2}\left( {2\over in\pi} -{1\over in\pi}(e^{in\pi} +e^{-in\pi})\right) \\={1\over in\pi}(1-(-1)^n) ,n=0,\pm 1,\pm 2,\dots\Rightarrow \bbox[red, 2pt]{f(t)= \sum_{n=-\infty}^\infty  {1\over in\pi}(1-(-1)^n)e^{in \pi t}}$$

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解題僅供參考，碩士班歷年試題及詳解