1 files changed, 50 insertions, 11 deletions
diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 04fa307..3d06e20 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -225,14 +225,6 @@ Integrating by parts,
 \end{align*}
 for $\lambda$ concave, monotonic,  $\lambda(1)=0$, and $\lambda'(1)=-1$
 \[
-  0=\frac{\partial\Sigma}{\partial R_d}
-  =\frac12\hat\epsilon f'(1)-\frac12\frac1{\hat\epsilon}\int_0^1dq\,\frac{\lambda(q)}{[\lambda(q)+R_d/\hat\epsilon]^2}
-\]
-\[
-  0=\frac{\partial\Sigma}{\partial\hat\epsilon}
-  =-\epsilon+\frac12R_d f'(1)+\hat\epsilon\int_0^1dq\,\lambda(q)f''(q)+\frac12\frac{R_d}{\hat\epsilon^2}\int_0^1dq\,\frac{\lambda(q)}{[\lambda(q)+R_d/\hat\epsilon]^2}
-\]
-\[
   0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2}
 \]
 \[
@@ -248,10 +240,57 @@ We suppose that solutions are given by
 \end{equation}
 where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac-Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.
 
-The two saddle points equations for $R_d$ and $\hat\epsilon$ give
+Inserting this into the complexity, we find
+\begin{align*}
+  \Sigma
+  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_df'(1)
+  +\frac12\int_0^{q^*}dq\left[
+    \hat\epsilon(f''(q)^{-1/2}-R_d)f''(q)+\hat\epsilon f''(q)^{1/2}
+  \right]
+  +\frac12\int_{q^*}^1dq\left[
+    \hat\epsilon^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\epsilon}
+  \right] \\
+  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_d\left[f'(1)-f'(q^*)\right]
+  +\hat\epsilon\int_0^{q^*}dq\,f''(q)^{1/2}
+  +\frac12\hat\epsilon^2\int_{q^*}^1dq\,
+    (1-q)f''(q)
+    -\log\left[1-(1-q^*)\hat\epsilon/R_d\right]
+\end{align*}
+$R_d$ can be extremized now, with
+\[
+  R_d=\frac12\left(
+    (1-q^*)\hat\epsilon\pm\sqrt{
+      (1-q^*)\left(
+        (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*)]
+      \right)
+    }
+  \right)
+\]
+
+This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives
 \[
-  0=\frac12\hat\epsilon f'(1)-\frac12\int_0^{q_*}dq\,\left[f''(q)^{1/2}-R_df''(q)\right]
-  -\frac12\frac1{\hat\epsilon}\int_{q^*}^1dq\,\frac{1-q}{(1-q+R_d/\hat\epsilon)^2}
+  F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right]
+\]
+and
+\begin{align*}
+  \Sigma
+  =-\epsilon\hat\epsilon+
+    \frac12\hat\epsilon R_df'(1)
+    +\frac12\int_0^1dq\,\left[
+      \hat\epsilon^2\lambda(q)f''(q)
+      +\frac1{\lambda(q)+R_d/\hat\epsilon}
+    \right]\\
+    -\mu R_d-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)+\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)
+\end{align*}
+All that changes is now
+\[
+  R_d=\frac12\left(
+    (1-q^*)\hat\epsilon\pm\sqrt{
+      (1-q^*)\left(
+        (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*) - 2\mu/\hat\epsilon]
+      \right)
+    }
+  \right)
 \]
 
 \end{document}