Lots of confusion.

1 files changed, 24 insertions, 0 deletions

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index c1f0bd8..e1ac91d 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -419,6 +419,30 @@ Diagonal ansatz requires that
   0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+R_dQ^{-1}f'(Q)
 \end{equation}
 or $\hat\beta f'(Q_{ab})=-R_d[Q^{-1}f'(Q)]_{ab}$ for $a\neq b$.
+We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)$. Inserting the diagonal, we get $Q^{-1}f'(Q)=\frac1{R_d^2}(I-D_df'(Q))$, and
+\begin{equation}
+  0=-\mu I+\hat\beta f'(Q)+R_df''(1)I+\frac1{R_d}(I-D_df'(Q))
+  =(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q)
+\end{equation}
+
+\begin{equation}
+  0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q
+\end{equation}
+
+\begin{equation}
+  \begin{aligned}
+    &\Sigma(\epsilon,\mu)
+    =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\
+    &\lim_{n\to0}\frac1n\left(
+      -\frac12\mu\sum_a^nR_{aa}
+      +\frac12\sum_{ab}\left[
+        \hat\beta^2f(Q_{ab})+\hat\beta R_{ab}f'(Q_{ab})
+      \right]
+      +\frac12\log\det(f'(Q)^{-1}Q)
+    \right)
+  \end{aligned}
+\end{equation}
+
 
 \subsection{Solution}