More computations than needed in Htilde
We are computing one order too many in V
for H_tilde
, as \tilde{H}^{AA}_n
does not require V_n
.
See thread.
Edited by Isidora Araya
We are computing one order too many in V
for H_tilde
, as \tilde{H}^{AA}_n
does not require V_n
.
See thread.