자연계열 I 1번

[1-1]$\displaystyle \mathrm{P}(-\pi\le X\le\pi)=\mathrm{P}\left(-\pi\le X\le\frac\pi2\right)+\mathrm{P}\left(\frac\pi2\le X\le\pi\right)$$\displaystyle \begin{aligned}\mathrm{P}\left(-\pi\le X\le\frac\pi2\right)&=\frac32\pi\times\frac32\pi a\times\frac12\\&=\frac98\pi^2a\end{aligned}$$\displaystyle \begin{aligned}\mathrm{P}\left(\frac\pi2\le X\le\pi\right)&=\frac12\pi\times\frac12\pi a\times\frac12\\&=\frac18\pi^2a\end{aligned}$따라서 $\displaystyle \mathrm{P}(-\pi\le X\le\pi)=\frac98\pi^2a+\frac18\pi^2a=1$이므로$\displaystyle a=\frac4{5\pi^2}$이다.별해) $\displaystyle \begin{aligned}\int_{-\pi}^\pi a\left|x-\frac\pi2\right|dx&=a\int_{-\pi}^{\frac\pi2}\left(\frac\pi2-x\right)dx+a\int_{\frac\pi2}^\pi\left(x-\frac\pi2\right)\\&=a\Biggl[\frac\pi2x-\frac12x^2\Biggr]_{-\pi}^{\frac\pi2}+a\Biggl[\frac12x^2-\frac\pi2x\Biggr]_{\frac\pi2}^\pi\\&=\frac98\pi^2a+\frac18\pi^2a\end{aligned}$Correction. 절댓값을 풀어 적은 둘째 정적분의 끝에 미분기호 $\displaystyle dx$가 빠졌다. 이 항은 $\displaystyle a\int_{\pi/2}^{\pi}(x-\pi/2)\,dx$이다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.따라서 $\displaystyle \mathrm{P}(-\pi\le X\le\pi)=\frac98\pi^2a+\frac18\pi^2a=1$이므로$\displaystyle a=\frac4{5\pi^2}$이다.

[1-2]$\displaystyle \begin{aligned}\mathrm{P}\left(\left|X-\frac\pi2\right|\le\pi\right)&=\mathrm{P}\left(-\frac\pi2\le X\le\frac32\pi\right)\\&=\mathrm{P}\left(-\frac\pi2\le X\le\pi\right)\\&=\mathrm{P}\left(-\frac\pi2\le X\le\frac\pi2\right)+\mathrm{P}\left(\frac\pi2\le X\le\pi\right)\\&=\frac25+\frac1{10}=\frac12\end{aligned}$따라서 $\displaystyle \mathrm{P}\left(\left|X-\frac\pi2\right|\le\pi\right)=\frac12$이다.별해) $\displaystyle \begin{aligned}\mathrm{P}\left(\left|X-\frac\pi2\right|\le\pi\right)&=\int_{-\pi}^\pi a\left|x-\frac\pi2\right|dx\\&=\frac4{5\pi^2}\left\{\int_{-\frac\pi2}^{\frac\pi2}\left(\frac\pi2-x\right)dx+\int_{\frac\pi2}^\pi\left(x-\frac\pi2\right)dx\right\}\\&=\frac12\end{aligned}$Correction. 별해의 첫 적분식에서 하한 $\displaystyle -\pi$는 $\displaystyle -\pi/2$의 오기이다. 다음 계산줄과 정답 $\displaystyle 1/2$는 맞다. 원문 표기는 그대로 실었다.

[1-3]$\displaystyle \begin{aligned}b_n&=\int_{-n\pi}^{n\pi}f\left(\frac xn\right)\cos x\,dx\\&=n\int_{-\pi}^\pi f(t)\cos nt\,dt\\&=\frac{4n}{5\pi^2}\left[\int_{\frac\pi2}^{\pi}\left(t-\frac\pi2\right)\cos nt\,dt-\int_{-\pi}^{\frac\pi2}\left(t-\frac\pi2\right)\cos nt\,dt\right]\end{aligned}$$\displaystyle =\frac{4n}{5\pi^2}\left\{\Biggl[\frac1n\left(t-\frac\pi2\right)\sin nt\Biggr]_{\frac\pi2}^{\pi}-\frac1n\int_{\frac\pi2}^{\pi}\sin nt\,dt+\Biggl[\frac1n\left(\frac\pi2-t\right)\sin nt\Biggr]_{-\pi}^{\frac\pi2}+\frac1n\int_{-\pi}^{\frac\pi2}\sin nt\,dt\right\}$$\displaystyle \begin{aligned}&=\frac{4n}{5\pi^2}\left(\frac2{n^2}\cos n\pi-\frac2{n^2}\cos\frac n2\pi\right)\\&=\frac8{5\pi^2n}\left(\cos n\pi-\cos\frac n2\pi\right)\end{aligned}$따라서 $\displaystyle b_2=\frac8{5\pi^2},b_5=-\frac8{25\pi^2}$이므로,$\displaystyle \frac{b_2}{b_5}=-5$이다.

별해 1)$\displaystyle \begin{aligned}b_n&=\int_{-n\pi}^{n\pi}f\left(\frac xn\right)\cos x\,dx\\&=n\int_{-\pi}^\pi f(t)\cos nt\,dt\\&=\frac{4n}{5\pi^2}\left[\int_{\frac\pi2}^\pi\left(t-\frac\pi2\right)\cos nt\,dt-\int_{-\pi}^{\frac\pi2}\left(t-\frac\pi2\right)\cos nt\,dt\right]\\&=\frac{4n}{5\pi^2}\left[\int_{\frac\pi2}^\pi t\cos nt\,dt-\frac\pi2\int_{\frac\pi2}^\pi\cos nt\,dt-\int_{-\pi}^{\frac\pi2}t\cos nt\,dt+\frac\pi2\int_{-\pi}^{\frac\pi2}\cos nt\,dt\right]\end{aligned}$$\displaystyle =\frac{4n}{5\pi^2}\left\{\Biggl[\frac tn\sin nt\Biggr]_{\frac\pi2}^{\pi}-\frac1n\int_{\frac\pi2}^{\pi}\sin nt\,dt-\frac\pi{2n}\Biggl[\sin nt\Biggr]_{\frac\pi2}^{\pi}-\Biggl[\frac tn\sin nt\Biggr]_{-\pi}^{\frac\pi2}+\frac1n\int_{-\pi}^{\frac\pi2}\sin nt\,dt+\frac\pi{2n}\Biggl[\sin nt\Biggr]_{-\pi}^{\frac\pi2}\right\}$$\displaystyle \begin{aligned}&=\frac{4n}{5\pi^2}\left\{-\frac\pi{2n}\sin\frac n2\pi+\frac1{n^2}\Biggl[\cos nt\Biggr]_{\frac\pi2}^\pi+\frac\pi{2n}\sin\frac n2\pi-\frac\pi{2n}\sin\frac n2\pi-\frac1{n^2}\Biggl[\cos nt\Biggr]_{-\pi}^{\frac\pi2}+\frac\pi{2n}\sin\frac n2\pi\right\}\\&=\frac{4n}{5\pi^2}\left(\frac2{n^2}\cos n\pi-\frac2{n^2}\cos\frac n2\pi\right)\\&=\frac8{5\pi^2n}\left(\cos n\pi-\cos\frac n2\pi\right)\end{aligned}$따라서 $\displaystyle b_2=\frac8{5\pi^2},\ b_5=-\frac8{25\pi^2}$이므로,$\displaystyle \frac{b_2}{b_5}=-5$이다.

별해 2)$\displaystyle \begin{aligned}b_2&=\int_{-2\pi}^{2\pi}f\left(\frac x2\right)\cos x\,dx=2\int_{-\pi}^\pi f(t)\cos2t\,dt\\&=\frac8{5\pi^2}\left[\int_{\frac\pi2}^\pi\left(t-\frac\pi2\right)\cos2t\,dt-\int_{-\pi}^{\frac\pi2}\left(t-\frac\pi2\right)\cos2t\,dt\right]\end{aligned}$$\displaystyle =\frac8{5\pi^2}\left\{\Biggl[\frac12\left(t-\frac\pi2\right)\sin2t\Biggr]_{\frac\pi2}^{\pi}-\frac12\int_{\frac\pi2}^{\pi}\sin2t\,dt+\Biggl[\frac12\left(\frac\pi2-t\right)\sin2t\Biggr]_{-\pi}^{\frac\pi2}+\frac12\int_{-\pi}^{\frac\pi2}\sin2t\,dt\right\}$$\displaystyle =\frac8{5\pi^2}$$\displaystyle \begin{aligned}b_5&=\int_{-5\pi}^{5\pi}f\left(\frac x5\right)\cos x\,dx=5\int_{-\pi}^\pi f(t)\cos5t\,dt\\&=\frac4{\pi^2}\left[\int_{\frac\pi2}^\pi\left(t-\frac\pi2\right)\cos5t\,dt-\int_{-\pi}^{\frac\pi2}\left(t-\frac\pi2\right)\cos5t\,dt\right]\end{aligned}$$\displaystyle =\frac4{\pi^2}\left\{\Biggl[\frac15\left(t-\frac\pi2\right)\sin5t\Biggr]_{\frac\pi2}^{\pi}-\frac15\int_{\frac\pi2}^{\pi}\sin5t\,dt+\Biggl[\frac15\left(\frac\pi2-t\right)\sin5t\Biggr]_{-\pi}^{\frac\pi2}+\frac15\int_{-\pi}^{\frac\pi2}\sin5t\,dt\right\}$$\displaystyle =-\frac8{25\pi^2}$　따라서 $\displaystyle \frac{b_2}{b_5}=-5$이다.

별해 3)$\displaystyle \begin{aligned}b_2&=\int_{-2\pi}^{2\pi}f\left(\frac x2\right)\cos x\,dx\\&=\frac a2\int_{-2\pi}^\pi(\pi-x)\cos x\,dx+\frac a2\int_\pi^{2\pi}(x-\pi)\cos x\,dx\end{aligned}$　: 10점$\displaystyle =\frac2{5\pi^2}\left\{\Biggl[(\pi-x)\sin x\Biggr]_{-2\pi}^\pi+\int_{-2\pi}^\pi\sin x\,dx+\Biggl[(x-\pi)\sin x\Biggr]_\pi^{2\pi}-\int_\pi^{2\pi}\sin x\,dx\right\}$$\displaystyle =\frac8{5\pi^2}$　: 10점$\displaystyle \begin{aligned}b_5&=\int_{-5\pi}^{5\pi}f\left(\frac x5\right)\cos x\,dx\\&=\frac a5\int_{-5\pi}^{\frac52\pi}\left(\frac52\pi-x\right)\cos x\,dx+\frac a5\int_{\frac52\pi}^{5\pi}\left(x-\frac52\pi\right)\cos x\,dx\end{aligned}$　: 10점$\displaystyle =\frac4{25\pi^2}\left\{\Biggl[\left(\frac52\pi-x\right)\sin x\Biggr]_{-5\pi}^{\frac52\pi}+\int_{-5\pi}^{\frac52\pi}\sin x\,dx+\Biggl[\left(x-\frac52\pi\right)\sin x\Biggr]_{\frac52\pi}^{5\pi}-\int_{\frac52\pi}^{5\pi}\sin x\,dx\right\}$$\displaystyle =-\frac8{25\pi^2}$　: 10점따라서 $\displaystyle \frac{b_2}{b_5}=-5$이다. : 10점