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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} |