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update num and theo

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christian
2020-05-20 11:22:14 +02:00
parent a822783306
commit 77a16d71f2
5 changed files with 47 additions and 14 deletions
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sose2020/num/uebungen/num3.pdf
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sose2020/num/uebungen/num3.tex
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@@ -46,17 +46,25 @@
&\Vert x \Vert_1 = \sum_{k=1}^{n} |x_k|
= \Vert x \Vert_2 \left( \sum_{k=1}^{n} \underbrace{\frac{|x_k|}{\Vert x \Vert_2}}_{\le 1} \right)
\ge \Vert x \Vert_2 \left( \sum_{k=1}^{n} \frac{|x_k|^{2}}{\Vert x \Vert_2^2} \right)
= \Vert x \Vert_2 \frac{\Vert x \Vert_2^2}{\Vert_x \Vert_2^2} = \Vert x \Vert_2 \\
= \Vert x \Vert_2 \frac{\Vert x \Vert_2^2}{\Vert x \Vert_2^2} = \Vert x \Vert_2 \\
&\Vert x \Vert_2 = \sqrt{\sum_{k=1}^{n} |x_k|^2} \ge \sqrt{\Vert x \Vert_{\infty}^2}
= \Vert x \Vert_{\infty}
.\end{align*}
Damit folgt
\begin{align*}
&\Vert x \Vert_2 \le \Vert x \Vert_1 \le \sqrt{n} \Vert x \Vert_2 \\
&\Vert x \Vert_\infty \le \Vert x \Vert_2 \le \sqrt{n} \Vert x \Vert_\infty
\frac{1}{\sqrt{n}} \Vert x \Vert_1 \le &\Vert x \Vert_2 \le \Vert x \Vert_1 \\
\frac{1}{\sqrt{n}} \Vert x \Vert_1 \le &\Vert x \Vert_{\infty} \le \Vert x \Vert_1 \\
\Vert x \Vert_2 \le &\Vert x \Vert_1 \le \sqrt{n} \Vert x \Vert_2 \\
\frac{1}{\sqrt{n}} \Vert x \Vert_2 \le &\Vert x \Vert_{\infty} \le \Vert x \Vert_2 \\
\Vert x \Vert_\infty \le \Vert x \Vert_2 \le &\Vert x \Vert_1 \le \sqrt{n} \Vert x \Vert_2
\le n \Vert x \Vert_\infty \\
\Vert x \Vert_\infty \le &\Vert x \Vert_2 \le \sqrt{n} \Vert x \Vert_\infty
.\end{align*}
Die anderen Kombinationen folgen durch Multiplikation mit $\frac{1}{\sqrt{n}}$.
Für $n \to \infty$ ist $\sqrt{n} \to \infty$.
Für $n = 1$ sind alle Abschätzungen scharf, denn dann ist
$\Vert x \Vert_1 = \Vert x \Vert_2 = \Vert x \Vert_{\infty}$ und $\sqrt{n} = \frac{1}{\sqrt{n}} = 1$.
Es gilt $n \xrightarrow{n \to \infty} \infty$, $\sqrt{n} \xrightarrow{n \to \infty} \infty$ und
$\frac{1}{\sqrt{n}} \xrightarrow{n \to \infty} 0$.
\end{enumerate}
\end{aufgabe}