國立成功大學115學年度碩士班招生考試試題
系所:土木工程學系
科目:工程數學
解答:$$\textbf{(a) }\det(A-\lambda I) = \begin{vmatrix} 3-\lambda& 1& 1\\1& 3-\lambda& 1\\1& 1& 3-\lambda \end{vmatrix} = 0 \Rightarrow (\lambda-5)(\lambda-2)^2 =0 \Rightarrow \lambda=2,5 \\ \qquad \Rightarrow \text{characteristic equation: }\bbox[red, 2pt]{(\lambda-5)(\lambda-2)^2=0}, \text{ eigenvalues: }\bbox[red, 2pt]{2,5}\\ \textbf{(b) }\lambda_1=5 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-2& 1& 1\\ 1&-2& 1\\ 1& 1& -2 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow x_1=x_2=x_3 \Rightarrow v=k \begin{bmatrix}1\\1\\1 \end{bmatrix} \\ \qquad \text{choosing }v_1= \begin{bmatrix}1\\1\\1 \end{bmatrix} \Rightarrow \text{normalized }v_1= e_1= \begin{bmatrix}1/\sqrt 3\\ 1/\sqrt 3\\ 1/\sqrt 3 \end{bmatrix}\\ \quad \lambda_2=2 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}1& 1& 1\\1& 1& 1\\1& 1& 1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_2 \end{bmatrix} =0 \Rightarrow x_1+x_2+x_3=0 \Rightarrow v= \begin{bmatrix}x_1\\ -x_1-x_3\\ x_3 \end{bmatrix} \\ \quad \Rightarrow v=x_1 \begin{bmatrix}1\\ -1\\0 \end{bmatrix} +x_3 \begin{bmatrix}0\\ -1\\1 \end{bmatrix}, \text{choosing } v_2=\begin{bmatrix}1\\ -1\\0 \end{bmatrix} \Rightarrow e_2=v_2/|v_2|= \begin{bmatrix}1/\sqrt 2\\ -1/\sqrt 2\\0 \end{bmatrix} \\ \quad \Rightarrow e_3=v_3/|v_3|= e_1\times e_2 = \begin{bmatrix}1/\sqrt 6\\ 1/\sqrt 6\\ -2/\sqrt 6 \end{bmatrix} \Rightarrow P=[v_1\; v_2\; v_3] \Rightarrow \bbox[red, 2pt]{P= \begin{bmatrix}1/\sqrt 3& 1/\sqrt 2& 1/\sqrt 6\\ 1/\sqrt 3& -1/\sqrt 2& 1/\sqrt 6\\ 1/\sqrt 3& 0& -2/\sqrt 6 \end{bmatrix}}\\ D= \begin{bmatrix}\lambda_1& 0& 0\\ 0&\lambda_2 & 0\\ 0&0& \lambda_2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{D= \begin{bmatrix}5&0&0\\0& 2& 0\\ 0&0& 2 \end{bmatrix}}$$
解答:$$\textbf{(a) }f(x)=x^3-2x-5 \Rightarrow f'(x)=3x^2-2 \Rightarrow x_{n+1} =x_n-{f(x) \over f'(x)}=x_n-{x_n^3-2x_n-5\over 3x_n^2-2} \\\qquad \Rightarrow \bbox[red, 2pt]{x_{n+1} ={2x_n^3 +5\over 3x_n^2-2}} \\\textbf{(b) }x_0=2 \Rightarrow f(x_0)=f(2)=-1 \Rightarrow f'(x_0)=10 \Rightarrow x_1={16+5\over 3\cdot 2^2-2} = \bbox[red, 2pt]{2.1}$$
解答:$$y''+2y'+5y=0 \Rightarrow r^2+2r+5=0 \Rightarrow r=-1\pm 2i \Rightarrow y_h= e^{-x}(c_1\cos 2x+ c_2\sin 2x) \\ y_p=A\cos x+B\sin x \Rightarrow y_p'=-A\sin x+B\cos x \Rightarrow y_p''=-A\cos x-B\sin x \\ \Rightarrow y_p''+2y_p'+5y_p = (4A+2B) \cos x+(-2A+4B) \sin x=10\cos x \Rightarrow \cases{4A+2B=10\\ -2A+4B=0} \\ \Rightarrow \cases{A=2\\ B=1} \Rightarrow y_p=2\cos x+\sin x \Rightarrow y=y_h+y_p =e^{-x}(c_1\cos 2x+ c_2\sin 2x)+ 2\cos x+\sin x \\ \Rightarrow y'=e^{-x}((-c_1+2c_2) \cos x+ (-c_2-2c_1) \sin x)-2\sin x+\cos x \\ \Rightarrow \cases{y(0)=c_1+2=0\\ y'(0)=-c_1+2c_2+1=0} \Rightarrow \cases{c_1=-2\\c_2=-3/2}\\ \Rightarrow\bbox[red, 2pt]{ y=e^{-x} \left( -2\cos 2x-{3\over 2}\sin x \right)+ 2\cos x+\sin x}$$
解答:$$\textbf{(a) }\mathbf F=x^2 \mathbf i+y^2 \mathbf j+(z+3)\mathbf k \Rightarrow \nabla\cdot \mathbf F = {\partial \over \partial x}(x^2) +{\partial \over \partial y}(y^2) +{\partial \over \partial z}(z+3) = \bbox[red, 2pt]{2x+2y+1} \\\textbf{(b) }\iint_S \mathbf F\cdot\mathbf n\, dA = \iiint_V (\nabla \cdot \mathbf F)\,dV=\bbox[red, 2pt]{ \iiint_V (2x+2y+ 1)\,dV} \\\textbf{(c) }\cases{x=r\cos \theta\\ y= r\sin \theta\\ z=z } \Rightarrow dV=rdzdrd\theta \\\Rightarrow \iiint_V (2x+2y+z)\,dV= \int_0^{2\pi} \int_0^2 \int_{r^2}^4 (2r\cos \theta+ 2r\sin \theta +1)r\,dz dr d\theta =\int_0^{2\pi} \int_0^2 \int_{r^2}^4 r\,dz dr d\theta \\= 2\pi \int_0^2 (4-r^2)r\,dr =2\pi \left. \left[ 2r^2-{1\over 4}r^4 \right] \right|_0^2= \bbox[red, 2pt]{8\pi}$$
解答:$$u(z,t)= Z(z)T(t) \Rightarrow {\partial u\over \partial t}=c_v{\partial^2 u\over \partial z^2} \Rightarrow ZT'=c_vZ''T \Rightarrow {T'\over c_vT}={Z''\over Z}=-\lambda \Rightarrow \cases{Z''+\lambda Z=0\\ T'+ c_v\lambda T=0}\\ \text{boundary conditions:}\cases{u(0,t)=Z(0)T(t) =0\\ u(H,t)=Z(H)T(t)=0} \Rightarrow \cases{Z(0)=0\\ Z(H)=0} \\\text{Solving for }Z(z):Z''+\lambda Z=0 \Rightarrow Z=A\cos \sqrt{\lambda }z +B\sin \sqrt \lambda z \Rightarrow Z(0)=A=0 \\ \qquad \Rightarrow Z(H) =B\sin \sqrt \lambda H=0 \Rightarrow \sqrt \lambda H=n\pi \Rightarrow \lambda_n={n^2\pi^2\over H^2} \Rightarrow Z_n= \sin {n\pi z\over H},n=1,2,\dots \\ \text{Solving for }T(t): T'+c_v\lambda T=0 \Rightarrow T_n=e^{-c_v(n^2\pi^2/H^2)t} \\ \Rightarrow u(z,t) = \sum_{n=1}^\infty B_n \sin{n\pi z\over H} e^{-c_v(n^2\pi^2/H^2)t} \Rightarrow u(z,0) =\sum_{n=1}^\infty B_n \sin{n\pi z\over H} =u_0\sin{\pi z\over H} \\ \Rightarrow \cases{n=1 \Rightarrow B_1=u_0\\ n\ne1 \Rightarrow B_n=0} \Rightarrow \bbox[red, 2pt]{u(z,t)=u_0\sin{\pi z\over H} e^{-c_v(\pi^2/H^2)t}}$$
解答:$$\textbf{(a) }f(x) =\int_0^\infty [A(\omega) \cos(\omega x)+ B(\omega) \sin(\omega)] \,d\omega\\\qquad f(x)= \begin{cases} 2,& -1\lt x\lt 1 \\0, & \text{otherwise}\end{cases} \Rightarrow f(-x)=f(x) \Rightarrow f(x)\text{ is even} \Rightarrow B(\omega )=0\\ \qquad \Rightarrow A(\omega) ={1\over \pi} \int_{-\infty}^\infty f(v)\cos(\omega v)\,dv ={1\over \pi} \int_{-1}^1 2\cos(\omega v)\,dv = {2\over \pi } \left. \left[ {\sin(\omega v)\over \omega} \right] \right|_{-1}^1 ={4\sin \omega\over \pi\omega} \\ \Rightarrow \bbox[red, 2pt]{f(x)={4\over \pi} \int_0^\infty {\sin \omega \cos \omega x\over \omega}\,d\omega} \\\textbf{(b) } \hat f(\omega) = \mathcal F\{f(x)\} =\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx =\int_{-1}^1 2e^{-i\omega x} \,dx =2 \left. \left[ {e^{-i\omega} x\over -i\omega} \right] \right|_{-1}^1 ={2\over -i\omega} \left( e^{-i\omega}-e^{i\omega} \right) \\\qquad = {2\over i\omega}(2i \sin(\omega)) ={4\sin \omega\over \omega} \Rightarrow \bbox[red, 2pt]{\hat f(\omega) =4 \text{ sinc}(\omega)}$$
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解題僅供參考，碩士班歷年試題及詳解
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