% % Presentation pour banff. % % cw june 07 % \documentclass{beamer} % [handout]{beamer} \usepackage{times} %\usepackage{graphicx} \usepackage{amsmath,amssymb,amsthm,bbm} %,calc,picins} \usepackage[mathcal]{euscript} \usepackage{mathrsfs} \setlength{\parindent}{0pt} %-- cyrillic \DeclareFontEncoding{OT2}{}{} % to enable usage of cyrillic fonts \newcommand{\textcyr}[1]{% {\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}% \selectfont #1}} \newcommand{\Sha}{{\mbox{\textcyr{Sh}}}} \newcommand{\Deltabar}{\overline{\Delta}} %-- nicer Greek letters \usepackage{pifont} \renewcommand{\Lambda}{\mbox{{\Pifont{psy}L}}} \renewcommand{\Gamma}{\mbox{\Pifont{psy}G}} %------ arrows %\newarrow{MapsTo}|---> %\newarrow{CorrTo}<---> % --------------------------------- \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant} \newcommand{\QQ}{\mathbb Q} \newcommand{\FF}{\mathbb F} \newcommand{\ZZ}{\mathbb Z} \newcommand{\CC}{\mathbb C} \newcommand{\PP}{\mathbb P} \newcommand{\Fp}{\mathbb F_{\! p}} \newcommand\vu\upsilon \newcommand{\sss}{\scriptscriptstyle} \newcommand{\fE}{f_{\sss E}} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Reg}{Reg} \DeclareMathOperator{\an}{an} %\DeclareMathOperator{\rank}{rank} %-- Colours \definecolor{dgreen}{rgb}{0,0.3,0} \definecolor{dblue}{rgb}{0,0,0.7} \definecolor{dred}{rgb}{0.7,0,0} \definecolor{lightgrey}{rgb}{0.9,0.9,0.9} \setbeamercolor{sagecolors}{fg=dblue,bg=lightgrey} \newcommand{\dblue}{\color{dblue}} \newcommand{\dred}{\color{dred}} \newcommand{\dgreen}{\color{dgreen}} % sage env ----- \newcommand{\s}{\underline{ }} \newcommand{\p}{\^{ }} \newcommand{\sage}[2]{% \begin{center} \begin{minipage}[h]{10cm} \begin{beamercolorbox}[rounded=true,shadow=true]{sagecolors}% \begin{flushleft}% \vspace{-1.1cm} \texttt{\small\color{dred}% \begin{semiverbatim}sage : #1\end{semiverbatim}}% \end{flushleft}% \vspace{-5mm} \begin{gather*}% #2% \end{gather*}% \vspace{-5mm} \end{beamercolorbox}% \end{minipage} \end{center} } \newcommand{\sageinput}[1]{% \begin{center} \begin{minipage}[h]{10cm} \begin{beamercolorbox}[rounded=true,shadow=true]{sagecolors}% \begin{flushleft}% \vspace{-1cm} \texttt{\tiny\color{dred}% \begin{semiverbatim}sage : #1\end{semiverbatim}}% \end{flushleft}% \end{beamercolorbox}% \end{minipage} \end{center} } % titlepage ----------------------- \title{\Large\bf Elliptic Curves and Convergence:\\Sato-Tate, GRH, and BSD} \author{\large William Stein (joint work with Barry Mazur, with input from Andrew Granville,Chris Swierczewski and Tom Boothby)} \date{2007-10-16} % -------------------------------- \begin{document} \maketitle \begin{frame} \frametitle{Purpose} \vfill \Large Find a possible {\dred ``next question to ask''}, now that so much is understood {\dblue about the Sato-Tate conjecture} due to work of Taylor, Haris, et al. \vfill More generally study the general notion of {\dblue rate of convergence} in the context of elliptic curves. \end{frame} \begin{frame} \frametitle{Hecke Eigenvalues} \large Let $E$ be a {\bf non-CM} elliptic curve over $\QQ$, and $$ a_p = p+1-\#E(\mathbf{F}_p). $$ \vfill {\bf Theorem (Hasse):} $\displaystyle -1 < \frac{a_p}{2\sqrt{p}} < 1$. \vfill {\bf\dred Sato and Tate:} How are these numbers distributed? A conjecture... \begin{center} \includegraphics[height=1.25in]{graphics/mikio_sato} \hspace{1in}\includegraphics[height=1.25in]{graphics/tate_on_slide} \end{center} \end{frame} \begin{frame} \frametitle{Convergence to the semicircle distribution} \large The following slides each contain 8 plots. Each plot displays the distribution of normalized $a_p$ for the lowest conductor elliptic curves of different rank and all $a_p$ for $p < C$, for $C=10^3, 10^4, 10^5, 10^6$. \begin{center} \begin{tabular}{ccc} Rank 0 & Rank 1 & Rank 2 \\ Rank 3 & Rank 4 & Rank 5 \\ Rank 6 & Rank 7 & Rank 8 \\ \end{tabular} \end{center} \end{frame} \begin{frame}[fragile] \frametitle{Sato-Tate Frequency Histograms: $C=10^3$} \includegraphics[width=4in]{graphics/hist-1000.png} \end{frame} \begin{frame} \frametitle{Sato-Tate Frequence Histograms: $C=10^4$} \includegraphics[width=4in]{graphics/hist-10000.png} \end{frame} \begin{frame} \frametitle{Sato-Tate Frequence Histograms: $C=10^5$} \includegraphics[width=4in]{graphics/hist-100000.png} \end{frame} \begin{frame} \frametitle{Sato-Tate Frequence Histograms: $C=10^6$} \includegraphics[width=4in]{graphics/hist-1000000.png} \end{frame} \begin{frame} \frametitle{Quantify the convergence?} \LARGE Barry Mazur: ``How can we precisely quantify the convergence of the {\dblue blue data} to the {\dred red semicircle} theoretical distribution?'' \end{frame} \begin{frame} \frametitle{Some Functions (copy on blackboard)} $E$ an elliptic curve; $a_p = p+1-\#E(\mathbf{F}_p)$ \vfill \begin{itemize} \item $\displaystyle X(T) = \frac{\int_{-1}^T \sqrt{1-x^2}dx}{\int_{-1}^1 \sqrt{1-x^2}dx}= \text{ area under arc of semicircle }$ \vfill \item $\displaystyle Y_{C}(T) = \frac{\#\left\{\text{primes }p < C : -1 < \frac{a_p}{2\sqrt{p}} < T\right\}} {\#\left\{\text{primes }p < C \right\}}$. \vfill \item $\displaystyle \Delta(C) = \sqrt{\int_{-1}^1 (X(T) - Y_{C}(T))^2 dT} =$ the $L_2$-norm of the difference of $X(T)$ and $Y_{C}(T)$, and $\Delta(C)_{\infty}$ the $L_{\infty}$-norm. \end{itemize} \end{frame} \begin{frame} \frametitle{The Sato-Tate Conjecture} Let $\Delta(C)_{\infty}$ be the max of the difference between the theoretical semicircle distribution and actual data using primes up to $C$. \Huge {\bf Sato-Tate Conjecture:} $$ \lim_{C\to\infty} \Delta(C)_{\infty} = 0 $$ \vfill \Large {\bf\dblue Theorem (Taylor, M.~Harris, et al.):} If $E$ has multiplicative reduction at some prime, then the Sato-Tate conjecture is true. [Key part of proof is to establish certain analytic properties of symmetric power $L$-functions.] \end{frame} \begin{frame}[fragile] \frametitle{Plotting $\Delta$ (up to $10^3$)} {\dblue \begin{verbatim} sage: e37a = SatoTate(EllipticCurve('37a'), 10^6) sage: show(e37a.plot_Delta(10^3, plot_points=400, max_points=100), ymax=0.1, ymin=0, figsize=[10,3]) \end{verbatim} } \includegraphics[width=1.2\textwidth]{graphics/37a-plot_Delta-1000} The {\dred red line is $\Delta(C)_{\infty}$} and the {\dblue blue line is $\Delta(C)$}. By Sato-Tate, they both go to $0$ as $C\to\infty$. \end{frame} \begin{frame}[fragile] \frametitle{Plotting $\Delta$ (up to $10^4$)} {\dblue \begin{verbatim} sage: e37a = SatoTate(EllipticCurve('37a'), 10^6) sage: show(e37a.plot_Delta(10^4, plot_points=200, max_points=100), ymax=0.1, ymin=0, figsize=[10,3]) \end{verbatim} } \includegraphics[width=1.2\textwidth]{graphics/37a-plot_Delta-10000} The {\dred red line is $\Delta(C)_{\infty}$} and the {\dblue blue line is $\Delta(C)$}. By Sato-Tate, they both go to $0$ as $C\to\infty$. \end{frame} \begin{frame}[fragile] \frametitle{Plotting $\Delta$ (up to $10^5$)} {\dblue \begin{verbatim} sage: e37a = SatoTate(EllipticCurve('37a'), 10^6) sage: show(e37a.plot_Delta(10^5, plot_points=200, max_points=100), ymax=0.1, ymin=0, figsize=[10,3]) \end{verbatim} } \includegraphics[width=1.2\textwidth]{graphics/37a-plot_Delta-100000} The {\dred red line is $\Delta(C)_{\infty}$} and the {\dblue blue line is $\Delta(C)$}. By Sato-Tate, they both go to $0$ as $C\to\infty$. \end{frame} \begin{frame}[fragile] \frametitle{Plotting $\Delta$ (up to $10^6$)} {\dblue \begin{verbatim} sage: e37a = SatoTate(EllipticCurve('37a'), 10^6) sage: show(e37a.plot_Delta(10^6, plot_points=200, max_points=100), ymax=0.1, ymin=0, figsize=[10,3]) \end{verbatim} } \includegraphics[width=1.2\textwidth]{graphics/37a-plot_Delta-1000000} The {\dred red line is $\Delta(C)_{\infty}$} and the {\dblue blue line is $\Delta(C)$}. By Sato-Tate, they both go to $0$ as $C\to\infty$. \end{frame} \begin{frame} \frametitle{``The next question to ask...''} \LARGE {\bf\dred QUESTION:} What about the speed of convergence? I.e., {\em how} does $\Delta(C)$ or $\Delta(C)_{\infty}$ converge to 0? \vfill \includegraphics[width=\textwidth]{graphics/37a-plot_Delta-1000000} \end{frame} \begin{frame} \frametitle{The Akiyama-Tanigawa Conjecture} \Large {\dred \bf Conjecture (Akiyama-Tanigawa [Math Comp., 1999]):} For every $\epsilon>0$, for $C\gg0$ we have $$\Delta(C)_{\infty} \leq \frac{1}{C^{1/2-\epsilon}}.$$ \vfill {\dblue Theorem (A-T):} This conjecture implies the {\dgreen Generalized Riemann Hypothesis} for $L(E,s)$. \vfill See Barry Mazur's forthcoming Notices paper for more discussion, references, and pretty pictures. \end{frame} \begin{frame}[fragile] \frametitle{Converse} {\dred Possibly GRH implies the above conjecture}: \small \begin{verbatim} From: Shigeki Akiyama Date: Sun, 30 Sep 2007 08:17:02 +0900 Dear Professor Mazur I feel very honored to have your comments on our old experimental paper. I was very pleased to read your expository paper itself, of course including subsections you wrote us. I did not consider the error term problem in this comprehensive manner, My only comment is that a partial converse is true. If we assume Riemann hypothesis for all symmetric L, then the conjecture is valid for L_{0,1}. This is a claim from H. Nagoshi and basically comes from Erdos-Turan inequility as far as I remember... We did not explore nor publish this observation. \end{verbatim} \end{frame} \begin{frame} \frametitle{Log Plots} \Large Let's test out Akiyama-Tanigawa, instead of plotting $\Delta(C)$ which just goes to $0$ quickly, {\dblue we instead plot $-\log_C(\Delta(C))$}. \begin{enumerate} \item How does this function {\dred compare} to $\displaystyle \frac{1}{2}$? I.e., does it eventually get within $\epsilon$ of $\frac{1}{2}$. \item Can we find a simple function that conjecturally nicely {\dred approximates} $-\log_C(\Delta(C))$? \end{enumerate} \end{frame} \begin{frame}[fragile] \frametitle{Rank 0 curve 11a; $p < 10^6$; with 300 sample points} \includegraphics[width=\textwidth]{graphics/11a-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 1 curve 37a; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/37a-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 2 curve 389a; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/389a-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 3 curve 5077a; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/5077a-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 4 curve [1,-1,0,-79,289]; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank4-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 5 curve [0, 0, 1, -79, 342]; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank5-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 6 curve [1, 1, 0, -2582, 48720]; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank6-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 7 curve [0, 0, 0, -10012, 346900]; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank7-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Rank 8 curve [0, 0, 1, -23737, 960366]; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank8-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Elkies rank $\geq 28$ curve; $p < 10^6$} \includegraphics[width=\textwidth]{graphics/rank28-theta-1000000} \vfill \begin{itemize} \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item {\dred Red line is $1/2$.} \end{itemize} {\bf OK, are those lines really going up to $1/2$???} \end{frame} \begin{frame} \frametitle{Understanding the Data Better?} \Large Can one {\dred predict} the asymptotic shape of the curve $\Delta(C)$, say, in terms of either arithmetic invariants of the curve or perhaps in terms of zeros of $L(E,s)$ on the critical strip? \vfill For some curves $\Delta(C)$ is quickly very close to $1/2$, e.g., the curves of rank $0$ and $1$ above. \end{frame} \begin{frame}[fragile] \frametitle{Fitting the ``random'' Rank 0 curve $y^2 = x^3 +19x + 234$} \includegraphics[width=\textwidth]{graphics/rank0-large_cond-fit} \vfill \begin{itemize} \item The black curve is $$ \frac{1}{2} - \frac{1}{\log(X)}. $$ \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \item Conductor = $24093568 = 2^7 \cdot 41 \cdot 4591$ \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Low zeros?} \small {\dblue \begin{verbatim} sage: EllipticCurve('11a').Lseries_zeros(10) [6.36261389, 8.60353962, 10.0355091, 11.4512586, 13.5686391, 15.9140726, 17.0336103, 17.9414336, 19.1857250, 20.3792605] \end{verbatim} } {\dred \begin{verbatim} sage: EllipticCurve([19,234]).Lseries_zeros(10) [0.255961213, 0.739839807, 1.03144159, 1.78804887, 2.11227980, 2.42762599, 3.11102036, 3.26810134, 3.68155235, 4.13888170] \end{verbatim} } \end{frame} \begin{frame}[fragile] \frametitle{Fitting the Rank 3 Curve 5077a} \includegraphics[width=\textwidth]{graphics/rank3-fit-1000000} \vfill \begin{itemize} \item The black curve is $$ \frac{1}{2} - \frac{3/3}{\log(X)}. $$ \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Fitting the Rank 4 [1,-1,0,-79,289]; $p< 10^6$} \includegraphics[width=\textwidth]{graphics/rank4-fit} \vfill \begin{itemize} \item The black curve is $$ \frac{1}{2} - \frac{4/3}{\log(X)}. $$ \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Fitting Rank 8 [0, 0, 1, -23737, 960366]; $p< 10^6$} \includegraphics[width=\textwidth]{graphics/rank8-fit} \vfill \begin{itemize} \item The black curve is $$ \frac{1}{2} - \frac{19/9}{\log(X)}. $$ \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \end{itemize} \end{frame} \begin{frame}[fragile] \frametitle{Fitting Rank 28 curve; $p< 10^6$} \includegraphics[width=\textwidth]{graphics/rank28-fit} \vfill \begin{itemize} \item The black curve is $$ \frac{1}{2} - \frac{28/9}{\log(X)}. $$ \item {\dgreen Green line is $-\log_C(\Delta(C)_\infty)$.} \item {\dblue Blue line is $-\log_C(\Delta(C))$}, with a grey tubular {\em numerical integration error bound}. \end{itemize} \end{frame} \begin{frame} \frametitle{Conjectural convergence of the measure of convergence} {\bf\dblue Conjecture (Stein):} For any $E$ there is a constant $\alpha$ such that $$ -\log_C(\Delta(C)) \geq \frac{1}{2} - \frac{\alpha}{\log(C)} $$ {\bf for all} $C$. \vfill For comparison, recall the {\dred Akiyama-Tanigawa conjecture} asserts that for all $\epsilon>0$, we have that $$ \Delta(C) \leq O\left(\frac{1}{C^{1/2-\epsilon}}\right) $$ Equivalently, $$ -\log_C( \Delta(C)) \gg 1/2 - \epsilon $$ \end{frame} \begin{frame} \frametitle{The Sato-Tate convergence parameter} \vfill For an elliptic curve $E$ let $\alpha(C)$ be the infimum of all constants that minimizes the $L_2$ norm of this (i.e. the distance between the black and blue curves above!): $$ -\log_C(\Delta(C)) - \left( \frac{1}{2} - \frac{\alpha(C)}{\log(C)} \right). $$ Thus $\alpha(C)$ is a function of $C$ (and the fixed curve $E$). \vfill \vfill {\bf \dblue Definition:} The {\em\dred Sate-Tate convergence parameter} of $E$ is $$ \alpha_E = \lim_{C\to\infty} \alpha(C). $$ \vfill (I don't know if this exists. replace by limsup and liminf?) \vfill {\bf \dred Challenge:} Find a conjectural formula for $k_E$ in terms the critical zeros of $L(E,s)$? \end{frame} \begin{frame} \frametitle{Another future direction...} We have {\dblue $$ X^{1/2 - 1/\log(X)} = \frac{X^{1/2}}{X^{1/\log(X)}} = e \cdot X^{1/2}. $$} We thus entertain the possibility (following the format of the people who work with random matrices etc.) that the true distribution is well approximated by something like {\dblue $$ a \cdot (\log X)^b \cdot X^c $$ } \noindent for appropriate constants $a,b,c$. So for the rank $3$ example above we might choose $$ a= e, \qquad b= 0, \qquad c = 1/2, $$ but there may be better choices? \end{frame} \begin{frame} \frametitle{More future direction...} \Large \vfill \begin{enumerate} \item Restrict to intervals $[a,b]\subset (-1,1)$. (This seems to have little to know impact.) \vfill \item Push computations much further (next slide). \end{enumerate} \vfill \end{frame} \begin{frame} \frametitle{Pushing Computations Further} \begin{enumerate} \item {\dred Drew Sutherland} (of MIT) has some amazingly fast {\em parallel} C code for computing all $a_p$ for $p 0$. \item So study the rate of {\em divergence} of the sum $$F(0) = \sum \frac{a_{p^n}}{p^n} \log(p).$$ \end{enumerate} \end{frame} \begin{frame} \frametitle{A Numerical Experiment} Let $$R_E(C) = \sum_{p^n\leq C} \frac{a_{p^n}}{p^n} \log(p).$$ \vfill {\dblue Guess:} $R_E(C) \sim \alpha \log(C)^\beta$, where $\beta$ depends only on the rank of $E$ and $\alpha$ depends on the arithmetic invariants of $E$. \vfill {\dblue Experiment:} Compute $\log(R_E(C)) / \log(\log(C))$. Does this depend only on rank of $E$? \vfill Next we give some {\dred data}. In each case I give several curves with a given rank, along with the value of the above quantity for $C = 10^6$: \vfill \end{frame} \begin{frame}[fragile] \begin{verbatim} curve rank log(R_E(C)) / log(log(C)) for C=10^6 37a1 1 0.622551283326 43a1 1 0.664628966956 53a1 1 0.64056834932 57a1 1 0.743607790253 58a1 1 0.639927175062 61a1 1 0.776549775927 65a1 1 0.717652219993 389a1 2 1.00758391471 433a1 2 0.988605592917 446d1 2 1.0273311084 563a1 2 0.987041109677 571b1 2 0.919099487872 643a1 2 0.889281143176 655a1 2 0.925749865705 664a1 2 0.957156816404 \end{verbatim} \end{frame} \begin{frame}[fragile] \begin{verbatim} curve rank log(R_E(C)) / log(log(C)) for C=10^6 5077a1 3 1.16071903587 11197a1 3 1.14902783005 11642a1 3 1.16976814614 12279a1 3 1.13108926023 13766a1 3 1.14886584781 16811a1 3 1.04598722161 18097b1 3 1.13427759105 18562c1 3 1.12453834551 234446a1 4 1.20905312451 19047851a1 5 1.29409998755 5187563742a1 6 1.34691224576 \end{verbatim} \end{frame} \begin{frame} Because of the log-log's, etc., I'm probably getting 2 digits correct above usually, from the huge sum. In pictures though, the lines are quickly fairly horizontal (so the true limit is likely close to the above numbers). The striking thing about the above clumps of numbers (for each rank), is they all lie in disjoint intervals. \end{frame} \end{document} = Another perhaps related question =