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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} |