In der oben abgebildeten Lemniskate gilt folgende Formel:
arcsl
(
r
)
=
s
(
r
)
=
∫
0
r
1
1
−
ρ
4
d
ρ
{\displaystyle \operatorname {arcsl} (r)=s(r)=\int _{0}^{r}{\frac {1}{\sqrt {1-\rho ^{4}}}}d\rho }
Das doppelte des Integrals von 0 bis 1 und liefert die Lemniskatische Konstante :
2
arcsl
(
1
)
=
2
∫
0
1
1
1
−
ρ
4
d
ρ
=
ϖ
{\displaystyle 2\operatorname {arcsl} (1)=2\int _{0}^{1}{\frac {1}{\sqrt {1-\rho ^{4}}}}d\rho =\varpi }
Beweis
Es gilt folgende Parametrisierung für die oben abgebildete Lemniskate:
x
(
t
)
=
sin
(
t
)
cos
(
t
)
2
+
1
{\displaystyle x(t)={\frac {\sin(t)}{\cos(t)^{2}+1}}}
y
(
t
)
=
sin
(
t
)
cos
(
t
)
cos
(
t
)
2
+
1
{\displaystyle y(t)={\frac {\sin(t)\cos(t)}{\cos(t)^{2}+1}}}
Daraus folgt für r:
r
(
t
)
=
x
(
t
)
2
+
y
(
t
)
2
=
sin
(
t
)
cos
(
t
)
2
+
1
{\displaystyle r(t)={\sqrt {x(t)^{2}+y(t)^{2}}}={\frac {\sin(t)}{\sqrt {\cos(t)^{2}+1}}}}
In Abhängigkeit von r ergeben sich folgende Formeln:
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 {d}{dr}}x(r)(r=\rho )\right]^{2}+\left[{\frac {d}{dr}}y(r)(r=\rho )\right]^{2}}}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 {d}{d\rho }}\rho {\sqrt {1+\rho ^{2}}}/{\sqrt {2}}\right]^{2}+\left[{\frac {d}{d\rho }}\rho {\sqrt {1-\rho ^{2}}}/{\sqrt {2}}\right]^{2}}}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})}}}}d\rho =\int _{0}^{r}{\frac {1}{\sqrt {1-\rho ^{4}}}}d\rho }
Reihenentwicklung
Die Taylorreihe des lemniskatischen Arkussinus mit dem Entwicklungspunkt erhält man durch Entwickeln der Ableitung in eine binomische Reihe und anschließende Integration:
arcsl
(
x
)
=
∫
0
x
1
1
−
y
4
d
y
=
∫
0
x
∑
k
=
0
∞
(
2
k
k
)
y
4
k
4
k
d
y
=
∑
k
=
0
∞
∫
0
x
(
2
k
k
)
y
4
k
4
k
d
y
=
∑
k
=
0
∞
(
2
k
k
)
x
4
k
+
1
4
k
(
4
k
+
1
)
{\displaystyle \operatorname {arcsl} (x)=\int _{0}^{x}{\frac {1}{\sqrt {1-y^{4}}}}dy=\int _{0}^{x}\sum _{k=0}^{\infty }{\binom {2k}{k}}{\frac {y^{4k}}{4^{k}}}dy=\sum _{k=0}^{\infty }\int _{0}^{x}{\binom {2k}{k}}{\frac {y^{4k}}{4^{k}}}dy=\sum _{k=0}^{\infty }{\binom {2k}{k}}{\frac {x^{4k+1}}{4^{k}(4k+1)}}}
Daraus folgt:
ϖ
=
2
∑
k
=
0
∞
(
2
k
k
)
1
4
k
(
4
k
+
1
)
{\displaystyle \varpi =2\sum _{k=0}^{\infty }{\binom {2k}{k}}{\frac {1}{4^{k}(4k+1)}}}
Diese Reihe konvergiert genau dann, wenn
|
x
|
≤
1
{\displaystyle |x|\leq 1}
Mit folgender Gleichung können noch schärfere Näherungen erzielt werden:
sl
[
arcsl
(
x
)
/
2
]
=
1
+
x
−
1
−
x
2
1
+
x
2
+
2
=
sin
[
arcsin
(
x
)
/
2
]
sech
[
arsinh
(
x
)
/
2
]
{\displaystyle \operatorname {sl} [\operatorname {arcsl} (x)/2]={\frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{\sqrt {2{\sqrt {1+x^{2}}}+2}}}=\sin[\arcsin(x)/2]\operatorname {sech} [\operatorname {arsinh} (x)/2]}
Dabei ist sl der lemniskatische Sinus.
Weitere Darstellungen
Der Arcussinus lemniscatus hat als elliptisches Integral erster Art ebenso folgende Darstellungen:
arcsl
(
x
)
=
1
2
F
[
arcsin
(
2
x
x
2
+
1
)
;
1
2
]
{\displaystyle \operatorname {arcsl} (x)={\frac {1}{\sqrt {2}}}F\left[\arcsin \left({\frac {{\sqrt {2}}x}{\sqrt {x^{2}+1}}}\right);{\frac {1}{\sqrt {2}}}\right]}
arcsl
(
x
)
=
ϖ
2
−
1
2
F
[
arccos
(
x
)
;
1
2
]
{\displaystyle \operatorname {arcsl} (x)={\frac {\varpi }{2}}-{\frac {1}{\sqrt {2}}}F\left[\arccos(x);{\frac {1}{\sqrt {2}}}\right]}
arcsl
(
x
)
=
(
2
2
−
2
)
F
[
arcsin
[
(
2
+
1
)
x
x
2
+
1
+
1
]
;
(
2
−
1
)
2
]
{\displaystyle \operatorname {arcsl} (x)=(2{\sqrt {2}}-2)F\left[\arcsin \left[{\frac {({\sqrt {2}}+1)x}{{\sqrt {x^{2}+1}}+1}}\right];({\sqrt {2}}-1)^{2}\right]}
Additionstheorem
Das Additionstheorem sieht so aus:
sl
[
arcsl
(
x
)
+
arcsl
(
y
)
]
=
x
1
−
y
4
+
y
1
−
x
4
1
+
x
2
y
2
{\displaystyle \operatorname {sl} [\operatorname {arcsl} (x)+\operatorname {arcsl} (y)]={\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}}
Denn es gilt folgender Zusammenhang:
(
d
d
x
x
1
−
y
4
+
y
1
−
x
4
1
+
x
2
y
2
)
[
1
−
(
x
1
−
y
4
+
y
1
−
x
4
1
+
x
2
y
2
)
4
]
−
1
/
2
=
1
1
−
x
4
{\displaystyle \left({\frac {d}{dx}}{\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}\right)\left[1-\left({\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}\right)^{4}\right]^{-1/2}={\frac {1}{\sqrt {1-x^{4}}}}}
Werte und Ableitungen
Werte des lemniskatischen Arkussinus:
arcsl
(
1
)
=
ϖ
/
2
{\displaystyle \operatorname {arcsl} (1)=\varpi /2}
arcsl
(
2
−
1
)
=
ϖ
/
4
{\displaystyle \operatorname {arcsl} ({\sqrt {{\sqrt {2}}-1}})=\varpi /4}
arcsl
[
(
3
+
1
−
12
4
)
/
2
]
=
ϖ
/
6
{\displaystyle \operatorname {arcsl} [({\sqrt {3}}+1-{\sqrt[{4}]{12}})/2]=\varpi /6}
arcsl
(
2
3
−
3
4
)
=
ϖ
/
3
{\displaystyle \operatorname {arcsl} ({\sqrt[{4}]{2{\sqrt {3}}-3}})=\varpi /3}
Ableitungen des lemniskatischen Arkussinus:
d
d
x
arcsl
(
x
)
=
1
1
−
x
4
{\displaystyle {\frac {d}{dx}}\operatorname {arcsl} (x)={\frac {1}{\sqrt {1-x^{4}}}}}
d
d
x
arcsl
(
x
x
2
+
1
)
=
1
(
x
2
+
1
)
(
2
x
2
+
1
)
{\displaystyle {\frac {d}{dx}}\operatorname {arcsl} \left({\frac {x}{\sqrt {x^{2}+1}}}\right)={\frac {1}{\sqrt {(x^{2}+1)(2x^{2}+1)}}}}
d
d
x
arcsl
(
2
x
x
4
+
6
x
2
+
1
+
x
2
+
1
)
=
1
x
4
+
6
x
2
+
1
{\displaystyle {\frac {d}{dx}}\operatorname {arcsl} \left({\frac {{\sqrt {2}}x}{\sqrt {{\sqrt {x^{4}+6x^{2}+1}}+x^{2}+1}}}\right)={\frac {1}{\sqrt {x^{4}+6x^{2}+1}}}}
d
d
x
2
arcsl
(
x
x
4
+
1
+
1
)
=
1
x
4
+
1
{\displaystyle {\frac {d}{dx}}{\sqrt {2}}\operatorname {arcsl} \left({\frac {x}{\sqrt {{\sqrt {x^{4}+1}}+1}}}\right)={\frac {1}{\sqrt {x^{4}+1}}}}
d
d
x
2
arcsl
(
x
1
+
1
−
x
4
)
=
1
(
1
−
x
4
)
3
4
{\displaystyle {\frac {d}{dx}}{\sqrt {2}}\operatorname {arcsl} \left({\frac {x}{\sqrt {1+{\sqrt {1-x^{4}}}}}}\right)={\frac {1}{\sqrt[{4}]{(1-x^{4})^{3}}}}}
d
d
x
arcsl
(
x
x
4
+
1
4
)
=
1
(
x
4
+
1
)
3
4
{\displaystyle {\frac {d}{dx}}\operatorname {arcsl} \left({\frac {x}{\sqrt[{4}]{x^{4}+1}}}\right)={\frac {1}{\sqrt[{4}]{(x^{4}+1)^{3}}}}}
d
d
x
2
arcsl
(
x
1
+
1
−
x
2
)
=
1
(
1
−
x
2
)
3
4
{\displaystyle {\frac {d}{dx}}2\operatorname {arcsl} \left({\frac {x}{1+{\sqrt {1-x^{2}}}}}\right)={\frac {1}{\sqrt[{4}]{(1-x^{2})^{3}}}}}
d
d
x
2
arcsl
(
x
x
2
+
1
+
1
)
=
1
(
x
2
+
1
)
3
4
{\displaystyle {\frac {d}{dx}}2\operatorname {arcsl} \left({\frac {x}{{\sqrt {x^{2}+1}}+1}}\right)={\frac {1}{\sqrt[{4}]{(x^{2}+1)^{3}}}}}
d
d
x
2
2
4
a
2
c
−
a
b
2
4
arcsl
[
2
a
x
+
b
4
a
(
a
x
2
+
b
x
+
c
)
+
4
a
c
−
b
2
]
=
1
(
a
x
2
+
b
x
+
c
)
3
4
{\displaystyle {\frac {d}{dx}}{\frac {2{\sqrt {2}}}{\sqrt[{4}]{4a^{2}c-ab^{2}}}}\operatorname {arcsl} \left[{\frac {2ax+b}{{\sqrt {4a(ax^{2}+bx+c)}}+{\sqrt {4ac-b^{2}}}}}\right]={\frac {1}{\sqrt[{4}]{(ax^{2}+bx+c)^{3}}}}}
Weblinks 
![{\displaystyle s(r)=\int _{0}^{r}{\sqrt {\left[{\frac {d}{dr}}x(r)(r=\rho )\right]^{2}+\left[{\frac {d}{dr}}y(r)(r=\rho )\right]^{2}}}d\rho =}](/media/api/rest_v1/media/math/render/svg/c2425124e0fb97f48fb40c9e35e58cb1cad87956)
![{\displaystyle =\int _{0}^{r}{\sqrt {\left[{\frac {d}{d\rho }}\rho {\sqrt {1+\rho ^{2}}}/{\sqrt {2}}\right]^{2}+\left[{\frac {d}{d\rho }}\rho {\sqrt {1-\rho ^{2}}}/{\sqrt {2}}\right]^{2}}}d\rho =}](/media/api/rest_v1/media/math/render/svg/77dfa2d1864c4362b35b3cd769d346e647ffdd2f)
![{\displaystyle \operatorname {sl} [\operatorname {arcsl} (x)/2]={\frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{\sqrt {2{\sqrt {1+x^{2}}}+2}}}=\sin[\arcsin(x)/2]\operatorname {sech} [\operatorname {arsinh} (x)/2]}](/media/api/rest_v1/media/math/render/svg/5d57e309d195478e52ee4c796559fad515ee858a)
![{\displaystyle \operatorname {arcsl} (x)={\frac {1}{\sqrt {2}}}F\left[\arcsin \left({\frac {{\sqrt {2}}x}{\sqrt {x^{2}+1}}}\right);{\frac {1}{\sqrt {2}}}\right]}](/media/api/rest_v1/media/math/render/svg/bd5af1a858299195395e2e7308eada3a8801514a)
![{\displaystyle \operatorname {arcsl} (x)={\frac {\varpi }{2}}-{\frac {1}{\sqrt {2}}}F\left[\arccos(x);{\frac {1}{\sqrt {2}}}\right]}](/media/api/rest_v1/media/math/render/svg/cf23521e95ad143da051a88c4a612f7980e95c3f)
![{\displaystyle \operatorname {arcsl} (x)=(2{\sqrt {2}}-2)F\left[\arcsin \left[{\frac {({\sqrt {2}}+1)x}{{\sqrt {x^{2}+1}}+1}}\right];({\sqrt {2}}-1)^{2}\right]}](/media/api/rest_v1/media/math/render/svg/5b6ab9168b56001b8ff77c45377f9abf1acfde53)
![{\displaystyle \operatorname {sl} [\operatorname {arcsl} (x)+\operatorname {arcsl} (y)]={\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}}](/media/api/rest_v1/media/math/render/svg/76abf576d89977d973ece30772bf0647f8ed205b)
![{\displaystyle \left({\frac {d}{dx}}{\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}\right)\left[1-\left({\frac {x{\sqrt {1-y^{4}}}+y{\sqrt {1-x^{4}}}}{1+x^{2}y^{2}}}\right)^{4}\right]^{-1/2}={\frac {1}{\sqrt {1-x^{4}}}}}](/media/api/rest_v1/media/math/render/svg/fa0ab51eae267acdb7c20e61d2beea89166df05b)
![{\displaystyle \operatorname {arcsl} [({\sqrt {3}}+1-{\sqrt[{4}]{12}})/2]=\varpi /6}](/media/api/rest_v1/media/math/render/svg/d139ce6a5a74a64db55b52482a6ff6106864c2dc)
![{\displaystyle \operatorname {arcsl} ({\sqrt[{4}]{2{\sqrt {3}}-3}})=\varpi /3}](/media/api/rest_v1/media/math/render/svg/421084da629c9a1b8c62004c5f0648d32ed59f2d)
![{\displaystyle {\frac {d}{dx}}{\sqrt {2}}\operatorname {arcsl} \left({\frac {x}{\sqrt {1+{\sqrt {1-x^{4}}}}}}\right)={\frac {1}{\sqrt[{4}]{(1-x^{4})^{3}}}}}](/media/api/rest_v1/media/math/render/svg/dd3fe0cafaa5a71242a19a5b07d9196c45b028f6)
![{\displaystyle {\frac {d}{dx}}\operatorname {arcsl} \left({\frac {x}{\sqrt[{4}]{x^{4}+1}}}\right)={\frac {1}{\sqrt[{4}]{(x^{4}+1)^{3}}}}}](/media/api/rest_v1/media/math/render/svg/e532ca7296c92b3efdff654369f25afecf2ca83b)
![{\displaystyle {\frac {d}{dx}}2\operatorname {arcsl} \left({\frac {x}{1+{\sqrt {1-x^{2}}}}}\right)={\frac {1}{\sqrt[{4}]{(1-x^{2})^{3}}}}}](/media/api/rest_v1/media/math/render/svg/bace243989b7e59cc5ce4f820110dd33b35957f3)
![{\displaystyle {\frac {d}{dx}}2\operatorname {arcsl} \left({\frac {x}{{\sqrt {x^{2}+1}}+1}}\right)={\frac {1}{\sqrt[{4}]{(x^{2}+1)^{3}}}}}](/media/api/rest_v1/media/math/render/svg/d8f10f7880e7b39ad7414feb1c5351b6ae98ef99)
![{\displaystyle {\frac {d}{dx}}{\frac {2{\sqrt {2}}}{\sqrt[{4}]{4a^{2}c-ab^{2}}}}\operatorname {arcsl} \left[{\frac {2ax+b}{{\sqrt {4a(ax^{2}+bx+c)}}+{\sqrt {4ac-b^{2}}}}}\right]={\frac {1}{\sqrt[{4}]{(ax^{2}+bx+c)^{3}}}}}](/media/api/rest_v1/media/math/render/svg/419d5e0dc99040fe005e42efdd16504e21293622)