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authorJaron Kent-Dobias <jaron@kent-dobias.com>2022-06-20 15:57:24 +0200
committerJaron Kent-Dobias <jaron@kent-dobias.com>2022-06-20 15:57:24 +0200
commit55dcf34e0283712a453ba0461c5b6ffd9364a7ea (patch)
tree8133436dc559d895c05fbac802c6f078aa11dd53
parentf670ae696e496ecf687cd5555b127e53b00fd912 (diff)
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Some unpolished work notes.
-rw-r--r--frsb_kac_new.tex10
1 files changed, 10 insertions, 0 deletions
diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index e1ac91d..e7b65a9 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -425,6 +425,16 @@ We also have from the first (before the diagonal ansatz) $Df'(Q)=I-RQ^{-1}Rf'(Q)
=(R_df''(1)+R_d^{-1}-\mu)I+(\hat\beta-D_d/R_d)f'(Q)
\end{equation}
+Always true:
+\begin{equation}
+ 0=-\mu R+\hat\beta Rf'(Q)+R(R\odot f''(Q))+I-Df'(Q)
+\end{equation}
+Insert $D=D_dI+\delta D$, $R=R_dI+\delta R$, and $D_d=\hat\beta R_d$
+\begin{equation}
+ 0=-\hat\beta \mu R_d I +\hat\beta R_df'(Q)+R_d^2f''(1)I+I-\hat\beta R_df'(Q)
+ -\mu\delta R
+\end{equation}
+
\begin{equation}
0=(\hat\beta Q+R) f'(Q)+(R\odot f''(Q)-\mu I)Q
\end{equation}