\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{amssymb} \usepackage{pstricks} \usepackage{graphicx} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.5in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\cC}{\mathcal{C}} \newcommand{\cS}{\mathcal{S}} \newcommand{\ie}{\emph{i.e.}} \newcommand{\al}{\alpha} \newcommand{\hs}{h^{(s)}} \newcommand{\Ren}{R{\'e}nyi} \newcommand{\con}{\operatorname{con}} \newcommand{\card}{\operatorname{card}} \newcommand{\id}{\operatorname{id}} \renewcommand{\ge}{\geqslant} \renewcommand{\le}{\leqslant} \newcommand{\eps}{\varepsilon} \newcommand{\Fsl}{F_{s,\lambda}} \newcommand{\Fs}{F_{s}} \newcommand{\Fu}{F_{u}} \newcommand{\G}{\Gamma} \newcommand{\Gu}{G_{u}} \begin{center} \vskip 1cm{\LARGE\bf On the Entropy of Curves } \vskip 1cm \large Michael Maurice Dodson\\ University of York\\ Heslington \\ York YO10 5DD \\ United Kingdom \\ \href{mailto:mmd1@york.ac.uk}{\tt mmd1@york.ac.uk} \\ \ \\ Michel Mend\`es France \\ Universit\'e Bordeaux I \\ 351 cours de la Lib\'eration\\ 33405 Talence Cedex \\ F-33405 France \\ \href{mailto:Michel.Mendes-France@math.u-bordeaux1.fr}{\tt Michel.Mendes-France@math.u-bordeaux1.fr} \end{center} \centerline{\it To Jean-Paul Allouche for his 60th birthday} \vskip .2 in \begin{abstract} Using geometric probability, we apply the formal definitions of Shannon entropy and \Ren's generalization to study the complexity of planar curves of finite length within a convex set. The bounds for the Shannon and \Ren{} entropies depend on the arc length of the curve and on that of the boundary of the convex set; they involve a Gibbs distribution and a power law distribution, respectively. We also obtain explicit formulae for the two entropies and determine convex sets that maximize the entropy of curves. \end{abstract} \section{Introduction} Planar curves range from a simple straight line segment or a circle to a complicated tangle. Their complexity has been analysed using both the thermodynamic concept of Shannon entropy~\cite{mmfNTDS,mmf91} and the geometrical concept of dimension~\cite{dmf81,dmft83}. Examples can be found in various other domains~\cite{amt2011,bcc09,cauchy1841,cpmmfjpbp94,cd02,crofton1868,mmf06,mmftt}. A brief overview of the fundamental ideas is given and the Shannon and the more general Shannon-\Ren{} entropies are defined. The question of bounds for these entropies is considered; the analysis of the general \Ren{} entropy turns out to be quite different from that of the Shannon entropy and involves a modified Hurwitz zeta function. A more precise viewpoint which gives exact formulae for the entropies is then developed. \section{Geometric probability} \label{sec:geomprob} Given a plane curve $\Gamma$ of finite length $|\Gamma|$, we recall the classical definition of the probability that a straight line $D$ intersects $\Gamma$ in exactly $n$ points~\cite{mmfNTDS,Santalo06}. In the plane, a straight line $D$ that does not pass through the origin $O$ is determined by the polar coordinates $(\omega, \rho)$, where $\omega$, $0\le \omega< 2\pi$, is the angle the normal to $D$ makes with the $x$-axis and $\rho>0$ is the distance from $O$. Straight lines that pass through the origin are not counted since, as we will see, the family of such lines has measure 0. \begin{figure} \begin{center} \begin{pspicture}(-2,-1)(2,3.5) %\showgrid \psset{unit=0.6cm} \psline(-4.25,0)(4.3,0) %x axis \rput(4.2,-0.3){$x$} \rput(-2.6,-0.5){$0$} %ORIGIN \psline(-2.5,-.2)(-2.5,6) %y axis \rput(-2.75,5.9){$y$} \rput(-1.5,0.4){$\omega$} % %\rput(-3,5){$(0,1)$} \psline(-3.2,5.8)(3.9,-0.2) %DROITE D \rput(2.5,2){$D$} %DROITE \rput(-1.4,1.72){$\rho$} \psline(-2.55,0)(.1,3.0) %orthogonal to D \rput(1.2,3.1){$(\omega,\rho)$} \end{pspicture} \end{center} \caption{A straight line $D$ determined by polar coordinates $(\omega, \rho)$ } \end{figure} In the $(\omega, \rho)$ plane, a point in the strip $\cS:= [0,2\pi)\times (0,\infty)$ represents a straight line not passing through the origin in the usual $(x,y)$-plane. The measure $\mu$ in the strip is defined by \begin{equation*} d\mu := d\rho\, d\omega, \end{equation*} i.e., the usual Lebesgue measure. Now let $K$ be a bounded convex set in the $(x,y)$-plane and let $F(K)$ be the set of straight lines that intersect $K$. The $\mu$-measure of $F(K)$ is known to be equal to the length $|\partial K|$ of the boundary $\partial K$ of $K$, i.e., \begin{equation*} \mu(F(K)) = |\partial K|. \end{equation*} It follows that the measure given by \begin{equation*} dp = \frac{d\mu}{|\partial K|} = \frac{ d\rho\, d\omega}{|\partial K|} \end{equation*} is a probability measure defined on $F(K)$. Suppose $K_0\subseteq K$ is a convex subset of $K$. Then $|\partial K_0|\le |\partial K|$ and the probability that a straight line meeting $K$ also meets $K_0$ is \begin{equation*} \label{eq:prob} \frac{|\partial K_0|}{|\partial K|}. \end{equation*} Consequently the probability that a straight line meeting $K$ misses $K_0$ is \begin{equation} \label{eq:p0} 1- \frac{|\partial K_0|}{|\partial K|}. \end{equation} Consider a curve $\Gamma_0$ of finite length lying in the convex set $K$. Denote by $p_n(\Gamma_0,K)$ the probability that a straight line $D$ meeting $K$ intersects $\Gamma_0$ in exactly $n$ points. Observe that if $K=K_0=\con(\Gamma_0)$, the convex hull of $\Gamma_0$, then $D$ must hit $\Gamma_0$, so that $p_0(\Gamma_0,K_0)=0$. By the definition of probability, \begin{equation} \label{eq:sumpn1} \sum_{n=0}^\infty p_n(\G_0,K) =1. \end{equation} A classical result of H.~Steinhaus~\cite{Steinhaus54} states that the expected number $\lambda$ say of intersection points of $\Gamma_0$ with random straight lines is \begin{equation} \label{eq:expectation} \lambda :=\sum_{n=1}^\infty n\, p_n(\G_0,K) = \frac{2|\Gamma_0|}{|\partial K|}. \end{equation} The next moment $ \sum_{n=1}^\infty n^2\, p_n(\G_0,K) $ corresponds to the ``energy'' of the curve but is much less well-behaved and will not be considered. \section{Entropy} \label{sec:entropy} The celebrated Shannon entropy of thermodynamics and information theory has a generalization due to \Ren~\cite{RenyiWahrs,RenyiPT}. Their definitions are used in conjunction with geometric probability to define formally notions of entropy for curves of finite length within a convex set $K$. Note that unlike the thermodynamic setting, there is no underlying mechanism or dynamic here that leads to the entropies increasing. \subsection{The Shannon entropy} The sequence $( p_n(\G_0,K)\colon n=0,1,2,\dots)$ of probabilities of the number of intersection points can be substituted into the formula for Shannon entropy. To be precise, the {\em{Shannon entropy}} $h(\Gamma_0,K)$ of the curve $\Gamma_0$ relative to a convex set $K$ containing $\G_0$ is defined by the formula \begin{equation} \label{eq:shent} h(\Gamma_0,K)=\sum_{n=0}^\infty p_n(\Gamma_0,K) \log \frac1{p_n(\Gamma_0,K)} = \sum_{n=0}^\infty p_n(K) \log\frac1{p_n(K)}, \end{equation} where as usual we put $ p \log (1/p)=0$ if $p=0$ and where for simplicity here and elsewhere whenever the argument is clear, we suppress $\G_0$ or $K$ and write the probability \begin{equation*} p_n(\Gamma_0,K) = p_n(K) {\text{ or }} p_n. \end{equation*} Similarly we will write the Shannon entropy \begin{equation*} h(\Gamma_0,K) = h(K) = h. \end{equation*} \subsection{The R{\'e}nyi entropy} \Ren~\cite{RenyiWahrs,RenyiPT} has defined a form of entropy $h^{(s)}$, where $s\ge 0$, by \begin{equation} \label{eq:rents} h^{(s)} := \frac1{1-s} \log\, \sum_{n=0}^\infty p_n^{s}, \end{equation} which tends to the Shannon entropy $h$ as $s\to 1$: \begin{equation} \label{eq:rent1} h^{(1)} := \lim_{s\to 1} h^{(s)} = \lim_{s\to 1} \left(\frac1{1-s} \log\, \sum_{n=0}^\infty p_n^{s}\right) = \sum_{n=0}^\infty p_n \log 1/p_n=h\,. \end{equation} In addition it is readily verified that $d\,h^{(s)}/ds=0$ at $s=1$ and that $h^{(s)}$ is minimal for $s=1$. The \Ren{} entropy can also be applied formally to curves. We will write \begin{equation} \label{eq:rentcurve} \hs(\G_0,K) = \frac1{1-s} \log\, \sum_{n=0}^\infty p_n^{s}(\G_0,K) \end{equation} for the \Ren{} entropy for a curve $\G_0$ relative to a convex set $K\supseteq \G_0$ and, as with the Shannon entropy, we will write \begin{equation*} h^{(s)}(\G_0,K) = h^{(s)}(K) = h^{(s)} \end{equation*} when the meaning is clear. The two series have finitely many terms for an algebraic curve, since $p_n=0$ for all $n$ larger than its degree, but they converge (absolutely) in general by the following results. A separate argument is needed for the two entropies. The Shannon entropy ($s=1$) has been treated in this context~\cite{mmfNTDS,mmf91} and is now discussed for completeness. \subsection{A bound for the Shannon entropy of curves} \label{ssec:h1curves} \begin{theorem} \label{thm:h(K)} The Shannon entropy $h(\Gamma_0,K)$ of a curve $\Gamma_0$ of finite length lying in the convex set $K$ satisfies \begin{equation} \label{eq:h(K)est1} h(\G_0,K)\le \log \left(\frac{2|\Gamma_0|}{|\partial K|} + 1\right) + \frac{2|\Gamma_0|}{|\partial K|} \log\left(1+\frac{|\partial K|}{2|\Gamma_0|}\right) \le \log \left(\frac{2|\Gamma_0|}{|\partial K|} + 1\right) + 1 \end{equation} and more precisely in the case $K=K_0=\con(\G_0)$ (so that $p_0=0$), \begin{eqnarray} \label{eq:h(K)est2} h(\G_0,K_0)&\le& \log \left(\frac{2|\Gamma_0|}{|\partial K_0|}\right) + \left(\frac{2|\Gamma_0|}{|\partial K_0|}-1\right) \log \left(\frac{2|\Gamma_0|}{2|\Gamma_0|-|\partial K_0|}\right) \notag \\ & \le &\log \left(\frac{2|\Gamma_0|}{|\partial K_0|}\right) + 1. \end{eqnarray} \end{theorem} \begin{proof} By~\eqref{eq:shent}, it suffices to bound the sum $ \sum_{n=k}^\infty p_n \log 1/p_n$, where $p_n\in [0,1]$ and $k=0$ in~\eqref{eq:h(K)est1} and $k=1$ in~\eqref{eq:h(K)est2}, subject to the two constraints~\eqref{eq:sumpn1} and~\eqref{eq:expectation}, i.e., \begin{equation*} \sum_{n=k}^\infty p_n =1 \end{equation*} and \begin{equation*} \sum_{n=k}^\infty n\,p_n= \frac{2|\Gamma_0|}{|\partial K|} = \lambda, \end{equation*} where $k=0$ or $1$. This is done using Lagrange multipliers. Let $\alpha,\beta\in\R$ and consider \begin{equation*} U=\sum_{n= k}^\infty \widehat{p}_n \log 1/\widehat{p}_n - \alpha \sum_{n=k}^\infty \widehat{p}_n - \beta\sum_{n=k}^\infty n\widehat{p}_n. \end{equation*} Then for each $n=k,k+1,\dots$, $\partial U/\partial \widehat{p}_n = 0$ implies \begin{equation*} -\log \widehat{p}_n-1 - \alpha - n\beta = 0, \end{equation*} whence the `maximal' probability distribution $\widehat{p}_n$ is a negative exponential with constant factor. More precisely, \begin{equation} \label{eq:sentdist} \widehat{p}_n = e^{-1-\alpha - n\beta} =Ce^{-\beta n}, \end{equation} where $C=e^{-1-\alpha}$ and $\widehat{p}_n$ is a Gibbs distribution~\cite{Schrod52}. Recall that Gibbs measure is a probability measure of importance in thermodynamics: it is the unique measure that maximizes the entropy for a given expected energy. It underlies maximum entropy methods and related algorithms and its appearance here is accordingly natural. The constant $C=e^{-1-\alpha}$ (or $\alpha$) is determined from the power series \begin{equation*} \sum_{n=k}^\infty \widehat{p}_n = C\sum_{n=k}^\infty e^{-\beta n} = 1 \end{equation*} which gives $C=C(\beta)=e^{\beta k}(1-e^{-\beta})$. In statistical mechanics, $C(\beta)$ is the reciprocal of the partition function and in physics $\beta$ corresponds to the inverse temperature~\cite{Schrod52}. Moreover $C\sum_{n=k}^\infty n\, e^{-\beta n} = \lambda$ whence \begin{equation*} \lambda=\frac1{1-e^{-\beta} }\left(k-(k-1)\,e^{-\beta }\right)= \frac1{e^{\beta} -1 }\left(k\,e^{\beta }-k+1\right). \end{equation*} Thus \begin{equation*} e^{\beta }=\frac{\lambda-k+1}{\lambda-k} \ {\text{and}} \ e^{\beta}- 1=\frac{1}{\lambda-k} . \end{equation*} Hence the entropy $h$ satisfies \begin{equation*} h\le C\sum_{n=k}^\infty e^{-\beta n}\log \frac{e^{\beta n}}{C} = H \end{equation*} say, so that \begin{eqnarray*} H&=&\frac{\beta}{1-e^{-\beta}}\left(k-(k-1)e^{-\beta}\right) + \log \frac{e^{-k\beta}}{1-e^{-\beta}} \\ &=& \frac{\beta}{e^{\beta}-1}\left(ke^{\beta}-k+1\right) -\beta\,k + \log \frac{e^{\beta}}{e^{\beta}-1}\\ &=& \log(\lambda-k+1) + (\lambda-k)\log \frac{\lambda-k+1}{\lambda-k}\\ &\le& \log(\lambda-k+1) + 1. \end{eqnarray*} \vspace{0.1in} {\bf{Case 1}} $k=0$, corresponds to the inequality~\eqref{eq:h(K)est1}. \vspace{0.05in} {\bf{Case 2}} $k=1$ corresponds to the inequality~\eqref{eq:h(K)est2}. \end{proof} \vspace{0.05in} \theoremstyle{remark} \begin{remark} The first inequality~\eqref{eq:h(K)est1} in the theorem shows that the entropy vanishes as $|\partial K|$ increases to infinity, a fact we shall rediscover in~\S\ref{sec:perspective}. \end{remark} \theoremstyle{remark} \begin{remark} In the above calculation, \begin{equation*} \lambda= \sum_{n=k}^\infty n\, \widehat{p}_n \ge k\sum_{n=k}^\infty \, \widehat{p}_n = k. \end{equation*} Thus obviously if $\lambda\to k$ from above, then $H=0$. Hence \end{remark} \begin{remark} In the case $k=1$, if $2|\Gamma|/|\partial K| \to 1$ from above, then $h=0$; which is otherwise obvious since $2|\Gamma|=|\partial K|$ implies $\Gamma$ is a segment of straight line. \end{remark} \subsection{ Bounds for the R\'enyi entropy of curves} \label{ssec:hscurves} Recall from equation~\eqref{eq:rentcurve} that the \Ren{} entropy $h^{(s)}(\Gamma_0,K)$ of $\G_0$ relative to the convex set $K\supset\G_0$ is given by \begin{equation*} h^{(s)}(\G_0,K) := \frac1{1-s} \log\, \sum_{n=0}^\infty p_n^{s}(\G_0,K). \end{equation*} For simplicity, we let $K=K_0=\con(\G_0)$, so that $p_0(\G_0,K_0)=0$, and denote the \Ren{} entropy $h^{(s)}(\Gamma_0,K_0)$ of $\G_0$ relative to its convex hull $K_0$ by $h^{(s)}_0$. Two bounds for the R{\'e}nyi entropy $ h^{(s)}_0$ are obtained, one for the range $01$. For convenience, the function $\Fs$ is simplified when $1/2\le s < 1$ by putting $u=1/(1-s)\in [2,\infty)$, so that $s/(1-s)=u-1$ and \begin{equation*} \label{eq:Fu} \Fs(\gamma)= \frac{ \zeta_1 (u-1;\gamma)}{ \zeta_1(u;\gamma)} := \Gu(\gamma)= G_{1/(1-s)}(\gamma). \end{equation*} Then for $u>2$ and $\gamma\ge 0$, \begin{eqnarray*} \label{eq:z>gz} (\gamma +1)\zeta_1(u;\gamma) &=& \sum_{k=1}^\infty \frac{\gamma+1}{(k+\gamma)^{u}} = \sum_{k=1}^\infty\left( \frac{k+\gamma}{(k+\gamma)^{u}} - \frac{k-1}{(k+\gamma)^{u}}\right) \\ &<& \sum_{k=1}^\infty \frac1{(k+\gamma)^{u-1}} =\zeta_1(u-1;\gamma), \end{eqnarray*} and $\Gu(\gamma):=\zeta_1(u-1;\gamma)/\zeta_1(u;\gamma) >1+\gamma$. A more precise estimate is possible using the familiar inequality \begin{equation} \label{eq:fests} \int_1^\infty f(x) dx \le \sum_{k=1}^\infty f(k) \le f(1) + \int_1^\infty f(x) dx \end{equation} for a positive increasing integrable function $f\colon\R\to [0,\infty)$. It follows from~\eqref{eq:fests} that \begin{equation} \label{eq:zests} \frac{(1+\gamma)^{-u+1}}{u-1} < \zeta_1(u,\gamma) := \sum_{k=1}^\infty \frac1{(k+\gamma)^u} < \frac1{(1+\gamma)^u}+ \frac1{(u-1)(1+\gamma)^{u-1}}. \end{equation} Hence for any fixed $u>2$, \begin{eqnarray*} \label{eq:lbG1} \Gu(\gamma) &=& \frac{\zeta_1(u-1,\gamma)}{\zeta_1(u,\gamma)} >\frac1{(u-2) (1+\gamma)^{u-2}}\, \left(\frac1{(1+\gamma)^u} + \frac1{(u-1)(1+\gamma)^{u-1}}\right)^{-1} \\ &>& \frac{(u-1)}{(u-2)}\, \frac{(1+\gamma)^2}{(u+\gamma)}. \end{eqnarray*} Thus given any $u>2$, for all sufficiently large $\gamma$, there exists an $\eps_u>0$ such that \begin{equation*} \Gu(\gamma) >(1+\eps_u)(1+\gamma). \label{eq:lbG} \end{equation*} It follows that given any real $\kappa$, $\eps_u(1+\gamma)>\kappa $ for all sufficiently large $\gamma$. Hence for fixed $u$ the inequality \begin{equation} \label{eq:G>g+k} \Gu(\gamma) > \gamma + \kappa \end{equation} holds for any $\kappa$ and all sufficiently large $\gamma$. This lower bound for the function $F_s= \Gu$ is used to show that the equation $F_s(\gamma)= \gamma +\lambda $, where $\lambda=2|\Gamma_0|/|\partial K_0|\in (0,\infty)$ (see~\eqref{eq:expectation}), is soluble when $\lambda\ge \Fs(0)=\Gu(0)$, i.e., when $\lambda\ge \zeta(u-1)/\zeta(u)> 1$. \begin{lemma} \label{lem:Flg} Suppose $\lambda\ge \Gu(0)$. Then there exists a unique $\gamma_0 \ge 0$ such that \begin{equation*} \label{eq:Flg} \Gu(\gamma_0) = \gamma_0 + \lambda. \end{equation*} \end{lemma} \begin{proof} Suppose without loss of generality that $\lambda>\Gu(0)$. By definition \begin{equation*} \Gu(0) - \lambda < 0. \end{equation*} But by equation~\eqref{eq:G>g+k}, for fixed $u$ and sufficiently large $\gamma$, \begin{equation*} \Gu(\gamma) >\gamma +\lambda \end{equation*} and by continuity, there exists a $\gamma_0>0$ such that $\Gu(\gamma_0) =\gamma_0 +\lambda $. The uniqueness of the root $\gamma_0$ follows from considering the derivative $\partial \Gu(\gamma)/\partial \gamma$ for $\gamma> 0$. The partial derivative $\partial\zeta_1(u;\gamma)/\partial\gamma = - u\,\zeta_1(u+1;\gamma)$. Thus \begin{eqnarray*} \label{eq:dFdg} \frac{\partial \Gu(\gamma)}{\partial \gamma} &=& \frac{\partial{}}{\partial\gamma} \frac{\zeta_1(u-1,\gamma)}{\zeta_1(u;\gamma)} = \frac{-(u-1)\, \zeta_1(u;\gamma) \zeta_1(u;\gamma) +u\,\zeta_1(u-1;\gamma)\zeta_1(u+1;\gamma)}{ \zeta_1(u;\gamma)^2}\\ &=& u\left(\frac{ \zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma)} { \zeta_1(u;\gamma)^2} - 1\right) +1>1 \end{eqnarray*} if $\zeta_1(u;\gamma)^2 < \zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma) $. Now by the Cauchy-Schwarz inequality, \begin{eqnarray*} \zeta_1(u;\gamma) &=& \sum_{k=1}^\infty \frac1{(k+\gamma)^{(u-1)/2}} \frac1{(k+\gamma)^{(u+1)/2}}\\ &\le& \left(\sum_{k=1}^\infty \frac1{(k+\gamma)^{u-1}} \right)^{1/2} \, \left(\sum_{k=1}^\infty \frac1{(k+\gamma)^{u+1}} \right)^{1/2} = \left(\zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma) \right)^{1/2} \end{eqnarray*} and $\zeta_1(u;\gamma)^2 <\zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma)$, since the inequality is readily seen to be strict. Hence $\partial \Gu(\gamma)/\partial \gamma > 1$, so that the graph of $\Gu$ crosses that of $\gamma\mapsto \gamma + \lambda$ no more than once. \end{proof} \begin{theorem} \label{thm:hal(K)} Let $\Gamma_0$ be a curve of finite length with convex hull $K_0$. Then for $0\frac{\zeta(s/(1-s))}{\zeta(1/(1-s))}>1, \end{equation} there exists a unique $\gamma_0$ such that \begin{equation} \label{eq:hs0bound2} h^{(s)}_0 \le \frac1{1-s} \log\left(\frac{2|\Gamma_0|}{|\partial K_0|} + \gamma_0\right) +\log\zeta_1\left(\frac1{1-s};\gamma_0\right). \end{equation} \end{theorem} \begin{proof} As with Shannon entropy, we use Lagrange multipliers under the same two constraints~\eqref{eq:sumpn1} and~\eqref{eq:expectation} to find a bound for $h^{(s)}_{0}$. Consider \begin{equation*} V=\frac1{1-s} \log \sum_{n= 1}^\infty \widehat{p}_n^{\,s} - \alpha \sum_{n=1}^\infty \widehat{p}_n - \beta\sum_{n=1}^\infty n\widehat{p}_n. \end{equation*} Then for each $n=1,2,\dots$, \begin{equation*} \frac{\partial V}{\partial \widehat{\,p}_n}= \frac{s \widehat{p}_n^{\,s-1} }{(1-s)\sum_{n= 1}^\infty \widehat{p}_n^{\,s}} -\al - \beta n. \end{equation*} Hence $\partial V/\partial \widehat{p}_n = 0$ implies \begin{equation*} \widehat{p}_n^{\, s-1} = \beta \left(\frac{1-s}{s}\right) \left(\sum_{n= 1}^\infty \widehat{p}_n^{\,s}\right) (\gamma+n), \end{equation*} where $\gamma=\al/\beta$. Hence the `maximal' distribution is given by \begin{equation} \label{eq:pna} \widehat{p}_n = \left(\frac{s}{\beta(1-s)}\right)^{1/(1-s)} \left(\sum_{n= 1}^\infty \widehat{p}_n^{\,s}\right)^{- 1/(1-s)} \left(\frac1{\gamma+n}\right)^{1/(1-s)}. \end{equation} We compute $\sum_{n= 1}^\infty \widehat{p}_n^{\,s} $ in two ways. First using~\eqref{eq:sumpn1}, \begin{eqnarray*} 1&=&\sum_{n=1}^\infty \widehat{p}_n = \left(\frac{s}{\beta(1-s)}\right)^{1/(1-s)} \left(\sum_{n= 1}^\infty \widehat{p}_n^{\,s}\right)^{- 1/(1-s)} \sum_{n=1}^\infty \left(\frac1{\gamma+n}\right)^{1/(1-s)}, \\ \end{eqnarray*} whence \begin{equation*} \left(\sum_{n= 1}^\infty \widehat{p}_n^{\,s}\right)^{ 1/(1-s)} = \left(\frac{s}{\beta(1-s)}\right)^{1/(1-s)} \zeta_1\left(\frac{1}{1-s};\gamma\right). \end{equation*} i.e., \begin{equation} \label{eq:pns3} \sum_{n= 1}^\infty \widehat{p}_n^{\,s} = \left(\frac{s}{\beta(1-s)}\right) \zeta_1\left(\frac{1}{1-s};\gamma\right)^{1-s}, \end{equation} where the right hand side converges for $s\in (0,1)$ and~\eqref{eq:hs0bound1} follows. \vspace{0.05in} Secondly, substitute~\eqref{eq:pns3} in~\eqref{eq:pna} to get \begin{equation*} \label{eq:pn2} \widehat{p}_n= \zeta_1\left(\frac{1}{1-s};\gamma\right)^{-1} \, \left(\frac1{\gamma+n}\right)^{1/(1-s)}, \end{equation*} so that, by contrast with the Shannon entropy case~\eqref{eq:sentdist}, the distribution $\widehat{p}_n$ obeys an inverse power law, with factor the reciprocal of a modified Hurwitz zeta function. Hence \begin{equation} \label{eq:pn3} \sum_{n=1}^\infty \widehat{p}_n^{\,s}= \zeta_1\left(\frac{1}{1-s}; \gamma\right)^{-s} \sum_{n=1}^\infty \left(\frac1{\gamma+n}\right)^{s/(1-s)} = \frac{\zeta_1\left(\frac{s}{1-s}; \gamma\right)}{\zeta_1\left(\frac{1}{1-s};\gamma\right)^s}\,, \end{equation} which converges when $1/21$. This and the values $\al,\beta$ do not have an obvious interpretation. Nor does the bound for the Shannon entropy appear to follow from the limit of the \Ren{} entropy case as $s\to 1$. \end{remark} \section{Changing the viewpoint: the parameter $t$} \label{sec:perspective} In this section, we discuss the Shannon-R\'enyi entropy $h^{(s)}$ of the curve $\Gamma_0$ relative to a compact subset $ K$ containing $K_0$; this involves a parameter $t$. Consider the curve $\Gamma_0$, its convex hull $K_0$ and a bounded convex set $K\supseteq K_0$. The set $K$ can be thought of as a variable set that ``increases'' and within which the curve can increase its entropy. \subsection{The parameter $t$} \label{ssec:t} Let $t\ge 1$ be the ratio of the lengths of the boundaries of $K$ and $K_0$, i.e., let \begin{equation*} \label{eq:t} t=\frac{|\partial K|}{|\partial K_0|}\,\in [1,\infty). \end{equation*} In this construction, the length $|\partial K|$ of the boundary of the convex set $K$ increases. \vspace{0.05in} \begin{lemma} \label{lem:pnKG0} The probability $p_n(\G_0,K)$ that a straight line $D$ meeting $K$ intersects $\Gamma_0$ in exactly $n$ points is given by \begin{equation} \label{eq:seqpn} p_n(\G_0,K) =p_n(K)= \begin{cases} \frac1{t}\,p_n(K_0), &\text{if }n\ge 1; \\ 1-\frac1{t}, &\text{if } n = 0. \end{cases} \end{equation} \end{lemma} \begin{proof} If $n\ge 1$, \begin{eqnarray*} p_n(\G_0,K)&=&\frac{\mu\{D\colon \card(D\cap \Gamma_0)=n\}} {\mu\{D\colon D\cap K\ne \emptyset\}}\\ &=& \frac{\mu\{D\colon D\cap K_0\ne \emptyset\}} {\mu\{D\colon D\cap K\ne \emptyset\}} \, \frac{\mu\{D\colon \card(D\cap \Gamma_0)=n\}} {\mu\{D\colon D\cap K_0\ne \emptyset\}}\\ &=&\frac{|\partial K_0|}{|\partial K|} \frac{\mu\{D\colon \card(D\cap \Gamma_0)=n\}} {\mu\{D\colon D\cap K_0\ne \emptyset\}} = \frac1{t} \, p_n(K_0,\G_0). \end{eqnarray*} If $n=0$, \begin{equation*} p_0(\G_0,K) = 1-\frac{|\partial K_0|}{|\partial K|} = 1 - \frac1{t}\,, \end{equation*} as in~\eqref{eq:p0} in~\S\ref{sec:geomprob}. \end{proof} \begin{figure} \begin{center} \begin{pspicture}(-3,-2)(4,2) %\showgrid \psline[linestyle=dashed](.44,-1)(1.06,1.1) \psline[linestyle=dashed](1.03,1.54)(-1.95,1.54) \psline[linestyle=dashed](-2.5,0)(.4,-1) \psellipse(0,.5)(3,2.5) \pscurve(-2.5,0)(-2,1.5)(-.5,.5)(-1.5,.5)(-.2,.7)(0,.6) (1,1)(1.1,1.5)(.5,.75)(.4,-1) \rput(-1,0){$\Gamma_0$} \rput(-.4,-1){$K_0$} \rput(2,-1.75){$K$} \end{pspicture} \end{center} \caption{The curve $\Gamma_0$ inside a bounded convex set $K$.} \end{figure} Let $h_0^{(s)}:=h^{(s)}(\G_0,K_0)$, $0h_0$. \end{theorem} \begin{proof} Let $h=h(K)$ be the entropy of $\Gamma_0$ relative to $K$ and let $h_0=h(K_0)$ be the entropy of $\Gamma_0$ relative to $K_0$. Then by~\eqref{eq:shent} and Lemma~\ref{lem:pnKG0} (and writing $p_n(K)=p_n(\G_0,K)$), \begin{eqnarray*} h&=&\sum_{n= 0}^\infty p_n(K)\,\log \frac1{p_n(K)}\\ &=&\left(1-\frac1{t}\right)\log \left(1-\frac1{t}\right)^{-1} + \sum_{n=1}^\infty \frac1{t}\, p_n(K_0)\,\log \frac{t}{p_n(K_0)}\\ &=&\left(1-\frac1{t}\right)\log \frac{t}{t-1} + \frac1{t} \sum_{n=1}^\infty p_n(K_0)\,\log t +\frac1{t} \sum_{n=1}^\infty p_n(K_0) \log \frac1{p_n(K_0)}\\ &=& \frac{h_0}{t} + \log t - \frac{t-1}{t} \,\log (t-1). \end{eqnarray*} We wish to find the maximum value of $h$ when $t\in (1,\infty)$. The derivative \begin{equation*} \frac{dh}{dt}= - \frac1{t^2}(h_0+\log(t-1))=0 \end{equation*} for $t_1=|\partial K_1|/|\partial K_0|=1+e^{-h_0}$, with entropy $h$ a maximum at $t=t_1$ and corresponding value \begin{equation*} \label{eq:h1} h_1:=\log(1+e^{h_0})>h_0. \end{equation*} Here $K_1$ is any convex set whose boundary has length $t_1 |\partial K_0|$ and which contains $K_0$. \end{proof} \begin{figure} \begin{pspicture}(-6,-2)(2,2) %\showgrid \psline(-2.25,-2)(7,-2) %x axis \rput(7,-2.5){$t$} \rput(-2.5,-2.5){$0$} \rput(0,-2.5){$1$} \rput(2,-2.5){$t_1=1+e^{-h_0}$} \rput(-2.75,0){$h_0$} \rput(-2.75,2.45){$h_1$} \psline(-2.25,-2)(-2.25,2.8) %y axis \psline[linestyle=dashed](0,-2)(0,0) %vertical through 1 \psline[linestyle=dashed](1.5,-2)(1.5,2.5) %vertical through t_1 \psline[linestyle=dashed](0.5,2.5)(2.5,2.5) %vhorizontal through h_1 \pscurve(0,0)(0.04,0.2)(0.2,.91)(1.5,2.5)(2.88,1.1)(3.22,.55)(3.26,.5) (3.78,0)(5.5,-.9)(6,-1.05)(7,-1.2) \end{pspicture} \vskip .2in \caption{The graph of $h$ with maximum at $t_1$} \end{figure} \subsection{The R\'enyi entropy case} \label{ssec:hst} For convenience $h^{(s)}(\Gamma_0,K)$, the \Ren{} entropy with respect to the convex set $K\supset\G_0$ will be written $h^{(s)}$, $p_n(\G_0,K)$ will be written $p_n(K)$ and $h^{(s)}(\Gamma_0,K_0)$, the \Ren{} entropy with respect to the convex hull $K_0$ of $\G_0$ will be written $h^{(s)}_0$. The dependence of $h^{(s)}$ on the parameter $t$ will be indicated where helpful; thus when $t=1$, $h^{(s)}(t) =h^{(s)}(1) = h^{(s)}_0$. Note that by Theorem~\ref{thm:hal(K)}, the quantity $h^{(s)}_0$ is finite for $s\in (0,1)$. \begin{theorem} The \Ren{} entropy $h^{(s)}(\Gamma_0,K)$, $0\hs(1)$. \end{proof} \begin{remark} For a certain `dilution' corresponding to $|\partial K|/|\partial K_0| =t_1$, the \Ren{} $s$-entropy attains a maximal value and then decreases to 0. Suppose now $s \to 1$. Then it is easily seen that $t_1\to 1 + e^{-h_0}$ and $\hs(t_1)\to \log (1 + e^{h_0})$, which is consistent with the results above in~\S\ref{ssec:h1t}. If we agree to identify entropy and complexity, we see that the complexity of a curve depends on the point of observation. 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Published in {\it Journal of Integer Sequences}, March 2 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}