國立中山大學 115學年度碩士班考試入學招生考試試題
科目名稱:工程數學【光電系碩士班】
Part 1: True or False (20 points, 2 points each question). Select A for True; select B for False.
解答:$$Ax=b \Rightarrow x=A^{-1}b \text{ is one unique vector} \Rightarrow \bbox[red, 2pt]{A}$$
解答:$$(A+B)^2 =A^2+AB+BA +B^2 \Rightarrow \bbox[red, 2pt]{B}$$
解答:$$\det(A) \ne 0 \text{ iff A is invertible }\Rightarrow \bbox[red, 2pt]{A}$$
解答:$$\text{a counter example: }\cases{A= \begin{bmatrix}1&0\\ 0&0 \end{bmatrix} \ne 0\\ B= \begin{bmatrix}0& 0\\0& 1 \end{bmatrix} \ne 0} \Rightarrow AB=0 \Rightarrow \bbox[red, 2pt]{B}$$
解答:$$\cases{v_1=[1,0]^T \\v_2= [0,1]^T} \Rightarrow v_1,v_2 \text{ span }\mathbb R^2, \text{ but } v_1 \text{ and }v_2 \text{ are linearly independent} \Rightarrow \bbox[red, 2pt]{B}$$
解答:$$\cases{U=\{(x,0), x\in R\} \\ V=\{(0,y), y\in R\}} \Rightarrow U \text{ and }V \text{ are subspace of }R^2 \Rightarrow \cases{u=(1,0) \in U\\ v=(0,1)\in V} \\\Rightarrow u+v =(1,1) \not \in U\cup V \Rightarrow U\cup V \text{ is not a subspace} \Rightarrow \bbox[red, 2pt]{B}$$
Part 2: Calculation and Proof (80 points)
解答:$$\mathbf F=(2xy+z^2,x^2+2yz, y^2+2xz) \Rightarrow \nabla\times \mathbf F= \begin{vmatrix} \vec i& \vec j& \vec k \\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\ 2xy+z^2& x^2+2yz& y^2+2xz\end{vmatrix} \\=2y\vec i+2x\vec k+2z \vec j-2x\vec k-2z\vec j-2y\vec i=0 \\ \text{From the Stoke's Theorem, } \oint_C \mathbf F\cdot d\mathbf l = \iint_S (\nabla \times \mathbf F)\,dS = \bbox[red, 2pt]{0}$$
解答:$$\textbf{(a) }F(s) ={1\over \sqrt{s^2+1}} ={1\over s\sqrt{1+{1\over s^2}}} ={1\over s} \left( 1+{1\over s^2} \right)^{-1/2} ={1\over s} \sum_{n=0}^\infty {-1/2\choose n} \left( -{1\over s^2} \right)^n \\= \sum_{n=0}^\infty {(-1)^n (2n)!\over 2^{2n}(n!)^2} \cdot {1\over s^{2n+1}} \Rightarrow f(t)=L^{-1}\{F(s)\} = \sum_{n=0}^\infty {(-1)^n (2n)!\over 2^{2n}(n!)^2} \cdot L^{-1} \left\{{1\over s^{2n+1}} \right \}\\ =\sum_{n=0}^\infty {(-1)^n (2n)!\over 2^{2n}(n!)^2} \cdot {t^{2n}\over (2n!)} =\sum_{n=0}^\infty {(-1)^n \over (n!)^2} \left( {t\over 2} \right)^{2n} =\bbox[red, 2pt]{J_0(t)} \quad \text{(Bessel function )} \\\textbf{(b) }L\left\{ {1\over \sqrt{\pi t}} e^{-1\over 4t}\right\}=I = \int_0^\infty {1\over \sqrt{\pi t}} e^{-1\over 4t} \cdot e^{-st}\,dt ={1\over \sqrt \pi} \int_0^\infty {1\over \sqrt t} e^{-(st+{1\over 4t})} \,dt \\ \text{Let }t=x^2 , \text{ then }dt =2x\,dx =2\sqrt t\,dx \Rightarrow I= {2\over \sqrt \pi} \int_0^\infty e^{-(sx^2+{1\over 4x^2})} \,dx \\ \text{Let }u=\sqrt s x \Rightarrow dx={du\over \sqrt s} \Rightarrow I={2\over \sqrt{\pi}} \int_0^\infty e^{-(u^2+{s\over 4u^2})} \cdot {du\over \sqrt s} ={2\over \sqrt{\pi s}} \int_0^\infty e^{-(u^2+{s\over 4u^2})} \,du \\={2\over \sqrt{\pi s}} \cdot {\sqrt \pi\over 2}e^{-\sqrt s} ={e^{-\sqrt s} \over \sqrt s} \Rightarrow L^{-1} \left \{ {e^{-\sqrt s} \over \sqrt s} \right\} = \bbox[red, 2pt]{{1\over \sqrt{\pi t}} e^{-1\over 4t}}$$
解答:$$\int_{-1}^1 [g_0(x)]^2\,dx =1 \Rightarrow \int_{-1}^1 a^2 \,dx =2a^2 =1 \Rightarrow a={\sqrt 2\over 2} \\ \int_{-1}^1 g_0(x)g_1(x)\,dx=0 \Rightarrow \int_{-1}^1 a(b_1+b_2 x) \,dx = a \left. \left[ b_1x+{1\over 2}b_2x^2 \right] \right|_{-1}^1=2ab_1= 0 \Rightarrow b_1=0 \Rightarrow g_1(x)=b_2x \\ \int_{-1}^1 [g_1(x)]^2\,dx =1 \Rightarrow \int_{-1}^1b_2^2 x^2 \,dx = b_2^2 \left. \left[ {1\over 3}x^3 \right] \right|_{-1}^1= {2\over 3}b_2^2=1 \Rightarrow b_2={\sqrt 6\over 2} \Rightarrow g_1(x) ={\sqrt 6\over 2}x \\\text{At the same way, } \int_{-1}^1 g_0(x)g_2(x)\,dx=0 \Rightarrow c_1=-{1\over 3}c_3, \int_{-1}^1 g_1(x)g_2(x)\,dx =0 \Rightarrow c_2=0 \\ \Rightarrow g_2(x)=c_3(x^2-{1\over 3}) \Rightarrow \int_{-1}^1 [g_2(x)]^2=1 \Rightarrow c_3={3\sqrt{10} \over 4} \Rightarrow c_1=-{1\over 3} \cdot {3\sqrt{10} \over 4}=-{\sqrt{10} \over 4} \\ \Rightarrow \bbox[red, 2pt]{\cases{a=\sqrt 2/2\\ b_1=0 \\ b_2=\sqrt 6/2\\ c_1=-\sqrt{10}/4\\c_2=0\\ c_3=3\sqrt{10}/4}}$$
解答:$$\textbf{(a) }xy''-y'=x^2e^{3x} \Rightarrow {1\over x}y''-{1\over x^2}y'=e^{3x} \Rightarrow \left( {1\over x}y' \right)'=e^{3x} \Rightarrow {1\over x}y'= \int e^{3x}\,dx ={1\over 3}e^{3x}+c_1 \\\quad \Rightarrow y'={1\over 3}xe^{3x}+c_1 x \Rightarrow y= \int \left( {1\over 3}xe^{3x}+c_1 x \right)\,dx = {1\over 9}xe^{3x}-{1\over 27}e^{3x}+{1\over 2}c_1x^2 +c_2 \\\quad \Rightarrow \bbox[red, 2pt]{y={1\over 9}xe^{3x}-{1\over 27}e^{3x}+c_3x^2+c_2} \\\textbf{(b) } x=t^2 \Rightarrow {dx\over dt}=2t \Rightarrow y'={dy\over dt} ={dy\over dx}{dx\over dt} =2t{dy\over dx} \Rightarrow y''=2{dy\over dx}+ 4t^2{d^2y\over dx^2} \\ y''+(4t-t^{-1})y'+8t^2 y= 2{dy\over dx}+ 4t^2{d^2y\over dx^2}+ (4t-t^{-1})(2t{dy\over dx})+8t^2y=0 \\ \quad \Rightarrow 4t^2 {d^2y\over dx^2} +8t^2 {dy\over dx}+8t^2y =0 \Rightarrow {d^2 y\over dx^2}+2{dy\over dx}+2y=0 \Rightarrow r^2+2r+2=0 \\ \quad \Rightarrow r=-1\pm i \Rightarrow y=e^{-x}(c_1\cos x+c_2 \sin x) \Rightarrow \bbox[red, 2pt]{y=e^{-t^2} (c_1 \cos t^2+ c_2 \sin t^2)}$$
解答:$$\text{The general solution for Laplace's equation in a full circular domain is:} \\ u(r,\theta)= A_0+ B_0\ln r+ \sum_{n=1}^\infty [(A_nr^n+B_nr^{-n}) \cos(n\theta)+(C_nr^n+ D_nr^{-n}) \sin(n\theta)] \\ u(1,\theta)=0 \Rightarrow A_0+ \sum_{n=1}^\infty [(A_n +B_n ) \cos(n\theta)+(C_n + D_n ) \sin(n\theta)] =0 \\ \Rightarrow \cases{A_0=0\\ A_n+B_n=0 \Rightarrow B_n=-A_n\\ C_n+D_n =0 \Rightarrow D_n=-C_n} \\ \Rightarrow u(r,\theta)= B_0\ln r+ \sum_{n=1}^\infty [A_n(r^n-r^{-n}) \cos(n\theta)+ C_n(r^n-r^{-n}) \sin(n\theta)] \\ u(0.5, \theta)= \cos^2 \theta={1\over 2}+{1\over 2}\cos(2\theta) \\ \Rightarrow B_0\ln (0.5)+ \sum_{n=1}^\infty [A_n(0.5^n-0.5^{-n}) \cos(n\theta)+ C_n(0.5^n-0.5^{-n}) \sin(n\theta)] ={1\over 2}+{1\over 2}\cos(2\theta) \\ n=0 \Rightarrow B_0 \ln(0.5)={1\over 2 } \Rightarrow B_0=-{1\over 2\ln 2} \\ n=2 \Rightarrow A_2(0.5^2-0.5^{-2}) ={1\over 2} \Rightarrow A_2=-{2\over 15} \\ \Rightarrow \bbox[red, 2pt]{u(r,\theta)= \left( -{1\over 2\ln 2} \right) \ln r+ \left( -{1\over 15} \right)(r^2-r^{-2}) \cos (2\theta)}$$
解答:$$ \cases{\nabla \times A = B + iA \\\nabla \times B = -A + iB} \Rightarrow \nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B)= B\cdot(B+iA)-A\cdot(-A+iB) \\=(B \cdot B) + i(B \cdot A) - (-A \cdot A + i(A \cdot B)) = |B|^2 + i(A \cdot B) + |A|^2 - i(A \cdot B) = |A|^2 + |B|^2\\|A - B|^2 \ge 0 \Rightarrow (A - B) \cdot (A - B) \ge 0 \Rightarrow |A|^2 - 2(A \cdot B) + |B|^2 \ge 0 \Rightarrow |A|^2 + |B|^2 \ge 2(A \cdot B) \\ \Rightarrow 2(|A|^2 + |B|^2) \ge |A|^2 + 2(A \cdot B) + |B|^2 =|A+B|^2 \Rightarrow |A|^2 + |B|^2 \ge \frac{1}{2} |A + B|^2 \\\int_V \nabla \cdot (A \times B) dV = \int_V (|A|^2 + |B|^2) dV \Rightarrow \int_V \nabla \cdot (A \times B) dV \ge \frac{1}{2} \int_V |A + B|^2 dV \\\text{From the Divergence Theorem,}\int_V \nabla \cdot (A \times B) dV = \int_S (A \times B) \cdot ds \\ \Rightarrow \int_S (A \times B) \cdot ds \ge \frac{1}{2} \int_V |A + B|^2 dV\qquad \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(a) }\text{Assume }a\gt b\gt 0 \Rightarrow I= \int_0^\pi {d\theta\over (a+b\cos \theta)^2} ={1\over 2} \int_0^{2\pi} {d\theta\over (a+b\cos \theta)^2} \\ ={1\over 2} \oint_{|z|=1} {1\over \left( a+b \left( {z^2+1\over 2z} \right) \right)^2} {dz\over iz} ={2\over i} \oint_{|z|=1} {z\over (bz^2+2az+b)^2}\,dz \quad \left( z=e^{i\theta} \Rightarrow \cos \theta={z^2+1\over 2z} \right)\\ bz^2+2az+b^2=0 \Rightarrow z= {-a\pm \sqrt{a^2-b^2}\over b} \Rightarrow \alpha= {-a+ \sqrt{a^2-b^2}\over b}, \beta={-a- \sqrt{a^2-b^2}\over b}\\ \text{Let }f(z) ={z\over (bz^2+2az+b)^2} \Rightarrow \text{Res}(f,\alpha) ={1\over b^2}\cdot {-(\alpha+\beta) \over (\alpha-\beta)^3} = {a\over 4(a^2-b^2)^{3/2}} \\ \Rightarrow \oint_{|z|=1} f(z)\,dz = 2\pi i \cdot \text{Res}(f,\alpha) =2\pi i \left( {a\over 4(a^2-b^2)^{3/2}} \right) ={\pi a i\over 2(a^2-b^2)^{3/2}} \\ \Rightarrow I={2\over i} \left( {\pi a i\over 2(a^2-b^2)^{3/2}} \right) = \bbox[red, 2pt]{\pi a\over (a^2-b^2)^{3/2}} \\\textbf{(b) }\text{Let }I=\int_0^\infty {x^{1/p} \over x^2+1}\,dx \text{ and }f(z) ={z^{1/p} \over z^2+1}.\\ \text{We integrate over a "keyhole" contour $C$ consisting of:} \\\quad \cases{C_1: \text{A path along the top of the positive real axis from }x=\epsilon\text{ to }R\\ C_R: \text{A large circle of radius }R\\ C_2:\text{A path along the bottom of the positive real axis from $R$ back to $\epsilon$} \\ C_\epsilon: \text{A small circle of radius $\epsilon$ around the origin, connecting back to }C_1} \\\text{As }R\to \infty \text{ and }\epsilon \to 0, \text{ the integrals over }C_R\text{ and }C_\epsilon \text{ vanish. Then} \\ \oint_C f(z)\,dz = \int_{C_1}f(z)\,dz +\int_{C_2} f(z)dz =\int_0^\infty {x^{1/p} \over x^2+1}-\int_0^\infty {x^{1/p}e^{2\pi i/p}\over x^2+1} \,dx =I(1-e^{2\pi i/p}) \\ \cases{\text{Res}(f,i)= {e^{\pi i/(2p)} \over 2i} \\\text{Res}(f,-i)= -{e^{3\pi i/(2p)}\over 2i}} \Rightarrow \oint_C f(z)\,dz =2\pi i \left( {e^{\pi i/(2p)} \over 2i}- {e^{3\pi i/(2p)}\over 2i}\right) =I(1-e^{2\pi i/p}) \\ \Rightarrow I= \bbox[red, 2pt]{{\pi\over 2} \sec {\pi\over 2p}}$$
解答:$$\text{Using Fourier transform to solve this problem. }\\ f(x)=e^{i\omega_1 x}+e^{i\omega_2 x} \Rightarrow \mathcal F\{f(x)\} =F(k)=2\pi \delta(k-\omega_1)+2\pi \delta(k-\omega_2\\ g(x) ={1\over x^2+a^2} \Rightarrow \mathcal F\{g(x)\} =G(k)={\pi\over a}e^{-a|k|} \\ H(k) =F(k)\cdot G(k) ={2\pi^2\over a}e^{-a|k|} \delta(k-\omega_1) +{2\pi^2\over a}e^{-a|k|} \delta(k-\omega_2) \\={2\pi^2\over a}e^{-a|\omega_1|} \delta(k-\omega_1) +{2\pi^2\over a}e^{-a|\omega_2|} \delta(k-\omega_2) \\ \Rightarrow h(x)= \mathcal F^{-1}\{H(k)\} = {1\over 2\pi} \int_{-\infty}^\infty H(k)e^{ikx}\,dk ={1\over 2\pi} \left( {2\pi^2\over a}e^{-a|\omega_1| }e^{i\omega_1x} +{2\pi^2\over a} e^{-a|\omega_2|} e^{i\omega_2 x}\right) \\=\bbox[red, 2pt]{{\pi\over a} \left( e^{i\omega_1x-a|\omega_1|} +e^{i \omega_2 x -a|\omega_2|} \right)}$$
解題僅供參考，其他碩士班題及詳解
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