【数学I】例題1.2.8：式の値（One More）★★★

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

(1) $\alpha^2+2\alpha-1$

(2) $\alpha^4+\alpha^3+\alpha^2+\alpha+1$

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% 例題I1.2.8の解答（One More）★★★
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(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{align*}
&\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{align*}

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*}

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

(1) $2\alpha^2-2\alpha-1$

(2) $\alpha^8$

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% 問題I1.2.8の解答（One More）★★★
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(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，$\alpha^2=\alpha+\frac{1}{2}$

$\alpha^8=(\alpha^4)^2$であるから，

\[
\alpha^4=(\alpha^2)^2=(\alpha+\frac{1}{2})^2=\alpha^2+\alpha+\frac{1}{4}=(\alpha+\frac{1}{2})+\alpha+\frac{1}{4}=2\alpha+\frac{3}{4}
\]

したがって，

\[
\alpha^8=(\alpha^4)^2=(2\alpha+\frac{3}{4})^2=4\alpha^2+3\alpha+\frac{9}{16}=4(\alpha+\frac{1}{2})+3\alpha+\frac{9}{16}=7\alpha+\frac{41}{16}
\]

よって，

\[
\alpha^8=7\cdot\frac{1+\sqrt{3}}{2}+\frac{41}{16}=\frac{56(1+\sqrt{3})+41}{16}=\frac{97+56\sqrt{3}}{16}
\]

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% 側注1
右辺を根号のみの形にする．

% 側注2
$\alpha^3$は，$\alpha^2=-2\alpha+1$を用いて，次数を下げる．$\alpha^4$は，$\alpha^3$の結果を用いて次数を下げる．なお，$\alpha^4=(\alpha^2)^2=(-2\alpha+1)^2$を用いてもよい．

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% 問題I1.2.8の側注：式の値（One More）★★★
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% 側注1
右辺を根号のみの形にする．

% 側注2
$\alpha^2=\alpha+\frac{1}{2}$を用いて，次数を下げる．なお，解答では$\alpha^8$に$\alpha=\frac{1+\sqrt{3}}{2}$を代入しているが，$\alpha^4$に$\alpha=\frac{1+\sqrt{3}}{2}$を代入して，その結果を2乗することで$\alpha^8$を求めてもよい．

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(1) 直接$\alpha$を代入して求めることもできるが，計算に手間が掛かる．計算を簡単にするために，$\alpha=\sqrt{2}-1$を根号を含まない形に変形することを考える．$\alpha=\sqrt{2}-1$を移項すると，$\alpha+1=\sqrt{2}$となる．この式の両辺を2乗すると，根号を含まない形が得られる．

(2) $\alpha$の値を$\alpha^4+\alpha^3+\alpha^2+\alpha+1$に直接代入しようとすると，計算に手間が掛かる．そこで，(1)の結果を利用することを考える．(1)より，$\alpha^2+2\alpha-1=0$であり，移項すると，$\alpha^2=-2\alpha+1\cdots(\mathrm{i})$が得られる．これより，$\alpha^2$は$\alpha$の1次式で表されるので，$\alpha^3=\alpha\cdot\alpha^2$は，(i)を用いることで次数を下げることができる．$\alpha^4$も同様の操作で次数を下げ，$\alpha$の1次式で表す．

$\underbrace{\alpha\cdot\alpha^2}_{3\text{次式}}=\underbrace{\alpha(-2\alpha+1)}_{2\text{次式}}$

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次数の高い式の値を求めるには，次数を下げることを考える．

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数学IIで学習する，多項式の除法を用いると，$\alpha^4+\alpha^3+\alpha^2+\alpha+1$は次のように変形できる．

\[
\alpha^4+\alpha^3+\alpha^2+\alpha+1=(\alpha^2+2\alpha-1)(\alpha^2-\alpha+4)-8\alpha+5
\]

ここで，$\alpha^2+2\alpha-1=0$より，$\alpha^4+\alpha^3+\alpha^2+\alpha+1$の値は$-8\alpha+5$の値と等しいことがわかる．