diff --git a/main.tex b/main.tex index ad2fceda11b5d96a0547774519dd5262a0390ed7..231d55c6441435019445a98b725c7077f304265f 100644 --- a/main.tex +++ b/main.tex @@ -24,6 +24,7 @@ \newcommand{\minorheading}[1]{{\noindent\normalfont\normalsize\bfseries #1}} \newtheorem{theorem}{Theorem}[section] +\newtheorem{lemmadfn}{Lemma/Definition}[section] \begin{document} @@ -985,7 +986,36 @@ rhs_numerator = ( \noindent Let $\aa^{'}$ be an integer representative of $\aa^{-1}$ in $\ZZ/n\ZZ$. -Considering the following tautology: +Next, we seek to find a larger $\epsilon$ to use in place of $\epsilon_F$ in the +proof of theorem \ref{thm:rmax_with_uniform_eps}: + +\begin{lemmadfn}[ + Finding better alternatives to $\epsilon_F$: + $\epsilon_q^1$ and $\epsilon_q^1$ +] +Suppose $d \in \frac{1}{m}\ZZ$ is satisfies the condition in +eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}. +That is: + +\begin{equation*} + \sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()} +\end{equation*} + +\noindent +Then we have: + +\begin{equation*} + d - \frac{(\aa r + 2\bb)\aa}{2n^2} + \geq \epsilon_q^2 \geq \epsilon_q^1 > 0 +\end{equation*} + +Where $\epsilon_q^1$ and $\epsilon_q^2$ are defined as follows: + +\end{lemmadfn} + +\begin{proof} + +Consider the following tautology: \begin{align} &\frac{ x }{ m } @@ -1021,12 +1051,12 @@ Considering the following tautology: \label{eqn:better_eps_problem_k_mod_n} \end{align} -In our situation, we want to find the gap between the right-hand-side of eqn +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. -Since such a $k$ satisfies eqn \ref{eqn:better_eps_problem_k_mod_n}, +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 $\frac{k}{2mn^2}$. @@ -1036,6 +1066,7 @@ $k \in \ZZ_{>0}$ 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. +\end{proof}