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author | Andreas Müller <andreas.mueller@ost.ch> | 2022-06-02 23:01:38 +0200 |
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committer | Andreas Müller <andreas.mueller@ost.ch> | 2022-06-02 23:01:38 +0200 |
commit | 83d215597b5df724022de2a08ae1dfa1e8d59497 (patch) | |
tree | 4d491a966ee8c3fa06e5f66df6e0a700f2eb98f6 /vorlesungen/slides/hermite/loesung.tex | |
parent | fix participant lists (diff) | |
download | SeminarSpezielleFunktionen-83d215597b5df724022de2a08ae1dfa1e8d59497.tar.gz SeminarSpezielleFunktionen-83d215597b5df724022de2a08ae1dfa1e8d59497.zip |
phases
Diffstat (limited to '')
-rw-r--r-- | vorlesungen/slides/hermite/loesung.tex | 19 |
1 files changed, 14 insertions, 5 deletions
diff --git a/vorlesungen/slides/hermite/loesung.tex b/vorlesungen/slides/hermite/loesung.tex index 7d4741f..68ee32e 100644 --- a/vorlesungen/slides/hermite/loesung.tex +++ b/vorlesungen/slides/hermite/loesung.tex @@ -20,36 +20,45 @@ P(t)e^{-\frac{t^2}2} \] in geschlossener Form angeben? \end{block} +\uncover<2->{% \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{block}} \end{column} \begin{column}{0.48\textwidth} +\uncover<3->{% \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 +&\uncover<4->{= +\int \sum_{k=0}^n a_kH_k(x)e^{-x^2}\,dx} \\ +\uncover<5->{ &= \sum_{k=0}^n a_k \int H_k(x)e^{-x^2}\,dx +} \\ +\uncover<6->{ &= a_0\operatorname{erf}(x) + C +} \\ +\uncover<6->{ &\hspace*{2mm} + \sum_{k=1}^n a_k\int H_k(x)e^{-x^2}\,dx +} \end{align*} -\end{block} +\end{block}} +\uncover<7->{% \begin{theorem} Das Integral von $P(x)e^{-x^2}$ ist genau dann elementar darstellbar, wenn $a_0=0$ -\end{theorem} +\end{theorem}} \end{column} \end{columns} \end{frame} |