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%
% polarformel.tex
%
% (c) 2021 Prod Dr Andreas Müller, OST Ostschweizer Fachhochschule
%
\bgroup
\definecolor{darkcolor}{rgb}{0,0.6,0}
\def\yone{-2.1}
\def\ytwo{-3.55}
\def\ythree{-5.0}
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
\frametitle{Polarformel}
\vspace{-5pt}
\begin{block}{Aufgabe}
$\langle x,y\rangle$ aus Werten von $\|\cdot\|_2$ rekonstruieren:
\end{block}
\begin{center}
\begin{tikzpicture}[>=latex,thick]
\node at (0,0) {$
\begin{aligned}
\uncover<2->{
\|x+ty\|_2^2
&=
\|x\|_2^2
+t\langle x,y\rangle
+\overline{t}\langle y,x\rangle
+ \|y\|_2^2}
\\
\uncover<3->{
&=
\|x\|_2^2
+t\langle x,y\rangle
+\overline{t\langle x,y\rangle}
+ \|y\|_2^2}
\\
\uncover<4->{
&=
\|x\|_2^2
+2\operatorname{Re}(t\langle x,y\rangle)
+ \|y\|_2^2}
\end{aligned}$};
\uncover<5->{
\draw[->] (-1,-0.9) -- (-3.3,{\yone+0.25});
\node at (-3.5,\yone) {$
\|x\pm y\|_2^2
=
\|x\|_2^2
\pm2\operatorname{Re}\langle x,y\rangle
+
\|y\|_2^2
$};
}
\uncover<8->{
\draw[->] (1,-0.9) -- (3.3,{\yone+0.25});
\node at (3.5,\yone) {$
\|x\pm iy\|_2^2
=
\|x\|_2^2
\pm2i\operatorname{Im}\langle x,y\rangle
+
\|y\|_2^2
$};
}
\uncover<6->{
\draw[->] (-3.5,{\yone-0.2}) -- (-3.5,{\ytwo+0.2});
\node at (-3.5,\ytwo) {$\operatorname{Re}\langle x,y\rangle
=
\frac12\bigl(
\|x+y\|_2^2-\|x-y\|_2^2
\bigr)
$};
}
\uncover<9->{
\draw[->] (3.5,{\yone-0.2}) -- (3.5,{\ytwo+0.2});
\node at (3.5,\ytwo) {$
\operatorname{Im}\langle x,y\rangle
=
\frac1{2i}\bigl(
\|x+iy\|_2^2-\|x-iy\|_2^2
\bigr)
$};
}
\uncover<7->{
\draw[->] (-3.3,{\ytwo-0.25}) -- (-1.5,{\ythree+0.25});
\node at (0,\ythree) {$
\langle x,y\rangle
=
\frac12\bigl(
\|x+y\|_2^2-\|x-y\|_2^2
\uncover<10->{
+
\|x+iy\|_2^2-\|x-iy\|_2^2
}
\bigr)$};
}
\uncover<10->{
\draw[->] (3.3,{\ytwo-0.25}) -- (1.5,{\ythree+0.25});
}
\end{tikzpicture}
\end{center}
\end{frame}
\egroup
|
