109年國立台北大學轉學考-工程數學詳解

國立臺北大學109學年度日間學士班轉學生招生

學制系級：通訊工程學系日間學士班3年級
科目：工程數學

解答： $$\mathbf{(1)}\;\cases{z_1=-2-10i\\ z_2= 5-i} \Rightarrow z=\bar z_1/\bar z_2 = {-2+10i\over 5+i} ={(-2+10i)(5-i) \over (5+i) (5-i)} = { 52i\over 26} =0+2i \Rightarrow \bbox[red,2pt]{\cases{x=0\\ y=2}} \\\mathbf{(2)}\; \cases{z_1= -2-10i \\ z_2=5-2}  \Rightarrow z_1/z_2 = {-2-10i\over 5-i} = {(-2-10i)(5+i) \over (5-i)(5+i)} ={-52i\over 26} =-2i \\\quad \Rightarrow \overline{z_1/z_2}= 0+2i \Rightarrow \bbox[red,2pt]{\cases{x=0\\ y=2}} \\\mathbf{(3)}\; \ln z=e+\pi i \Rightarrow z=e^{e+\pi i} =e^e\cdot e^{\pi i} =e^e(\cos \pi +i\sin \pi)= -e^e \Rightarrow  \bbox[red,2pt]{\cases{x=-e^e\\ y=0}} \\\mathbf {(4)}\; \ln z= 0.4+0.2i \Rightarrow z=e^{0.4+ 0.2i} =e^{0.4} (\cos 0.2+i\sin 0.2) \Rightarrow \bbox[red,2pt]{\cases{x=e^{0.4}\cos 0.2\\ y=e^{0.4}\sin 0.2}}$$

解答： $$\mathbf{(1)}\;\cases{u=e^x\cos y\\ v= e^x\sin y} \Rightarrow \cases{u_x=e^x\cos y\\ u_y= -e^x\sin y\\ v_x=e^x\sin y\\ v_y=e^x\cos y} \Rightarrow \cases{u_x=v_y\\ u_y=-v_x} \Rightarrow f(z) \bbox[red,2pt]{\text{analytic}} \\\mathbf{(2)}\; f(z)=z^2=(x+iy)^2 = x^2-y^2+2xyi \Rightarrow \cases{u=x^2-y^2\\ v=2xy} \Rightarrow \cases{u_x=2x\\ u_y=-2y\\ v_x=2y\\ v_y=2x}\Rightarrow \cases{u_x=v_y\\ u_y=-v_x} \\\qquad \Rightarrow f(z) \bbox[red,2pt]{\text{analytic}}$$

解答： $$\mathbf{(1)}\; r^2+6r+5=0 \Rightarrow (r+5)(r+1)=0 \Rightarrow r=-5,-1 \Rightarrow \cases{y_1=e^{-5x}\\ y_2=e^{-x}} \\ \qquad \Rightarrow \text{Wronskian }=\begin{vmatrix} y_1 & y_2\\ y_1' & y_2'\end{vmatrix} =\begin{vmatrix} e^{-5x} & e^{-x}\\ -5e^{-5x} & -e^{-x} \end{vmatrix} = \bbox[red, 2pt]{4e^{-6x}} \\\mathbf{(2)}\;  y=C_1e^{-5x}+ C_2e^{-x}  \Rightarrow y'=-5C_1e^{-5x}-C_2 e^{-x} \Rightarrow y''=25C_1e^{-5x} +C_2e^{-x}\\ \qquad 因此y'(0)=y''(0)=4 \Rightarrow \cases{-5C_1-C_2= 4\\ 25C_1+C_2= 4} \Rightarrow \cases{C_2 =2/5\\ C_2=-6} \Rightarrow \bbox[red,2pt]{y={2\over 5}e^{-5x}-6e^{-x}}$$

解答： $$先求齊次解，即y''-4y'+4y=0 \Rightarrow r^2-4r+4=0 \Rightarrow r=2 \Rightarrow y_h=C_1e^{2x}+ C_2xe^{2x}\\ 令\cases{y_1= e^{2x}\\ y_2=xe^{2x}} \Rightarrow w(y_1,y_2)= y_1y_2'-y_2y_1'= e^{4x} \Rightarrow y_p = -y_1 \int {y_2 \cdot r(x)\over w}\,dx + y_2 \int{ y_1\cdot r(x)\over w}\,dx \\=-e^{2x} \int{1\over x^3}\,dx +xe^{2x} \int  {1\over x^4}\,dx ={e^{2x}\over 2x^2} -{1\over 3x^2}e^{2x} ={1\over 6x^2}e^{2x} \Rightarrow y=y_h+y_p\\ \Rightarrow \bbox[red,2pt]{y= C_1e^{2x}+ C_2xe^{2x}+{1\over 6x^2}e^{2x}}$$

解答： $$\mathbf{(1)}\;r(t)=\begin{cases}1,& 0\lt t\lt 1\\ 0, &\text{otherwise} \end{cases} \Rightarrow r(t)=u(t)-u(t-1) \Rightarrow L\{r(t)\}= L\{u(t)\}-L\{u(t-1)\} \\ \qquad ={1\over s}-{e^{-s}\over s} =\bbox[red,2pt]{{1\over s}(1-e^{-s})} \\\mathbf{(2)}\; L\{ y''\}+L\{y' \}= L\{r(t) \} \Rightarrow s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0) ={1\over s}(1-e^{-s}) \\\qquad \Rightarrow (s^2+1)Y(s)={1\over s}(1-e^{-s}) \Rightarrow Y(s)={1\over s(s^2+1)}(1-e^{-s})  =({1\over s}-{s\over s^2+1})(1-e^{-s}) \\ \qquad \Rightarrow y=L^{-1}\left( ({1\over s}-{s\over s^2+1})(1-e^{-s}) \right) =L^{-1}\{ {1\over s}(1-e^{-s})\}  -L^{-1}\{{s\over s^2+1} \}+L^{-1}\{{s\over s^2+1}e^{-s} \} \\\qquad \Rightarrow \bbox[red,2pt]{y(t)=r(t)-\cos(s)+u(t-1)\cos(t-1)}$$

解答： $$f為奇函數且週期為2\pi \Rightarrow f(x)=\begin{cases} \pi/2, & 0\lt x\le \pi/2\\ \pi-x, & \pi/2\lt x\lt \pi \\ -\pi/2, & -\pi/2 \lt x\le 0\\ -x-\pi,& -\pi\lt x\lt -\pi/2\end{cases}\\ \Rightarrow a_n=0且b_n={1\over \pi }\int_{-\pi}^\pi f(x)\sin nx \,dx= {1\over \pi }\left(\int_{-\pi}^{-\pi/2} (-x-\pi)\sin nx\,dx +\int_{-\pi/2}^0 (-{\pi\over 2}\sin nx\,dx \right)\\ \quad +{1\over \pi }\left(\int_0^{\pi/2} {\pi\over 2}\sin nx\,dx +\int_{\pi/2}^\pi (\pi-x)\sin nx\,dx \right) \\ = {1\over \pi }\left({2\over n^2}\sin(n\pi/2) +{\pi\over n}\cos(n\pi/2)+{2\pi \over n}\sin^2(n\pi/4)   \right) = {1\over \pi}\left( {2\over n^2} \sin(n\pi/2) +{\pi\over n}\right) \\={1\over n}+{2\over n^2\pi} \sin(n\pi/2) \Rightarrow \text{Fourier series} = \bbox[red,2pt]{\sum_{n=1}^\infty \left({1\over n}+{2\over n^2\pi} \sin(n\pi/2)\right)\sin(nx)}$$

解答： $$\mathbf{f}=[x^2+y^2,3xy+z^2, 4z^3] \Rightarrow \cases{f_x= 2x\\ f_y=3x\\ f_z= 12z^2} \Rightarrow \text{div }\mathbf f= f_x+ f_y +f_z= 5x+12z^2\\ \text{curl }\mathbf f=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\{\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\ x^2+y^2 & 3xy+z^2 & 4z^3\end{vmatrix} =0\mathbf i +0\mathbf j+3y\mathbf k-2y \mathbf k-0\mathbf j-2z\mathbf i = [-2z,0,y] \\ 因此\bbox[red,2pt]{\cases{\text{div }\mathbf f= 5x+12z^2\\ \text{curl }\mathbf f= [-2z,0,y]} }$$
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