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% Author, Title, etc.

\Large
\title[Continued Fraction Digit Averages and Maclaurin's Inequalities]
{
Continued Fraction Digit Averages and Maclaurin's Inequalities
}

\author[Steven J. Miller]
{
\textcolor{violet}{Steven J. Miller, Williams College} \newline \footnotesize
  \textbf{\textcolor{violet}{\texttt{sjm1@williams.edu,\ Steven.Miller.MC.96@aya.yale.edu}}}\newline \small
  \textcolor{violet}{Joint with Francesco Cellarosi, Doug Hensley and Jake Wellens} \newline \normalsize
  \footnotesize \textcolor{blue}{\url{http://web.williams.edu/Mathematics/sjmiller/public_html/}}\normalsize
%\inst{1} %\and
%  Tzvika~Hartman\inst{2} \and
%  Till~Nierhoff\inst{3} \and
%  Roded~Sharan\inst{4} \and
%  \textcolor{green!50!black}{Till~Tantau}\inst{5}
}
\date[March 2013]
{\alert{AMS Special Session on Continued Fractions, January 12, 2015}
 \vspace{-0.01in}
%  \rightline{\scalebox{.9}{\includegraphics{CookieMonsterAndBlueMandM.eps}}}
}


\begin{document}

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%\begin{frame}{Outline}
%  \tableofcontents
%\end{frame}




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\section{Intro}

\begin{frame}

\begin{block}{}
\begin{center}
\alert{\ \\ Introduction\\ \ }
\end{center}
\end{block}

\end{frame}



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\subsection{}
\begin{frame}{Plan of the talk}
\begin{itemize}
\item Classical ergodic theory of continued fractions. \\
\noindent $\alert{\diamond}$ Almost surely geometric mean $\sqrt[n]{a_1\cdots a_n} \to K_0$.\\
\noindent $\alert{\diamond}$ Almost surely arithmetic mean $(a_1+\cdots+a_n)/n \to \infty$. \\ \

\item Symmetric averages and Maclaurin's inequalities.\\
\noindent   $\alert{\diamond}$ $S(x,n,k) := \ncr{n}{k}^{-1} \sum_{1\leq i_1<i_2<\cdots<i_k\leq n}x_{i_1}x_{i_2}\cdots x_{i_k}$.\\ \small
\noindent  $\alert{\diamond}$ ${\rm AM} = S(x,n,1)^{1/1} \geq S(x,n,2)^{1/2}\geq\cdots\geq S(x,n,n)^{1/n}  ={\rm GM}$. \\ \ \normalsize

\item Results / conjectures on typical / periodic continued fraction averages.\\ \

\item Elementary proofs of weak results, sketch of stronger results.
%\item<6->Some open questions
\end{itemize}

\ \\

\noindent  To appear in Exp. Math.:\small \textcolor{blue}{\url{http://arxiv.org/abs/1402.0208}}. \normalsize



\end{frame}



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%\subsection{}
%\begin{frame}{Continued Fractions}
%\begin{itemize}
%\item Every real number $\alpha\in(0,1)$ can be expressed as
%$$x \ = \ \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ldots}}}}\ = \ [a_1,a_2,a_3,\ldots], \ \ a_i \in \{1, 2, \dots\}.$$ \pause \ \\ \

%\only<2>{
%\item The sequence $\{a_i\}_i$ is finite iff $\alpha\in\mathbb{Q}$. \pause  \\ \
%}

%\only<3>{
%\item $x=\frac{p}{q}\in\mathbb{Q}$ then $a_i$'s the partial quotients of Euclidean Alg.
%\begin{align}&&\mathbf{333}&\ =\ \textcolor{red}{3}\cdot\mathbf{106}+15\nonumber\\
%\frac{\mathbf{106}}{\mathbf{333}}\ = \ [\textcolor{red}{3},\textcolor{blue}{7},\textcolor{purple}{15}]&&106& \ = \ \textcolor{blue}{7}\cdot %15+1\nonumber\\
%&&15&\ = \ \textcolor{purple}{15}\cdot 1+0.\nonumber
%\end{align}
%}

%\only<4>{
%\item $\{a_i\}_i$ preperiodic iff $\alpha$ a quadratic irrational;\\ ex: $\sqrt{3}-1=[1,2,1,2,1,2,\dots]$.
%}

%\end{itemize}
%\end{frame}



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%\subsection{}
%\begin{frame}{Gauss Map: Definition}
%\begin{itemize}
%\item The Gauss map $T:(0,1]\to(0,1]$, $T(x)=\{\frac{1}{x}\}=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor$
%generates the continued fraction digits
%$$a_1\ = \ \lfloor 1/T^0(\alpha)\rfloor,\ \ \ \ a_{i+1}\ = \ \lfloor 1/T^i(\alpha)\rfloor, \ \ \ \  \ldots$$
%corresponding to the Markov partition $$(0,1]\ =\ \bigsqcup_{k=1}^\infty\left(\frac{1}{k+1},\frac{1}{k}\right].$$
%\item $T$  preserves the measure $d\mu=\frac{1}{\log 2}\frac{1}{1+x}dx$ and it is mixing.
%\end{itemize}
%\end{frame}


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%\subsection{}
%\begin{frame}{Gauss Map: Example: $\sqrt{3}-1=[1,2,1,2,1,2,\dots]$}

%\noindent $T:(0,1]\to(0,1]$, $T(x)=\{\frac{1}{x}\}=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor$
%generates digits
%$$a_1\ = \ \lfloor 1/T^0(\alpha)\rfloor,\ \ \ \ a_{i+1}\ = \ \lfloor 1/T^i(\alpha)\rfloor, \ \ \ \  \ldots$$

%\ \\

%\noindent $\alpha = \sqrt{3} -1 =[1,2,1,2,\dots]$: Note $a_1 = \lfloor \frac{1}{\sqrt{3}-1}\rfloor = 1$   and  \bea T^1(\sqrt{3}-1) & \ = \ &  \frac1{\sqrt{3}-1} - \left\lfloor\frac1{\sqrt{3}-1}\right\rfloor \ = \ \frac{\sqrt{3}+1}{3-1} - 1 \ = \ \frac{\sqrt{3}-1}{2} \nonumber\\ a_2 & \ = \ & \left\lfloor \frac{2}{\sqrt{3}-1}\right\rfloor \ = \ 2.  \nonumber\\ T^2(\sqrt{3}-1) & \ = \ & \frac{2}{\sqrt{3}-1} - \left\lfloor \frac{2}{\sqrt{3}-1}\right\rfloor \ = \ \frac{2\sqrt{3}+2}{2} - 2 \ = \ \sqrt{3} - 1 \nonumber\\ a_3 & \ = \ & \left\lfloor\frac1{\sqrt{3}-1}\right\rfloor \ = \ 1. \nonumber
%\eea

%\end{frame}



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%\subsection{}
%\begin{frame}{Gauss Map: Example: $\sqrt{3}-1=[1,2,1,2,1,2,\dots]$}

%\noindent $T:(0,1]\to(0,1]$, $T(x)=\{\frac{1}{x}\}=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor$
%generates digits
%$$a_1\ = \ \lfloor 1/T^0(\alpha)\rfloor,\ \ \ \ a_{i+1}\ = \ \lfloor 1/T^i(\alpha)\rfloor, \ \ \ \  \ldots$$

%\ \\

%\noindent $\alert{\alpha = \sqrt{3} -1} =[\textcolor{blue}{1},\textcolor{blue}{2},1,2,\dots]$: Note $a_1 = \lfloor \frac{1}{\sqrt{3}-1}\rfloor = 1$ and  \bea T^1(\sqrt{3}-1) & \ = \ &  \frac1{\sqrt{3}-1} - \left\lfloor\frac1{\sqrt{3}-1}\right\rfloor \ = \ \frac{\sqrt{3}+1}{3-1} - 1 \ = \ \frac{\sqrt{3}-1}{2} \nonumber\\ a_2 & \ = \ & \left\lfloor \frac{2}{\sqrt{3}-1}\right\rfloor \ = \ \textcolor{blue}{2}. \nonumber\\ T^2(\sqrt{3}-1) & \ = \ & \frac{2}{\sqrt{3}-1} - \left\lfloor \frac{2}{\sqrt{3}-1}\right\rfloor \ = \ \frac{2\sqrt{3}+2}{2} - 2 \ = \ \alert{\sqrt{3} - 1} \nonumber\\ a_3 & \ = \ & \left\lfloor\frac1{\sqrt{3}-1}\right\rfloor \ = \ \textcolor{blue}{1}. \nonumber
%\eea

%\end{frame}




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\subsection{}
\begin{frame}{Statistics of Continued Fraction Digits 1/3}
\begin{itemize}
\item<1->The digits $a_i$ follow the Gauss-Kuzmin distribution:
$$\lim_{n\to\infty}\mathbb{P}(a_n=k)=\log_2\left(1+\frac{1}{k(k+2)}\right)$$
(note the expectation is infinite). \\ \

\item<2->The function $x\mapsto f(x)=\lfloor 1/T(x) \rfloor$ on $(0,1]$ is not integrable wrt $\mu$. However, $\log f\in L^1(\mu)$. \\ \

\item<3-> Pointwise ergodic theorem (applied to $f$ and $\log f$) reads
\begin{align}&\lim_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n} \ = \
\infty\hspace{.5cm}\mbox{almost surely}\nonumber\\
&\lim_{n\to\infty}\left(a_1 a_2\cdots a_n\right)^{1/n} \ = \ e^{\int \log f\, d\mu}\hspace{.5cm}\mbox{almost surely.}\nonumber
\end{align}
\end{itemize}




\end{frame}



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\subsection{}
\begin{frame}{Statistics of Continued Fraction Digits 2/3}
\begin{itemize}
\item<1-> Geometric mean converges a.s. to Khinchin's constant:
\begin{align}
\lim_{n\to\infty}\left(a_1 a_2\cdots a_n\right)^{1/n} = \prod_{k=1}^\infty\left(1+\frac{1}{k(k+2)}\right)^{\log_2 k} =  K_0 \approx 2.6854.\nonumber
\end{align}

\ \\

\item<2->H\"{o}lder means:  For $p<1$, almost surely
\begin{align}&\lim_{n\to\infty}\left(\frac{1}{n}\sum_{i=1}^n a_i^p\right)^{1/p} = K_p=\left(\sum_{k=1}^{\infty}-k^p \log_2\left(1-\frac{1}{(k+1)^2}\right)\right)^{1/p}. \ \ \ \
\nonumber
\end{align}

 \ \\

\item<3-> Example: The harmonic mean $K_{-1}=1.74540566\dots$.

\ \\

\item<4-> $\lim_{p\to 0} K_p=K_0$.

\end{itemize}



\end{frame}



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\subsection{}
\begin{frame}{Statistics of Continued Fraction Digits 3/3}
\begin{itemize}
\item<1->Khinchin also proved: For $a_m' = a_m$ if $a_m < m(\log m)^{4/3}$ and 0 otherwise:
$$
\lim_{n\to\infty}\frac{\sum_{i=1}^n a_i'}{n\log n}\ = \
\frac{1}{\log 2}\hspace{.5cm}\mbox{in measure.}
$$

\ \\

\item<2->Diamond and Vaaler (1986) showed that
$$
\lim_{n\to\infty}\frac{\sum_{i=1}^n a_i-\max_{1\leq i\leq n}a_i}{n\log n}\ = \
\frac{1}{\log 2}\hspace{.5cm}\mbox{almost surely.}
$$%\item<3-> For example, the harmonic mean $K_{-1}=1.74540566\ldots$
%\item<4->$\lim_{p\to 0} K_p=K_0$.
\end{itemize}





\end{frame}



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\section{Maclaurin Inequalities}

\begin{frame}

\begin{block}{}
\begin{center}
\alert{\ \\ Maclaurin Inequalities\\ \ }
\end{center}
\end{block}

\end{frame}



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\subsection{}
\begin{frame}{Definitions and Maclaurin's Inequalities}
\begin{itemize}
\item Both $\frac{1}{n}\sum_{i=1}^n x_i$ and $\left(\prod_{i=1}^{n}x_i\right)^{1/n}$ are defined in terms of elementary symmetric polynomials in $x_1,\ldots, x_n$.

\item Define \alert{\textbf{$k$\textsuperscript{th} elementary symmetric mean of $x_1,\ldots,x_n$}} by
$$
S(x,n,k)\ := \ \frac{1}{{n\choose k}} \sum_{1\leq i_1<i_2<\cdots<i_k\leq n}x_{i_1}x_{i_2}\cdots x_{i_k}.
$$
\end{itemize} \pause \ \\

\begin{block}{Maclaurin's Inequalities}
For positive $x_1,\ldots, x_n$ we have
$$
{\rm AM} := S(x,n,1)^{1/1} \geq S(x,n,2)^{1/2}\geq\cdots\geq S(x,n,n)^{1/n}  =: {\rm GM}
$$
(and equalities hold iff $x_1=\cdots =x_n$). \end{block}
%\item<4->When $(x_1,x_2,\ldots)=(a_1,a_2,\ldots)$, Khinchin's results read
%\begin{align}
%S(1,1)^{1/1}\to\infty\hspace{.4cm}\mbox{and}\hspace{.4cm}S(n,n)^{1/n}\to K_0\nonumber
%\end{align}
%almost surely as $n\to\infty$.

\end{frame}



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\subsection{}
\begin{frame}{Maclaurin's work}
\vspace{-0.5cm}
\begin{figure}
%\hspace{-.5cm}
\hspace{-5.5cm}
\includegraphics[angle=0,width=5.3cm]{maclaurin-1.eps}
%\vspace{-3cm}
\end{figure}
%...
\vspace{-5.5cm}
%...
\begin{figure}
\hspace{5cm}
\includegraphics[angle=0,width=4.0cm]{maclaurin-2b.eps}
\end{figure}



\end{frame}



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\subsection{}
\begin{frame}{Proof}

\noindent Standard proof through Newton's inequalities.\\ \ \\

\noindent Define the \alert{\textbf{$k$\textsuperscript{th} elementary symmetric function}} by $$s_k(x) \ = \ \sum_{1\le i_1 < i_2 < \cdots < i_k \le n} x_{i_1} x_{i_2} \cdots x_{i_k},$$  and the \alert{\textbf{$k$\textsuperscript{th} elementary symmetric mean}} by $$E_k(x) \ = \ s_k(x) \Big/ \ncr{n}{k}.$$\ \\

\noindent Newton's inequality: $E_k(x)^2 \ge E_{k-1}(x) E_{k+1}(x)$. \\ \ \\

\noindent New proof by Iddo Ben-Ari and Keith Conrad: \scriptsize \begin{center} \burl{http://homepages.uconn.edu/benari/pdf/maclaurinMathMagFinal.pdf}.\normalsize \end{center}

\end{frame}

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\subsection{}
\begin{frame}{Sketch of Ben-Ari and Conrad's Proof}

\noindent \alert{Bernoulli's inequality:} $t > -1$: $(1+t)^n \ge 1 + nt$ or $1+\frac1{n}x \ge (1+x)^{1/n}$. \ \\ \

\noindent \alert{Generalized Bernoulli:} $x > -1$: $$1 + \frac1{n}x  \ge \left(1+\frac{2}{n}x\right)^{1/2} \ge \left(1+\frac{3}{n}x\right)^{1/3} \ge \cdots \ge \left(1+\frac{n}{n}x\right)^{1/n}.$$

\noindent \textbf{Proof:} Equivalent to $\frac1{k}\log\left(1+\frac{k}{n}x\right) \ge \frac1{k+1}\log\left(1+\frac{k+1}{n}x\right)$, which follows by $\log t$ is strictly concave:

\noindent  \ \ \ \ \ \ $\lambda = \frac1{k+1}$, $1 + \frac{k}{n}x = \lambda \cdot 1 + (1-\lambda) \cdot \left(1+\frac{k+1}{n}x\right)$.


\end{frame}



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\subsection{}
\begin{frame}{Sketch of Ben-Ari and Conrad's Proof}

\noindent \textbf{Proof of Maclaurin's Inequalities:}\\ \

\noindent Trivial for $n \in \{1,2\}$, wlog assume $x_1 \le x_2 \le \cdots \le x_n$.\\ \ \\  

\noindent Set $E_k := s_k(x)/\ncr{n}{k}$, $\epsilon_k := E_k(x_1,\dots,x_{n-1})$. \\ \ \\  

\noindent Have $E_k(x_1,\dots,x_n) = \left(1-\frac{k}{n}\right)E_k(x_1,\dots,x_{n-1}) + \frac{k}{n}E_k(x_1,\dots,x_{n-1})x_n$.\\ \ \\  

\noindent Proceed by induction in number of variables, use Generalized Bernoulli.

\end{frame}

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\section{Main Results (Elementary)}

\begin{frame}

\begin{block}{}
\begin{center}
\alert{\ \\ Main Results\\ \ (Elementary Techniques) }
\end{center}
\end{block}

\end{frame}



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\subsection{}
\begin{frame}{Symmetric Averages and Maclaurin's Inequalities}
\begin{itemize}
%\item<1->Both $\frac{1}{n}\sum_{i=1}^n x_i$ and $\left(\prod_{i=1}^{n}x_i\right)^{1/n}$ are defined in terms of elementary symmetric polynomials in $x_1,\ldots, x_n$.
\item Recall: $\displaystyle{S(x,n,k)=\frac{1}{{n\choose k}} \sum_{1\leq i_1<\cdots<i_k\leq n}x_{i_1}\cdots x_{i_k}}$\\ and $S(x,n,1)^{1/1}\geq S(x,n,2)^{1/2}\geq\cdots\geq S(x,n,n)^{1/n}$. \pause \ \\ \ \\

\item  Khinchin's results: almost surely as $n\to\infty$
\begin{align}
S(\alpha,1,1)^{1/1}\to\infty\hspace{.4cm}\mbox{and}\hspace{.4cm}S(\alpha,n,n)^{1/n}\to K_0.\nonumber
\end{align}

\ \\

\item We study the intermediate means $S(\alpha,n,k)^{1/k}$ as $n\to\infty$ when $k=k(n)$, with
$$S(\alpha,n,k(n))^{1/k(n)}\ =  S(\alpha,n,\lceil k(n)\rceil)^{1/\lceil k(n)\rceil}.$$
\end{itemize}




\end{frame}



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\subsection{}
\begin{frame}{Our results on typical continued fraction averages}

%\item<1->Both $\frac{1}{n}\sum_{i=1}^n x_i$ and $\left(\prod_{i=1}^{n}x_i\right)^{1/n}$ are defined in terms of elementary symmetric polynomials in $x_1,\ldots, x_n$.
\noindent Recall: $\displaystyle{S(\alpha,n,k)=\frac{1}{{n\choose k}} \sum_{1\leq i_1<\cdots<i_k\leq n}a_{i_1}\cdots a_{i_k}}$\\ and $S(\alpha,n,1)^{1/1}\geq S(\alpha, n,2)^{1/2}\geq\cdots\geq S(\alpha, n,n)^{1/n}$.


\begin{block}{Theorem 1}
Let $f(n)=o(\log\log n)$ as $n\to\infty$. Then, almost surely, $$\lim_{n\to\infty}S(\alpha,n,f(n))^{1/f(n)}\ = \ \infty.$$
\end{block}

\begin{block}{Theorem 2}
Let $f(n)=o(n)$ as $n\to\infty$. Then, almost surely, $$\lim_{n\to\infty}S(\alpha,n,n-f(n))^{1/(n-f(n))}\ = \ K_0.$$
\end{block}

Note: Theorems do not cover the case $f(n)=cn$ for $0<c<1$.



\end{frame}



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\subsection{}
\begin{frame}{Sketch of Proofs of Theorems 1 and 2}

\begin{block}{Theorem 1: For $f(n)=o(\log\log n)$ as $n\to\infty$:}
$${\rm Almost\ surely\ }\lim_{n\to\infty}S(\alpha,n,f(n))^{1/f(n)}\ = \ \infty.$$
\end{block}

\noindent Uses Niculescu's strengthening of Maclaurin (2000):
$$S(n,tj+(1-t)k)\ \ge\ S(n,j)^t\cdot S(n,k)^{1-t}.$$

\pause

\begin{block}{Theorem 2: For $f(n)=o(n)$ as $n\to\infty$:}
$${\rm Almost\ surely\ }\lim_{n\to\infty}S(\alpha,n,n-f(n))^{1/(n-f(n))}\ = \ K_0.$$
\end{block}

\bea {\rm Use\ (a.s.)}\ K_0\leq \limsup_{n\to\infty}S(\alpha,n,cn)^{1/cn}\leq K_0^{1/c}<\infty, 0<c<1. \nonumber \eea

\end{frame}



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\subsection{}
\begin{frame}{Proof of Theorem 1: Preliminaries}

\begin{block}{Lemma}  Let $X$ be a sequence of positive real numbers. Suppose $\lim_{n\to\infty}S(X,n, k(n))^{1/k(n)}$ exists. Then, for any $f(n) = o(k(n))$ as $n\to\infty$, we have
\be \lim_{n\to\infty}S(X,n, k(n)+f(n))^{1/(k(n)+f(n))} \ = \  \lim_{n\to\infty}S(n, k(n))^{1/k(n)}.\nonumber\ee
\end{block}

\ \\

\noindent \textbf{Proof:} Assume $f(n) \geq 0$ for large enough $n$, and for display purposes write $k$ and $f$ for $k(n)$ and $f(n)$. \\ \

\noindent From Newton's inequalities and Maclaurin's inequalities, we get \footnotesize
\be \left(S(X,n, k)^{1/k}\right)^{\frac{k}{k+f}}\ =\ S(X,n, k)^{1/(k+f)}\ \leq\ S(X,n, k+f)^{1/(k+f)}\ \leq\ S(X,n, k)^{1/k}. \nonumber\ee \normalsize





\end{frame}



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\subsection{}
\begin{frame}{Proof of Theorem 1: $f(n) = o(\log \log n)$}

\noindent Each entry of $\alpha$ is at least 1.

\noindent Let $f(n) = o(\log{\log{n}})$. Set $t=1/2$ and $(j, k) = (1, 2f(n) -1)$, so that $tj+(1-t)k = f(n)$. Niculescu's result yields
\be S(\alpha, n, f(n)) \ \ge \  \sqrt{S(\alpha, n, 1)\cdot S(\alpha, n, 2f(n) - 1)}\ >\ \sqrt{S(\alpha, n, 1)}. \nonumber \ee

\noindent Square both sides, raise to the power $1/f(n)$:
\be S(\alpha, n, f(n))^{2/f(n)} \ \ge \  S(\alpha, n, 1)^{1/f(n)}.\nonumber\ee

\noindent From Khinchin almost surely if  $g(n)=o(\log n)$
\be \lim_{n \to \infty} \frac{S(\alpha, n, 1)}{g(n)} \ = \  \infty. \nonumber\ee

\noindent Let $g(n) = \log{n}/\log{\log{n}}$. Taking logs:
\be\log\left( S(\alpha,n,1)^{1/f(n)}\right)\ > \ \frac{\log g(n)}{f(n)}\ >\ \frac{ \log{\log{n}}}{2f(n)}.\nonumber\ee


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\subsection{}
\begin{frame}{Proof of Theorem 2}

\noindent \textbf{Theorem 2:} Let $f(n)=o(n)$ as $n\to\infty$. Then, almost surely, $$\lim_{n\to\infty}S(\alpha,n,n-f(n))^{1/(n-f(n))}\ = \ K_0.$$ \ \\

\noindent \textbf{Proof:} Follows immediately from:\\ \noindent For any constant $0 < c < 1$ and almost all $\alpha$ have
\be K_0 \ \le \  \limsup_{n \to \infty}S(\alpha, n, cn)^{1/cn} \ \le \   K_0^{1/c} \ < \  \infty. \nonumber \ee

\ \\

\noindent To see this, note \footnotesize
\be S(\alpha, n, cn)^{1/cn} \ = \  \left( \prod_{i=1}^{n} a_i(\alpha)^{1/n} \right)^{n/cn} \left( \dfrac{ \displaystyle\sum_{i_1 < \cdots < i_{(1-c)n} \leq n} 1/(a_{i_1}(\alpha) \cdots a_{i_{(1-c)n}}(\alpha))}{\dbinom{n}{cn}}\right)^{1/cn}. \nonumber \ee
\normalsize





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\subsection{}
\begin{frame}{Evidence for Limiting Behavior}
\begin{itemize}
\item<1->$n\mapsto S(\alpha,n,cn)^{1/cn}$  for $c=\tfrac{1}{4},\tfrac{1}{2},\tfrac{3}{4}$ and $\textcolor{red}{\alpha=\pi-3}$, $\textcolor{blue}{\gamma}$, $\textcolor{green}{\sin(1)}$ .
\begin{figure}
\includegraphics[width=10cm]{sixgraphs4000.eps}
\end{figure}
\end{itemize}



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\subsection{}
\begin{frame}{Our results on periodic continued fraction averages 1/2}
\begin{itemize}
\item<1-> For $\alpha=\sqrt3-1=[1,2,1,2,1,2,\ldots]$,
\begin{align}
&\lim_{n\to\infty}S(\alpha,n,1)^{1/1}=\frac{3}{2}\neq \infty\nonumber\\
&\lim_{n\to\infty}S(\alpha,n,n)^{1/n}=\sqrt 2\neq K_0\nonumber
\end{align}
\item<2-> What can we say about $\lim_{n\to\infty}S(\alpha,n,cn)^{1/cn}$?
\item<3->Consider the quadratic irrational $\alpha=[x,y,x,y,x,y,\ldots]$.
\item<4-> Let us look at $S(\alpha,n,cn)^{1/cn}$ for $c=1/2$.\pause
$$S(\alpha,n,\lceil \tfrac{n}{2}\rceil)=\begin{cases}S(\alpha, n, \tfrac{n}{2})&\mbox{if $n\equiv 0\bmod 2$;}\\ S(\alpha, n,\tfrac{n+1}{2})&\mbox{if $n\equiv 1\bmod 2$.}\end{cases}$$
\item<5-> We find the limit $\lim_{n\to\infty}S(\alpha,n,\lceil \tfrac{n}{2}\rceil)^{1/\lceil \tfrac{n}{2}\rceil}$ in terms of $x, y$. %\pause For $\alpha=\sqrt3-1=[\overline{1,2}]$, $S(\alpha,2n,n)^{1/n}=\frac{1}{4}(3+2\sqrt2)$.
%\item<4->\textbf{Theorem 3}. $S(\alpha,2n,n)^{1/n}$ converges monotonically as $n\to\infty$ to the $\frac{1}{2}$-H\"{o}lder mean of $x$ and $y$: $$\lim_{n\to\infty}S(\alpha,2n,n)^{1/n}=\left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2.$$
%\item<5->For $\alpha=\sqrt3-1=[\overline{1,2}]$, $S(\alpha,2n,n)^{1/n}=\frac{1}{4}(3+2\sqrt2)$.
\end{itemize}



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\subsection{}
\begin{frame}{Our results on periodic continued fraction averages 2/2}

\begin{block}{Theorem 3}
Let $\alpha=[\overline{x,y}]$. Then $S(\alpha,n,\lceil\tfrac{n}{2}\rceil)^{1/\lceil\tfrac{n}{2}\rceil}$ converges as $n\to\infty$ to the $\frac{1}{2}$-H\"{o}lder mean of $x$ and $y$: $$\lim_{n\to\infty}S(\alpha,n,\lceil\tfrac{n}{2}\rceil)^{1/\lceil\tfrac{n}{2}\rceil} = \left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2.$$
\end{block}

\ \\  

\noindent Suffices to show for $n\equiv 0\bmod2$, say $n=2k$.

\noindent In this case we have that $S(\alpha,2k,k)^{1/k}\to\left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2$ \textcolor{red}{monotonically} as $k\to\infty$.



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\subsection{}
\begin{frame}{On the proof of Theorem 3, 1/2}
\alert{
$${\rm Goal:}\ \alpha = [\overline{x,y}] \Rightarrow \lim_{n\to\infty}S(\alpha,n,\lceil\tfrac{n}{2}\rceil)^{1/\lceil\tfrac{n}{2}\rceil} =\left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2.$$
}

\noindent The proof uses an asymptotic formula for Legendre polynomials $P_k$ (with $t=\frac{x}{y}<1$ and $u=\frac{1+t}{1-t}>1$):
\begin{align}
P_k(u)&=\frac{1}{2^k}\sum_{j\ = \ 0}^k{k\choose j}^2(u-1)^{k-j}(u+1)^j\nonumber\\
S(\alpha,2k,k)&=\frac{1}{{2k\choose k}}\sum_{j=0}^k{k\choose j}^2x^j y^{k-j}= \frac{y^k}{{2k\choose k}}\sum_{j=0}^k{k\choose j}^2 t^j\nonumber\\
&=\frac{y^k}{{2k\choose k}}(1-t)^k P_k(u).\nonumber
\end{align}





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\subsection{}
\begin{frame}{On the proof of Theorem 3, 2/2}
$$\alert{
{\rm Goal:}\ \alpha = [\overline{x,y}] \Rightarrow \lim_{n\to\infty}S(\alpha,n,\lceil\tfrac{n}{2}\rceil)^{1/\lceil\tfrac{n}{2}\rceil} =\left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2.}$$
\noindent Using the \emph{generalized Laplace-Heine asymptotic formula} for $P_k(u)$ for $u>1$ and $t=\frac{x}{y}<1$ and $u=\frac{1+t}{1-t}>1$ gives
\begin{align}
S(\alpha,2k,k)^{1/k}&\ = \ y(1-t)\left(\frac{P_k(u)}{{2k\choose k}}\right)^{1/k} \nonumber\\
&\ \longrightarrow\ y(1-t)\frac{u+\sqrt{u^2-1}}{4} \ = \ &y\left(\frac{1+\sqrt t}{2}\right)^2\nonumber\\
&\ = \ \left(\frac{x^{1/2}+y^{1/2}}{2}\right)^2.\nonumber
\end{align}

\ \\

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\subsection{}
\begin{frame}{A conjecture on periodic continued fraction averages 1/3}

Expect the same result of Theorem 3 to hold for every quadratic irrational $\alpha$ and for every $c$. \ \\

\begin{block}{Conjecture 2} For every $\alpha=[\overline{x_1,\ldots, x_L}]$ and every $0\leq c\leq 1$ the limit
$$\lim_{n\to\infty}S(\alpha,n,\lceil cn\rceil)^{1/\lceil cn\rceil}=:F(\alpha, c)$$
exists and it is a continuous function of $c$.
\end{block}

\ \\

Notice $c\mapsto F(\alpha,c)$ is automatically decreasing by Maclaurin's inequalities.





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\subsection{}
\begin{frame}{A conjecture on periodic continued fraction averages 2/3}
\textbf{Conjecture 2} for period 2 and period 3, $0\leq c\leq1$.
\vspace{-0.2cm}
\begin{figure}
\includegraphics[width=10cm]{threeFXkc.eps}
\end{figure}
\vspace{-1.38cm}
\begin{figure}
\includegraphics[width=11cm]{threeFXkcL3.eps}
\end{figure}
%\end{itemize}



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\section{Main Results (Technical)}

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\begin{block}{}
\begin{center}
\alert{\ \\ Main Results\\ \ (Sketch of More Technical Arguments)}
\end{center}
\end{block}

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\subsection{}
\begin{frame}{Explicit Formula for $F(c)$}

\noindent Result of Hal\'asz and Sz\'ekely yields conjecture and $F(c)$. \\ \

\small
\begin{block}{Theorem 4}\small If $\lim_{n \to \infty} \frac{k}{n} = c \in (0,1]$, then for almost all $\alpha \in [0,1]$
\be
\lim_{n \to \infty} S(\alpha, n, k)^{1/k}\ =:\ F(c)\nonumber
\ee exists, and $F(c)$ is continuous and given explicitly by \footnotesize
\be
c(1-c)^{\frac{1-c}{c}}\exp\left\{{\frac{1}{c}\left( (c-1)\log{r_c} - \sum_{k=1}^{\infty}\log{(r_c + k)\log_2\left({1-\frac{1}{(k+1)^2}}\right) } \right) }\right\},\nonumber \normalsize
\ee \small
where $r_c$ is the unique nonnegative solution of the equation
\be
\sum_{k=1}^{\infty} \frac{r}{r+k}\log_2\left({1-\frac{1}{(k+1)^2}}\right)\ =\ c - 1.\nonumber
\ee \normalsize
\end{block}

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\subsection{}
\begin{frame}{Proof: Work of Hal\'asz and Sz\'ekely }

\begin{itemize}

\item Hal\'asz and Sz\'ekely calculate asymptotic properties of iidrv $\xi_1, \dots, \xi_n$ when

\noindent\ \ \ \  $\diamond$ $c = \lim_{n\to\infty} k/n \in [0,1]$.

\noindent\ \ \ \  $\diamond$ $\xi_j$ non-negative.

\noindent\ \ \ \  $\diamond$ $\mathbb{E}[\log \xi_j] < \infty$ if $c=1$.

\noindent\ \ \ \  $\diamond$ $\mathbb{E}[\log (1 + \xi_j) < \infty$ if $0<c<1$.

\noindent\ \ \ \  $\diamond$ $\mathbb{E}[\xi_j] < \infty$ if $c=0$.\\ \

\item Prove $\lim_{n\to\infty} \sqrt[k]{S(\xi,n,k)/\ncr{n}{k}}$ exists with probability 1 and determine it.\\ \

\end{itemize}

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\subsection{}
\begin{frame}{Proof: Work of Hal\'asz and Sz\'ekely}

\noindent Random variables $a_i(\alpha)$ not independent, but Hal\'asz and Sz\'ekely only use independence to conclude sum of the form
\be
\frac{1}{n} \sum_{k=1}^{n} f(T^k(\alpha))\nonumber
\ee
(where $T$ is the Gauss map and $f$ is some function integrable with respect to the Gauss measure) converges a.e. to $\mathbb{E}f$ as $n \to \infty$.\\ \

\noindent Arrive at the same conclusion by appealing to the pointwise ergodic theorem.

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\section{Refs}

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\begin{block}{}
\begin{center}
\alert{\ \\ References\\ \ }
\end{center}
\end{block}

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\subsection{}
\begin{frame}{References}

\footnotesize

\begin{thebibliography}{1234}

\bibitem[BC]{BC}
I. Ben-Ari and K. Conrad, \emph{Maclaurin's inequality and a generalized Bernoulli inequality}, Math. Mag. \textbf{87} (2014), 14--24.%\\ \

\bibitem[C]{C}
F. Cellarosi, D. Hensley, S. J. Miller and J. Wellens, \emph{Continued Fraction Digit Averages and Maclaurin's Inequalities}, to appear in Experimental Mathematics. \textcolor{blue}{\url{http://arxiv.org/abs/1402.0208}}.%\\ \

\bibitem[DV]{DV}
H. G. Diamond and J. D. Vaaler, \emph{Estimates for Partial Sums of Continued Fraction
Partial Quotients}, Pacific Journal of Mathematics \textbf{122} (1986), 73--82.%\\ \

\bibitem[HS]{HS}
G. Hal\'asz and G. J. Sz\'ekely, \emph{On the elementary symmetric polynomials of independent random variables},
Acta Math. Acad. Sci. Hungar. \textbf{28} (1976), no. 3-4, 397--400.%\\ \

\bibitem[Kh]{Kh}
A. Y. Khinchin, \emph{Continued Fractions},  3rd edition, University of Chicago Press, Chicago, 1964.%\\ \

\end{thebibliography}

\ \\

%\normalsize

Work supported by AMS-Simons Travel grant, NSF grants DMS0850577, DMS0970067, DMS1265673 and DMS1363227, and Williams College.

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\end{document}

