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author | enezerdem <105669082+enezerdem@users.noreply.github.com> | 2022-05-22 15:35:23 +0200 |
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committer | GitHub <noreply@github.com> | 2022-05-22 15:35:23 +0200 |
commit | 9a4d65ec7e6c5e5041d2904128b6bd202f66594b (patch) | |
tree | f460bc3b8f45399004b4a07d8d3eac13006193c9 /vorlesungen/slides/hermite/skalarprodukt.tex | |
parent | Korrektur 21.05 (diff) | |
parent | Merge pull request #14 from enezerdem/master (diff) | |
download | SeminarSpezielleFunktionen-9a4d65ec7e6c5e5041d2904128b6bd202f66594b.tar.gz SeminarSpezielleFunktionen-9a4d65ec7e6c5e5041d2904128b6bd202f66594b.zip |
Merge pull request #4 from AndreasFMueller/master
update
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-rw-r--r-- | vorlesungen/slides/hermite/skalarprodukt.tex | 72 |
1 files changed, 72 insertions, 0 deletions
diff --git a/vorlesungen/slides/hermite/skalarprodukt.tex b/vorlesungen/slides/hermite/skalarprodukt.tex new file mode 100644 index 0000000..32b933f --- /dev/null +++ b/vorlesungen/slides/hermite/skalarprodukt.tex @@ -0,0 +1,72 @@ +% +% skalarprodukt.tex -- slide template +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\bgroup +\begin{frame}[t] +\setlength{\abovedisplayskip}{5pt} +\setlength{\belowdisplayskip}{5pt} +\frametitle{Skalarprodukt} +\vspace{-20pt} +\begin{columns}[t,onlytextwidth] +\begin{column}{0.48\textwidth} +\begin{block}{Orthogonale Zerlegung} +Orthogonale $H_k$ normalisieren: +\[ +\tilde{H}_k(x) = \frac{1}{\|H_k\|_w} H_k(x) +\] +mit Gewichtsfunktion $w(x)=e^{-x^2}$ +\end{block} +\begin{block}{``Hermite''-Analyse} +\begin{align*} +P(x) +&= +\sum_{k=1}^\infty a_k H_k(x) += +\sum_{k=1}^\infty \tilde{a}_k \tilde{H}_k(x) +\\ +\tilde{a}_k +&= +\| H_k\|_w\, a_k +\\ +a_k +&= +\frac{1}{\|H_k\|} +\langle \tilde{H}_k, P\rangle_w += +\frac{1}{\|H_k\|^2} +\langle H_k, P\rangle_w +\end{align*} +\end{block} +\end{column} +\begin{column}{0.48\textwidth} +\begin{block}{Integrationsproblem} +Bedingung: +\begin{align*} +a_0=0 +\qquad\Leftrightarrow\qquad +\langle H_0,P\rangle_w +&= +0 +\\ +\int_{-\infty}^\infty +P(t) w(t) \,dt += +\int_{-\infty}^\infty +P(t) e^{-t^2} \,dt +&= +0 +\end{align*} +\end{block} +\begin{theorem} +Das Integral von $P(t)e^{-t^2}$ ist in geschlossener Form darstellbar +genau dann, wenn +\[ +\int_{-\infty}^\infty P(t)e^{-t^2}\,dt = 0 +\] +\end{theorem} +\end{column} +\end{columns} +\end{frame} +\egroup |