114年中央光電碩士班-工程數學詳解

國立中央大學114學年度碩士班考試入學

系所：光電類
科目：工程數學
本科考試可使用計算機，廠牌、功能不拘

解答:$$U(\xi, \eta) =\cases{1, \text{ if } \left(-{a\over 2}\le \xi \le {a\over 2} \right) \cap \left( -{b\over 2} \le \eta\le {b\over 2}\right);\\ 0, \text{ otherwise.}} \Rightarrow U(\xi, \eta) = \text{rect} \left({\xi\over a} \right) \text{rect} \left({\eta\over b} \right) \\ \Rightarrow U(x,y)=C \int_{-b/2}^{b/2} \int_{-a/2}^{a/2} 1\cdot e^{-2\pi j(x \xi+ y\eta)/\lambda z} \,d\xi\,d\eta, C={e^{kj(x^2+y^2)/2+jkz} \over j\lambda z} \\\qquad = C \left( \int_{-a/2}^{a/2} e^{-2\pi j x\xi/\lambda z} \,d\xi\right) \times \left( \int_{-b/2}^{b/2} e^{-2\pi j y\eta/\lambda z}\, d\eta\right) = C \left( a\cdot \text{sinc}{ax\over \lambda z} \right) \left(b\cdot \text{sinc}{by\over \lambda z} \right) \\= {e^{kj(x^2+y^2)/2+jkz} \over j\lambda z}(ab)\text{sinc} \cdot{ax\over \lambda z} \cdot \text{sinc}{by\over \lambda z} \Rightarrow I(x,y) = \left| {e^{kj(x^2+y^2)/2+jkz} \over j\lambda z}(ab)\text{sinc} \cdot{ax\over \lambda z} \cdot \text{sinc}{by\over \lambda z} \right|^2 \\= \bbox[red, 2pt]{\left({ab\over \lambda z} \right)^2 \text{sinc}^2 \left({ax\over \lambda z} \right)  \text{sinc}^2\left({by\over \lambda z} \right)}$$

解答:$$y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=1,2 \\ \Rightarrow y= c_1e^t+c_2e^{2t} \Rightarrow y'=c_1e^t+ 2c_2e^{2t} \Rightarrow \cases{y(0) =c_1+c_2=1\\ y'(0)=c_1+ 2c_2=0} \Rightarrow \cases{c_1=2\\ c_2=-1} \\ \Rightarrow y= 2e^t-e^{2t} =-e^{2t}(1-2e^{-t}), 故選\bbox[red, 2pt]{(b)}$$

解答:$$y''-2y'+2y=0 \Rightarrow \lambda^2-2\lambda+2=0 \Rightarrow \lambda=1\pm i \Rightarrow y=e^t(c_1\cos t+c_2\sin t) \\ \Rightarrow y'=e^t((c_1+c_2)\cos t+(c_2-c_1)\sin t)  \Rightarrow \cases{y(0) =c_1=1\\ y'(0)=c_1+c_2=0} \Rightarrow \cases{c_1=1 \\ c_2=-1} \\ \Rightarrow y=e^t(\cos t -\sin t)，故選\bbox[red, 2pt]{(b)}$$

解答:$$y''+5y'+6y=0 \Rightarrow \lambda^2+5 \lambda+6=0 \Rightarrow (\lambda+2)( \lambda+3) =0 \Rightarrow \lambda=-2,-3 \\ \Rightarrow y_h=c_1e^{-2t} +c_2e^{-3t} \\ y_p=Ae^{-t} \Rightarrow y_p'=-Ae^{-t}  \Rightarrow y_p''=Ae^{-t} \Rightarrow y_p''+5y_p'+6y_p= 2Ae^{-t} =2e^{-t} \Rightarrow A=1 \\ \Rightarrow y_p=e^{-t} \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{-2t} +c_2e^{-3t}+e^{-t}}$$

解答:$$y= c_1e^{-2t} +c_2e^{-3t}+e^{-t} \Rightarrow y'=-2c_1e^{-2t}-3c_2e^{-3t}-e^{-t} \Rightarrow \cases{y(0)=c_1+c_2+1= 0\\ y'(0)= -2c_1-3c_2-1=0}  \\ \Rightarrow \cases{c_1= -2\\ c_2= 1} \Rightarrow y= -2e^{-2t}+e^{-3t}+e^{-t} =e^{-t}(-2e^{-t}+e^{-2t}+1), 故選\bbox[red, 2pt]{(a)}$$

解答:$$y= c_1e^{-2t} +c_2e^{-3t}+e^{-t} \Rightarrow y'=-2c_1e^{-2t}-3c_2e^{-3t}-e^{-t} \Rightarrow \cases{y(0)=c_1+c_2+1= 1\\ y'(0)= -2c_1-3c_2-1=0}  \\ \Rightarrow \cases{c_1= 1\\ c_2= -1} \Rightarrow y= e^{-2t}-e^{-3t}+e^{-t} =e^{-t}(e^{-t}-e^{-2t}+1), 故選\bbox[red, 2pt]{(a)}$$

解答:$$aL\{y''\}+ bL\{y'\}+ cL\{y\}=a \left( s^2Y(s)-sy(0)-y'(0)\right) +b (sY(s)-y(0)) +cY(s)=L\{g(t)\} \\ \Rightarrow (as^2+bs +c)Y(s)-ay_0s-au_0-by_0 = L\{g(t)\} \\\Rightarrow \bbox[red, 2pt]{Y(s) ={1\over as^2+bs+c} \left( L\{g(t)\}+ay_0s+ au_0+by_0\right)}$$

解答:$$y= \sum_{n=0}^\infty a_n x^n \Rightarrow y'= \sum_{n=0}^\infty na_n x^{n-1} \Rightarrow y''= \sum_{n=0}^\infty n(n-1)a_n x^{n-2} \\ \Rightarrow y''-y=\sum_{n=0}^\infty n(n-1)a_n x^{n-2} -\sum_{n=0}^\infty a_n x^n =\sum_{n=0}^\infty \left((n+2)(n+1)a_{n+2}-a_n \right) x^n =0 \\ \Rightarrow (n+2)(n+1)a_{n+2}-a_n  =0 \Rightarrow a_{n+2} ={a_n\over (n+2) (n+1)}, n=0,1,2, \dots \\ \textbf{Case I }n=2k \Rightarrow \cases{n=0 \Rightarrow a_2=a_0/2!\\ n=2 \Rightarrow a_4=a_2/(4\cdot 3) =a_0/4!\\ n=4 \Rightarrow a_6= a_4/(6\cdot 5) =a_0/6! \\ \cdots} \Rightarrow a_{2k} ={a_0\over (2k)!}, k=0,1,2,\dots \\\textbf{Case II }n=2k+1 \Rightarrow \cases{n=1 \Rightarrow a_3=a_1/3!\\ n=3 \Rightarrow a_5=a_1/5!\\ n=5 \Rightarrow a_7=a_1/7!\\ \cdots} \Rightarrow a_{2k+1}={a_1\over (2k+1)!}, k=0,1,2,\dots \\ \Rightarrow y(x) = \sum_{k=0}^\infty {a_0\over (2k)!}x^{2k} +\sum_{k=0}^\infty {a_1\over (2k+1)!} x^{2k+1} =a_0 \cosh x+ a_1 \sinh x \quad \bbox[red, 2pt]{QED.}$$

解答:$$\cases{A(2,6,1) \\B(3,1,7) \\C(4,3,9) \\D(5,2,8)} \Rightarrow \cases{ \vec u= \overrightarrow{AB} =(1,-5,6) \\\vec v= \overrightarrow{AC} =(2,-3,8) \\ \vec w=\overrightarrow{AD} =(3,-4,7)} \Rightarrow V={1\over 6}|(\vec u\times \vec v) \cdot \vec w|= {1\over 6}|(-22,4,7) \cdot (3,-4,7)| = \bbox[red, 2pt]{11\over 2}$$

解答:$$\text{Divergence Theorem: } \iint_S \vec F\cdot \vec n\,dA = \iiint \nabla \cdot \vec f \,dV = \iiint (-3x^2-3y^2-3z^2) \,dV \\= \int_0^{2\pi} \int_0^{\pi/2} \int_0^3 -3\rho^2 \cdot \rho^2 \sin \phi \,d\rho d\phi d\theta =-3 \left(\int_0^{2\pi}\,d\theta \right) \left( \int_0^{\pi/2}\sin \rho\right) \left( \int_0^3 \rho^4\, d\rho\right) \\=-3\cdot 2\pi\cdot 1\cdot {243\over 5} = \bbox[red, 2pt]{-{1458\pi \over 5}}$$

解答:$$\textbf{(a) }A_1 \begin{bmatrix} x_1\\ x_2\end{bmatrix} =\begin{bmatrix} x_1\\ 3x_2\end{bmatrix} \Rightarrow A_1 = \bbox[red, 2pt]{\begin{bmatrix} 1&0\\ 0& 3\end{bmatrix}} \\\textbf{(b) } A_2= \begin{bmatrix} \cos \pi/6 &-\sin \pi/6\\ \sin \pi/6 & \cos \pi/6\end{bmatrix} = \bbox[red, 2pt] {\begin{bmatrix} {\sqrt 3\over 2}& -{1\over 2} \\{1\over 2}& {\sqrt 3\over 2}\end{bmatrix}} \\\textbf{(c) }A_3= A_2 A_1 =\begin{bmatrix} 1&0\\ 0& 3\end{bmatrix} \begin{bmatrix} {\sqrt 3\over 2}& -{1\over 2} \\{1\over 2}& {\sqrt 3\over 2}\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} {\sqrt 3\over 2}& -{1\over 2} \\{3 \over 2}& {3\sqrt 3\over 2}\end{bmatrix}} \\\textbf{(d) } P\in \Gamma:x_1^2+x_2^2=1 \Rightarrow P=[\cos \theta \; \sin \theta]^T \Rightarrow A_3 P = \begin{bmatrix} {\sqrt 3\over 2}& -{1\over 2} \\{3 \over 2}& {3\sqrt 3\over 2}\end{bmatrix} \begin{bmatrix} \cos \theta \\ \sin \theta\end{bmatrix}\\\quad = \begin{bmatrix}{1\over 2}(\sqrt 3 \cos \theta-\sin \theta) \\{3\over 2}(\cos \theta+ \sqrt 3\sin \theta)  \end{bmatrix} = \begin{bmatrix} \cos(\pi/6) \cos \theta-\sin \pi/6 \sin \theta\\ 3(\sin \pi/6 \cos \theta+ \cos \pi/6 \sin \theta) \end{bmatrix} = \begin{bmatrix} \cos \left( {\pi\over 6}+\theta \right)\\ 3 \sin \left({\pi\over 6}+\theta  \right) \end{bmatrix} = \begin{bmatrix} y_1\\ y_2\end{bmatrix} \\ \quad \Rightarrow y_1^2+{y_2^2 \over 9} =1 \Rightarrow \text{ resulting figure is an ellipse }$$

解答:$$\textbf{(a) } \det(A) = 20+24+84-28-16-90= \bbox[red, 2pt]{-6} \\\textbf{(b) } \det(A-\lambda I) =-\lambda^3+7\lambda -6=-(\lambda-1) (\lambda-2) (\lambda+3) =0 \Rightarrow \lambda=1, -3,2\\\quad \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}4 & 4 & -4 \\-1 & -2 & 3 \\7 & 6 & -5 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2=2x_3} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} -1\\ 2\\ 1\end{pmatrix}, \text{ choose }v_1= \begin{pmatrix} -1\\ 2\\ 1\end{pmatrix} \\\quad \lambda_2=-3  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}8 & 4 & -4 \\-1 & 2 & 3 \\7 & 6 & -1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2+ x_3=0} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 1\\ -1\\ 1\end{pmatrix}, \text{ choose }v_2= \begin{pmatrix} 1\\ -1\\ 1\end{pmatrix} \\\quad \lambda_3=2  \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix}3 & 4 & -4 \\-1 & -3 & 3 \\7 & 6 & -6 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1= 0\\ x_2= x_3 } \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 0\\  1\\ 1\end{pmatrix}, \text{ choose }v_3= \begin{pmatrix} 0\\ 1\\ 1\end{pmatrix} \\ \Rightarrow \text{ eigenvalues: } \bbox[red, 2pt]{1,-3,2}, \text{ and the corresponding eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix} -1\\ 2\\ 1\end{pmatrix}, \begin{pmatrix} 1\\ -1\\ 1\end{pmatrix}, \begin{pmatrix} 0\\ 1\\ 1\end{pmatrix}} \\\textbf{(c) }  \mathbf X= [v_1 v_2 v_3] \Rightarrow \bbox[red, 2pt]{\mathbf X= \begin{bmatrix}-1 & 1 & 0 \\2 & -1 & 1 \\1 & 1 & 1 \end{bmatrix}}, \mathbf D= \begin{bmatrix} \lambda_1 & 0 & 0\\ 0& \lambda_2 & 0\\ 0& 0& \lambda_3\end{bmatrix} \Rightarrow  \bbox[red, 2pt]{\mathbf {D= \begin{bmatrix} 1 & 0 & 0 \\0 & -3 & 0 \\0 & 0 & 2\end{bmatrix}}} \\\textbf{(d) } \mathbf [X\mid I]=\left[ \begin{array}{rrr|rrr}-1 & 1 & 0&1& 0& 0 \\2 & -1 & 1&0& 1& 0 \\1 & 1 & 1 &0&0& 1\end{array}  \right]\xrightarrow{2R_1+R_2 \to R_2, R_1+R_3\to R_3} \left[ \begin{array}{rrr|rrr}-1 & 1 & 0 & 1 & 0 & 0\\0 & 1 & 1 & 2 & 1 & 0\\0 & 2 & 1 & 1 & 0 & 1\end{array}  \right] \\\quad \xrightarrow{R_1-R_2 \to R_1, R_3-2R_2 \to R_3} \left[ \begin{array}{rrr|rrr}-1 & 0 & -1 & -1 & -1 & 0\\0 & 1 & 1 & 2 & 1 & 0\\0 & 0 & -1 & -3 & -2 & 1\end{array}  \right] \xrightarrow{-R_1\to R_1, -R_3\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 1 & 0\\0 & 1 & 1 & 2 & 1 & 0\\0 & 0 & 1 & 3 & 2 & -1\end{array}  \right] \\ \quad \xrightarrow{R_1-R_3\to R_1, R_2-R_3\to R_2}  \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -2 & -1 & 1\\0 & 1 & 0 & -1 & -1 & 1\\0 & 0 & 1 & 3 & 2 & -1\end{array}  \right]  \Rightarrow \bbox[red, 2pt]{\mathbf X^{-1} = \begin{bmatrix} -2&-1&1\\ -1&-1& 1\\3&2 & -1\end{bmatrix}} \\\textbf{(e) } \mathbf A= \mathbf {XDX^{-1}} \Rightarrow \bbox[red, 2pt]{A^8 =XD^8X^{-1}}$$

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