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Riemannsche Summen
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Stückweise stetige Funktion ist Riemann-integrierbar  
Beitrag No.4 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2021-04-12 19:02
LukasNiessen
J

Danke, ich habe das wohl übersehen, ja!

Und es ist noch keine Mail raus ;D

Riemannsche Summen
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Stückweise stetige Funktion ist Riemann-integrierbar  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2021-04-12 16:59
LukasNiessen
J
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Hallo,

ich wiederhole gerade den Stoff aus dem letzten Semester. Nach Vorlesung ist eine stückweise stetige, beschränkte Funktion Riemann-integrierbar.

Ich sehe allerdings nicht ganz an einem Beispiel warum das so ist, also mir geht es hier eher um das intuitive Verständnis. Daher folgendes sehr einfaches Beispiel:

$f: [0,2] \rightarrow \IR$ mit $f(x) = 0$ für alle x ungleich 1 und $f(1) = 1$.

Offenbar ist f beschränkt und stückweise stetig.
Wenn ich mir die Wikipedia Definition anschaue (hier), dann sehe ich direkt, dass f auch integrierbar ist.
Man wähle nämlich als Zerlegung Z einfach $(0,1,2)$.

Das Problem ist aber, dass wir in der Vorlesungen die Ober- und Untersummen ein wenig anders definiert haben:
$O(Z):=\sum _{k=1}^{n}{\Big (}(x_{k}-x_{k-1})\cdot \sup _{x \in [x_{k-1},x_k]}f(x)\Big )$

Der Unterschied ist unter dem supremums Zeichen: Wir haben das äußere Intervall genommen, auf Wikipedia wird aber strikte Ordnung benutzt.
Also bei unserer Definition kann insbesondere auch $x = x_k$ oder $x = x_{k+1}$ gelten.

Ich sehe unter den Umständen aber nicht, wie man dann eine Zerlegung Z wählen könnte, sodass die Untersumme und die Obersumme nicht betragsmäßig Differenz 1 haben.

Kann mir jemand helfen?
Danke!
\(\endgroup\)

Numerik & Optimierung
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Matrixnorm < 1, dann A+E invertierbar  
Beitrag No.3 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2021-03-08
LukasNiessen
J

Jetzt seh ich's! Danke sehr!

Grüße

Numerik & Optimierung
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Matrixnorm < 1, dann A+E invertierbar  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2021-03-08
LukasNiessen
J
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Hallo,

ich soll zeigen:

Sei $||.||$ eine Vektornorm auf dem $\IR^n$ und $|||.|||$ die von ihr induzierte Matrixnorm. Sei nun: Für $A \in \IR^{\text{n x n}}: |||A||| < 1$. Dann ist $E+A$ regulär.

Mein Ansatz ist:
Nach der Def. einer induzierten Matrixnorm haben wir also:
Für alle $x \in \IR^n - {0}: \frac{||Ax||}{||x||} < 1$, also $||Ax|| < ||x||$.

Man kann das dann noch umformen zu:
$||Ax||-||Ex|| < 0$

Ich habe versucht zu nutzen, dass $|||.|||$ mit $||.||$ verträglich ist, oder die umgekehrte Dreiecksungl., aber das bringt ja beides nichts, weil wir dann die Abschätzung < 0 verlieren würden.

Ich sehe aber insb. nicht wie man hier auf irgendwas von Invertierbarkeit kommen kann, zB, dass A vollen Rang haben muss, oder die Determinante nicht 0 sein kann.

Vielen Dank!
\(\endgroup\)

Körper und Galois-Theorie
  
Thema eröffnet von: riemann05
nicht archimedisch angeordneter Körper  
Beitrag No.5 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-12-25
LukasNiessen
 

Hallo,

ich hoffe es ist in Ordnung, dass ich nochmal auf dieses alte Thema antworte und nachfrage. (Oder sollte ich lieber ein neues eröffnen und auf dieses hier verlinken?)

Der geordnete Körper aus Fabis Beitrag, habe ich mir überlegt, ist doch sogar totalgeordnet, richtig?

Das heißt insb, dass ein totalgeordneter Körper nicht archimedisch geordnet sein muss, korrekt?

Danke!

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Allgemeine Gleichung 5ten Grades nicht auflösbar  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-16
LukasNiessen
 
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Danke!

Verstehe es richtig, dass dann der Körper K eigentlich garkeine Rolle spielt? Also ich meine welcher Körper K nun ist, ist doch eig. egal, oder?

Denn wenn wir die allgemeine Gleichung n-ten Grades jetzt so wie ich anfangs als $X5+aX4+bX3+cX2+dX+e$ interpretieren, dann betrachtet man $a,b,c,d,e$ ja nicht als Elemente von K sondern als Elemente von $K(X_1,...,X_5)$. Also quasi als "feste Variablen".

Und die Art wie wir "mit ihnen rechnen" können ist dann aber dennoch nicht ganz unabhängig von K. Bsplw. wäre für $K=\{0,1\}$ folgendes der Fall: $a + a = 2a = 0$. Nicht aber für $K = \IQ$.

Trotzdem bleibt aber das Resultat, dass die Gleichung nicht auflösbar ist für alle Körper K bestehen, richtig?
Also auch für zb so "einfache" Körper wie $\{0,1\}$.

Verstehe ich das richtig?
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Allgemeine Gleichung 5ten Grades nicht auflösbar  
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Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-16
LukasNiessen
 
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Hallo,

mit der Galois-Theorie kann man leicht zeigen, dass die n-te allgemeine Gleichung für n größer 4 nicht auflösbar ist.

Ich habe aber mein Problem beim genauen Verständnis von was mit der n-ten allgemeinen Gleichung nun gemeint ist.

Die intuitivste "Erkenntnis" obiger Aussage wäre wohl: $aX^5 + bX^4+cX^3+dX^2+eX+f = 0$ hat keine allgemeine Lösungsmenge (durch Radikale), mit $a,b,c,d,e,f \in K$ mit K Körper.

---

Aber die n-te allgemeine Gleichung wird ganz anders definiert:

Sei k ein Körper und $L = Q(k(T_1,...,T_n))$ mit $T_1,...,T_n$ Variablen.
Dann ist der Körper der symmetrischen rationalen Funktionen K definiert als der Fixkörper $L^{S(n)}$ wobei $S(n)$ die symmetrische Gruppe sei.
Die Gleichung der Erweiterung $L/K$ ist:
$f(X) = \prod_{i=1}^{n} (X-T_i) \in k[T_1,...,T_n][X]$.

Man kann nun zeigen, dass $L/K$ galoissch mit Galois-Gruppe $S(n)$ ist und, dass L ein Zerfällungskörper von f über K ist.

---

Wendet man darauf die Theorie von Radikalerweiterungen an, folgt sofort, dass die allg. Gleichung n-ten Grades für n größer 4 nicht auflösbar ist, denn $S(n)$ ist es nicht.

Aber ich sehe nicht die Parallelen von dieser Definition der allg. Gleichung n-ten Grades und meinem anfangs erwähnten Polynom $aX^5 + bX^4+cX^3+dX^2+eX+f$.

Ich schätze $T_1,...,T_n$ sollen hier die allgemeinen Variablen sein (also quasi $a,...,f$).
Aber wieso entspricht dann $f(X) = \prod_{i=1}^{n} (X-T_i)$ dem Polynom $aX^5 + bX^4+cX^3+dX^2+eX+f$?

Wir erhalten doch bei $\prod_{i=1}^{n} (X-T_i)$ nur bestimmte Summen und Produkte von den $T_i$ als Koeffizienten vor X wenn man ausmultipliziert?
Und zwar ein elementarsymmetrisches Polynom aus K.
Außerdem haben wir noch einen alternierenden Faktor $(-1)^i$ davor.

Ich sehe also nicht, wie diese Definition von der allgemeinen n-ten Gleichung einen Schluss auf meine anfangs gennante intuitive "Erkenntnis" zulässt.

Ich hoffe man versteht, was ich meine.

Könnte vielleicht jemand helfen? Danke!
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Reelle Zahlen nicht eindeutig bestimmt?!  
Beitrag No.6 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-15
LukasNiessen
 

Dankeschön!

[Die Antwort wurde nach Beitrag No.4 begonnen.]

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Reelle Zahlen nicht eindeutig bestimmt?!  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-15
LukasNiessen
 
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Danke!

Wie sähe denn eine solche Definition von $\IC$ etwa aus?

Konnte dazu nichts im Netz finden, die deutsche und englische Wikipedia scheinen es nicht zu beinhalten.
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Reelle Zahlen nicht eindeutig bestimmt?!  
Themenstart
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LukasNiessen
 
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Hey,

was ist mit folgendem Satz gemeint?

"Der Körper $\IR$ der reellen Zahlen als Teilkörper von $\IC$ ist aus rein algebraischer Sicht keineswegs eindeutig bestimmt, da es Automorphismen von $\IC$ gibt, die $\IR$ nicht invariant lassen."

Mir geht es nicht um einen Beweis oder Formales, sondern nur um die Intuition.

Ich dachte $\IC$ wird erst über $\IR$ definiert, und zwar durch Adjunktion einer Lösung von $X^2+1 = 0$ über einem algebraischen Abschluss von $\IR$ bzw. als Zerfällungskörper jenes Polynoms.

Inwiefern kann man dann von $\IC$ ausgehen und überhaupt von Eindeutigkeit von $\IR$ als Teilkörper sprechen?

Danke!
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Erweiterung endlich oder algebraisch  
Beitrag No.1 im Thread
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LukasNiessen
J

Ok habs selbst raus!

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Erweiterung endlich oder algebraisch  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-15
LukasNiessen
J
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Hey,

es mag sein, dass ich hier auf dem Schlauch stehe.
Es geht um einen Beweis zum Fundamentalsatz der Algebra, mittels Galois-Theorie.

Es heisst:

Wir betrachten eine Körperkette $\IR \subset \IC \subset L$, wobei $L/\IC $ endlich sei. Zu zeigen ist $L = \IC$.

Ich sehe aber nicht, wie daraus die algebraische Abgeschlossenheit folgen soll.

Es gilt, das haben wir auch bewiesen:
Ein Körper L ist genau dann algebraisch abgeschlossen, wenn er keine echten algebraischen Erweiterungen zulässt.

Daher müsste man doch annehmen, dass $L/\IC$ algebraisch sei, und das ist ja allgemeiner als, dass die Erweiterung endlich ist.

Wieso also reicht es hier schon anzunehmen, die Erweiterung sei endlich?

(In Skripten im Netz konnte ich auch keine Erklärung finden, dort wurde auch einfach Endlichkeit angenommen.)

Danke!
\(\endgroup\)

Strukturen und Algebra
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Nur eine p-Sylow-Gruppe  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-13
LukasNiessen
J

Danke sehr!

Strukturen und Algebra
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Nur eine p-Sylow-Gruppe  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-11-13
LukasNiessen
J
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Hey,

ich hänge gerade an folgendem Problem. Ich soll zeigen, dass für eine Gruppe G der Ordnung $pq$ (verschiedene Primzahlen) gilt:
Es gibt nur eine p-Sylow-Gruppe

Nach den Sylowsätzen gibt es eine, sei die Anzahl der p-Sylow-Gruppen mit $n$ bezeichnet.

Auch nach den Sätzen gilt:
$n | pq$ und $n \mod p = 1 \mod p$

Damit ist n entweder 1 oder q.

Aber wieso kann n nicht gleich q sein? Dann würde $p | q-1$ gelten, wieso aber kann das nicht sein?

---

Was man aber dennoch folgen können müsste ist, dass es nur 1 p-Sylow-Gruppe gibt oder es nur 1 q-Sylow-Gruppe gibt. Denn Wenn $p | q-1$, Dann kann nicht auch $q | p-1$ gelten, also gibt es genau eine q-Sylow-Gruppe in dem Fall.

Ist das korrekt?

Danke!
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Galois-Gruppe bestimmen  
Beitrag No.5 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-10-26
LukasNiessen
J

Perfekt, danke euch!


Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Galois-Gruppe bestimmen  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-10-23
LukasNiessen
J
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Hallo!

Es geht nochmal darum eine Galois-Gruppe zu bestimmen, und zwar folgenden Polynoms:

$X^8 - 2 \in \IQ[X]$.

---

Der Zerfällungskörper ist $\IQ(\sqrt[8]{2}, i)$ und die Erweiterung hat den Grad 16.

Es gilt:

$\text{Gal}(\IQ(\sqrt[8]{2}, i) / \IQ) \subset S_8$.

---

Ich komme hier aber nicht weiter, wie kann ich nun die 16 Automorphismen der Galois-Gruppe bestimmen?

Danke!
\(\endgroup\)

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Thema eröffnet von: LukasNiessen
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Beitrag No.5 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-10-21
LukasNiessen
J
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Okay, ich glaube ich hab's!

Man darf x auf die Länge 1 skalieren, denn es geht lediglich um die Existenz eines x ungleich 0, welches $z = \frac{x}{\bar{x}}$ erfüllt.

Und angenommen, ein solches x erfüllt obiges, dann tut es auch die auf Länge 1 skalierte "Version" $x'$, denn:

$\frac{x}{\bar{x}}\bar{(\frac{x}{\bar{x}})}^{-1} = \frac{x}{\bar{x}} \cdot (\bar{x} \frac{1}{|x|})^{-1} = \frac{x}{\bar{x}}$

Ich sehe gerade, obige Formel sieht unleserlich aus.

Ich habe auf jeden Fall nur $\frac{x'}{\bar{x'}}$ berechnet und gesehen, dass $\frac{x}{\bar{x}}$ rauskommt, also damit auch gleich z ist.

Danke!
\(\endgroup\)

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Komplexe Zahlen Betrag Rechnung  
Beitrag No.4 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-10-21
LukasNiessen
J
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Danke euch!

Ich dachte der Kontext wäre irrelevant, da die Folgerung allgemein gültig wäre.

Hier ist der Kontext:
Man soll Theorem 90 von Hilbert (hier) bezüglich $\IC / \IR$ untersuchen.

---

Man sieht schnell, dass die Erweiterung zyklisch von der Ordnung 2 ist und die Galois-Gruppe von der komplexen Konjugation erzeugt wird.
Außerdem gilt für die Norm:

$N_{\IC / \IR}(z) = z \bar{z} = |z|^2$ wobei $z \in \IC$.

Wenn nun $N_{\IC / \IR}(z) = 1$ gilt, dann folgt mit Hilberts Satz:

Es gibt $x \in \IC^*: z = \frac{x}{\bar{x}}$.

Hier wird nun gesagt: Man dürfe $x \bar{x} = |x|^2 = 1$ annehmen.
Gefolgert wird dann:

$z = x^2$, also, dass x eine Quadratwurzel von z ist.

---

Ich verstehe alles, außer, dass man $x \bar{x} = |x|^2 = 1$ annehmen darf.

Danke!
\(\endgroup\)

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Thema eröffnet von: LukasNiessen
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Hallo!

Ich verstehe folgenden Schritt in einer Rechnung nicht:

Sei $z \in \IC$ und $x \in \IC-\{0\}$.
Weiter sei $z \bar{z} = |z|^2 = 1$ und $z = \frac{x}{\bar{x}}$.

Es heißt nun, man dürfe annehmen, dass $x \bar{x} = |x|^2 = 1$ gilt.

Das sehe ich aber nicht, wieso muss das gelten?

Danke!
\(\endgroup\)

Körper und Galois-Theorie
Universität/Hochschule 
Thema eröffnet von: LukasNiessen
Unterkörper mindestens Grad p  
Beitrag No.2 im Thread
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LukasNiessen
J

Ach klar! Es ist also in der Tat trivial, ich habe es nur übersehen. :D

Danke dir!
 

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