【数学I】例題1.2.8：式の値（One More）★★★

% 例題I1.2.8：式の値 （One More）★★★
$\alpha=\sqrt{2}-1$のとき,次の式の値を求めよ. (1)$\alpha^2+2\alpha-1$(2)$\alpha^4+\alpha^3+\alpha^2+\alpha+1$

% 解答（例題I1.2.8）
(1)$\alpha=\sqrt{2}-1$より,$\alpha+1=\sqrt{2}$両辺を2乗すると,$(\alpha+1)^2=(\sqrt{2})^2$したがって,$\alpha^2+2\alpha+1=2$よって,$\alpha^2+2\alpha-1=0$(2)(1)より,$\alpha^2=-2\alpha+1$$\alpha^3=\alpha \cdot \alpha^2,\alpha^4=\alpha \cdot \alpha^3$であるから,$$\begin{aligned} & \alpha^3=\alpha \cdot \alpha^2=\alpha(-2 \alpha+1)=-2 \alpha^2+\alpha=-2(-2 \alpha+1)+\alpha=5 \alpha-2,\\ & \alpha^4=\alpha \cdot \alpha^3=\alpha(5 \alpha-2)=5 \alpha^2-2 \alpha=5(-2 \alpha+1)-2 \alpha=-12 \alpha+5\end{aligned}$$よって, \begin{align*} \alpha^4+\alpha^3+\alpha^2+\alpha+1&=(-12\alpha+5)+(5\alpha-2)+(-2\alpha+1)+\alpha+1\\ &=-8\alpha+5=-8(\sqrt{2}-1)+5=13-8\sqrt{2} \end{align*}

% 問題I1.2.8
$\alpha=\frac{1+\sqrt{3}}{2}$のとき,次の式の値を求めよ. (1)$2 \alpha^2-2 \alpha-1$(2)$\alpha^8$

% 解答I1.2.8
(1)$\alpha=\frac{1+\sqrt{3}}{2}$より,$2\alpha-1=\sqrt{3}$両辺を2乗すると,$(2\alpha-1)^2=(\sqrt{3})^2$したがって,$4\alpha^2-4\alpha-2=0$よって,$2\alpha^2-2\alpha-1=0$(2)(1)より,$\alpha^2=\alpha+\frac{1}{2}$$\alpha^8=\left(\alpha^4\right)^2$であるから,$$\alpha^4 =\left(\alpha^2\right)^2=\left(\alpha+\frac{1}{2}\right)^2 =\alpha^2+\alpha+\frac{1}{4} =\left(\alpha+\frac{1}{2}\right)+\alpha+\frac{1}{4} =2\alpha+\frac{3}{4}$$したがって,$$\alpha^8=\left(\alpha^4\right)^2=\left(2\alpha+\frac{3}{4}\right)^2=4\alpha^2+3\alpha+\frac{9}{16} =4\left(\alpha+\frac{1}{2}\right)+3\alpha+\frac{9}{16}=7\alpha+\frac{41}{16}$$よって,$$\alpha^8=7 \cdot \frac{1+\sqrt{3}}{2}+\frac{41}{16}=\frac{56(1+\sqrt{3})+41}{16}=\frac{97+56\sqrt{3}}{16}$$