108年高考三級應用數學詳解

108年公務人員高等考試三級考試

類 科 ：氣象
科 目：應用數學（包括微積分、微分方程與向量分析）

解： $$先求齊次解，即x^2y''+xy'-y=0\\ 令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow m(m-1)x^m+mx^m-x^m=0\\\Rightarrow (m^2-1)x^m=0\Rightarrow m^2=1\Rightarrow m=\pm 1\Rightarrow y_h=c_1x+c_2\cdot\frac{1}{x} \\ y_h=c_1x+c_2\frac{1}{x} \Rightarrow y_p=\phi_1x+\phi_2\frac{1}{x} \Rightarrow\left[ \begin{matrix}  x & {1\over x} \\ {d\over dx}x &{d\over dx}{1\over x} \end{matrix} \right] \left[ \begin{matrix}  \phi_1' \\ \phi_2' \end{matrix} \right] = \left[ \begin{matrix}  0 \\ {16x^3\over x^2} \end{matrix} \right]\\ \Rightarrow \left[ \begin{matrix}  x & {1\over x} \\ 1 &-{1\over x^2} \end{matrix} \right] \left[ \begin{matrix}  \phi_1' \\ \phi_2' \end{matrix} \right] = \left[ \begin{matrix}  0 \\ {16x} \end{matrix} \right] \Rightarrow \begin{cases}x\phi_1'+{\phi_2'\over x}=0\\ \phi_1'-{\phi_2'\over x^2}=16x\end{cases} \Rightarrow \begin{cases}\phi_1'=8x\\ \phi_2'=-8x^3\end{cases} \\ \Rightarrow \begin{cases}\phi_1=\int{8x\,dx}=4x^2\\ \phi_2=\int{-8x^3\,dx}=-2x^4\end{cases} \Rightarrow y_p= \phi_1x+\phi_2\frac{1}{x}=4x^3-2x^3=2x^3 \\ \Rightarrow y=y_h+y_p=c_1x+c_2{1\over x}+2x^3\Rightarrow y'=c_1-c_2{1\over x^2}+6x^2   \\再加上初始條件\cases{y(1)=-1\\ y'(1)=1} \Rightarrow \cases{c_1+c_2+2=-1\\ c_1-c_2+6=1} \Rightarrow \cases{c_1=-4\\ c_2=1} \Rightarrow \bbox[red,2pt]{y=-4x+{1\over x}+2x^3}$$

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解：
$$F(\omega)={1\over \sqrt{2\pi}}\int_{-\infty}^\infty{f(x)e^{-i\omega x}\,dx}= {1\over \sqrt{2\pi}}\int_{a}^b {kxe^{-i\omega x}\,dx}= {k\over \sqrt{2\pi}}\left.\left[ \left({1\over \omega^2} -{x\over i\omega}\right) e^{-i\omega x}\right] \right|_a^b \\= \bbox[red,2pt]{{k\over \sqrt{2\pi}}\left[ \left({1\over \omega^2} -{b\over i\omega}\right) e^{-i\omega b}- \left({1\over \omega^2} -{a\over i\omega}\right) e^{-i\omega a}\right] }$$

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解：
$$\begin{cases}3x-2y+z=13\\ -2x+y-4z=11 \\ x+4y-5z=-31\end{cases} \Rightarrow \left[\begin{array}{rrr|r}3&-2&1&13\\ -2& 1&4&11 \\1&4&-5& -31 \end{array}\right] \xrightarrow{-3r_3+r_1,2r_3+r_2}\left[\begin{array}{rrr|r}0&-14&16&106\\ 0& 9 &-6 &-51 \\1&4&-5& -31 \end{array}\right]\\ \xrightarrow{r_1/2+r_1,r_2/3}  \left[\begin{array}{rrr|r}0&-7&8&53\\ 0& 3 &-2 &-17 \\1&4&-5& -31 \end{array}\right]  \xrightarrow{2r_2+r_1}\left[\begin{array}{rrr|r} 0&-1&4 &19\\ 0& 3 & -2 &-17 \\1&4&-5& -31 \end{array}\right] \\ \xrightarrow{3r_1+r_2,4r_1+r_3}\left[\begin{array}{rrr|r} 0&-1&4&19\\ 0& 0 &10 &40 \\1&0&11& 45 \end{array}\right] \Rightarrow \left\{\begin{aligned}-y+4z=19\\ 10z=40 \\ x+11z= 45\end{aligned}\right.  \Rightarrow \bbox[red,2pt]{\left\{\begin{aligned}x&=1\\ y&=-3 \\ z&= 4\end{aligned}\right.}$$

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解：
$$令v=u_x 則 u_{xy}=u_x \Rightarrow v_y=v\Rightarrow v_y-v=0 \Rightarrow v=A(x)e^y\Rightarrow u=\int{v\,dx} = B(x)e^y+C(y)\\ \Rightarrow \bbox[red,2pt]{u(x,y)=B(x)e^y+C(y)},其中B及C為x的函數$$

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解：

(一)泰勒級數及其收斂半徑 $$(1)f(z)={z^8\over 1-z^4}=z^8\left(1+z^4+z^8+\cdots\right)=z^8+z^{12}+z^{16}+\cdots\\ \Rightarrow \bbox[red,2pt]{ f(z)=\sum_{n=0}^\infty{z^{4(n+2)}}} \Rightarrow |z|<1\Rightarrow \bbox[red,2pt]{收斂半徑R=1}$$ (二)羅蘭級數及其收斂半徑 $$A_1:0<|z|<1 \Rightarrow z\in A_1 \Rightarrow f(z)的羅倫級數與泰勒級數相同，即\bbox[red,2pt]{f(z)=\sum_{n=0}^\infty{z^{4(n+2)}} }\Rightarrow \bbox[red,2pt]{收斂半徑R=1}\\A_2:1<|z|<\infty \Rightarrow z\in A_2 \Rightarrow f(z)={z^8\over 1-z^4}=-{z^4\over 1-{1\over z^4}} = -z^4\left(1+ {1\over z^4}+{1\over z^8}+{1\over z^{12}}+\cdots \right)\\ \Rightarrow \bbox[red,2pt]{f(z)=-z^4-1-{1\over z^4} -{1\over z^8}-\cdots}\Rightarrow \bbox[red,2pt]{收斂半徑1< R<\infty}$$

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