Commit da8698d6 by Sander, Oliver

### Während der Vorlesung gefundene Tippfehler

parent a054ce98
 ... ... @@ -7736,7 +7736,7 @@ und (Produktregel) Daraus erhält man die Formel für die partielle Integration \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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