113年中央光電碩士班-工程數學詳解

國立中央大學113學年度碩士班考試入學

所別：光電類
科目：工程數學

解答： $$\textbf{(a)}\; W[\sin(x),e^x]= \begin{vmatrix}\sin x & e^x \\\cos x & e^x \end{vmatrix} =e^x(\sin x-\cos x) \ne 0, \text{ if }x=0 \Rightarrow \bbox[red, 2pt]{\text{ linearly independent}} \\ \textbf{(b)}\; W[e^x,e^{x+2}] = \begin{vmatrix} e^x & e^{x+2} \\e^x & e^{x+2} \end{vmatrix} =0, \forall x \Rightarrow  \bbox[red, 2pt]{\text{ linearly dependent}}$$

解答： $$y''+5y'+6y=0 \Rightarrow \lambda^2+5\lambda+6=0 \Rightarrow (\lambda+3)( \lambda+2)=0 \Rightarrow \lambda=-3,-2 \\ \Rightarrow y_h=c_1e^{-2x}+c_2 e^{-3x} \\ y_p= Ax^2+Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \\ \Rightarrow y_p''+5y_p'+6y_p==6Ax^2+ (10A+6B)x+ 2A+5B+6C =2x+1 \\ \Rightarrow \cases{A=0\\ 10A+6B=2\\ 2A+5B+6C=1} \Rightarrow \cases{A=2\\ B=1/3\\ C=-1/9} \Rightarrow y_p={1\over 3}x-{1\over 9} \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-2x}+c_2 e^{-3x} +{1\over 3}x-{1\over 9}}$$

解答： $$xy''+y'=0 \Rightarrow (xy')'=0 \Rightarrow xy'=c_1 \Rightarrow y'={c_1\over x} \Rightarrow \bbox[red, 2pt]{ y=c_1\ln x+c_2}$$

解答： $$L\{x''(t)\}+L\{x(t) \}=L\{ f(t)\} =s^2X(s)-sx(0)-x'(0)+X(s)= \int_1^5 e^{-st}\,dt \\ \Rightarrow (s^2+1)X(s)={1\over s}e^{-s}-{1\over s} e^{-5s} \Rightarrow X(s)= {1\over s(s^2+1)}e^{-s}-{1\over s(s^2+1)} e^{-5s} \\ \Rightarrow x(t)= L^{-1}\{ X(s)\} \Rightarrow \bbox[red, 2pt]{x(t) =u(t-1)(1- \cos(t-1))+ u(t-5)(\cos(t-5)-1)}$$

解答： $$f(t)=|t| \Rightarrow f(-t)=f(t) \Rightarrow f(t) \text{ is even }\Rightarrow b_n=0\\ a_0= {1\over 2}\int_{-1}^1 |t|\,dt =\int_0^1 t\,dt ={1\over 2} \\ a_n = \int_{-1}^1 |t|\cos (n\pi t)\,dt  = 2  \int_0^1 t\cos(n\pi t)\,dt ={2\over n^2\pi^2}((-1)^n-1) \\ \Rightarrow \mathcal F(f(t))= \bbox[red, 2pt]{{1\over 2} +\sum_{n=1}^\infty {2\over n^2\pi^2}((-1)^n-1) \cos(n \pi t)\,dt}$$

解答： $$\text{The boundary curve will be the circle of radius 2 that is in the plane  z=0}\\ \text{The parameterization of this curve is, } \vec r=[2\cos t,2\sin t,0], 0\le t\le 2\pi \\ \text{Using Stokes' theorem, we have} \iint_S (\nabla \times \vec F) \cdot \vec n\,dA= \int_0^{2\pi} \vec F(\vec r(t))\cdot \vec r'(t)\,dt \\ =\int_0^{2\pi} [8\cos^2 t\sin t,-8\cos t\sin^2 t,0] \cdot [-2\sin t,2\cos t,0]\,dt = \int_0^{2\pi} -32\cos^2t \sin^2 t\,dt \\= \int_0^{2\pi} -8\sin^2(2t)\,dt  = \int_0^{2\pi} -4(1-\cos(4t))\,dt = \bbox[red, 2pt]{-8\pi}$$

解答： $$A=\begin{bmatrix}-1 & 1 & 1 \\2 & -6 & 3 \\ 2 & 6 & -5 \end{bmatrix} \Rightarrow A^{-1}=\begin{bmatrix}\frac{3}{7} & \frac{11}{28} & \frac{9}{28} \\\frac{4}{7} & \frac{3}{28} & \frac{5}{28} \\\frac{6}{7} & \frac{2}{7} & \frac{1}{7}\end{bmatrix}\\ \Rightarrow \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =A^{-1}  \begin{bmatrix}-5 \\ 4\\ 8 \end{bmatrix} =\begin{bmatrix}\frac{3}{7} & \frac{11}{28} & \frac{9}{28} \\\frac{4}{7} & \frac{3}{28} & \frac{5}{28} \\\frac{6}{7} & \frac{2}{7} & \frac{1}{7}\end{bmatrix} \begin{bmatrix}-5 \\ 4\\ 8 \end{bmatrix} =\begin{bmatrix}2 \\-1 \\-2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{ \cases{x_1= 2\\ x_2=-1\\ x_3=-2} }$$

解答： $$A=\begin{bmatrix}1 & -1 \\2 & 4 \end{bmatrix} \Rightarrow \det(A-\lambda I)= (\lambda-2)(\lambda -3) \\ \Rightarrow f(\lambda)=\lambda^{17} =c_1+c_2 \lambda \Rightarrow \cases{f(2)=2^{17}= c_1+ 2c_2\\ f(3)=3^{17}=c_1+3c_2} \Rightarrow \cases{c_1=3\cdot 2^{17}-2\cdot 3^{17}\\ c_2=3^{17}-2^{17}} \\ \Rightarrow A^{17}= f(A)=c_1I+c_2A = (3\cdot 2^{17}-2\cdot 3^{17})I+(3^{17}-2^{17})A \\=\begin{bmatrix}3\cdot 2^{17}-2\cdot 3^{17} & 0 \\0 & 3\cdot 2^{17}-2\cdot 3^{17} \end{bmatrix} +\begin{bmatrix}(3^{17}-2^{17}) & -(3^{17}-2^{17}) \\2(3^{17}-2^{17}) & 4 (3^{17}-2^{17}) \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix} 2^{18}-3^{17} & 2^{17}-3^{17} \\2\cdot 3^{17}-2^{18} & 2\cdot 3^{17}-2^{17} \end{bmatrix}}$$

解答： $$\textbf{(a)}\; f(t)=\cos^2(2\pi t)= {1\over 2} \cos(4\pi t) +{1\over 2} = {1\over 4}\left( e^{4 \pi t i} +e^{-4\pi t i}\right) +{1\over 2} \\ \Rightarrow F(\omega)={1\over \sqrt{2\pi}} \int_{-\infty}^\infty \left( {1\over 4}\left( e^{4 \pi t i} +e^{-4\pi t i}\right) +{1\over 2} \right) e^{-i\omega t}\,dt \\\qquad ={1\over \sqrt{2\pi}} \int_{-\infty}^\infty \left({1\over 4}e^{-i(\omega-4\pi)t} +{1\over 4}e^{-i(\omega + 4\pi)t} +{1\over 2}e^{-i\omega t}\right)\,dt \\ \qquad ={1\over \sqrt{2\pi}} \left({2\pi \over 4}  ( \delta(\omega-4\pi) +\delta(\omega+ 4\pi)  +{1 \over 2}\cdot 2\pi \delta(\omega) \right)\\\qquad  = \bbox[red, 2pt]{{\sqrt{2\pi} \over 4} (\delta(\omega-4\pi)+ \delta(\omega+4\pi)+2 \delta(\omega)) } \\\textbf{(b)}\; g(t)=\begin{cases} 1 & 0\le t\le 1\\ 0& \text{otherwise }\end{cases} \Rightarrow G(\omega) ={1\over \sqrt{2\pi}} \int_0^1 e^{-i\omega t}\,dt={1\over \sqrt{2\pi}} \left. \left[ {1\over -i\omega} e^{-i\omega t}\right] \right|_0^1 \\\qquad ={1\over \sqrt{2\pi}}\cdot {1\over -i\omega} \left( e^{-i\omega} -1\right) \Rightarrow\bbox[red, 2pt]{ G(\omega)={i\over \omega \sqrt{2\pi}}\left( e^{-i\omega} -1\right)} \\ \quad h(t)= \begin{cases} 1 & 0\le t\le 1/2\\ 0& \text{ otherwise }\end{cases} \Rightarrow H(\omega) ={1\over \sqrt{2\pi}} \int_0^{1/2} e^{-i\omega t}\,dt={1\over \sqrt{2\pi}} \left. \left[ {1\over -i\omega} e^{-i\omega t}\right] \right|_0^{1/2} \\\qquad = {1\over \sqrt{2\pi}}\cdot {1\over -i\omega} \left( e^{-i\omega/2} -1\right) \Rightarrow \bbox[red, 2pt]{H(\omega) = {i\over \omega \sqrt{2\pi}} \left( e^{-i\omega/2} -1\right)}$$

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