Added rough discussion of the saddle requirements of the diagonal ansatz.

1 files changed, 31 insertions, 0 deletions

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index dcf6dd8..c1f0bd8 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -389,6 +389,37 @@ Similarly, ....  one shows that
 D_d = \hat \beta R_d
 \end{equation}
 
+
+Is it a saddle?
+
+\begin{equation}
+  0=\frac{\partial\Sigma}{\partial R}
+  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
+    +(QD+R^2)^{-1}R
+  \right)
+\end{equation}
+\begin{equation}
+  0=\frac{\partial\Sigma}{\partial D}
+  =\lim_{n\to0}\frac1n\left(-\frac12f'(Q)
+    +\frac12(QD+R^2)^{-1}Q
+  \right)
+\end{equation}
+\begin{equation}
+  D=f'(Q)^{-1}-RQ^{-1}R
+\end{equation}
+Diagonal anstaz requires that $f'(Q_{ab})^{-1}=R_d^2Q_{ab}^{-1}$ for $a\neq b$.
+\begin{equation}
+  0
+  =\lim_{n\to0}\frac1n\left(-\mu I+\hat\beta f'(Q)+R\odot f''(Q)
+    +Q^{-1}Rf'(Q)
+  \right)
+\end{equation}
+Diagonal ansatz requires that
+\begin{equation}
+  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 $a\neq b$.
+
 \subsection{Solution}