n n n を正の整数とする.
複素数を係数とする
n n n 次の多項式
f ( x ) f(x) f ( x ) に対して, 方程式
f ( x ) = 0 f(x) = 0 f ( x ) = 0 の相異なる解の個数は
n n n 以下である.
証明
α 1 , \alpha _1, α 1 , ⋯ , \cdots, ⋯ , α r \alpha _r α r を
f ( x ) = 0 f(x) = 0 f ( x ) = 0 の相異なる解とする.
因数定理により,
f ( x ) f(x) f ( x ) は
x − α 1 , x-\alpha _1, x − α 1 , ⋯ , \cdots, ⋯ , x − α r x-\alpha _r x − α r で割り切れる.
よって,
f ( x ) f(x) f ( x ) は
( x − α 1 ) ⋯ ( x − α r ) (x-\alpha _1)\cdots (x-\alpha _r) ( x − α 1 ) ⋯ ( x − α r ) で割り切れ,
0 0 0 でない多項式
q ( x ) q(x) q ( x ) を用いて
f ( x ) = ( x − α 1 ) ⋯ ( x − α r ) q ( x ) f(x) = (x-\alpha _1)\cdots (x-\alpha _r)q(x) f ( x ) = ( x − α 1 ) ⋯ ( x − α r ) q ( x )
と書けるから, 両辺の次数を比較すると,
n = r + deg q ( x ) ≧ r n = r+\deg q(x) \geqq r n = r + deg q ( x ) ≧ r
が得られる.
累乗根は, 多くの教科書で指数関数の章で定義されるが, 高次方程式を解くうえで基本的な概念であるため (理論的には累乗根を使わずに解ける方程式は珍しい), この章で導入されることが望ましい.
(1) 正の整数 n , n, n , a a a x n = a x^n = a x n = a n n n a a a a a a n n n n n n 2 2 2 3 3 3 ⋯ , \cdots, ⋯ , 累乗根 (radical root) と呼ぶ.
(2) 正の整数 n , n, n , a a a x n = a x^n = a x n = a x x x 1 1 1 a n \sqrt[n]{a} n a
m , m, m , n n n を正の整数,
a , a, a , b b b を正の数とする.
(A)
a n b n = a b n \sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab} n a n b = n a b (A)'
a n b n = a b n \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}} n b n a = n b a (A)''
a n b n = a b n \sqrt[n]{a^nb} = a\sqrt[n]{b} n a n b = a n b
(B) ( a n ) m = a m n (\sqrt[n]{a})^m = \sqrt[n]{a^m} ( n a ) m = n a m (C) a m n = a m n \sqrt[n]{\sqrt[m]{a}} = \sqrt[mn]{a} n m a = m n a
が成り立つことを示せ.
2022 / 11 / 23 2022/11/23 2 0 2 2 / 1 1 / 2 3 2022 / 11 / 25 2022/11/25 2 0 2 2 / 1 1 / 2 5 解答例
c n \sqrt[n]{c} n c ( c > 0 ) (c > 0) ( c > 0 ) は
x n = c x^n = c x n = c を満たすただ
1 1 1 つの正の数であることを使う.
(A) 指数法則により
( a n b n ) n = ( a n ) n ( b n ) n = a b (\sqrt[n]{a}\sqrt[n]{b})^n = (\sqrt[n]{a})^n(\sqrt[n]{b})^n = ab ( n a n b ) n = ( n a ) n ( n b ) n = a b a n b n = a b n \sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab} n a n b = n a b (A)' 指数法則により
( a n b n ) n = ( a n ) n ( b n ) n = a b \left(\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\right) ^n = \frac{(\sqrt[n]{a})^n}{(\sqrt[n]{b})^n} = \frac{a}{b} ( n b n a ) n = ( n b ) n ( n a ) n = b a a n b n = a b n \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}} n b n a = n b a (A)'' (A) において a a a a n a^n a n a n b n = a n n b n = a b n \sqrt[n]{a^nb} = \sqrt[n]{a^n}\sqrt[n]{b} = a\sqrt[n]{b} n a n b = n a n n b = a n b (B) 指数法則により
{ ( a n ) m } n = ( a n ) m n = { ( a n ) n } m = a m \{ (\sqrt[n]{a})^m\} ^n = (\sqrt[n]{a})^{mn} = \{ (\sqrt[n]{a})^n\} ^m = a^m { ( n a ) m } n = ( n a ) m n = { ( n a ) n } m = a m ( a n ) m = a m n (\sqrt[n]{a})^m = \sqrt[n]{a^m} ( n a ) m = n a m (C) 指数法則により
( a m n ) m n = { ( a m n ) n } m = ( a m ) m = a (\sqrt[n]{\sqrt[m]{a}})^{mn} = \{ (\sqrt[n]{\sqrt[m]{a}})^n\} ^m = (\sqrt[m]{a})^m = a ( n m a ) m n = { ( n m a ) n } m = ( m a ) m = a a m n = a m n \sqrt[n]{\sqrt[m]{a}} = \sqrt[mn]{a} n m a = m n a
n n n を正の整数,
a a a を
n n n 乗数 (ある整数の
n n n 乗の形に表せる整数) でない正の整数とする.
(1) a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0 a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0 = 0 a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0 a k a_k a k a 0 a n ≠ 0 a_0a_n \neq 0 a 0 a n = 0 α \alpha α α \alpha α α = p q \alpha = \dfrac{p}{q} α = q p p p p a 0 a_0 a 0 q q q a n a_n a n (2) (1) の結果を使って, a n \sqrt[n]{a} n a
2019 / 05 / 01 2019/05/01 2 0 1 9 / 0 5 / 0 1 2024 / 02 / 29 2024/02/29 2 0 2 4 / 0 2 / 2 9 解答例
(1) α = p q \alpha = \dfrac{p}{q} α = q p p , p, p , q q q q > 0 q > 0 q > 0 a n α n + a n − 1 α n − 1 + ⋯ + a 1 α + a 0 = 0 a n ( p q ) n + a n − 1 ( p q ) n − 1 + ⋯ + a 1 ( p q ) + a 0 = 0 a n p n + a n − 1 p n − 1 q + ⋯ + a 1 p q n − 1 + a 0 q n = 0 \begin{aligned}
a_n\alpha ^n+a_{n-1}\alpha ^{n-1}+\cdots +a_1\alpha +a_0 &= 0 \\
a_n\left(\frac{p}{q}\right) ^n\!+a_{n-1}\left(\frac{p}{q}\right) ^{n-1}\!+\cdots +a_1\left(\frac{p}{q}\right) +a_0 &= 0 \\
a_np^n+a_{n-1}p^{n-1}q+\cdots +a_1pq^{n-1}+a_0q^n &= 0
\end{aligned} a n α n + a n − 1 α n − 1 + ⋯ + a 1 α + a 0 a n ( q p ) n + a n − 1 ( q p ) n − 1 + ⋯ + a 1 ( q p ) + a 0 a n p n + a n − 1 p n − 1 q + ⋯ + a 1 p q n − 1 + a 0 q n = 0 = 0 = 0 p ( a n p n − 1 + a n − 1 p n − 2 q + ⋯ + a 1 q n − 1 ) = − a 0 q n p(a_np^{n-1}+a_{n-1}p^{n-2}q+\cdots +a_1q^{n-1}) = -a_0q^n p ( a n p n − 1 + a n − 1 p n − 2 q + ⋯ + a 1 q n − 1 ) = − a 0 q n p p p − a 0 q n -a_0q^n − a 0 q n p , p, p , q q q p p p a 0 a_0 a 0 q ( a n − 1 p n − 1 + ⋯ + a 1 p q n − 2 + a 0 q n − 1 ) = − a n p n q(a_{n-1}p^{n-1}+\cdots +a_1pq^{n-2}+a_0q^{n-1}) = -a_np^n q ( a n − 1 p n − 1 + ⋯ + a 1 p q n − 2 + a 0 q n − 1 ) = − a n p n q q q − a n p n -a_np^n − a n p n p , p, p , q q q q q q a n a_n a n α \alpha α α = p q \alpha = \dfrac{p}{q} α = q p p p p a 0 a_0 a 0 q q q a n a_n a n (2) x n − a = 0 x^n-a = 0 x n − a = 0 a a a a a a d d d a a a n n n a ≠ d n a \neq d^n a = d n d n − a ≠ 0 d^n-a \neq 0 d n − a = 0 d d d x n − a = 0 x^n-a = 0 x n − a = 0 x n − a = 0 x^n-a = 0 x n − a = 0 x n − a = 0 x^n-a = 0 x n − a = 0 x = a n x = \sqrt[n]{a} x = n a
別解
(2) a n \sqrt[n]{a} n a a a a n n n a n \sqrt[n]{a} n a c , c, c , d d d ( d ≠ 0 ) (d \neq 0) ( d = 0 ) a n = c d \sqrt[n]{a} = \frac{c}{d} n a = d c n n n a = c n d n a = \dfrac{c^n}{d^n} a = d n c n a d n = c n ad^n = c^n a d n = c n
となる.
各素数 p p p x ( ≠ 0 ) x\,(\neq 0) x ( = 0 ) p p p o r d p ( x ) \mathrm{ord}_p(x) o r d p ( x ) o r d p ( x y ) = o r d p ( x ) + o r d p ( y ) ( x y ≠ 0 ) \mathrm{ord}_p(xy) = \mathrm{ord}_p(x)+\mathrm{ord}_p(y) \quad (xy \neq 0) o r d p ( x y ) = o r d p ( x ) + o r d p ( y ) ( x y = 0 ) o r d p ( a ) + n o r d p ( d ) = n o r d p ( c ) \mathrm{ord}_p(a)+n\,\mathrm{ord}_p(d) = n\,\mathrm{ord}_p(c) o r d p ( a ) + n o r d p ( d ) = n o r d p ( c ) o r d p ( a ) = n o r d p ( c ) − n o r d p ( d ) \mathrm{ord}_p(a) = n\,\mathrm{ord}_p(c)-n\,\mathrm{ord}_p(d) o r d p ( a ) = n o r d p ( c ) − n o r d p ( d ) n n n a a a n n n a a a n n n
参考
多項式 f ( x ) f(x) f ( x ) f ( x ) = 0 f(x) = 0 f ( x ) = 0 f ( x ) f(x) f ( x ) 「根」 (root) と呼ぶ.
(1) の結果は「有理根定理」 (rational root theorem) として知られており,
因数分解, 因数定理を使って解くことができる整数係数の高次方程式の解を探すのに有用である.
1 1 1 辺の長さが
1 1 1 の正三角形
A B C \mathrm{ABC} A B C において, 辺
A B \mathrm{AB} A B の
B \mathrm B B を越える延長上に
B D = 1 \mathrm{BD} = 1 B D = 1 なる点
D \mathrm D D をとる.
また, 辺
B C \mathrm{BC} B C の
C \mathrm C C を越える延長上の点
E \mathrm E E を,
A E \mathrm{AE} A E と
C D \mathrm{CD} C D の交点
F \mathrm F F について
E F = 1 \mathrm{EF} = 1 E F = 1 となるようにとり,
x = A F x = \mathrm{AF} x = A F とおく.
さらに, 点
F \mathrm F F を通って
A C \mathrm{AC} A C に平行な直線と
B E \mathrm{BE} B E の交点を
G \mathrm G G とおく.
(1) x x x F G , \mathrm{FG}, F G , C F \mathrm{CF} C F (2) x 4 + 2 x 3 − 2 x − 4 = 0 x^4+2x^3-2x-4 = 0 x 4 + 2 x 3 − 2 x − 4 = 0 (3) x x x
2018 / 06 / 09 2018/06/09 2 0 1 8 / 0 6 / 0 9 2022 / 05 / 18 2022/05/18 2 0 2 2 / 0 5 / 1 8 解答例
(1) △ E F G \triangle\mathrm{EFG} △ E F G △ E A C \triangle\mathrm{EAC} △ E A C E F : F G = E A : A C 1 : F G = ( x + 1 ) : 1 F G = 1 x + 1 \begin{aligned}
\mathrm{EF}:\mathrm{FG} &= \mathrm{EA}:\mathrm{AC} \\
1:\mathrm{FG} &= (x+1):1 \\
\mathrm{FG} &= \frac{1}{x+1}
\end{aligned} E F : F G 1 : F G F G = E A : A C = ( x + 1 ) : 1 = x + 1 1 B C = B D = 1 \mathrm{BC} = \mathrm{BD} = 1 B C = B D = 1 ∠ B C D = 6 0 ∘ ÷ 2 = 3 0 ∘ \angle\mathrm{BCD} = 60^\circ\div 2 = 30^\circ ∠ B C D = 6 0 ∘ ÷ 2 = 3 0 ∘ ∠ A C F = 18 0 ∘ − ∠ A C B − ∠ B C D = 18 0 ∘ − 6 0 ∘ − 3 0 ∘ = 9 0 ∘ \begin{aligned}
\angle\mathrm{ACF} &= 180^\circ -\angle\mathrm{ACB}-\angle\mathrm{BCD} \\
&= 180^\circ -60^\circ -30^\circ = 90^\circ
\end{aligned} ∠ A C F = 1 8 0 ∘ − ∠ A C B − ∠ B C D = 1 8 0 ∘ − 6 0 ∘ − 3 0 ∘ = 9 0 ∘ A C \mathrm{AC} A C F G \mathrm{FG} F G ∠ C F G = ∠ A C F = 9 0 ∘ , ∠ F G C = ∠ A C B = 6 0 ∘ \begin{aligned}
\angle\mathrm{CFG} &= \angle\mathrm{ACF} = 90^\circ, \\
\angle\mathrm{FGC} &= \angle\mathrm{ACB} = 60^\circ
\end{aligned} ∠ C F G ∠ F G C = ∠ A C F = 9 0 ∘ , = ∠ A C B = 6 0 ∘ C F = 3 F G = 3 x + 1 \mathrm{CF} = \sqrt 3\mathrm{FG} = \frac{\sqrt 3}{x+1} C F = 3 F G = x + 1 3 (2) △ A C F \triangle\mathrm{ACF} △ A C F x 2 = 1 2 + ( 3 x + 1 ) 2 x 2 ( x + 1 ) 2 = ( x + 1 ) 2 + 3 x 2 ( x 2 + 2 x + 1 ) = ( x 2 + 2 x + 1 ) + 3 \begin{aligned}
x^2 &= 1^2+\left(\frac{\sqrt 3}{x+1}\right) ^2 \\
x^2(x+1)^2 &= (x+1)^2+3 \\
x^2(x^2+2x+1) &= (x^2+2x+1)+3
\end{aligned} x 2 x 2 ( x + 1 ) 2 x 2 ( x 2 + 2 x + 1 ) = 1 2 + ( x + 1 3 ) 2 = ( x + 1 ) 2 + 3 = ( x 2 + 2 x + 1 ) + 3 x 4 + 2 x 3 − 2 x − 4 = 0 x^4+2x^3-2x-4 = 0 x 4 + 2 x 3 − 2 x − 4 = 0 (3) 左辺を変形すると
x 3 ( x + 2 ) − 2 ( x + 2 ) = 0 ( x 3 − 2 ) ( x + 2 ) = 0 \begin{aligned}
x^3(x+2)-2(x+2) &= 0 \\
(x^3-2)(x+2) &= 0
\end{aligned} x 3 ( x + 2 ) − 2 ( x + 2 ) ( x 3 − 2 ) ( x + 2 ) = 0 = 0 x x x x = 2 3 x = \sqrt[3]{2} x = 3 2
参考
ギリシアの三大作図問題として, 与えられた立方体の
2 2 2 倍の体積をもつ立方体の
1 1 1 辺の長さは作図可能であるか, つまり
2 3 \sqrt[3]{2} 3 2 は定規とコンパスのみで作図できるか, という問題がある.
この問題は,
「立方体倍積問題」 (cube-doubling problem) または
「デロスの問題」 (Delian problem) などと呼ばれ, 長い間未解決であったが,
19 19 1 9 世紀に不可能であることが証明された (ピエール・ヴァンツェル,
1837 1837 1 8 3 7 年).
しかし, 本問で見たように,
2 3 \sqrt[3]{2} 3 2 は目盛付きの定規とコンパスを使えば作図できる (
E F = 1 \mathrm{EF} = 1 E F = 1 を測るために定規の目盛を使う).
実数係数方程式
f ( x ) = 0 f(x) = 0 f ( x ) = 0 が複素数
α \alpha α を解にもつとき,
f ( x ) = 0 f(x) = 0 f ( x ) = 0 は共役複素数
α ˉ \bar\alpha α ˉ も解にもつ.
証明
次のことを示せ.
(1) 複素数 α , \alpha, α , β \beta β α + β ‾ = α ˉ + β ˉ , α β ‾ = α ˉ β ˉ \overline{\alpha +\beta} = \bar\alpha +\bar\beta, \quad \overline{\alpha\beta} = \bar\alpha\bar\beta α + β = α ˉ + β ˉ , α β = α ˉ β ˉ (2) 実数係数方程式 f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α f ( x ) = 0 f(x) = 0 f ( x ) = 0 α ˉ \bar\alpha α ˉ
2022 / 08 / 05 2022/08/05 2 0 2 2 / 0 8 / 0 5 2022 / 08 / 06 2022/08/06 2 0 2 2 / 0 8 / 0 6 解答例
(1) 複素数 α = a 1 + a 2 i , \alpha = a_1+a_2i, α = a 1 + a 2 i , β = b 1 + b 2 i \beta = b_1+b_2i β = b 1 + b 2 i a 1 , a_1, a 1 , a 2 , a_2, a 2 , b 1 , b_1, b 1 , b 2 b_2 b 2 α + β ‾ = ( a 1 + a 2 i ) + ( b 1 + b 2 i ) ‾ = ( a 1 + b 1 ) + ( a 2 + b 2 ) i ‾ = ( a 1 + b 1 ) − ( a 2 + b 2 ) i = ( a 1 − a 2 i ) + ( b 1 − b 2 i ) = α ˉ + β ˉ , α β ‾ = ( a 1 + a 2 i ) ( b 1 + b 2 i ) ‾ = ( a 1 b 1 − a 2 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i ‾ = ( a 1 b 1 − a 2 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i = ( a 1 − a 2 i ) ( b 1 − b 2 i ) = α ˉ β ˉ \begin{aligned}
\overline{\alpha +\beta} &= \overline{(a_1+a_2i)+(b_1+b_2i)} \\
&= \overline{(a_1+b_1)+(a_2+b_2)i} \\
&= (a_1+b_1)-(a_2+b_2)i \\
&= (a_1-a_2i)+(b_1-b_2i) \\
&= \bar\alpha +\bar\beta, \\
\overline{\alpha\beta} &= \overline{(a_1+a_2i)(b_1+b_2i)} \\
&= \overline{(a_1b_1-a_2b_2)+(a_1b_2+a_2b_1)i} \\
&= (a_1b_1-a_2b_2)-(a_1b_2+a_2b_1)i \\
&= (a_1-a_2i)(b_1-b_2i) \\
&= \bar\alpha\bar\beta
\end{aligned} α + β α β = ( a 1 + a 2 i ) + ( b 1 + b 2 i ) = ( a 1 + b 1 ) + ( a 2 + b 2 ) i = ( a 1 + b 1 ) − ( a 2 + b 2 ) i = ( a 1 − a 2 i ) + ( b 1 − b 2 i ) = α ˉ + β ˉ , = ( a 1 + a 2 i ) ( b 1 + b 2 i ) = ( a 1 b 1 − a 2 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i = ( a 1 b 1 − a 2 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i = ( a 1 − a 2 i ) ( b 1 − b 2 i ) = α ˉ β ˉ (2) 実数係数多項式 f ( x ) = a n x n + ⋯ + a 1 x + a 0 f(x) = a_nx^n+\cdots +a_1x+a_0 f ( x ) = a n x n + ⋯ + a 1 x + a 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 α \alpha α f ( α ) = 0 f(\alpha ) = 0 f ( α ) = 0 f ( α ) ‾ = 0 ˉ a n α n + ⋯ + a 1 α + a 0 ‾ = 0 a n α n ‾ + ⋯ + a 1 α ‾ + a 0 ‾ = 0 a n ‾ α ˉ n + ⋯ + a 1 ‾ α ˉ + a 0 ‾ = 0 a n α ˉ n + ⋯ + a 1 α ˉ + a 0 = 0 f ( α ˉ ) = 0 \begin{aligned}
\overline{f(\alpha )} &= \bar 0 \\
\overline{a_n\alpha ^n+\cdots +a_1\alpha +a_0} &= 0 \\
\overline{a_n\alpha ^n}+\cdots +\overline{a_1\alpha}+\overline{a_0} &= 0 \\
\overline{a_n}\bar\alpha ^n+\cdots +\overline{a_1}\bar\alpha +\overline{a_0} &= 0 \\
a_n\bar\alpha ^n+\cdots +a_1\bar\alpha +a_0 &= 0 \\
f(\bar\alpha ) &= 0
\end{aligned} f ( α ) a n α n + ⋯ + a 1 α + a 0 a n α n + ⋯ + a 1 α + a 0 a n α ˉ n + ⋯ + a 1 α ˉ + a 0 a n α ˉ n + ⋯ + a 1 α ˉ + a 0 f ( α ˉ ) = 0 ˉ = 0 = 0 = 0 = 0 = 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 α ˉ \bar\alpha α ˉ
参考
集合 A A A x x x A ′ A' A ′ 1 1 1 f ( x ) f(x) f ( x ) A A A A ′ A' A ′ 「写像」 (map) と呼ぶ.
集合によっては, これを関数と呼ぶ.
四則演算が定義され, 結合法則, 交換法則, 分配法則を満たす, 少なくとも複数の要素をもつ集合を「体」 (field) と呼ぶ.
有理数全体の集合 Q , \mathbb Q, Q , R , \mathbb R, R , C \mathbb C C 「有理数体」 ,「実数体」 ,「複素数体」 と呼ばれる.
「体」K K K K ′ K' K ′ σ \sigma σ σ ( a + b ) = σ ( a ) + σ ( b ) , σ ( a b ) = σ ( a ) σ ( b ) , σ ( 1 ) = 1 \sigma (a+b) = \sigma (a)+\sigma (b), \quad \sigma (ab) = \sigma (a)\sigma (b), \quad \sigma (1) = 1 σ ( a + b ) = σ ( a ) + σ ( b ) , σ ( a b ) = σ ( a ) σ ( b ) , σ ( 1 ) = 1 「準同型写像」 (homomorphism) と呼ぶ.
C \mathbb C C C \mathbb C C f ( α ) = α ˉ f(\alpha ) = \bar\alpha f ( α ) = α ˉ
d d d を平方数でない整数とし, 有理数
a 1 , a_1, a 1 , a 2 a_2 a 2 を用いて
α = a 1 + a 2 d ⋯ [ 1 ] \alpha = a_1+a_2\sqrt d \quad \cdots [1] α = a 1 + a 2 d ⋯ [ 1 ]
の形に表される複素数
α \alpha α について, その全体の集合を
K K K とおき,
σ ( α ) = a 1 − a 2 d \sigma (\alpha ) = a_1-a_2\sqrt d σ ( α ) = a 1 − a 2 d
と定める (
K K K の要素は
[ 1 ] [1] [ 1 ] の形にただ
1 1 1 通りに表されることに注意,
こちら を参照).
次のことを示せ.
(1) 集合 K K K α , \alpha, α , β \beta β σ ( α + β ) = σ ( α ) + σ ( β ) , σ ( α β ) = σ ( α ) σ ( β ) \sigma (\alpha +\beta ) = \sigma (\alpha )+\sigma (\beta ), \quad \sigma (\alpha\beta ) = \sigma (\alpha )\sigma (\beta) σ ( α + β ) = σ ( α ) + σ ( β ) , σ ( α β ) = σ ( α ) σ ( β ) (2) 有理数係数方程式 f ( x ) = 0 f(x) = 0 f ( x ) = 0 K K K α \alpha α f ( x ) = 0 f(x) = 0 f ( x ) = 0 σ ( α ) \sigma (\alpha ) σ ( α )
2022 / 08 / 06 2022/08/06 2 0 2 2 / 0 8 / 0 6 2022 / 08 / 06 2022/08/06 2 0 2 2 / 0 8 / 0 6 解答例
(1) K K K α = a 1 + a 2 d , \alpha = a_1+a_2\sqrt d, α = a 1 + a 2 d , β = b 1 + b 2 d \beta = b_1+b_2\sqrt d β = b 1 + b 2 d a 1 , a_1, a 1 , a 2 , a_2, a 2 , b 1 , b_1, b 1 , b 2 b_2 b 2 σ ( α + β ) = σ ( ( a 1 + a 2 d ) + ( b 1 + b 2 d ) ) = σ ( ( a 1 + b 1 ) + ( a 2 + b 2 ) d ) = ( a 1 + b 1 ) − ( a 2 + b 2 ) d = ( a 1 − a 2 d ) + ( b 1 − b 2 d ) = σ ( α ) + σ ( β ) , σ ( α β ) = σ ( ( a 1 + a 2 d ) ( b 1 + b 2 d ) ) = σ ( ( a 1 b 1 + d a 2 b 2 ) + ( a 1 b 2 + a 2 b 1 ) d ) = ( a 1 b 1 + d a 2 b 2 ) − ( a 1 b 2 + a 2 b 1 ) d = ( a 1 − a 2 d ) ( b 1 − b 2 d ) = σ ( α ) σ ( β ) \begin{aligned}
\sigma (\alpha +\beta) &= \sigma ((a_1+a_2\sqrt d)+(b_1+b_2\sqrt d)) \\
&= \sigma ((a_1+b_1)+(a_2+b_2)\sqrt d) \\
&= (a_1+b_1)-(a_2+b_2)\sqrt d \\
&= (a_1-a_2\sqrt d)+(b_1-b_2\sqrt d) \\
&= \sigma (\alpha )+\sigma (\beta ), \\
\sigma (\alpha\beta ) &= \sigma ((a_1+a_2\sqrt d)(b_1+b_2\sqrt d)) \\
&= \sigma ((a_1b_1+da_2b_2)+(a_1b_2+a_2b_1)\sqrt d) \\
&= (a_1b_1+da_2b_2)-(a_1b_2+a_2b_1)\sqrt d \\
&= (a_1-a_2\sqrt d)(b_1-b_2\sqrt d) \\
&= \sigma (\alpha )\sigma (\beta )
\end{aligned} σ ( α + β ) σ ( α β ) = σ ( ( a 1 + a 2 d ) + ( b 1 + b 2 d ) ) = σ ( ( a 1 + b 1 ) + ( a 2 + b 2 ) d ) = ( a 1 + b 1 ) − ( a 2 + b 2 ) d = ( a 1 − a 2 d ) + ( b 1 − b 2 d ) = σ ( α ) + σ ( β ) , = σ ( ( a 1 + a 2 d ) ( b 1 + b 2 d ) ) = σ ( ( a 1 b 1 + d a 2 b 2 ) + ( a 1 b 2 + a 2 b 1 ) d ) = ( a 1 b 1 + d a 2 b 2 ) − ( a 1 b 2 + a 2 b 1 ) d = ( a 1 − a 2 d ) ( b 1 − b 2 d ) = σ ( α ) σ ( β ) (2) 有理数係数多項式 f ( x ) = a n x n + ⋯ + a 1 x + a 0 f(x) = a_nx^n+\cdots +a_1x+a_0 f ( x ) = a n x n + ⋯ + a 1 x + a 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 K K K α \alpha α f ( α ) = 0 f(\alpha ) = 0 f ( α ) = 0 σ \sigma σ σ ( f ( α ) ) = σ ( 0 ) σ ( a n α n + ⋯ + a 1 α + a 0 ) = 0 σ ( a n α n ) + ⋯ + σ ( a 1 α ) + σ ( a 0 ) = 0 σ ( a n ) σ ( α ) n + ⋯ + σ ( a 1 ) σ ( α ) + σ ( a 0 ) = 0 a n σ ( α ) n + ⋯ + a 1 σ ( α ) + a 0 = 0 f ( σ ( α ) ) = 0 \begin{aligned}
\sigma (f(\alpha )) &= \sigma (0) \\
\sigma (a_n\alpha ^n+\cdots +a_1\alpha +a_0) &= 0 \\
\sigma (a_n\alpha ^n)+\cdots +\sigma (a_1\alpha )+\sigma (a_0) &= 0 \\
\sigma (a_n)\sigma (\alpha )^n+\cdots +\sigma (a_1)\sigma (\alpha )+\sigma (a_0) &= 0 \\
a_n\sigma (\alpha )^n+\cdots +a_1\sigma (\alpha )+a_0 &= 0 \\
f(\sigma (\alpha )) &= 0
\end{aligned} σ ( f ( α ) ) σ ( a n α n + ⋯ + a 1 α + a 0 ) σ ( a n α n ) + ⋯ + σ ( a 1 α ) + σ ( a 0 ) σ ( a n ) σ ( α ) n + ⋯ + σ ( a 1 ) σ ( α ) + σ ( a 0 ) a n σ ( α ) n + ⋯ + a 1 σ ( α ) + a 0 f ( σ ( α ) ) = σ ( 0 ) = 0 = 0 = 0 = 0 = 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 σ ( α ) \sigma (\alpha ) σ ( α )
参考
「体」K K K K K K K K K f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x ) K K K f ( x ) f(x) f ( x ) K K K 「既約」 (irreducible) であるという.
また, 多項式 f ( x ) f(x) f ( x ) f ( x ) = 0 f(x) = 0 f ( x ) = 0 f ( x ) f(x) f ( x ) 「根」 (root) と呼ぶ.
本問において, α = a 1 + a 2 d , \alpha = a_1+a_2\sqrt d, α = a 1 + a 2 d , σ ( α ) = a 1 − a 2 d \sigma (\alpha ) = a_1-a_2\sqrt d σ ( α ) = a 1 − a 2 d Q \mathbb Q Q x 2 − 2 a 1 x + ( a 1 2 − d a 2 2 ) x^2-2a_1x+(a_1{}^2-da_2{}^2) x 2 − 2 a 1 x + ( a 1 2 − d a 2 2 ) α , \alpha, α , β \beta β K K K α , \alpha, α , β \beta β K K K K K K K K K σ ( α ) \sigma (\alpha ) σ ( α ) 2 2 2 こちら を参照) の理論において複素共役のような役割を果たす.
ω \omega ω を
x 3 = 1 x^3 = 1 x 3 = 1 の虚数解の
1 1 1 つとする.
(1) ω 2 + ω + 1 \omega ^2+\omega +1 ω 2 + ω + 1 (2) a , a, a , b b b ( a ω + b ω 2 ) ( a ω 2 + b ω ) (a\omega +b\omega ^2)(a\omega ^2+b\omega ) ( a ω + b ω 2 ) ( a ω 2 + b ω ) (3) ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) (x+y+z)(x+\omega y+\omega ^2z)(x+\omega ^2y+\omega z) ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z )
2022 / 11 / 01 2022/11/01 2 0 2 2 / 1 1 / 0 1 2022 / 11 / 03 2022/11/03 2 0 2 2 / 1 1 / 0 3 解答例
(1) ω 3 = 1 \omega ^3 = 1 ω 3 = 1 ω 3 − 1 = 0 \omega ^3-1 = 0 ω 3 − 1 = 0 ( ω − 1 ) ( ω 2 + ω + 1 ) = 0 (\omega -1)(\omega ^2+\omega +1) = 0 ( ω − 1 ) ( ω 2 + ω + 1 ) = 0 ω ≠ 1 \omega \neq 1 ω = 1 ω 2 + ω + 1 = 0 \omega ^2+\omega +1 = 0 ω 2 + ω + 1 = 0 (2) (1) により, 与式は
( a ω + b ω 2 ) ( a ω 2 + b ω ) = ( a 2 + b 2 ) ω 3 + a b ( ω 2 + ω 4 ) = ( a 2 + b 2 ) ⋅ 1 + a b ( ω 2 + ω ) = a 2 + b 2 + a b ⋅ ( − 1 ) = a 2 − a b + b 2 \begin{aligned}
(a\omega +b\omega ^2)(a\omega ^2+b\omega ) &= (a^2+b^2)\omega ^3+ab(\omega ^2+\omega ^4) \\
&= (a^2+b^2)\cdot 1+ab (\omega ^2+\omega ) \\
&= a^2+b^2+ab\cdot (-1) \\
&= a^2-ab+b^2
\end{aligned} ( a ω + b ω 2 ) ( a ω 2 + b ω ) = ( a 2 + b 2 ) ω 3 + a b ( ω 2 + ω 4 ) = ( a 2 + b 2 ) ⋅ 1 + a b ( ω 2 + ω ) = a 2 + b 2 + a b ⋅ ( − 1 ) = a 2 − a b + b 2 (3) (2) により
( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) = x 2 + ( ω y + ω 2 z + ω 2 y + ω z ) x + ( ω y + ω 2 z ) ( ω 2 y + ω z ) = x 2 + ( ω + ω 2 ) ( y + z ) x + ( y 2 − y z + z 2 ) = x 2 − ( y + z ) x + ( y 2 − y z + z 2 ) \begin{aligned}
&(x+\omega y+\omega ^2z)(x+\omega ^2y+\omega z) \\
&= x^2+(\omega y+\omega ^2z+\omega ^2y+\omega z)x+(\omega y+\omega ^2z)(\omega ^2y+\omega z) \\
&= x^2+(\omega +\omega ^2)(y+z)x+(y^2-yz+z^2) \\
&= x^2-(y+z)x+(y^2-yz+z^2) \\
\end{aligned} ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) = x 2 + ( ω y + ω 2 z + ω 2 y + ω z ) x + ( ω y + ω 2 z ) ( ω 2 y + ω z ) = x 2 + ( ω + ω 2 ) ( y + z ) x + ( y 2 − y z + z 2 ) = x 2 − ( y + z ) x + ( y 2 − y z + z 2 ) ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) = ( x + y + z ) { x 2 − ( y + z ) x + ( y 2 − y z + z 2 ) } = x 3 − ( y + z ) x 2 + ( y 2 − y z + z 2 ) x + ( y + z ) x 2 − ( y + z ) 2 x + ( y + z ) ( y 2 − y z + z 2 ) = x 3 + ( y 2 − y z + z 2 ) x − ( y 2 + 2 y z + z 2 ) x + ( y 3 + z 3 ) = x 3 + y 3 + z 3 − 3 x y z \begin{aligned}
&(x+y+z)(x+\omega y+\omega ^2z)(x+\omega ^2y+\omega z) \\
&= (x+y+z)\{ x^2-(y+z)x+(y^2-yz+z^2)\} \\
&= x^3-(y+z)x^2+(y^2-yz+z^2)x \\
&\quad +(y+z)x^2-(y+z)^2x+(y+z)(y^2-yz+z^2) \\
&= x^3+(y^2-yz+z^2)x-(y^2+2yz+z^2)x+(y^3+z^3) \\
&= x^3+y^3+z^3-3xyz
\end{aligned} ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) = ( x + y + z ) { x 2 − ( y + z ) x + ( y 2 − y z + z 2 ) } = x 3 − ( y + z ) x 2 + ( y 2 − y z + z 2 ) x + ( y + z ) x 2 − ( y + z ) 2 x + ( y + z ) ( y 2 − y z + z 2 ) = x 3 + ( y 2 − y z + z 2 ) x − ( y 2 + 2 y z + z 2 ) x + ( y 3 + z 3 ) = x 3 + y 3 + z 3 − 3 x y z
参考
等式
x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z ) x^3\!+\!y^3\!+\!z^3\!-\!3xyz = (x\!+\!y\!+\!z)(x\!+\!\omega y\!+\!\omega ^2z)(x\!+\!\omega ^2y\!+\!\omega z) x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x + ω y + ω 2 z ) ( x + ω 2 y + ω z )
は
3 3 3 次方程式の
「オイラーの解法」 に利用される (
こちら を参照).
3 3 3 次方程式は, 解と係数の関係を利用して,
2 2 2 次方程式に帰着させて解くことができる.
3 3 3 次方程式
x 3 − x − 1 = 0 ⋯ [ 1 ] x^3-x-1 = 0\ \cdots [1] x 3 − x − 1 = 0 ⋯ [ 1 ] について, 次の問に答えよ.
(1) x = u + v x = u+v x = u + v u 3 + v 3 − 1 = 3 u v − 1 = 0 ⋯ [ 2 ] u^3+v^3-1 = 3uv-1 = 0\ \cdots [2] u 3 + v 3 − 1 = 3 u v − 1 = 0 ⋯ [ 2 ] [ 1 ] [1] [ 1 ] (2) 複素数 u , u, u , v v v [ 2 ] [2] [ 2 ] u 3 , u^3, u 3 , v 3 v^3 v 3 2 2 2 1 1 1 (3) 1 1 1 ω = − 1 + 3 i 2 \omega = \dfrac{-1+\sqrt 3i}{2} ω = 2 − 1 + 3 i [ 1 ] [1] [ 1 ]
2018 / 02 / 12 2018/02/12 2 0 1 8 / 0 2 / 1 2 2024 / 04 / 13 2024/04/13 2 0 2 4 / 0 4 / 1 3 解答例
(1) x = u + v x = u+v x = u + v u 3 + v 3 − 1 = 3 u v − 1 = 0 ⋯ [ 2 ] u^3+v^3-1 = 3uv-1 = 0\ \cdots [2] u 3 + v 3 − 1 = 3 u v − 1 = 0 ⋯ [ 2 ] x 3 − x − 1 = ( u + v ) 3 − ( u + v ) − 1 = u 3 + v 3 + 3 u v ( u + v ) − ( u + v ) − 1 = ( u 3 + v 3 − 1 ) + ( 3 u v − 1 ) ( u + v ) = 0 + 0 ⋅ ( u + v ) = 0 ⋯ [ 1 ] \begin{aligned}
x^3-x-1 &= (u+v)^3-(u+v)-1 \\
&= u^3+v^3+3uv(u+v)-(u+v)-1 \\
&= (u^3+v^3-1)+(3uv-1)(u+v) \\
&= 0+0\cdot (u+v) = 0 \quad \cdots [1]
\end{aligned} x 3 − x − 1 = ( u + v ) 3 − ( u + v ) − 1 = u 3 + v 3 + 3 u v ( u + v ) − ( u + v ) − 1 = ( u 3 + v 3 − 1 ) + ( 3 u v − 1 ) ( u + v ) = 0 + 0 ⋅ ( u + v ) = 0 ⋯ [ 1 ] (2) 複素数 u , u, u , v v v [ 2 ] [2] [ 2 ] u 3 + v 3 = 1 , u 3 v 3 = ( u v ) 3 = ( 1 3 ) 3 = 1 27 u^3+v^3 = 1, \quad u^3v^3 = (uv)^3 = \left(\frac{1}{3}\right) ^3 = \dfrac{1}{27} u 3 + v 3 = 1 , u 3 v 3 = ( u v ) 3 = ( 3 1 ) 3 = 2 7 1 2 2 2 u 3 , u^3, u 3 , v 3 v^3 v 3 2 2 2 t 2 − t + 1 27 = 0 ⋯ [ 3 ] t^2-t+\dfrac{1}{27} = 0 \quad \cdots [3] t 2 − t + 2 7 1 = 0 ⋯ [ 3 ] (3) [ 3 ] [3] [ 3 ] 2 2 2 t = 1 2 ( 1 ± 1 − 4 27 ) = 1 2 ± 1 6 23 3 t = \frac{1}{2}\left( 1\pm\sqrt{1-\frac{4}{27}}\right) = \frac{1}{2}\pm\frac{1}{6}\sqrt{\frac{23}{3}} t = 2 1 ( 1 ± 1 − 2 7 4 ) = 2 1 ± 6 1 3 2 3 u 3 = 1 2 + 1 6 23 3 , u^3 = \dfrac{1}{2}+\dfrac{1}{6}\sqrt{\dfrac{23}{3}}, u 3 = 2 1 + 6 1 3 2 3 , v 3 = 1 2 − 1 6 23 3 v^3 = \dfrac{1}{2}-\dfrac{1}{6}\sqrt{\dfrac{23}{3}} v 3 = 2 1 − 6 1 3 2 3 u = 1 2 + 1 6 23 3 3 ω k ( k = 0 , 1 , 2 ) , v = 1 2 − 1 6 23 3 3 ω l ( l = 0 , 1 , 2 ) \begin{aligned}
u &= \sqrt[3]{\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}}\omega ^k \quad (k = 0,1,2), \\
v &= \sqrt[3]{\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}}\omega ^l \quad (l = 0,1,2)
\end{aligned} u v = 3 2 1 + 6 1 3 2 3 ω k ( k = 0 , 1 , 2 ) , = 3 2 1 − 6 1 3 2 3 ω l ( l = 0 , 1 , 2 ) u , u, u , v v v u 3 + v 3 − 1 = 0 u^3+v^3-1 = 0 u 3 + v 3 − 1 = 0 u v = ( 1 2 + 1 6 23 3 ) ( 1 2 − 1 6 23 3 ) 3 ω k + l = 1 4 − 23 4 ⋅ 27 3 ω k + l = 1 27 3 ω k + l = 1 3 ω k + l \begin{aligned}
uv &= \sqrt[3]{\left(\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}\right)\left(\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}\right)}\omega ^{k+l} \\
&= \sqrt[3]{\frac{1}{4}-\frac{23}{4\cdot 27}}\omega ^{k+l} = \sqrt[3]{\frac{1}{27}}\omega ^{k+l} = \frac{1}{3}\omega ^{k+l}
\end{aligned} u v = 3 ( 2 1 + 6 1 3 2 3 ) ( 2 1 − 6 1 3 2 3 ) ω k + l = 3 4 1 − 4 ⋅ 2 7 2 3 ω k + l = 3 2 7 1 ω k + l = 3 1 ω k + l 3 u v − 1 = 0 3uv-1 = 0 3 u v − 1 = 0 ( k , l ) = ( 0 , 0 ) , (k,l) = (0,0), ( k , l ) = ( 0 , 0 ) , ( 1 , 2 ) , (1,2), ( 1 , 2 ) , ( 2 , 1 ) (2,1) ( 2 , 1 ) x = u + v x = u+v x = u + v [ 1 ] [1] [ 1 ] 3 3 3 3 3 3 [ 1 ] [1] [ 1 ] x = 1 2 + 1 6 23 3 3 + 1 2 − 1 6 23 3 3 , 1 2 + 1 6 23 3 3 ω + 1 2 − 1 6 23 3 3 ω 2 , 1 2 + 1 6 23 3 3 ω 2 + 1 2 − 1 6 23 3 3 ω \begin{aligned}
x = &\sqrt[3]{\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}}+\sqrt[3]{\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}}, \\
&\sqrt[3]{\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}}\omega +\sqrt[3]{\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}}\omega ^2, \\
&\sqrt[3]{\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}}\omega ^2+\sqrt[3]{\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}}\omega
\end{aligned} x = 3 2 1 + 6 1 3 2 3 + 3 2 1 − 6 1 3 2 3 , 3 2 1 + 6 1 3 2 3 ω + 3 2 1 − 6 1 3 2 3 ω 2 , 3 2 1 + 6 1 3 2 3 ω 2 + 3 2 1 − 6 1 3 2 3 ω
参考
2 2 2 3 3 3 4 4 4 3 3 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0 x^3+a_2x^2+a_1x+a_0 = 0 x 3 + a 2 x 2 + a 1 x + a 0 = 0 a k a_k a k x = X − a 2 3 x = X-\dfrac{a_2}{3} x = X − 3 a 2 X 3 + 3 p X + 2 q = 0 X^3+3pX+2q = 0 X 3 + 3 p X + 2 q = 0 p , p, p , q q q ω \omega ω 1 1 1 1 1 1 X = − q + q 2 + p 3 3 ω k + − q − q 2 + p 3 3 ω 3 − k ( k = 0 , 1 , 2 ) X = \sqrt[3]{-q+\sqrt{q^2+p^3}}\omega ^k+\sqrt[3]{-q-\sqrt{q^2+p^3}}\omega ^{3-k}\ (k = 0,1,2) X = 3 − q + q 2 + p 3 ω k + 3 − q − q 2 + p 3 ω 3 − k ( k = 0 , 1 , 2 ) 「カルダーノの解法」 (Cardano's method) と呼ばれることが多いが, 先に S・デル・フェッロと N・フォンタナ (タルタリア) によって発見された.
3 3 3 x 3 = x + 1 x^3 = x+1 x 3 = x + 1 「プラスチック数」 (plastic number) と呼ばれ,
P 1 = P 2 = 1 , P 3 = 2 , P n + 3 = P n + P n + 1 P_1 = P_2 = 1,\ P_3 = 2, \quad P_{n+3} = P_n+P_{n+1} P 1 = P 2 = 1 , P 3 = 2 , P n + 3 = P n + P n + 1 「パドヴァン数列」 (Padovan sequence) { P n } \{ P_n\} { P n } 2 2 2 { P n + 1 P n } \left\{\dfrac{P_{n+1}}{P_n}\right\} { P n P n + 1 } x x x P n + 3 P n + 2 ⋅ P n + 2 P n + 1 ⋅ P n + 1 P n = P n + 3 P n = 1 + P n + 1 P n \frac{P_{n+3}}{P_{n+2}}\cdot\frac{P_{n+2}}{P_{n+1}}\cdot\frac{P_{n+1}}{P_n} = \frac{P_{n+3}}{P_n} = 1+\frac{P_{n+1}}{P_n} P n + 2 P n + 3 ⋅ P n + 1 P n + 2 ⋅ P n P n + 1 = P n P n + 3 = 1 + P n P n + 1 x 3 = 1 + x x^3 = 1+x x 3 = 1 + x 「プラスチック数」は
P 1 = 0 , P 2 = 2 , P 3 = 3 , P n + 3 = P n + P n + 1 P_1 = 0,\ P_2 = 2,\ P_3 = 3, \quad P_{n+3} = P_n+P_{n+1} P 1 = 0 , P 2 = 2 , P 3 = 3 , P n + 3 = P n + P n + 1 「ペラン数列」 の隣り合う 2 2 2
(A) x 4 − 5 x 2 + 5 = 0 x^4-5x^2+5 = 0 x 4 − 5 x 2 + 5 = 0 (B) 80 x 4 − 100 x 2 + 1 = 0 80x^4-100x^2+1 = 0 8 0 x 4 − 1 0 0 x 2 + 1 = 0
2020 / 01 / 01 2020/01/01 2 0 2 0 / 0 1 / 0 1 2022 / 12 / 06 2022/12/06 2 0 2 2 / 1 2 / 0 6 解答例
(A) 与式を変形すると
x 4 − 2 ⋅ 5 2 x 2 + ( 5 2 ) 2 = − 5 + ( 5 2 ) 2 ( x 2 − 5 2 ) 2 = 5 4 x 2 − 5 2 = ± 5 2 x 2 = 5 ± 5 2 \begin{aligned}
x^4-2\cdot\frac{5}{2}x^2+\left(\frac{5}{2}\right) ^2 &= -5+\left(\frac{5}{2}\right) ^2 \\
\left( x^2-\frac{5}{2}\right) ^2 &= \frac{5}{4} \\
x^2-\frac{5}{2} &= \pm\frac{\sqrt 5}{2} \\
x^2 &= \frac{5\pm\sqrt 5}{2}
\end{aligned} x 4 − 2 ⋅ 2 5 x 2 + ( 2 5 ) 2 ( x 2 − 2 5 ) 2 x 2 − 2 5 x 2 = − 5 + ( 2 5 ) 2 = 4 5 = ± 2 5 = 2 5 ± 5 x = ± 5 ± 5 2 = ± 10 ± 2 5 2 x = \pm\sqrt{\frac{5\pm\sqrt 5}{2}} = \pm\frac{\sqrt{10\pm 2\sqrt 5}}{2} x = ± 2 5 ± 5 = ± 2 1 0 ± 2 5 (B) 与式を変形すると
x 4 − 5 4 x 2 + 1 80 = 0 x 4 − 2 ⋅ 5 8 x 2 + ( 5 8 ) 2 = − 1 80 + ( 5 8 ) 2 ( x 2 − 5 8 ) 2 = 121 320 x 2 − 5 8 = ± 11 8 5 x 2 = 25 ± 11 5 40 \begin{aligned}
x^4-\frac{5}{4}x^2+\frac{1}{80} &= 0 \\
x^4-2\cdot\frac{5}{8}x^2+\left(\frac{5}{8}\right) ^2 &= -\frac{1}{80}+\left(\frac{5}{8}\right) ^2 \\
\left( x^2-\frac{5}{8}\right) ^2 &= \frac{121}{320} \\
x^2-\frac{5}{8} &= \pm\frac{11}{8\sqrt 5} \\
x^2 &= \frac{25\pm 11\sqrt 5}{40}
\end{aligned} x 4 − 4 5 x 2 + 8 0 1 x 4 − 2 ⋅ 8 5 x 2 + ( 8 5 ) 2 ( x 2 − 8 5 ) 2 x 2 − 8 5 x 2 = 0 = − 8 0 1 + ( 8 5 ) 2 = 3 2 0 1 2 1 = ± 8 5 1 1 = 4 0 2 5 ± 1 1 5 x = ± 25 ± 11 5 40 = ± 1 2 25 ± 11 5 10 x = \pm\sqrt{\frac{25\pm 11\sqrt 5}{40}} = \pm\frac{1}{2}\sqrt{\frac{25\pm 11\sqrt 5}{10}} x = ± 4 0 2 5 ± 1 1 5 = ± 2 1 1 0 2 5 ± 1 1 5
参考
a x 4 + b x 2 + c ax^4+bx^2+c a x 4 + b x 2 + c a , a, a , b , b, b , c c c a ≠ 0 a \neq 0 a = 0 「複 2 2 2 (biquadratic polynomial) と呼ぶ.
1 1 1 1 1 1 1 , 1, 1 , 1 + 5 2 \dfrac{1+\sqrt 5}{2} 2 1 + 5 x x x x 2 = 1 2 + ( 1 + 5 2 ) 2 = 5 + 5 2 2 x 2 − 5 = 5 ( 2 x 2 − 5 ) 2 = 5 4 x 4 − 20 x 2 + 20 = 0 \begin{aligned}
&x^2 = 1^2+\left(\frac{1+\sqrt 5}{2}\right) ^2 = \frac{5+\sqrt 5}{2} \\
&2x^2-5 = \sqrt 5 \\
&(2x^2-5)^2 = 5 \\
&4x^4-20x^2+20 = 0
\end{aligned} x 2 = 1 2 + ( 2 1 + 5 ) 2 = 2 5 + 5 2 x 2 − 5 = 5 ( 2 x 2 − 5 ) 2 = 5 4 x 4 − 2 0 x 2 + 2 0 = 0 4 4 4 (B) の 4 4 4 1 2 25 + 11 5 10 \dfrac{1}{2}\sqrt{\dfrac{25+11\sqrt 5}{10}} 2 1 1 0 2 5 + 1 1 5 1 1 1 1 1 1 こちら を参照).
4 4 4 次方程式
x 4 + 4 x 3 + 16 x 2 + 64 x + 256 = 0 ⋯ [ ∗ ] x^4+4x^3+16x^2+64x+256 = 0\ \cdots [*] x 4 + 4 x 3 + 1 6 x 2 + 6 4 x + 2 5 6 = 0 ⋯ [ ∗ ] について考える.
(1) [ ∗ ] [*] [ ∗ ] x = X − 1 x = X-1 x = X − 1 4 4 4 1 1 1 3 3 3 X X X 4 4 4 f ( X ) = 0 f(X) = 0 f ( X ) = 0 (2) f ( X ) = ( X 2 + t ) 2 − ( p X + q ) 2 f(X) = (X^2+t)^2-(pX+q)^2 f ( X ) = ( X 2 + t ) 2 − ( p X + q ) 2 t , t, t , p , p, p , q q q 1 1 1 (3) [ ∗ ] [*] [ ∗ ]
2020 / 03 / 27 2020/03/27 2 0 2 0 / 0 3 / 2 7 2022 / 05 / 18 2022/05/18 2 0 2 2 / 0 5 / 1 8 解答例
(1) [ ∗ ] [*] [ ∗ ] x = X − 1 x = X-1 x = X − 1 [ ∗ ] ⟺ x 4 + 4 x 3 + 16 x 2 + 64 x + 256 = 0 ⟺ ( X − 1 ) 4 + 4 ( X − 1 ) 3 + 16 ( X − 1 ) 2 + 64 ( X − 1 ) + 256 = 0 ⟺ ( X 4 − 4 X 3 + 6 X 2 − 4 X + 1 ) + 4 ( X 3 − 3 X 2 + 3 X − 1 ) + 16 ( X 2 − 2 X + 1 ) + 64 ( X − 1 ) + 256 = 0 ⟺ X 4 + 10 X 2 + 40 X + 205 = 0 ⋯ [ 1 ] \begin{aligned}
[*] &\iff x^4+4x^3+16x^2+64x+256 = 0 \\
&\iff (X-1)^4+4(X-1)^3 \\
&\qquad\qquad +16(X-1)^2+64(X-1)+256 = 0 \\
&\iff (X^4\!-\!4X^3\!+\!6X^2\!-\!4X\!+\!1)\!+\!4(X^3\!-\!3X^2\!+\!3X\!-\!1) \\
&\qquad\qquad +16(X^2-2X+1)+64(X-1)+256 = 0 \\
&\iff X^4+10X^2+40X+205 = 0 \quad \cdots [1]
\end{aligned} [ ∗ ] ⟺ x 4 + 4 x 3 + 1 6 x 2 + 6 4 x + 2 5 6 = 0 ⟺ ( X − 1 ) 4 + 4 ( X − 1 ) 3 + 1 6 ( X − 1 ) 2 + 6 4 ( X − 1 ) + 2 5 6 = 0 ⟺ ( X 4 − 4 X 3 + 6 X 2 − 4 X + 1 ) + 4 ( X 3 − 3 X 2 + 3 X − 1 ) + 1 6 ( X 2 − 2 X + 1 ) + 6 4 ( X − 1 ) + 2 5 6 = 0 ⟺ X 4 + 1 0 X 2 + 4 0 X + 2 0 5 = 0 ⋯ [ 1 ] (2) X 4 + 10 X 2 + 40 X + 205 = ( X 2 + t ) 2 − ( p X + q ) 2 = X 4 + ( 2 t − p 2 ) X 2 − 2 p q X + ( t 2 − q 2 ) \begin{aligned}
&X^4+10X^2+40X+205 = (X^2+t)^2-(pX+q)^2 \\
&= X^4+(2t-p^2)X^2-2pqX+(t^2-q^2)
\end{aligned} X 4 + 1 0 X 2 + 4 0 X + 2 0 5 = ( X 2 + t ) 2 − ( p X + q ) 2 = X 4 + ( 2 t − p 2 ) X 2 − 2 p q X + ( t 2 − q 2 ) 10 = 2 t − p 2 , 10 = 2t-p^2, 1 0 = 2 t − p 2 , 40 = − 2 p q , 40 = -2pq, 4 0 = − 2 p q , 205 = t 2 − q 2 205 = t^2-q^2 2 0 5 = t 2 − q 2 p 2 = 2 ( t − 5 ) , p 2 q 2 = 400 , q 2 = t 2 − 205 p^2 = 2(t-5), \quad p^2q^2 = 400, \quad q^2 = t^2-205 p 2 = 2 ( t − 5 ) , p 2 q 2 = 4 0 0 , q 2 = t 2 − 2 0 5 2 ( t − 5 ) ( t 2 − 205 ) = 400 t 3 − 5 t 2 − 205 t + 825 = 0 \begin{aligned}
2(t-5)(t^2-205) &= 400 \\
t^3-5t^2-205t+825 &= 0
\end{aligned} 2 ( t − 5 ) ( t 2 − 2 0 5 ) t 3 − 5 t 2 − 2 0 5 t + 8 2 5 = 4 0 0 = 0 t = 15 , t = 15, t = 1 5 , p = 2 5 , p = 2\sqrt 5, p = 2 5 , q = − 2 5 q = -2\sqrt 5 q = − 2 5 (3) [ 1 ] [1] [ 1 ] ( X 2 + 15 ) 2 − ( 2 5 X − 2 5 ) 2 = 0 ( X 2 − 2 5 X + 15 + 2 5 ) ( X 2 + 2 5 X + 15 − 2 5 ) = 0 \begin{aligned}
&(X^2+15)^2-(2\sqrt 5X-2\sqrt 5)^2 = 0 \\
&(X^2-2\sqrt 5X+15+2\sqrt 5)(X^2+2\sqrt 5X+15-2\sqrt 5) = 0
\end{aligned} ( X 2 + 1 5 ) 2 − ( 2 5 X − 2 5 ) 2 = 0 ( X 2 − 2 5 X + 1 5 + 2 5 ) ( X 2 + 2 5 X + 1 5 − 2 5 ) = 0 X = 5 ± 10 + 2 5 i , − 5 ± 10 − 2 5 i \begin{aligned}
X = \sqrt 5\pm\sqrt{10+2\sqrt 5}i,\ -\sqrt 5\pm\sqrt{10-2\sqrt 5}i
\end{aligned} X = 5 ± 1 0 + 2 5 i , − 5 ± 1 0 − 2 5 i x = X − 1 x = X-1 x = X − 1 [ ∗ ] [*] [ ∗ ] x = − 1 + 5 ± 10 + 2 5 i , − 1 − 5 ± 10 − 2 5 i \begin{aligned}
x = &-1+\sqrt 5\pm\sqrt{10+2\sqrt 5}i, \\
&-1-\sqrt 5\pm\sqrt{10-2\sqrt 5}i
\end{aligned} x = − 1 + 5 ± 1 0 + 2 5 i , − 1 − 5 ± 1 0 − 2 5 i 4 4 4
参考
4 4 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0 x^4+a_3x^3+a_2x^2+a_1x+a_0 = 0 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0 a k a_k a k 「フェラーリの解法」 (Ferrari's method) と呼ばれる方法で解ける:
与式に x = X − a 3 4 x = X-\dfrac{a_3}{4} x = X − 4 a 3 3 3 3 X 4 + A 2 X 2 + A 1 X + A 0 = 0 X^4+A_2X^2+A_1X+A_0 = 0 X 4 + A 2 X 2 + A 1 X + A 0 = 0 A k A_k A k A 1 ≠ 0 A_1 \neq 0 A 1 = 0 ( X の 2 次式 ) 2 − ( X の 1 次式 ) 2 = 0 (X\text{ の }2\text{ 次式})^2-(X\text{ の}1\text{ 次式})^2 = 0 ( X の 2 次式 ) 2 − ( X の 1 次式 ) 2 = 0 3 3 3 本問では, z 5 = 1 z^5 = 1 z 5 = 1 cos ( ± 2 π 5 ) + i sin ( ± 2 π 5 ) , cos ( ± 4 π 5 ) + i sin ( ± 4 π 5 ) \cos\left(\pm\dfrac{2\pi}{5}\right) +i\sin\left(\pm\dfrac{2\pi}{5}\right),\ \cos\left(\pm\dfrac{4\pi}{5}\right) +i\sin\left(\pm\dfrac{4\pi}{5}\right) cos ( ± 5 2 π ) + i sin ( ± 5 2 π ) , cos ( ± 5 4 π ) + i sin ( ± 5 4 π ) 4 4 4 x 5 = 2 10 x^5 = 2^{10} x 5 = 2 1 0 z 5 = 1 z^5 = 1 z 5 = 1 z 4 + z 3 + z 2 + z + 1 = 0 z^4+z^3+z^2+z+1 = 0 z 4 + z 3 + z 2 + z + 1 = 0 t = z + 1 z t = z+\dfrac{1}{z} t = z + z 1 2 2 2 こちら を参照).
