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\begin{document}
\parskip \dist
\title{Distribution of Powers Modulo 1 and Related Topics}
\author{Miguel A. Lerma}
\address{Department of Mathematics, RLM 8.100,
University of Texas at Austin, Austin, Texas 78712}
\email{mlerma@@math.utexas.edu}
\date{June 19, 1995}
\thanks{I am grateful to Prof.~Jeffrey Vaaler for his
support and advice. I am also grateful to Prof.~David Boyd
and Prof.~Alf~van der Poorten for their valuable help.}
\maketitle
\begin{abstract}
This is a review of several results related to distribution
of powers and combinations of powers modulo~1.
We include a proof that given any sequence
of real numbers $\theta_n$, it is possible to get an
$\alpha$ (given $\lambda \neq 0$), or a $\lambda$
(given $\alpha > 1$) such that $\lambda\,\alpha^n$
is close to $\theta_n$ modulo~1. We also prove that
in a number field, if a combination of powers
$\lambda_1\,\alpha_1^n + \dots + \lambda_m\,\alpha_m^n$
has bounded $v$-adic absolute value (where $v$ is any
non-Archimedean place) for $n \geq n_0$, then the
$\alpha_i$'s are $v$-adic algebraic integers.
Finally we present several open problems
and topics for further research.
\end{abstract}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction} \label{s:intro}
The study of the behaviour of sequences of the form
$\alpha^n$ modulo 1 has some interesting connections
with subjects such as Waring's problem. Let $g$ be
the function:
\dis
{
g(k) = \min\,\{\,s \in \mathbb N\,:
\,a = n_1^k + \dots + n_s^k \text{ for all } a \in \mathbb N\,\}
}
and $\|x\| =$ distance from x to the nearest integer.
Then it is well known that for $k \geq 5$, if
\dis
{
\|(3/2)^k\|\ >\ (3/4)^k
}
then
\dis
{
g(k)\ =\ 2^k + [(3/2)^k] - 2
}
so that the rate to which $(3/2)^k \pmod 1$
accumulates near 0 determines the value of $g(k)$.
In a slightly more general setting, the study of sequences
of the form $\lambda\,\alpha^n$ also has interesting
applications. As an example, it is known
(\cite{kuip74}, theorem~8.1) that a number
$\lambda$ is normal to the base $b$, i.e., all its
finite sequences of k digits to base b occur with
the same relative frequency $1/b^k$, if and only if the
sequence $\lambda\,b^n$ is u.d.~mod~1 (uniformly
distributed modulo 1, see definition~\ref{d:unidis}
below), so that techniques from the theory of distribution
modulo~1 apply to this problem.
The study of slowly growing sequences such as
$\{n\omega\}_{n=1}^\infty$ is not hard,
but when the sequence grows very rapidly, as with
$x_n = \alpha^n$ for $\alpha >1$, the fractional part
of $x_n$ becomes almost negligible compared to its integer
part. So, special techniques have had to be developed
to tackle the problem in that case.
Several results related to distribution of powers modulo~1
are presented below.
\section{Overview of results} \label{over}
A natural question is if a sequence of the form $\alpha^n$,
or more generally, of the form $\lambda\,\alpha^n$ for some
fixed $\lambda \neq 0$, is u.d.~mod~1.
Koksma's metric theorem~\ref{t:koksma}
shows that in fact, $\lambda\,\alpha^n$
is u.d.~mod~1 for almost every $\alpha > 1$, i.e., the
exceptional set of $\alpha$'s for which $\lambda\,\alpha^n$
is not u.d.~mod~1 has Lebesge measure zero. It is surprising,
however, that there is no known concrete example of a real
number $\alpha > 1$ for which $\alpha^n$ is u.d.~mod~1;
only members of the exceptional set are known. For instance,
the following classes of numbers are known to be in the
exceptional set:
\begin{enumerate}
\item Integers ($>1$), since $\alpha^n = 0 \pmod 1$
for every $\alpha \in \mathbb Z$.
\vskip \dist
\item Pisot-Vijayaraghavan (or P.V., or Thue) numbers.
A P.V. number is a real algebraic integer $\alpha > 1$
whose conjugates lie inside the open unit disc
$\{\,z~\in~\mathbb C\,:\, |z| < 1\,\}$ (the rational
integers greater than 1 are P.V. numbers).
If $\alpha$ is a P.V number then
$\lim_{n \to \infty} \alpha^n = 0 \pmod 1$ geometrically
(\cite{sale63}, p.~3; \cite{bert92}, theorem~5.3.1).
\vskip \dist
\item Salem numbers. A Salem number is a real algebraic
integer $\alpha > 1$ whose conjugates lie all in the closed
unit disc $\{\,z \in \mathbb C\,:\, |z| \leq 1\,\}$,
and at least one of them is in the border of the disc
(actually it can be readily seen that all of them except one
will be in the border). If $\alpha$ is a Salem number,
then $\{\alpha^n\}_{n=1}^\infty$ is dense modulo 1,
i.e., the fractional parts of $\alpha^n$ are dense in
the interval $[0,1)$, but it is not u.d.~mod~1
(\cite{sale63}, p.~33; \cite{bert92}, theorem~5.3.2).
\end{enumerate}
All those examples are algebraic numbers, so it is natural
to ask if there is any transcendental number $\alpha$ such
that $\alpha^n$ is not u.d.~mod~1. A result in this direction
is the following theorem of Boyd (\cite{boyd69}):
\begin{theorem} \label{t:baddis}
Let $A$, $B$ be real numbers with $3 < A < B$,
and let $a_0$ be an integer satisfying
$a_0 > (A + 1)\,(A - 1)^{-1}\,(B - A)^{-1}$.
Then there is an uncountable set $S \subset [A,B]$,
such that for each $\alpha \in S$, there is a real number
$\lambda = \lambda(\alpha) > 0$ for which
\dis
{ \label{e:baddis}
\|\lambda\,\alpha^n\|\ \leq\ (A - 1)^{-1}\,(\alpha - 1)^{-1}
\qquad \text{\rm for } n = 0,1,\dots
}
The integer $a_0$ is the nearest integer to $\lambda(\alpha)$
for all $\alpha \in S$.
\end{theorem}
Since $S$ is uncountable, it will contain transcendental
numbers. On the other hand, (\ref{e:baddis}) shows that
$\lambda\,\alpha^n$ will be in a small interval
around zero modulo 1, so it cannot be u.d.~mod~1.
The idea behind the proof of theorem~\ref{t:baddis}
is to get a sequence of positive integers $a_n$ that
will play the role of nearest integers to
$\lambda\,\alpha^n$. Then $\lambda$ and
$\alpha$ will be obtained as the following
limits:
\dis
{
\alpha\ =\ \lim_{n \to \infty} \frac {a_{n+1}} {a_n}
\qquad\text{ and }\qquad
\lambda\ =\ \lim_{n \to \infty} a_{n} \alpha^{-n}
}
Here, $a_0$ is given in the hypothesis of the theorem,
$a_1$ will be any integer such that
\dis
{
a_0\,A + (A-1)^{-1} < a_1 < a_0\,B - (A-1)^{-1}
}
and for $n \geq 1$:
\dis
{
a_{n+1}\ =\ [a_n^2/a_{n-1}] + f(n)
}
where $[x]$ is the integer part of x, and
$f \in \mathfrak J =$ the set of functions
$\mathbb Z^{+} \to \{0,1\}$. Since $\mathfrak J$
is uncountable, and each function $f \in \mathfrak J$
gives a different $\alpha$, the set of $\alpha$'s
that can be found this way is uncountable. Furthermore,
it can be proved that
\dis
{
|a_n - \lambda\,\alpha^n|\ \leq
\ (A - 1)^{-1}\,(\alpha - 1)^{-1}
\qquad \text{\rm for } n = 0,1,\dots
}
which gives (\ref{e:baddis}) and ensures that $a_n$ is, in fact,
the nearest integer to $\lambda\,\alpha^n$.
However, this result does not say anything about $\lambda = 1$
or any fixed value of $\lambda \neq 0$.
The answer to the question is actually positive,
as shown in theorem~\ref{t:seq}.
Given $\lambda > 0$, and given any sequence
$\theta_n$, we can get an $\alpha$ such that
$\lambda\,\alpha^n$ is close to $\theta_n$ mod~1.
The idea is to start with an "initial" value
$\alpha_1 > A > 2$ (A large) such that
$\lambda\,\alpha_1 = \theta_1 \pmod 1$.
Next, slightly perturb the value of $\alpha_1$, i.e.,
find a slightly greater $\alpha_2 \geq \alpha_1$, such
that now $\lambda\,\alpha^2 = \theta_2$.
After doing so, we will not have
$\lambda\,\alpha_2 = \theta_1 \pmod 1$ any more, but the difference
$\|\lambda\,\alpha_2 - \theta_1\|$ can be made less than
$1/\alpha_1 < A^{-1}$ just by taking
$0 \leq \alpha_2 - \alpha_1 \leq \frac 1 {\lambda\,\alpha_1}$.
The process can be repeated making
$\lambda\,\alpha_n^n = \theta_n \pmod 1$ by perturbing the value
of $\alpha_{n-1}$ less than
$\lambda^{-1}\,\alpha_{n-1}^{1-n} < \lambda^{-1}\,A^{1-n}$,
in such a way that the sequence $\alpha_n$ will converge
to a limit $\alpha$ such that
\dis
{
\|\lambda\,\alpha^n - \theta_n \|\ \leq\
\sum_{k=1}^\infty A^{-k}\ =\ \frac 1 {A-1}
}
for $n=1,2,\dots$ (actually, a little trick at
the end of the proof allows us to halve the bound).
>From here, the existence of uncountably many real numbers,
so uncountably many transcendental numbers, whose powers are not
u.d.~mod~1, follows easily (corollary~\ref{c:trans}).
In Pisot's thesis (\cite{piso38}) there are some
general results related to the previous one.
In particular, the application of his theorem~II,
on p.~215 of \cite{piso38}, would show that there are
arbitrarily large numbers $\alpha$ such that:
\dis
{
\limsup_{n \to \infty} \|\alpha^n - \theta_n\|
\ \leq\ \frac 1 {2\,(\alpha-1)}
}
Also, his theorem on p.~225 of \cite{piso38} implies
that for every $\varepsilon > 0$ there is some arbitrarily
large $\alpha$ such that:
\dis
{
\|\alpha^n - \theta_n\|
\ <\ \frac {1 + \varepsilon} {2\,(\alpha-1)}
}
for $n \geq n_0(\varepsilon)$.
The next question that arises is if
the sequence $\langle \alpha^n \rangle =$ fractional
part of $\alpha^n$ uniquely determines $\alpha$.
This can be stated as if $\alpha^n - \beta^n \in \mathbb Z$
for every $n \geq n_0$, for some $n_0$, implies that
$\alpha = \beta$. The answer is yes, except in the trivial
case when $\alpha$ and $\beta$ are both integers.
The techniques used to solve this problem are easy to
generalize to the study of combinations
$S_n = \lambda\,\alpha^n + \mu\,\beta^n$ (with
$\lambda,\mu,\alpha,\beta \neq 0$ and $\alpha \neq \beta$)
such that $S_n \in \mathbb Z$ for every $n \geq n_0$ for some
$n_0$. The result now is that $\alpha$ and $\beta$ are rational
integers or conjugate algebraic integers of degree two.
Furthermore, $\lambda \in \mathbb Q(\alpha)$ and
$\mu \in \mathbb Q(\beta)$. In general we do not have
that $\mathbb Q(\lambda) = \mathbb Q(\alpha)$ and
$\mathbb Q(\mu) = \mathbb Q(\beta)$.
Of course, if $\alpha$ is a quadratic integer
and $\beta = \bar \alpha$ denotes its conjugate,
then $\alpha^n + \beta^n \in \mathbb Z$,
whilst $\lambda = 1$ does not generate
$\mathbb Q(\alpha)$.
However, the equality does
hold whenever $\lambda \neq \mu$.
Pisot's thesis (\cite{piso38}) also contains similar
results, applicable to sequences of the form
\dis
{
S_n\ =\ \lambda_1(n)\,\alpha_1^n +
\lambda_2(n)\,\alpha_2^n + \dots + \lambda_m(n)\,\alpha_m^n
%S_n\ =\ \sum_{i=1}^{m} \lambda_i(n)\,\alpha_i^n
}
where $\alpha_i \in \mathbb C$ and the coefficients
are polynomials with complex coefficients.
A generalization of those results to local fields
is given in theorem~\ref{t:compow}. Note that
$S_n \in \mathbb Z$ can be expressed as $S_n \in \mathbb Q$
and $|S_n|_v \leq 1$ for every non-Archimedean place
$v$ in $\mathbb Q$. The generalization given in
theorem~\ref{t:compow} allows us to study what happens
at each individual non-Archimedean place $v$. In that theorem,
$\mathbb Q$ is replaced by some subfield $k'$ of a finite
extension of a local field $k_v$, where $k_v =$ completion of
a number field $k$ at a non-Archimedean place $v$,
the $\alpha_i$'s and the coefficients of the
$\lambda_i$'s are in an arbitrary
extension of $k'$, and for $n \geq n_0$,
$S_n$ is assumed to be in $k'$ and to have
bounded $v$-adic absolute value.
Under those hypothesis, the main conclusion
is that $\alpha_i \in \overline {k'}$ and
$|\alpha_i|_v \leq 1$ for every $i$, which
is the local version of being algebraic integers.
The proof runs along the following lines:
Since $|S_n|_v$ is bounded for $n \geq n_0$, it is possible
to multiply it by some $b \in k'$ so that $c_n = b\,S_n$ verifies
that $|c_n|_v \leq 1$, i.e.,
\dis
{
c_n\ \in\ \mathcal O_{k',v}\ =
\ \{\,x \in k'\,:\,|x|_v \leq 1\,\}
}
for $n \geq n_0$.
Furthermore, it is well known that if
\dis
{
f(X)\ =
\ \prod_{i=1}^{m} (X-\alpha_i)^{d_i} =
X^M - r_1\,X^{M-1} - r_2\,X^{M-2} - \dots - r_M
}
where $d_i = 1 + \deg\,\lambda_i$, $i=1,2,\dots,m$, and
$M = \sum_{i=1}^m d_i$,
then $c_n$ verifies a recurrence relation of the form
\dis
{
c_n\ =\ r_1\,c_{n-1} + r_2\,c_{n-2} + \dots + r_M\,c_{n-M}
}
Next, define the formal power series
\dis
{
F_t(X)\ =
\ c_t + c_{t+1}\,X + c_{t+2}\,X^2 + c_{t+3}\,X^3 + \dots
}
which is in $\mathcal O_{k',v}[[X]]$.
We have $F_t(X) = p_t(X)/q(X)$,
where $p_t(X),q(X) \in k'[X]$ and
$q(X) = X^M\,f(1/X)$ is the reciprocal of $f(X)$.
Using the generalized Fatou's lemma (lemma~\ref{l:fatou}),
we conclude that $q(X) \in \mathcal O_{k',v}[X]$, so
$f(X) \in \mathcal O_{k',v}[X]$. Since $f(X)$ is monic,
we get that its roots verify $|\alpha_i|_v \leq 1$.
Theorem~\ref{t:compow} also shows that the coefficients
of each $\lambda_i(X)$ are in $k'(\alpha_i)$. Even more,
any $k'$-automorphism $\sigma$ in $k'(\alpha_1,\dots,\alpha_m)$,
which acts as a permutation over the $\alpha_i$'s,
induces the same permutation over the $\lambda_i$'s.
Finally, proposition~\ref{p:grow} shows that the condition
$|S-n|_v = O(1)$ can be weakened to a subexponential
growth condition of the form $|S-n|_v = O(A^n)$ for
every $A > 1$.
In the next sections we give details about the above results.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Distribution of powers modulo 1} \label{s:dispow}
\begin{definition}
For any real number $x$, we define:
\begin{enumerate}
\item Integer part of $x$:
$[x] = \max\,\{\,n \in \mathbb Z\,:\,n \leq x\,\}$.
\vskip \dist
\item Fractional part of $x$: $\langle x \rangle = x - [x]$.
\vskip \dist
\item Nearest integer to $x$:
$E(x) = \max \{\,n \in \mathbb Z\,:\,n \leq x+1/2\,\}$.
\vskip \dist
\item Residue of $x$ modulo~1:
$\varepsilon(x) = x - E(x)$.
\vskip \dist
\item Distance from $x$ the the nearest integer:
\[
\|x\| = |\varepsilon(x)| =
\min\,\{\,|x-n|\,:\,n \in \mathbb Z\,\}
\]
\end{enumerate}
\end{definition}
\begin{definition} \label{d:unidis}
A sequence of real numbers $\{x_n\}_{n=1}^\infty$
is said to be uniformly distributed modulo 1,
abbreviated u.d.~mod~1, if
\dis
{
\lim_{N \to \infty} \frac 1 N \, \sum_{n=1}^N \chi_{s,t}(x_n)
\ =\ t - s
}
whenever $s < t < s+1$. Here:
\dis
{
\chi_{s,t}(x)\ =
\ \begin{cases}
1 &\text{ if } \quad s < x-n < t
\quad \text{ for some } \quad n \in \mathbb Z \\
1/2 &\text{ if } \quad s-x \in \mathbb Z
\quad \text{ or } \quad t-x \in \mathbb Z \\
0 &\text{ otherwise }
\end{cases}
}
\end{definition}
\begin{theorem}[Weyl criterion]
The sequence $\{x_n\}_{n=1}^\infty$ is u.d.~mod~1
if and only if for every $h \in \mathbb Z \setminus \{0\}$:
\dis
{
\lim_{N \to \infty} \frac 1 N\,
\sum_{n=1}^N \exp(2\pi i\,h\,x_n)\ =\ 0
}
\end{theorem}
\begin{pf} See theorem~2.1 in \cite{kuip74}. \end{pf}
\begin{theorem}[Weyl-Koksma's Metric Theorem] \label{t:koksma}
\item{(a)} Let $\alpha > 1$ be a real number; the sequence
$\{\lambda\,\alpha^n\}_{n=1}^\infty$ is uniformly distributed
modulo 1 for almost all real $\lambda$ (Weyl).
\item{(b)} Let $\lambda$ be a non zero real number; the sequence
$\{\lambda\,\alpha^n\}_{n=1}^\infty$ is uniformly distributed
modulo 1 for almost all real $\alpha > 1$ (Koksma).
\end{theorem}
\begin{pf} See \cite{weyl16} and \cite{koks35}.
See also \cite{bert92}, p.~71.
\end{pf}
\begin{theorem} \label{t:seq}
Let $\{\theta_n\}$ be any sequence of
real numbers. Then:
\item[(1)] Given any $\lambda \neq 0$ and $A > 1$, there exists
an $\alpha$ such that:
\dis
{ \label{e:seq1a}
A\ \leq\ \alpha\ \leq A + \frac A {|\lambda|\,(A-1)}
}
and for every $n \geq 1$
\dis
{ \label{e:seq1b}
\| \lambda\,\alpha^n - \theta_n \|\ \leq\ \frac 1 {2\,(A-1)}
}
\item[(2)] Given any $\alpha > 1$ and $L \neq 0$, there exists
a $\lambda$ (with the same sign as L) such that:
\dis
{ \label{e:seq2a}
|L|\ \leq\ |\lambda|\ \leq\ |L| + \frac 1 {\alpha - 1}
}
and for every $n \geq 1$
\dis
{ \label{e:seq2b}
\| \lambda\,\alpha^n - \theta_n \|\ \leq\ \frac 1 {2\,(\alpha - 1)}
}
\end{theorem}
\begin{pf}
\item{(1)} There is no loss of generality in assuming
that $\lambda > 0$. Otherwise, make
$\lambda' = -\lambda$ and $\theta^{\prime}_n = -\theta_n$,
and use that $\|-x\| = \|x\|$.
Construct an increasing sequence $A = \alpha_0
\leq \alpha_1 \leq \alpha_2 \dots$, by the following recursive
rule ($n \geq 0$):
\dis
{ \label{e:recrul}
\begin{aligned}
\alpha_{n+1}\ &=\ \lambda^{-1/(n+1)}\,(\lambda\,\alpha_n^{n+1}
+ \langle \theta_{n+1} - \lambda\,\alpha_n^{n+1} \rangle)^{1/(n+1)} \\
&=\ \lambda^{-1/(n+1)}\,(\theta_{n+1}
- [\theta_{n+1} - \lambda\,\alpha_n^{n+1}])^{1/(n+1)}
\end{aligned}
}
where $[x]=$ integer part of $x$, and
$\langle x \rangle = x - [x]$.
Since $\langle \theta_{n+1} - \lambda\,\alpha_n^{n+1} \rangle
\geq 0$ we have that $\alpha_{n+1} \geq \alpha_n$, so the
sequence is non decreasing. Also, by construction
$\langle \lambda\,\alpha_n^n \rangle = \langle \theta_n \rangle$
for every $n \geq 1$.
Now, for every $1 \leq m \leq n$:
\dis
{
\begin{aligned}
\alpha_{n+1}^m - \alpha_n^m
\ &=\ \lambda^{-m/(n+1)}\,(\lambda\,\alpha_n^{n+1}
+ \langle \theta_{n+1} - \lambda\,\alpha_n^{n+1} \rangle)^{m/(n+1)}
- \alpha_n^m \\
&\leq\ \lambda^{-m/(n+1)}\,\left( (\lambda\,\alpha_n^{n+1} + 1)^{m/(n+1)}
- (\lambda\,\alpha_n^{n+1})^{m/(n+1)} \right) \\
&=\ \lambda^{-m/(n+1)}\,\frac m {n+1}\,
\int_0^1 (\lambda\,\alpha_n^{n+1} + x)^{(m-n-1)/(n+1)}\,dx \\
&\leq\ \lambda^{-m/(n+1)}\,\frac m {n+1}\,
(\lambda\,\alpha_n^{n+1})^{(m-n-1)/(n+1)} \\
&<\ \lambda^{-1}\,\alpha_n^{m-n-1}\ \leq\ \lambda^{-1}\,A^{m-n-1}
\end{aligned}
}
For $m=1$ we have $\alpha_{n+1} - \alpha_n < \lambda^{-1}\,A^{-n}$,
thus:
\begin{multline}
\alpha_N - \alpha_0\ =\ \sum_{n=0}^{N-1} (\alpha_{n+1} - \alpha_n) \\
\ =
\ \sum_{n=0}^{N-1} \lambda^{-1}\,A^{-n}\ <
\ \sum_{n=0}^{\infty} \lambda^{-1}\,A^{-n}\ =
\ \frac A {\lambda\,(A-1)}
\end{multline}
So, for every $N \geq 0$ we have
$A \leq \alpha_N < A + \frac A {\lambda\,(A-1)}$.
Hence, $\alpha = \lim_{N \to \infty} \alpha_N$ exists, and
\[
A\ \leq\ \alpha\ \leq\ A + \frac A {\lambda\,(A-1)}
\]
which is (\ref{e:seq1a}).
Now, for $N > m \geq 1$:
\begin{multline}
0\ \leq \lambda\,\alpha_N^m - \lambda\,\alpha_m^m\ =
\ \sum_{n=m}^{N-1} \lambda\,(\alpha_{n+1}^m - \alpha_n^m) \\
\leq\ \sum_{n=m}^{N-1} A^{m-n-1}\ <
\sum_{n=m}^{\infty} A^{m-n-1}\ =\ \frac 1 {A-1}
\end{multline}
Now let $N \to \infty$:
\dis
{
0\ \leq \lambda\,\alpha^m - \lambda\,\alpha_m^m
\ \leq\ \frac 1 {A-1}
}
Since
$\langle \lambda\,\alpha_m^m \rangle = \langle \theta_m \rangle$,
we get:
\dis
{
\langle \lambda\,\alpha^m - \theta_m \rangle
\ \leq\ \frac 1 {A-1}
}
To get the desired result, apply the previous result
to the sequence
$\theta_n^\prime = \theta_n - \frac 1 {2\,(A-1)}$,
i.e., for $n \geq 1$:
\dis
{
0\ \leq
\ \langle \lambda\,\alpha^n - \theta_n + \frac 1 {2\,(A-1)} \rangle
\ \leq\ \frac 1 {A-1}
}
Subtracting $\frac 1 {2\,(A-1)}$ we get:
\dis
{
\left|
\langle \lambda\,\alpha^n - \theta_n + \frac 1 {2\,(A-1)} \rangle
- \frac 1 {2\,(A-1)}
\right|
\ \leq\ \frac 1 {2\,(A-1)}
}
or, using $\langle x \rangle = x - [x]$:
\dis
{
\left|
\lambda\,\alpha^n - \theta_n
- \left[ \lambda\,\alpha^n - \theta_n + \frac 1 {2\,(A-1)} \right]
\right|
\ \leq\ \frac 1 {2\,(A-1)}
}
Hence:
\[
\| \lambda\,\alpha^n - \theta_n \|
\ \leq\ \frac 1 {2\,(A-1)}
\]
which is (\ref{e:seq1b}).
\item{(2)} There is no loss of generality in assuming
that $L > 0$. Otherwise, make $L' = -L$,
$\lambda' = -\lambda$ and $\theta^{\prime}_n = -\theta_n$,
and use that $\|-x\| = \|x\|$.
Construct an increasing sequence
$L = \lambda_0 \leq \lambda_1 \leq \lambda_2 \leq \dots$,
by the following recursive rule for $n \geq 0$:
\dis
{
\begin{aligned}
\lambda_{n+1}\ &=
\ \lambda_n +
\alpha^{-n-1}\,\langle \theta_{n+1} - \lambda_n\,\alpha^{n+1} \rangle \\
&=
\ \alpha^{-n-1}\,(\theta_{n+1} - [\theta_{n+1} - \lambda\,\alpha^{n+1}])
\end{aligned}
}
Since
$\langle \theta_{n+1} - \lambda_n\,\alpha^{n+1} \rangle \geq 0$,
we have that $\lambda_{n+1} \geq \lambda_n$ ($n \geq 0$),
so the sequence is actually increasing. Also, by construction
$\langle \lambda_n\,\alpha^n \rangle = \langle \theta_n \rangle$
for every $n \geq 1$. Furthermore, since
$\langle \theta_{n+1} - \lambda_n\,\alpha^{n+1} \rangle < 1$:
\dis
{
0\ \leq\ \lambda_{n+1} - \lambda_n\ <\ \alpha^{-n-1}
}
so, for $N \geq 1$:
\begin{multline}
0\ \leq\ \lambda_N - \lambda_0\ =
\ \sum_{n=0}^{N-1} (\lambda_{n+1} - \lambda_n) \\
\ <\ \sum_{n=0}^{N-1} \alpha^{-n-1}\ <
\ \sum_{n=0}^\infty \alpha^{-n-1}\ =\ \frac 1 {\alpha -1 }
\end{multline}
Hence,
$L \leq \lambda_N < L + \frac 1 {\alpha - 1}$,
so $\lambda = \lim_{N \to \infty} \lambda_N$
exists, and:
\[
L\ \leq\ \lambda\ \leq L + \frac 1 {\alpha - 1}
\]
which is (\ref{e:seq2a}).
Now, for $N > m \geq 1$:
\begin{multline}
0\ \leq\ \lambda_N\,\alpha^m - \lambda_m\,\alpha^m\ =
\ \alpha^m\,\sum_{n=m}^{N-1} (\lambda_{n+1} - \lambda_n) \\
<\ \alpha^m\,\sum_{n=m}^{N-1} \alpha^{-n-1}\ <
\ \sum_{n=m}^\infty \alpha^{m-n-1}\ =
\ \frac 1 {\alpha - 1}
\end{multline}
Letting $N \to \infty$:
\dis
{
0\ \leq\ \lambda\,\alpha^m - \lambda_m\,\alpha^m\ \leq
\ \frac 1 {\alpha - 1}
}
Since
$\langle \lambda_m\,\alpha^m \rangle = \langle \theta_m \rangle$,
we get:
\dis
{
\langle \lambda\,\alpha^m - \theta_m \rangle\ \leq
\ \frac 1 {\alpha - 1}
}
To get the desired result, apply the previous result to
the sequence
$\theta^\prime_n = \theta_n - \frac 1 {\alpha - 1}$,
and apply the same reasoning as in part (1).
\end{pf}
\begin{corollary} \label{c:trans}
For any $\lambda \neq 0$ there are uncountably many numbers
$\alpha > 1$ such that $\lambda\,\alpha^n$ is not uniformly
distributed modulo 1. Hence, there are transcendental numbers
with that property.
\end{corollary}
\begin{pf}
For every subset $I \subseteq \mathbb Z^{+}$ take an
$\alpha_I > 1$ such that
$\|\lambda\,\alpha_I^n - 1/4\| \leq 1/5$
if $n \in I$ and $\|\lambda\,\alpha_I^n - 3/4\| \leq 1/5$ if
$n \notin I$. Such $\alpha_I$ exists by theorem~\ref{t:seq}.
We have that $I \neq I' \implies \alpha_I \neq \alpha_{I'}$,
so the map $I \mapsto \alpha_I$ is injective.
Since there are uncountably many subsets in $\mathbb Z^{+}$,
the result follows.
\end{pf}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Combinations of powers} \label{s:compow}
\begin{remark}[Notation] In this section,
$k$ will be a number field,
$v$ a non-trivial, non-Archimedean place in $k$,
$k_v$ the completion of $k$ at $v$,
$\overline k_v$ an algebraic closure of $k_v$,
$\Omega_v$ the completion of $\overline k_v$,
$k' \subseteq k_v(\gamma) \subseteq \overline k_v$
any subfield of a finite extension
$k_v(\gamma)$ of $k_v$ ($\gamma \in \overline k_v$),
$\overline {k'} \subseteq \overline k_v$ the set
of elements of $\overline k_v$
that are algebraic over $k'$,
$K$ any field extension of $\overline {k'}$, and
$\mathcal O_{k',v}$ the ring
$\mathcal O_{k',v} =
\{\,x \in k'\,:\,|x|_v \leq 1\,\}$.
\end{remark}
\begin{theorem} \label{t:compow}
Let $\lambda_1(X), \lambda_2(X), \dots, \lambda_m(X)$
be polynomials in
$K[X] \setminus \{0\}$. Assume that
$\alpha_1, \alpha_2, \dots, \alpha_m$ are
distinct elements from $K \setminus \{0\}$. Let $S_n$ be
the following sum:
\dis
{
S_n\ =\ \sum_{i=1}^{m} \lambda_i(n)\,\alpha_i^n
}
Also assume that there is an integer $n_0$ such that
for every $n \geq n_0$, $S_n \in k'$. Then
for every $n \in \mathbb Z$, $S_n \in k'$, and:
\item{(i)} $\alpha_1,\alpha_2,\dots,\alpha_m$ are
the roots of a polynomial $f(X) \in k'[X]$,
so they are in $\overline {k'}$.
Moreover, if $|S_n|_v = O(1)$,
then $|\alpha_i|_v \leq 1$, $i=1,2,\dots,m$.
\item{(ii)} For every $i=1,2,\dots,m$:
$k'(\lambda_i) \subseteq k'(\alpha_i)$,
where $k'(\lambda_i)$ is the field generated
by the coefficients of $\lambda_i(X)$.
\item{(iii)} For every automorphism $\sigma \in
\Gal(k'(\alpha_1,\dots,\alpha_m)/k')$ such that
$\sigma(\alpha_i)=\alpha_{\pi(i)}$, $i=1,\dots,m$,
where $\pi$ is a permutation of $\{1,\dots,m\}$, we have
that $\sigma(\lambda_i(X)) = \lambda_{\pi(i)}(X)$,
where by definition $\sigma(\sum_i a_i\,X^i) =
\sum_i \sigma(a_i)\,X^i$.
\item{(iv)} If $\lambda_1(X),\dots,\lambda_m(X)$
are distinct polynomials, then
$k'(\alpha_i) = k'(\lambda_i)$ for every $i$.
\end{theorem}
The proof of this theorem requires several lemmas.
\begin{lemma}[Generalized Fatou's Lemma] \label{l:fatou}
Let $A$ be a Dedekind ring and $F$ a rational series
in $A[[X]]$, i.e., $F = p/q$ for some
$p, q \in A[X]$. Then there exist two polynomials
$P, Q \in A[X]$ such that $F = P/Q$,
where $P$ and $Q$ are relatively prime and
$Q(0) = 1$.
\end{lemma}
\begin{pf}
See \cite{bert92}, p.~15, theorem~1.3.
\end{pf}
\begin{lemma} \label{l:req}
Let $\{c_n\}_{n=-\infty}^{\infty}$ a set of
elements from $K$ such that $c_n \in k'$ for every
$n \geq n_0$, and verifying the following recurrence
relation of order M:
\dis
{
c_n\ =\ r_1\,c_{n-1} + r_2\,c_{n-2} + \dots + r_M\,c_{n-M}
}
for every $n \in \mathbb Z$, where $r_1,r_2,\dots,r_M$ are in
$K$, $r_M \neq 0$.
Then:
\item{(i)} The coefficients $r_1,r_2,\dots,r_M$ are in
$k'$, and for every $n \in \mathbb Z$, $c_n \in k'$.
\item{(ii)} If $c_n \in \mathcal O_{k',v}$
for every $n \geq n_0$, then the coefficients
$r_1,r_2,\dots,r_M$ are all in
$\mathcal O_{k',v}$.
\end{lemma}
\begin{pf}
\item{(i)} Let $C_n$ and $R$ be the matrices:
\dis
{
C_n\ =
\ \left(
\begin{array}{llll}
% \format\l&\quad\l&\quad\l&\quad\l\\
c_n & c_{n+1} & \hdots & c_{n+M-1} \\
c_{n+1} & c_{n+2} & \hdots & c_{n+M} \\
\vdots & \vdots & \ddots & \vdots \\
c_{n+M-1} & c_{n+M} & \hdots & c_{n+2M-2}
\end{array}
\right)
}
and
\dis
{
R\ =
\ \left(
\begin{array}{lllll}
% \format\l&\quad\l&\quad\l&\quad\l&\quad\l\\
0 & 1 & 0 & \hdots & 0 \\
0 & 0 & 1 & \hdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \hdots & 1 \\
r_M & r_{M-1} & r_{M-2} & \hdots & r_1
\end{array}
\right)
}
We have that $C_{n+1} = R\,C_n$. Since the recurrence relation
is of order M, $C_n$ is non singular.
On the other hand, $R = C_{n+1}\,C_{n}^{-1}$. Since the elements of
$C_n$ are in $k'$ for $n \geq n_0$, the entries of $R$, and those
of $R^{-1}$, will be in $k'$. Since $C_{n-1} = R^{-1}\,C_n$,
we get that the entries of $C_n$ will be in $k'$ also for
$n < n_0$.
\item{(ii)} For each $t \geq n_0$ define the formal
power series
\dis
{
F_t(X)\ =\ \sum_{n=0}^{\infty} c_{t+n}\,X^n
}
which is in $\mathcal O_{k',v}[[X]]$.
We have $F_t(X) = p_t(X)/q(X)$,
where $p_t(X),q(X) \in k'[X]$ are the following:
\dis
{
p_t(X)\ =\ \sum_{j=0}^{M-1} \Bigl( c_{t+j} -
\sum_{i=1}^{j} r_i\,c_{t+j-i} \Bigr)\,X^j
}
\dis
{
q(X)\ =\ 1 - r_1\,X - r_2\,X^2 - \dots - r_M\,X^M
}
This can be checked by multiplying $F_t(X)$ by $q_t(X)$
and using the recurrence relation, which gives
$F_t(X)\,q(X) = p_t(X)$ (see \cite{poor81} and \cite{poor88}).
Now we will prove that $p_t(X)$ and $q(X)$ are relatively prime.
To do so, we will see that they cannot have any common root
(in $\overline {k'}$). In fact, assume
that $\alpha$ is a common root of $p_{t_0}(X)$ and $q(X)$
for some $t_0 \geq n_0$, i.e.: $p_{t_0}(\alpha) = q(\alpha) = 0$.
Since $q(0)=1$, then $\alpha \neq 0$. Now we have:
\dis
{
X\,F_{t_0+1}(X) = F_{t_0}(X) - c_{t_0}
}
so:
\begin{multline}
X\,p_{t_0+1}(X) = X\,q(X)\,F_{t_0+1}(X) \\
= q(X)\,(F_{t_0}(X) - c_{t_0}) = p_{t_0}(X) - c_{t_0}\,q(X)
\end{multline}
Hence $p_{t_0+1}(\alpha) = 0$, which means that $\alpha$ is
also a root of $p_{t_0+1}(X)$. By induction we get that
$p_t(\alpha) = 0$ for every $t \geq t_0$. Grouping the
terms of $p_t(X)$ with respect to $c_t,c_{t+1},\dots,c_{t+M-1}$,
we get:
\dis
{
p_t(X) = \sum_{j=0}^{M-1} a_j(X)\,c_{t+j}
}
where
\dis
{
a_j(X) = X^j\,\Bigl( 1 - \sum_{i=1}^{M-j-1} r_i\,X^i \Bigr)
}
Note that $a_0(X),a_1(X),\dots,a_{M-1}(X)$ do not depend on t.
On the other hand $p_t(\alpha)=0$ implies
\dis
{ \label{e:coldep}
\sum_{j=0}^{M-1} a_j(\alpha)\,c_{t+j} = 0
}
for every $t \geq t_0$. Note that $a_{M-1}(\alpha)=\alpha^{M-1}
\neq 0$, so $a_0(\alpha),a_1(\alpha),\dots,a_{M-1}(\alpha)$
are not all zero, and (\ref{e:coldep}) means that the columns of
the matrix $C_{t_0}$ are linearly dependent, so $\det C_{t_0}=0$,
which contradicts the fact that $C_{t_0}$ is non singular.
Hence, the hypothesis that $p_t(X)$ and
$q(X)$ have a common root has to be false. This proves that
$p_t(X)$ and $q(X)$ are relatively prime.
By (generalized Fatou's) lemma~\ref{l:fatou},
and taking into account that
$\mathcal O_{k',v}$ is a Dedekind ring (see, for instance,
\cite{mcca76}, chap.~5),
we get that there exist two relatively prime
polynomials $P_t(X)$ and $Q_t(X)$ in
$\mathcal O_{k',v}[X]$ such that
$F_t(X) = P_t(X)/Q_t(X)$ and $Q_t(0)=1$. Hence:
$p_t(X)\,Q_t(X) = q(X)\,P_t(X)$. By unique factorization
of polynomials in $k'[X]$, there is a $u \in k'$ such that
$P_t(X) = u\,p_t(X)$ and $Q_t(X) = u\,q_t(X)$. Since
$Q_t(0)=q(0)=1$, we get that $u=1$, so
$P_t(X) = p_t(X)$ and $Q_t(X) = q(X)$.
Hence, the coefficients of $q(X)$ are in
$\mathcal O_{k',v}$.
\end{pf}
\begin{pf*}{Proof of Theorem~\ref{t:compow}}
Let $f(X) \in K[X]$ be the polynomial
\dis
{
f(X) = \prod_{i=1}^{m} (X-\alpha_i)^{d_i} =
X^M - r_1\,X^{M-1} - r_2\,X^{M-2} - \dots - r_M
}
where $d_i = 1 + \deg\,\lambda_i$, $i=1,2,\dots,m$, and
$M = \sum_{i=1}^m d_i$. Note that this is just
the reciprocal of the polynomial $q(X)$ defined
in the proof lemma~\ref{l:req}. It is well known
(see \cite{poor81} and \cite{poor88}) that $S_n$ verifies
a recurrence relation of order $M$, of the form:
\dis
{
S_n\ =\ r_1\,S_{n-1} + r_2\,S_{n-2} + \dots + r_M\,S_{n-M}
}
By part (i) of lemma~\ref{l:req} we get that $S_n \in k'$ for every
$n \in \mathbb Z$.
Now we prove the remaining results.
\item{(i)} Since $\alpha_1,\dots,\alpha_m$ are the roots of
$f(X) \in k'[X]$, it is clear that they are
in $\overline {k'}$.
Assume that $|S_n|_v = O(1)$.
We have that there are integers $B > 0$ and $n_0$ such that
for every $n \geq n_0$, $|S_n|_v \leq B$. Let $b \in k'$ be
any element such that $|b|_v \leq 1/B$.
Then for every $n \geq n_0$,
$|b\,S_n|_v \leq 1$. Putting $c_n = b\,S_n$ we get that
$c_n \in \mathcal O_{k',v}$ for every $n \geq n_0$, and also
verifies the recurrence:
\dis
{
c_n\ =\ r_1\,c_{n-1} + r_2\,c_{n-2} + \dots + r_M\,c_{n-M}
}
By part (ii) of lemma~\ref{l:req} we get that $r_1,r_2,\dots,r_M$
are in $\mathcal O_{k',v}$, i.e., $|r_i|_v \leq 1$
for $i=1,2,\dots,M$.
We already know that $\alpha_i \in \overline {k'}$, and
using the well known equality:
\dis
{
|r_0|_v\,\prod_{l=1}^m \max\,\{1,|\alpha_l|_v\}^{d_l}\ =
\ \max\,\{\,|r_l|_v\,:\,0 \leq l \leq M\,\}
}
($r_0=1$), we get that actually $|\alpha_i|_v \leq 1$.
\item{(ii)} For $i=1,2,\dots,m$, we have
$\lambda_i(X)\ =\ \sum_{j=0}^{d_j-1} a_{ij}\,X^j$,
where the coefficients $a_{ij}$ are in $K$.
If $d = \max \{d_1,d_2,\dots,d_m\}$, then we can write
$\lambda_i(X)\ =\ \sum_{j=0}^{d-1} a_{ij}\,X^j$,
where $a_{ij} = 0$ for $d_i \leq j < d$.
Now consider the following matrices:
\dis
{
\vec S\ =
\ \begin{pmatrix} S_0 \\ S_1 \\ \vdots \\ S_{md-1} \end{pmatrix},
\qquad
\vec L\ =
\ \begin{pmatrix} \vec L_1 \\ \vec L_2 \\
\vdots \\ \vec L_{m} \end{pmatrix}
}
where the $\vec L_i$'s are blocks of the form:
\dis
{
\vec L_i\ =
\ \begin{pmatrix}
a_{i\,0} \\ a_{i\,1} \\ \vdots \\ a_{i\,d-1}
\end{pmatrix}
}
and:
\dis
{
A\ =\ (A_1\,A_2\,\hdots\,A_m)
}
where the $A_i$'s are blocks of the form:
{\newcommand{\ai}{\alpha_i{}}
\dis
{
A_i\ =
\ \begin{pmatrix}
1 & 0 & \hdots & 0 \\
\ai & \ai & \hdots & \ai \\
\ai^2 & 2\ai^2 & \hdots & 2^{d-1}\,\ai^2 \\
\ai^3 & 3\ai^3 & \hdots & 3^{d-1}\,\ai^3 \\
\vdots & \vdots & \ddots & \vdots \\
\ai^{(md-1)} & (md-1)\,\ai^{(md-1)}
& \hdots & (md-1)^{(d-1)}\,\ai^{(md-1)}
\end{pmatrix}
}}
i.e., the element of $A_i$ in row $n$, column $j$, is
$n^j\,\alpha_i^n$, for $0 \leq n \leq md-1$ and
$0 \leq j \leq d-1$ (by convention, $0^0=1$).
Note that $A$ is a square matrix.
It is easy to check that $\vec S = A\,\vec L$. Furthermore,
it is known
that $\det A \neq 0$ (see lemma~8.5.1 in \cite{stol74}, 177-182).
Next, considering the coefficients $a_{ij}$ as unknowns of
a system of $md$ linear equations and $md$ unknowns, and using
Cramer's rule, we get
\dis
{
a_{ij} = \det(A^\prime_{ij})/\det(A)
}
where $A^\prime_{ij}$ is matrix $A$ with the $j$-th column
of block $A_i$ substituted by $\vec S$. Considering $a_{ij}$
as a function of $\alpha_1, \alpha_2, \dots, \alpha_{i-1},
\alpha_{i+1}, \dots, \alpha_m$, we note that interchanging
any two of them, say $\alpha_l$ and $\alpha_r$ ($l,r \neq i$),
does not change the value of $a_{ij}$, so
$a_{ij}$ is a symmetric rational function of the
$\alpha_l$'s ($l \neq 1$), with coefficients
in $k'(\alpha_i)$. Hence, $a_{ij}$ is a rational function in
\dis
{
k'(\alpha_i)(\phi_{i1},\phi_{i2},
\dots,\phi_{i,i-1},\phi_{i,i+1},\dots,\phi_{i,m})
}
where $\phi_{il}$ for $1 \leq l \leq m$, $l \neq i$,
are the elementary symmetric functions of
$\{\alpha_l\}_{l \neq i}$.
In other words, the $\phi_{il}$'s are $\pm$
the coefficients of the polynomial
\dis
{ \label{e:symm}
\prod \Sb l=1 \\ l \neq i \endSb \Sp m \endSp
(X - \alpha_l)\ =
\ \frac {\prod_{l=1}^{m}(X - \alpha_l)} {X - \alpha_i}
}
Since $f(X) = \prod_{l=1}^{m} (X-\alpha_l)^{d_l} \in k'[X]$,
and in characteristic zero every irreducible polynomial
is separable, it is clear that the polynomial in the numerator
of (\ref{e:symm}) is in $k'[X]$.
Hence, the quotient is in $k'(\alpha_i)[[X]]$,
and actually in $k'(\alpha_i)[X]$, since it equals a polynomial.
This implies that the $\phi_{il}$'s are in $k'(\alpha_i)$,
hence $a_{ij} \in k'(\alpha_i)$. This proves the desired
result.
\item{(iii)} We use again the expression
$a_{ij} = \det(A^\prime_{ij})/\det(A)$.
If $\sigma(\alpha_i) = \alpha_{\pi(i)}$, then:
\dis
{
\sigma(\det (A_1\,A_2\,\dots\,A_m)) =
\det (A_{\pi(1)}\,A_{\pi(2)}\,\dots\,A_{\pi(m)})
}
The effect of $\sigma$ on $\det(A^\prime_{ij})$
is similar, but now $S$ will be in the $j$-th column
of block $A_{\pi(i)}$.
Hence, $\sigma(\det(A^\prime_{ij}))/\sigma(\det(A))
= a_{\pi(i),j}$, i.e., $\sigma(a_{ij}) = a_{\pi(i),j}$,
which proves the result.
\item{(iv)} From (ii) we have that
$k'(\lambda_i) \subseteq k'(\alpha_i)$,
so we will prove the other containment.
If $\lambda_1(X),\dots,\lambda_m(X)$ are distinct
non zero polynomials, then for some $n \in \mathbb Z$
the numbers $\lambda_1(n),\dots,\lambda_m(n)$
are distinct and non zero.
For $j \in \mathbb Z$, consider the following sums:
\dis
{
T_j\ =\ \sum_{i=1}^{m} \lambda_i(n)^j\,\alpha_i
}
It is easy to see that for every $j \in \mathbb Z$,
$T_j \in k'$. In fact, take any automorphism
$\sigma \in \Gal(k'(\alpha_1,\dots,\alpha_m)/k')$ such
that $\sigma(\alpha_i) = \alpha_{\pi(i)}$ for
$i=1,2,\dots,m$, where $\pi$ is some permutation
of $\{1,2,\dots,m\}$. Then:
\dis
{
\sigma(T_j)\ =
\ \sum_{i=1}^{m} \lambda_{\pi(i)}(n)^j\,\alpha_{\pi(i)}\ =
\ \sum_{i=1}^{m} \lambda_i(n)^j\,\alpha_i\ =\ T_j
}
So, $T_j$ is invariant for $\sigma$. This implies that
$T_j \in k'$.
Now, consider each $\alpha_i$ as a polynomial of degree
zero. Reasoning as in the proof of (ii) with the roles
of $\lambda_i$ and $\alpha_i$ interchanged, we get that
$k'(\alpha_i) \subseteq k'(\lambda_i(n))$. Since
$\lambda_i(n) \in k'(\lambda_i)$, we get the desired
result.
\end{pf*}
Next corollary is the global version of part~(i)
of theorem~\ref{t:compow}.
Here, we take $k' = k$, and
$\displaystyle {\mathcal O_k =
{\bigcap_{\text{finite } v}} \mathcal O_{k',v}}$.
\begin{corollary} \label{c:glob}
Under the same hypothesis as in theorem~\ref{t:compow}
with $k' = k$, if there is an element $b \in k$ such that
$b\,S_n \in \mathcal O_k$ for every $n \geq n_0$,
then $\alpha_i$ ($i=1,2,\dots,m$) are
algebraic integers.
\end{corollary}
\begin{pf}
$b\,S_n \in \mathcal O_k$ for every $n \geq n_0$
implies that $S_n \in k$ and
$|S_n|_v = O(1)$ at every finite place $v$, so that
by theorem~\ref{t:compow} we get $\alpha \in \overline k$
and $|\alpha_i|_v \leq 1$ for every finite $v$,
hence, the $\alpha_i$'s are algebraic integers.
\end{pf}
Next proposition shows that the condition $|S_n|_v = O(1)$
in theorem~\ref{t:compow} can be weakened to a growth
condition.
\begin{proposition} \label{p:grow}
Under the hypothesis of theorem~\ref{t:compow}, if
$|S_n|_v = O(A^n)$ for every $A > 1$, then $|S_n|_v = O(1)$.
\end{proposition}
\begin{pf} Let $\pi$ be any element in $\overline {k'}$
such that $\pi^2$ is a prime generator of the ideal
\dis
{
\mathcal M\ =
\ \{\,x \in k'(\alpha_1,\dots,\alpha_m)\,:\,|x|_v < 1\,\}
}
Take $A = |\pi|_v^{-1}$. Let $\alpha^\prime_i = \alpha_i\,\pi$,
and let $S^\prime_n$ be:
\dis
{
S^\prime_n\ =
\ \sum_{i=1}^m \lambda_i(n)\,(\alpha^\prime_i)^n\ =
\ \sum_{i=1}^m \lambda_i(n)\,(\alpha_i\,\pi)^n\ =
\ S_n\,\pi^n
}
By hypothesis,
$|S^\prime_n|_v = |S_n|_v\,|\pi|_v^n = |S_n|_v/A^n = O(1)$.
Furthermore, we have that
$S^\prime_n \in k'(\pi) \subseteq k_v(\gamma,\pi)$.
Since $k_v(\gamma,\pi)$ is a finite extension of $k_v$,
theorem~\ref{t:compow} applies,
hence, by part (i) of that theorem,
for every $i$, $|\alpha^\prime_i|_v = |\alpha_i\,\pi|_v \leq 1$,
thus $|\alpha_i|_v \leq |\pi|_v^{-1}$, which is strictly less than
$|\pi^2|_v^{-1} =
\min\{\,|x|_v^{-1}\,:\,x \in \mathcal M \setminus \{0\}\,\}$,
so $|\alpha_i|_v \leq 1$, and from here the result follows.
\end{pf}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Other topics} \label{s:other}
\subsection{Asymptotic results} \label{ss:asym}
We mentioned in section~\ref{s:intro} that P.V.~numbers
belong to the exceptional set of Koksma's theorem, because
if $\alpha$ is a P.V.~number, then
$\lim_{n \to \infty} \|\alpha^n\| = 0$ (geometrically).
We could ask if there are other real numbers ($> 1$),
besides the P.V.~numbers, with this property.
In other words, given a real number $\alpha > 1$, does
$\lim_{n \to \infty} \|\alpha^n\| = 0$
imply that $\alpha$ is a P.V.~number?
The answer is not known so far, however it is possible
to prove that $\alpha$ is actually a P.V.~number
when any of the following (increasingly weak)
conditions is added (\cite{bert92}, section 5.4):
\begin{enumerate}
\item $\alpha$ is algebraic, and there exists a
real $\lambda \neq 0$ such that:
\dis
{
\lim_{n \to \infty} \|\lambda\,\alpha^n\|\ =\ 0
}
\vskip \dist
\item There exists a real $\lambda \neq 0$ such that:
\dis
{
\sum_{n=1}^\infty \|\lambda\,\alpha^n\|^2\ <\ +\infty
}
\vskip \dist
\item There exists a real $\lambda \neq 0$ such that:
\dis
{
\|\lambda\,\alpha^n\|^2\ =\ o(n^{-1/2})
}
\vskip \dist
\item There exists a real $\lambda \neq 0$ such that:
\dis
{
\|\lambda\,\alpha^n\|^2\ \leq\ \frac a {\sqrt n}
\quad (\forall n \geq n_0)
}
for some $0 < a < 1/2\sqrt2 (\alpha + 1)^2$ and some
integer $n_0$.
\end{enumerate}
The proofs rest on the rationality of the power series
$f(X) = \sum_{n=0}^\infty u_n\,X^n$, where $u_n =$ nearest
integer to $\lambda\,\alpha^n$. Next, by Fatou's lemma,
$f(X) = A(X)/Q(X)$ for some relatively prime
$A,Q \in \mathbb Z[X]$, such that $Q(0) = 1$.
On the other hand we have that:
\dis
{
f(X)\ =\ \frac {A(X)} {Q(X)}\ =
\ \frac \lambda {1 - \alpha\,X} + \varepsilon(X)
}
where $\varepsilon(X) = \sum_{n=0}^\infty \varepsilon_n\,X^n$;
$\varepsilon_n = u_n - \lambda\,\alpha^n \in [-1/2,1/2)$.
Since $\varepsilon_n \to 0$, we get that the meromorphic function
$\varepsilon(z)$ has no pole on the close unit disk $|z| \leq 1$,
so $Q(z)$ has a single zero in the close unit disk.
>From here it is easy to see that $\alpha$ is a P.V.~number
(\cite{bert92}, sec.~5.4.; \cite{sale63}, 4-10).
\subsection{Generalizations to adeles}
Let $\mathbb A$ be the ring of adeles of $\mathbb Q$,
and $I$ a finite set of places in $\mathbb Q$.
The $I$-adele ring of $\mathbb Q$ is defined as:
\dis
{
\mathbb A_I\ =
\ \{\,x \in \mathbb A\,:
\,x_p = 0 \text{ for } p \notin I\,\}
}
Which is isomorphic to $\prod_{p \in I} \mathbb Q_p$ and
contains a field canonically isomorphic to $\mathbb Q$
(which will also be designed $\mathbb Q$). Let $\mathbb Q^I$ be:
\dis
{
\mathbb Q^I\ =
\ \{\,x \in \mathbb Q\,:
\,|x|_p \leq 1 \text{ for } p \notin I \cup \{\infty\}\,\}
}
Then we have that $\mathbb Q^I$ is a Dedekind ring, and:
\vskip \dist
\begin{enumerate}
\item The field $\mathbb Q$ is dense in $\mathbb A_I$.
\vskip \dist
\item $\mathbb Q^I$ is a discrete subring of
$\mathbb A_I$, and the quotient
$\mathbb A_I/\mathbb Q^I$ is locally compact.
\end{enumerate}
Let $F_I$ be the set:
\dis
{
F_I\ =
\ [-\frac 1 2 , \frac 1 2) \times
\prod_{p \in I \setminus \{\infty\}} \mathcal O_p
\ \cong\ \mathbb A_I/\mathbb Q^I
}
then every element $x \in \mathbb A_I$ can be expressed
in one and only one way as $x = E(x) + \varepsilon(x)$,
with $E(x) \in \mathbb Q^I$ and $\varepsilon(x) \in F_I$
(Artin decomposition, see \cite{bert92}, theorem 10.1.2).
Here, $E(x)$ plays the role of
"nearest integer" to $x$, and $\varepsilon(x)$ is
the residue of $x$ modulo $\mathbb Q^I$.
Note that this theory becomes the usual theory
of distribution modulo~1 when $I = \{\infty\}$.
In this setting we may define the concept of uniform
distribution by using Weyl's criterion:
\begin{definition} A sequence $\{x_n\}_{n=1}^\infty$
in $\mathbb A_I$ is uniformly distributed modulo
$\mathbb Q^I$ if for all $a \in \mathbb Q^I$:
\dis
{
\lim_{N \to \infty}
\,\frac 1 N\,\sum_{n=1}^N \exp(2\pi i\,\varepsilon_\infty(a\,x_n))
\ =\ 0
}
\end{definition}
Other possible definitions of uniform distribution
over rings of adeles can be found in \cite{gran72}.
The version of Koksma's theorem here is the following
(\cite{bert92}, theorem~10.1.6):
\begin{theorem}
\item{(i)} Let $\alpha \in \mathbb A_I$, with $|\alpha|_p > 1$
for every $p \in I$. Then the sequence
$\{\lambda\,\alpha^n\}_{n=1}^\infty$ is uniformly distributed
modulo $\mathbb Q^I$ for almost every invertable element
$\lambda$ in $\mathbb A_I$.
\item{(ii)} Let $\lambda$ be an invertable element of $\mathbb A_I$.
Then the sequence
$\{\lambda\,\alpha^n\}_{n=1}^\infty$ is uniformly distributed
modulo $\mathbb Q^I$ for almost all $\alpha \in \mathbb A_I$
with $|\alpha|_p > 1$ for every $p \in I$.
\end{theorem}
\subsection{Generalization to fields of formal power series}
Let $k$ be any finite field, $\mathcal Z = k[X]$ and
$\mathcal F = k(X)$. If $|k| = q$, then the following
are absolute values on $\mathcal F$ (\cite{bert92}, chap.~12):
\begin{enumerate}
\item If $f,g \in \mathcal Z \setminus \{0\}$,
define $|f/g|_\infty = q^{(\deg f - \deg g)}$.
\vskip \dist
\item If $v$ is a prime polynomial in $\mathcal Z$ and
$f,g \in \mathcal Z \setminus \{0\}$ are relatively prime
to $v$, then $|v^h\,f/g|_v = q^{-h}$.
\end{enumerate}
Let $\mathcal F_v$ be the completion of $\mathcal F$ at place $v$,
and $\mathcal Z_v = \{\,x\,:\,|x|_v \leq 1\,\}$ the valuation
ring of $\mathcal F_v$. Then we have that
$\mathcal F_\infty = k\{X^{-1}\} =$ formal Laurent series of
the form $\sum_{n=-\infty}^h a_n\,X^n$, $a_n \in k$.
Furthermore, every element $x \in \mathcal F_\infty$ can be
written in a unique way as $x = E(x) + \varepsilon(x)$,
with $E(x) \in \mathcal Z$ and $|\varepsilon(x)|_\infty < 1$
(Artin decomposition; \cite{bert92}, theorem 12.0.3).
The uniform distribution can be defined like this
(\cite{bert92}, definition~12.0.1):
\begin{definition}
A sequence $\{x_n\}_{n=1}^\infty$ in $\mathcal F_\infty$
is said to be uniformly distributed modulo $\mathcal Z$
if, for every $h \in \mathbb N$ and every
$\beta \in \mathcal F_\infty$, we have:
\dis
{
\lim_{N \to \infty} \frac 1 N\,A(N;h,\beta) = q^{-h}
}
where $A(N;h,\beta) =$ number of terms $x_n$ of the series
such that $n \leq N$ and $|x_n~-~\beta|_\infty < q^{-h}$.
\end{definition}
Here the version of Weyl's metric theorem is the following
(\cite{bert92}, theorem 12.0.4):
\begin{theorem}
Given any $\alpha \in \mathcal F_\infty \ \mathcal Z_\infty$,
the sequence $\{\lambda\,\alpha^n\}_{n=1}^\infty$ is uniformly
distributed modulo $\mathcal Z$ for almost all
$\lambda \in \mathcal F_\infty$
(in the sense of a Haar measure).
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conclusions and future research} \label{s:conclu}
Most of the results presented here have been known for
several decades, although I found theorems~\ref{t:seq}
and~\ref{t:compow} independently.
Theorem~\ref{t:seq} can be considered as a direct consequence
of very general results in Pisot's thesis (\cite{piso38}).
On the other hand, the proof presented here gives more details
for the specific case of a sequence of the form $\lambda\,\alpha^n$.
Expression~(\ref{e:recrul}) gives a recursive rule that could be
used to design an algorithm that approximates the desired
value of $\alpha$.
In Pisot's thesis there are also global versions of
theorem~\ref{t:compow}. So far, however, I have not found
in any of the papers and works I have read the kind of
local version given here.
Several topics that deserve further study are the following:
\begin{enumerate}
\item Find a concrete (computable?) real number $\alpha > 1$
such that $\alpha^n$ is u.d.~mod~1. To this end, an answer
to the following question might be helpful.
\vskip \dist
\item In section~\ref{s:dispow} we saw a method to get a
number $\alpha$ such that $\lambda\,\alpha^n$ is close
modulo~1 to a given sequence $\theta_n$.
If the sequence $\theta_n$ is u.d.~mod~1,
does that imply that $\lambda\,\alpha^n$ is u.d.~mod~1?
\vskip \dist
\item Find a concrete transcendental number $\alpha > 1$
such that $\alpha^n$ is not u.d.~mod~1. The results in
section~\ref{s:dispow} prove that there are transcendental
numbers with this property, but no explicit example is given.
\vskip \dist
\item Is there any transcendental number $\alpha > 1$
such that $\lim_{n \to \infty} \|\alpha^n\| = 0$?
According to the results in section~\ref{ss:asym},
the convergence $\|\alpha^n\| \to 0$ should be
rather slow.
\vskip \dist
\item Generalize theorem~\ref{s:compow}
to Dedekind fields (see \cite{mcca76}, chap.~5,
for a definition of Dedekind field).
\vskip \dist
\item Further asymptotic results:
we already know (in $\mathbb R$) that if
$\|\alpha^n - \beta^n\| = 0$
for every $n \geq n_0$ then $\alpha = \beta$ or
$\alpha, \beta \in \mathbb Z$, but what conclusions
can be drawn from
$\lim_{n \to \infty} \|\alpha^n - \beta^n\| = 0$? And from
$\lim_{n \to \infty} \|\sum_{i=1}^m \lambda_i\,\alpha_i^n\| = 0$?
How do the results extend to local fields?
\vskip \dist
\item Extend results to distribution over $\mathcal O_v$
for some ring of $v$-adic integers
$\mathcal O_v = \{\,x \in k_v\,:\,|x|_v \leq 1\,\}$, to
rings of adeles, and to fields of formal power series.
\end{enumerate}
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\end{document}