109年調查三等-工程數學詳解

109年法務部調查局調查人員考試試題

考試別：調查人員
等  別：三等
類科組：電子科學組
科  目：工程數學

解：

(一) $$依傅立葉級數定義，\omega_0=\bbox[red, 2pt]{\pi}$$ (二) $$A=\frac{1}{2}\int_{-1}^1t^2\;dt =\frac{1}{2}\left. \left[ {1\over 3}t^3\right] \right|_{-1}^1 ={1\over 2}\times {2\over 3}= \bbox[red, 2pt]{1\over 3}$$ (三) $$x(t)為偶函數\Rightarrow b_n=0 \Rightarrow \bbox[red,2pt]{b_1=b_2=b_3=0}$$

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解：

(一) $$y''+2y'-3y=0 \Rightarrow \lambda^2+2\lambda-3=0 \Rightarrow (\lambda+3)(\lambda-1)=0 \Rightarrow \lambda=-3,1\\ \Rightarrow 齊次解: \bbox[red, 2pt]{y_h=C_1e^{-3x}+ C_2e^{x}}$$ (二) $$r(x)=2e^{-x} \Rightarrow 非齊次解y_p=ke^{-x}代回原式\Rightarrow (ke^{-x})''+2(ke^{-x})'-3ke^{-x}=2e^{-x} \\ \Rightarrow ke^{-x}-2ke^{-x}-3ke^{-x}=2e^{-x} \Rightarrow -4k=2 \Rightarrow k=-{1\over 2} \Rightarrow y_p=-{1\over 2} e^{-x}\\ \Rightarrow 特定解y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=C_1e^{-3x}+ C_2e^{x}-{1\over 2} e^{-x}}$$ (三) $$y=C_1e^{-3x}+ C_2e^{x}-{1\over 2} e^{-x} \Rightarrow y'=-3C_1e^{-3x}+C_2e^x+ {1\over 2}e^{-x}，將初始值\cases{y(0)=1\\ y'(0)=-1}代入 \\ \Rightarrow \cases{C_1+C_2-{1\over 2}=1 \\ -3C_1+C_2+{1\over 2}=-1}  \Rightarrow C_1=C_2={3\over 4} \Rightarrow 精確解\bbox[red, 2pt]{y={3\over 4}e^{-3x}+ {3\over 4}e^{x}-{1\over 2} e^{-x}}$$

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解：

$$由\cases{3x+2y=1\\ 2x-y=5\\ x+4y=-2 \\} \Rightarrow \begin{bmatrix}3 & 2\\ 2& -1 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 1\\ 5 \\ -2\end{bmatrix} \equiv Ax=b，即 \cases{A= \begin{bmatrix}3 & 2\\ 2& -1 \\ 1 & 4 \end{bmatrix}\\b=\begin{bmatrix} 1\\ 5 \\ -2\end{bmatrix}} \\\Rightarrow A^T= \begin{bmatrix}3 & 2 &1\\ 2& -1  & 4 \end{bmatrix} \Rightarrow  \cases{A^TA= \begin{bmatrix}14 & 8\\8& 21 \end{bmatrix}\\A^Tb=\begin{bmatrix} 11\\ -11 \end{bmatrix}} \Rightarrow \begin{bmatrix}14 & 8\\8& 21 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 11\\-11\end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{ \cases{ x={319\over 230}\\ y=-{121\over 115} }}$$

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解：

(一) $$A=\begin{bmatrix} 2& 6 \\4 & 7\end{bmatrix},\det(A-\lambda I)=0 \Rightarrow \begin{vmatrix} 2-\lambda& 6 \\4 & 7-\lambda\end{vmatrix}=0 \Rightarrow (\lambda-2)(\lambda-7)-24=0 \\ \Rightarrow (\lambda+1)(\lambda-10)=0 \Rightarrow 特徵值\bbox[red, 2pt]{\lambda =-1,10}$$ (二) $$\lambda_1=-1 \Rightarrow (A-\lambda_1 I)=  \begin{bmatrix} 2+1& 6 \\4 & 7+1\end{bmatrix} =  \begin{bmatrix} 3& 6 \\4 & 8\end{bmatrix}，\\因此(A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} 3& 6 \\4 & 8\end{bmatrix}\begin{bmatrix} x_1 \\x_2\end{bmatrix}=0  \Rightarrow x_1+2x_2=0，取\vec v_1=\begin{bmatrix} 2 \\-1\end{bmatrix};\\\lambda_2=10 \Rightarrow (A-\lambda_2 I)=  \begin{bmatrix} 2-10& 6 \\4 & 7-10\end{bmatrix} =  \begin{bmatrix} -8& 6 \\4 & -3\end{bmatrix}，\\因此(A-\lambda_2 I)X=0 \Rightarrow \begin{bmatrix} -8 & 6 \\4 & -3\end{bmatrix}\begin{bmatrix} x_1 \\x_2\end{bmatrix}=0  \Rightarrow 4x_1=3x_2，取\vec v_2=\begin{bmatrix} 3 \\4\end{bmatrix};\\Ans:\bbox[red, 2pt]{\lambda_1=-1對應的特徵向量\vec v_1=\begin{bmatrix} 2 \\-1\end{bmatrix}，\lambda_2=10對應的特徵向量\vec v_2=\begin{bmatrix} 3 \\4\end{bmatrix}}$$

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解：

(一) $$\iint f(x,y)\;dxdy =1  \Rightarrow \int_0^1 \int_0^1 A\cdot x\cdot (1+y^2)\;dxdy=1\\ \int_0^1 \int_0^1 A\cdot x\cdot (1+y^2)\;dxdy= A\int_0^1 \left.\left[{1\over 2} x^2\cdot (1+y^2)\right] \right|_0^1dy= {1\over 2}A \int_0^1 (1+y^2)\;dy \\ = {1\over 2}A \left. \left[ y+{1\over 3}y^3\right] \right|_0^1 ={2\over 3}A=1 \Rightarrow A=\bbox[red,2pt]{3\over 2}$$ (二) $$f_X(x)=  \int_0^1 A\cdot x\cdot (1+y^2)\;dy= {3\over 2}x\int_0^1 (1+y^2)\;dy = {3\over 2}x \left. \left[y+{1\over 3}y^3 \right] \right|_0^1 =2x\\ \Rightarrow \bbox[red,2pt]{f_X(x)=2x}$$ (三) $$f_Y(y)=  \int_0^1 A\cdot x\cdot (1+y^2)\;dx= {3\over 2}(1+y^2)\int_0^1 x\;dx = {3\over 4}(1+y^2) \\ \Rightarrow E(Y) = \int_0^1 yf_Y(y)\;dy = {3\over 4}\int_0^1 (y+y^3)\;dy = {3\over 4}\left .\left[ {1\over 2}y^2+{1\over 4}y^4 \right] \right|_0^1 =\bbox[red, 2pt]{9\over 16}$$ (四) $$E(Z)=E(XY) = \int_0^1 \int_0^1 xy\cdot {3\over 2}\cdot x(1+y^2)\;dxdy = \int_0^1 \int_0^1 {3\over 2}\cdot x^2(y+y^3)\;dxdy \\ =\int_0^1{1\over 2}(y+y^3)\;dy = \bbox[red, 2pt]{3\over 8}$$

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解：

(一) $$f(z) =f(x+yi) = (x+yi)^2 = x^2+2xyi-y^2 = (x^2-y^2)+2xyi = u(x,y)+i\cdot v(x,y)\\ \Rightarrow \cases{\bbox[red, 2pt] {u(x,y)=x^2-y^2} \\ \bbox[red, 2pt]{v(x,y) =2xy}}$$ (二) $$\cases{ u(x,y)=x^2-y^2 \\ v(x,y) =2xy} \Rightarrow \cases{{\partial \over \partial x}u(x,y)=2x \\{\partial \over \partial y} u(x,y)= -2y \\ {\partial \over \partial x}v(x,y)=2y \\ {\partial \over \partial y}v(x,y)=2x } \Rightarrow \cases{{\partial \over \partial x}u(x,y) ={\partial \over \partial y}v(x,y) \\ {\partial \over \partial y}u(x,y) = -{\partial \over \partial x}v(x,y)} \\ \Rightarrow u(x,y)及v(x,y)滿足柯西黎曼方程式$$ (三) $$滿足柯西-黎曼方程式且{\partial \over \partial x}u,{\partial \over \partial y}u,{\partial \over \partial x}v,{\partial \over \partial y}v均存在且連續，所以\bbox[red,2pt]{f(z)可解析}$$ (四) $$f(z)可解析\Rightarrow 線積分與路徑無關，因此我們將路徑改為一條水直線及一條垂直線:\\1+i\cdot 0\xrightarrow{C_1}0+i\cdot 0 \xrightarrow{C_2}0+i\cdot 1;\\\int_\Gamma f(z)\;dz = \int_{C_1}f(z)\;dz + \int_{C_2}f(z)\;dz,其中C_1為水平線,C_2為垂直線;\\ C_1:z=x+iy = x \Rightarrow dz= dx \Rightarrow \int_{C_1}z^2\;dz = \int_1^0x^2\;dx= -{1\over 3}\\ C_2:z=x+iy=iy \Rightarrow dz=idy \Rightarrow \int_{C_2}z^2\;dz = \int_0^1(iy)^2i\;dy= \int_0^1-iy^2\;dy=-{1\over 3}i\\ \Rightarrow \int_\Gamma f(z)\;dz =\bbox[red, 2pt]{-{1\over 3}-{1\over 3}i}$$

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