자연계열 II 1번

【1-1】 조건 (1)로부터 $\displaystyle f(x+1/2)^2+f(x)^2=1\Rightarrow f(x+1)^2+f(x+1/2)^2=1$이므로 $\displaystyle f(x)^2=f(x+1)^2$이다. 그러므로 정수 $\displaystyle n$에 대하여 $\displaystyle |f(n)|=|f(0)|\ne0$이다.한편, $\displaystyle 2f(x)f^{\prime}(x)=2f(x+1)f^{\prime}(x+1)$이므로$\displaystyle |f(n)|\ |f^{\prime}(n)|=|f(n+1)|\ |f^{\prime}(n+1)|\Rightarrow|f^{\prime}(n)|=|f^{\prime}(n+1)|$.따라서, $\displaystyle |f^{\prime}(n)|=|f^{\prime}(0)|$

【1-2】 $\displaystyle g(-1/12)=a$. $\displaystyle f(-1/12)f(5/12)=3a-16a^3$이고$\displaystyle f(-1/12)^2+f(5/12)^2=1\Rightarrow f(-1/12)=\pm\sqrt2/2$이므로 $\displaystyle 16a^3-3a\pm1/2=0$이다.조건 (3)에 의해 $\displaystyle a=-1/4$이다.

【1-3】 $\displaystyle h(x)=1/2f(x)^2-1/2f(x)^4=1/2f(x)^2f(x+1/2)^2$ (조건 (1)).준적분$\displaystyle \begin{aligned}{}&=\int_{-1/12}^0g(x)h^{\prime}(x)\,dx=\Biggl[g(x)h(x)\Biggr]_{-1/12}^0-\int_{-1/12}^0g^{\prime}(x)h(x)\,dx\\&=-g(-1/12)h(-1/12)-1/2\int_{-1/12}^0(f(x)f(x+1/2))^2g^{\prime}(x)\,dx\\&=-g(-1/12)h(-1/12)-1/2\int_{-1/12}^0(3g(x)-16g(x)^3)^2g^{\prime}(x)\,dx\\&=-g(-1/12)h(-1/12)-1/2\int_{-1/4}^0(3u-16u^3)^2\,du\end{aligned}$

$\displaystyle =1/32-1/2\times17/560=9/560$

$\displaystyle p+q=569$.