diff --git a/main.tex b/main.tex index 31376b69684f59385589f5c5efce8144f7332cd8..77c2cedef6e7ec8f85f4634143eae8afb53e304c 100644 --- a/main.tex +++ b/main.tex @@ -1032,6 +1032,10 @@ Where $\epsilon_q^1$ and $\epsilon_q^2$ are defined as follows: \end{lemmadfn} +It's worth noting that $\epsilon_q^2$ is potentially larger than $\epsilon_q^2$ +but calculating it involves a $\gcd$, a modulo reduction, and a modulo $n$ +inverse, for each $q$ considered. + \begin{proof} Consider the following tautology: @@ -1070,20 +1074,16 @@ Consider the following tautology: \label{eqn:better_eps_problem_k_mod_n} \end{align} -In our situation, we want to find a gap between the right-hand-side of eqn -\ref{eqn:positive_rad_condition_in_terms_of_q_beta}, -and the least element of $\frac{1}{m}\ZZ$ which is strictly greater. -This amounts to finding the least $k \in \ZZ_{>0}$ for which -eqn \ref{eqn:finding_better_eps_problem} holds. +In our situation, we want to find the least $k$ satisfying +eqn \ref{eqn:finding_better_eps_problem}. Since such a $k$ must also satisfy eqn \ref{eqn:better_eps_problem_k_mod_n}, -we can pick the smallest $k \in \ZZ_{>0}$ which satisfies this new condition -(a computation only depending on $q$ and $\beta$, but not $r$) -and we are then guaranteed that the gap is at least $\frac{k}{2mn^2}$. - -A potentially larger gap can also be guaranteed if we choose the least -$k \in \ZZ_{>0}$ satisfying eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn} -instead, but at the cost of computing several $\gcd$'s and modulo reductions -for each $q$ considered. +we can pick the smallest $k_q^1 \in \ZZ_{>0}$ which satisfies this new condition +(a computation only depending on $q$ and $\beta$, but not $r$). +We are then guaranteed that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^1$. +Furthermore, $k$ also satisfies +eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn} +so we can also pick the smallest $k_q^2 \in \ZZ_{>0}$ satisfying this condition, +which also guarantees that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^2$. \end{proof}