다중근호 (多重根號, nested radical sign )는 루트 근호 안에 1개이상의 루트 근호를 포함하는 루트를 말한다.
다중근호는 고차방정식 의 해, 수학 상수 등을 표현할 때 중요하게 사용된다.
다중근호의 종류로는 이중근호, 다중근호, 중첩근호 등이 있다.
루트 근호안에 1개의 루트 근호를 포함한다.
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{\displaystyle {\sqrt {(a+b)+2{\sqrt {ab}}}}}
루트 근호안에 2개이상의 루트 근호를 포함한다.
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{\displaystyle {\sqrt {({\sqrt {a}}+{\sqrt {b}})^{2}}},{\sqrt {({\sqrt {a}})^{2}+({\sqrt {b}})^{2}+2{\sqrt {ab}}}}}
루트 근호안에 루트 근호를 갖는 수식이 반복해서 중첩되게 내재한다.
1
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1
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{\displaystyle {\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1...}}}}}}}}}}}
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⋯
{\displaystyle {\sqrt {1+2{\sqrt {1+3{\sqrt {1+4{\sqrt {1+5{\sqrt {1+6{\sqrt {1+\cdots }}}}}}}}}}}}}
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{\displaystyle {\sqrt {{\sqrt {(a)^{2}}}+{\sqrt {(b)^{2}}}+2{\sqrt {ab}}}}={\sqrt {\left({\sqrt {a}}+{\sqrt {b}}\right)+2{\sqrt {ab}}}}={\sqrt {\left({\sqrt {a}}+{\sqrt {b}}\right)^{2}}}={\sqrt {a}}+{\sqrt {b}}}
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125
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)
{\displaystyle {\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right)}
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1
3
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9
3
−
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9
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+
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{\displaystyle {\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}}
[ 1]
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{\displaystyle {\sqrt {a}}+{\sqrt {a}}={\sqrt {({\sqrt {a}}+{\sqrt {a}})^{2}}}={\sqrt {({\sqrt {a}})^{2}+({\sqrt {a}})^{2}+2{\sqrt {aa}}}}={\sqrt {(a+a)+2{\sqrt {a^{2}}}}}={\sqrt {(a+a)+2{a}}}={\sqrt {2a+2a}}={\sqrt {4a}}=2{\sqrt {a}}}
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{\displaystyle \;\;{\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}={\sqrt {(a+b+c)+2\left({\sqrt {ab}}+{\sqrt {ac}}+{\sqrt {bc}}\right)}}={\sqrt {{\sqrt {a}}^{2}+{\sqrt {b}}^{2}+{\sqrt {c}}^{2}+2{\sqrt {ab}}+2{\sqrt {ac}}+2{\sqrt {bc}}}}={\sqrt {\left({\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\right)^{2}}}={\sqrt {\left({\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\right)}}{\sqrt {\left({\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\right)}}}
3차방정식 이상의 일반적인 근 에서 다중근호가 사용된다.
다음은 4차방정식 의 근의 공식이다.
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b
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e
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{\displaystyle \textstyle ax^{4}+bx^{3}+cx^{2}+dx+e=0\ }
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{\displaystyle x=-{b \over 4a}+\left({{{\sqrt {p+2t_{1}}}+{\sqrt {-3p-2{t_{1}}-{2q \over {\sqrt {p+2t_{1}}}}}}} \over {2a}}\right),-{b \over 4a}+\left({{{\sqrt {p+2t_{1}}}-{\sqrt {-3p-2{t_{1}}-{2q \over {\sqrt {p+2t_{1}}}}}}} \over {2a}}\right)}
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{\displaystyle ,-{b \over 4a}-\left({{{\sqrt {p+2t_{1}}}+{\sqrt {-3p-2{t_{1}}-{2q \over {\sqrt {p+2t_{1}}}}}}} \over {2a}}\right),-{b \over 4a}-\left({{{\sqrt {p+2t_{1}}}-{\sqrt {-3p-2{t_{1}}-{2q \over {\sqrt {p+2t_{1}}}}}}} \over {2a}}\right)}
무한한 중첩근호는 일반식으로 표현될 수 있다.
1
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1
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1
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φ
{\displaystyle {\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1...}}}}}}}}}}={{{\sqrt {5}}+1} \over {2}}=\varphi }
황금비
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{\displaystyle {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2...}}}}}}}}}}={{{\sqrt {9}}+1} \over {2}}=2}
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{\displaystyle {\sqrt {3+{\sqrt {3+{\sqrt {3+{\sqrt {3+{\sqrt {3...}}}}}}}}}}={{{\sqrt {13}}+1} \over {2}}}
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{\displaystyle {\sqrt {4+{\sqrt {4+{\sqrt {4+{\sqrt {4+{\sqrt {4...}}}}}}}}}}={{{\sqrt {17}}+1} \over {2}}}
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{\displaystyle {\sqrt {5+{\sqrt {5+{\sqrt {5+{\sqrt {5+{\sqrt {5...}}}}}}}}}}={{{\sqrt {21}}+1} \over {2}}}
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{\displaystyle {\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6...}}}}}}}}}}={{{\sqrt {25}}+1} \over {2}}}
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{\displaystyle {\sqrt {7+{\sqrt {7+{\sqrt {7+{\sqrt {7+{\sqrt {7...}}}}}}}}}}={{{\sqrt {29}}+1} \over {2}}}
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{\displaystyle {\sqrt {8+{\sqrt {8+{\sqrt {8+{\sqrt {8+{\sqrt {8...}}}}}}}}}}={{{\sqrt {33}}+1} \over {2}}}
따라서 이것을 일반화하면,
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{\displaystyle {\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n...}}}}}}}}}}={{{\sqrt {((4\cdot n)+1)}}+1} \over {2}}}
한편,
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{\displaystyle x^{2}-x-1=0}
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{\displaystyle {{-b\pm {\sqrt {b^{2}-4ac}}} \over {2a}}}
근의 공식 )
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{\displaystyle {{-(-1)\pm {\sqrt {(-1)^{2}-4\cdot 1\cdot (-1)}}} \over {2}}}
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{\displaystyle {{1\pm {\sqrt {1+4}}} \over {2}}}
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{\displaystyle {{1\pm {\sqrt {5}}} \over {2}}}
이고,
따라서,
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{\displaystyle {\sqrt {c+{\sqrt {c+{\sqrt {c+{\sqrt {c+{\sqrt {c...}}}}}}}}}}}
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{\displaystyle ={{{\sqrt {((4\cdot c)+1)}}+1} \over {2}}}
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{\displaystyle x^{2}-x-c=0}
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1
1
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1
1
1
…
3
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{\displaystyle {\sqrt {1}}={\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1\dots }}}}}}}}}}}}}}}}}
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2
…
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{\displaystyle {\sqrt {2}}={\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2\dots }}}}}}}}}}}}}}}}}
3
=
3
3
3
3
3
3
3
3
…
3
3
3
3
3
3
3
3
{\displaystyle {\sqrt {3}}={\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3\dots }}}}}}}}}}}}}}}}}
⋮
{\displaystyle \qquad \qquad \qquad \vdots }
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7
…
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{\displaystyle {\sqrt {7}}={\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7\dots }}}}}}}}}}}}}}}}}
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8
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{\displaystyle {\sqrt {8}}={\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8\dots }}}}}}}}}}}}}}}}}
⋮
{\displaystyle \qquad \qquad \qquad \vdots }
따라서 이것을 일반화 하면,
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=
x
x
x
x
…
3
3
3
3
{\displaystyle {\sqrt {x}}={\sqrt[{3}]{x{\sqrt[{3}]{x{\sqrt[{3}]{x{\sqrt[{3}]{x\dots }}}}}}}}}
그리고 이것을 확장하면,[ 2]
x
1
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=
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x
x
x
…
n
n
n
n
{\displaystyle x^{{1} \over {(n-1)}}={\sqrt[{n}]{x{\sqrt[{n}]{x{\sqrt[{n}]{x{\sqrt[{n}]{x\dots }}}}}}}}}
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2
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2
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+
…
{\displaystyle 2={\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+\dots }}}}}}}}}}}}}}}}}
3
=
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+
6
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6
+
6
+
…
{\displaystyle 3={\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+{\sqrt {6+\dots }}}}}}}}}}}}}}}}}
4
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12
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{\displaystyle 4={\sqrt {12+{\sqrt {12+{\sqrt {12+{\sqrt {12+{\sqrt {12+{\sqrt {12+{\sqrt {12+{\sqrt {12+\dots }}}}}}}}}}}}}}}}}
⋮
{\displaystyle \qquad \qquad \qquad \vdots }
8
=
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56
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56
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56
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56
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…
{\displaystyle 8={\sqrt {56+{\sqrt {56+{\sqrt {56+{\sqrt {56+{\sqrt {56+{\sqrt {56+{\sqrt {56+{\sqrt {56+\dots }}}}}}}}}}}}}}}}}
⋮
{\displaystyle \qquad \qquad \qquad \vdots }
따라서, 이것을 일반화하면,
x
=
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x
⋅
(
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−
1
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)
+
(
x
⋅
(
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(
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)
+
(
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)
+
…
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,
x
=
i
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t
e
g
e
r
)
{\displaystyle x={\sqrt {(x\cdot (x-1))+{\sqrt {(x\cdot (x-1))+{\sqrt {(x\cdot (x-1))+{\sqrt {(x\cdot (x-1))+\dots }}}}}}}}\qquad (x>1,x=integer)}
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+
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+
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+
…
{\displaystyle x={\sqrt {(x^{2}-x)+{\sqrt {(x^{2}-x)+{\sqrt {(x^{2}-x)+{\sqrt {(x^{2}-x)+\dots }}}}}}}}}
한편, 이것을 일반식으로 표현하면,
(
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)
2
−
(
x
)
{\displaystyle \left({x}\right)^{2}-\left({x}\right)}
x
{\displaystyle x}
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{\displaystyle {\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n...}}}}}}}}}}={{{\sqrt {((4\cdot n)+1)}}+1} \over {2}}}
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{\displaystyle x={\sqrt {\left({x}\right)^{2}-\left({x}\right)+{\sqrt {\left({x}\right)^{2}-\left({x}\right)+{\sqrt {\left({x}\right)^{2}-\left({x}\right)+{\sqrt {\left({x}\right)^{2}-\left({x}\right)+{\sqrt {\left({x}\right)^{2}-\left({x}\right)...}}}}}}}}}}}
=
(
(
4
⋅
(
x
2
−
x
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)
+
1
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+
1
2
{\displaystyle ={{{\sqrt {\left(\left(4\cdot \left({x}^{2}-{x}\right)\right)+1\right)}}+1} \over {2}}}
=
4
x
2
−
4
x
+
1
+
1
2
{\displaystyle ={{{\sqrt {4x^{2}-4x+1}}+1} \over {2}}}
=
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2
x
−
1
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2
+
1
2
{\displaystyle ={{{\sqrt {(2x-1)^{2}}}+1} \over {2}}}
=
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2
x
−
1
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+
1
2
=
x
{\displaystyle ={{(2x-1)+1} \over {2}}=x}
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a
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a
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⋯
{\displaystyle {x+n+a}={\sqrt {ax+(n+a)^{2}+x{\sqrt {a(x+1n)+(n+a)^{2}+(x+1n){\sqrt {a(x+2n)+(n+a)^{2}+(x+2n){\sqrt {a(x+3n)+(n+a)^{2}+(x+3n){\sqrt {\cdots }}}}}}}}}}}
a
=
0
,
n
=
1
{\displaystyle a=0,n=1}
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x
1
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⋯
{\displaystyle x+1={\sqrt {1+x{\sqrt {1+(x+1){\sqrt {1+(x+2){\sqrt {1+\cdots }}}}}}}}}
3
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1
=
1
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⋯
{\displaystyle 3=2+1={\sqrt {1+2{\sqrt {1+3{\sqrt {1+4{\sqrt {1+5{\sqrt {1+6{\sqrt {1+\cdots }}}}}}}}}}}}}
매스월드 (Ramanujan 1911)Ramanujan, S. Question No. 298. J. Indian Math. Soc. 1911.
(Ramanujan 2000)Ramanujan, S. Collected Papers of Srinivasa Ramanujan (Ed. G. H. Hardy, P. V. S. Aiyar, and B. M. Wilson). Providence, RI: Amer. Math. Soc., p. 327, 2000. 
![{\displaystyle {\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81539ceeadda96b7dd53a8f6c15f727563430906)
![{\displaystyle {\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55d6701f2c4aae4e261edfa19bd8711347f10257)
![{\displaystyle {\sqrt {1}}={\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1{\sqrt[{3}]{1\dots }}}}}}}}}}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae2ce87264a9a93143204518e475fd8990c3ba3b)
![{\displaystyle {\sqrt {2}}={\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2{\sqrt[{3}]{2\dots }}}}}}}}}}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef7074d33136ffc8bb457e2a94bb422eedeeeda8)
![{\displaystyle {\sqrt {3}}={\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3{\sqrt[{3}]{3\dots }}}}}}}}}}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5702b9519980a32c963a8397ad3470c6a67ffbae)
![{\displaystyle {\sqrt {7}}={\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7{\sqrt[{3}]{7\dots }}}}}}}}}}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd631aaf4aa01e0d997c32a9ecdf97a39973d0dc)
![{\displaystyle {\sqrt {8}}={\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8{\sqrt[{3}]{8\dots }}}}}}}}}}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd8930c09c89053fc9e53009a409cda09a9f5462)
![{\displaystyle {\sqrt {x}}={\sqrt[{3}]{x{\sqrt[{3}]{x{\sqrt[{3}]{x{\sqrt[{3}]{x\dots }}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35b76fc4a298e84200ee7f4e673178ca88ac94d0)
![{\displaystyle x^{{1} \over {(n-1)}}={\sqrt[{n}]{x{\sqrt[{n}]{x{\sqrt[{n}]{x{\sqrt[{n}]{x\dots }}}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9053c95daab79e3736097ac3ade6132399f6a706)
