자연계열 I (오전) 2번

[문제 2-1]주어진 식에 $\displaystyle b=\frac12$을 대입하면$\displaystyle g\left(a+\frac12\right)+g\left(a-\frac12\right)=2g(a)\cos\frac\pi2=0$$\displaystyle g\left(a+\frac12\right)=-g\left(a-\frac12\right)$$\displaystyle a$에 $\displaystyle a+\frac12$를 대입하면 $\displaystyle g(a+1)=-g(a)$$\displaystyle g(a+2)=-g(a+1)=g(a)$따라서 $\displaystyle g(0)=g(2)=\cdots=g(2020)=1$주어진 식에 $\displaystyle a=0$, $\displaystyle b=x$를 대입하면 $\displaystyle g(x)+g(-x)=2g(0)\cos\pi x=2\cos\pi x$위 식을 $\displaystyle \left[-\frac12,\frac12\right]$에서 적분하면,$\displaystyle \int_{-\frac12}^{\frac12}g(x)\,dx+\int_{-\frac12}^{\frac12}g(-x)\,dx=2\int_{-\frac12}^{\frac12}\cos\pi x\,dx=\frac4\pi$$\displaystyle -x=t$로 치환적분법을 이용하면 $\displaystyle \int_{-\frac12}^{\frac12}g(-x)\,dx=\int_{-\frac12}^{\frac12}g(t)\,dt$따라서 $\displaystyle \int_{-\frac12}^{\frac12}g(x)\,dx=\frac2\pi$가 된다.

[문제 2-2]$\displaystyle a=0$, $\displaystyle b=x$를 대입하면$\displaystyle g(x)+g(-x)=2\cos\pi x$ --- ①$\displaystyle a=x+\frac12$, $\displaystyle b=\frac12$를 대입하면$\displaystyle g(x+1)+g(x)=0$ --- ②$\displaystyle a=\frac12$, $\displaystyle b=x+\frac12$을 대입하면$\displaystyle g(x+1)+g(-x)=2g\left(\frac12\right)\cos\pi\left(x+\frac12\right)=-2g\left(\frac12\right)\sin\pi x$ --- ③①과 ②을 더한 식에 ③을 빼면$\displaystyle g(x)=\cos\pi x+g\left(\frac12\right)\sin\pi x$

$\displaystyle x$에 각각 $\displaystyle \frac13$, $\displaystyle -\frac13$을 대입하면$\displaystyle g\left(\frac13\right)=\cos\frac\pi3+g\left(\frac12\right)\sin\frac\pi3=\frac12+\frac{\sqrt3}2g\left(\frac12\right)$$\displaystyle g\left(-\frac13\right)=\cos\left(-\frac\pi3\right)+g\left(\frac12\right)\sin\left(-\frac\pi3\right)=\frac12-\frac{\sqrt3}2g\left(\frac12\right)$$\displaystyle g\left(\frac13\right)g\left(-\frac13\right)=\frac14-\frac34\left\{g\left(\frac12\right)\right\}^2=\frac1{10}$따라서, $\displaystyle \left\{g\left(\frac12\right)\right\}^2=\frac15$이다.