114年專技高考電機工程技師-工程數學詳解

 114年專門職業及技術人員高等考試

等 別：高等考試
類 科：電機工程技師
科 目：工程數學（包括線性代數、微分方程、複變函數與機率）

解答:$$70\%\times 1\% +20\%\times 2\% +10\%\times 5\% = \bbox[red, 2pt]{0.016}$$

解答:$$e^x =1+x+{ x^2\over 2!} +{x^3\over 3!}+ \cdots \Rightarrow e^{1/z^2}=1+ {1\over z^2}+{1\over 2 z^4}+ {1\over 6z^6}+\cdots \Rightarrow \text{Res}(e^{1/z^2},0) =0 \\ \Rightarrow \int_C e^{1/z^2}\,dz = 2\pi i\cdot \text{Res}(e^{1/z^2},0) = \bbox[red, 2pt]0$$

解答:$$\lambda^2+4\lambda+ 4=0 \Rightarrow (\lambda+2)^2=0 \Rightarrow \lambda=-2 \Rightarrow y= c_1e^{-2x}+ c_2xe^{-2x}  \\\Rightarrow y'=(-2c_1+c_2)e^{-2x}- 2c_2xe^{-2x} \Rightarrow \cases{y(0) =c_1 =1\\ y'(0)=-2c_1 +c_2=3} \Rightarrow \cases{c_1 =1\\ c_2= 5}\\ \Rightarrow \bbox[red, 2pt]{y= e^{-2x}+ 5xe^{-2x}} \\ 題目\bbox[cyan,2pt]{有疑義}，y^{(1)}=3,也許是y^{(1)}(0)=3$$

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

解答:$$\textbf{(一) } Ax=b \Rightarrow \begin{bmatrix}0&-1&0& 1\\ 0&1& -1&0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} = \begin{bmatrix}0\\ 1 \end{bmatrix} \Rightarrow [A \mid b] = \left[\begin{matrix}0 & -1 & 0 & 1 & 0\\0 & 1 & -1 & 0 & 1\end{matrix}\right] \\ \quad \Rightarrow \text{RREF} \left( \left[\begin{matrix}0 & -1 & 0 & 1 & 0\\0 & 1 & -1 & 0 & 1\end{matrix}\right] \right) = \left[\begin{matrix}0 & 1 & 0 & -1 & 0\\0 & 0 & 1 & -1 & -1\end{matrix}\right] \Rightarrow \cases{x_2-x_4=0\\ x_3-x_4=-1} \\ \Rightarrow \bbox[red, 2pt]{x = \left\{  \left. \begin{bmatrix}s\\ t\\ t-1\\ t \end{bmatrix} \right|  s,t \in \mathbb R\right\}} \\ \textbf{(二) } \text{RREF}(A)= \left[\begin{matrix}0 & 1 & 0 & -1\\0 & 0 & 1 & -1\end{matrix}\right] \Rightarrow \cases{x_2-x_4=0\\ x_3-x_4=0} \Rightarrow \bbox[red, 2pt]{N(A) =\left\{  \left. \begin{bmatrix}s\\ t\\ t \\ t \end{bmatrix} \right|  s,t \in \mathbb R\right\}}$$

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