Als Probe berechnen wir
3
2
x
1
2
+
2
x
2
2
+
2
x
1
x
2
−
2
x
2
x
3
=
3
2
(
2
6
y
1
+
2
14
y
2
−
1
4
21
y
3
)
2
+
2
(
−
1
6
y
1
+
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14
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2
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21
y
3
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2
+
2
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2
6
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14
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21
y
3
)
(
−
1
6
y
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+
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14
y
2
+
1
2
21
y
3
)
−
2
(
−
1
6
y
1
+
3
14
y
2
+
1
2
21
y
3
)
(
1
6
y
1
−
1
14
y
2
+
1
21
y
3
)
{\displaystyle {}{\begin{aligned}{\frac {3}{2}}x_{1}^{2}+2x_{2}^{2}+2x_{1}x_{2}-2x_{2}x_{3}&={\frac {3}{2}}{\left({\frac {2}{\sqrt {6}}}y_{1}+{\frac {2}{\sqrt {14}}}y_{2}-{\frac {1}{4{\sqrt {21}}}}y_{3}\right)}^{2}+2{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}^{2}+2{\left({\frac {2}{\sqrt {6}}}y_{1}+{\frac {2}{\sqrt {14}}}y_{2}-{\frac {1}{4{\sqrt {21}}}}y_{3}\right)}{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}-2{\left(-{\frac {1}{\sqrt {6}}}y_{1}+{\frac {3}{\sqrt {14}}}y_{2}+{\frac {1}{2{\sqrt {21}}}}y_{3}\right)}{\left({\frac {1}{\sqrt {6}}}y_{1}-{\frac {1}{\sqrt {14}}}y_{2}+{\frac {1}{\sqrt {21}}}y_{3}\right)}\\\,\end{aligned}}}
euklidischer Raum
ist insbesondere ein Hilbert-Raum. Diese beiden Begriffe werden vor allem für unendlich dimensionale Vektorräume eingesetzt.
Es sei
(
M
,
d
)
{\displaystyle {}(M,d)}
metrischer Raum , sei
T
⊆
M
{\displaystyle {}T\subseteq M}
a
∈
M
{\displaystyle {}a\in M}
Berührpunkt von
T
{\displaystyle {}T}
f
:
T
⟶
V
{\displaystyle f\colon T\longrightarrow V}
eine
Abbildung
in einen
endlichdimensionalen
normierten Vektorraum
V
{\displaystyle {}V}
Komponentenfunktionen
f
1
,
…
,
f
n
:
M
⟶
R
{\displaystyle f_{1},\ldots ,f_{n}\colon M\longrightarrow \mathbb {R} }
bezüglich einer Basis von
V
{\displaystyle {}V}
Limes
lim
x
→
a
f
(
x
)
{\displaystyle \operatorname {lim} _{x\rightarrow a}\,f(x)}
genau dann existiert, wenn sämtliche Limiten
lim
x
→
a
f
j
(
x
)
{\displaystyle \operatorname {lim} _{x\rightarrow a}\,f_{j}(x)}
existieren.
Wie hoch muss ein Spiegel mindestens sein, damit man sich in ihm vollständig sehen kann
(ohne sich zu verrenken)?
Fakt
lässt sich der Abstand
d
1
{\displaystyle {}d_{1}}
A
B
→
{\displaystyle {}{\overrightarrow {AB}}}
d
2
{\displaystyle {}d_{2}}
d
=
d
1
−
d
2
{\displaystyle {}d=d_{1}-d_{2}}
d
2
=
⟨
(
a
1
a
2
)
,
1
c
(
b
2
−
a
2
−
b
1
+
a
1
)
⟩
=
1
c
(
a
1
b
2
−
a
1
a
2
−
a
2
b
1
+
a
1
a
2
)
=
1
c
(
a
1
b
2
−
a
2
b
1
)
,
{\displaystyle {}{\begin{aligned}d_{2}=\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\frac {1}{c}}{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}}\right\rangle &={\frac {1}{c}}{\left(a_{1}b_{2}-a_{1}a_{2}-a_{2}b_{1}+a_{1}a_{2}\right)}\\&={\frac {1}{c}}{\left(a_{1}b_{2}-a_{2}b_{1}\right)},\,\end{aligned}}}
d
1
=
⟨
1
a
+
b
+
c
(
a
a
1
+
b
b
1
+
c
c
1
a
a
2
+
b
b
2
+
c
c
2
)
,
1
c
(
b
2
−
a
2
−
b
1
+
a
1
)
⟩
=
1
a
+
b
+
c
1
c
(
a
a
1
b
2
−
b
a
2
b
1
+
c
c
1
b
2
−
c
c
1
a
2
−
a
a
2
b
1
+
b
a
1
b
2
−
c
b
1
c
2
+
c
a
1
c
2
)
=
1
a
+
b
+
c
1
c
(
a
(
a
1
b
2
−
a
2
b
1
)
+
b
(
a
1
b
2
−
a
2
b
1
)
+
c
(
c
1
b
2
−
b
1
c
2
)
+
c
(
−
c
1
a
2
+
a
1
c
2
)
)
=
1
a
+
b
+
c
1
c
(
(
a
+
b
+
c
)
(
a
1
b
2
−
a
2
b
1
)
+
c
(
−
a
1
b
2
+
a
2
b
1
+
c
1
b
2
−
b
1
c
2
−
c
1
a
2
+
a
1
c
2
)
)
=
1
c
(
a
1
b
2
−
a
2
b
1
)
+
1
a
+
b
+
c
(
⟨
(
a
1
a
2
)
,
(
−
b
2
b
1
)
⟩
+
⟨
(
b
1
b
2
)
,
(
−
c
2
c
1
)
⟩
+
⟨
(
c
1
c
2
)
,
(
−
a
2
a
1
)
⟩
)
=
d
2
+
1
a
+
b
+
c
(
⟨
(
a
1
a
2
)
,
(
−
b
2
b
1
)
⟩
+
⟨
(
b
1
b
2
)
,
(
−
c
2
c
1
)
⟩
+
⟨
(
c
1
c
2
)
,
(
−
a
2
a
1
)
⟩
)
.
{\displaystyle {}{\begin{aligned}d_{1}=\left\langle {\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}},{\frac {1}{c}}{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}}\right\rangle &={\frac {1}{a+b+c}}{\frac {1}{c}}{\left(aa_{1}b_{2}-ba_{2}b_{1}+cc_{1}b_{2}-cc_{1}a_{2}-aa_{2}b_{1}+ba_{1}b_{2}-cb_{1}c_{2}+ca_{1}c_{2}\right)}\\&={\frac {1}{a+b+c}}{\frac {1}{c}}{\left(a(a_{1}b_{2}-a_{2}b_{1})+b(a_{1}b_{2}-a_{2}b_{1})+c(c_{1}b_{2}-b_{1}c_{2})+c(-c_{1}a_{2}+a_{1}c_{2})\right)}\\&={\frac {1}{a+b+c}}{\frac {1}{c}}{\left((a+b+c)(a_{1}b_{2}-a_{2}b_{1})+c(-a_{1}b_{2}+a_{2}b_{1}+c_{1}b_{2}-b_{1}c_{2}-c_{1}a_{2}+a_{1}c_{2})\right)}\\&={\frac {1}{c}}{\left(a_{1}b_{2}-a_{2}b_{1}\right)}+{\frac {1}{a+b+c}}{\left(\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\begin{pmatrix}-b_{2}\\b_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}},{\begin{pmatrix}-c_{2}\\c_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}},{\begin{pmatrix}-a_{2}\\a_{1}\end{pmatrix}}\right\rangle \right)}\\&=d_{2}+{\frac {1}{a+b+c}}{\left(\left\langle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},{\begin{pmatrix}-b_{2}\\b_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}},{\begin{pmatrix}-c_{2}\\c_{1}\end{pmatrix}}\right\rangle +\left\langle {\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}},{\begin{pmatrix}-a_{2}\\a_{1}\end{pmatrix}}\right\rangle \right)}.\,\end{aligned}}}
Der rechte Summand ist
d
{\displaystyle {}d}
Man kann auch zur Berechnung des Schnittpunkts die obige Bedingung mit
a
b
c
{\displaystyle {}abc}
s
(
a
b
(
b
1
−
a
1
)
+
a
c
(
c
1
−
a
1
)
)
−
t
(
a
b
(
a
1
−
b
1
)
+
b
c
(
c
1
−
b
1
)
)
=
a
b
c
(
b
1
−
a
1
)
{\displaystyle s(ab(b_{1}-a_{1})+ac(c_{1}-a_{1}))-t(ab(a_{1}-b_{1})+bc(c_{1}-b_{1}))=abc(b_{1}-a_{1})}
s
(
a
b
(
b
2
−
a
2
)
+
a
c
(
c
2
−
a
2
)
)
−
t
(
a
b
(
a
2
−
b
2
)
+
b
c
(
c
2
−
b
2
)
)
=
a
b
c
(
b
2
−
a
2
)
{\displaystyle s(ab(b_{2}-a_{2})+ac(c_{2}-a_{2}))-t(ab(a_{2}-b_{2})+bc(c_{2}-b_{2}))=abc(b_{2}-a_{2})}
schreiben.
Mit Cramer:
s
=
[
[
:
V
o
r
l
a
g
e
:
O
p
:
d
e
t
]
]
[
[
:
V
o
r
l
a
g
e
:
O
p
:
d
e
t
]
]
=
b
c
[
[
:
V
o
r
l
a
g
e
:
O
p
:
d
e
t
]
]
[
[
:
V
o
r
l
a
g
e
:
O
p
:
d
e
t
]
]
=
b
c
(
b
1
−
a
1
)
(
a
(
a
2
−
b
2
)
+
c
(
c
2
−
b
2
)
)
−
(
b
2
−
a
2
)
(
a
(
a
1
−
b
1
)
+
c
(
c
1
−
b
1
)
)
(
b
(
b
1
−
a
1
)
+
c
(
c
1
−
a
1
)
)
(
a
(
a
2
−
b
2
)
+
c
(
c
2
−
b
2
)
)
−
(
a
(
a
1
−
b
1
)
+
c
(
c
1
−
b
1
)
)
(
b
(
b
2
−
a
2
)
+
c
(
c
2
−
a
2
)
)
=
b
c
a
(
b
1
−
a
1
)
(
a
2
−
b
2
)
+
c
(
b
1
−
a
1
)
(
c
2
−
b
2
)
−
a
(
b
2
−
a
2
)
(
a
1
−
b
1
)
−
c
(
b
2
−
a
2
)
(
c
1
−
b
1
)
a
b
(
b
1
−
a
1
)
(
a
2
−
b
2
)
+
b
c
(
b
1
−
a
1
)
(
c
2
−
b
2
)
+
a
c
(
c
1
−
a
1
)
(
a
2
−
b
2
)
+
c
2
(
c
1
−
a
1
)
(
c
2
−
b
2
)
−
a
b
(
a
1
−
b
1
)
(
b
2
−
a
2
)
−
a
c
(
a
1
−
b
1
)
(
c
2
−
a
2
)
+
b
c
(
c
1
−
b
1
)
(
b
2
−
a
2
)
−
a
c
(
c
1
−
b
1
)
(
c
2
−
a
2
)
=
b
c
(
b
1
−
a
1
)
(
c
2
−
b
2
)
−
(
b
2
−
a
2
)
(
c
1
−
b
1
)
b
(
b
1
−
a
1
)
(
c
2
−
b
2
)
+
a
(
c
1
−
a
1
)
(
a
2
−
b
2
)
+
c
(
c
1
−
a
1
)
(
c
2
−
b
2
)
−
a
(
a
1
−
b
1
)
(
c
2
−
a
2
)
+
b
(
c
1
−
b
1
)
(
b
2
−
a
2
)
−
a
(
c
1
−
b
1
)
(
c
2
−
a
2
)
=
b
c
(
b
1
−
a
1
)
(
c
2
−
b
2
)
−
(
b
2
−
a
2
)
(
c
1
−
b
1
)
(
a
+
b
+
c
)
(
(
b
1
−
a
1
)
(
c
2
−
b
2
)
−
(
b
2
−
a
2
)
(
c
1
−
b
1
)
)
=
b
c
a
+
b
+
c
.
{\displaystyle {}{\begin{aligned}s&={\frac {[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&={\frac {bc[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&=bc{\frac {(b_{1}-a_{1})(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(b_{2}-a_{2})(a(a_{1}-b_{1})+c(c_{1}-b_{1}))}{(b(b_{1}-a_{1})+c(c_{1}-a_{1}))(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(a(a_{1}-b_{1})+c(c_{1}-b_{1}))(b(b_{2}-a_{2})+c(c_{2}-a_{2}))}}\\&=bc{\frac {a(b_{1}-a_{1})(a_{2}-b_{2})+c(b_{1}-a_{1})(c_{2}-b_{2})-a(b_{2}-a_{2})(a_{1}-b_{1})-c(b_{2}-a_{2})(c_{1}-b_{1})}{ab(b_{1}-a_{1})(a_{2}-b_{2})+bc(b_{1}-a_{1})(c_{2}-b_{2})+ac(c_{1}-a_{1})(a_{2}-b_{2})+c^{2}(c_{1}-a_{1})(c_{2}-b_{2})-ab(a_{1}-b_{1})(b_{2}-a_{2})-ac(a_{1}-b_{1})(c_{2}-a_{2})+bc(c_{1}-b_{1})(b_{2}-a_{2})-ac(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{b(b_{1}-a_{1})(c_{2}-b_{2})+a(c_{1}-a_{1})(a_{2}-b_{2})+c(c_{1}-a_{1})(c_{2}-b_{2})-a(a_{1}-b_{1})(c_{2}-a_{2})+b(c_{1}-b_{1})(b_{2}-a_{2})-a(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{(a+b+c)((b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1}))}}\\&={\frac {bc}{a+b+c}}.\end{aligned}}}
![{\displaystyle {}{\begin{aligned}s&={\frac {[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&={\frac {bc[[:Vorlage:Op:det]]}{[[:Vorlage:Op:det]]}}\\&=bc{\frac {(b_{1}-a_{1})(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(b_{2}-a_{2})(a(a_{1}-b_{1})+c(c_{1}-b_{1}))}{(b(b_{1}-a_{1})+c(c_{1}-a_{1}))(a(a_{2}-b_{2})+c(c_{2}-b_{2}))-(a(a_{1}-b_{1})+c(c_{1}-b_{1}))(b(b_{2}-a_{2})+c(c_{2}-a_{2}))}}\\&=bc{\frac {a(b_{1}-a_{1})(a_{2}-b_{2})+c(b_{1}-a_{1})(c_{2}-b_{2})-a(b_{2}-a_{2})(a_{1}-b_{1})-c(b_{2}-a_{2})(c_{1}-b_{1})}{ab(b_{1}-a_{1})(a_{2}-b_{2})+bc(b_{1}-a_{1})(c_{2}-b_{2})+ac(c_{1}-a_{1})(a_{2}-b_{2})+c^{2}(c_{1}-a_{1})(c_{2}-b_{2})-ab(a_{1}-b_{1})(b_{2}-a_{2})-ac(a_{1}-b_{1})(c_{2}-a_{2})+bc(c_{1}-b_{1})(b_{2}-a_{2})-ac(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{b(b_{1}-a_{1})(c_{2}-b_{2})+a(c_{1}-a_{1})(a_{2}-b_{2})+c(c_{1}-a_{1})(c_{2}-b_{2})-a(a_{1}-b_{1})(c_{2}-a_{2})+b(c_{1}-b_{1})(b_{2}-a_{2})-a(c_{1}-b_{1})(c_{2}-a_{2})}}\\&=bc{\frac {(b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1})}{(a+b+c)((b_{1}-a_{1})(c_{2}-b_{2})-(b_{2}-a_{2})(c_{1}-b_{1}))}}\\&={\frac {bc}{a+b+c}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64fa5881ed8c015928ff5f6a98696155d69ce100)