最初に２群に分けてから3次方程式を解いた方が、きれいな解が得られる。

正13角形のベキ根表示

2COS(2π/13)=(-1+√13)/6+{(26-5√13+i3√39)/54}^1/3+{(26-5√13-i3√(39)/54}^1/3

が成り立つ。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

最初に３群に分けてから２次方程式を解くと、

ω=(-1+i√3)/2

B1=-1/3+ω^2{(-65+39i√3)/54}^1/3+ω{(-65-39i√3)/54}^1/3

B4=-1/3+ω{(-65+39i√3)/54}^1/3+ω^2{(-65-39i√3)/54}^1/3とおくと、

x=2COS(2π/13)=(B1+√(B1*B1-4*B4))/2

実数解で比較したい。

a+bi=(a^2+b^2)^1/2{a/(a^2+b^2)^1/2+ib/(a^2+b^2)^1/2}として、虚部を消すことを考える。

|-65+39i√3|/54=(65^2+3・39^2)^1/2/54＝(13^3・4)^1/2/54=r1

-65/54=r1cost1,39√3/54=r1sint1

B1+1/3=ω^2(r1^1/3cos(t1/3)+ir1^1/3sin(t1/3))+ω(r1^1/3cos(t1/3)-ir1^1/3sin(t1/3))

=(ω^2+ω)r1^1/3cos(t1/3)+i(ω^2-ω)r1^1/3sin(t1/3)

=-r1^1/3cos(t1/3)-i√3r1^1/3sin(t1/3)

B4+1/3=ω(r1^1/3cos(t1/3)+ir1^1/3sin(t1/3))+ω^2(r1^1/3cos(t1/3)-ir1^1/3sin(t1/3))

=(ω+ω^2)r1^1/3cos(t1/3)+i(ω-ω^2)r1^1/3sin(t1/3)

=-r1^1/3cos(t1/3)+i√3r1^1/3sin(t1/3)

B1^2=1/9+(r1^1/3cos(t1/3))^2-3(r1^1/3sin(t1/3))^2+2/3r1^1/3cos(t1/3)+2i√3/3r1^1/3sin(t1/3)+2i√3(r1^1/3cos(t1/3))(r1^1/3sin(t1/3))

B1^2-4B4=1/9+(r1^1/3cos(t1/3))^2-3(r1^1/3sin(t1/3))^2+2/3r1^1/3cos(t1/3)+2i√3/3r1^1/3sin(t1/3)+2i√3(r1^1/3cos(t1/3))(r1^1/3sin(t1/3)) +4/3+4r1^1/3cos(t1/3)-4i√3r1^1/3sin(t1/3)

=13/9+(r1^1/3cos(t1/3))^2-3(r1^1/3sin(t1/3))^2+2/3r1^1/3cos(t1/3)+4r1^1/3cos(t1/3) +2i√3/3r1^1/3sin(t1/3)+2i√3(r1^1/3cos(t1/3))(r1^1/3sin(t1/3))-4i√3r1^1/3sin(t1/3)

=13/9+(r1^1/3cos(t1/3))^2-3(r1^1/3sin(t1/3))^2+14/3r1^1/3cos(t1/3) -10i√3/3r1^1/3sin(t1/3)+2i√3(r1^1/3cos(t1/3))(r1^1/3sin(t1/3)) (√13/3+r1^1/3cos(t1/3)+i√3(r1^1/3sin(t1/3))^2

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

x=(-1+√13)/6+{(26-5√13+i3√39)/54}^1/3+{(26-5√13-i3√(39)/54}^1/3

|26-5√13+i3√39|/54→{(1352-260√13)^1/2}/54={52(26-5√13)}^1/2/54=r1

(26-5√13)/54=r1cost1,3√39/54=r1sint1,3√39/54=r1sin(-t1)

x=(-1+√13)/6+2r1^1/3cos(t1/3) と簡単な形になる

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝