Commit da8698d6 authored by Sander, Oliver's avatar Sander, Oliver
Browse files

Während der Vorlesung gefundene Tippfehler

parent a054ce98
......@@ -7736,7 +7736,7 @@ und (Produktregel)
Daraus erhält man die Formel für die partielle Integration
\begin{equation}
\label{eq:partielle_integration_differentialformen}
\int_\Sigma \bd \omega \wedge \eta + (-1)^l(\omega \wedge \eta)
\int_\Sigma \bd \omega \wedge \eta + (-1)^l(\omega \wedge \bd \eta)
=
\int_\Sigma \bd(\omega \wedge \eta)
=
......@@ -7760,13 +7760,13 @@ Daraus erhält man die Formel für die partielle Integration
%
\Upsilon_3(\omega \wedge \eta)
& =
(\Upsilon_2 \omega) \times (\Upsilon_1 \eta)
(\Upsilon_2 \omega) \cdot (\Upsilon_1 \eta)
& \qquad &
\forall \omega \in \mathcal{DF}^2(\Omega) , \eta \in \mathcal{DF}^1(\Omega) \\
%
\Upsilon_l(\omega \wedge \eta)
& =
(\Upsilon_l \omega) \times (\Upsilon_0 \eta)
(\Upsilon_l \omega) \cdot (\Upsilon_0 \eta)
& \qquad &
\forall \omega \in \mathcal{DF}^l(\Omega) , \eta \in \mathcal{DF}^0(\Omega)
\end{alignat*}
......@@ -7877,8 +7877,12 @@ $(\ba_0,\dots,\widehat{\ba_i},\dots,\ba_l)$ in $F$.
Es gilt $\partial \circ \partial = 0$.
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (2,0) -- (4,2) -- (1.5,2.1) -- cycle;
\draw (2,0) -- (1.5,2.1);
\coordinate[label=below:$\ba_0$] (a0) at (0,0);
\coordinate[label=below:$\ba_1$] (a1) at (2,0) {};
\coordinate[label=above:$\ba_2$] (a2) at (1.5,2.1) {};
\coordinate[label=above:$\ba_3$] (a3) at (3.5,2) {};
\draw (a0) -- (a1) -- (a3) -- (a2) -- (a0);
\draw (a1) -- (a2);
\end{tikzpicture}
\end{center}
......@@ -7892,13 +7896,13 @@ $(\ba_0,\dots,\widehat{\ba_i},\dots,\ba_l)$ in $F$.
& =
(\ba_1,\ba_2) - (\ba_0, \ba_2) + (\ba_0, \ba_1) + (\ba_3, \ba_2) - (\ba_1, \ba_2) + (\ba_1, \ba_3)\\
& =
- (\ba_0, \ba_2) + (\ba_0, \ba_1) + (\ba_3, \ba_2) + (\ba_1, \ba_3)
- (\ba_0, \ba_2) + (\ba_0, \ba_1) + (\ba_3, \ba_2) + (\ba_1, \ba_3) \\
%
\partial \partial \mathcal{S}_3(\mathcal{T}) \\
\partial \partial \mathcal{S}_3(\mathcal{T})
& =
- \partial (\ba_0, \ba_2) + \partial (\ba_0, \ba_1) + \partial(\ba_3, \ba_2) + \partial (\ba_1, \ba_3) \\
& =
\dots \\
-(\ba_2 - \ba_0) + (\ba_1 - \ba_0) + (\ba_2 - \ba_3) + (\ba_3 - \ba_1) \\
& =
0.
\end{align*}
......@@ -8129,14 +8133,13 @@ $\bx, \by \in T$ als Linearkombination der Kanten von $T$ dar:
\sum_i \sum_j \lambda_i(\bx) \lambda_j(\by) (t\ba_i + (1-t) \ba_j)
\; : \; 0 \le t \le 1 \bigg\}.
\end{align*}
Hence, taking into account orientation, we require that the interpolating
differential $1$-form $\mathsf{W}^1 \vec \omega$ satisfies
Wir fordern deshalb für die interpolierende
$1$-Form $\mathsf{W}^1 \vec \omega$
\begin{equation*}
\int_{(\bx,\by)} \mathsf{W}^1|_T \vec \omega
\colonequals
\sum_{i,j} \lambda_i(\bx) \lambda_j(\by) \vec \omega_{(i,j)}
=
\sum_{i < j} (\lambda_i(\bx)\lambda_j(\by) - \lambda_i(\by) \lambda_j(\bx)) \vec \omega_{(i,j)}.
\sum_{i,j} \lambda_i(\bx) \lambda_j(\by) \vec \omega_{(i,j)}.
\end{equation*}
Hier ist $\vec \omega_{(i,j)}$ ist der Wert den die $1$-Kokette $\vec \omega$
der orientierten Kante $(\ba_i,\ba_j)$ zuweist.
......
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