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| -rw-r--r-- | .gitignore | 6 | ||||
| -rw-r--r-- | frsb_kac-rice.tex | 61 |
2 files changed, 56 insertions, 11 deletions
diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..da5f7f8 --- /dev/null +++ b/.gitignore @@ -0,0 +1,6 @@ +frsb_kac-rice.aux +frsb_kac-rice.fdb_latexmk +frsb_kac-rice.fls +frsb_kac-rice.log +frsb_kac-rice.pdf +frsb_kac-rice.synctex.gz 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} |