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S Jul 15, 2023 at 19:23 history unlocked CommunityBot
S Jul 15, 2023 at 19:23 history locked CommunityBot
S Jul 15, 2023 at 19:23 history closed M.G.
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Jun 19, 2023 at 14:53 vote accept Gauge
Jun 18, 2023 at 23:45 answer added Michael Engelhardt timeline score: 0
Jun 18, 2023 at 23:18 comment added Michael Engelhardt The $\nabla \cdot \nabla $ term doesn't correspond to what's on the Wikipedia page and just seems simply wrong to me. The correct form on the Wikipedia page refers to the case that $g$, in your notation, contains $\nabla f$ (which it doesn't in your concrete example), and then one has to take into account $\partial g/\partial (\nabla f)$.
Jun 18, 2023 at 20:55 comment added Gauge Yea that's the one
Jun 18, 2023 at 20:49 review Close votes
Jul 15, 2023 at 19:23
Jun 18, 2023 at 19:33 comment added LSpice I didn't see your exact formula on the Wikipedia article, but it seemed quite close to the formula linked at "Functional derivative > Formula", so I changed the link to point there (while also fixing up some TeX). I hope that was correct.
Jun 18, 2023 at 19:32 history edited LSpice CC BY-SA 4.0
TeX
Jun 18, 2023 at 19:18 history edited Gauge CC BY-SA 4.0
added 5 characters in body
S Jun 18, 2023 at 19:18 review First questions
Jun 19, 2023 at 5:15
S Jun 18, 2023 at 19:18 history asked Gauge CC BY-SA 4.0