수학 에서, 특히 타원 함수 의 이론인 놈 (nome)은 특별한 함수이며 다음과 같이 주어진다.
q
=
e
−
π
K
′
K
=
e
i
π
ω
2
ω
1
=
e
i
π
τ
{\displaystyle q=e^{-{\frac {\pi K'}{K}}}=e^{\frac {{\rm {i}}\pi \omega _{2}}{\omega _{1}}}=e^{{\rm {i}}\pi \tau }\,}
여기서
K
{\displaystyle K}
i
K
′
{\displaystyle iK'}
1
/
4
{\displaystyle 1/4}
ω
1
{\displaystyle \omega _{1}}
ω
2
{\displaystyle \omega _{2}}
1
/
4
{\displaystyle 1/4}
K
{\displaystyle K}
i
K
′
{\displaystyle iK'}
야코비 타원 함수 의 문맥에서만 일반적으로 사용되는 반면, 절반 기간(반기 ,
1
/
2
{\displaystyle 1/2}
ω
1
{\displaystyle \omega _{1}}
ω
2
{\displaystyle \omega _{2}}
바이어슈트라스(Weierstrass) 타원 함수 의 맥락에서만 사용된다. 일부 저자들은
ω
1
{\displaystyle \omega _{1}}
ω
2
{\displaystyle \omega _{2}}
기본 설명 [ 편집 ] 놈(nome)은 타원 함수와 모듈 형식을 설명 할 수 있는 값으로 자주 사용된다. 반면에
1
/
4
{\displaystyle 1/4}
1
/
4
{\displaystyle 1/4}
함수
τ
=
i
K
′
K
=
ω
2
ω
1
{\displaystyle \tau ={{iK'} \over {K}}={{\omega _{2}} \over {\omega _{1}}}}
ω
1
{\displaystyle \omega _{1}}
ω
2
{\displaystyle \omega _{2}}
보완적인
n
1
{\displaystyle n1}
q
1
=
e
−
π
K
K
′
{\displaystyle q_{1}=e^{-{\frac {\pi K}{K'}}}}
그러나 일부 출처는 관습에 따라 다음을 그대로 사용하기도 한다.
q
=
e
2
i
π
τ
{\displaystyle q=e^{{2{\rm {i}}}\pi \tau }}
q
=
e
2
i
π
z
{\displaystyle q=e^{{2{\rm {i}}}\pi z}}
놈(nome)에 대한 추가 정의 및 관계에 대해서는 분기 (
1
/
4
{\displaystyle 1/4}
타원 적분 에 대한 항목을 참조할 수 있다.
변수에 대한 함수로 타원 놈은 다음과 같이 정의된다:
q
(
x
)
=
exp
[
−
π
K
(
1
−
x
2
)
K
(
x
)
−
1
]
{\displaystyle q(x)=\exp {\bigl [}-\pi \,K({\sqrt {1-x^{2}}})\,K(x)^{-1}{\bigr ]}}
그리고 제1종 완전 타원 적분은 다음과 같이 정의된다:
K
(
ε
)
=
∫
0
π
/
2
1
1
−
ε
2
sin
(
φ
)
2
d
φ
{\displaystyle K(\varepsilon )=\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-\varepsilon ^{2}\sin(\varphi )^{2}}}}\,\mathrm {d} \varphi }
K
(
ε
)
=
∫
0
1
2
(
w
2
+
1
)
2
−
4
ε
2
w
2
d
w
{\displaystyle K(\varepsilon )=\int _{0}^{1}{\frac {2}{\sqrt {(w^{2}+1)^{2}-4\,\varepsilon ^{2}w^{2}}}}\,\mathrm {d} w}
두 공식은 서로 일치하며 동일한 결과를 가져온다.
무한급수 [ 편집 ] 놈 기능의 무한 시리즈는 다음과 같이 표시될 수 있다:
q
(
x
)
=
∑
n
=
1
∞
Kt
(
n
)
16
n
x
2
n
{\displaystyle q(x)=\sum _{n=1}^{\infty }{\frac {{\text{Kt}}(n)}{16^{n}}}\,x^{2n}}
이 무한급수의 수렴 반경은 1이다. 여기서
Kt
(
n
)
{\displaystyle {\text{Kt}}(n)}
Kt
(
n
)
∈
N
{\displaystyle {\text{Kt}}(n)\in \mathbb {N} }
n
∈
N
{\displaystyle n\in \mathbb {N} }
Kt
(
1
)
=
1
{\displaystyle {\text{Kt}}(1)=1}
Kt
(
n
+
1
)
=
1
n
∑
k
=
1
n
k
Kt
(
k
)
[
16
Ap
(
n
+
1
−
k
)
−
Ap
(
n
+
2
−
k
)
]
{\displaystyle {\text{Kt}}(n+1)={\frac {1}{n}}\sum _{k=1}^{n}k\,{\text{Kt}}(k)[16\,{\text{Ap}}(n+1-k)-{\text{Ap}}(n+2-k)]}
Ap
(
n
)
=
∑
a
=
0
n
−
1
(
2
a
a
)
2
(
2
n
−
2
−
2
a
n
−
1
−
a
)
2
{\displaystyle {\text{Ap}}(n)=\sum _{a=0}^{n-1}{\binom {2a}{a}}^{2}{\binom {2n-2-2a}{n-1-a}}^{2}}
이 정수열
Kt
(
n
)
{\displaystyle {\text{Kt}}(n)}
Ap
(
n
)
{\displaystyle {\text{Ap}}(n)}
Kt
(
n
)
{\displaystyle {\text{Kt}}(n)}
Kt
(
n
)
{\displaystyle {\text{Kt}}(n)}
Kt
(
1
)
=
1
{\displaystyle {\text{Kt}}(1)=1}
n
∈
N
{\displaystyle n\in \mathbb {N} }
위치 n
정수 시퀀스의 수 Ap(n)
정수 시퀀스의 수 Kt(n)
1
1
1
2
8
8
3
88
84
4
1088
992
5
14296
12514
6
195008
164688
7
2728384
2232200
8
38879744
30920128
9
561787864
435506703
10
8206324928
6215660600
11
120929313088
89668182220
12
1794924383744
1305109502496
13
26802975999424
19138260194422
14
402298219288064
282441672732656
15
6064992788397568
4191287776164504
16
91786654611673088
62496081197436736
17
1393772628452578264
935823746406530603
놈의 경우 두 번째 시리즈를 개발할 수 있다:
q
(
x
)
=
∑
n
=
1
∞
Sw
(
n
)
2
4
n
−
3
(
1
−
1
−
x
2
4
1
+
1
−
x
2
4
)
4
n
−
3
=
x
2
{
1
2
+
[
∑
n
=
1
∞
Sw
(
n
+
1
)
2
4
n
+
1
x
2
n
]
}
4
{\displaystyle q(x)=\sum _{n=1}^{\infty }{\frac {{\text{Sw}}(n)}{2^{4n-3}}}{\biggl (}{\frac {1-{\sqrt[{4}]{1-x^{2}}}}{1+{\sqrt[{4}]{1-x^{2}}}}}{\biggr )}^{4n-3}=x^{2}{\biggl \{}{\frac {1}{2}}+{\biggl [}\sum _{n=1}^{\infty }{\frac {{\text{Sw}}(n+1)}{2^{4n+1}}}x^{2n}{\biggr ]}{\biggr \}}^{4}}
정수 시퀀스
S
w
(
n
)
{\displaystyle Sw(n)}
Hermann Amandus Schwarz 는 54-56페이지의 "Berechnung der Grösse k" 장에서 "Formeln und Lehrsätze zum Gebrauche der elliptischen Funktionen" 이라는 저서에서 정수 수열을 썼다. 이 슈바르츠 수열
S
w
(
n
)
{\displaystyle Sw(n)}
Karl Theodor Wilhelm Weierstrass 와 Louis Melville Milne-Thomson 에 의해 분석되었다. 수학자 Adolf Kneser 는 다음 패턴을 기반으로 이 시퀀스에 대한 합성 방법을 결정했다.
Sw
(
1
)
=
1
{\displaystyle {\text{Sw}}(1)=1}
Sw
(
n
+
1
)
=
2
n
∑
m
=
1
n
Sw
(
m
)
Kn
(
n
+
1
−
m
)
{\displaystyle {\text{Sw}}(n+1)={\frac {2}{n}}\sum _{m=1}^{n}{\text{Sw}}(m){\text{Kn}}(n+1-m)}
Schwarz 시퀀스
S
w
(
n
)
{\displaystyle Sw(n)}
K
n
(
n
)
{\displaystyle Kn(n)}
K
n
(
n
)
{\displaystyle Kn(n)}
Apéry 시퀀스
A
p
(
n
)
{\displaystyle Ap(n)}
Ap
(
n
)
=
∑
a
=
0
n
−
1
(
2
a
a
)
2
(
2
n
−
2
−
2
a
n
−
1
−
a
)
2
{\displaystyle {\text{Ap}}(n)=\sum _{a=0}^{n-1}{\binom {2a}{a}}^{2}{\binom {2n-2-2a}{n-1-a}}^{2}}
이러한 방식으로 모든 자연수 n에 대해 Kneser 수열을 정의할 수 있다.
Kn
(
n
+
1
)
=
2
4
n
+
1
−
1
8
Ap
(
n
+
2
)
−
∑
b
=
1
n
Kn
(
b
)
Ap
(
n
+
2
−
b
)
{\displaystyle {\text{Kn}}(n+1)=2^{4n+1}-{\tfrac {1}{8}}{\text{Ap}}(n+2)-\sum _{b=1}^{n}{\text{Kn}}(b){\text{Ap}}(n+2-b)}
Kneser 시퀀스는 생성 함수 에 의해 다음과 같이 생성될 수도 있다.
π
2
8
x
(
1
−
x
2
)
K
(
x
)
2
−
1
2
x
=
∑
n
=
1
∞
Kn
(
n
)
2
4
n
−
2
x
2
n
−
1
{\displaystyle {\frac {\pi ^{2}}{8x(1-x^{2})K(x)^{2}}}-{\frac {1}{2x}}=\sum _{n=1}^{\infty }{\frac {{\text{Kn}}(n)}{2^{4n-2}}}x^{2n-1}}
다음 표에는 슈바르츠 수와 크네저 수와 아페리 수가 포함되어 있다.
Kneser에 따른 시퀀스 구성 방법
인덱스 n
Ap(n) (A036917)
Kn(n) (A227503)
Sw(n) (A002103)
1
1
1
1
2
8
13
2
3
88
184
15
4
1088
2701
150
5
14296
40456
1707
6
195008
613720
20910
7
2728384
9391936
268616
8
38879744
144644749
3567400
값 목록 [ 편집 ] 다음에서 놈의 일부 기능 값이 제공된다. 다음은 렘니스케이트 값이다.
q
(
0
)
=
0
{\displaystyle q(0)=0}
q
(
1
)
=
1
{\displaystyle q(1)=1}
q
(
−
1
)
=
1
{\displaystyle q(-1)=1}
q
(
1
2
2
)
=
e
−
π
{\displaystyle q({\tfrac {1}{2}}{\sqrt {2}})={\text{e}}^{-\pi }}
q
[
(
2
−
1
)
2
]
=
e
−
2
π
{\displaystyle q[({\sqrt {2}}-1)^{2}]={\text{e}}^{-2\pi }}
q
[
2
2
4
(
2
−
1
)
]
=
exp
(
−
1
2
π
)
{\displaystyle q[2{\sqrt[{4}]{2}}({\sqrt {2}}-1)]=\exp(-{\tfrac {1}{2}}\pi )}
q
[
1
2
(
3
−
1
)
(
2
−
3
4
)
]
=
e
−
3
π
{\displaystyle q[{\tfrac {1}{2}}({\sqrt {3}}-1)({\sqrt {2}}-{\sqrt[{4}]{3}})]={\text{e}}^{-3\pi }}
q
[
1
2
(
3
−
1
)
(
2
+
3
4
)
]
=
exp
(
−
1
3
π
)
{\displaystyle q[{\tfrac {1}{2}}({\sqrt {3}}-1)({\sqrt {2}}+{\sqrt[{4}]{3}})]=\exp(-{\tfrac {1}{3}}\pi )}
q
[
1
2
(
10
−
2
2
)
(
3
−
2
5
4
)
]
=
e
−
5
π
{\displaystyle q[{\tfrac {1}{2}}({\sqrt {10}}-2{\sqrt {2}})(3-2{\sqrt[{4}]{5}})]={\text{e}}^{-5\pi }}
q
[
1
2
(
10
−
2
2
)
(
3
+
2
5
4
)
]
=
exp
(
−
1
5
π
)
{\displaystyle q[{\tfrac {1}{2}}({\sqrt {10}}-2{\sqrt {2}})(3+2{\sqrt[{4}]{5}})]=\exp(-{\tfrac {1}{5}}\pi )}
일부 비 렘니스케이트 값은 다음과 같다:
q
[
1
4
(
6
−
2
)
]
=
e
−
3
π
{\displaystyle q[{\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})]={\text{e}}^{-{\sqrt {3}}\pi }}
q
[
1
4
(
6
+
2
)
]
=
exp
(
−
1
3
3
π
)
{\displaystyle q[{\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})]=\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi )}
q
{
sin
[
1
2
arcsin
(
5
−
2
)
]
}
=
e
−
5
π
{\displaystyle q{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}{\bigr \}}={\text{e}}^{-{\sqrt {5}}\pi }}
q
{
cos
[
1
2
arcsin
(
5
−
2
)
]
}
=
exp
(
−
1
5
5
π
)
{\displaystyle q{\bigl \{}\cos {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}{\bigr \}}=\exp(-{\tfrac {1}{5}}{\sqrt {5}}\,\pi )}
q
[
1
8
(
3
2
−
14
)
]
=
e
−
7
π
{\displaystyle q[{\tfrac {1}{8}}(3{\sqrt {2}}-{\sqrt {14}})]={\text{e}}^{-{\sqrt {7}}\pi }}
q
[
1
8
(
3
2
+
14
)
]
=
exp
(
−
1
7
7
π
)
{\displaystyle q[{\tfrac {1}{8}}(3{\sqrt {2}}+{\sqrt {14}})]=\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi )}
q
[
1
16
(
22
+
3
2
)
(
1
3
6
3
+
2
11
3
−
1
3
6
3
−
2
11
3
+
1
3
11
−
1
)
4
]
=
e
−
11
π
{\displaystyle q{\bigl [}{\tfrac {1}{16}}{\bigl (}{\sqrt {22}}+3{\sqrt {2}}{\bigr )}{\bigl (}{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\tfrac {1}{3}}{\sqrt {11}}-1{\bigr )}^{4}{\bigr ]}={\text{e}}^{-{\sqrt {11}}\pi }}
q
[
1
16
(
22
−
3
2
)
(
1
3
6
3
+
2
11
3
−
1
3
6
3
−
2
11
3
+
1
3
11
+
1
)
4
]
=
exp
(
−
1
11
11
π
)
{\displaystyle q{\bigl [}{\tfrac {1}{16}}{\bigl (}{\sqrt {22}}-3{\sqrt {2}}{\bigr )}{\bigl (}{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\tfrac {1}{3}}{\sqrt {11}}+1{\bigr )}^{4}{\bigr ]}=\exp(-{\tfrac {1}{11}}{\sqrt {11}}\,\pi )}
q
{
sin
[
1
2
arcsin
(
5
13
−
18
)
]
}
=
e
−
13
π
{\displaystyle q{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}{\bigr \}}={\text{e}}^{-{\sqrt {13}}\pi }}
q
{
cos
[
1
2
arcsin
(
5
13
−
18
)
]
}
=
exp
(
−
1
13
13
π
)
{\displaystyle q{\bigl \{}\cos {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}{\bigr \}}=\exp(-{\tfrac {1}{13}}{\sqrt {13}}\,\pi )}
q
⟨
sin
{
1
2
arcsin
[
(
1
4
17
+
1
4
−
1
4
2
17
+
2
)
6
]
}
⟩
=
e
−
17
π
{\displaystyle q{\bigl \langle }\sin {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {1}{4}}-{\tfrac {1}{4}}{\sqrt {2{\sqrt {17}}+2}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {17}}\pi }}
q
⟨
cos
{
1
2
arcsin
[
(
1
4
17
+
1
4
−
1
4
2
17
+
2
)
6
]
}
⟩
=
exp
(
−
1
17
17
π
)
{\displaystyle q{\bigl \langle }\cos {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {1}{4}}-{\tfrac {1}{4}}{\sqrt {2{\sqrt {17}}+2}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{17}}{\sqrt {17}}\,\pi )}
q
⟨
sin
{
1
2
arcsin
[
(
37
−
6
)
3
]
}
⟩
=
e
−
37
π
{\displaystyle q{\bigl \langle }\sin {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}({\sqrt {37}}-6)^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {37}}\pi }}
q
⟨
cos
{
1
2
arcsin
[
(
37
−
6
)
3
]
}
⟩
=
exp
(
−
1
37
37
π
)
{\displaystyle q{\bigl \langle }\cos {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}({\sqrt {37}}-6)^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{37}}{\sqrt {37}}\,\pi )}
다음 값은 홀수의 제곱근과 함께 Gelfold 상수의 역수의 거듭제곱으로 발생한다:
q
(
2
−
1
)
=
e
−
2
π
{\displaystyle q({\sqrt {2}}-1)={\text{e}}^{-{\sqrt {2}}\pi }}
q
(
2
2
−
2
)
=
exp
(
−
1
2
2
π
)
{\displaystyle q({\sqrt {2{\sqrt {2}}-2}})=\exp(-{\tfrac {1}{2}}{\sqrt {2}}\,\pi )}
q
[
(
2
−
3
)
(
3
−
2
)
]
=
e
−
6
π
{\displaystyle q[(2-{\sqrt {3}})({\sqrt {3}}-{\sqrt {2}})]={\text{e}}^{-{\sqrt {6}}\pi }}
q
[
(
2
−
3
)
(
3
+
2
)
]
=
exp
(
−
1
3
6
π
)
{\displaystyle q[(2-{\sqrt {3}})({\sqrt {3}}+{\sqrt {2}})]=\exp(-{\tfrac {1}{3}}{\sqrt {6}}\,\pi )}
q
[
(
10
−
3
)
(
2
−
1
)
2
]
=
e
−
10
π
{\displaystyle q[({\sqrt {10}}-3)({\sqrt {2}}-1)^{2}]={\text{e}}^{-{\sqrt {10}}\pi }}
q
[
(
10
−
3
)
(
2
+
1
)
2
]
=
exp
(
−
1
5
10
π
)
{\displaystyle q[({\sqrt {10}}-3)({\sqrt {2}}+1)^{2}]=\exp(-{\tfrac {1}{5}}{\sqrt {10}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
2
+
1
2
−
1
2
4
2
+
5
)
3
]
}
⟩
=
e
−
14
π
{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\sqrt {2}}+{\tfrac {1}{2}}-{\tfrac {1}{2}}{\sqrt {4{\sqrt {2}}+5}}{\bigr )}^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {14}}\pi }}
q
⟨
tan
{
1
2
arctan
[
(
2
+
1
2
+
1
2
4
2
+
5
)
3
]
}
⟩
=
exp
(
−
1
7
14
π
)
{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\sqrt {2}}+{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {4{\sqrt {2}}+5}}{\bigr )}^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {14}}\,\pi )}
q
[
(
10
−
3
11
)
(
3
11
−
7
2
)
]
=
e
−
22
π
{\displaystyle q[(10-3{\sqrt {11}})(3{\sqrt {11}}-7{\sqrt {2}})]={\text{e}}^{-{\sqrt {22}}\pi }}
q
[
(
10
−
3
11
)
(
3
11
+
7
2
)
]
=
exp
(
−
1
11
22
π
)
{\displaystyle q[(10-3{\sqrt {11}})(3{\sqrt {11}}+7{\sqrt {2}})]=\exp(-{\tfrac {1}{11}}{\sqrt {22}}\,\pi )}
q
{
(
26
+
5
)
(
2
−
1
)
2
tan
[
1
4
π
−
arctan
(
1
3
3
3
+
26
3
−
1
3
3
3
−
26
3
+
1
6
26
−
1
2
2
)
]
4
}
=
e
−
26
π
{\displaystyle q{\bigl \{}({\sqrt {26}}+5)({\sqrt {2}}-1)^{2}\tan {\bigl [}{\tfrac {1}{4}}\pi -\arctan({\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}+{\sqrt {26}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}-{\sqrt {26}}}}+{\tfrac {1}{6}}{\sqrt {26}}-{\tfrac {1}{2}}{\sqrt {2}}){\bigr ]}^{4}{\bigr \}}={\text{e}}^{-{\sqrt {26}}\pi }}
q
{
(
26
+
5
)
(
2
+
1
)
2
tan
[
arctan
(
1
3
3
3
+
26
3
−
1
3
3
3
−
26
3
+
1
6
26
+
1
2
2
)
−
1
4
π
]
4
}
=
exp
(
−
1
13
26
π
)
{\displaystyle q{\bigl \{}({\sqrt {26}}+5)({\sqrt {2}}+1)^{2}\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}+{\sqrt {26}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}-{\sqrt {26}}}}+{\tfrac {1}{6}}{\sqrt {26}}+{\tfrac {1}{2}}{\sqrt {2}})-{\tfrac {1}{4}}\pi {\bigr ]}^{4}{\bigr \}}=\exp(-{\tfrac {1}{13}}{\sqrt {26}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
1
4
17
+
3
4
−
1
4
6
17
+
10
)
6
]
}
⟩
=
e
−
34
π
{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {3}{4}}-{\tfrac {1}{4}}{\sqrt {6{\sqrt {17}}+10}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {34}}\pi }}
q
⟨
tan
{
1
2
arctan
[
(
1
4
17
+
3
4
+
1
4
6
17
+
10
)
6
]
}
⟩
=
exp
(
−
1
17
34
π
)
{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {3}{4}}+{\tfrac {1}{4}}{\sqrt {6{\sqrt {17}}+10}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{17}}{\sqrt {34}}\,\pi )}
q
[
(
13
58
−
99
)
(
2
−
1
)
6
]
=
e
−
58
π
{\displaystyle q[(13{\sqrt {58}}-99)({\sqrt {2}}-1)^{6}]={\text{e}}^{-{\sqrt {58}}\pi }}
q
[
(
13
58
−
99
)
(
2
+
1
)
6
]
=
exp
(
−
1
29
58
π
)
{\displaystyle q[(13{\sqrt {58}}-99)({\sqrt {2}}+1)^{6}]=\exp(-{\tfrac {1}{29}}{\sqrt {58}}\,\pi )}
놈를 제곱하면 다음 값이 된다:
q
{
tan
[
1
2
arcsin
(
2
−
1
)
]
2
}
=
e
−
2
2
π
{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {2}}-1{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {2}}\pi }}
q
[
(
3
−
2
)
2
(
2
−
1
)
2
]
=
e
−
2
3
π
{\displaystyle q[({\sqrt {3}}-{\sqrt {2}})^{2}({\sqrt {2}}-1)^{2}]={\text{e}}^{-2{\sqrt {3}}\pi }}
q
[
(
3
−
2
)
2
(
2
+
1
)
2
]
=
exp
(
−
2
3
3
π
)
{\displaystyle q[({\sqrt {3}}-{\sqrt {2}})^{2}({\sqrt {2}}+1)^{2}]=\exp(-{\tfrac {2}{3}}{\sqrt {3}}\,\pi )}
q
{
tan
[
1
4
arcsin
(
5
−
2
)
]
2
}
=
e
−
2
5
π
{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {5}}\pi }}
q
{
tan
[
1
4
π
−
1
4
arcsin
(
5
−
2
)
]
2
}
=
exp
(
−
2
5
5
π
)
{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}^{2}{\bigr \}}=\exp(-{\tfrac {2}{5}}{\sqrt {5}}\,\pi )}
q
[
(
2
−
1
)
4
(
2
2
−
7
)
2
]
=
e
−
2
7
π
{\displaystyle q[({\sqrt {2}}-1)^{4}(2{\sqrt {2}}-{\sqrt {7}})^{2}]={\text{e}}^{-2{\sqrt {7}}\pi }}
q
[
(
2
−
1
)
4
(
2
2
+
7
)
2
]
=
exp
(
−
2
7
7
π
)
{\displaystyle q[({\sqrt {2}}-1)^{4}(2{\sqrt {2}}+{\sqrt {7}})^{2}]=\exp(-{\tfrac {2}{7}}{\sqrt {7}}\,\pi )}
q
{
tan
[
1
4
arcsin
(
5
13
−
18
)
]
2
}
=
e
−
2
13
π
{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {13}}\pi }}
q
{
tan
[
1
4
π
−
1
4
arcsin
(
5
13
−
18
)
]
2
}
=
exp
(
−
2
13
13
π
)
{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}^{2}{\bigr \}}=\exp(-{\tfrac {2}{13}}{\sqrt {13}}\,\pi )}
가치의 특정 사중주가 아래에 나열되어 있다:
q
⟨
tan
{
1
2
arctan
[
(
10
−
3
)
2
(
5
−
2
)
2
]
}
⟩
=
exp
(
−
30
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}-3)^{2}({\sqrt {5}}-2)^{2}]\}{\bigr \rangle }=\exp(-{\sqrt {30}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
10
−
3
)
2
(
5
+
2
)
2
]
}
⟩
=
exp
(
−
1
3
30
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}-3)^{2}({\sqrt {5}}+2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{3}}{\sqrt {30}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
10
+
3
)
2
(
5
−
2
)
2
]
}
⟩
=
exp
(
−
1
5
30
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}+3)^{2}({\sqrt {5}}-2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{5}}{\sqrt {30}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
10
+
3
)
2
(
5
+
2
)
2
]
}
⟩
=
exp
(
−
1
15
30
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}+3)^{2}({\sqrt {5}}+2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{15}}{\sqrt {30}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
2
7
−
3
3
)
2
(
2
2
−
7
)
2
]
}
⟩
=
exp
(
−
42
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}-3{\sqrt {3}})^{2}(2{\sqrt {2}}-{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\sqrt {42}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
2
7
−
3
3
)
2
(
2
2
+
7
)
2
]
}
⟩
=
exp
(
−
1
3
42
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}-3{\sqrt {3}})^{2}(2{\sqrt {2}}+{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{3}}{\sqrt {42}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
2
7
+
3
3
)
2
(
2
2
−
7
)
2
]
}
⟩
=
exp
(
−
1
7
42
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}+3{\sqrt {3}})^{2}(2{\sqrt {2}}-{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {42}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
2
7
+
3
3
)
2
(
2
2
+
7
)
2
]
}
⟩
=
exp
(
−
1
21
42
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}+3{\sqrt {3}})^{2}(2{\sqrt {2}}+{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{21}}{\sqrt {42}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
5
−
2
)
4
(
2
−
1
)
6
]
}
⟩
=
exp
(
−
70
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}-2)^{4}({\sqrt {2}}-1)^{6}]\}{\bigr \rangle }=\exp(-{\sqrt {70}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
5
−
2
)
4
(
2
+
1
)
6
]
}
⟩
=
exp
(
−
1
5
70
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}-2)^{4}({\sqrt {2}}+1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{5}}{\sqrt {70}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
5
+
2
)
4
(
2
−
1
)
6
]
}
⟩
=
exp
(
−
1
7
70
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}+2)^{4}({\sqrt {2}}-1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {70}}\,\pi )}
q
⟨
tan
{
1
2
arctan
[
(
5
+
2
)
4
(
2
+
1
)
6
]
}
⟩
=
exp
(
−
1
35
70
π
)
{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}+2)^{4}({\sqrt {2}}+1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{35}}{\sqrt {70}}\,\pi )}
지수 정리 [ 편집 ] 대수 숫자의 놈를 밑수로 하고 양의 유리수를 지수로 사용하는 모든 거듭제곱은 대수 숫자의 놈이 된다:
q
(
x
1
∈
A
)
w
∈
Q
+
=
q
(
x
2
∈
A
)
{\displaystyle q(x_{1}\in \mathbb {A} )^{w\in \mathbb {Q^{+}} }=q(x_{2}\in \mathbb {A} )^{}}
야코비 타원 함수 를 사용하여 정리를 설명하는 경우 정리에 대해 다음 공식을 설정할 수 있다:
q
(
x
)
2
=
q
[
x
2
(
1
+
1
−
x
2
)
−
2
]
{\displaystyle q(x)^{2}=q[x^{2}(1+{\sqrt {1-x^{2}}})^{-2}]}
q
(
x
)
3
=
q
{
x
3
sn
[
1
3
K
(
x
)
;
x
]
4
}
{\displaystyle q(x)^{3}=q\{x^{3}\operatorname {sn} [{\tfrac {1}{3}}K(x);x]^{4}\}}
q
(
x
)
5
=
q
{
x
5
sn
[
1
5
K
(
x
)
;
x
]
4
sn
[
3
5
K
(
x
)
;
x
]
4
}
{\displaystyle q(x)^{5}=q\{x^{5}\operatorname {sn} [{\tfrac {1}{5}}K(x);x]^{4}\operatorname {sn} [{\tfrac {3}{5}}K(x);x]^{4}\}}
q
(
x
)
7
=
q
{
x
7
sn
[
1
7
K
(
x
)
;
x
]
4
sn
[
3
7
K
(
x
)
;
x
]
4
sn
[
5
7
K
(
x
)
;
x
]
4
}
{\displaystyle q(x)^{7}=q\{x^{7}\operatorname {sn} [{\tfrac {1}{7}}K(x);x]^{4}\operatorname {sn} [{\tfrac {3}{7}}K(x);x]^{4}\operatorname {sn} [{\tfrac {5}{7}}K(x);x]^{4}\}}
실수 값 구간 [-1;1]의 대수적 x 값의 경우 표시된 야코비 사인 진폭 표현식은 항상 대수적이다. 일반적으로 모든 자연수 n에 대해:
q
(
x
)
2
n
+
1
=
q
{
x
2
n
+
1
∏
k
=
1
n
sn
[
2
k
−
1
2
n
+
1
K
(
x
)
;
x
]
4
}
{\displaystyle q(x)^{2n+1}=q{\biggl \{}x^{2n+1}\prod _{k=1}^{n}\operatorname {sn} {\bigl [}{\tfrac {2k-1}{2n+1}}K(x);x{\bigr ]}^{4}{\biggr \}}}
입방체에 대한 정리는 단순화된 방식으로 매개변수화할 수도 있다:
q
[
u
(
u
4
−
u
2
+
1
−
u
2
+
1
)
]
3
=
q
[
u
(
u
4
−
u
2
+
1
+
u
2
−
1
)
]
{\displaystyle q[u({\sqrt {u^{4}-u^{2}+1}}-u^{2}+1)]^{3}=q[u({\sqrt {u^{4}-u^{2}+1}}+u^{2}-1)]}
이 공식은 모든 값
−
1
<
u
<
1
{\displaystyle -1<u<1}
ln
⟨
q
{
sin
[
1
4
π
−
1
2
arctan
(
x
)
]
}
⟩
ln
⟨
q
{
sin
[
1
4
π
+
1
2
arctan
(
x
)
]
}
⟩
=
π
2
{\displaystyle \ln {\biggl \langle }q{\bigl \{}\sin {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{2}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }\ln {\biggl \langle }q{\bigl \{}\sin {\bigl [}{\tfrac {1}{4}}\pi +{\tfrac {1}{2}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }=\pi ^{2}}
ln
{
q
[
1
2
2
−
2
x
(
x
2
+
1
)
−
1
/
2
]
}
ln
{
q
[
1
2
2
+
2
x
(
x
2
+
1
)
−
1
/
2
]
}
=
π
2
{\displaystyle \ln {\bigl \{}q{\bigl [}{\tfrac {1}{2}}{\sqrt {2-2x(x^{2}+1)^{-1/2}}}{\bigr ]}{\bigr \}}\ln {\bigl \{}q{\bigl [}{\tfrac {1}{2}}{\sqrt {2+2x(x^{2}+1)^{-1/2}}}{\bigr ]}{\bigr \}}=\pi ^{2}}
ln
⟨
q
{
tan
[
1
8
π
−
1
4
arctan
(
x
)
]
}
⟩
ln
⟨
q
{
tan
[
1
8
π
+
1
4
arctan
(
x
)
]
}
⟩
=
2
π
2
{\displaystyle \ln {\biggl \langle }q{\bigl \{}\tan {\bigl [}{\tfrac {1}{8}}\pi -{\tfrac {1}{4}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }\ln {\biggl \langle }q{\bigl \{}\tan {\bigl [}{\tfrac {1}{8}}\pi +{\tfrac {1}{4}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }=2\pi ^{2}}
ln
{
q
[
(
x
2
+
1
+
x
)
2
+
1
−
x
2
+
1
−
x
]
}
ln
{
q
[
(
x
2
+
1
−
x
)
2
+
1
−
x
2
+
1
+
x
]
}
=
2
π
2
{\displaystyle \ln {\bigl \{}q{\bigl [}{\sqrt {({\sqrt {x^{2}+1}}+x)^{2}+1}}-{\sqrt {x^{2}+1}}-x{\bigr ]}{\bigr \}}\ln {\bigl \{}q{\bigl [}{\sqrt {({\sqrt {x^{2}+1}}-x)^{2}+1}}-{\sqrt {x^{2}+1}}+x{\bigr ]}{\bigr \}}=2\pi ^{2}}
미분법 [ 편집 ] 타원 놈 함수는 다음과 같이 파생된다:
d
d
x
q
(
x
)
=
π
2
2
x
(
1
−
x
2
)
K
(
x
)
2
q
(
x
)
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}q(x)={\frac {\pi ^{2}}{2x(1-x^{2})K(x)^{2}}}q(x)}
2차 도함수는 다음과 같다:
d
2
d
x
2
q
(
x
)
=
π
4
+
2
π
2
(
1
+
x
2
)
K
(
x
)
2
−
4
π
2
K
(
x
)
E
(
x
)
4
x
2
(
1
−
x
2
)
2
K
(
x
)
4
q
(
x
)
{\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}q(x)={\frac {\pi ^{4}+2\pi ^{2}(1+x^{2})K(x)^{2}-4\pi ^{2}K(x)E(x)}{4x^{2}(1-x^{2})^{2}K(x)^{4}}}q(x)}
그리고 3차 도함수는 다음과 같은 형식을 취한다:
d
3
d
x
3
q
(
x
)
=
π
6
+
6
π
4
(
1
+
x
2
)
K
(
x
)
2
+
8
π
2
(
1
+
x
2
)
2
K
(
x
)
4
+
12
π
2
K
(
x
)
E
(
x
)
[
−
π
2
−
2
(
1
+
x
2
)
K
(
x
)
2
+
2
K
(
x
)
E
(
x
)
]
8
x
3
(
1
−
x
2
)
3
K
(
x
)
6
q
(
x
)
{\displaystyle {\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}q(x)={\frac {\pi ^{6}+6\pi ^{4}(1+x^{2})K(x)^{2}+8\pi ^{2}(1+x^{2})^{2}K(x)^{4}+12\pi ^{2}K(x)E(x)[-\pi ^{2}-2(1+x^{2})K(x)^{2}+2K(x)E(x)]}{8x^{3}(1-x^{2})^{3}K(x)^{6}}}q(x)}
두 번째 종류의 완전 타원 적분은 다음과 같이 정의된다:
E
(
ε
)
=
∫
0
π
/
2
1
−
ε
2
sin
(
φ
)
2
d
φ
{\displaystyle E(\varepsilon )=\int _{0}^{\pi /2}{\sqrt {1-\varepsilon ^{2}\sin(\varphi )^{2}}}\mathrm {d} \varphi }
E
(
ε
)
=
2
∫
0
1
(
w
2
+
1
)
2
−
4
ε
2
w
2
(
w
2
+
1
)
2
d
w
{\displaystyle E(\varepsilon )=2\int _{0}^{1}{\frac {\sqrt {(w^{2}+1)^{2}-4\,\varepsilon ^{2}w^{2}}}{(w^{2}+1)^{2}}}\mathrm {d} w}
두 공식은 서로 일치하며 동일한 결과를 가져온다.
제2종 완전 타원 적분을 지우면 다음과 같은 결과를 얻을 수 있다:
3
[
d
2
d
x
2
q
(
x
)
]
2
−
2
[
d
d
x
q
(
x
)
]
[
d
3
d
x
3
q
(
x
)
]
=
π
8
−
4
π
4
(
1
+
x
2
)
2
K
(
x
)
4
16
x
4
(
1
−
x
2
)
4
K
(
x
)
8
q
(
x
)
2
{\displaystyle 3{\biggl [}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}q(x){\biggr ]}^{2}-2{\biggl [}{\frac {\mathrm {d} }{\mathrm {d} x}}q(x){\biggr ]}{\biggl [}{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}q(x){\biggr ]}={\frac {\pi ^{8}-4\pi ^{4}(1+x^{2})^{2}K(x)^{4}}{16x^{4}(1-x^{2})^{4}K(x)^{8}}}q(x)^{2}}
따라서 이 3차 미분방정식은 유효하다:
x
2
(
1
−
x
2
)
2
[
2
q
(
x
)
2
q
′
(
x
)
q
‴
(
x
)
−
3
q
(
x
)
2
q
″
(
x
)
2
+
q
′
(
x
)
4
]
=
(
1
+
x
2
)
2
q
(
x
)
2
q
′
(
x
)
2
{\displaystyle x^{2}(1-x^{2})^{2}[2q(x)^{2}q'(x)q'''(x)-3q(x)^{2}q''(x)^{2}+q'(x)^{4}]=(1+x^{2})^{2}q(x)^{2}q'(x)^{2}}
무한 합과 무한 곱 [ 편집 ] 타원 놈는 Richard Dedekind에 의해 연구되었으며 이것은 그의 에타 함수 이론의 기초를 형성한다. 타원형 놈는 Lambert 급수의 구성에서 시작점을 형성하고 산술 기하 평균 의 대수적 조합에 대한 Carl Gustav Jacobi의 테타 함수 에서 가로 좌표로 지정된다. 일반적으로 많은 시리즈 확장은 타원 놈로 설명된다:
∑
n
=
1
∞
q
(
x
)
n
2
=
1
2
ϑ
00
[
q
(
x
)
]
−
1
2
=
1
2
2
π
−
1
K
(
x
)
−
1
2
=
1
2
agm
(
1
−
x
;
1
+
x
)
−
1
/
2
−
1
2
{\displaystyle \sum _{n=1}^{\infty }q(x)^{n^{2}}={\tfrac {1}{2}}\vartheta _{00}[q(x)]-{\tfrac {1}{2}}={\tfrac {1}{2}}{\sqrt {2\pi ^{-1}K(x)}}-{\tfrac {1}{2}}={\tfrac {1}{2}}\operatorname {agm} (1-x;1+x)^{-1/2}-{\tfrac {1}{2}}}
∑
n
=
1
∞
q
(
x
)
(
2
n
−
1
)
2
=
1
4
ϑ
00
[
q
(
x
)
]
−
1
4
ϑ
01
[
q
(
x
)
]
=
1
4
(
1
−
1
−
x
2
4
)
2
π
−
1
K
(
x
)
{\displaystyle \sum _{n=1}^{\infty }q(x)^{(2n-1)^{2}}={\tfrac {1}{4}}\vartheta _{00}[q(x)]-{\tfrac {1}{4}}\vartheta _{01}[q(x)]={\tfrac {1}{4}}(1-{\sqrt[{4}]{1-x^{2}}}){\sqrt {2\pi ^{-1}K(x)}}}
∑
n
=
1
∞
2
q
(
x
)
n
q
(
x
)
2
n
+
1
=
1
2
ϑ
00
[
q
(
x
)
]
2
−
1
2
=
π
−
1
K
(
x
)
−
1
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {2q(x)^{n}}{q(x)^{2n}+1}}={\tfrac {1}{2}}\vartheta _{00}[q(x)]^{2}-{\tfrac {1}{2}}=\pi ^{-1}K(x)-{\tfrac {1}{2}}}
∑
n
=
1
∞
2
q
(
x
)
2
n
−
1
q
(
x
)
4
n
−
2
+
1
=
1
4
ϑ
00
[
q
(
x
)
]
2
−
1
4
ϑ
01
[
q
(
x
)
]
2
=
1
2
(
1
−
1
−
x
2
)
π
−
1
K
(
x
)
{\displaystyle \sum _{n=1}^{\infty }{\frac {2q(x)^{2n-1}}{q(x)^{4n-2}+1}}={\tfrac {1}{4}}\vartheta _{00}[q(x)]^{2}-{\tfrac {1}{4}}\vartheta _{01}[q(x)]^{2}={\tfrac {1}{2}}(1-{\sqrt {1-x^{2}}})\,\pi ^{-1}K(x)}
∑
n
=
1
∞
n
2
q
(
x
)
n
2
=
2
−
1
/
2
π
−
5
/
2
K
(
x
)
3
/
2
[
E
(
x
)
−
(
1
−
x
2
)
K
(
x
)
]
{\displaystyle \sum _{n=1}^{\infty }n^{2}q(x)^{n^{2}}=2^{-1/2}\pi ^{-5/2}K(x)^{3/2}[E(x)-(1-x^{2})K(x)]}
∑
n
=
1
∞
[
2
q
(
x
)
n
1
+
q
(
x
)
2
n
]
2
=
2
π
−
2
E
(
x
)
K
(
x
)
−
1
2
{\displaystyle \sum _{n=1}^{\infty }{\biggl [}{\frac {2q(x)^{n}}{1+q(x)^{2n}}}{\biggr ]}^{2}=2\pi ^{-2}E(x)K(x)-{\tfrac {1}{2}}}
∑
n
=
1
∞
[
2
q
(
x
)
n
1
−
q
(
x
)
2
n
]
2
=
2
3
π
−
2
(
2
−
x
2
)
K
(
x
)
2
−
2
π
−
2
K
(
x
)
E
(
x
)
+
1
6
{\displaystyle \sum _{n=1}^{\infty }{\biggl [}{\frac {2q(x)^{n}}{1-q(x)^{2n}}}{\biggr ]}^{2}={\tfrac {2}{3}}\pi ^{-2}(2-x^{2})K(x)^{2}-2\pi ^{-2}K(x)E(x)+{\tfrac {1}{6}}}
∏
n
=
1
∞
[
1
−
q
(
x
)
2
n
]
[
1
+
q
(
x
)
2
n
−
1
]
2
=
2
π
−
1
K
(
x
)
{\displaystyle \prod _{n=1}^{\infty }[1-q(x)^{2n}][1+q(x)^{2n-1}]^{2}={\sqrt {2\pi ^{-1}K(x)}}}
∏
n
=
1
∞
[
1
−
q
(
x
)
2
n
]
[
1
−
q
(
x
)
2
n
−
1
]
2
1
−
x
2
4
2
π
−
1
K
(
x
)
{\displaystyle \prod _{n=1}^{\infty }[1-q(x)^{2n}][1-q(x)^{2n-1}]^{2}{\sqrt[{4}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}
테타 함수 [ 편집 ] 세 가지 주요 테타 함수는 다음과 같이 정의된다:
ϑ
00
(
y
)
=
1
+
2
(
∑
n
=
1
∞
y
n
2
)
{\displaystyle \vartheta _{00}(y)=1+2{\biggl (}\sum _{n=1}^{\infty }y^{n^{2}}{\biggr )}}
ϑ
01
(
y
)
=
1
+
2
[
∑
n
=
1
∞
(
−
1
)
n
y
n
2
]
{\displaystyle \vartheta _{01}(y)=1+2{\biggl [}\sum _{n=1}^{\infty }(-1)^{n}y^{n^{2}}{\biggr ]}}
ϑ
10
(
y
)
=
2
[
∑
n
=
1
∞
y
(
n
−
1
/
2
)
2
]
{\displaystyle \vartheta _{10}(y)=2{\biggl [}\sum _{n=1}^{\infty }y^{(n-1/2)^{2}}{\biggr ]}}
야코비 정체성는 이 세 가지 함수을 수학적 조합으로 제공한다:
ϑ
10
(
y
)
=
ϑ
00
(
y
)
4
−
ϑ
01
(
y
)
4
4
{\displaystyle \vartheta _{10}(y)={\sqrt[{4}]{\vartheta _{00}(y)^{4}-\vartheta _{01}(y)^{4}}}}
테타 함수와 타원형 놈 함수는 서로 다음과 같은 관계를 갖는다:
ϑ
00
[
q
(
x
)
]
=
2
π
−
1
K
(
x
)
{\displaystyle \vartheta _{00}[q(x)]={\sqrt {2\pi ^{-1}K(x)}}}
ϑ
01
[
q
(
x
)
]
=
1
−
x
2
4
2
π
−
1
K
(
x
)
{\displaystyle \vartheta _{01}[q(x)]={\sqrt[{4}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}
ϑ
00
[
q
(
x
)
2
]
=
cos
[
1
2
arcsin
(
x
)
]
2
π
−
1
K
(
x
)
{\displaystyle \vartheta _{00}[q(x)^{2}]=\cos {\bigl [}{\tfrac {1}{2}}\arcsin(x){\bigr ]}{\sqrt {2\pi ^{-1}K(x)}}}
ϑ
01
[
q
(
x
)
2
]
=
1
−
x
2
8
2
π
−
1
K
(
x
)
{\displaystyle \vartheta _{01}[q(x)^{2}]={\sqrt[{8}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}
오차 방정식 [ 편집 ] Abel-Ruffini 정리에 따르면 5차 방정식의 일반적인 경우는 기본적으로 풀 수 없다. 그러나 타원 명사와 세타 함수의 조합으로 모든 오차 방정식 을 풀 수 있다. 다음의 브링-제라드 정규형의 5차 다항식에 대해 언급된 타원 함수가 있는 실제 솔루션이 이제 표현된다:
x
5
+
5
x
=
4
c
{\displaystyle x^{5}+5\,x=4\,c}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}}
x
=
[
ϑ
00
(
Q
1
/
5
)
2
−
5
ϑ
00
(
Q
5
)
2
]
ϑ
00
(
Q
1
/
5
)
2
+
5
ϑ
00
(
Q
5
)
2
−
4
ϑ
00
(
Q
)
2
−
2
ϑ
00
(
Q
1
/
5
)
ϑ
00
(
Q
5
)
4
ϑ
10
(
Q
)
ϑ
01
(
Q
)
ϑ
00
(
Q
)
{\displaystyle x={\frac {{\bigl [}\vartheta _{00}(Q^{1/5})^{2}-5\,\vartheta _{00}(Q^{5})^{2}{\bigr ]}{\sqrt {\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}-2\,\vartheta _{00}(Q^{1/5})\,\vartheta _{00}(Q^{5})}}}{4\,\vartheta _{10}(Q)\,\vartheta _{01}(Q)\,\vartheta _{00}(Q)}}}
기본 함수로 풀 수 있는 방정식의 예:
첫 번째 계산 예:
x
5
+
5
x
=
2
3
{\displaystyle x^{5}+5\,x=2{\sqrt {3}}}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
(
c
=
1
2
3
)
=
exp
(
−
1
3
3
π
)
{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c={\tfrac {1}{2}}{\sqrt {3}})=\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi )}
x
=
ϑ
00
[
exp
(
−
1
15
3
π
)
]
2
−
5
ϑ
00
[
exp
(
−
5
3
3
π
)
]
2
4
ϑ
10
[
exp
(
−
1
3
3
π
)
]
ϑ
01
[
exp
(
−
1
3
3
π
)
]
ϑ
00
[
exp
(
−
1
3
3
π
)
]
×
{\displaystyle {\color {JungleGreen}x={\frac {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}}{4\,\vartheta _{10}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{01}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}}\times }}
×
ϑ
00
[
exp
(
−
1
15
3
π
)
]
2
+
5
ϑ
00
[
exp
(
−
5
3
3
π
)
]
2
−
4
ϑ
00
[
exp
(
−
1
3
3
π
)
]
2
−
2
ϑ
00
[
exp
(
−
1
15
3
π
)
]
ϑ
00
[
exp
(
−
5
3
3
π
)
]
{\displaystyle {\color {JungleGreen}\times {\sqrt {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}+5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-4\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-2\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}}}}
ϑ
00
[
exp
(
−
1
15
3
π
)
]
=
[
4
3
sin
(
1
5
π
)
10
3
+
1
80
6
+
1
3
cot
(
1
10
π
)
]
ϑ
00
[
exp
(
−
1
3
3
π
)
]
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}={\biggl [}{\frac {4}{3}}\sin {\bigl (}{\tfrac {1}{5}}\pi {\bigr )}{\frac {{\sqrt[{3}]{10}}+1}{\sqrt[{6}]{80}}}+{\frac {1}{3}}\cot({\tfrac {1}{10}}\pi ){\biggr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}
5
ϑ
00
[
exp
(
−
5
3
3
π
)
]
=
[
4
3
cos
(
1
10
π
)
10
3
+
1
80
6
−
1
3
tan
(
1
5
π
)
]
ϑ
00
[
exp
(
−
1
3
3
π
)
]
{\displaystyle {\sqrt {5}}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}={\biggl [}{\frac {4}{3}}\cos {\bigl (}{\tfrac {1}{10}}\pi {\bigr )}{\frac {{\sqrt[{3}]{10}}+1}{\sqrt[{6}]{80}}}-{\frac {1}{3}}\tan({\tfrac {1}{5}}\pi ){\biggr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}
x
=
1
3
3
(
10
3
−
1
)
{\displaystyle {\color {ForestGreen}x={\tfrac {1}{3}}{\sqrt {3}}\,({\sqrt[{3}]{10}}-1)}}
두 번째 계산 예:
x
5
+
5
x
=
3
7
{\displaystyle x^{5}+5\,x=3{\sqrt {7}}}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
(
c
=
3
4
7
)
=
exp
(
−
1
7
7
π
)
{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c={\tfrac {3}{4}}{\sqrt {7}})=\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi )}
x
=
ϑ
00
[
exp
(
−
1
35
7
π
)
]
2
−
5
ϑ
00
[
exp
(
−
5
7
7
π
)
]
2
4
ϑ
10
[
exp
(
−
1
7
7
π
)
]
ϑ
01
[
exp
(
−
1
7
7
π
)
]
ϑ
00
[
exp
(
−
1
7
7
π
)
]
×
{\displaystyle {\color {JungleGreen}x={\frac {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}}{4\,\vartheta _{10}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{01}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}}\times }}
×
ϑ
00
[
exp
(
−
1
35
7
π
)
]
2
+
5
ϑ
00
[
exp
(
−
5
7
7
π
)
]
2
−
4
ϑ
00
[
exp
(
−
1
7
7
π
)
]
2
−
2
ϑ
00
[
exp
(
−
1
35
7
π
)
]
ϑ
00
[
exp
(
−
5
7
7
π
)
]
{\displaystyle {\color {JungleGreen}\times {\sqrt {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}+5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-4\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-2\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}}}}
ϑ
00
[
exp
(
−
1
35
7
π
)
]
=
4
3
3
cos
(
1
10
π
)
cosh
[
1
3
artanh
(
1
9
21
)
]
ϑ
00
[
exp
(
−
1
7
7
π
)
]
{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}={\tfrac {4}{3}}{\sqrt {3}}\cos({\tfrac {1}{10}}\pi )\cosh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}
5
ϑ
00
[
exp
(
−
5
7
7
π
)
]
=
4
3
3
sin
(
1
5
π
)
cosh
[
1
3
artanh
(
1
9
21
)
]
ϑ
00
[
exp
(
−
1
7
7
π
)
]
{\displaystyle {\sqrt {5}}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}={\tfrac {4}{3}}{\sqrt {3}}\sin({\tfrac {1}{5}}\pi )\cosh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}
x
=
1
2
7
−
1
2
3
tanh
[
1
3
artanh
(
1
9
21
)
]
{\displaystyle {\color {ForestGreen}x={\tfrac {1}{2}}{\sqrt {7}}-{\tfrac {1}{2}}{\sqrt {3}}\tanh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}}}
타원 함수로만 풀 수 있는 계산 예:
x
5
+
5
x
=
2
2
4
{\displaystyle x^{5}+5\,x=2{\sqrt[{4}]{2}}}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
(
c
=
2
−
3
/
4
)
=
q
[
cos
(
1
8
π
)
]
{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c=2^{-3/4})=q{\bigl [}\cos {\bigl (}{\tfrac {1}{8}}\pi {\bigr )}{\bigr ]}}
Q
≈
0.11785793531185771914155254110648923544545944879394196130445
{\displaystyle Q\approx 0.11785793531185771914155254110648923544545944879394196130445}
x
=
ϑ
00
{
q
[
cos
(
1
8
π
)
]
1
/
5
}
2
−
5
ϑ
00
{
q
[
cos
(
1
8
π
)
]
5
}
2
4
ϑ
10
{
q
[
cos
(
1
8
π
)
]
}
ϑ
01
{
q
[
cos
(
1
8
π
)
]
}
ϑ
00
{
q
[
cos
(
1
8
π
)
]
}
×
{\displaystyle x={\frac {\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}^{2}-5\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}^{2}}{4\,\vartheta _{10}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}\,\vartheta _{01}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}}}\times }
×
ϑ
00
{
q
[
cos
(
1
8
π
)
]
1
/
5
}
2
+
5
ϑ
00
{
q
[
cos
(
1
8
π
)
]
5
}
2
−
4
ϑ
00
{
q
[
cos
(
1
8
π
)
]
}
2
−
2
ϑ
00
{
q
[
cos
(
1
8
π
)
]
1
/
5
}
ϑ
00
{
q
[
cos
(
1
8
π
)
]
5
}
{\displaystyle \times {\sqrt {\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}^{2}+5\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}^{2}-4\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}^{2}-2\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}}}}
x
≈
0.4710447387949811740242434591671230417409496435087081512857
{\displaystyle x\approx 0.4710447387949811740242434591671230417409496435087081512857}
이제 솔루션이 타원 함수의 도움으로만 표현될 수 있는 두 번째 예가 표시된다:
x
5
+
5
x
=
4
{\displaystyle x^{5}+5\,x=4}
Q
=
q
[
(
2
c
2
+
2
+
2
c
4
+
1
)
−
1
/
2
(
c
4
+
1
+
1
+
c
)
]
(
c
=
1
)
=
q
[
2
4
2
+
sin
(
π
8
)
]
{\displaystyle Q=q{\bigl [}{\bigl (}2c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}{\bigl (}c=1{\bigr )}=q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}{\bigr )}{\bigr ]}}
Q
≈
0.18520287008030014142515182307361246060360377625
{\displaystyle Q\approx 0.18520287008030014142515182307361246060360377625}
x
=
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
1
/
5
}
2
−
5
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
5
}
2
4
ϑ
10
{
q
[
2
4
2
+
sin
(
π
8
)
]
}
ϑ
01
{
q
[
2
4
2
+
sin
(
π
8
)
]
}
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
}
×
{\displaystyle x={\frac {\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}^{1/5}\}^{2}-5\,\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}^{5}\}^{2}}{4\,\vartheta _{10}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}\,\vartheta _{01}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}\,\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}}}\times }
×
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
1
/
5
}
2
+
5
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
5
}
2
−
4
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
}
2
−
2
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
1
/
5
}
ϑ
00
{
q
[
2
4
2
+
sin
(
π
8
)
]
5
}
{\displaystyle \times {\sqrt {\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{1/5}\}^{2}+5\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{5}\}^{2}-4\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}\}^{2}-2\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{1/5}\}\,\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{5}\}}}}
x
≈
0.75192639869405948026865366345020738740978383913
{\displaystyle x\approx 0.75192639869405948026865366345020738740978383913}
같이 보기 [ 편집 ] Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions OCLC 1097832 . See sections 16.27.4 and 17.3.17. 1972 edition: ISBN 0-486-61272-4
Tom M. Apostol , Modular Functions and Dirichlet Series in Number Theory, Second Edition (1990), Springer, New York ISBN 0-387-97127-0 Folkmar Bornemann, Dirk Laurie, Stan Wagon and Jörg Waldvogel, Vom Lösen numerischer Probleme , page 275
Edmund Taylor Whittaker and George Neville Watson : A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, 1990. page 469–470.Toshio Fukushima: Fast Computation of Complete Elliptic Integrals and Jacobian Elliptic Functions . 2012, National Astronomical Observatory of Japan (国立天文台)
Lowan, Blanch and Horenstein: On the Inversion of the q-Series Associated with Jacobian Elliptic Functions . Bull. Amer. Math. Soc. 48, 1942
H. Ferguson, D. E. Nielsen, G. Cook: A partition formula for the integer coefficients of the theta function nome . Mathematics of computation, Volume 29, number 131, Juli 1975
J. D. Fenton and R. S. Gardiner-Garden: Rapidly-convergent methods for evaluating elliptic integrals and theta and elliptic functions . J. Austral. Math. Soc. (Series B) 24, 1982, page 57
Charles Hermite: Sur la résolution de l'Équation du cinquiéme degré Comptes rendus . Acad. Sci. Paris, Nr. 11, 1858
Nikolaos Bagis: On the solution of the general quintic using the Rogers-Ramanujan continued fraction . Pella, Makedonien, Griechenland, 2015
Nikolaos Bagis: Solution of Polynomial Equations with Nested Radicals . Pella, Makedonien, Griechenland, 2020
Viktor Prasolov (Прасолов) und Yuri Solovyev (Соловьёв): Elliptic Functions and Elliptic Integrals . Volume 170, Rhode Island, 1991. pages 149 – 159
Sun Zhi-Hong: New congruences involving Apery-like numbers . Huaiyin Normal University, Huaian (淮安), China, 2020. page 2
Robert Fricke: Die elliptischen Funktionen und ihre Anwendungen: Dritter Teil . Springer-Verlag Berlin Heidelberg, 2012. ISBN 978-3-642-20953-6 , ISBN 978-3-642-20954-3 (eBook)
Adolf Kneser: Neue Untersuchung einer Reihe aus der Theorie der elliptischen Funktionen . J. reine u. angew. Math. 157, 1927. pages 209 – 218
G. P. Young: Solution of Solvable Irreducible Quintic Equations, Without the Aid of a Resolvent Sextic . In: Amer. J. Math. Band 7, pages 170–177, 1885.
C. Runge: Über die auflösbaren Gleichungen von der Form x 5 + u x + v = 0 {\displaystyle x^{5}+ux+v=0} x^{5}+ux+v=0 . In: Acta Math. Band 7, pages 173–186, 1885, doi:10.1007/BF02402200.
Edward Neuman: Two-sided inequalitites for the lemniscate functions. Volume 1, Southern Illinois University Carbondale , USA, 2014.
Ji-en Deng und Chao-ping Chen: Sharp Shafer-Fink type inequalities for Gauss lemniscate functions. Universität Henan (河南大学), China, 2014.
Jun-Ling Sun und Chao-ping Chen: Shafer-type inequalities for inverse trigonometric functions and Gauss lemniscate functions. Universität Henan, China, 2016.
Minjie Wei, Yue He and Gendi Wang: Shafer–Fink type inequalities for arc lemniscate functions . Zhejiang Sci-Tech University, Hangzhou, China, 2019 
![{\displaystyle q(x)=\exp {\bigl [}-\pi \,K({\sqrt {1-x^{2}}})\,K(x)^{-1}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e48085d7b295c4ae69291931925d980ac9e6c5f7)
![{\displaystyle {\text{Kt}}(n+1)={\frac {1}{n}}\sum _{k=1}^{n}k\,{\text{Kt}}(k)[16\,{\text{Ap}}(n+1-k)-{\text{Ap}}(n+2-k)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68bc7a99e16b0c56e6f0c08740c23278d4376567)
![{\displaystyle q(x)=\sum _{n=1}^{\infty }{\frac {{\text{Sw}}(n)}{2^{4n-3}}}{\biggl (}{\frac {1-{\sqrt[{4}]{1-x^{2}}}}{1+{\sqrt[{4}]{1-x^{2}}}}}{\biggr )}^{4n-3}=x^{2}{\biggl \{}{\frac {1}{2}}+{\biggl [}\sum _{n=1}^{\infty }{\frac {{\text{Sw}}(n+1)}{2^{4n+1}}}x^{2n}{\biggr ]}{\biggr \}}^{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd0a7560444b5ca057221a2bff2bff27fda2c10)
![{\displaystyle q[({\sqrt {2}}-1)^{2}]={\text{e}}^{-2\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/079402741cc0fc41dc58f0e23a62abbca26d77b9)
![{\displaystyle q[2{\sqrt[{4}]{2}}({\sqrt {2}}-1)]=\exp(-{\tfrac {1}{2}}\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11984d99c16031886f0383b494ffa10ab429a23f)
![{\displaystyle q[{\tfrac {1}{2}}({\sqrt {3}}-1)({\sqrt {2}}-{\sqrt[{4}]{3}})]={\text{e}}^{-3\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6dcd2c89e0738a9106d96eaa5390dfc9b49b7f4)
![{\displaystyle q[{\tfrac {1}{2}}({\sqrt {3}}-1)({\sqrt {2}}+{\sqrt[{4}]{3}})]=\exp(-{\tfrac {1}{3}}\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2676715f1a31a3992484d9be1e5b65b8b8a515b6)
![{\displaystyle q[{\tfrac {1}{2}}({\sqrt {10}}-2{\sqrt {2}})(3-2{\sqrt[{4}]{5}})]={\text{e}}^{-5\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32af9a31804139cd85901a4c2f8e18f9f425f3ad)
![{\displaystyle q[{\tfrac {1}{2}}({\sqrt {10}}-2{\sqrt {2}})(3+2{\sqrt[{4}]{5}})]=\exp(-{\tfrac {1}{5}}\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f1efb5d6ef4d5179909c56dc432fb23a0abfcef)
![{\displaystyle q[{\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})]={\text{e}}^{-{\sqrt {3}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b699f110a1d772eea046ff5e2964f7bf2ed90e90)
![{\displaystyle q[{\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})]=\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65439422ce60fa3d5fc6fe179c42e1009ed2d75f)
![{\displaystyle q{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}{\bigr \}}={\text{e}}^{-{\sqrt {5}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40d9b92fcbab137058bee0caef4a6190e2b80260)
![{\displaystyle q{\bigl \{}\cos {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}{\bigr \}}=\exp(-{\tfrac {1}{5}}{\sqrt {5}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbb7640deb60ac3f5183ed7491180f875efdf9fc)
![{\displaystyle q[{\tfrac {1}{8}}(3{\sqrt {2}}-{\sqrt {14}})]={\text{e}}^{-{\sqrt {7}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/683e33b54167148807e57e6ed6e2e6880862fd16)
![{\displaystyle q[{\tfrac {1}{8}}(3{\sqrt {2}}+{\sqrt {14}})]=\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1876f432bebf77e540f1f05ed861a3063a377e5d)
![{\displaystyle q{\bigl [}{\tfrac {1}{16}}{\bigl (}{\sqrt {22}}+3{\sqrt {2}}{\bigr )}{\bigl (}{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\tfrac {1}{3}}{\sqrt {11}}-1{\bigr )}^{4}{\bigr ]}={\text{e}}^{-{\sqrt {11}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b445a9d25425ea9900ac5c744d5a8a4c246dabce)
![{\displaystyle q{\bigl [}{\tfrac {1}{16}}{\bigl (}{\sqrt {22}}-3{\sqrt {2}}{\bigr )}{\bigl (}{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}+2{\sqrt {11}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {3}}-2{\sqrt {11}}}}+{\tfrac {1}{3}}{\sqrt {11}}+1{\bigr )}^{4}{\bigr ]}=\exp(-{\tfrac {1}{11}}{\sqrt {11}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d693754f74e8d7112c827dcd59404b39dc24e73)
![{\displaystyle q{\bigl \{}\sin {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}{\bigr \}}={\text{e}}^{-{\sqrt {13}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3d941e94004f3933d2d98ab81075247415755ac)
![{\displaystyle q{\bigl \{}\cos {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}{\bigr \}}=\exp(-{\tfrac {1}{13}}{\sqrt {13}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ffff40a1205dc4ad0013ac9e82751ba6371f2be)
![{\displaystyle q{\bigl \langle }\sin {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {1}{4}}-{\tfrac {1}{4}}{\sqrt {2{\sqrt {17}}+2}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {17}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1972a86fafd8375849fa04e6da51d1362cbce40c)
![{\displaystyle q{\bigl \langle }\cos {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {1}{4}}-{\tfrac {1}{4}}{\sqrt {2{\sqrt {17}}+2}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{17}}{\sqrt {17}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5eed9291d917f428fa715ce05fdc123fedfbd28b)
![{\displaystyle q{\bigl \langle }\sin {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}({\sqrt {37}}-6)^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {37}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11a02c5ca5b39f1f2deb7f1989052ad6031b1ae4)
![{\displaystyle q{\bigl \langle }\cos {\bigl \{}{\tfrac {1}{2}}\arcsin {\bigl [}({\sqrt {37}}-6)^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{37}}{\sqrt {37}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1576fa186b7e2426abe96311f8afa456ed9aac1e)
![{\displaystyle q[(2-{\sqrt {3}})({\sqrt {3}}-{\sqrt {2}})]={\text{e}}^{-{\sqrt {6}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2723be025d3fed8e2362fe5b1854fa2fff6e4c25)
![{\displaystyle q[(2-{\sqrt {3}})({\sqrt {3}}+{\sqrt {2}})]=\exp(-{\tfrac {1}{3}}{\sqrt {6}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75700ebac9ec6effff134f031941de5df9346d04)
![{\displaystyle q[({\sqrt {10}}-3)({\sqrt {2}}-1)^{2}]={\text{e}}^{-{\sqrt {10}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9e47ec37e6c0ad20953bf71aba4aaa8c47d1c11)
![{\displaystyle q[({\sqrt {10}}-3)({\sqrt {2}}+1)^{2}]=\exp(-{\tfrac {1}{5}}{\sqrt {10}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45461a275c98757ce2a36099ac9967977bc6e24c)
![{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\sqrt {2}}+{\tfrac {1}{2}}-{\tfrac {1}{2}}{\sqrt {4{\sqrt {2}}+5}}{\bigr )}^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {14}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a065a560aa4567c63e6672a9e8e61beb10f5447)
![{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\sqrt {2}}+{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {4{\sqrt {2}}+5}}{\bigr )}^{3}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {14}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/125bb65a58f927d35c4b58bcf037cd8ba013e12b)
![{\displaystyle q[(10-3{\sqrt {11}})(3{\sqrt {11}}-7{\sqrt {2}})]={\text{e}}^{-{\sqrt {22}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb04a02c149b69c2bff965da54d9c1c71d237ee9)
![{\displaystyle q[(10-3{\sqrt {11}})(3{\sqrt {11}}+7{\sqrt {2}})]=\exp(-{\tfrac {1}{11}}{\sqrt {22}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec5febe724b4a827262e9b8e2f6d626fea510e84)
![{\displaystyle q{\bigl \{}({\sqrt {26}}+5)({\sqrt {2}}-1)^{2}\tan {\bigl [}{\tfrac {1}{4}}\pi -\arctan({\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}+{\sqrt {26}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}-{\sqrt {26}}}}+{\tfrac {1}{6}}{\sqrt {26}}-{\tfrac {1}{2}}{\sqrt {2}}){\bigr ]}^{4}{\bigr \}}={\text{e}}^{-{\sqrt {26}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f847a405c8ccecc10b7b9b8cddfb607cb7ba7b1f)
![{\displaystyle q{\bigl \{}({\sqrt {26}}+5)({\sqrt {2}}+1)^{2}\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}+{\sqrt {26}}}}-{\tfrac {1}{3}}{\sqrt[{3}]{3{\sqrt {3}}-{\sqrt {26}}}}+{\tfrac {1}{6}}{\sqrt {26}}+{\tfrac {1}{2}}{\sqrt {2}})-{\tfrac {1}{4}}\pi {\bigr ]}^{4}{\bigr \}}=\exp(-{\tfrac {1}{13}}{\sqrt {26}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/282659359e4e2b4141082d4da6f772e0f2981726)
![{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {3}{4}}-{\tfrac {1}{4}}{\sqrt {6{\sqrt {17}}+10}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }={\text{e}}^{-{\sqrt {34}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/970438766acf02a4ec663c1d0bc64eb386627be3)
![{\displaystyle q{\bigl \langle }\tan {\bigl \{}{\tfrac {1}{2}}\arctan {\bigl [}{\bigl (}{\tfrac {1}{4}}{\sqrt {17}}+{\tfrac {3}{4}}+{\tfrac {1}{4}}{\sqrt {6{\sqrt {17}}+10}}{\bigr )}^{6}{\bigr ]}{\bigr \}}{\bigr \rangle }=\exp(-{\tfrac {1}{17}}{\sqrt {34}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bfb268f194f330c1913350d1ed3ee38de9ce9ae)
![{\displaystyle q[(13{\sqrt {58}}-99)({\sqrt {2}}-1)^{6}]={\text{e}}^{-{\sqrt {58}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8546165868a164a8088d715833a99e17770ad80)
![{\displaystyle q[(13{\sqrt {58}}-99)({\sqrt {2}}+1)^{6}]=\exp(-{\tfrac {1}{29}}{\sqrt {58}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b66236db93f7eb9d8f9596ff99272b207805a216)
![{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\sqrt {2}}-1{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {2}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9312b4fb437a305198c59a64fd6314d6b82c417c)
![{\displaystyle q[({\sqrt {3}}-{\sqrt {2}})^{2}({\sqrt {2}}-1)^{2}]={\text{e}}^{-2{\sqrt {3}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a22dd2e071883694b796ae5eb1e00d9c853a55f)
![{\displaystyle q[({\sqrt {3}}-{\sqrt {2}})^{2}({\sqrt {2}}+1)^{2}]=\exp(-{\tfrac {2}{3}}{\sqrt {3}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/110e1efd7bdf2e134b66cd42bc53cc67e7989c7e)
![{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {5}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe124b242628b6945ffade888a1d26b4729a3497)
![{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arcsin {\bigl (}{\sqrt {5}}-2{\bigr )}{\bigr ]}^{2}{\bigr \}}=\exp(-{\tfrac {2}{5}}{\sqrt {5}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24cce58b069acbef7ba14c371cdd8ee4aab230c0)
![{\displaystyle q[({\sqrt {2}}-1)^{4}(2{\sqrt {2}}-{\sqrt {7}})^{2}]={\text{e}}^{-2{\sqrt {7}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edb165dd56071527301c66fec25f1bb7b447ae4a)
![{\displaystyle q[({\sqrt {2}}-1)^{4}(2{\sqrt {2}}+{\sqrt {7}})^{2}]=\exp(-{\tfrac {2}{7}}{\sqrt {7}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0d1eb94543a6bb49f95423ff985878d4ebe68c9)
![{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}^{2}{\bigr \}}={\text{e}}^{-2{\sqrt {13}}\pi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cb9a8423956a3c68561399635a91610ddec99c8)
![{\displaystyle q{\bigl \{}\tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arcsin {\bigl (}5{\sqrt {13}}-18{\bigr )}{\bigr ]}^{2}{\bigr \}}=\exp(-{\tfrac {2}{13}}{\sqrt {13}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e4eac78e8b80497a54b1ddeb23f2b9fd864784f)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}-3)^{2}({\sqrt {5}}-2)^{2}]\}{\bigr \rangle }=\exp(-{\sqrt {30}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d69eb4b6d26661dc5cd4fd371661e5fccf829167)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}-3)^{2}({\sqrt {5}}+2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{3}}{\sqrt {30}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be2757b8d402e4f74852616257103ca5dc6f65e9)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}+3)^{2}({\sqrt {5}}-2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{5}}{\sqrt {30}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fab313381b6bcb8f7177009041ada5d0340793fc)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {10}}+3)^{2}({\sqrt {5}}+2)^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{15}}{\sqrt {30}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/790a8b7e2ae017b7c4724b827f4143b28264fba9)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}-3{\sqrt {3}})^{2}(2{\sqrt {2}}-{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\sqrt {42}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/812460a9a3377d3092a95784540d6383ca8fa9ed)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}-3{\sqrt {3}})^{2}(2{\sqrt {2}}+{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{3}}{\sqrt {42}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eeff4d809ee740f2337cbd638e87eae69c4ec73b)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}+3{\sqrt {3}})^{2}(2{\sqrt {2}}-{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {42}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d02a19ff96d8c7e93a671943c99e571ee094e21b)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[(2{\sqrt {7}}+3{\sqrt {3}})^{2}(2{\sqrt {2}}+{\sqrt {7}})^{2}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{21}}{\sqrt {42}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91cc6acb4d9cc60a1dd23541595e895d4bd6abd7)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}-2)^{4}({\sqrt {2}}-1)^{6}]\}{\bigr \rangle }=\exp(-{\sqrt {70}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99e7cf8530c26b9c60d13ffff4cf3aa9a68bd44d)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}-2)^{4}({\sqrt {2}}+1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{5}}{\sqrt {70}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b98aa9218cb41de881cc81c73987a7c09da2f550)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}+2)^{4}({\sqrt {2}}-1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{7}}{\sqrt {70}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fbf5d1f54881a796c734b0db1df201764a1abed)
![{\displaystyle q{\bigl \langle }\tan\{{\tfrac {1}{2}}\arctan[({\sqrt {5}}+2)^{4}({\sqrt {2}}+1)^{6}]\}{\bigr \rangle }=\exp(-{\tfrac {1}{35}}{\sqrt {70}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17dcb93c4a745a60ea3dfff87ef5861b62a58964)
![{\displaystyle q(x)^{2}=q[x^{2}(1+{\sqrt {1-x^{2}}})^{-2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81d4e2f22d84b78a189b6ec0e811866e009d0340)
![{\displaystyle q(x)^{3}=q\{x^{3}\operatorname {sn} [{\tfrac {1}{3}}K(x);x]^{4}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1aaa2a20204c337d524e60c96f5990d649bb4d2d)
![{\displaystyle q(x)^{5}=q\{x^{5}\operatorname {sn} [{\tfrac {1}{5}}K(x);x]^{4}\operatorname {sn} [{\tfrac {3}{5}}K(x);x]^{4}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21c632a335335d04a3202551ddef5b1687703263)
![{\displaystyle q(x)^{7}=q\{x^{7}\operatorname {sn} [{\tfrac {1}{7}}K(x);x]^{4}\operatorname {sn} [{\tfrac {3}{7}}K(x);x]^{4}\operatorname {sn} [{\tfrac {5}{7}}K(x);x]^{4}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51ea4a64c7e6c5dd7935d06f7e62975ae57cd0c4)
![{\displaystyle q(x)^{2n+1}=q{\biggl \{}x^{2n+1}\prod _{k=1}^{n}\operatorname {sn} {\bigl [}{\tfrac {2k-1}{2n+1}}K(x);x{\bigr ]}^{4}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/179540b77a63ca6859837bb8624ada84e7c749bf)
![{\displaystyle q[u({\sqrt {u^{4}-u^{2}+1}}-u^{2}+1)]^{3}=q[u({\sqrt {u^{4}-u^{2}+1}}+u^{2}-1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77c196e70df24473cb42f48397b7b7de4ed10e92)
![{\displaystyle \ln {\biggl \langle }q{\bigl \{}\sin {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{2}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }\ln {\biggl \langle }q{\bigl \{}\sin {\bigl [}{\tfrac {1}{4}}\pi +{\tfrac {1}{2}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }=\pi ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3ef0bae7362a1536667682c41d165230f55d0f7)
![{\displaystyle \ln {\bigl \{}q{\bigl [}{\tfrac {1}{2}}{\sqrt {2-2x(x^{2}+1)^{-1/2}}}{\bigr ]}{\bigr \}}\ln {\bigl \{}q{\bigl [}{\tfrac {1}{2}}{\sqrt {2+2x(x^{2}+1)^{-1/2}}}{\bigr ]}{\bigr \}}=\pi ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b3ce9b22a1c61c56aee8371f913b7d983fac4bf)
![{\displaystyle \ln {\biggl \langle }q{\bigl \{}\tan {\bigl [}{\tfrac {1}{8}}\pi -{\tfrac {1}{4}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }\ln {\biggl \langle }q{\bigl \{}\tan {\bigl [}{\tfrac {1}{8}}\pi +{\tfrac {1}{4}}\arctan(x){\bigr ]}{\bigr \}}{\biggr \rangle }=2\pi ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/296d0a4815c0f995f2ab32997b2c52db559f8f31)
![{\displaystyle \ln {\bigl \{}q{\bigl [}{\sqrt {({\sqrt {x^{2}+1}}+x)^{2}+1}}-{\sqrt {x^{2}+1}}-x{\bigr ]}{\bigr \}}\ln {\bigl \{}q{\bigl [}{\sqrt {({\sqrt {x^{2}+1}}-x)^{2}+1}}-{\sqrt {x^{2}+1}}+x{\bigr ]}{\bigr \}}=2\pi ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc886987b4daaa465cbb16636aef29774116d3e0)
![{\displaystyle {\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}q(x)={\frac {\pi ^{6}+6\pi ^{4}(1+x^{2})K(x)^{2}+8\pi ^{2}(1+x^{2})^{2}K(x)^{4}+12\pi ^{2}K(x)E(x)[-\pi ^{2}-2(1+x^{2})K(x)^{2}+2K(x)E(x)]}{8x^{3}(1-x^{2})^{3}K(x)^{6}}}q(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fbdc67eecc2dc36f2bb759f6e06cb01a159d3ed)
![{\displaystyle 3{\biggl [}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}q(x){\biggr ]}^{2}-2{\biggl [}{\frac {\mathrm {d} }{\mathrm {d} x}}q(x){\biggr ]}{\biggl [}{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}q(x){\biggr ]}={\frac {\pi ^{8}-4\pi ^{4}(1+x^{2})^{2}K(x)^{4}}{16x^{4}(1-x^{2})^{4}K(x)^{8}}}q(x)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3abf79eb163d0b918b655dfbfb2e7939bf5038b)
![{\displaystyle x^{2}(1-x^{2})^{2}[2q(x)^{2}q'(x)q'''(x)-3q(x)^{2}q''(x)^{2}+q'(x)^{4}]=(1+x^{2})^{2}q(x)^{2}q'(x)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d20a96bbd628efdd6806277004869a4047b3497c)
![{\displaystyle \sum _{n=1}^{\infty }q(x)^{n^{2}}={\tfrac {1}{2}}\vartheta _{00}[q(x)]-{\tfrac {1}{2}}={\tfrac {1}{2}}{\sqrt {2\pi ^{-1}K(x)}}-{\tfrac {1}{2}}={\tfrac {1}{2}}\operatorname {agm} (1-x;1+x)^{-1/2}-{\tfrac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e085950d228d9179c621c398544247f600bff16)
![{\displaystyle \sum _{n=1}^{\infty }q(x)^{(2n-1)^{2}}={\tfrac {1}{4}}\vartheta _{00}[q(x)]-{\tfrac {1}{4}}\vartheta _{01}[q(x)]={\tfrac {1}{4}}(1-{\sqrt[{4}]{1-x^{2}}}){\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5a62222de595b13f9cdbe4bf9fe33d33273cfcf)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {2q(x)^{n}}{q(x)^{2n}+1}}={\tfrac {1}{2}}\vartheta _{00}[q(x)]^{2}-{\tfrac {1}{2}}=\pi ^{-1}K(x)-{\tfrac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18262c129c6083740984bc3f7a60b2d312b0f3f8)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {2q(x)^{2n-1}}{q(x)^{4n-2}+1}}={\tfrac {1}{4}}\vartheta _{00}[q(x)]^{2}-{\tfrac {1}{4}}\vartheta _{01}[q(x)]^{2}={\tfrac {1}{2}}(1-{\sqrt {1-x^{2}}})\,\pi ^{-1}K(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30949bc6516abb4054272fec3fa64717fd19ad98)
![{\displaystyle \sum _{n=1}^{\infty }n^{2}q(x)^{n^{2}}=2^{-1/2}\pi ^{-5/2}K(x)^{3/2}[E(x)-(1-x^{2})K(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c021e24fcffd4932fa6c67f169c08f43709ec89a)
![{\displaystyle \sum _{n=1}^{\infty }{\biggl [}{\frac {2q(x)^{n}}{1+q(x)^{2n}}}{\biggr ]}^{2}=2\pi ^{-2}E(x)K(x)-{\tfrac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d21f2ef968a158473bf752d5b75003715e1142c)
![{\displaystyle \sum _{n=1}^{\infty }{\biggl [}{\frac {2q(x)^{n}}{1-q(x)^{2n}}}{\biggr ]}^{2}={\tfrac {2}{3}}\pi ^{-2}(2-x^{2})K(x)^{2}-2\pi ^{-2}K(x)E(x)+{\tfrac {1}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b51ce7017fc886ab22c947d7604999cd321ea527)
![{\displaystyle \prod _{n=1}^{\infty }[1-q(x)^{2n}][1+q(x)^{2n-1}]^{2}={\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a8cbaa0386c57b053f1224d8c1d71edaf020d2a)
![{\displaystyle \prod _{n=1}^{\infty }[1-q(x)^{2n}][1-q(x)^{2n-1}]^{2}{\sqrt[{4}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20b2823c8fa6c1f395845fef897c3e043d5f2a35)
![{\displaystyle \vartheta _{01}(y)=1+2{\biggl [}\sum _{n=1}^{\infty }(-1)^{n}y^{n^{2}}{\biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d905f4c6825af4295cc710543708bd2d3e83cca8)
![{\displaystyle \vartheta _{10}(y)=2{\biggl [}\sum _{n=1}^{\infty }y^{(n-1/2)^{2}}{\biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13fff24d030f1595f256c708bfe4d87ad8b0dced)
![{\displaystyle \vartheta _{10}(y)={\sqrt[{4}]{\vartheta _{00}(y)^{4}-\vartheta _{01}(y)^{4}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee7006330b7da303d51e23a393a0463fc60f3309)
![{\displaystyle \vartheta _{00}[q(x)]={\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ec62244207921549ce46631dc9c28b0c06f3a98)
![{\displaystyle \vartheta _{01}[q(x)]={\sqrt[{4}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13d5e158ff5af8b967d896938815a2f36fb02fad)
![{\displaystyle \vartheta _{00}[q(x)^{2}]=\cos {\bigl [}{\tfrac {1}{2}}\arcsin(x){\bigr ]}{\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb370e9f95c0eca3c627360f5cc557a67d995699)
![{\displaystyle \vartheta _{01}[q(x)^{2}]={\sqrt[{8}]{1-x^{2}}}\,{\sqrt {2\pi ^{-1}K(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11ca8f8d576d3ac0ac9e204a8b662589a906887b)
![{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15032b7bc8a5111463de7d0f27666af4ec3b00f9)
![{\displaystyle x={\frac {{\bigl [}\vartheta _{00}(Q^{1/5})^{2}-5\,\vartheta _{00}(Q^{5})^{2}{\bigr ]}{\sqrt {\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}-2\,\vartheta _{00}(Q^{1/5})\,\vartheta _{00}(Q^{5})}}}{4\,\vartheta _{10}(Q)\,\vartheta _{01}(Q)\,\vartheta _{00}(Q)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bfa954d820b6bcc9a441b962c15b6259e3ac23f)
![{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c={\tfrac {1}{2}}{\sqrt {3}})=\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef3d221a7456ec7e0e9b2dc5c23db200768f816d)
![{\displaystyle {\color {JungleGreen}x={\frac {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}}{4\,\vartheta _{10}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{01}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}}\times }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9de4ac75573b42f1370ad353535abfc992cc4eec)
![{\displaystyle {\color {JungleGreen}\times {\sqrt {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}+5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-4\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}^{2}-2\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f9b234f9affca9a5897c3281d08faf850c893f4)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{15}}{\sqrt {3}}\,\pi ){\bigr ]}={\biggl [}{\frac {4}{3}}\sin {\bigl (}{\tfrac {1}{5}}\pi {\bigr )}{\frac {{\sqrt[{3}]{10}}+1}{\sqrt[{6}]{80}}}+{\frac {1}{3}}\cot({\tfrac {1}{10}}\pi ){\biggr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9afe8543f48d650bf8453c19cf43b1e895a39fbf)
![{\displaystyle {\sqrt {5}}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{3}}{\sqrt {3}}\,\pi ){\bigr ]}={\biggl [}{\frac {4}{3}}\cos {\bigl (}{\tfrac {1}{10}}\pi {\bigr )}{\frac {{\sqrt[{3}]{10}}+1}{\sqrt[{6}]{80}}}-{\frac {1}{3}}\tan({\tfrac {1}{5}}\pi ){\biggr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{3}}{\sqrt {3}}\,\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06b51c8b162df8285e30dfe1ae9b112316ee686e)
![{\displaystyle {\color {ForestGreen}x={\tfrac {1}{3}}{\sqrt {3}}\,({\sqrt[{3}]{10}}-1)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12384887bf98126a12ceb6e8f4b86c0a04e3f448)
![{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c={\tfrac {3}{4}}{\sqrt {7}})=\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1b67cf5888ba13ed6def615f121075b283c18f8)
![{\displaystyle {\color {JungleGreen}x={\frac {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}}{4\,\vartheta _{10}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{01}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}}\times }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ba2617fd9105f7797695b5f999935517e8e4aa2)
![{\displaystyle {\color {JungleGreen}\times {\sqrt {\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}+5\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-4\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}^{2}-2\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38303418eccc98052839fa2450c28f57611f0d08)
![{\displaystyle \vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{35}}{\sqrt {7}}\,\pi ){\bigr ]}={\tfrac {4}{3}}{\sqrt {3}}\cos({\tfrac {1}{10}}\pi )\cosh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/187bb72fa99abdb66f494074feaf93c66d61b945)
![{\displaystyle {\sqrt {5}}\,\vartheta _{00}{\bigl [}\exp(-{\tfrac {5}{7}}{\sqrt {7}}\,\pi ){\bigr ]}={\tfrac {4}{3}}{\sqrt {3}}\sin({\tfrac {1}{5}}\pi )\cosh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}\vartheta _{00}{\bigl [}\exp(-{\tfrac {1}{7}}{\sqrt {7}}\,\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d99046d5c05aa72c8885d68f9bfa7324799807c8)
![{\displaystyle {\color {ForestGreen}x={\tfrac {1}{2}}{\sqrt {7}}-{\tfrac {1}{2}}{\sqrt {3}}\tanh {\bigl [}{\tfrac {1}{3}}{\text{artanh}}({\tfrac {1}{9}}{\sqrt {21}}){\bigr ]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e88e8cf567a70cf74f78e25436a1c3ffeeb8537)
![{\displaystyle x^{5}+5\,x=2{\sqrt[{4}]{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05faad68a2e97cbb9c635ec1b229037c49241dec)
![{\displaystyle Q=q{\bigl [}{\bigl (}2\,c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}(c=2^{-3/4})=q{\bigl [}\cos {\bigl (}{\tfrac {1}{8}}\pi {\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e27f9dca9633971974e96ea07218ba844534c57)
![{\displaystyle x={\frac {\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}^{2}-5\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}^{2}}{4\,\vartheta _{10}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}\,\vartheta _{01}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/15753ed06fa600e6a845ff2a1e7dbc398b77e5b5)
![{\displaystyle \times {\sqrt {\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}^{2}+5\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}^{2}-4\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}{\bigr \}}^{2}-2\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{1/5}{\bigr \}}\,\vartheta _{00}{\bigl \{}q{\bigl [}\cos({\tfrac {1}{8}}\pi ){\bigr ]}^{5}{\bigr \}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10299bf9c98b4e3065db8a43dfdaeaa9760ee467)
![{\displaystyle Q=q{\bigl [}{\bigl (}2c^{2}+2+2{\sqrt {c^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {c^{4}+1}}+1}}+c{\bigr )}{\bigr ]}{\bigl (}c=1{\bigr )}=q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}{\bigr )}{\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a0c37522775c49cdf66ecd87a5ae1814ae729c1)
![{\displaystyle x={\frac {\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}^{1/5}\}^{2}-5\,\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}^{5}\}^{2}}{4\,\vartheta _{10}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}\,\vartheta _{01}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}\,\vartheta _{00}\{q{\bigl [}{\frac {\sqrt[{4}]{2}}{2}}+\sin({\frac {\pi }{8}}){\bigr ]}\}}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/27325fca4eafb52c217a4b22aead1ab9c2904cb4)
![{\displaystyle \times {\sqrt {\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{1/5}\}^{2}+5\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{5}\}^{2}-4\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}\}^{2}-2\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{1/5}\}\,\vartheta _{00}\{q{\bigl [}{\tfrac {\sqrt[{4}]{2}}{2}}+\sin({\tfrac {\pi }{8}}){\bigr ]}^{5}\}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2d46f674c8953f7305327bda2ecdcf44b320ce1)
