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Unterseiten:
/Algebra
/Analysis
/Differentialgeometrie
/Physik
https://de.wikipedia.org/wiki/Kategorie:Wikipedia:Formatvorlage
https://de.wikipedia.org/wiki/Kategorie:Vorlage:Infobox
https://de.wikipedia.org/wiki/Wikipedia:Starthilfe
Quellen:
van der Waerden, Algebra I, Springer-Verlag
3
x
1
+
2
x
2
−
x
3
=
1
2
x
1
−
2
x
2
+
4
x
3
=
−
2
−
x
1
+
1
2
x
2
−
x
3
=
0
{\displaystyle {\begin{matrix}3x_{1}&+&2x_{2}&-&x_{3}&=&1\\2x_{1}&-&2x_{2}&+&4x_{3}&=&-2\\-x_{1}&+&{1 \over 2}x_{2}&-&x_{3}&=&0\end{matrix}}}
a
11
x
1
+
a
12
x
2
+
⋯
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
⋯
+
a
2
n
x
n
=
b
2
⋮
a
m
1
x
1
+
a
m
2
x
2
+
⋯
+
a
m
n
x
n
=
b
m
{\displaystyle {\begin{matrix}a_{11}x_{1}+a_{12}x_{2}\,+&\cdots &+\,a_{1n}x_{n}&=&b_{1}\\a_{21}x_{1}+a_{22}x_{2}\,+&\cdots &+\,a_{2n}x_{n}&=&b_{2}\\&&&\vdots &\\a_{m1}x_{1}+a_{m2}x_{2}\,+&\cdots &+\,a_{mn}x_{n}&=&b_{m}\\\end{matrix}}}
x
1
{\displaystyle x_{1}}
p1
(
x
2
−
5
)
2
−
24
{\displaystyle \left(x^{2}-5\right)^{2}-24}
(erscheint richtig in der Überschrift)
2D:
∫
0
R
∫
0
2
π
r
d
ϕ
1
d
r
=
π
R
2
{\displaystyle \int _{0}^{R}\int _{0}^{2\pi }r\mathrm {d} \phi _{1}\mathrm {d} r=\pi R^{2}}
3D:
∫
0
R
∫
0
2
π
∫
0
π
r
2
sin
(
ϕ
1
)
d
ϕ
1
d
ϕ
2
d
r
=
4
π
R
3
3
{\displaystyle \int _{0}^{R}\int _{0}^{2\pi }\int _{0}^{\pi }r^{2}\sin(\phi _{1}){\text{d}}\phi _{1}{\text{d}}\phi _{2}{\text{d}}r={\frac {4\pi R^{3}}{3}}}
4D:
∫
0
R
∫
0
2
π
∫
0
π
∫
0
π
r
3
sin
2
(
ϕ
1
)
sin
(
ϕ
2
)
d
ϕ
1
d
ϕ
2
d
ϕ
3
d
r
=
π
2
R
4
2
{\displaystyle \int _{0}^{R}\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{\pi }r^{3}\sin ^{2}(\phi _{1})\sin(\phi _{2}){\text{d}}\phi _{1}{\text{d}}\phi _{2}{\text{d}}\phi _{3}{\text{d}}r={\frac {\pi ^{2}R^{4}}{2}}}
Vaughan F. R. Jones: A polynomial invariant for knots via von Neumann algebras . In: Hyman Bass, Meyer Jerison, Calvin C. Moore (Hrsg.): Bulletin of the American Mathematical Society (New Series) . Vol. 12, Nr. 1 . American Mathematical Society, 1985, ISSN 0273-0979 S. 103–111 , doi :10.1090/S0273-0979-1985-15304-2 ams.org [PDF; abgerufen am 2. Dezember 2012]).
Q
{\displaystyle \textstyle \mathbb {Q} }
Q
(
2
)
=
{
a
+
b
2
∣
a
,
b
∈
Q
}
{\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}})=\{a+b{\sqrt {2}}\mid a,b\in \mathbb {Q} \}}
{
1
,
2
{\displaystyle \textstyle 1,{\sqrt {2}}}
Q
(
2
,
3
)
=
{
a
+
b
2
+
c
3
+
d
6
∣
a
,
b
,
c
,
d
∈
Q
}
{\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})=\{a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mid a,b,c,d\in \mathbb {Q} \}}
{
1
,
2
,
3
,
6
{\displaystyle \textstyle 1,{\sqrt {2}},{\sqrt {3}},{\sqrt {6}}}
5
±
2
6
=
(
2
±
3
)
2
{\displaystyle 5\pm 2{\sqrt {6}}=({\sqrt {2}}\pm {\sqrt {3}})^{2}}
x
1
=
2
+
3
{\displaystyle x_{1}={\sqrt {2}}+{\sqrt {3}}}
2
=
1
2
(
x
1
3
−
9
x
1
)
{\displaystyle {\sqrt {2}}={\tfrac {1}{2}}(x_{1}^{3}-9x_{1})}
3
=
−
1
2
(
x
1
3
−
11
x
1
)
{\displaystyle {\sqrt {3}}=-{\tfrac {1}{2}}(x_{1}^{3}-11x_{1})}
1
a
+
b
2
=
(
a
−
b
2
)
(
a
+
b
2
)
⋅
(
a
−
b
2
)
=
(
a
−
b
2
)
(
a
2
−
2
b
2
)
=
a
(
a
2
−
2
b
2
)
+
−
b
(
a
2
−
2
b
2
)
2
{\displaystyle {\frac {1}{a+b{\sqrt {2}}}}={\frac {(a-b{\sqrt {2}})}{(a+b{\sqrt {2}})\cdot (a-b{\sqrt {2}})}}={\frac {(a-b{\sqrt {2}})}{(a^{2}-2b^{2})}}={\frac {a}{(a^{2}-2b^{2})}}+{\frac {-b}{(a^{2}-2b^{2})}}{\sqrt {2}}}
'kursiv' (???)
Beispiel (???)
fett
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
.
.
.
{\displaystyle \left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto ...}
Nr.
Permutation
Nr.
Permutation
Nr.
Permutation
Nr.
Permutation
1
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle \left(x_{1},x_{2},x_{3},x_{4}\right)}
7
(
x
2
,
x
1
,
x
3
,
x
4
)
{\displaystyle \left(x_{2},x_{1},x_{3},x_{4}\right)}
13
(
x
3
,
x
1
,
x
2
,
x
4
)
{\displaystyle \left(x_{3},x_{1},x_{2},x_{4}\right)}
19
(
x
4
,
x
1
,
x
2
,
x
3
)
{\displaystyle \left(x_{4},x_{1},x_{2},x_{3}\right)}
2
(
x
1
,
x
2
,
x
4
,
x
3
)
{\displaystyle \left(x_{1},x_{2},x_{4},x_{3}\right)}
8
(
x
2
,
x
1
,
x
4
,
x
3
)
{\displaystyle \left(x_{2},x_{1},x_{4},x_{3}\right)}
14
(
x
3
,
x
1
,
x
4
,
x
2
)
{\displaystyle \left(x_{3},x_{1},x_{4},x_{2}\right)}
20
(
x
4
,
x
1
,
x
3
,
x
2
)
{\displaystyle \left(x_{4},x_{1},x_{3},x_{2}\right)}
3
(
x
1
,
x
3
,
x
2
,
x
4
)
{\displaystyle \left(x_{1},x_{3},x_{2},x_{4}\right)}
9
(
x
2
,
x
3
,
x
1
,
x
4
)
{\displaystyle \left(x_{2},x_{3},x_{1},x_{4}\right)}
15
(
x
3
,
x
2
,
x
1
,
x
4
)
{\displaystyle \left(x_{3},x_{2},x_{1},x_{4}\right)}
21
(
x
4
,
x
2
,
x
1
,
x
3
)
{\displaystyle \left(x_{4},x_{2},x_{1},x_{3}\right)}
4
(
x
1
,
x
3
,
x
4
,
x
2
)
{\displaystyle \left(x_{1},x_{3},x_{4},x_{2}\right)}
10
(
x
2
,
x
3
,
x
4
,
x
1
)
{\displaystyle \left(x_{2},x_{3},x_{4},x_{1}\right)}
16
(
x
3
,
x
2
,
x
4
,
x
1
)
{\displaystyle \left(x_{3},x_{2},x_{4},x_{1}\right)}
22
(
x
4
,
x
2
,
x
3
,
x
1
)
{\displaystyle \left(x_{4},x_{2},x_{3},x_{1}\right)}
5
(
x
1
,
x
4
,
x
2
,
x
3
)
{\displaystyle \left(x_{1},x_{4},x_{2},x_{3}\right)}
11
(
x
2
,
x
4
,
x
1
,
x
3
)
{\displaystyle \left(x_{2},x_{4},x_{1},x_{3}\right)}
17
(
x
3
,
x
4
,
x
1
,
x
2
)
{\displaystyle \left(x_{3},x_{4},x_{1},x_{2}\right)}
23
(
x
4
,
x
3
,
x
1
,
x
2
)
{\displaystyle \left(x_{4},x_{3},x_{1},x_{2}\right)}
6
(
x
1
,
x
4
,
x
3
,
x
2
)
{\displaystyle \left(x_{1},x_{4},x_{3},x_{2}\right)}
12
(
x
2
,
x
4
,
x
3
,
x
1
)
{\displaystyle \left(x_{2},x_{4},x_{3},x_{1}\right)}
18
(
x
3
,
x
4
,
x
2
,
x
1
)
{\displaystyle \left(x_{3},x_{4},x_{2},x_{1}\right)}
24
(
x
4
,
x
3
,
x
2
,
x
1
)
{\displaystyle \left(x_{4},x_{3},x_{2},x_{1}\right)}
Ohne Textstyle:
{
1
,
2
{\displaystyle 1,{\sqrt {2}}}
1
,
2
{\displaystyle \textstyle 1,{\sqrt {2}}}
n
→
i
⋅
x
→
=
b
i
{\displaystyle {\vec {n}}_{i}\cdot {\vec {x}}=b_{i}}
n
→
i
=
(
a
i
1
a
i
2
⋮
a
i
n
)
{\displaystyle {\vec {n}}_{i}={\begin{pmatrix}a_{i1}\\a_{i2}\\\vdots \\a_{in}\end{pmatrix}}\qquad }
(
2
−
1
)
⋅
(
x
1
x
2
)
=
4
{\displaystyle {\begin{pmatrix}2\\-1\end{pmatrix}}\cdot {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}=4}
Fehler beim Parsen (Syntaxfehler): {\displaystyle \begin{pmatrix} 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = == Zwischenräume == :<math>\sigma_1:\left(x_1,x_2,x_3,x_4\right)\mapsto\left(x_1,x_2,x_3,x_4\right)}
σ
2
:
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
(
x
2
,
x
1
,
x
4
,
x
3
)
{\displaystyle \sigma _{2}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{2},x_{1},x_{4},x_{3}\right)}
Q
(
2
,
3
)
{\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}
p
1
(
x
1
)
=
x
1
,
p
2
(
x
1
)
=
x
2
,
p
3
(
x
1
)
=
x
3
,
p
4
(
x
1
)
=
x
4
,
⟹
σ
1
:
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle p_{1}(x_{1})=x_{1},\quad p_{2}(x_{1})=x_{2},\quad p_{3}(x_{1})=x_{3},\quad p_{4}(x_{1})=x_{4},\quad \Longrightarrow \quad \sigma _{1}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{3},x_{4}\right)}
p
1
(
x
1
)
=
x
1
,
p
2
(
x
1
)
=
x
2
,
p
3
(
x
1
)
=
x
3
,
p
4
(
x
1
)
=
x
4
,
⟹
σ
1
:
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle p_{1}(x_{1})=x_{1},\quad p_{2}(x_{1})=x_{2},\quad p_{3}(x_{1})=x_{3},\quad p_{4}(x_{1})=x_{4},\quad \Longrightarrow \quad \sigma _{1}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{3},x_{4}\right)}
Beispiel zum Satz vom primitiven Element
Galois-Resolvente
Nachweis der Stetigkeit der Funktion
f
(
x
)
=
2
x
+
3
{\displaystyle f(x)=2x+3}
an der Stelle
x
0
{\displaystyle x_{0}}
Herleitung
Mithilfe der Formel von Moivre-Binet lässt sich eine einfach Herleitung angeben. Denn für die Zahlen
Φ
,
Ψ
{\displaystyle \Phi ,\Psi }
n
>
0
{\displaystyle n>0}
−
Φ
<
0
<
−
Ψ
|
⋅
Ψ
2
n
>
0
{\displaystyle -\Phi <0<-\Psi \qquad |\cdot \Psi ^{2n}>0}
−
Φ
Ψ
2
n
<
−
Ψ
2
n
+
1
|
+
Φ
2
n
+
1
{\displaystyle -\Phi \Psi ^{2n}<-\Psi ^{2n+1}\qquad |+\Phi ^{2n+1}}
Φ
2
n
+
1
−
Φ
Ψ
2
n
=
Φ
(
Φ
2
n
−
Ψ
2
n
)
<
Φ
2
n
+
1
−
Ψ
2
n
+
1
|
:
(
Φ
2
n
−
Ψ
2
n
)
>
0
{\displaystyle \Phi ^{2n+1}-\Phi \Psi ^{2n}=\Phi (\Phi ^{2n}-\Psi ^{2n})<\Phi ^{2n+1}-\Psi ^{2n+1}\qquad |:(\Phi ^{2n}-\Psi ^{2n})>0\quad }
(1)
Φ
<
Φ
2
n
+
1
−
Ψ
2
n
+
1
Φ
2
n
−
Ψ
2
n
=
f
2
n
+
1
f
2
n
{\displaystyle \Phi <{\frac {\Phi ^{2n+1}-\Psi ^{2n+1}}{\Phi ^{2n}-\Psi ^{2n}}}={\frac {f_{2n+1}}{f_{2n}}}}
Φ
−
Ψ
{\displaystyle \Phi -\Psi }
−
Φ
<
0
<
−
Ψ
|
⋅
Ψ
2
n
−
1
<
0
{\displaystyle -\Phi <0<-\Psi \qquad |\cdot \Psi ^{2n-1}<0}
−
Φ
Ψ
2
n
−
1
>
−
Ψ
2
n
|
+
Φ
2
n
{\displaystyle -\Phi \Psi ^{2n-1}>-\Psi ^{2n}\qquad |+\Phi ^{2n}}
Φ
2
n
−
Φ
Ψ
2
n
−
1
=
Φ
(
Φ
2
n
−
1
−
Ψ
2
n
−
1
)
>
Φ
2
n
−
Ψ
2
n
|
:
(
Φ
2
n
−
1
−
Ψ
2
n
−
1
)
>
0
{\displaystyle \Phi ^{2n}-\Phi \Psi ^{2n-1}=\Phi (\Phi ^{2n-1}-\Psi ^{2n-1})>\Phi ^{2n}-\Psi ^{2n}\qquad |:(\Phi ^{2n-1}-\Psi ^{2n-1})>0}
Φ
>
Φ
2
n
−
Ψ
2
n
Φ
2
n
−
1
−
Ψ
2
n
−
1
=
f
2
n
f
2
n
−
1
{\displaystyle \Phi >{\frac {\Phi ^{2n}-\Psi ^{2n}}{\Phi ^{2n-1}-\Psi ^{2n-1}}}={\frac {f_{2n}}{f_{2n-1}}}\quad }
(2)
Die Ungleichungen (1) und (2) ergeben zusammen die Behauptung.
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle \left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{3},x_{4}\right)}
p
1
(
x
1
)
=
x
1
,
p
2
(
x
1
)
=
x
2
,
p
3
(
x
1
)
=
x
3
,
p
4
(
x
1
)
=
x
4
,
⟹
σ
1
:
(
x
1
,
x
2
,
x
3
,
x
4
)
↦
(
x
1
,
x
2
,
x
3
,
x
4
)
{\displaystyle p_{1}(x_{1})=x_{1},\quad p_{2}(x_{1})=x_{2},\quad p_{3}(x_{1})=x_{3},\quad p_{4}(x_{1})=x_{4},\quad \Longrightarrow \quad \sigma _{1}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{3},x_{4}\right)}
id
{\displaystyle \operatorname {id} }
[1]
https://de.wikiversity.org/wiki/Kurs:K%C3%B6rper-_und_Galoistheorie_(Osnabr%C3%BCck_2018-2019)/Vorlesung_17
↑ Nieper-Wißkirchen, Universität Augsburg: "Galoissche Theorie", S. 126, Proposition 4.8 und Beispiel, [1]
