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Martin Sleziak
Martin Sleziak
12:51 PM
This is a bit embarrassing - I have edited this answer before, yet I missed that still more macros are missing there: math.stackexchange.com/posts/1192632/revisions
 
Martin Sleziak
Martin Sleziak
1:08 PM
There is only one comment where the problem is caused by the change in software:
That fixes both expressions of $\ast F$ and tells me the form of $\ast\mathrm{d}\ast F$ is right except for index height, which I decidedly like because it fixes the sign problems with turning it into an equations involving $J$. I get $\ast\mathrm{d}\ast F=-\pd_\mu F^{\mu\nu}\mathrm{d}x^\nu=\pd_\mu F^{\nu\mu}\mathrm{d}x^\nu=\mu_0J^\nu\mathrm{d}x^\nu$, as I wanted. Now I have to check the field form, but since I have dedicated too much time to this I will not. Besides, I will probably change the sign of the $E$ part, getting the sign alternance right. — MickG Mar 11 '15 at 16:12
The other comments found with these queries seem to be false positives:
$\newcommand{\pd}{\frac{\partial}{\partial\mu}}$ $\pd$ is linear, so $\pd\sum_{i = 1}^nf_i(\mu) = \sum_{i=1}^n\pd f_i(\mu)$, so you just need to calculate $\pd f_i(\mu)$. — Mark Aug 23 '16 at 2:21
@mrf I guess for this case, I am considering $\pdv{f}{z}$ is still bounded when $z_j \to z$ for $z_j \in W$, where $W$ is a domain for $f$. — The Hong Kong Journalist Nov 21 '18 at 6:18
It might help to first solve the quadratic equations of the symbol to then recognize $\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}$ $(\pd{}{x}-2\pd{}{y})(\pd{}x-\pd{}y)z=3x^2y+9xy^2$ and $(\pd{}{x}+4\pd{}{y})(\pd{}x-15\pd{}y)z=3x^2+4x^2y+9y$. — LutzL Oct 2 at 20:34
search posts \et is prefix of \eta and \eth - this can be solved by the regex query above or: data.stackexchange.com/math/revision/1071496/1323755/…
All posts found by this query seem to be false positives. (Typos, \etc, \\etc...)
search comments Comments with \et, but neither \eta nor \eth: data.stackexchange.com/math/revision/1071502/1323761/…
Again, only false positives.
 

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