國立高雄大學 115 學年度轉學招生考試試題(轉三年級)
科目： 工程數學
系所：電機工程學系(無組別)
解答:$$\textbf{(a) }\bigcirc: \bar a_1+\bar a_2+\bar a_3= \bar 0 \Rightarrow \bar a_1,\bar a_2, \text{ and }\bar a_3 \text{ are linearly dependent }\Rightarrow A^{-1} \text{ does not exist}\\\qquad \Rightarrow \det(A)=0 \\ \textbf{(b) }\times: \text{no solution }\ne \text{ at least one solution} \\ \textbf{(c) }\times: A= \begin{bmatrix}0& 1\\0&0 \end{bmatrix} \Rightarrow N(A)=R(A)= \text{span} \left\{ \begin{bmatrix}1\\0 \end{bmatrix} \right\} \Rightarrow N(A) \not \bot R(A) \\ \textbf{(d) }\times: AA^T \text{ is a projection matrix } \Rightarrow rank(AA^T)=3 \Rightarrow AA^T \ne I_5\\ \text{Correct statements: }\bbox[red, 2pt]{(a)}$$
解答:$$ A = \begin{bmatrix} 1 & 0 & -3 \\ -2 & 2 & 6 \\ -1 & -6 & 1 \end{bmatrix} \xrightarrow{R_2 \gets R_2+2R_1\\ R_3 \gets R_3+R_1} \begin{bmatrix} 1 & 0 & -3 \\ 0 & 2 & 0 \\ 0 & -6 & -2 \end{bmatrix} \xrightarrow{R_3\gets R_3+3R_2} \begin{bmatrix} 1 & 0 & -3 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{U = \begin{bmatrix} 1 & 0 & -3 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix} }, L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{L= \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & -3 & 1 \end{bmatrix} }$$
解答:$$ \mathbf{u}_1 = \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 2 \\ 4 \end{bmatrix} \Rightarrow \mathbf{u}_2 = \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix} - \frac{10}{25} \begin{bmatrix} 1 \\ 2 \\ 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -2/5 \\ 1/5 \\ -4/5 \\ 2/5 \end{bmatrix} \\ \text{let } \mathbf{u}_2' = 5\mathbf{u}_2 = \begin{bmatrix} -2 \\ 1 \\ -4 \\ 2 \end{bmatrix} \Rightarrow \mathbf{u}_3 = \mathbf{v}_3 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_2'}{\mathbf{u}_2' \cdot \mathbf{u}_2'} \mathbf{u}_2' = \begin{bmatrix} -1 \\ 2 \\ 2 \\ -4 \end{bmatrix} - \frac{-9}{25} \begin{bmatrix} 1 \\ 2 \\ 2 \\ 4 \end{bmatrix} - \frac{-12}{25} \begin{bmatrix} -2 \\ 1 \\ -4 \\ 2 \end{bmatrix} \\ = \begin{bmatrix}-8/5\\ 16/5\\ 4/5\\ -8/5 \end{bmatrix}, \text{ let } \mathbf{u}_3' = \frac{5}{4}\mathbf{u}_3 = \begin{bmatrix} -2 \\ 4 \\ 1 \\ -2 \end{bmatrix} \Rightarrow \mathbf{q}_1 = \frac{\mathbf{u}_1}{\Vert{}\mathbf{u}_1\Vert{}} = \frac{1}{5} \begin{bmatrix} 1 \\ 2 \\ 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 1/5 \\ 2/5 \\ 2/5 \\ 4/5 \end{bmatrix} ,\\ \mathbf{q}_2 = \frac{\mathbf{u}_2'}{\Vert{}\mathbf{u}_2'\Vert{}} = \frac{1}{5} \begin{bmatrix} -2 \\ 1 \\ -4 \\ 2 \end{bmatrix} = \begin{bmatrix} -2/5 \\ 1/5 \\ -4/5 \\ 2/5 \end{bmatrix}, \mathbf{q}_3 = \frac{\mathbf{u}_3'}{\Vert{}\mathbf{u}_3'\Vert{}} = \frac{1}{5} \begin{bmatrix} -2 \\ 4 \\ 1 \\ -2 \end{bmatrix} = \begin{bmatrix} -2/5 \\ 4/5 \\ 1/5 \\ -2/5 \end{bmatrix} \\ \Rightarrow \text{an orthonormal basis: }\{\mathbf q_1, \mathbf q_2, \mathbf q_3 \} = \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1/5 \\ 2/5 \\ 2/5 \\ 4/5 \end{bmatrix}, \begin{bmatrix} -2/5 \\ 1/5 \\ -4/5 \\ 2/5 \end{bmatrix}, \begin{bmatrix} -2/5 \\ 4/5 \\ 1/5 \\ -2/5 \end{bmatrix} \right\}}$$
解答:$$A= \begin{bmatrix}3& 1& 4\\ -1& 1& -1 \\ 0& 0& -1\end{bmatrix} \Rightarrow \det(A-\lambda I) = -(\lambda + 1)(\lambda - 2)^2 = 0 \Rightarrow \lambda=-1,2 \\ \lambda_1= -1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 4 & 1 & 4 \\ -1 & 2 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_2=0\\ x_1+x_3=0} \\\qquad \Rightarrow v= x_1 \begin{bmatrix}1\\0\\ -1 \end{bmatrix} \Rightarrow \text{ choosing }v_1= \begin{bmatrix}1\\0\\-1 \end{bmatrix} \\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} 1 & 1 & 4 \\ -1 & -1 & -1 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1+x_2=0\\ x_3=0 }\\ \qquad \Rightarrow v = x_1 \begin{bmatrix}-1\\ 1\\0 \end{bmatrix} \Rightarrow \text{ choosing }v_2= \begin{bmatrix} -1\\1\\0 \end{bmatrix} \\ \text{$A$ is $3\times 3$. but there are only two independent eigenvectors. Thus, $A$ is not diagonlizable.}\\ \text{Now , we want to find $v_3$ such that }(A-\lambda_2 I)v_3=v_2, \begin{bmatrix} 1 & 1 & 4 \\ -1 & -1 & -1 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\\ \Rightarrow \cases{x_3=0 \\ x_1+x_2=-1} \Rightarrow \text{ choosing }v_3= \begin{bmatrix}-1 \\0\\0 \end{bmatrix} \Rightarrow X= [v_1\; v_2\; v_3] \\ \Rightarrow \bbox[red, 2pt]{X = \begin{bmatrix} 1 & -1 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{bmatrix}, \quad J = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} }$$
解答:$$v=y^{-1} \Rightarrow {dv\over dx}=-y^{-2}{dy\over dx} \Rightarrow x\frac{dy}{dx} - (1+x)y = xy^2 \Rightarrow \frac{dy}{dx} - \left(\frac{1}{x} + 1\right)y = y^2 \\\Rightarrow -y^{-2}\frac{dy}{dx} + \left(\frac{1}{x} + 1\right)y^{-1} = -1 \Rightarrow \frac{dv}{dx} + \left(\frac{1}{x} + 1\right)v = -1 \\ \text{integrating factor }I(x) = e^{\int \left(\frac{1}{x} + 1\right) dx} = e^{\ln x + x} = x e^x \Rightarrow x e^x \frac{dv}{dx} + x e^x \left(\frac{1}{x} + 1\right)v = -x e^x \\ \Rightarrow \frac{d}{dx} (v \cdot x e^x) = -x e^x \Rightarrow v \cdot x e^x = \int -x e^x \, dx= -x e^x + e^x + C \\ \Rightarrow v=y^{-1} = \frac{-x e^x + e^x + C}{x e^x} \Rightarrow y={xe^x \over -xe^x+e^x+C} \Rightarrow y(1)={e\over C}=1 \Rightarrow C=e \\ \Rightarrow \bbox[red, 2pt]{y = \frac{x e^x}{(1-x)e^x + e} }$$
解答:$$y = x^m\Rightarrow y' = mx^{m-1} \Rightarrow y'' = m(m-1)x^{m-2} \Rightarrow x^2(m(m-1)x^{m-2}) - 2x(mx^{m-1}) + 2x^m = 0 \\ \Rightarrow (m^2-3m+2)x^m=0 \Rightarrow m^2-3m+2=0 \Rightarrow (m-2)(m-1)=0 \Rightarrow m=1,2\\ \Rightarrow y_h = c_1 x + c_2 x^2 \Rightarrow \cases{y_1=x\\ y_2=x^2} \Rightarrow W= \begin{vmatrix} y_1& y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} x& x^2\\ 1& 2x \end{vmatrix}=x^2 \\ \text{Using variation of parameters, }y_p =-x \int{x^2 \ln x\over x\cdot x^2} \,dx+ x^2 \int {x\ln x\over x\cdot x^2}\,dx =-{x\over 2}(\ln x)^2-x\ln x-x \\ \Rightarrow y= y_h+y_p = c_1 x + c_2 x^2 - \frac{x}{2}(\ln x)^2 - x\ln x \Rightarrow y'= c_1 + 2c_2 x - \frac{1}{2}(\ln x)^2 - 2\ln x - 1 \\ \Rightarrow \cases{y(1)=c_1+c_2=1\\ y'(1)= c_1+2c_2-1=0} \Rightarrow \cases{c_1=1\\ c_2=0} \Rightarrow \bbox[red, 2pt]{y = x - \frac{x}{2}(\ln x)^2 - x\ln x}$$
解答:$$ L\{y'' + 6y' + 5y\} = L\{x - xU(x - 2)\}\\ \Rightarrow [s^2Y(s) - sy(0) - y'(0)] + 6[sY(s) - y(0)] + 5Y(s) = L\{x\} - L\{xU(x - 2)\} \\ \Rightarrow [s^2Y(s) - s] + 6[sY(s) - 1] + 5Y(s) = \frac{1}{s^2} - L\{(x-2+2)U(x - 2)\} \\ \Rightarrow Y(s)(s^2 + 6s + 5) - s - 6 = \frac{1}{s^2} - \left[ L\{(x-2)U(x - 2)\} + L\{2U(x - 2)\} \right] \\ \Rightarrow Y(s)(s + 1)(s + 5) - (s + 6) = \frac{1}{s^2} - e^{-2s}\left(\frac{1}{s^2}\right) - e^{-2s}\left(\frac{2}{s}\right) \\ \Rightarrow Y(s) = \frac{s + 6}{(s + 1)(s + 5)} + \frac{1}{s^2(s + 1)(s + 5)} - e^{-2s}\frac{2s + 1}{s^2(s + 1)(s + 5)} \\= \left( \frac{5/4}{s + 1} - \frac{1/4}{s + 5} \right)+ \left( -\frac{6/25}{s} + \frac{1/5}{s^2} + \frac{1/4}{s + 1} - \frac{1/100}{s + 5} \right)- e^{-2s} \left( \frac{4/25}{s} + \frac{1/5}{s^2} - \frac{1/4}{s + 1} + \frac{9/100}{s + 5} \right) \\= -\frac{6}{25} \cdot \frac{1}{s} + \frac{1}{5} \cdot \frac{1}{s^2} + \frac{3}{2} \cdot \frac{1}{s + 1} - \frac{13}{50} \cdot \frac{1}{s + 5} - e^{-2s} \left( \frac{4/25}{s} + \frac{1/5}{s^2} - \frac{1/4}{s + 1} + \frac{9/100}{s + 5} \right) \\ \Rightarrow y(x) =L^{-1} \{ Y(s)\} \\ \Rightarrow \bbox[red, 2pt]{y(x) = -\frac{6}{25} + \frac{1}{5}x + \frac{3}{2}e^{-x} - \frac{13}{50}e^{-5x} - \left[ -\frac{6}{25} + \frac{1}{5}x - \frac{1}{4}e^{-(x-2)} + \frac{9}{100}e^{-5(x-2)} \right] U(x-2) }$$
解答:$$a_0 = \frac{1}{L} \int_{0}^{2L} f(x) \, dx = \int_{0}^{2} (3 - x^2) \, dx={10\over 3} \\ a_n = \frac{1}{L} \int_{0}^{2L} f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx = {1\over 1} \int_{0}^{2} (3 - x^2) \cos(n\pi x) \, dx = -{4\over n^2\pi^2 } \\ b_n = \frac{1}{L} \int_{0}^{2L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx = \int_{0}^{2} (3 - x^2) \sin(n\pi x) \, dx ={4\over n\pi} \\ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right] = \bbox[red, 2pt]{\frac{5}{3} + \sum_{n=1}^{\infty} \left[ -\frac{4}{n^2\pi^2} \cos(n\pi x) + \frac{4}{n\pi} \sin(n\pi x) \right] }$$
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