Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

consider also =: #20

Open
ouboub opened this issue Jan 9, 2018 · 0 comments
Open

consider also =: #20

ouboub opened this issue Jan 9, 2018 · 0 comments

Comments

@ouboub
Copy link

ouboub commented Jan 9, 2018

Hi
the following works nicely

\begin{dmath}
\label{eq:sec4-proof:17}
H^n-H^0 = \frac{\partial A^a}{\partial U}(U^n)DU^n \partial_aU^{n}-\frac{\partial A^a}{\partial U}(U^0)DU^0 \partial_aU^{0}
= DU^n \partial_a \left(\frac{\partial A^a}{\partial U}(U^n)-\frac{\partial A^a}{\partial U}(U^0)\right)
+ \frac{\partial A^a}{\partial U} \left( DU^n \partial_a U^n - DU^0 \partial_aU^0
\right)
= H_1+H_{2}
\end{dmath}

but that one not

\begin{dmath}
\label{eq:sec4-proof:17}
H^n-H^0 = \frac{\partial A^a}{\partial U}(U^n)DU^n \partial_aU^{n}-\frac{\partial A^a}{\partial U}(U^0)DU^0 \partial_aU^{0}
= DU^n \partial_a \left(\frac{\partial A^a}{\partial U}(U^n)-\frac{\partial A^a}{\partial U}(U^0)\right)
+ \frac{\partial A^a}{\partial U} \left( DU^n \partial_a U^n - DU^0 \partial_aU^0
\right)
=: H_1+H_{2}
\end{dmath}

the issue is to consider =: in a similar way as =
I tried {=:} but it did not work as expected.

thanks for the nice package

Uwe Brauer

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Labels
None yet
Projects
None yet
Development

No branches or pull requests

1 participant