vault backup: 2026-08-04 13:31:58

00 Inbox/29622985 - 2026-04-28.md

@@ -16,3 +16,30 @@ tags:
 ---
 
 ## 📝 Content
+
+A containment hierarchy of classes of formal grammars. Grammars are classified into four types with different limitations.
+
+### Type 0
+no restrictions
+
+### Type 1
+Each rule $sans(w)_1 -> sans(w_2)$ satisfies $sans(abs(w_1) <= abs(w_2))$ with the exception that $sans(S) -> epsilon$ is allowed if $sans(S)$ does not occur on any right hand side of rules.
+$sans(abs(w_1) <= abs(w_2))$ implies that no shortening rules are allowed.
+
+### Type 2 (context free)
+Same restriction as [[#Type 1|type 1]] and additionally for each rule $sans(w_1 -> w_2)$ the string $sans(w_1)$ contains only a single variable (i.e. $sans(w_1 in V)$)
+
+### Type 3 (regular)
+Same restriction as [[#Type 2|type 2]] and additionally $sans(w_2 in Sigma union Sigma V)$), i.e. the right hand side is either a single terminal symbol or a terminal symbol followed by a variable.
+
+> [!REMARK]
+> The order of the variable and the terminal symbol is chosen arbitrarily.
+> Important is just that within a regular grammar only a single order occurs. If for all rules $sans(w_2 in Sigma union V Sigma)$ it would also be a regular grammar.
+
+## Definition
+A language $sans(L subset Sigma^*)$ is said to be of type 0 (1,2,3) if there exist a type 0 (1,2,3) grammar $sans(G)$ for which $sans(L(G) = L)$.
+
+> [!REMARK]
+> In order to show that a language is op type 0 (1,2,3) one thus just has to find a corresponding grammar.
+> In order to show that a grammar is not of type 0 (1,2,3) one has to prove that no type 0 (1,2,3) grammar can exist generating the corresponding language.
+

10 Courses/02 - SoSe 2026/Automatentheorie und formale Sprachen/29593852 - Strings.md

@@ -23,7 +23,7 @@ A word (or _string_) is a finite sequence $w = a_1 a_2 ... a_n$ if characters fr
 > We will use small letters to describe strings that are part of a language.
 
 > [!EXAMPLE]
-> $"aa", "ab", "bba"$ and $"baab"$ are strings over $Sigma = {a, b}.
+> $"aa", "ab", "bba"$ and $"baab"$ are strings over $Sigma = {a, b}$.
 
 #### Length of a string
 The _length_ $abs(x)$ of a string $x = a_1 ... a_n$ is its number $abs(x) = n$ of characters.

10 Courses/02 - SoSe 2026/Automatentheorie und formale Sprachen/29615600 - Chomsky Hierarchy.md

@@ -0,0 +1,45 @@
+---
+created: 2026-04-28 14:25
+course:
+topic:
+related:
+type: lecture
+status: 🔴
+tags:
+  - university
+---
+## 📌 Summary
+
+> [!abstract]
+> 
+
+---
+
+## 📝 Content
+
+A containment hierarchy of classes of formal grammars. Grammars are classified into four types with different limitations.
+
+### Type 0
+no restrictions
+
+### Type 1
+Each rule $sans(w)_1 -> sans(w_2)$ satisfies $sans(abs(w_1) <= abs(w_2))$ with the exception that $sans(S) -> epsilon$ is allowed if $sans(S)$ does not occur on any right hand side of rules.
+$sans(abs(w_1) <= abs(w_2))$ implies that no shortening rules are allowed.
+
+### Type 2 (context free)
+Same restriction as [[#Type 1|type 1]] and additionally for each rule $sans(w_1 -> w_2)$ the string $sans(w_1)$ contains only a single variable (i.e. $sans(w_1 in V)$)
+
+### Type 3 (regular)
+Same restriction as [[#Type 2|type 2]] and additionally $sans(w_2 in Sigma union Sigma V)$), i.e. the right hand side is either a single terminal symbol or a terminal symbol followed by a variable.
+
+> [!REMARK]
+> The order of the variable and the terminal symbol is chosen arbitrarily.
+> Important is just that within a regular grammar only a single order occurs. If for all rules $sans(w_2 in Sigma union V Sigma)$ it would also be a regular grammar.
+
+## Definition
+A language $sans(L subset Sigma^*)$ is said to be of type 0 (1,2,3) if there exist a type 0 (1,2,3) grammar $sans(G)$ for which $sans(L(G) = L)$.
+
+> [!REMARK]
+> In order to show that a language is op type 0 (1,2,3) one thus just has to find a corresponding grammar.
+> In order to show that a grammar is not of type 0 (1,2,3) one has to prove that no type 0 (1,2,3) grammar can exist generating the corresponding language.
+

10 Courses/02 - SoSe 2026/Mathe II/29595454 - Komplexe Zahlen und Eigenwertaufgaben.md

@@ -0,0 +1,40 @@
+---
+created: 2026-04-09 11:34
+course: "[[29595454 - Mathematik II]]"
+topic: komplexe Zahlen, Eigenwert
+related:
+type: lecture
+status: 🔴
+tags:
+  - university
+---
+## 📌 Summary
+
+> [!abstract]
+> 
+
+---
+
+## 📝 Content
+
+## Imaginäre Einheit
+Die imaginäre Einheit $i$ ist definiert über $i^2 = -1$.
+
+## Menge der komplexen Zahlen
+Die Elemente der menge $CC := {x + i y : x, y in RR}$ nennt man _komplexe Zahlen_.
+
+## Komplexe Zahlen
+Zu einer komplexen Zahl $z = x + i y in CC$ mit $x, y in RR$ heißt
+- $"Re" z := x$ _Realteil_ von $z$
+- $"Im" z := y$ _Imaginärteil_ von $z$
+- $overline(z) := x - i y$ die _konjugiert komplexe Zahl_ zu $z$
+- $abs(z) := sqrt(x^2 + y^2)$ der _Betrag_ von $z$
+- $phi in [0, 2pi]$ _Argument_ oder _Phase_ von $z$
+
+### Operationen
+
+#### Addition
+Die Addition wird (komponentenweise) wie für Vektoren definiert: 
+$a = vec(x, y), b = vec(x_2, y_2)$
+
+#### Multiplikation

11 Homework/Mathe II/Homework 6.typ

@@ -0,0 +1,42 @@
+#set page(
+    header: align(right, [
+      #set text(size: 9pt)
+      #set par(leading: .3em)
+      #pad(y: -.5cm, [
+        Mathe II \
+        Hausaufgabe 02 \
+        Jan Meyer \
+        664237
+      ])
+    ])
+  )
+
+= Problem 2
+
+== (a)
+$f_n (x) = (sin(n x) + n) / (3n + 1) = (sin(n x) / n + 1) / (3 + 1/n)$
+Da $lim_(n -> oo) sin(n x) / n = 0$, ist der punktweise Grenzwert:
+$f(x) = lim_(n -> oo) f_n (x) = (0 + 1) / (3 + 0) = 1/3$
+
+== (b)
+Zu zeigen: $f_n (x) = (n x) / (n x^2 + 1)$ konvergiert auf $[r, 1)$ mit $r > 0$ gleichmäßig gegen $f(x) = 1/x$.
+$|f_n (x) - f(x)| = |(n x) / (n x^2 + 1) - 1/x| = |(n x^2 - (n x^2 + 1)) / (x(n x^2 + 1))| = 1 / (n x^3 + x)$
+Da $x >= r > 0$, gilt $n x^3 + x > n r^3$. Daraus folgt für das Supremum:
+$sup_(x in [r, 1)) |f_n (x) - f(x)| <= 1 / (n r^3)$
+Wegen $lim_(n -> oo) 1 / (n r^3) = 0$ konvergiert die Funktionenfolge gleichmäßig.
+
+= Problem 3
+
+Sei $cal(E) = (bold(m)_0, bold(m)_1)$ die Standardbasis. Die Basiswechselmatrizen von $cal(B)$ und $cal(C)$ in $cal(E)$ sind:
+$M_cal(B) = mat(3, 2; 3, 1), quad M_cal(C) = mat(4, 1; 0, 1)$
+Der Koordinatenvektor von $bold(p) = bold(m)_0 + bold(m)_1$ in $cal(E)$ ist $bold(p)^cal(E) = vec(1, 1)$.
+
+== (a)
+$bold(p)^cal(B) = M_cal(B)^(-1) bold(p)^cal(E) = 1/(3-6) mat(1, -2; -3, 3) vec(1, 1) = mat(-1/3, 2/3; 1, -1) vec(1, 1) = vec(1/3, 0)$
+
+== (b)
+$T_(cal(C) <- cal(B)) = M_cal(C)^(-1) M_cal(B) = 1/4 mat(1, -1; 0, 4) mat(3, 2; 3, 1) = mat(1/4, -1/4; 0, 1) mat(3, 2; 3, 1) = mat(0, 1/4; 3, 1)$
+$T_(cal(B) <- cal(C)) = (T_(cal(C) <- cal(B)))^(-1) = 1/(0 - 3/4) mat(1, -1/4; -3, 0) = -4/3 mat(1, -1/4; -3, 0) = mat(-4/3, 1/3; 4, 0)$
+
+== (c)
+$bold(p)^cal(C) = T_(cal(C) <- cal(B)) bold(p)^cal(B) = mat(0, 1/4; 3, 1) vec(1/3, 0) = vec(0, 1)$

11 Homework/Mathe II/Homework 7.typ

@@ -0,0 +1,52 @@
+#set page(header: align(right, [
+  #set text(size: 9pt)
+  #set par(leading: .3em)
+  #pad(y: -.5cm, [
+    Mathe II \
+    Hausaufgabe 07 \
+    Jan Meyer \
+    664237
+  ])
+]))
+
+= Problem 1
+
+== (a)
+Punktweiser Grenzwert für $x in (0, oo)$:
+$ f(x) = lim_(n -> oo) n / (1 + n x) = lim_(n -> oo) 1 / (1/n + x) = 1/x $
+
+== (b)
+Zu prüfen ist, ob $lim_(n -> oo) sup_(x in RR) |f_n (x) - f(x)| = 0$ gilt.
+Die Differenz lautet:
+$ |f_n (x) - f(x)| = |(x^2 + n^2 x)/n^2 - x| = x^2/n^2 $
+Das Supremum auf ganz $RR$ ist für jedes $n$ unendlich:
+$ sup_(x in RR) x^2/n^2 = oo $
+Die Konvergenz ist somit nicht gleichmäßig.
+
+= Problem 4
+
+== (a)
+$A$ ist symmetrisch ($A^T = A$). Für die positive Definitheit prüfen wir die Hauptminoren nach Sylvester:
+- $D_1 = 1 > 0$
+- $D_2 = det mat(1, 0; 0, 1) = 1 > 0$
+- $D_3 = 1 dot (5 - 4) = 1 > 0$
+- $D_4 = 3 dot D_3 = 3 > 0$
+Da alle Hauptminoren positiv sind, ist $A$ positiv definit und definiert ein Skalarprodukt.
+
+== (b)
+Ansatz für die Projektion $p$: $p = X c$ mit $X = (u_1, u_2)$ und $c = vec(c_1, c_2)$.
+Normalengleichung: $X^T A X c = X^T A v$.
+
+Berechnung der Gramschen Matrix $X^T A X$:
+$ X^T A X = mat(u_1^T A u_1, u_1^T A u_2; u_2^T A u_1, u_2^T A u_2) = mat(4, 1; 1, 2) $
+
+Berechnung der rechten Seite $X^T A v$:
+$ A v = mat(1, 0, 0, 0; 0, 1, -2, 0; 0, -2, 5, 0; 0, 0, 0, 3) mat(2; 2; 2; 2) = mat(2; -2; 6; 6) $
+$ X^T A v = mat(u_1^T A v; u_2^T A v) = mat(8; 0) $
+
+Lösen des LGS:
+$ mat(4, 1; 1, 2) mat(c_1; c_2) = mat(8; 0) $
+Aus der 2. Zeile folgt $c_1 = -2 c_2$. Eingesetzt in die 1. Zeile ergibt $4(-2 c_2) + c_2 = 8 <=> -7 c_2 = 8 <=> c_2 = -8/7$. Damit ist $c_1 = 16/7$.
+
+Einsetzen in $p$:
+$ p = c_1 u_1 + c_2 u_2 = 16/7 mat(1; 0; 0; 1) - 8/7 mat(1; 1; 0; 0) = mat(8/7; -8/7; 0; 16/7) $