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--- a/bezout.tex
+++ b/bezout.tex
@@ -164,6 +164,41 @@ not be greater than the product over all singular values \cite{Weyl_1912_Das}.
Therefore, the absence of zero eigenvalues implies the absence of zero singular
values.
+% This is kind of a boring definition...
+\begin{equation} \label{eq:count.def.marginal}
+ \overline{\mathcal N}(\kappa,\epsilon)
+ =\int da\,\overline{\mathcal N}(\kappa,\epsilon,a)
+\end{equation}
+
+\begin{equation} \label{eq:count.zero.energy}
+ \overline{\mathcal N}(\kappa,0,a)
+ =\left[(p-1)a^{p-1}\sqrt{\frac{1-a^{-2}}{a^{2(p-1)}-|\kappa|^2}}\right]^N
+\end{equation}
+
+\begin{equation}
+ \overline{\mathcal N}(\kappa,\epsilon)
+ =\lim_{a\to\infty}\overline{\mathcal N}(\kappa,\epsilon,a)
+ =(p-1)^N
+\end{equation}
+
+For $|\kappa|<1$,
+\begin{equation}
+ \lim_{a\to1}\overline{\mathcal N}(\kappa,\epsilon,a)
+ =0
+\end{equation}
+
+\begin{equation}
+ \lim_{a\to1}\overline{\mathcal N}(1,0,a)
+ =(p-1)^{N/2}
+\end{equation}
+
+\begin{equation} \label{eq:threshold.energy}
+ |\epsilon_{\mathrm{th}}|^2
+ =\frac{p-1}{2p}\frac{(1-|\delta|^2)^2a^{p-2}}
+ {1+|\delta|^2-2|\delta|\cos(\arg\kappa+2\arg\epsilon)}
+\end{equation}
+for $\delta=\kappa a^{-(p-2)}$.
+
\bibliographystyle{apsrev4-2}
\bibliography{bezout}