Sinus lemniscatus sl (schwarz) und Cosinus lemniscatus cl (blau), zum Vergleich der auf sl normierte Sinus (hellgrau) Die Länge s des Lemniskatenbogens vom Ursprung korreliert mit dem Abstand r des Kurvenpunktes zum Ursprung.Quadrant enthält einen Viertelbogen (der Länge
ϖ
2
{\displaystyle {\tfrac {\varpi }{2}}}
Brennpunkte liegen hier bei
(
±
1
2
∣
0
)
{\displaystyle \left(\pm {\tfrac {1}{\sqrt {2}}}\mid 0\right)}
Der Lemniskatische Sinus und der Lemniskatische Cosinus (kurz sinlemn und coslemn oder
sl
{\displaystyle \operatorname {sl} }
cl
{\displaystyle \operatorname {cl} }
Carl Friedrich Gauß eingeführte mathematische Funktionen aus der Gruppe der elliptischen Funktionen. Der Sinus lemniscatus und Cosinus lemniscatus entsprechen denjenigen Funktionen für die Lemniskate , welche der Sinus und Cosinus für den Kreis sind. Der lemniskatische Cosinus leitet sich direkt vom lemniskatischen Sinus ab. Denn indem die Funktion
sl
{\displaystyle \operatorname {sl} }
ϖ
÷
2
{\displaystyle \varpi \div 2}
cl
{\displaystyle \operatorname {cl} }
elliptischen Funktionen . Nach der Definition durch Jacobi ist der Kehrwert der Quadratwurzel aus Zwei der elliptische Modul der lemniskatischen Funktionen.
Der 19-jährige Gauß beschäftigte sich 1796 (in erst nach seinem Tod veröffentlichten Notizen) mit der Frage, wie man aus einer gegebenen Bogenlänge
s
{\displaystyle s}
r
∈
(
−
1
,
1
)
{\displaystyle r\in (-1,1)}
r
=
0
{\displaystyle r=0}
Umkehrfunktion
r
=
r
(
s
)
{\displaystyle r=r(s)}
elliptischen Integrals
s
(
r
)
=
∫
0
r
d
ρ
1
−
ρ
4
.
{\displaystyle s(r)=\int _{0}^{r}{\frac {\mathrm {d} \,\rho }{\sqrt {1-\rho ^{4}}}}.}
Beweis:
Für den ersten und dritten Quadranten kann die Lemniskate von Bernoulli auf folgende Weise parametrisiert werden:
x und y als Koordinaten eines Punktes auf der Kurve im Abstand r vom Ursprung (Pythagoras) erfüllen die Lemniskatengleichung. Aus diesen zwei Gleichungen ergeben sich
x
(
r
)
=
r
1
+
r
2
/
2
{\displaystyle x(r)=r{\sqrt {1+r^{2}}}/{\sqrt {2}}}
y
(
r
)
=
r
1
−
r
2
/
2
{\displaystyle y(r)=r{\sqrt {1-r^{2}}}/{\sqrt {2}}}
Für die Berechnung der vom Ursprung ausgehenden Kurvenlänge s wird der Pythagoras der ersten Ableitungen von x und y gebildet und dieser integriert:
s
(
r
)
=
∫
0
r
[
d
d
r
x
(
r
)
(
r
=
ρ
)
]
2
+
[
d
d
r
y
(
r
)
(
r
=
ρ
)
]
2
d
ρ
=
{\displaystyle s(r)=\int _{0}^{r}{\sqrt {\left[{\frac {\mathrm {d} }{\mathrm {d} r}}x(r)(r=\rho )\right]^{2}+\left[{\frac {\mathrm {d} }{\mathrm {d} r}}y(r)(r=\rho )\right]^{2}}}\mathrm {d} \rho =}
=
∫
0
r
[
d
d
ρ
ρ
1
+
ρ
2
/
2
]
2
+
[
d
d
ρ
ρ
1
−
ρ
2
/
2
]
2
d
ρ
=
{\displaystyle =\int _{0}^{r}{\sqrt {\left[{\frac {\mathrm {d} }{\mathrm {d} \rho }}\rho {\sqrt {1+\rho ^{2}}}/{\sqrt {2}}\right]^{2}+\left[{\frac {\mathrm {d} }{\mathrm {d} \rho }}\rho {\sqrt {1-\rho ^{2}}}/{\sqrt {2}}\right]^{2}}}\mathrm {d} \rho =}
=
∫
0
r
(
1
+
2
ρ
2
)
2
2
(
1
+
ρ
2
)
+
(
1
−
2
ρ
2
)
2
2
(
1
−
ρ
2
)
d
ρ
=
∫
0
r
1
1
−
ρ
4
d
ρ
{\displaystyle =\int _{0}^{r}{\sqrt {{\frac {(1+2\rho ^{2})^{2}}{2(1+\rho ^{2})}}+{\frac {(1-2\rho ^{2})^{2}}{2(1-\rho ^{2})}}}}\mathrm {d} \rho =\int _{0}^{r}{\frac {1}{\sqrt {1-\rho ^{4}}}}\mathrm {d} \rho }
Gauß nannte diese Umkehrfunktion Sinus lemniscatus und bezeichnete sie mit
sl
{\displaystyle \operatorname {sl} }
r
=
sl
s
{\displaystyle r=\operatorname {sl} \,s}
Entsprechend definierte er den Cosinus lemniscatus
cl
s
=
sl
(
ϖ
2
−
s
)
{\displaystyle \operatorname {cl} \,s=\operatorname {sl} \,({\tfrac {\varpi }{2}}-s)}
ϖ
{\displaystyle \varpi }
ϖ
=
2
∫
0
1
d
ρ
1
−
ρ
4
≈
2,622
05
75542
92119
81046
48395
89891
…
{\displaystyle \varpi =2\int _{0}^{1}{\frac {\mathrm {d} \,\rho }{\sqrt {1-\rho ^{4}}}}\approx 2{,}62205\ 75542\ 92119\ 81046\,48395\ 89891\ldots }
A062539 in OEIS )Gauß ließ sich bei diesen Bezeichnungen von der Analogie zu den Kreisfunktionen leiten, denn der Sinus ist die Umkehrfunktion des Integrals
s
(
r
)
=
∫
0
r
d
ρ
1
−
ρ
2
,
und
2
∫
0
1
d
ρ
1
−
ρ
2
=
π
.
{\displaystyle s(r)=\int _{0}^{r}{\frac {\mathrm {d} \,\rho }{\sqrt {1-\rho ^{2}}}},\qquad {\mbox{und}}\qquad 2\int _{0}^{1}{\frac {\mathrm {d} \,\rho }{\sqrt {1-\rho ^{2}}}}=\pi .}
also
r
=
sin
s
{\displaystyle r=\sin s}
cos
s
=
sin
(
π
2
−
s
)
{\displaystyle \cos s=\sin({\tfrac {\pi }{2}}-s)}
sl
{\displaystyle \operatorname {sl} }
cl
{\displaystyle \operatorname {cl} }
Komplexe fortzusetzen. Er bewies dann die Periodizitätsrelationen
sl
(
s
+
2
ϖ
)
=
sl
s
,
sl
(
s
+
2
i
ϖ
)
=
sl
s
.
{\displaystyle \operatorname {sl} \,(s+2\varpi )=\operatorname {sl} \,s,\qquad \operatorname {sl} \,(s+2\mathrm {i} \varpi )=\operatorname {sl} \,s.}
Im Gegensatz zum Sinus hat also der lemniskatische Sinus
sl
{\displaystyle \operatorname {sl} }
zwei Perioden
2
ϖ
{\displaystyle 2\varpi }
2
i
ϖ
{\displaystyle 2\mathrm {i} \varpi }
cl
{\displaystyle \operatorname {cl} }
elliptisch . Carl Gustav Jacobi führte um 1830 die jacobischen elliptischen Funktionen ein und verallgemeinerte damit die beiden lemniskatischen Funktionen. Diese lassen sich auf folgende Weise durch die Jacobi-Funktionen mit dem Modul λ*(1) = 1/sqrt(2) ausdrücken:
sl
(
s
)
=
sd
(
2
s
;
1
/
2
)
/
2
{\displaystyle \operatorname {sl} \,(s)=\operatorname {sd} ({\sqrt {2}}s;1/{\sqrt {2}})/{\sqrt {2}}}
cl
(
s
)
=
cn
(
2
s
;
1
/
2
)
{\displaystyle \operatorname {cl} \,(s)=\operatorname {cn} ({\sqrt {2}}s;1/{\sqrt {2}})}
Somit sind der lemniskatische Sinus und der lemniskatische Cosinus auch über die Thetafunktionen auf folgende Weise[ 1]
sl
(
s
)
=
ϑ
10
(
π
/
2
−
π
s
/
ϖ
;
e
−
π
)
ϑ
01
(
π
/
2
−
π
s
/
ϖ
;
e
−
π
)
{\displaystyle \operatorname {sl} \,(s)={\frac {\vartheta _{10}(\pi /2-\pi s/\varpi ;\operatorname {e} ^{-\pi })}{\vartheta _{01}(\pi /2-\pi s/\varpi ;\operatorname {e} ^{-\pi })}}}
cl
(
s
)
=
ϑ
10
(
π
s
/
ϖ
;
e
−
π
)
ϑ
01
(
π
s
/
ϖ
;
e
−
π
)
{\displaystyle \operatorname {cl} \,(s)={\frac {\vartheta _{10}(\pi s/\varpi ;\operatorname {e} ^{-\pi })}{\vartheta _{01}(\pi s/\varpi ;\operatorname {e} ^{-\pi })}}}
Folgende algebraische Beziehung gilt für die lemniskatischen Funktionen:
[
1
+
sl
(
x
)
2
]
⋅
[
1
+
cl
(
x
)
2
]
=
2
{\displaystyle [1+\operatorname {sl} (x)^{2}]\cdot [1+\operatorname {cl} (x)^{2}]=2}
Die Additionstheoreme für die lemniskatischen Funktionen lauten wie folgt:
sl
(
a
+
b
)
=
sl
(
a
)
⋅
cl
(
b
)
+
cl
(
a
)
⋅
sl
(
b
)
1
−
sl
(
a
)
⋅
cl
(
a
)
⋅
sl
(
b
)
⋅
cl
(
b
)
{\displaystyle \operatorname {sl} (a+b)={\frac {\operatorname {sl} (a)\cdot \operatorname {cl} (b)+\operatorname {cl} (a)\cdot \operatorname {sl} (b)}{1-\operatorname {sl} (a)\cdot \operatorname {cl} (a)\cdot \operatorname {sl} (b)\cdot \operatorname {cl} (b)}}}
cl
(
a
+
b
)
=
cl
(
a
)
⋅
cl
(
b
)
−
sl
(
a
)
⋅
sl
(
b
)
1
+
sl
(
a
)
⋅
cl
(
a
)
⋅
sl
(
b
)
⋅
cl
(
b
)
{\displaystyle \operatorname {cl} (a+b)={\frac {\operatorname {cl} (a)\cdot \operatorname {cl} (b)-\operatorname {sl} (a)\cdot \operatorname {sl} (b)}{1+\operatorname {sl} (a)\cdot \operatorname {cl} (a)\cdot \operatorname {sl} (b)\cdot \operatorname {cl} (b)}}}
Alternative Darstellungen für die Additionstheoreme:
sl
(
a
+
b
)
=
sl
(
a
)
⋅
sl
′
(
b
)
+
sl
′
(
a
)
⋅
sl
(
b
)
1
+
sl
(
a
)
2
⋅
sl
(
b
)
2
{\displaystyle \operatorname {sl} (a+b)={\frac {\operatorname {sl} (a)\cdot \operatorname {sl} '(b)+\operatorname {sl} '(a)\cdot \operatorname {sl} (b)}{1+\operatorname {sl} (a)^{2}\cdot \operatorname {sl} (b)^{2}}}}
cl
(
a
+
b
)
=
sl
′
(
a
)
⋅
sl
′
(
b
)
−
2
⋅
sl
(
a
)
⋅
sl
(
b
)
1
+
sl
(
a
)
2
+
sl
(
b
)
2
−
sl
(
a
)
2
⋅
sl
(
b
)
2
{\displaystyle \operatorname {cl} (a+b)={\frac {\operatorname {sl} '(a)\cdot \operatorname {sl} '(b)-2\cdot \operatorname {sl} (a)\cdot \operatorname {sl} (b)}{1+\operatorname {sl} (a)^{2}+\operatorname {sl} (b)^{2}-\operatorname {sl} (a)^{2}\cdot \operatorname {sl} (b)^{2}}}}
Dabei gilt die Beziehung sl' = cl*(1+sl^2).
Darstellung über den Arkustangens:
arctan
[
sl
(
a
+
b
)
]
=
arctan
[
sl
(
a
)
⋅
cl
(
b
)
]
+
arctan
[
cl
(
a
)
⋅
sl
(
b
)
]
{\displaystyle \arctan[\operatorname {sl} (a+b)]=\arctan[\operatorname {sl} (a)\cdot \operatorname {cl} (b)]+\arctan[\operatorname {cl} (a)\cdot \operatorname {sl} (b)]}
arctan
[
cl
(
a
+
b
)
]
=
arctan
[
cl
(
a
)
⋅
cl
(
b
)
]
−
arctan
[
sl
(
a
)
⋅
sl
(
b
)
]
{\displaystyle \arctan[\operatorname {cl} (a+b)]=\arctan[\operatorname {cl} (a)\cdot \operatorname {cl} (b)]-\arctan[\operatorname {sl} (a)\cdot \operatorname {sl} (b)]}
Für die Verdopplung gelten diese Formeln:
sl
(
2
x
)
=
2
sl
(
x
)
cl
(
x
)
1
+
sl
(
x
)
2
1
+
sl
(
x
)
4
{\displaystyle \operatorname {sl} (2x)=2\,\operatorname {sl} (x)\operatorname {cl} (x){\frac {1+\operatorname {sl} (x)^{2}}{1+\operatorname {sl} (x)^{4}}}}
cl
(
2
x
)
=
−
1
+
2
cl
(
x
)
2
+
cl
(
x
)
4
1
+
2
cl
(
x
)
2
−
cl
(
x
)
4
{\displaystyle \operatorname {cl} (2x)={\frac {-1+2\,\operatorname {cl} (x)^{2}+\operatorname {cl} (x)^{4}}{1+2\,\operatorname {cl} (x)^{2}-\operatorname {cl} (x)^{4}}}}
Dementsprechend gelten folgende Formeln für die Halbierung:
sl
(
x
2
)
2
=
1
−
cl
(
x
)
1
+
sl
(
x
)
2
1
+
sl
(
x
)
2
+
1
{\displaystyle \operatorname {sl} \left({\frac {x}{2}}\right)^{2}={\frac {1-\operatorname {cl} (x){\sqrt {1+\operatorname {sl} (x)^{2}}}}{{\sqrt {1+\operatorname {sl} (x)^{2}}}+1}}}
cl
(
x
2
)
2
=
1
+
cl
(
x
)
1
+
sl
(
x
)
2
1
+
sl
(
x
)
2
+
1
{\displaystyle \operatorname {cl} \left({\frac {x}{2}}\right)^{2}={\frac {1+\operatorname {cl} (x){\sqrt {1+\operatorname {sl} (x)^{2}}}}{{\sqrt {1+\operatorname {sl} (x)^{2}}}+1}}}
Für die Verdreifachung gilt Folgendes:
sl
(
3
x
)
=
3
sl
(
x
)
−
6
sl
(
x
)
5
−
sl
(
x
)
9
1
+
6
sl
(
x
)
4
−
3
sl
(
x
)
8
{\displaystyle \operatorname {sl} (3x)={\frac {3\,\operatorname {sl} (x)-6\,\operatorname {sl} (x)^{5}-\operatorname {sl} (x)^{9}}{1+6\,\operatorname {sl} (x)^{4}-3\,\operatorname {sl} (x)^{8}}}}
Diese alternativen Darstellungen ermöglichen eine Umkehrung durch Lösen kubischer Gleichungen:
sl
(
3
x
)
=
27
4
(
3
+
1
)
y
−
2
y
3
2
+
27
4
(
3
+
1
)
y
2
[
y
=
3
4
(
3
−
1
)
sl
(
x
)
+
2
sl
(
x
)
3
2
−
3
4
(
3
−
1
)
sl
(
x
)
2
]
{\displaystyle \operatorname {sl} (3x)={\frac {{\sqrt[{4}]{27}}({\sqrt {3}}+1)y-{\sqrt {2}}y^{3}}{{\sqrt {2}}+{\sqrt[{4}]{27}}({\sqrt {3}}+1)y^{2}}}\left[y={\frac {{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)+{\sqrt {2}}\operatorname {sl} (x)^{3}}{{\sqrt {2}}-{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)^{2}}}\right]}
sl
(
3
x
)
=
27
4
(
3
+
1
)
z
+
2
z
3
2
−
27
4
(
3
+
1
)
z
2
[
z
=
3
4
(
3
−
1
)
sl
(
x
)
−
2
sl
(
x
)
3
2
+
3
4
(
3
−
1
)
sl
(
x
)
2
]
{\displaystyle \operatorname {sl} (3x)={\frac {{\sqrt[{4}]{27}}({\sqrt {3}}+1)z+{\sqrt {2}}z^{3}}{{\sqrt {2}}-{\sqrt[{4}]{27}}({\sqrt {3}}+1)z^{2}}}\left[z={\frac {{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)-{\sqrt {2}}\operatorname {sl} (x)^{3}}{{\sqrt {2}}+{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)^{2}}}\right]}
Der Cosinus Lemniscatus ergibt sich als negatives Analogon zum Sinus Lemniscatus:
cl
(
3
x
)
=
−
3
cl
(
x
)
+
6
cl
(
x
)
5
+
cl
(
x
)
9
1
+
6
cl
(
x
)
4
−
3
cl
(
x
)
8
{\displaystyle \operatorname {cl} (3x)={\frac {-3\,\operatorname {cl} (x)+6\,\operatorname {cl} (x)^{5}+\operatorname {cl} (x)^{9}}{1+6\,\operatorname {cl} (x)^{4}-3\,\operatorname {cl} (x)^{8}}}}
Die lemniskatischen Funktionen haben folgende Ableitungen:
d
d
x
sl
(
x
)
=
cl
(
x
)
⋅
[
1
+
sl
(
x
)
2
]
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {sl} (x)=\operatorname {cl} (x)\cdot [1+\operatorname {sl} (x)^{2}]}
d
d
x
cl
(
x
)
=
−
sl
(
x
)
⋅
[
1
+
cl
(
x
)
2
]
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {cl} (x)=-\operatorname {sl} (x)\cdot [1+\operatorname {cl} (x)^{2}]}
Daraus folgt die Tatsache, dass die zweite Ableitung das negative doppelte vom Kubus ist.
d
d
x
d
d
x
sl
(
x
)
=
−
2
⋅
sl
(
x
)
3
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {sl} (x)=-2\cdot \operatorname {sl} (x)^{3}}
d
d
x
d
d
x
cl
(
x
)
=
−
2
⋅
cl
(
x
)
3
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {cl} (x)=-2\cdot \operatorname {cl} (x)^{3}}
Über die Formeln der Ableitungen lassen sich ebenso die Stammfunktionen von Sinus Lemniscatus und Cosinus lemniscatus ermitteln.
cl
(
x
)
=
d
d
x
arctan
[
sl
(
x
)
]
{\displaystyle \operatorname {cl} (x)={\frac {\mathrm {d} }{\mathrm {d} x}}\arctan[\operatorname {sl} (x)]}
sl
(
x
)
=
−
d
d
x
arctan
[
cl
(
x
)
]
{\displaystyle \operatorname {sl} (x)=-{\frac {\mathrm {d} }{\mathrm {d} x}}\arctan[\operatorname {cl} (x)]}
Einzelne Funktionswerte für die lemniskatischen Funktionen:
sl
(
0
)
=
0
=
cl
(
ϖ
2
)
{\displaystyle \operatorname {sl} \left(0\right)=0=\operatorname {cl} \left({\frac {\varpi }{2}}\right)}
sl
(
ϖ
2
)
=
1
=
cl
(
0
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{2}}\right)=1=\operatorname {cl} \left(0\right)}
sl
(
ϖ
4
)
=
2
−
1
=
cl
(
ϖ
4
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{4}}\right)={\sqrt {{\sqrt {2}}-1}}=\operatorname {cl} \left({\frac {\varpi }{4}}\right)}
sl
(
ϖ
6
)
=
1
2
⋅
(
3
+
1
−
12
4
)
=
cl
(
ϖ
3
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{6}}\right)={\frac {1}{2}}\cdot \left({\sqrt {3}}+1-{\sqrt[{4}]{12}}\right)=\operatorname {cl} \left({\frac {\varpi }{3}}\right)}
sl
(
ϖ
3
)
=
3
8
2
4
⋅
3
−
1
=
cl
(
ϖ
6
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{3}}\right)={\frac {\sqrt[{8}]{3}}{\sqrt[{4}]{2}}}\cdot {\sqrt {{\sqrt {3}}-1}}=\operatorname {cl} \left({\frac {\varpi }{6}}\right)}
sl
(
ϖ
8
)
=
(
2
4
−
1
)
⋅
(
2
+
1
−
2
+
2
)
=
cl
(
3
⋅
ϖ
8
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{8}}\right)={\sqrt {\left({\sqrt[{4}]{2}}-1\right)\cdot \left({\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}\right)}}=\operatorname {cl} \left({\frac {3\cdot \varpi }{8}}\right)}
sl
(
3
⋅
ϖ
8
)
=
(
2
4
−
1
)
⋅
(
2
+
1
+
2
+
2
)
=
cl
(
ϖ
8
)
{\displaystyle \operatorname {sl} \left({\frac {3\cdot \varpi }{8}}\right)={\sqrt {\left({\sqrt[{4}]{2}}-1\right)\cdot \left({\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}\right)}}=\operatorname {cl} \left({\frac {\varpi }{8}}\right)}
sl
(
ϖ
5
)
=
1
2
⋅
2
4
⋅
(
5
−
1
)
⋅
20
4
−
5
−
1
=
cl
(
3
⋅
ϖ
10
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{5}}\right)={\frac {1}{2\cdot {\sqrt[{4}]{2}}}}\cdot ({\sqrt {5}}-1)\cdot {\sqrt {{\sqrt[{4}]{20}}-{\sqrt {{\sqrt {5}}-1}}}}=\operatorname {cl} \left({\frac {3\cdot \varpi }{10}}\right)}
sl
(
2
⋅
ϖ
5
)
=
1
2
⋅
2
4
⋅
(
5
−
1
)
⋅
20
4
+
5
−
1
=
cl
(
ϖ
10
)
{\displaystyle \operatorname {sl} \left({\frac {2\cdot \varpi }{5}}\right)={\frac {1}{2\cdot {\sqrt[{4}]{2}}}}\cdot ({\sqrt {5}}-1)\cdot {\sqrt {{\sqrt[{4}]{20}}+{\sqrt {{\sqrt {5}}-1}}}}=\operatorname {cl} \left({\frac {\varpi }{10}}\right)}
sl
(
ϖ
10
)
=
1
2
⋅
(
5
4
−
1
)
⋅
(
5
+
2
−
1
)
=
cl
(
2
⋅
ϖ
5
)
{\displaystyle \operatorname {sl} \left({\frac {\varpi }{10}}\right)={\frac {1}{2}}\cdot \left({\sqrt[{4}]{5}}-1\right)\cdot \left({\sqrt {{\sqrt {5}}+2}}-1\right)=\operatorname {cl} \left({\frac {2\cdot \varpi }{5}}\right)}
sl
(
3
⋅
ϖ
10
)
=
1
2
⋅
(
5
4
−
1
)
⋅
(
5
+
2
+
1
)
=
cl
(
ϖ
5
)
{\displaystyle \operatorname {sl} \left({\frac {3\cdot \varpi }{10}}\right)={\frac {1}{2}}\cdot \left({\sqrt[{4}]{5}}-1\right)\cdot \left({\sqrt {{\sqrt {5}}+2}}+1\right)=\operatorname {cl} \left({\frac {\varpi }{5}}\right)}
Weitere lemniskatische Funktionswerte in trigonometrischer Darstellung:
sl
(
1
12
ϖ
)
=
1
2
8
4
[
sin
(
5
24
π
)
−
3
4
sin
(
1
24
π
)
]
(
2
3
+
3
4
−
1
)
=
cl
(
5
12
ϖ
)
{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {1}{12}}\varpi {\bigr )}={\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]\left({\sqrt[{4}]{2{\sqrt {3}}+3}}-1\right)=\operatorname {cl} {\bigl (}{\tfrac {5}{12}}\varpi {\bigr )}}
sl
(
5
12
ϖ
)
=
1
2
8
4
[
sin
(
5
24
π
)
−
3
4
sin
(
1
24
π
)
]
(
2
3
+
3
4
+
1
)
=
cl
(
1
12
ϖ
)
{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {5}{12}}\varpi {\bigr )}={\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]\left({\sqrt[{4}]{2{\sqrt {3}}+3}}+1\right)=\operatorname {cl} {\bigl (}{\tfrac {1}{12}}\varpi {\bigr )}}
sl
(
1
14
ϖ
)
=
tanh
{
1
2
arcoth
[
1
2
2
cos
(
3
14
π
)
cot
(
1
28
π
)
+
cos
(
1
7
π
)
]
}
=
cl
(
3
7
ϖ
)
{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {1}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {3}{14}}\pi )\cot({\tfrac {1}{28}}\pi )}}+\cos({\tfrac {1}{7}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {3}{7}}\varpi {\bigr )}}
sl
(
3
14
ϖ
)
=
tanh
{
1
2
arcoth
[
1
2
2
cos
(
1
14
π
)
tan
(
5
28
π
)
+
sin
(
3
14
π
)
]
}
=
cl
(
2
7
ϖ
)
{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {3}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {1}{14}}\pi )\tan({\tfrac {5}{28}}\pi )}}+\sin({\tfrac {3}{14}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {2}{7}}\varpi {\bigr )}}
sl
(
5
14
ϖ
)
=
tanh
{
1
2
arcoth
[
1
2
2
sin
(
1
7
π
)
cot
(
3
28
π
)
+
sin
(
1
14
π
)
]
}
=
cl
(
1
7
ϖ
)
{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {5}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\sin({\tfrac {1}{7}}\pi )\cot({\tfrac {3}{28}}\pi )}}+\sin({\tfrac {1}{14}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {1}{7}}\varpi {\bigr )}}
Folgende Produktreihen für die lemniskatischen Funktionen konvergieren schnell:
sl
(
x
)
=
2
exp
(
−
1
4
π
)
sin
(
π
x
/
ϖ
)
∏
k
=
1
∞
1
−
2
cos
(
2
π
x
/
ϖ
)
exp
(
−
2
k
π
)
+
exp
(
−
4
k
π
)
1
+
2
cos
(
2
π
x
/
ϖ
)
exp
[
−
(
2
k
−
1
)
π
]
+
exp
[
−
(
4
k
−
2
)
π
]
{\displaystyle \operatorname {sl} (x)=2\exp \left(-{\tfrac {1}{4}}\pi \right)\sin(\pi x/\varpi )\prod _{k=1}^{\infty }{\frac {1-2\cos(2\pi x/\varpi )\exp(-2k\pi )+\exp(-4k\pi )}{1+2\cos(2\pi x/\varpi )\exp[-(2k-1)\pi ]+\exp[-(4k-2)\pi ]}}}
cl
(
x
)
=
2
exp
(
−
1
4
π
)
cos
(
π
x
/
ϖ
)
∏
k
=
1
∞
1
+
2
cos
(
2
π
x
/
ϖ
)
exp
(
−
2
k
π
)
+
exp
(
−
4
k
π
)
1
−
2
cos
(
2
π
x
/
ϖ
)
exp
[
−
(
2
k
−
1
)
π
]
+
exp
[
−
(
4
k
−
2
)
π
]
{\displaystyle \operatorname {cl} (x)=2\exp \left(-{\tfrac {1}{4}}\pi \right)\cos(\pi x/\varpi )\prod _{k=1}^{\infty }{\frac {1+2\cos(2\pi x/\varpi )\exp(-2k\pi )+\exp(-4k\pi )}{1-2\cos(2\pi x/\varpi )\exp[-(2k-1)\pi ]+\exp[-(4k-2)\pi ]}}}
Auf den Forschungsresultaten Edmund Taylor Whittaker und George Neville Watson [ 2] [ 3] [ 4]
Diese Summen stellen schnell konvergierende Reihen zur numerischen Berechnung des lemniskatischen Sinus und Cosinus dar:[ 5]
sl
(
x
)
=
4
π
ϖ
sin
(
π
x
ϖ
)
∑
k
=
0
∞
(
−
1
)
k
sinh
[
(
k
+
1
/
2
)
π
]
cosh
[
(
2
k
+
1
)
π
]
+
cos
(
2
π
x
/
ϖ
)
,
{\displaystyle \operatorname {sl} (x)={\frac {4\pi }{\varpi }}\sin \left({\frac {\pi x}{\varpi }}\right)\sum _{k=0}^{\infty }{\frac {(-1)^{k}\sinh[(k+1/2)\pi ]}{\cosh[(2k+1)\pi ]+\cos \left(2\pi x/\varpi \right)}},}
cl
(
x
)
=
4
π
ϖ
cos
(
π
x
ϖ
)
∑
k
=
0
∞
(
−
1
)
k
sinh
[
(
k
+
1
/
2
)
π
]
cosh
[
(
2
k
+
1
)
π
]
−
cos
(
2
π
x
/
ϖ
)
,
{\displaystyle \operatorname {cl} (x)={\frac {4\pi }{\varpi }}\cos \left({\frac {\pi x}{\varpi }}\right)\sum _{k=0}^{\infty }{\frac {(-1)^{k}\sinh[(k+1/2)\pi ]}{\cosh[(2k+1)\pi ]-\cos(2\pi x/\varpi )}},}
Hierbei verläuft die Präzision der Annäherung mit endlichem oberen Index
m
{\displaystyle m}
10
−
3
m
/
2
{\displaystyle 10^{-3m/2}}
Beide Reihen zeigen deutlich den Zusammenhang mit den Kreisfunktionen , indem die nach der lemniskatischen Form ausgestreckten Kreisfunktionen als Summanden in den genannten Differenzen gezeigt werden.
Basierend auf der Summendefinition der Jacobischen Zetafunktion können diese nicht alternierenden Summen aufgestellt werden:
tan
{
1
2
arctan
[
sl
(
x
)
]
}
=
4
π
ϖ
sin
(
π
x
/
ϖ
)
∑
k
=
1
∞
cosh
[
(
2
k
−
1
)
π
]
cosh
[
(
2
k
−
1
)
π
]
2
−
cos
(
π
x
/
ϖ
)
2
{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}={\frac {4\pi }{\varpi }}\sin(\pi x/\varpi )\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\pi x/\varpi )^{2}}}}
tan
{
1
2
arctan
[
cl
(
x
)
]
}
=
4
π
ϖ
cos
(
π
x
/
ϖ
)
∑
k
=
1
∞
cosh
[
(
2
k
−
1
)
π
]
cosh
[
(
2
k
−
1
)
π
]
2
−
sin
(
π
x
/
ϖ
)
2
{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}={\frac {4\pi }{\varpi }}\cos(\pi x/\varpi )\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\pi x/\varpi )^{2}}}}
Zusatzinformation:
Die Tangenshalbierungen von Sinus lemniscatus und Cosinus lemniscatus führen zu den Jacobi-Funktionen mit dem Modul λ*(4):
tan
{
1
2
arctan
[
sl
(
x
)
]
}
=
(
2
−
1
)
sn
[
1
2
(
2
+
1
)
x
;
(
2
−
1
)
2
]
{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}=({\sqrt {2}}-1)\operatorname {sn} \left[{\tfrac {1}{2}}({\sqrt {2}}+1)x;({\sqrt {2}}-1)^{2}\right]}
tan
{
1
2
arctan
[
cl
(
x
)
]
}
=
(
2
−
1
)
cd
[
1
2
(
2
+
1
)
x
;
(
2
−
1
)
2
]
{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}=({\sqrt {2}}-1)\operatorname {cd} \left[{\tfrac {1}{2}}({\sqrt {2}}+1)x;({\sqrt {2}}-1)^{2}\right]}
Weitere Reihendarstellungen über alternierende Summen des Secans hyperbolicus lauten so:
ϖ
π
cl
(
ϖ
x
)
=
∑
k
=
−
∞
∞
(
−
1
)
k
sech
[
π
(
k
+
x
)
]
{\displaystyle {\frac {\varpi }{\pi }}\operatorname {cl} (\varpi x)=\sum _{k=-\infty }^{\infty }(-1)^{k}\operatorname {sech} [\pi (k+x)]}
ϖ
π
sl
(
ϖ
x
)
=
∑
k
=
−
∞
∞
(
−
1
)
k
sech
[
π
(
k
−
1
2
+
x
)
]
{\displaystyle {\frac {\varpi }{\pi }}\operatorname {sl} (\varpi x)=\sum _{k=-\infty }^{\infty }(-1)^{k}\operatorname {sech} [\pi (k-{\frac {1}{2}}+x)]}
Die Ramanujansche Thetafunktion hat diese Definition:
ϑ
R
(
v
;
w
)
=
∑
n
=
−
∞
∞
v
n
(
n
+
1
)
/
2
w
n
(
n
−
1
)
/
2
=
∑
n
=
−
∞
∞
v
△
(
n
)
w
△
(
n
−
1
)
{\displaystyle \vartheta _{R}(v;w)=\sum _{n=-\infty }^{\infty }v^{n(n+1)/2}\,w^{n(n-1)/2}=\sum _{n=-\infty }^{\infty }v^{\bigtriangleup (n)}\,w^{\bigtriangleup (n-1)}}
Der korrespondierende Ausdruck aus einer Summe mit dem Index Eins lautet demnach so:
ϑ
R
(
v
;
w
)
=
1
+
v
+
w
+
∑
n
=
1
∞
(
v
w
)
△
(
n
)
(
v
n
+
1
+
w
n
+
1
)
{\displaystyle \vartheta _{R}(v;w)=1+v+w+\sum _{n=1}^{\infty }(vw)^{\bigtriangleup (n)}(v^{n+1}+w^{n+1})}
So können darauf basierend folgende Identitäten hervorgebracht werden:
1
+
tan
{
1
8
π
−
1
2
arctan
[
s
l
(
x
)
]
}
=
2
exp
(
−
x
G
)
ϑ
R
[
exp
(
x
÷
G
−
1
2
π
)
;
exp
(
−
x
÷
G
−
3
2
π
)
]
2
ϑ
R
[
exp
(
x
÷
G
−
3
2
π
)
;
exp
(
−
x
÷
G
−
1
2
π
)
]
2
{\displaystyle 1+\tan {\biggl \{}{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\mathrm {sl} (x){\bigr ]}{\biggr \}}={\sqrt {2}}\exp {\bigl (}-{\frac {x}{G}}{\bigr )}\,{\frac {\vartheta _{R}{\bigl [}\exp(x\div G-{\tfrac {1}{2}}\pi );\exp(-x\div G-{\tfrac {3}{2}}\pi ){\bigr ]}^{2}}{\vartheta _{R}{\bigl [}\exp(x\div G-{\tfrac {3}{2}}\pi );\exp(-x\div G-{\tfrac {1}{2}}\pi ){\bigr ]}^{2}}}}
1
+
tan
{
1
8
π
−
1
2
arctan
[
c
l
(
x
)
]
}
=
2
exp
(
x
G
−
π
2
)
ϑ
R
[
exp
(
−
x
÷
G
)
;
exp
(
x
÷
G
−
2
π
)
]
2
ϑ
R
[
exp
(
−
x
÷
G
−
π
)
;
exp
(
x
÷
G
−
π
)
]
2
{\displaystyle 1+\tan {\biggl \{}{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\mathrm {cl} (x){\bigr ]}{\biggr \}}={\sqrt {2}}\exp {\bigl (}{\frac {x}{G}}-{\frac {\pi }{2}}{\bigr )}\,{\frac {\vartheta _{R}{\bigl [}\exp(-x\div G);\exp(x\div G-2\pi ){\bigr ]}^{2}}{\vartheta _{R}{\bigl [}\exp(-x\div G-\pi );\exp(x\div G-\pi ){\bigr ]}^{2}}}}
Noch viel schneller als die Reihen der vorherigen Abschnitte konvergieren somit folgende zwei Reihen für die lemniskatischen Funktionen, welche sich direkt aus den Formeln mit der Ramanujanschen Thetafunktion herleiten lassen:
1
+
tan
{
1
8
π
−
1
2
arctan
[
sl
(
x
)
]
}
=
2
{
∑
k
=
−
∞
∞
exp
[
−
π
(
k
+
1
4
+
x
2
ϖ
)
2
]
}
2
{
∑
k
=
−
∞
∞
exp
[
−
π
(
k
−
1
4
+
x
2
ϖ
)
2
]
}
−
2
{\displaystyle 1+\tan \left\{{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}={\sqrt {2}}\,\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {1}{4}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{2}\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k-{\frac {1}{4}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{-2}}
1
+
tan
{
1
8
π
−
1
2
arctan
[
cl
(
x
)
]
}
=
2
{
∑
k
=
−
∞
∞
exp
[
−
π
(
k
+
1
2
+
x
2
ϖ
)
2
]
}
2
{
∑
k
=
−
∞
∞
exp
[
−
π
(
k
+
x
2
ϖ
)
2
]
}
−
2
{\displaystyle 1+\tan \left\{{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}={\sqrt {2}}\,\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {1}{2}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{2}\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{-2}}
Diejenigen elliptischen Module, welche die Lambda-Stern-Funktionswerte von den Doppelten der ungeraden natürlichen Zahlen[ 6]
Weitere Werte:
λ
∗
(
14
)
=
sl
{
1
2
arcsl
[
(
8
2
+
11
)
−
1
/
2
]
}
2
{\displaystyle \lambda ^{*}(14)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[(8{\sqrt {2}}+11)^{-1/2}\right]\right\}^{2}}
λ
∗
(
26
)
=
sl
{
1
2
arcsl
[
1
33
(
2
132
78
+
837
3
−
2
132
78
−
837
3
−
9
)
]
}
2
{\displaystyle \lambda ^{*}(26)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{33}}\left(2{\sqrt[{3}]{132{\sqrt {78}}+837}}-2{\sqrt[{3}]{132{\sqrt {78}}-837}}-9\right)\right]\right\}^{2}}
λ
∗
(
38
)
=
sl
{
1
2
arcsl
[
1
627
19
(
2
3300
114
+
27323
3
−
2
3300
114
−
27323
3
−
35
)
]
}
2
{\displaystyle \lambda ^{*}(38)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{627}}{\sqrt {19}}\left(2{\sqrt[{3}]{3300{\sqrt {114}}+27323}}-2{\sqrt[{3}]{3300{\sqrt {114}}-27323}}-35\right)\right]\right\}^{2}}
λ
∗
(
46
)
=
sl
{
1
2
arcsl
[
1
3
(
104
2
+
147
)
−
1
/
2
]
}
2
{\displaystyle \lambda ^{*}(46)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{3}}(104{\sqrt {2}}+147)^{-1/2}\right]\right\}^{2}}
λ
∗
(
66
)
=
sl
{
1
2
arcsl
[
(
75
2
+
13
2
33
+
1
2
1842
33
+
10578
)
−
1
]
}
2
{\displaystyle \lambda ^{*}(66)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[\left({\tfrac {75}{2}}+{\tfrac {13}{2}}{\sqrt {33}}+{\tfrac {1}{2}}{\sqrt {1842{\sqrt {33}}+10578}}\right)^{-1}\right]\right\}^{2}}
↑ Derivative of the Jacobi theta function: Introduction to the Jacobi theta functions. Abgerufen am 1. August 2021 . ↑ Eric W. Weisstein : Jacobi Theta Functions .MathWorld ↑ http://wayback.cecm.sfu.ca/~pborwein/TEMP_PROTECTED/pi-agm.pdf ↑ DLMF: 20.5 Infinite Products and Related Results. Abgerufen am 13. August 2022 . ↑ https://www.mdpi.com/2073-8994/12/6/1040 ↑ Eric W. Weisstein: Elliptic Lambda Function. Abgerufen am 20. Februar 2022 (englisch). 
![{\displaystyle s(r)=\int _{0}^{r}{\sqrt {\left[{\frac {\mathrm {d} }{\mathrm {d} r}}x(r)(r=\rho )\right]^{2}+\left[{\frac {\mathrm {d} }{\mathrm {d} r}}y(r)(r=\rho )\right]^{2}}}\mathrm {d} \rho =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5fb508aa089a8396fb189536ad6dcc9da9a033b)
![{\displaystyle =\int _{0}^{r}{\sqrt {\left[{\frac {\mathrm {d} }{\mathrm {d} \rho }}\rho {\sqrt {1+\rho ^{2}}}/{\sqrt {2}}\right]^{2}+\left[{\frac {\mathrm {d} }{\mathrm {d} \rho }}\rho {\sqrt {1-\rho ^{2}}}/{\sqrt {2}}\right]^{2}}}\mathrm {d} \rho =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fb86ceeaaba5299a69ab89a5d4b4d0ab8a21d4b)
![{\displaystyle [1+\operatorname {sl} (x)^{2}]\cdot [1+\operatorname {cl} (x)^{2}]=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25bedbe8bc3cb82798417e7475131c227316f184)
![{\displaystyle \arctan[\operatorname {sl} (a+b)]=\arctan[\operatorname {sl} (a)\cdot \operatorname {cl} (b)]+\arctan[\operatorname {cl} (a)\cdot \operatorname {sl} (b)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/611b127f1a23968766906a7dc8c14da85f2d5d80)
![{\displaystyle \arctan[\operatorname {cl} (a+b)]=\arctan[\operatorname {cl} (a)\cdot \operatorname {cl} (b)]-\arctan[\operatorname {sl} (a)\cdot \operatorname {sl} (b)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff480ed1b05a16f17884dc9627a976d670417978)
![{\displaystyle \operatorname {sl} (3x)={\frac {{\sqrt[{4}]{27}}({\sqrt {3}}+1)y-{\sqrt {2}}y^{3}}{{\sqrt {2}}+{\sqrt[{4}]{27}}({\sqrt {3}}+1)y^{2}}}\left[y={\frac {{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)+{\sqrt {2}}\operatorname {sl} (x)^{3}}{{\sqrt {2}}-{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b691fe9318c20d68d6a99a3294c57e5d77ac492d)
![{\displaystyle \operatorname {sl} (3x)={\frac {{\sqrt[{4}]{27}}({\sqrt {3}}+1)z+{\sqrt {2}}z^{3}}{{\sqrt {2}}-{\sqrt[{4}]{27}}({\sqrt {3}}+1)z^{2}}}\left[z={\frac {{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)-{\sqrt {2}}\operatorname {sl} (x)^{3}}{{\sqrt {2}}+{\sqrt[{4}]{3}}({\sqrt {3}}-1)\operatorname {sl} (x)^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be4ade87163d1495f8510f49237e9b7f07702c88)
![{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {sl} (x)=\operatorname {cl} (x)\cdot [1+\operatorname {sl} (x)^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d67993f8ba30769274bd9f2f73cdc1d5e2aef90c)
![{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {cl} (x)=-\operatorname {sl} (x)\cdot [1+\operatorname {cl} (x)^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23833447f1b08c929192c159ead2b7fb3289aa14)
![{\displaystyle \operatorname {cl} (x)={\frac {\mathrm {d} }{\mathrm {d} x}}\arctan[\operatorname {sl} (x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12bfb1cc6596f3e2f46410b59c1a76de6896fe9f)
![{\displaystyle \operatorname {sl} (x)=-{\frac {\mathrm {d} }{\mathrm {d} x}}\arctan[\operatorname {cl} (x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1a3563e776a88cce215f9b6c311b4c0dc9bac4e)
![{\displaystyle \operatorname {sl} \left({\frac {\varpi }{6}}\right)={\frac {1}{2}}\cdot \left({\sqrt {3}}+1-{\sqrt[{4}]{12}}\right)=\operatorname {cl} \left({\frac {\varpi }{3}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4567e492022fcf047833fe6277d72d13296bcdb)
![{\displaystyle \operatorname {sl} \left({\frac {\varpi }{3}}\right)={\frac {\sqrt[{8}]{3}}{\sqrt[{4}]{2}}}\cdot {\sqrt {{\sqrt {3}}-1}}=\operatorname {cl} \left({\frac {\varpi }{6}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c75e22fbb228f468a3e09f2bad5de2cf37b23457)
![{\displaystyle \operatorname {sl} \left({\frac {\varpi }{8}}\right)={\sqrt {\left({\sqrt[{4}]{2}}-1\right)\cdot \left({\sqrt {2}}+1-{\sqrt {2+{\sqrt {2}}}}\right)}}=\operatorname {cl} \left({\frac {3\cdot \varpi }{8}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a0d34e79706a402b6b0e40feed9629a43f1ae86)
![{\displaystyle \operatorname {sl} \left({\frac {3\cdot \varpi }{8}}\right)={\sqrt {\left({\sqrt[{4}]{2}}-1\right)\cdot \left({\sqrt {2}}+1+{\sqrt {2+{\sqrt {2}}}}\right)}}=\operatorname {cl} \left({\frac {\varpi }{8}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07c211e5e6713e243e9ed1366386d0d8b9e06b5b)
![{\displaystyle \operatorname {sl} \left({\frac {\varpi }{5}}\right)={\frac {1}{2\cdot {\sqrt[{4}]{2}}}}\cdot ({\sqrt {5}}-1)\cdot {\sqrt {{\sqrt[{4}]{20}}-{\sqrt {{\sqrt {5}}-1}}}}=\operatorname {cl} \left({\frac {3\cdot \varpi }{10}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86657cb32824b96e71cbed628455658ed04f4a7d)
![{\displaystyle \operatorname {sl} \left({\frac {2\cdot \varpi }{5}}\right)={\frac {1}{2\cdot {\sqrt[{4}]{2}}}}\cdot ({\sqrt {5}}-1)\cdot {\sqrt {{\sqrt[{4}]{20}}+{\sqrt {{\sqrt {5}}-1}}}}=\operatorname {cl} \left({\frac {\varpi }{10}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/350ae080f58713ac7beb65c26a17cec253cb713c)
![{\displaystyle \operatorname {sl} \left({\frac {\varpi }{10}}\right)={\frac {1}{2}}\cdot \left({\sqrt[{4}]{5}}-1\right)\cdot \left({\sqrt {{\sqrt {5}}+2}}-1\right)=\operatorname {cl} \left({\frac {2\cdot \varpi }{5}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b5d827ddb42c56eb51d6b0b93dd558fd96e3725)
![{\displaystyle \operatorname {sl} \left({\frac {3\cdot \varpi }{10}}\right)={\frac {1}{2}}\cdot \left({\sqrt[{4}]{5}}-1\right)\cdot \left({\sqrt {{\sqrt {5}}+2}}+1\right)=\operatorname {cl} \left({\frac {\varpi }{5}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27bf6584ff8e8df753e156ecb8d6d068adcddc86)
![{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {1}{12}}\varpi {\bigr )}={\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]\left({\sqrt[{4}]{2{\sqrt {3}}+3}}-1\right)=\operatorname {cl} {\bigl (}{\tfrac {5}{12}}\varpi {\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9657d3fe04afc2b531958a8a8adb756471d857ff)
![{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {5}{12}}\varpi {\bigr )}={\tfrac {1}{2}}{\sqrt[{4}]{8}}\left[\sin \left({\tfrac {5}{24}}\pi \right)-{\sqrt[{4}]{3}}\sin \left({\tfrac {1}{24}}\pi \right)\right]\left({\sqrt[{4}]{2{\sqrt {3}}+3}}+1\right)=\operatorname {cl} {\bigl (}{\tfrac {1}{12}}\varpi {\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a656c9dfd0f204677aea8f3cd6668ef1b069fe4)
![{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {1}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {3}{14}}\pi )\cot({\tfrac {1}{28}}\pi )}}+\cos({\tfrac {1}{7}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {3}{7}}\varpi {\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29750d5d5b02c43a92f9f11906016353d0f981b3)
![{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {3}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\cos({\tfrac {1}{14}}\pi )\tan({\tfrac {5}{28}}\pi )}}+\sin({\tfrac {3}{14}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {2}{7}}\varpi {\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0cf31de1efc5ae59e952900560e134b97bcfa1a)
![{\displaystyle \operatorname {sl} {\bigl (}{\tfrac {5}{14}}\varpi {\bigr )}=\tanh {\bigl \{}{\tfrac {1}{2}}\operatorname {arcoth} {\bigl [}{\tfrac {1}{2}}{\sqrt {2\sin({\tfrac {1}{7}}\pi )\cot({\tfrac {3}{28}}\pi )}}+\sin({\tfrac {1}{14}}\pi ){\bigr ]}{\bigr \}}=\operatorname {cl} {\bigl (}{\tfrac {1}{7}}\varpi {\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80f6d75d10360b413779018571943e624276dfa0)
![{\displaystyle \operatorname {sl} (x)=2\exp \left(-{\tfrac {1}{4}}\pi \right)\sin(\pi x/\varpi )\prod _{k=1}^{\infty }{\frac {1-2\cos(2\pi x/\varpi )\exp(-2k\pi )+\exp(-4k\pi )}{1+2\cos(2\pi x/\varpi )\exp[-(2k-1)\pi ]+\exp[-(4k-2)\pi ]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a897cb10453fcf68d44c7115541aa1304a265d26)
![{\displaystyle \operatorname {cl} (x)=2\exp \left(-{\tfrac {1}{4}}\pi \right)\cos(\pi x/\varpi )\prod _{k=1}^{\infty }{\frac {1+2\cos(2\pi x/\varpi )\exp(-2k\pi )+\exp(-4k\pi )}{1-2\cos(2\pi x/\varpi )\exp[-(2k-1)\pi ]+\exp[-(4k-2)\pi ]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edbce9c95b4012cd7789d749bff8f841e101b737)
![{\displaystyle \operatorname {sl} (x)={\frac {4\pi }{\varpi }}\sin \left({\frac {\pi x}{\varpi }}\right)\sum _{k=0}^{\infty }{\frac {(-1)^{k}\sinh[(k+1/2)\pi ]}{\cosh[(2k+1)\pi ]+\cos \left(2\pi x/\varpi \right)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59eac2fa576e6c2c92661ab3e9f09756e0dfb202)
![{\displaystyle \operatorname {cl} (x)={\frac {4\pi }{\varpi }}\cos \left({\frac {\pi x}{\varpi }}\right)\sum _{k=0}^{\infty }{\frac {(-1)^{k}\sinh[(k+1/2)\pi ]}{\cosh[(2k+1)\pi ]-\cos(2\pi x/\varpi )}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc4e0f9abe55943ff9b80b09323654b7c4fa58c9)
![{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}={\frac {4\pi }{\varpi }}\sin(\pi x/\varpi )\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\pi x/\varpi )^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe62df0f57e4d6a3d839f6496abbf709605c3dd2)
![{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}={\frac {4\pi }{\varpi }}\cos(\pi x/\varpi )\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\pi x/\varpi )^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de3ea53c3c97f4abebf74abfdc25763b11f3ba3b)
![{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}=({\sqrt {2}}-1)\operatorname {sn} \left[{\tfrac {1}{2}}({\sqrt {2}}+1)x;({\sqrt {2}}-1)^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40d225e69b31192041d9249156b8fc780474f6bf)
![{\displaystyle \tan \left\{{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}=({\sqrt {2}}-1)\operatorname {cd} \left[{\tfrac {1}{2}}({\sqrt {2}}+1)x;({\sqrt {2}}-1)^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c779b4fce5f7b84fe7d82780bef94d5ecccd690)
![{\displaystyle {\frac {\varpi }{\pi }}\operatorname {cl} (\varpi x)=\sum _{k=-\infty }^{\infty }(-1)^{k}\operatorname {sech} [\pi (k+x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d9023c0cf7edeffdc9e5e077751b022dd02d07)
![{\displaystyle {\frac {\varpi }{\pi }}\operatorname {sl} (\varpi x)=\sum _{k=-\infty }^{\infty }(-1)^{k}\operatorname {sech} [\pi (k-{\frac {1}{2}}+x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/924341f927e4bc7ab354814eff730a7e449d3c75)
![{\displaystyle 1+\tan {\biggl \{}{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\mathrm {sl} (x){\bigr ]}{\biggr \}}={\sqrt {2}}\exp {\bigl (}-{\frac {x}{G}}{\bigr )}\,{\frac {\vartheta _{R}{\bigl [}\exp(x\div G-{\tfrac {1}{2}}\pi );\exp(-x\div G-{\tfrac {3}{2}}\pi ){\bigr ]}^{2}}{\vartheta _{R}{\bigl [}\exp(x\div G-{\tfrac {3}{2}}\pi );\exp(-x\div G-{\tfrac {1}{2}}\pi ){\bigr ]}^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/254d3ea11bff41778397bdd632574190a1614b7a)
![{\displaystyle 1+\tan {\biggl \{}{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\mathrm {cl} (x){\bigr ]}{\biggr \}}={\sqrt {2}}\exp {\bigl (}{\frac {x}{G}}-{\frac {\pi }{2}}{\bigr )}\,{\frac {\vartheta _{R}{\bigl [}\exp(-x\div G);\exp(x\div G-2\pi ){\bigr ]}^{2}}{\vartheta _{R}{\bigl [}\exp(-x\div G-\pi );\exp(x\div G-\pi ){\bigr ]}^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea48b4dbbba88a86f79781d152b87df400b0df3a)
![{\displaystyle 1+\tan \left\{{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\operatorname {sl} (x){\bigr ]}\right\}={\sqrt {2}}\,\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {1}{4}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{2}\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k-{\frac {1}{4}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/830801a2e6fff14e8c2eda81311eaa1b6e838ac2)
![{\displaystyle 1+\tan \left\{{\frac {1}{8}}\pi -{\frac {1}{2}}\arctan {\bigl [}\operatorname {cl} (x){\bigr ]}\right\}={\sqrt {2}}\,\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {1}{2}}+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{2}\left\{\sum _{k=-\infty }^{\infty }\exp \left[-\pi \left(k+{\frac {x}{2\varpi }}\right)^{2}\right]\right\}^{-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20fd40b6c87955f8318ac0309308ee56f83c4cfb)
![{\displaystyle \lambda ^{*}({\tfrac {3}{2}})=\operatorname {cl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}}{\sqrt {3}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c26bfb4ce5367b1df021a9ce84aa7e6b01469d1)
![{\displaystyle \lambda ^{*}(6)=\operatorname {sl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}}{\sqrt {3}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5bb7dce6bb8df2dea27420a83fdbb125bf49556)
![{\displaystyle \lambda ^{*}({\tfrac {1}{6}})=\operatorname {sl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}}{\sqrt {3}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/550d978d137a6b824c35132b64788844e58eb386)
![{\displaystyle \lambda ^{*}({\tfrac {2}{3}})=\operatorname {cl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}}{\sqrt {3}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99c88002bd8b838fbd77e3dee4040cc0d38ea4e0)
![{\displaystyle \lambda ^{*}({\tfrac {5}{2}})=\operatorname {cl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a720287a362386852a296cdd60ffb86cc7f5d5fc)
![{\displaystyle \lambda ^{*}(10)=\operatorname {sl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a99ba9a12a5012b0ab0edb3b6323e984aef5b23e)
![{\displaystyle \lambda ^{*}({\tfrac {1}{10}})=\operatorname {sl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/376245dc72ba174746588e68ab64eec5c34475be)
![{\displaystyle \lambda ^{*}({\tfrac {2}{5}})=\operatorname {cl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{3}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5416c50c110074595d2148df2bd5a2d0b2446c2)
![{\displaystyle \lambda ^{*}({\tfrac {11}{2}})=\operatorname {cl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{33}}{\sqrt {11}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe5aad1642270fb25ce1e1bd1ab290d917c8b6cf)
![{\displaystyle \lambda ^{*}(22)=\operatorname {sl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{33}}{\sqrt {11}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1239ddd5182e00a029b1fcff723382fb97f079ba)
![{\displaystyle \lambda ^{*}({\tfrac {1}{22}})=\operatorname {sl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{33}}{\sqrt {11}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3ac21951e2475e9017bb21331e06e50be3ef796)
![{\displaystyle \lambda ^{*}({\tfrac {2}{11}})=\operatorname {cl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{33}}{\sqrt {11}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2dfe23ef8fd8437259993e41975a8feb27b91ac)
![{\displaystyle \lambda ^{*}({\tfrac {17}{2}})=\operatorname {cl} '\{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{3}}({\sqrt {17}}-4)]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94ca5728716f5c7b313000d4da494e3d0c5ed763)
![{\displaystyle \lambda ^{*}(34)=\operatorname {sl} \{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{3}}({\sqrt {17}}-4)]\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c20c4c85beff448e366e813a5aedbc9e91130592)
![{\displaystyle \lambda ^{*}({\tfrac {1}{34}})=\operatorname {sl} '\{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{3}}({\sqrt {17}}-4)]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10a2bc68bea2054e9c027df53a2cbf70e58342ec)
![{\displaystyle \lambda ^{*}({\tfrac {2}{17}})=\operatorname {cl} \{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{3}}({\sqrt {17}}-4)]\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9f2d9768de50eb8c3a7dbfc01c20da196b6879)
![{\displaystyle \lambda ^{*}({\tfrac {29}{2}})=\operatorname {cl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{99}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51cca4c900fe922ff2b68f7a29619ef3d8c3e72d)
![{\displaystyle \lambda ^{*}(58)=\operatorname {sl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{99}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92909a6cd60c944a4792bef52c5e058491b3272d)
![{\displaystyle \lambda ^{*}({\tfrac {1}{58}})=\operatorname {sl} '[{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{99}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f66aa151eee0ec87c636ce2c1dc9cc26b17dc18)
![{\displaystyle \lambda ^{*}({\tfrac {2}{29}})=\operatorname {cl} [{\tfrac {1}{2}}\operatorname {arcsl} ({\tfrac {1}{99}})]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95c3c8745416e6536f110ee1db0f9c338d48d616)
![{\displaystyle \lambda ^{*}({\tfrac {41}{2}})=\operatorname {cl} '\{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{69}}(8{\sqrt {41}}-51)]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd3db69386580a67eee665d3d3b0be7e8152a981)
![{\displaystyle \lambda ^{*}(82)=\operatorname {sl} \{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{69}}(8{\sqrt {41}}-51)]\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c404d66e84023718bc1596240eb0446c2cdbda75)
![{\displaystyle \lambda ^{*}({\tfrac {1}{82}})=\operatorname {sl} '\{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{69}}(8{\sqrt {41}}-51)]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dce8870b61402b9f6b1eced8bdfb1963e7fde7c3)
![{\displaystyle \lambda ^{*}({\tfrac {2}{41}})=\operatorname {cl} \{{\tfrac {1}{2}}\operatorname {arcsl} [{\tfrac {1}{69}}(8{\sqrt {41}}-51)]\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/578ac501fc3495988dc128482e0f3cbe9d06fa1e)
![{\displaystyle \lambda ^{*}(14)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[(8{\sqrt {2}}+11)^{-1/2}\right]\right\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6a6a983f1e806472fd3648ea88d191f685f930e)
![{\displaystyle \lambda ^{*}(26)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{33}}\left(2{\sqrt[{3}]{132{\sqrt {78}}+837}}-2{\sqrt[{3}]{132{\sqrt {78}}-837}}-9\right)\right]\right\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d413f6ce2d19183c3737311a7997d8ef1ee80099)
![{\displaystyle \lambda ^{*}(38)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{627}}{\sqrt {19}}\left(2{\sqrt[{3}]{3300{\sqrt {114}}+27323}}-2{\sqrt[{3}]{3300{\sqrt {114}}-27323}}-35\right)\right]\right\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32a029bc66735723d996002c370544ee5d0f07eb)
![{\displaystyle \lambda ^{*}(46)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[{\tfrac {1}{3}}(104{\sqrt {2}}+147)^{-1/2}\right]\right\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71764dd984587472c6b073530e0f541ce53ac3f1)
![{\displaystyle \lambda ^{*}(66)=\operatorname {sl} \left\{{\tfrac {1}{2}}\operatorname {arcsl} \left[\left({\tfrac {75}{2}}+{\tfrac {13}{2}}{\sqrt {33}}+{\tfrac {1}{2}}{\sqrt {1842{\sqrt {33}}+10578}}\right)^{-1}\right]\right\}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0946fea179a483cbf4ff327800e91931c5c19dc6)