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authorRunterer <37069007+Runterer@users.noreply.github.com>2022-05-26 15:44:34 +0200
committerGitHub <noreply@github.com>2022-05-26 15:44:34 +0200
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+%
+% loesung.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Lösung}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Frage}
+Für welche Polynome $P(t)$ kann man eine Stammfunktion
+\[
+\int
+P(t)e^{-\frac{t^2}2}
+\,dt
+\]
+in geschlossener Form angeben?
+\end{block}
+\begin{block}{``Hermite-Antwort''}
+\[
+\int H_n(x)e^{-x^2}\,dx
+\]
+kann genau für $n>0$ in geschlossener Form angegeben werden.
+\end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+\begin{block}{Allgemein}
+\begin{align*}
+\int P(x)e^{-x^2}\,dx
+&=
+\int \sum_{k=0}^n a_kH_k(x)e^{-x^2}\,dx
+\\
+&=
+\sum_{k=0}^n
+a_k
+\int
+H_k(x)e^{-x^2}\,dx
+\\
+&=
+a_0\operatorname{erf}(x) + C
+\\
+&\hspace*{2mm} + \sum_{k=1}^n a_k\int H_k(x)e^{-x^2}\,dx
+\end{align*}
+\end{block}
+\begin{theorem}
+Das Integral von $P(x)e^{-x^2}$ ist genau dann elementar darstellbar, wenn
+$a_0=0$
+\end{theorem}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup