\documentstyle[12pt]{article}

\textwidth=17.0cm \textheight=630pt \evensidemargin=1mm \oddsidemargin=1mm

\renewcommand{\refname}{Bibliography}

\title{Continued Fraction of $\sqrt[N]{M}$}
%\title{• ®á ¨ ƒ ãáá}
\author{V.A.Malyshev}
\date{}

\begin{document}
\maketitle

\begin{abstract}
ø{\small\it $\!\!\!\!\!\!\!\!\!\!\!\!$ Let $\sqrt[N]{M}$ be an algebraic number of order
$N\ge 3$, and let $r_{0},r_{1},\ldots$ be the sequence of remainders of its continued
fraction. Hypothesis: The sequence
$$
ø\xi_{n}=\log_{2}(1+r_{n})
$$
has the uniform distribution.}
\end{abstract}

\begin{center}
\large{ø}
\end{center}
\bigskip

\noindent {Introduction}
\medskip

\noindent {1. Continued fractions}

\noindent {2. Gauss theorem}

\noindent {3. Remainders of the continued fraction}

\noindent {4. Elements of the continued fraction}


\medskip\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%
\section*{Introduction}

In the paper we present a reliable hypothesis about the remainders of the continued
fraction of the algebraic number $\sqrt[N]{M}$ of order $N\ge 3$. From the hypothesis it
follows that the continued fraction has an unbounded sequence of elements. In particular,
this gives the answer to the Kchinchin problem~\cite{Hi1} about the continued fractions
of algebraic numbers.
\bigskip
\bigskip
\bigskip

\begin{center}
 \begin{picture}(400,20)
 \put(-40,30){\line(1,0){70}}
 \put(-25,15){{\footnotesize{\it 1991 Mathematics Subject Classification.} Primary 11-04, 11K50.}}
 \put(-25,0){{\footnotesize{\it Key words and phrases.} Algebraic numbers, continued fractions.}}
\end{picture}
\end{center}

\newpage





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. Continued fractions}

Let $x>0$. Then
$$
x\;=\;a_{0}+r_{0}\;=\;a_{0}+\frac{1}{\displaystyle a_{1}+r_{1}}\;=\;
  a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}+r_{2}}}
\;=\;
 a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}
 +\frac{1}{\displaystyle a_{3}+r_{3}}}}\;=\;\cdots\;
$$
where
$$
ø\begin{array}{llr}
 a_{0}&=&[x]\\
 r_{0}&=&x-[x]
\end{array}
$$
and
$$
ø\begin{array}{llr}
 a_{n}&=&\left[1/r_{n-1}\right],\\
 r_{n}&=&1/r_{n-1}-\left[1/r_{n-1}\right].
\end{array}
$$
for $n=1,2\ldots\;$. The real numbers $r_{0},r_{1},\ldots$ are called the remainders, and
the integer numbers $a_{0},a_{1},\ldots$ are called the elements of the continued
fraction.

Continued fractions of rational numbers are finite.
$$
\frac{3}{5}\;=\;
  \frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 2}}}
$$
Continued fractions of algebraic numbers of order 2 are periodic.
$$
\sqrt{2}\;=\;1
 +\frac{1}{\displaystyle 2
 +\frac{1}{\displaystyle 2
 +\frac{1}{\displaystyle 2
 +\frac{1}{\displaystyle 2
+\cdots}}}}
$$
Continued fractions of algebraic numbers of order $\ge 3$ are chaotic.
$$
\sqrt[3]{2}\;=\;1
 +\frac{1}{\displaystyle 3
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 5
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 4
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 8
 +\frac{1}{\displaystyle 1
 +\frac{1}{\displaystyle 14+\cdots}}}}}}}}}}}
$$
The first $100\,000$ elements of the last continued fraction look like a stochastic
process between 1 and $532\,220$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Gauss theorem}

Let $x$ be a random variable with the uniform distribution on $[0,1]$. For each
$n=0,1,2,\ldots$ the remainder $r_{n}$ is a random variable on $[0,1]$. Let $f_{n}(t)$ be
the probability density of the random variable $r_{n}$. Since $r_{0}=x$, we have
$
f_{0}(t)=1.
$
Then
$$
f_{n}(t)=\sum_{k=1}^{\infty}\frac{1}{(t+k)^{2}}f_{n-1}\left(\frac{1}{t+k}\right).
$$
From
$$
ø\frac{1}{1+t}=
\sum_{k=1}^{\infty}\frac{1}{(t+k)^{2}}\frac{1}{1+{\displaystyle\frac{1}{t+k}}}
$$
and
$$
ø\int_{0}^{1}f_{n}(t)\,dt=1
$$
it may be proved that
$$
\lim_{n\to\infty}f_{n}(t)=\frac{1}{\ln2}\,\frac{1}{1+t}.
$$
The proof of the Gauss theorem is in~\cite{Hi1}.

Thus the random variables
$$
ø\xi_{n}=\log_{2}(1+r_{n})
$$
converge to the uniform distribution. This leads to a reliable hypothesis about the
remainders of the continued fraction of the algebraic number $\sqrt[N]{M}$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. Remainders of the continued fraction}


We need the notion of the density of a sequence~\cite{Pol1}. Let $x_{1},x_{2},\ldots\;$
be a sequence on $[0,1]$. Suppose that for each $0\le t\le 1$ there exists the limit
$$
F(t)=\lim_{n\to\infty}\frac{m_{n}(t)}{n},
$$
where $m_{n}(t)$ is the number of points $x_{1},\ldots, x_{n}$ such that $0\le x_{k}\le
t$. The function
$$
f(t)=\frac{d}{dt}F(t)
$$
is called the density of the sequence $x_{1},x_{2},\ldots\;$. In the case $f(t)=1$ we say
that the sequence has the uniform distribution.

For example, for any irrational $x$  the sequence
$$
x_{n}=nx-[nx]
$$
has the uniform distribution.

Suppose that the irrational algebraic number $\sqrt[N]{M}$ is expanded in the continued
fraction
$$
\sqrt[N]{M}\;=\;a_{0}+r_{0}\;=\;a_{0}+\frac{1}{\displaystyle a_{1}+r_{1}}\;=\;
  a_{0}+\frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}+r_{2}}}
\;=\;a_{0}+
 \frac{1}{\displaystyle a_{1}
 +\frac{1}{\displaystyle a_{2}
 +\frac{1}{\displaystyle a_{3}+r_{3}}}}\;=\;\cdots
$$
\medskip

\noindent{\bf Hypothesis.} If $N\ge 3$, then the sequence
$$
ø\xi_{n}=\log_{2}(1+r_{n})
$$
has the uniform distribution.
\medskip

\noindent{\bf Example.} Let us check the hypothesis for the number $\sqrt[3]{2}$. By
direct calculation, we obtain the first 1000 elements $\xi_{0},\xi_{1},\ldots,\xi_{999}$
and see the expected  identities
$$
ø\frac{1}{1000}\sum_{k=0}^{999}\xi_{k}=0.501912...\approx\frac{1}{2}
$$
and
$$
ø\frac{1}{1000}\sum_{k=0}^{999}\xi_{k}^{2}=0.332463...\approx\frac{1}{3}.
$$
In the table
\bigskip
\begin{center}
ø\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
  ø1 & ø2 & ø3 & ø4 & ø5 & ø6 & ø7 & ø8 & ø9 & 10\\\hline\hline
  ø90 & ø98 & ø109 & ø102 & ø101 & ø89 & ø115 & ø96 & ø112 & 88 \\
\end{tabular}
\end{center}
\bigskip
we see how many elements $\xi_{0},\xi_{1},\ldots,\xi_{999}$ are in the intervals
$$
 \left[\frac{k-1}{10},\frac{k}{10}\right]
$$
where $k=1,2,\ldots,10$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{4. Elements of the continued fraction}

The chaotic behavior of the remainders $r_{0},r_{1},\ldots$ leads to the chaotic behavior
of the elements $a_{0},a_{1},\ldots\;$. Nevertheless the elements may be calculated by
means of recurrence relations.

The numerator and denominator of the $k$th convergent are calculated as follows
\begin{eqnarray*} ø
 P_{k}&=&a_{k}P_{k-1}+P_{k-2},\\
 Q_{k}&=&a_{k}Q_{k-1}+Q_{k-2},\\
\end{eqnarray*}
where
$$
ø\begin{array}{c}
  øP_{-1}=1 \\
  øQ_{-1}=0
\end{array}
\;\;\;\;{\rm and}\;\;\;\; ø\begin{array}{c}
  øP_{0}=a_{0} \\
  øQ_{0}=1
\end{array}
$$
It seems to be true that for any $\sqrt[N]{M}$ there exists $n$ such that
$$
øa_{k}=\left[-\frac{(P^{N}_{k-2}-MQ^{N}_{k-2})P^{N-1}_{k-1}}
                   {(P^{N}_{k-1}-MQ^{N}_{k-1})P^{N-1}_{k-2}}\right].
$$
for all $k\ge n+1$. For $3\le N\le 5$ and $2\le M\le 10$ such $n$ may be taken from the
table.
\bigskip
\begin{center}
ø\begin{tabular}{c||c|c|c|c|c|c|c|c|c}
  ø  & ø2 & ø3 & ø4 & ø5 & ø6 & ø7 & ø8 & ø9 & ø10 \\\hline\hline
  ø3 & ø1 & ø2 & ø3 & ø3 & ø3 & ø3 & ø$\emptyset$ & ø1 & ø1 \\\hline
  ø4 & ø2 & ø2 & ø2 & ø3 & ø4 & ø3 & ø4 & ø2 & ø5 \\\hline
  ø5 & ø1 & ø2 & ø2 & ø2 & ø3 & ø2 & ø3 & ø3 & ø5 \\
\end{tabular}
\end{center}
\bigskip


\noindent{\bf Example.} For $\sqrt[3]{2}$ we have $n=1$. We put
$$
ø\begin{array}{c}
  øa_{0}=1 \\
  øP_{0}=1 \\
  øQ_{0}=1 \\
\end{array}
\;\;\;\;{\rm and}\;\;\;\; ø\begin{array}{c}
  øa_{1}=3 \\
  øP_{1}=4 \\
  øQ_{1}=3 \\
\end{array}
$$
Then
\begin{eqnarray*} ø
 a_{k}&=&\left[-\frac{(P^{3}_{k-2}-2Q^{3}_{k-2})P^{2}_{k-1}}
                   {(P^{3}_{k-1}-2Q^{3}_{k-1})P^{2}_{k-2}}\right],\medskip\\
 P_{k}&=&a_{k}P_{k-1}+P_{k-2},\medskip\\
 Q_{k}&=&a_{k}Q_{k-1}+Q_{k-2}\\
\end{eqnarray*}
for all $k=2,3\ldots\;$. The recurrence relations give the elements
$$
ø1\;\;3\;\;1\;\;5\;\;1\;\;1\;\;4\;\;1\;\;1\;\;8\;\;1\;\;14\;\;1\;\;10\;\;2\;\;1\;\;4\;\;12
 \;\;2\;\;3\;\;2\;\;1\;\;3\;\;4\;\;1\;\;1\;\;2\;\;14\;\;3\;\;12\;\;1\;\;\ldots
$$
of the continued fraction of the algebraic number $\sqrt[3]{2}$.




\begin{thebibliography}{9}


\bibitem{Hi1} A.J.Khinchin. Continued fractions. Nauka, Moscow, 1978.



\bibitem{Pol1}
G.Polya, G.Szeg$\ddot{o}$. Aufgaben und Lehrs$\ddot{a}$tze aus der Analysis.
Sprnger-Verlag, Berlin, 1964


\end{thebibliography}

Vladimir Malyshev

wmal@ryb.adm.yar.ru

\end{document}