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\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 $\omega$, $0\le \omega&lt; 2\pi$, is the angle the normal to $D$
makes with the $x$-axis and $\rho&gt;0$ 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)&amp;\le&amp; \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 \\
&amp; \le &amp;\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&amp;=&amp;\frac{\beta}{1-e^{-\beta}}\left(k-(k-1)e^{-\beta}\right) 
+ \log \frac{e^{-k\beta}}{1-e^{-\beta}}      \\
&amp;=&amp; \frac{\beta}{e^{\beta}-1}\left(ke^{\beta}-k+1\right) -\beta\,k
+ \log \frac{e^{\beta}}{e^{\beta}-1}\\
&amp;=&amp; \log(\lambda-k+1) + (\lambda-k)\log \frac{\lambda-k+1}{\lambda-k}\\
&amp;\le&amp; \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
$0&lt;s&lt;1$ and the other for $1/2&lt;s&lt;1$. Although the \Ren{} entropy
coincides with Shannon entropy at $s=1$, the bounds for the former may well
be infinite at $s=1$.

Two auxiliary functions are introduced. For each
 $\gamma\in
[0,\infty)$ and $s\in [1/2,1]$, 
we write 
 \begin{equation*}
   \label{eq:F}
  \Fs(\gamma)= \frac{ \zeta_1
\left(\frac{s}{1-s};
        \gamma\right)}{ \zeta_1\left(\frac{1}{1-s};\gamma\right)}, 
 \end{equation*}
 where $\zeta_1(u,c)$, $c\ge 0$, is the modified Hurwitz zeta function
 given by
\begin{equation*}
  \zeta_1(u,c) := \sum_{k=1}^\infty \frac1{(k+c)^u} =\sum_{k=0}^\infty
  \frac1{(k+c)^u}  - \frac1{c^u} = \zeta(u,c) - \frac1{c^u}.
\end{equation*} 
 Note that $\zeta_1(u,0)=\zeta(u)$ and that $ \Fs(0) &gt;1$.  
 
 
 For convenience, the function $\Fs$ is simplified when $1/2\le s &lt; 1$
 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 $u&gt;2$ and $\gamma\ge 0$, 
 \begin{eqnarray*}
   \label{eq:z&gt;gz}
  (\gamma +1)\zeta_1(u;\gamma) &amp;=&amp; \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) \\
&amp;&lt;&amp;  \sum_{k=1}^\infty
  \frac1{(k+\gamma)^{u-1}} =\zeta_1(u-1;\gamma),
 \end{eqnarray*}
and $\Gu(\gamma):=\zeta_1(u-1;\gamma)/\zeta_1(u;\gamma) &gt;1+\gamma$. 


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} &lt;  \zeta_1(u,\gamma) :=
\sum_{k=1}^\infty  \frac1{(k+\gamma)^u} &lt; \frac1{(1+\gamma)^u}+
 \frac1{(u-1)(1+\gamma)^{u-1}}.
\end{equation}
Hence for any fixed $u&gt;2$, 
\begin{eqnarray*}
  \label{eq:lbG1}
  \Gu(\gamma) &amp;=&amp; \frac{\zeta_1(u-1,\gamma)}{\zeta_1(u,\gamma)} 
  &gt;\frac1{(u-2) (1+\gamma)^{u-2}}\, 
  \left(\frac1{(1+\gamma)^u} +
    \frac1{(u-1)(1+\gamma)^{u-1}}\right)^{-1} \\
  &amp;&gt;&amp; \frac{(u-1)}{(u-2)}\, 
\frac{(1+\gamma)^2}{(u+\gamma)}.
\end{eqnarray*}
Thus given any $u&gt;2$, for all sufficiently large $\gamma$, there
exists an $\eps_u&gt;0$ such that
 \begin{equation*}
\Gu(\gamma) &gt;(1+\eps_u)(1+\gamma).
\label{eq:lbG}  
 \end{equation*}
 It follows that given any real $\kappa$, $\eps_u(1+\gamma)&gt;\kappa $
 for all sufficiently large $\gamma$.  Hence for fixed $u$ the
 inequality
 \begin{equation}
   \label{eq:G&gt;g+k}
 \Gu(\gamma)    &gt; \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 $\lambda\ge \zeta(u-1)/\zeta(u)&gt; 1$.
 \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 $\lambda&gt;\Gu(0)$.  By
   definition
 \begin{equation*}
   \Gu(0) - \lambda &lt; 0. 
 \end{equation*}
 But by equation~\eqref{eq:G&gt;g+k},  for fixed $u$
 and sufficiently large $\gamma$,
\begin{equation*}
  \Gu(\gamma) &gt;\gamma +\lambda  
\end{equation*}
and by continuity, there exists a $\gamma_0&gt;0$ 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 $\gamma&gt; 0$.
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} &amp;=&amp;
  \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}\\
  &amp;=&amp; u\left(\frac{ \zeta_1(u-1;\gamma)  \zeta_1(u+1;\gamma)}
    { \zeta_1(u;\gamma)^2}   - 1\right) +1&gt;1 
\end{eqnarray*}
if $\zeta_1(u;\gamma)^2 &lt; \zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma) $.
Now by the Cauchy-Schwarz inequality,
\begin{eqnarray*}
  \zeta_1(u;\gamma)  &amp;=&amp;  
  \sum_{k=1}^\infty \frac1{(k+\gamma)^{(u-1)/2}} \frac1{(k+\gamma)^{(u+1)/2}}\\
  &amp;\le&amp;  \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 $\zeta_1(u;\gamma)^2 &lt;\zeta_1(u-1;\gamma) \zeta_1(u+1;\gamma)$,
since the inequality is readily seen to be strict. Hence $\partial
\Gu(\gamma)/\partial \gamma &gt; 1$, 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 $0&lt;s&lt;1$,
\begin{equation}
  \label{eq:hs0bound1}
  h^{(s)}_0 \le 
  \frac1{1-s} \log  \left(\frac{s}{\beta(1-s)}\right) 
  + \log \zeta_1\left(\frac{1}{1-s};\gamma\right), 
\end{equation}
where $\beta=\beta(s,2|\Gamma_0|/|\partial K_0|)$,
$\gamma=\gamma(s,2|\Gamma_0|/|\partial K_0|)$.
Moreover, provided  $1/2&lt;s&lt;1$ and 
\begin{equation}
  \label{eq:fixptcondn}
\frac{2|\Gamma_0|}{|\partial K_0|}
&gt;\frac{\zeta(s/(1-s))}{\zeta(1/(1-s))}&gt;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&amp;=&amp;\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
$1/2&lt;s&lt;1$ but diverges for $s\le 1/2$. 


\vspace{0.05in}

To determine $\gamma$, use the constraint $2|\Gamma_0|/|\partial
K_0|=\lambda$ given by~\eqref{eq:expectation}:  
\begin{eqnarray}
\label{eq:l=F-g}
\frac{2|\Gamma_0|}{|\partial K_0|} = 
\lambda&amp;=&amp;\sum_{n=1}^\infty n \widehat{p}_n =
\zeta_1\left(\frac{1}{1-s};\gamma\right)^{-1} \,\sum_{n=1}^\infty n  
\left(\frac1{\gamma+n}\right)^{1/(1-s)} \notag\\
&amp;=&amp; \zeta_1\left(\frac{1}{1-s};\gamma\right)^{-1} 
\sum_{n=1}^\infty \left[\frac{\gamma+n}{(\gamma+
    n)^{1/(1-s)}} - \frac{\gamma}{(\gamma+n)^{1/(1-s)}} \right]\notag\\
&amp;=&amp; \zeta_1\left(\frac{1}{1-s};\gamma\right)^{-1} \left( \zeta_1
  \left(\frac{s}{1-s};
    \gamma\right) - \gamma \zeta_1\left(\frac{1}{1-s};
    \gamma\right) \right) \notag\\
&amp;=&amp; \frac{ \zeta_1
  \left(\frac{s}{1-s};
    \gamma\right)}{ \zeta_1\left(\frac{1}{1-s};\gamma\right)} -
\gamma.
\end{eqnarray}
By~\eqref{eq:fixptcondn} and Lemma~\ref{lem:Flg} with $u=1/(1-s)$,
$1/2&lt;s&lt;1$, there is a $\gamma_0=\gamma_0(s,\lambda)$
satisfying~\eqref{eq:l=F-g}, i.e., such that
\begin{equation*}
  \label{eq:fixpt}
  \Gu(\gamma_0)=\Fs(\gamma_0) = \frac{ \zeta_1
    \left(\frac{s}{1-s};
      \gamma_0\right)}{ \zeta_1\left(\frac{1}{1-s};\gamma_0\right)} =
  \lambda +\gamma_0.
\end{equation*}

Now by~\eqref{eq:pn3}, 
\begin{eqnarray*}
  h^{(s)} &amp;=&amp; \frac1{1-s}\log\left( \sum_{n= 1}^\infty
    p_n^s\right) \le \frac1{1-s}\log\left(\sum_{n= 1}^\infty
    \widehat{p}_n^{\,s}\right) = \widehat{h}^{\,(s)}
 \end{eqnarray*}
and for $u=1/(1-s)$, 
 \begin{eqnarray*}
 \widehat{h}^{(s)} &amp;=&amp; \frac1{1-s}\log \left[\frac{\zeta_1\left(\frac{s}{1-s};
        \gamma_0\right)}{\zeta_1\left(\frac{1}{1-s};\gamma_0\right)}\right]  +
  \log \zeta_1\left(\frac{1}{1-s};\gamma_0\right)   \\
  &amp;=&amp;  \frac{1}{1-s} \log (\lambda + \gamma_0) + 
  \log\zeta_1\left(\frac{1}{1-s};\gamma_0\right)  , 
\end{eqnarray*}
giving~\eqref{eq:hs0bound2}. 


The corresponding parameters $\beta_0$ and
$\alpha_0=\beta_0\,\gamma_0$, are fixed by the equation
\begin{equation*}
\label{eq:sbz}
  \frac{s}{\beta_0(1-s)} =
\frac{\zeta_1\left(\frac{s}{1-s};
\gamma_0\right)}{\zeta_1\left(\frac{1}{1-s};\gamma_0\right)}=
\Fs(\gamma_0) = 
\lambda +\gamma_0\, ,
\end{equation*}
obtained by dividing~\eqref{eq:pns3} by~\eqref{eq:pn3} and
using~\eqref{eq:l=F-g}.  This equation implies that as $s\to 1$ from
below, $\beta_0\to\infty$ and therefore $\al_0\to\infty$, while
$\Fs(\gamma_0) \to \infty$ as $s \to 1/2$ from above, which implies
that $\beta_0,\, \al_0\to 0$.
 \end{proof}


 \begin{remark} The bound for the \Ren{} entropy $h^{(s)}$ is finite
   for $1/2&lt;s&lt;1$.  The equation $\Fs(\gamma_0)=\gamma_0 + \lambda$
   only holds if $\lambda =2|\Gamma_0|/|\partial K_0|\ge \Fs(0)=
   \zeta(s/(s-1))/\zeta(1/(s-1))&gt;1$.  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), 
           &amp;\text{if }n\ge 1; \\
1-\frac1{t},
          &amp;\text{if }  n = 0.
                  \end{cases}
\end{equation} 
\end{lemma}
\begin{proof} 
If $n\ge 1$,
\begin{eqnarray*}
  p_n(\G_0,K)&amp;=&amp;\frac{\mu\{D\colon \card(D\cap \Gamma_0)=n\}}
{\mu\{D\colon D\cap K\ne \emptyset\}}\\
&amp;=&amp; \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\}}\\
&amp;=&amp;\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 $h_0^{(s)}:=h^{(s)}(\G_0,K_0)$, $0&lt;s\le 1$, be the Shannon-R\'enyi
entropy of $\Gamma_0$ with respect to its convex hull $K_0$. In the
preceding section~\S\ref{sec:entropy}, $h_0^{(s)}$ is written by $h$
or $h^{(1)}$ when $s=1$ (Shannon entropy) and $h^{(s)}$ when $0&lt;s&lt;1$.
We begin with the case $s=1$.



\subsection{The Shannon entropy case}
\label{ssec:h1t}
The Shannon entropy $h(\Gamma_0,K)=h$ of $\Gamma_0$ relative to $K$ is
now determined in terms of the Shannon entropy $h(\Gamma_0,K_0)=h_0$
of $\G_0$ relative to the convex hull $K_0$ of $\G_0$ ($h_0$ is finite
by Theorem~\ref{thm:h(K)}).

\begin{theorem} 
\label{thm:h1t}
\label{thm:h(K0)}
Let $h_0$ be the Shannon entropy of $\Gamma_0$ with respect to
$K_0$. The entropy $h(\Gamma_0,K)$ of $\Gamma_0$ with respect to $K$
is given by
    \begin{equation*}
      h(\Gamma_0,K)=\frac{h_0}{t}+ \log t - \frac{t-1}{t} \log (t-1) 
=\frac{h_0}{t} + (-1+\frac1{t})\log(1-\frac1{t}) + \frac{\log t}{t} \to 0
    \end{equation*}
    as $t\to\infty$. Moreover $ h(\Gamma_0,K)$ is maximal when
    $t=t_1=1+e^{-h_0}$, with value $h_1=\log (1+e^{h_0} ) &gt;h_0$.
\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&amp;=&amp;\sum_{n= 0}^\infty p_n(K)\,\log \frac1{p_n(K)}\\
  &amp;=&amp;\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)}\\
  &amp;=&amp;\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)}\\
  &amp;=&amp; \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})&gt;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 $h^{(s)}(\Gamma_0,K)$, $0&lt;s&lt;1$, of
  $\Gamma_0$ with respect to $K$ is given by
    \begin{eqnarray*}
      h^{(s)}(\Gamma_0,K)  
&amp;=&amp; \frac1{1-s} \log\left(\left(1-\frac1{t}\right)^s
+  \sum_{n=1}^\infty\frac1{t^s}\, p_n^s(K_0)  \right) 
\\
&amp;=&amp; \frac{1}{1-s} \log \left(1+(t-1)^{-s} e^{(1-s){h^{(s)}_0}}\right)
+ \frac{s}{1-s} \log\left(1-\frac1{t}\right),
    \end{eqnarray*}
so that $h^{(s)} \to 0$ as $t\to\infty$. 
 Moreover $ h^{(s)}(\Gamma_0,K)$ is maximal
    for 
    \begin{equation*}
t_1=1+ \left(\sum_{n=1}^\infty p_n^s (K_0) \right)^{-1/(1-s)}  =
1+e^{-\hs_0},     
    \end{equation*}
with value $h^{(s)}(t_1) 
=\log (1+e^{h_0^{(s)}})$.
  \end{theorem}

\begin{proof}
   Write $h^{(s)}(\G_0,K)=h^{(s)}$.  It follows from~\eqref{eq:seqpn}
that  
\begin{eqnarray*}
  h^{(s)}
&amp;=&amp;h^{(s)}(t) = \frac1{1-s} \log\, \sum_{n=0}^\infty p_n^{s}(K)
= \frac1{1-s} \log\left(\left(1-\frac1{t}\right)^s
+  \sum_{n=1}^\infty\frac1{t^s}\, p_n^s(K_0)
\right)\\
&amp;=&amp; \frac{s}{1-s} \log \left(1-\frac1{t}\right) +
\frac{1}{1-s} \log \left(1+\frac{e^{(1-s){h^{(s)}_0}}}{(t-1)^{s}}\right),
\end{eqnarray*}
as claimed. 
\vspace{0.05in}

We now ask which $t\ge 1$ maximizes $ h^{(s)}= h^{(s)}(t)$.  The
derivative
\begin{equation*}
 \frac{d\hs}{dt} 
= \frac{s}{(1-s)\left(\sum_{n=0}^\infty p_n^{s}(K_0)\right)} 
\left( (1-\frac1{t})^{s-1} \frac{1}{t^{2}}  - \frac1{t^{s+1}} 
\sum_{n=1}^\infty  p_n^s(K_0)\right)=0
\end{equation*}
iff 
\begin{equation*} 
\left(t-1\right)^{s-1}   =  \sum_{n=1}^\infty  p_n^s(K_0) =
e^{(1-s)\, h^{(s)}_0},
 \end{equation*} 
\ i.e., iff 
 \begin{equation}
\label{eq:tmax} 
t  = 1 + \left(\sum_{n=1}^\infty  p_n^s(K_0)\right)^{-1/(1-s)} 
=1 + e^{-h^{(s)}_0}.
\end{equation}
The entropy $ h^{(s)}(t)$ is maximal at the value $t_1$ given
by~\eqref{eq:tmax}, i.e.,
\begin{eqnarray}
  h^{(s)}(t_1) &amp;= &amp;\frac1{1-s}\,\log \sum_{n=1}^\infty p_n^s (K_0) +
 \log\left(1+\left(\sum_{n=1}^\infty p_n^s(K_0)\right)^{-\frac{1}{(1-s)}}\right)  \nonumber \\
  &amp;=&amp;  h^{(s)}_0 + \log\left(1+e^{-h_0^{(s)}}\right),
\label{eq:hmax}
\end{eqnarray}
whence $h^{(s)}(t_1) =\log\left(1+e^{h_0^{(s)}}\right)$.
The first term on the right hand side of~\eqref{eq:hmax} is the
$s$-entropy $\hs_0=\hs(1)$ of $\G_0$ relative to its convex hull
$K_0$, so that $\hs(t_1)&gt;\hs(1)$.
\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.  Seen from
a certain distance, the curve increases its complexity, while seen
from infinity the curves reduces to a point with entropy 0.
\end{remark}

 \section{Acknowledgment}
 We are grateful to Chris Hughes for his help with the Hurwitz zeta
 function.

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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 54C70; Secondary 52A22, 28D20, 11M35.

\noindent \emph{Keywords: } curves, geometric probability, entropy.

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\vspace*{+.1in}
\noindent
Received June 28 2012;
revised version received  August 31 2012.
Published in {\it Journal of Integer Sequences}, March 2 2013.

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