114年中山大學海下所碩士班-工程數學詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學【海下所碩士班】

解答: $$f(x)=e^x \Rightarrow f^{[n]}(x)=e^x \Rightarrow f^{[n]}(0)=1 \Rightarrow e^x = 1+x+{1\over 2!}x^2 +{1\over 3!}x^3+ {1\over 4!}x^4 + \cdots \\ \Rightarrow \bbox[red, 2pt]{e^x =1+x+{1\over 2}x^2+{1\over 6}x^3 +{1\over 24}x^4 +\cdots} \\ \Rightarrow e^{ix} =1+ix+ {1\over 2}(ix)^2 +{1\over 6}(ix)^3 +{1\over 24}(ix^4)+ \cdots \\ \Rightarrow \bbox[red, 2pt]{e^{ix} =1+ix -{1\over x^2}-{1\over 6}ix^3+{1\over 24}x^4 + \cdots}$$

解答: $$(AB)^{-1} =B^{-1}A^{-1} = \left( \begin{matrix}1 & 1 \\-1 & 1\end{matrix} \right) \left( \begin{matrix}4 & 3 \\2 & 1\end{matrix} \right) = \bbox[red, 2pt]{\left( \begin{matrix}6 & 4 \\-2 & -2\end{matrix} \right)}$$

解答: $$y'+{1\over x}y=3x \Rightarrow xy'+y=3x^2 \Rightarrow (xy)'=3x^2 \Rightarrow xy=x^3+c_1 \Rightarrow y=x^2+{c_1\over x} \\ \Rightarrow y(1)=1+c_1=2 \Rightarrow c_1=1 \Rightarrow \bbox[red, 2pt]{y=x^2+{1\over x}}$$

解答: $$L\{1\} = \int_0^\infty e^{-st}\,dt = \left. \left[ -{1\over s}e^{-st} \right] \right|_0^\infty ={1\over s} \\ L\{e^{at}\} = \int_0^\infty e^{at}\cdot e^{-st}\,dt= \int_0^\infty   e^{-(s-a)t}\,dt = \left. \left[ -{1\over s-a}e^{-(s-a)t} \right] \right|_0^\infty = {1\over s-a} \\ \Rightarrow \bbox[red, 2pt]{L\{1\}={1\over s}, L\{e^{at}\} ={1\over s-a}}$$

解答: $$x^2y''+xy'+(x^2-{1\over 11})y=0 \text{ is a Bessel function with }v^2={1\over 11} \\ \Rightarrow \bbox[red, 2pt]{y=c_1 J_{1/\sqrt{11}} (x)+ c_2J_{-1/ \sqrt{11}}(x), \text{ where}  J_{1/\sqrt{11}}(x) = \sum_{n=0}^\infty {(-1)^n \over n! \Gamma(1+{1\over \sqrt{11}}+n)}\left({x\over 2} \right)^{2n+{1/\sqrt{11}}},} \\ \bbox[red, 2pt]{\text{ and } J_{-1/\sqrt{11}}(x) = \sum_{n= 0}^\infty {(-1)^n \over n! \Gamma(1-{1\over \sqrt{11}}+n)}\left({x\over 2} \right)^{2n-{1/\sqrt{11}}}}$$

解答: $$f(t) =2t^2 \Rightarrow f(-t)=f(t) \Rightarrow f(t)\text{ is even } \Rightarrow b_n=0\\ a_0 ={1\over 2\pi} \int_0^{2\pi} 2t^2 \,dt ={8\over 3}\pi^2\\ a_n= {1\over \pi} \int_0^{2\pi} 2t^2 \cos(nt) \,dt ={1\over \pi} \cdot {8\pi\over n^3} ={8\over n^3} \\ \Rightarrow \bbox[red, 2pt]{f(t)={8\over 3}\pi^2 + \sum_{n=1}^\infty {8\over n^3} \cos (nt)}$$

解答: $${\partial^2 p\over \partial x^2} ={1\over c^2}\cdot {\partial^2 p\over \partial t^2} \Rightarrow {\partial^2 p\over \partial t^2}=c^2{\partial^2 p\over \partial x^2} \Rightarrow  \text{By d'Alembert's method: } \\p(x,t)= {1\over 2}\left( f(x+ct) +f(x-ct)\right)+{1\over 2c}\int_{x-ct}^{x+ct} g(s)\,ds, \\\cases{f(x)={1\over 1+4x^2}\\g(x)=0} \Rightarrow \bbox[red, 2pt]{p(x,t)={1\over 2}\left({1\over 1+4(x+ct)^2}+ {1\over 1+4(x-ct)^2} \right)}+0$$

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解題僅供參考，碩士班歷年試題及詳解