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:
李煌数学研究院/高次代数方程x^n=1的根式解的解析表达式
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School:李煌數學研究院
李煌公式
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方程
x
n
=1满足
n
=3k+2,
n
>3,n,k是整数,之李煌根式解为：
x
1
=
−
1
2
−
i
3
2
(
−
1
2
+
i
3
2
)
1
n
{\displaystyle x_{1}={\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}}
x
2
=
(
−
1
2
−
i
3
2
(
−
1
2
+
i
3
2
)
1
n
)
2
{\displaystyle x_{2}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{2}}
x
3
=
(
−
1
2
−
i
3
2
(
−
1
2
+
i
3
2
)
1
n
)
3
{\displaystyle x_{3}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{3}}
.
.
.
{\displaystyle ...}
x
n
=
(
−
1
2
−
i
3
2
(
−
1
2
+
i
3
2
)
1
n
)
n
=
1
{\displaystyle x_{n}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{n}=1}
方程
x
n
=1满足
n
=3k+1,
n
>3,n,k是整数,之李煌根式解为：
x
1
=
−
1
2
−
i
3
2
(
−
1
2
−
i
3
2
)
1
n
{\displaystyle x_{1}={\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}}
x
2
=
(
−
1
2
−
i
3
2
(
−
1
2
−
i
3
2
)
1
n
)
2
{\displaystyle x_{2}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{2}}
x
3
=
(
−
1
2
−
i
3
2
(
−
1
2
−
i
3
2
)
1
n
)
3
{\displaystyle x_{3}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{3}}
.
.
.
{\displaystyle ...}
x
n
=
(
−
1
2
−
i
3
2
(
−
1
2
−
i
3
2
)
1
n
)
n
=
1
{\displaystyle x_{n}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{n}=1}
来源
编辑
《南昌理工学院学报》.李煌
<<
School:李煌数学研究院
