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FGI2: Ausdruck in 1.3.2 verbessert

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Jim Martens
2014-10-18 12:49:15 +02:00
parent 22070e072b
commit 42962c3f93
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fgi2/Blatt1/Aufgabenblatt1.tex
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@ -42,7 +42,7 @@
\subsection{} %1.
\[\underbrace{aaa \cdots aaa}_{\text{n times}}d + d + \underbrace{\underbrace{aaa \cdots aaa}_{\text{2k - 1 times}} \cdot c \cdot \underbrace{bbb \cdots bbb}_{\text{2k - 1 times}}}_{\forall k \in \mathbb{N}| 0 < k \leq \frac{1}{2}n} + \underbrace{\underbrace{aaa \cdots aaa}_{\text{2k times}} \cdot d \cdot \underbrace{bbb \cdots bbb}_{\text{2k times}}}_{\forall k \in \mathbb{N}| 0 < k \leq \frac{1}{2}n}\]
\subsection{}
\[\bigcup\limits_{k = 1}^{\frac{1}{2}n} \left\lbrace \underbrace{aaa \cdots aaa}_{\text{2k -1 times}} c \underbrace{bbb \cdots bbb}_{\text{2k -1 times}}\right\rbrace \cup \bigcup\limits_{k = 1}^{\frac{1}{2}n} \left\lbrace \underbrace{aaa \cdots aaa}_{\text{2k times}} d \underbrace{bbb \cdots bbb}_{\text{2k times}}\right\rbrace \cup \left\lbrace \underbrace{aaa \cdots aaa}_{\text{n times}}d\right\rbrace \cup \{d\}\]
\[L(A_{n}) = \bigcup\limits_{k = 1}^{\frac{1}{2}n} \left\lbrace \underbrace{aaa \cdots aaa}_{\text{2k -1 times}} c \underbrace{bbb \cdots bbb}_{\text{2k -1 times}}\right\rbrace \cup \bigcup\limits_{k = 1}^{\frac{1}{2}n} \left\lbrace \underbrace{aaa \cdots aaa}_{\text{2k times}} d \underbrace{bbb \cdots bbb}_{\text{2k times}}\right\rbrace \cup \left\lbrace \underbrace{aaa \cdots aaa}_{\text{n times}}d\right\rbrace \cup \{d\}\]
\subsection{}
\subsection{}
\(L(A_{n})\) ist für ein beliebiges fest gewähltes \(n\) regulär, da die Sprache durch einen deterministischen endlichen Automaten akzeptiert wird (siehe Aufgabe 1.3).