115年成功大學船舶機電碩士班-工程數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:系統及船舶機電工程學系
科目:工程數學

解答:$$y''-8y'+16y=0 \Rightarrow r^2-8r+16=0 \Rightarrow (r-4)^2=0 \Rightarrow r=4\\ \Rightarrow y_h= c_1e^{4x} +c_2xe^{4x} \\ y_p=Ax^2e^{4x}+Bx+C \Rightarrow y_p'=Ae^{4x}(4x^2+2x)+B \Rightarrow y_p''= A^{4x}(16x^2+16x+2) \\ \Rightarrow y_p''-8y_p'+16y_p= 2Ae^{4x} +16Bx-8B+16C=6e^{4x}+12x-5 \\ \Rightarrow \cases{A=3\\ B=3/4\\ C=1/16} \Rightarrow y_p=3x^2e^{4x}+{3\over 4}x+{1\over 16} \Rightarrow y=y_h+y_p\\ \Rightarrow \bbox[red, 2pt]{y= c_1e^{4x} +c_2xe^{4x}+ 3x^2e^{4x}+{3\over 4}x+{1\over 16} }$$

解答:$$a_0={1\over \pi} \int_{-\pi}^\pi f(x)\,dx ={1\over \pi} \int_{0}^\pi x\,dx ={\pi\over 2} \\a_n={1\over \pi} \int_{-\pi}^\pi f(x) \cos(nx)\,dx ={1\over \pi} \int_{0}^\pi x\cos(nx)\,dx ={1\over \pi} \left. \left[ {x\sin(nx) \over n} +{\cos(nx)\over n^2}\right] \right|_0^\pi\\\qquad ={1\over n^2\pi}((-1)^n-1) \\b_n ={1\over \pi} \int_{-\pi}^\pi f(x) \sin(nx)\,dx ={1\over \pi} \int_{0}^\pi x\sin(nx)\,dx ={1\over \pi} \left. \left[ -{x\cos(nx) \over n} +{\sin(nx) \over n^2}\right] \right|_0^\pi \\\qquad = {1\over n}(-1)^{n+1} \\ \Rightarrow f(x)= \bbox[red, 2pt]{{\pi\over 4}+ \sum_{n=1}^\infty \left({1\over n^2\pi}((-1)^n-1)\cos(nx)+  {1\over n}(-1)^{n+1} \sin(nx)\right)}$$

解答:$$\Phi(X)={1\over 2}X^T A^{-1}X為二次型\text{(Quadratic form)} \Rightarrow \nabla \Phi(X)={1\over 2}(A^{-1} +(A^{-1})^T)X\\ {1\over 2}(A^{-1} +(A^{-1})^T)為對稱矩陣, 因此可假設\Phi(X)= \begin{bmatrix} c_{11} & c_{12}& c_{13} \\c_{21} &c_{22}& c_{23} \\c_{31}& c_{32}& c_{33}\end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} \\ = \begin{bmatrix}c_{11}x+ c_{12}y+ c_{13}z\\ c_{21}x+ c_{22}y+ c_{23}z \\ c_{31}x+ c_{32}y+ c_{33}z \end{bmatrix} \Rightarrow \nabla\cdot(\nabla \Phi)={\partial \over \partial x}(\nabla \Phi)_x +{\partial \over \partial y}(\nabla \Phi)_y +{\partial \over \partial z}(\nabla \Phi)_z =c_{11}+c_{22}+ c_{33} \\ =\text{tr} \left( {1\over 2}(A^{-1} +(A^{-1})^T) \right) = \text{tr}\left( A^{-1} \right) =\text{tr} \left( \begin{bmatrix}2 & 1 & 0 \\1 & 1 & 1 \\0 & 2 & 1 \end{bmatrix}^{-1} \right) =\text{tr} \left( \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{-1}{3} \\\frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\\frac{-2}{3} & \frac{4}{3} & \frac{-1}{3}\end{bmatrix} \right) \\={1\over 3}-{2\over 3}-{1\over 3} =\bbox[red, 2pt]{-{2\over 3}} \\ \nabla \Phi 為一純量\Rightarrow 純量的旋度為0,即 \nabla \times(\nabla \Phi) =\bbox[red, 2pt]0$$

解答:$$欲證M\subseteq \{s\}+Ker(L_A):\\ 已知s\in M \Rightarrow L_A(s)=b, 因此\forall x\in M \Rightarrow L_A(x)=b \Rightarrow L_A(x-s) =L_A(x)-L_A(s)=b-b =0 \\ \Rightarrow x-s\in Ker(L_A) \Rightarrow x=s+(x-s) \Rightarrow x\in \{s\}+Ker(L_A) \Rightarrow M\subseteq \{s\}+Ker(L_A) \\ 欲證\{s\}+Ker(L_A) \subseteq M:\\ \forall y\in \{s\}+Ker(L_A) \Rightarrow y=s+v, 其中v\in Ker(L_A),即L_A(v)=0 \\ 由於s\in M\Rightarrow L_A(s)=b \Rightarrow L_A(y)=L_A(s+v) =L_A(s)+L_A(v)=b+0=b\\ \Rightarrow y\in M \Rightarrow  \{s\}+Ker(L_A)\subseteq  M \\ 因此M\subseteq \{s\}+Ker(L_A) 且\{s\}+Ker(L_A)\subseteq  M,可得M=\{s\}+Ker(L_A) \;\bbox[red, 2pt]{故得證}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解