add ana2
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\documentclass{lecture}
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\begin{document}
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Jetzt: Fourier Analysis
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\subsection{Der Funktionen-Raum $R[a,b]$}
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\begin{definition}
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Eine $f\colon [a,b] \to \mathbb{C}$, $[a,b] \subset \R$ heißt
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Riemann-integrierbar auf $[a,b]$, falls $\text{Re}(f)$ und
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$\text{Im}(f)$ Riemann-integrierbar sind.
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Man setzt
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\[
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\int_{a}^{b} f(x) \d x := \int_{a}^{b} \text{Re} f(x) \d x + i \int_{a}^{b} \text{Im} f(x) \d x
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.\]
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\end{definition}
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\begin{bem}
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\begin{enumerate}
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\item Analog: Definitionen von uneigentlichen Riemann-integralen für
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komplexwertige Funktionen
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\item Die Rechenregeln f+r das reelle Riemann-integral übertragen sich auf komplexwertige
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Integrale, insbesondere gilt:
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\begin{align*}
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\int_{a}^{b} \overline{f(x)} \d x &= \int_{a}^{b} \left( \text{Re}f(x) - i \cdot \text{Im}f(x) \right) \d x \\
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&= \int_{a}^{b} \text{Re}f(x) \d x - i \int_{a}^{b} \text{Im}f(x) \d x \\
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&= \overline{\int_{a}^{b} f(x) \d x }
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.\end{align*}
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\end{enumerate}
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\end{bem}
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\begin{definition}
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Eine Funktion $f\colon [a,b] \to \mathbb{K}$ ($\mathbb{K} = \R$ oder $\mathbb{K} = \mathbb{C}$)
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heißt stückweise stetig, falls
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\begin{enumerate}[1)]
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\item $f$ in $[a,b]$ bis auf endlich viele Ausnahmestellen stetig und beschränkt ist.
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\item in jeder dieser Unstetigkeitsstellen $\xi \in [a,b]$ die links- bzw.
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rechtsseitigen Grenzwerte
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\[
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f(\xi_{\pm} := \lim_{h \searrow 0} f(\xi \pm h)
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.\] existieren. Für $\xi \in (a,b)$ wird
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\[
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f(\xi) := \frac{f(\xi_{-} + f(\xi_{+})}{2}
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.\] gesetzt.
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\end{enumerate}
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\end{definition}
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\begin{bem}
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Stückweise stetige Funktionen sind Riemann-integrierbar.
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Die Menge der in diesem Sinne auf $[a,b]$ stückweise stetigen (Riemann-integrierbaren)
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Funktionen bilden einen Vektorraum $R[a,b]$.
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\end{bem}
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\begin{definition}
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Wir definieren
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\[
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(f, g) := \int_{a}^{b} f(x) \overline{g(x)} \d x \qquad (\text{,,Sesquilinearform''})
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.\] Dies ist wohldefiniert da für $f, g \in R[a,b]$ das Produkt $f(x) \cdot \overline{g(x)} \in R[a,b]$ ist.
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\end{definition}
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\begin{definition}[Skalarprodukt]
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Sei $V$ Vektorraum über $\mathbb{K}$. Die Abbildung $<\cdot, \cdot >\colon V \times V \to \mathbb{K}$
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heißt Skalarprodukt auf $V$, falls $\forall u, v, w \in V$ und $\alpha \in \mathbb{K}$ gilt:
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\begin{enumerate}[(S1)]
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\item $\langle v, u\rangle = \overline{\langle u, v\rangle}$ (Symmetrie,
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hermitesch falls $\mathbb{K} = \mathbb{C}$
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symmetrisch falls $\mathbb{K} = \R$)
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\item $\langle\alpha v, u\rangle = \alpha \langle v, u\rangle$ \\
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$\langle v, \alpha u\rangle = \overline{\alpha}\langle v, u\rangle$ \\
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$\langle v, u + w\rangle = \langle v, u\rangle + \langle v, w\rangle$ \\
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$\langle v + u, w\rangle = \langle v, w\rangle + \langle u, w\rangle$
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\item Positivdefinitheit: $\langle v, v \rangle \ge 0$ \\
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$\langle v, v \rangle = 0 \iff v = 0$
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\end{enumerate}
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\end{definition}
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\begin{bem}
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Auf $R[a,b]$ besitzt $(\cdot , \cdot )$ die Eigenschaften eines Skalarprodukts, denn
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es gilt $\forall \alpha, \beta \in \mathbb{C}$, $\forall f, g \in R[a,b]$, $f_1, f_2 \in R[a,b]$,
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$g_1, g_2 \in R[a,b]$:
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\begin{enumerate}[(1)]
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\item $(\alpha f_1 + \beta f_2, g) = (\alpha f_1, g) + (\beta f_2, g) = \alpha (f_1, g) + \beta (f_2, g)$
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\item $(f, \alpha g_1 + \beta g_2) = (f, \alpha g_1) + (f, \beta g_2) = \overline{\alpha}(f, g_1) + \overline{\beta} (f, g_2)$
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\item $\displaystyle (f, g)
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= \int_{a}^{b} f \cdot \overline{g} \d x
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= \int_{a}^{b} \overline{\overline{f} g} \d x
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= \overline{\int_{a}^{b} \overline{f} g} \d x
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= \overline{\int_{a}^{b} g \overline{f} \d x }
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= \overline{(g, f)}$
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\item $(f,f) = \displaystyle \int_{a}^{b} f \overline{f} \d x = \int_{a}^{b} |f(x)|^2 \d x \ge 0$
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\item Aus (4) und der Definition von $R[a,b]$ folgt: $(f,f) = 0 \implies f \equiv 0$ auf $[a,b]$.
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\end{enumerate}
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$(\cdot , \cdot )$ wird auf $R[a,b]$ $L^2$-Skalarprodukt genannt.
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\end{bem}
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\begin{lemma}
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Für ein $L^2$-Skalarprodukt $(\cdot , \cdot )$ auf $R[a,b]$ gilt die Cauchy-Schwarz Ungleichung:
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\[
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|(f,g)|^2 \le (f,f)\cdot (g,g)
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.\]
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\end{lemma}
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\begin{proof}
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\begin{enumerate}[1)]
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\item Falls $g \equiv 0$ gilt trivialerweise
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\[
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|(f,g)|^2 = 0 = (f,f) \cdot (g,g)
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.\]
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\item Falls $g \not\equiv 0$, sei $\alpha \in \mathbb{K}$ beliebig
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\[
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0 \le (f + \alpha g, f + \alpha g) = (f,f) + \alpha(g,f) + \overline{\alpha}(f,g)
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+ \alpha \cdot \overline{\alpha}(g,g)
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.\] Setze $\alpha := - \frac{(f,g)}{(g,g)} = - \frac{\overline{(g,f)}}{(g,g)}$. Dann gilt
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\begin{align*}
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0 &\le (f,f) - \frac{(f,g) \cdot (g, f)}{(g,g)} - \frac{(g, f) \cdot (f,g)}{(g,g)}
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+ \frac{(f,g)(g,f)(g,g)}{(g,g)(g,g)} \\
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&= (f,f) - \frac{(f,g)(g, f)}{(g,g)} \\
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&= (f,f) - \frac{\overline{(f,g)}(f,g)}{(g,g)} \\
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&= (f,f) - \frac{|(f,g)|^2}{(g,g)} \\
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\implies 0 &\le (f,f)(g,g) - |(f,g)|^2
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.\end{align*}
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\end{enumerate}
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\end{proof}
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\begin{definition}[$L^2$-Norm]
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Das $L^2$-Skalarprodukt $(\cdot , \cdot )$ induziert die $L^2$-Norm auf $R[a,b]$ mit
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\[
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\Vert f \Vert = \Vert f \Vert_{L^2} := (f,f)^{\frac{1}{2}} = \left(\int_{a}^{b} f \cdot \overline{f} \d x\right)^{\frac{1}{2}}
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.\]
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\end{definition}
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\begin{bem}
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Normeigenschaften von $L^2$ auf $R[a,b]$ sind erfüllt:
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\begin{enumerate}[(N1)]
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\item Definitheit: $\Vert f \Vert = 0 \implies (f,f) = 0 \implies f = 0$ auf $[a,b]$
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\item Homogenität: $\Vert \alpha f \Vert = (\alpha f, \alpha f)^{\frac{1}{2}} = (|\alpha|^2 (f,f))^{\frac{1}{2}} = |\alpha| \cdot \Vert f \Vert$
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\item Dreiecksungleichung:
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\begin{align*}
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\Vert f + g \Vert &= (f + g, f + g)^{\frac{1}{2}} \\
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&= \left( \Vert f \Vert^2 + (f,g) + (g,f) + \Vert g \Vert^2 \right)^{\frac{1}{2}} \\
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&\stackrel{\text{CSU}}{\le} \left( \Vert f \Vert^2 + 2 \Vert f \Vert \Vert g \Vert
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+ \Vert g \Vert^2\right)^{\frac{1}{2}} \\
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&= \Vert f \Vert + \Vert g \Vert
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.\end{align*}
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\end{enumerate}
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\end{bem}
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\begin{definition}[Konvergenz im Quadratischen Mittel ($L^2$-Konvergenz)]
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Seien $f_n \in R[a,b], n \in \N, f \in R[a,b]$. $f_n$ konvergiert gegen $f$ im Quadratischen
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Mittel $f_n \xrightarrow[L^2]{n \to \infty}$, wenn gilt
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\[
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\Vert f_n - f \Vert_{L^2} \xrightarrow{n \to \infty} 0
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.\] Das heißt, dass die quadratische Abweichung zwischen $f_n$ und $f$ gegen Null konvergiert:
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\[
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\int_{a}^{b} |f_n(x) - f(x)|^2 \d x \xrightarrow{n \to \infty} 0
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.\]
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\end{definition}
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\begin{bem}
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\begin{enumerate}[(1)]
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\item Es gilt:
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\[
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\Vert f_n - f \Vert_{L^2}^2 = \int_{a}^{b} |f_n(x) - f(x)|^2 \d x
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\le \Vert f_n - f \Vert_{\infty}^2 (b-a)
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.\] Damit folgt
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\[
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\Vert f_n - f \Vert_{\infty} \xrightarrow{n \to \infty} 0
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\implies \Vert f_n - f \Vert_{L^2} \xrightarrow{n \to \infty} 0
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.\]
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Die Umkehrung gilt i.A. nicht! Beispiel: $f_n(x) := x^{n}$, $x \in [-1, 1]$
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\begin{figure}[h!]
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\centering
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\begin{tikzpicture}
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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grid style={line width=.1pt, draw=gray!10},
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major grid style={line width=.2pt,draw=gray!50},
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axis lines=middle,
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enlargelimits={abs=0.2},
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ymax=1,
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ymin=-1,
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]
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\addplot[domain=-1:1,samples=50,smooth,red] {x^1};
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\addplot[domain=-1:1,samples=50,smooth,purple] {x^2};
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\addplot[domain=-1:1,samples=50,smooth,green] {x^3};
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\legend{$n=1$, $n=2$, $n=3$}
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{axis}%
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[grid=none,
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minor tick num=4,
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grid style={line width=.1pt, draw=gray!10},
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major grid style={line width=.2pt,draw=gray!50},
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axis lines=middle,
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enlargelimits={abs=0.2},
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ymax=1,
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ymin=0,
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ytick={0},
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xtick = {0.2, 0.5, 0.9},
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xticklabels = {$a$, $\xi$, $b$}
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]
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\addplot[domain=0:1,samples=50,smooth,red] {0};
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\node[red,circle,fill,inner sep=0.5pt] at (axis cs:0.5,0.5) {};
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\node[black,circle,fill,inner sep=0.5pt] at (axis cs:0.5,0) {};
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\end{axis}
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\end{tikzpicture}
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\caption{Links: $f_n(x) = x^{n}$, Rechts: $f(x) \not\equiv 0$}
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\label{abb:nichtvollstaendig}
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\end{figure}
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\[
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\Vert f_n \Vert^2_{L^2} = \int_{-1}^{1} x^{2n} \d x
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= 2 \int_{0}^{1} x^{2n} \d x
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= 2 \frac{x^{2n+1}}{2n+1} \Big|_{0}^{1} = \frac{2}{2n+1} \xrightarrow{n \to \infty} 0
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.\] Damit folgt $f_n \xrightarrow[L^2]{n \to \infty} f \equiv 0$.
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Aber wegen $f_n(1) = 1$ für $x = 1$, $n \in \N$, konvergiert $f_n$ nicht punktweise gegen
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$f \equiv 0$ und wegen $f_n(-1) = (-1)^{n}, n \in \N, x = -1$ konvergiert $f_n$ nicht.
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\item Der Raum $R[a,b]$ mit $L^2$-Norm $\Vert \cdot \Vert$ ist \textbf{nicht vollständig}, d.h.
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es existieren Cauchy-Folgen in $R[a,b]$, die keinen Grenzwert in $R[a,b]$ haben. Beispiel: siehe
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Abb. \ref{abb:nichtvollstaendig} (Rechts). Hier ist
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$f(x) \not\equiv 0$, $x \in [a,b]$.
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\[
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\int_{a}^{b} |f(x)|^2 \d x = 0 = \Vert f \Vert_{L^2}
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,\] aber $f(x) \not\in R[a,b]$, denn
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\[
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f(\xi) \neq 0 = \frac{\lim_{h \searrow 0} f(\xi + h) - \lim_{h \searrow 0} f(\xi - h)}{2}
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.\]
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\end{enumerate}
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\end{bem}
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\begin{definition}[Orthogonalität]
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$f, g \in R[a,b]$ heißen orthogonal, wenn gilt $(f, g) = 0$.
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Eine Teilmenge $S \subset R[a,b]$ heißt Orthogonalsystem, wenn alle Elemente
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aus $S$ paarweise orthogonal sind, d.h.
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\[
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(f_i, f_j) = \begin{cases}
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\Vert f_i \Vert^2 & i = j \\
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0 & i \neq j
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\end{cases} \quad \forall f_i, f_j \in S
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.\]
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\end{definition}
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\begin{satz}
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Die trigonometrischen Funktionen, für $k, l \in \N$
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\begin{align*}
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c_k(x) &:= \begin{cases}
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1 & k = 0 \\
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\cos(k x) & \text{sonst}
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\end{cases} \\
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s_l(x) &:= \sin (l x)
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\end{align*} bilden auf $R[a,b]$ bezüglich des $L^2$-Skalarprodukts ein
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Orthogonalsystem und es gilt
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\begin{align*}
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&\int_{0}^{2 \pi} c_k(x) \d x = \int_{0}^{2 \pi} s_l(x) \d x = \int_{0}^{2\pi} c_k(x) s_l(x) \d x = 0 \\
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&\int_{0}^{2\pi} c_k(x) c_l(x) \d x = \pi \delta_{kl} \\
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&\int_{0}^{2\pi} s_k(x) s_l(x) \d x = \pi \delta_{kl}
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\intertext{Hier sei}
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&\delta_{kl} := \begin{cases}
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1 & k = l \\
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0 & k \neq l
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\end{cases} \qquad \text{Kroneckersymbol}
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.\end{align*}
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\end{satz}
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\begin{proof}
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\begin{align*}
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\int_{0}^{2\pi} c_k(x) \d x
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&= \int_{0}^{2\pi} \cos(k x) \d x
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= \frac{1}{k} \sin(k x) \Big|_{0}^{2\pi} = 0 \\
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\int_{0}^{2\pi} s_k(x) \d x &= \int_{0}^{2\pi} \sin(kx) \d x = - \frac{1}{k} \cos(k x) \Big|_{0}^{2\pi}
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= \frac{1}{k}(1-1) = 0
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\intertext{Damit folgt}
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\int_{0}^{2\pi} c_k(x) s_l(x) \d x \quad &= \quad
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\int_{0}^{2\pi} \underbrace{\cos(k x)}_{u'} \cdot \underbrace{\sin(l x)}_{v} \d x \\
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&\stackrel{\text{part. Int.}}{=}
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\quad
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\underbrace{\underbrace{\frac{1}{k} \sin(k x)}_{u} \cdot \underbrace{\sin(l x)}_{v} \Big|_{0}^{2\pi}}_{= 0}
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- \int_{0}^{2\pi} \underbrace{\frac{1}{k} \sin(k x)}_{u} \underbrace{l \cos(l x)}_{v'} \d x \\
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&= \quad - \frac{l}{k} \int_{0}^{2\pi} s_k(x) c_l(x) \d x
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\intertext{Für $l = k$ gilt}
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\int_{0}^{2\pi} c_k(x) s_k(x) \d x &= - \int_{0}^{2\pi} c_k(x) s_k(x) \d x \\
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\implies 2 \int_{0}^{2\pi} c_k(x) s_k(x) \d x &= 0 \\
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\implies \int_{0}^{2\pi} c_k(x) s_k(x) \d x &= 0
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\intertext{Analog folgt mit partieller Integration}
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\int_{0}^{2\pi} c_k(x) c_l(x) \d x
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&= \frac{l}{k} \int_{0}^{2\pi} s_k(x) s_l(x) \d x \\
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\stackrel{l = k}{\implies} \int_{0}^{2\pi} c_k^2 \d x
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&= \int_{0}^{2\pi} s_k^2 \d x = \int_{0}^{2\pi} (1- c_k^2(x)) \d x
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= 2\pi - \int_{0}^{2\pi} c_k^2(x) \d x \\
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\implies \int_{0}^{2\pi} c_k^2(x) \d x &= \pi = \int_{0}^{2\pi} s_k^2(x) \d x
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\intertext{Wenn $k \neq l$, dann folgt}
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\int_{0}^{2\pi} c_k(x) c_l(x) \d x \quad
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&= \quad \frac{l}{k} \int_{0}^{2\pi} s_k(x) s_l(x) \d x \\
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&\stackrel{\text{part. Int.}}{=} \quad \frac{l^2}{k^2} \int_{0}^{2\pi} c_k(x) c_l(x) \d x \\
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\implies \int_{0}^{2\pi} c_k(x) c_l(x) \d x \quad &= \quad 0
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\intertext{Analog}
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\int_{0}^{2\pi} s_k(x) s_l(x) \d x &= 0 \\
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\int_{0}^{2\pi} c_k(x) s_l(x) \d x &= 0
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.\end{align*}
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\end{proof}
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\end{document}
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\documentclass[titlepage]{../../../lecture}
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\documentclass[titlepage]{lecture}
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\usepackage{standalone}
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\usepackage{tikz}
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\newpage
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\input{ana1.tex}
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\input{ana2.tex}
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\end{document}
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+136
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\ProvidesClass{lecture}
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\LoadClass[a4paper, titlepage]{article}
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\RequirePackage[utf8]{inputenc}
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\RequirePackage[T1]{fontenc}
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\RequirePackage{textcomp}
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\RequirePackage[german]{babel}
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\RequirePackage{amsmath, amssymb, amsthm}
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\RequirePackage{mdframed}
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\RequirePackage{fancyhdr}
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\RequirePackage{geometry}
|
||||
\RequirePackage{import}
|
||||
\RequirePackage{pdfpages}
|
||||
\RequirePackage{transparent}
|
||||
\RequirePackage{xcolor}
|
||||
\RequirePackage{array}
|
||||
\RequirePackage[shortlabels]{enumitem}
|
||||
\RequirePackage{tikz}
|
||||
\RequirePackage{pgfplots}
|
||||
\RequirePackage[nobottomtitles]{titlesec}
|
||||
\RequirePackage{listings}
|
||||
\RequirePackage{mathtools}
|
||||
\RequirePackage{forloop}
|
||||
\RequirePackage{totcount}
|
||||
|
||||
\usetikzlibrary{quotes, angles}
|
||||
|
||||
\DeclareOption*{\PassOptionsToClass{\CurrentOption}{article}}
|
||||
\DeclareOption{uebung}{
|
||||
\makeatletter
|
||||
\lhead{\@title}
|
||||
\rhead{\@author}
|
||||
\makeatother
|
||||
}
|
||||
\ProcessOptions\relax
|
||||
|
||||
% PAGE GEOMETRY
|
||||
\geometry{
|
||||
left=15mm,
|
||||
right=40mm,
|
||||
top=20mm,
|
||||
bottom=20mm
|
||||
}
|
||||
|
||||
% PARAGRAPH no indent but skip
|
||||
\setlength{\parskip}{3mm}
|
||||
\setlength{\parindent}{0mm}
|
||||
|
||||
\theoremstyle{definition}
|
||||
\newmdtheoremenv{satz}{Satz}[section]
|
||||
\newmdtheoremenv{lemma}[satz]{Lemma}
|
||||
\newmdtheoremenv{korrolar}[satz]{Korrolar}
|
||||
\newmdtheoremenv{definition}[satz]{Definition}
|
||||
|
||||
\newtheorem{bsp}[satz]{Beispiel}
|
||||
\newtheorem{bem}[satz]{Bemerkung}
|
||||
\newtheorem{aufgabe}{Aufgabe}
|
||||
|
||||
% enable aufgaben counting
|
||||
\regtotcounter{aufgabe}
|
||||
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
|
||||
% HEADERS
|
||||
|
||||
\pagestyle{fancy}
|
||||
|
||||
\newcommand{\incfig}[1]{%
|
||||
\def\svgwidth{\columnwidth}
|
||||
\import{./figures/}{#1.pdf_tex}
|
||||
}
|
||||
\pdfsuppresswarningpagegroup=1
|
||||
|
||||
% horizontal rule
|
||||
\newcommand\hr{
|
||||
\noindent\rule[0.5ex]{\linewidth}{0.5pt}
|
||||
}
|
||||
|
||||
% punkte tabelle
|
||||
\newcommand{\punkte}{
|
||||
\@punkten{\totvalue{aufgabe}}
|
||||
}
|
||||
|
||||
\def\@punkten#1{
|
||||
\newcounter{n}
|
||||
\begin{tabular}{|c|*{#1}{m{1cm}|}m{1cm}|@{}m{0cm}@{}}
|
||||
\hline
|
||||
Aufgabe
|
||||
\forloop{n}{1}{\not{\value{n} > #1}}{
|
||||
& \centering A\then
|
||||
}
|
||||
& \centering $\sum$ & \\[5mm] \hline
|
||||
Punkte
|
||||
\forloop{n}{1}{\not{\value{n} > #1}}{
|
||||
&
|
||||
}
|
||||
& & \\[5mm] \hline
|
||||
\end{tabular}
|
||||
}
|
||||
|
||||
% code listings, define style
|
||||
\lstdefinestyle{mystyle}{
|
||||
commentstyle=\color{gray},
|
||||
keywordstyle=\color{blue},
|
||||
numberstyle=\tiny\color{gray},
|
||||
stringstyle=\color{black},
|
||||
basicstyle=\ttfamily\footnotesize,
|
||||
breakatwhitespace=false,
|
||||
breaklines=true,
|
||||
captionpos=b,
|
||||
keepspaces=true,
|
||||
numbers=left,
|
||||
numbersep=5pt,
|
||||
showspaces=false,
|
||||
showstringspaces=false,
|
||||
showtabs=false,
|
||||
tabsize=2
|
||||
}
|
||||
|
||||
% activate my colour style
|
||||
\lstset{style=mystyle}
|
||||
|
||||
% better stackrel
|
||||
\let\oldstackrel\stackrel
|
||||
\renewcommand{\stackrel}[2]{%
|
||||
\oldstackrel{\mathclap{#1}}{#2}
|
||||
}%
|
||||
|
||||
% integral d sign
|
||||
\makeatletter \renewcommand\d[1]{\ensuremath{%
|
||||
\;\mathrm{d}#1\@ifnextchar\d{\!}{}}}
|
||||
\makeatother
|
||||
Reference in New Issue
Block a user