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| author | Preston Pan <preston@nullring.xyz> | 2024-06-28 21:30:42 -0700 |
|---|---|---|
| committer | Preston Pan <preston@nullring.xyz> | 2024-06-28 21:30:42 -0700 |
| commit | e7dd5245c35d2794f59bcf700a6a92009ec8c478 (patch) | |
| tree | 0d0e81552f0426f8b715bd5bd3bdd0856058db2c /mindmap/del operator.org | |
| parent | 01ba01763b81a838dcbac4c08243804e068495b9 (diff) | |
stuff
Diffstat (limited to 'mindmap/del operator.org')
| -rw-r--r-- | mindmap/del operator.org | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/mindmap/del operator.org b/mindmap/del operator.org index 3d9e2f5..657ff69 100644 --- a/mindmap/del operator.org +++ b/mindmap/del operator.org @@ -107,6 +107,6 @@ It returns a scalar field and is the multivariable analogue to the second deriva and gradient have been described, I feel it is trivial to understand the Laplacian. ** Product Rules -The product rules pertaining to the del operator are consistent with that of linear algebra and single variable derivative rules. +The [[id:d1e245f4-0b04-450e-8465-a9c85fe57f7e][product rules]] pertaining to the del operator are consistent with that of linear algebra and single variable derivative rules. For example, \( \vec{\nabla} \times \vec{\nabla}f = 0\). You can show this yourself quite easily, so I find no need to go over it here. When in doubt, just assume the del works the same way as any old vector except you apply the [[id:d1e245f4-0b04-450e-8465-a9c85fe57f7e][product rule]], and you will usually be correct. |