<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">From f670ae696e496ecf687cd5555b127e53b00fd912 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias &lt;jaron@kent-dobias.com&gt;
Date: Fri, 17 Jun 2022 11:17:44 +0200
Subject: Lots of confusion.

---
 frsb_kac_new.tex | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)

(limited to 'frsb_kac_new.tex')

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}
+    &amp;\Sigma(\epsilon,\mu)
+    =\frac12+\mathcal D(\mu)+\hat\beta\epsilon+\\
+    &amp;\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}
 
-- 
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