ana7 done, fixed all bad breaks and overfull boxes
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@@ -32,8 +32,8 @@
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$x=0$: $f_n(0) = 0 = f(0)$\\
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$0 < x \le 2$: $\forall n \ge \frac{2}{x}$, $f_n(x) = 0 = f(x)$
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\item $f_n(x) = x^{n}$, $f_n\colon [0,1] \to \R$.
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\begin{figure}[h!]
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\begin{tikzpicture}
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\begin{figure}[ht!]
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\begin{tikzpicture}[scale = 0.97]
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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@@ -50,7 +50,7 @@
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\addplot[domain=0:1,samples=50,smooth,red] {x^4};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{tikzpicture}[scale = 0.97]
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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@@ -95,8 +95,8 @@
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\text{graph}(f_n) \subset \epsilon\text{-Umgebung von Graphen von } f
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:= \{ (x, y) \in D \times \R \mid | y - f(x)| < \epsilon\}
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.\]
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\begin{figure}[h!]
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\begin{tikzpicture}
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\begin{figure}[ht!]
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\begin{tikzpicture}[scale = 0.97]
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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@@ -113,7 +113,7 @@
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\addplot[domain=0:1,samples=50,smooth,dashed, blue] {0.3*sin(deg(8*x)) + 0.2*x + 0.3};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{tikzpicture}[scale = 0.97]
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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@@ -293,10 +293,12 @@ Wichtige Frage: Wenn $f_n \to f$, gilt dann auch $\int_{a}^{b} f_n \to \int_{a}^
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$f_n \xrightarrow[\text{gleichmäßig}]{n \to \infty} f \implies f$ stetig $\implies f$ Riemann-integrierbar.
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Es gilt
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\[
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\left| \int_{a}^{b} f_n(x) dx - \int_{a}^{b} f(x) dx\right| = \left| \int_{a}^{b} (f_n(x) - f(x))dx\right|
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\le \int_{a}^{b} |f_n(x) - f(x)| dx \le \underbrace{\max_{x \in [a,b]} |f_n(x) - f(x)|}_{= \underbrace{\Vert f_n - f \Vert_\infty}_{\xrightarrow{n \to \infty} 0}\underbrace{(b-a)}_{\text{beschränkt}}}
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.\]
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\begin{align*}
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\left| \int_{a}^{b} f_n(x) dx - \int_{a}^{b} f(x) dx\right| &= \left| \int_{a}^{b} (f_n(x) - f(x))dx\right|\\
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&\le \int_{a}^{b} |f_n(x) - f(x)| dx\\
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&\le \max_{x \in [a,b]} |f_n(x) - f(x)| \cdot (a-b)\\
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&=\underbrace{\norm{f_n - f}_\infty}_{\xrightarrow{n \to \infty} 0}\underbrace{(b-a)}_{\text{beschränkt}}.
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\end{align*}
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\end{proof}
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\begin{satz}\label{permutesumint}
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