Hi,
As usual \rho_{n+1}(in) = (1-\alpha)\rho_{n}(in) +
\alpha \rho_{n}(out).
So, it seems, yes, that is. But one have to be
careful, because the Broyden method is used by default
for charge mixing. In this case the above mentioned
mixing procedure is used just for a few initial
iterations.
Bests,
Eyvaz.
--- Aritz Leonardo <swblelia at sw.ehu.es> wrote:
> One simple question: A big value of beta means a
> small portion of the previous density?
>
> this is... n(i+1) = beta*n + (1-beta)*n(i)
>
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