Timeline for How to understand the combinatorial Laplacian $\Delta$ which is defined on the graph?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2020 at 22:36 | vote | accept | Hermi | ||
| Apr 20, 2020 at 17:58 | history | edited | ARG |
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| Apr 19, 2020 at 0:20 | vote | accept | Hermi | ||
| Apr 19, 2020 at 0:38 | |||||
| Apr 18, 2020 at 14:12 | answer | added | ARG | timeline score: 12 | |
| Apr 17, 2020 at 19:50 | answer | added | gmvh | timeline score: 2 | |
| Apr 17, 2020 at 14:27 | review | Close votes | |||
| Apr 22, 2020 at 3:02 | |||||
| Apr 17, 2020 at 14:26 | comment | added | Hermi | @kneidell The divergence is defined by $\nabla\cdot f(v)=\sum_{e} f(e).$ So $\nabla\cdot \nabla F(v)=\sum_{xy} c(x, y)(F(y)-F(x))$. | |
| Apr 17, 2020 at 14:24 | history | edited | Hermi | CC BY-SA 4.0 |
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| Apr 17, 2020 at 14:13 | comment | added | Abdelmalek Abdesselam | For the gradient the graph needs to be a digraph, whereas the Laplacian does not need oriented edges. | |
| Apr 17, 2020 at 13:41 | comment | added | kneidell | Given $F:V\to\mathbb R$, if I understand correctly, $\nabla F$ would be a function on the set of edges. How do you define $\nabla\cdot \nabla F$ then? | |
| Apr 17, 2020 at 13:23 | history | asked | Hermi | CC BY-SA 4.0 |