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
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,ω0=π(二)A=12∫1−1t2dt=12[13t3]|1−1=12×23=13(三)x(t)為偶函數⇒bn=0⇒b1=b2=b3=0
(一)y″+2y′−3y=0⇒λ2+2λ−3=0⇒(λ+3)(λ−1)=0⇒λ=−3,1⇒齊次解:yh=C1e−3x+C2ex(二)r(x)=2e−x⇒非齊次解yp=ke−x代回原式⇒(ke−x)″+2(ke−x)′−3ke−x=2e−x⇒ke−x−2ke−x−3ke−x=2e−x⇒−4k=2⇒k=−12⇒yp=−12e−x⇒特定解y=yh+yp⇒y=C1e−3x+C2ex−12e−x(三)y=C1e−3x+C2ex−12e−x⇒y′=−3C1e−3x+C2ex+12e−x,將初始值{y(0)=1y′(0)=−1代入⇒{C1+C2−12=1−3C1+C2+12=−1⇒C1=C2=34⇒精確解y=34e−3x+34ex−12e−x
解:
由{3x+2y=12x−y=5x+4y=−2⇒[322−114][xy]=[15−2]≡Ax=b,即{A=[322−114]b=[15−2]⇒AT=[3212−14]⇒{ATA=[148821]ATb=[11−11]⇒[148821][xy]=[11−11]⇒{x=319230y=−121115
解:
(一)A=[2647],det(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}}
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}
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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