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\begin{document}


\title{The Behavior of $\{n^2\alpha\}$ and $\{n^3\alpha\}$}

\author{Gloria Boyd\thanks{E-mail: \texttt{gboyd@.princeton.edu}} \\
\centerline{Mathematics Department} \\
 \centerline{Princeton University} \\
 \centerline{Princeton, NJ 08544}}

\date{January 17,2003}

\maketitle


\begin{abstract}
The aim of this project is to examine the behavior of
$\{n^2\alpha\}$ and $\{n^3\alpha\}$.
\end{abstract}


\tableofcontents

\chapter{Introduction}

\section{Background Information}

This paper is an expansion of the Robert Lipshitz paper, which
investigated the numerical results concerning the distribution of
the fractional part of $n^2\alpha$ for n from 1 to N.
Specifically, if we put the fractional part of $n^2\alpha$ for n
from 1 to N in increasing order, and let $\beta_j$ be the jth of
them, then the distribution of the differences between the
consecutive $\beta_j$ should resemble what would be expected for
numbers chosen at random in the interval [0,1] for almost all
$\alpha$.

Given a fixed irrational $\alpha$ and a positive integer n, let
$\{y\}$ denote the fractional part of y. Consider $\{n^2\alpha\}$
and $\{n^3\alpha\}$ for n from 1 to N throughout this paper.

The question of how the above behave is linked to many things,
which on the surface really do not appear to be of mathematical
interest. One such example is when a mathematician Von
Bortkewitsch, in 1898, analyzed the number of Prussian soldiers
killed each year as a result of being kicked by horses. He showed
that the Poisssonian distribution fitted his data very closely, by
tabulating the expected and observed frequencies and by comparing
the sample variance with the sample mean.(Barbour,1) As we have
studied in class the behavior of the fractional parts of n$\alpha$
are closely linked with the existence of Poissonian behavior.

We know that $\{n^2\alpha\}$ is equidistributed. We then extend
the treatment of the above to $\{n^3\alpha\}$ to show that it is also
equidistributed.

\section{Two: Lipshitz's Findings}

   We begin by stating a definition.

Definition: We say that a sequence $\{\alpha_k\}_{k=1}^\infty$ in
[0,1] is    equidistributed if for any interval $[a,b] \in [0,1]$,

\begin{equation} \lim_{n\to\infty} \frac{1}{n}( \#{\alpha_k \in [a,b], 1 \le k \le n}) \end{equation}

exists is equal to b-a.

The argument for real numbers is a result of the findings for the
sequence $\{\alpha_k\}^{\infty}_{k=1}$.

Definition: We say that a sequence of real numbers
$\{\alpha_k\}_{k=1}^{\infty}$ is equidistributed mod 1 if the
sequences of the fractional parts of $\alpha_k$ is
equididtributed.

\begin{thm} For $0 \le \alpha \le 1$, the fractional part of
$\alpha_n$ is equidistributed if and only if for any $C^{\infty}
function f : [0,1] \rightarrow C$, the following is true:

\begin{equation}\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} f(\alpha_k) = \int\limits_{x=0}^{x=1} f(x)\,dx \end{equation}.

\end{thm}

The proof of the above theorem is as follows: Let's suppose that:

\begin{equation}\lim_{n\to\infty}\frac {1}{n} \sum_{k=1}^{n} f(\alpha_k)= \int\limits_{x=0}^{x=1} f(x)\,dx \end{equation}

for any $C^{\infty}$ function $f : [0,1] \rightarrow C$. Fix
$[a,b] \in (0,1)$ and fix $\epsilon$ with 0<$\epsilon$<b-a. We
construct a $C^{\infty}$ function $f :[0,1] \rightarrow R$ with $0
\le f(x) \le 1$ for all $x \in [0,1]$. It is the simple function
below:

$f(x) = 0$ if $x< a-\epsilon$  or $x>b+\epsilon$ and $1$ if
$a<x<b$.


Then,
\begin{equation}\ b-a \le \int\limits_{x=0}^{x=1} f(x)\,dx \le b-a+2\epsilon \end{equation}

and , \begin{equation} \sum_{k=1}^{n} f(\alpha_k) \ge \#{\alpha_k
\in [a,b], 1 \le k \le n} \end{equation}

so,

\begin{equation} \lim_{n\to\infty} \frac{1}{n} \sum _{k=1}^{n} f(\alpha_k) \ge \limsup_{n\to\infty} \frac{1}{n} (\#{\alpha_k \in [a,b], 1 \le k \le n}) \end{equation}

Now we have $b-a+2\epsilon \ge\limsup_{n\to\infty} \frac{1}{n}
(\#\{\alpha_k \in [a,b], 1\le k \le n\})$. If we let $\epsilon$
tend to zero then we get:
\begin{equation} b-a \ge \limsup_{n\to\infty} \frac{1}{n} (\#{\alpha_k \in [a,b], 1 \le k \le n}) \end{equation}

Applying a similar argument to a function g(x) with the same boundaries as f(x) We obtain:

$g(x) = 0$ if $x<a$ or $a>b$ and $g(x) = 1$ if
$a+\epsilon<x<b-\epsilon$

If we let $\epsilon$ tend to zero then we have:

\begin{equation} b-a \le \liminf_{n\to\infty} \frac{1}{n} (\#{\alpha_k \in [a,b],1 \le k \le n}) \end{equation}.

So,

\begin{equation} \lim_{n\to\infty} \frac{1}{n}(\#{\alpha_k \in [a,b], 1\le k \le n}) \end{equation}

exists and is b-a.

Suppose the converse for any interval $[a,b] \in [0,1]$,

\begin{equation}\lim_{n\to\infty}\frac{1}{n}(\#{\alpha_k \in [a,b], 1 \le k \le n}) = b-a \end{equation}

Fix $\epsilon$>0 and $f: [0,1] \rightarrow C$. Since f is a
continuous function on a compact set, thus uniformly continuous,
there exists a $N  \in \N$  such that

\be |x-y|\le \frac{1}{N} \Rightarrow |f(x)-f(y)|<\epsilon \ee

Consider the partition of [0,1] into
$0<\frac{1}{N}<\frac{2}{N}<\frac{3}{N}<\ldots<1$.

There exists an M $\in$ N such that for m>M and any a between 1 and N-1,

\begin{equation} |\frac {\#\{\alpha_i \in [a/N,a+1/N], 1\le i \le m\}}{m} - \frac{1}{N}| < \frac{\epsilon}{N} \end{equation}

Fix m. Let $A_j = \#\{\alpha_i \in [j/N,(j+1)/N], 1 \le i \le
m\}$.

 Now,

\begin{eqnarray}
|\frac{1}{m}\sum_{i=1}^{m} f(\alpha_i)- \int_{x=0}^{x=1}f(x) dx| &
\le & |\frac{1}{m} \sum_{i=1}^{m} f(\alpha_i) -
|\frac{1}{N}\sum_{j=0}^{N-1} f(j/N)| \nonumber\\
& & +|\frac{1}{N}\sum_{j=0}^{N-1} f(j/N) - \int_{x=0}^{x=1}
f(x)dx| \nonumber\\ & < & |\sum_{j=0}^{N-1}[(\sum_{\alpha_i \in
[j/N,j+1/N]} \frac{f(\alpha_i)}{m} - \frac{f(j/N)}{N}]| \nonumber\\
& + & \frac{N\epsilon}{N} \le \sum_{j=1}^{N-1}(A_j
\frac{\epsilon}{m} \nonumber\\ & & + |\frac{A_j}{m}-\frac{1}{N}|)+
\epsilon \le \epsilon + \sum_{j=0}^{N-1}
\frac{\epsilon}{N}+\epsilon = 3\epsilon \nonumber\\
\end{eqnarray}

Since we found that $\epsilon$ was arbitrary, which implies the following result:

\begin{equation}
|\frac{1}{m}\sum_{i=1}^{m} f(\alpha_i) - \int\limits_{x=0}^{x=1}
f(x)dx|\rightarrow 0
\end{equation}

as m $\rightarrow \infty$.

\begin{thm} A sequence $\{\alpha_n\}^{\infty}_{n=1}$ is
equidistributed mod 1 if and only if for all integers t,
t$\not=$0,

\begin{equation} \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} e^{2\pi i \alpha_k}=0 \end{equation}.

\end{thm}


The proof follows from observing the periodicity of $e^{2\pi iz}$
and assuming that all sequences $\alpha_n$ lie between 0 and 1.
Thus equidistributed in this case is the same as being
equidistributed on the unit circle.

Suppose the latter part of the theorem, specifically, that for all
t$\not=$0, the

\begin{equation}\lim_{n\to\infty}\sum_{k=1}^{n}e^{2\pi i \alpha_k}=0 \end{equation}.

As we did earlier, we fix a $C^\infty$ function $f:[0,1]
\rightarrow C$. From Fourier analysis, we know that there exists a
sequence ${a_n}_{-\infty}^{\infty}$ with the

\begin{equation}\lim_{n\to\infty}n^2a_n=0 \end{equation} such that

\begin{equation} f(x) =\sum_{n=-\infty}^{\infty}a_ne^{2\pi i n x} \end{equation}.

Since the above limit converges to 0 the series converges and does so uniformly.So

\begin{equation}\int\limits_{x=0}^{x=1}f(x)\,dx = a_0 \end{equation}.

We now only need to show that

\begin{equation}\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}
f(\alpha_k) = a_0 \end{equation}

to show that the sequence is equidistributed.

\begin{equation}
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}  f(\alpha_k) =
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sum_{t=-\infty}^{\infty}
a_t e^{2\pi i t \alpha_k}
\end{equation}


\begin{equation}
= \sum_{t=-infty}^{\infty} \lim_{n\to\infty} \frac{a_t}{n}
\sum_{k=1}^{n} e^{2\pi i t \alpha_k} = a_0
\end{equation}.

The change in order of the summations is justified by uniform
convergence. By the previous theorem $\{\alpha_n\}_{n=1}^{\infty}$
is equidistributed because

\begin{equation}\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} f(\alpha_k) = \int\limits_{x=0}^{x=1} f(x)dx \end{equation}

\begin{thm}
Suppose $\{z_{q+h}-z_q\}$ is equi-distributed mod 1 for all
positive integers h. Then $\{z_q\}$ is equi-distributed mod 1
also.
\end{thm}

The proof begins by assuming the last theorem for any integer
$t\neq 0$ and a corollary, which states that for each positive h
and sequence $\{z_q\}_{q-1}^{\infty}$, if we are given

\begin{equation} \frac{1}{Q} \sum_{1\le q\le Q} e^{2\pi i (z_{q+h}-z_q)} \rightarrow \end{equation}

then

\begin{equation}\frac{1}{Q}\sum_{1\le q\le Q}e^{2\pi i
z_q}\rightarrow0,  Q\rightarrow \infty \end{equation}.

So by assuming the aforementioned corollary and the last theorem,

\begin{equation}\frac{1}{Q}\sum_{1\le q\le Q}e^{2\pi i t (z_{q+h}-z_q)}\rightarrow,  Q\rightarrow \infty \end{equation},

and

\begin{equation}\frac{1}{Q}\sum_{1\le q\le Q}e^{2\pi t i z_q}\rightarrow0, Q\rightarrow \infty \end{equation}.

Finally, by the last theorem ${z_q}_{q=1}^{\infty}$ is
equidistributed mod 1.

We now move to prove that $\{n\alpha\}$ is equidistributed mod 1.

\begin{thm}For irrational $\alpha$, $\{n\alpha\}$ is equidistributed mod
1.\end{thm}

The proof of this comes from knowledge of the behavior of
geometric series. We consider the following:

\begin{equation}\frac{1}{N}\sum_{k=0}^{N}e^{2\pi k i t \alpha} \end{equation}.

Since this series is geometric, and is equivalent to:

\begin{equation}\frac{1}{N} \frac{1-e^{2\pi i (N+1) t \alpha}}{1-e^{2\pi i t \alpha}} \end{equation},

$\{n\alpha\}$ is equidistributed because the above geometric series approaches 0 as N approaches infinity.

\begin{thm} For irrational $\alpha$, $\{n^2\alpha\}_{n=1}^{\infty}$ is equidistributed mod
1.\end{thm}

The proof of this follows from an aforementioned theorem.

For a fixed positive integer h, $(n+h)^2\alpha - n^2\alpha =
2h\alpha + h^2$, which is equidistributed by the penultimate
theorem.

So we have shown that $\{n^2\alpha\}$ is equidistributed mod 1,
but what does that imply about $\{n^3\alpha\}$?


\section{Three: The Equidistribution of $\{n^3\alpha\}$}

We learn that the above results can be extended to any polynomial
n with an irrational coefficient other than the constant
term.(Cassels,73) So we intend to use this knowledge to show that
$\{n^3\alpha\}$ is equidistributed mod 1, by similar means used to
show that $\{n^2\alpha\}$ is equidistributed.

Following the last proof, for a fixed positive integer h, we have
the polynomial, $(n+h)^3\alpha - n^3\alpha$, which is equal to
$3n^2h\alpha+3h^2n\alpha+h^3\alpha$. We notice that our equation
has more than one irrational coefficient. So by the same theorem
used to prove $\{n^2\alpha\}$ is equidistributed mod 1, we have,

\begin{thm} For irrational $\alpha$, $\{n^3\alpha\}$ is equidistributed mod
1.\end{thm}

What this also does is imply that for any degree k of n for a
fixed $\alpha$, the fractional parts are equidistributed mod 1.

\section{Four: Expectations for distribution of Consecutive $B_j$}

We found that $\{n^2\alpha\}$ and $\{n^3\alpha\}$ behave in the
same manner as $\{n\alpha\}$ with respect to equidistribution. we
now move to discuss the behavior of the consecutive events in both
with respect to the well studied case of $\{n\alpha\}$.

We know how $\{n\alpha\}$ behave. Now, we need to study how differently
$\{n^2\alpha\}$ and $\{n^3\alpha\}$ behave.

We know that a sequence of n uniformly distributed variables
splits into n+1 subintervals, when both $\in [0,1]$ with
probability 1. So what about the lengths of these subintervals.

We see that the results for $\{n^2\alpha\}$ show that the
consecutive spacings appear to behave as they should according to
Poissonian statistics. We expect more of the same for
$\{n^3\alpha\}$

\section{Five: Background Information to the Distribution and Spacing of Consecutive $\beta_j$}


In $1987$, three scientists studied an interesting stochastic
process, which models a common type of radiation damage  to
certain kinds of DNA in bacteria. The double-stranded DNA is
topologically a circle in this case.The damaging agent causes
single-stranded breaks called nicks. The nicks have independent
positions equally likely to occur on each of the two strands and
uniformly distributed around the strand.If two nicks on different
strands occur within a critical length of each other then the
circle of DNA breaks.


In the above study, the scientists used results from findings on
spacings to show that the binomial relation that described the
distribution of the consecutive nicks converges to the Poissonian.
This is similar to that done by Von Bortkewitsch, as mentioned in
section one.


The negative exponential characterizes the probability
distribution of distances between consecutive events of a Poisson
process. In other words, The probability of the distance of
consecutive events being greater than some value x is $e^{-\lambda
 x}$. [BHJ]

\begin{cor} Let $X_1,X_2,\ldots$ be a sequence of independent random variables in [0,1]. For fixed n,
$X_1,X_2,\ldots,X_n$ divide [0,1] into n+1 subintervals. Let $L_n$
denote the length of the first interval in this partition. Then,
for fixed t,the probability that $nL_n$ > t approaches $e^{-t}$ as
n approaches infinity. \end{cor}

 This corollary  leads us to the following definition. Given a sequence $\{\alpha_n\}^{\infty}_{n=1}$ in [0,1], for fixed N let ${\beta_{1,N},\beta_{2,N}, \ldots,\beta_{N,N}}$ denote the sequence ${\alpha_1,\alpha_2,\ldots,\alpha_N}$ reordered in increasing order. The first consecutive spacing measure is defined as

\begin{equation} \mu(N) = \frac{1}{N-1}\sum_{j=1}^{N-1}\delta_{N(\beta_j+1 - \beta_j)} \end{equation}

where $\delta$ is the Dirac delta-function. For the aforementioned corollary. we expect that if $\alpha_N$ act like random numbers then $\mu(N)$ approaches $e^{-x}dx$ as N approaches infinity.

We extend this to the kth consecutive spacing in this manner:

\begin{equation} \mu_{k}(N) = \frac{1}{N-k}\sum_{j=1}^{N-k}\delta_{N(\beta_j+1 - \beta_j)} \end{equation}

From more inspection we see that $\mu_{k}(N)$ approaches \begin{equation}\frac{x^{k}}{k!}e^{-x}dx \end{equation} as N approaches infinity.

We expect for both $\{n^2\alpha\}$ and $\{n^3\alpha\}$ that
$\mu_{k}(N,\alpha)\rightarrow\frac{x^{k}}{k!}e^{-x}dx$ and
$\mu_{k}(N,\alpha_{1},\alpha_{2})\rightarrow\frac{x^{k}}{k!}e^{-x}dx$
, where $\mu_{k}(N,\alpha)$ represents the kth consecutive spacing
measure of $\{n^2\alpha\}$ and $\mu_{k}(N,\alpha_{1},\alpha_{2})$
represents the kth consecutive spacing measure of $\{n^3\alpha\}$.



I have conducted similar data analysis as Robert Lipshitz has
done, but have also taken to examining the irrationals $\sqrt{3}$,
$2^{\sqrt{2}}$, and $\sum_{n=1}^{\infty} \frac{1}{2^{n}-1}$. He
was examining irrationals $\sqrt{2}$, $\pi$, and $e$.

Specifically, I will check convergence by graphing cumulative
distributions for $\mu_{k}(N,\alpha_{1},\alpha_{2})$ for the same
$\alpha$ as Lipshitz and comparing the two rates of convergence,
as convergence is expected. I will then check convergence for the
other irrationals chosen above for
$\mu_{k}(N,\alpha_{1},\alpha_{2})$. Then, I examine disc, which
measures the difference between the cumulative difference
functions at a point and our model of what we believe the behavior
resembles,$e^{-x}$

I believe that $\{n^3\alpha\}$ will behave similarly to  $\{n^2\alpha\}$ with respect to distribution. I believe that because we are keeping $\alpha$ fixed for each experimentation the differences in the behavior will be noticeable but not profound.

\section{Calculations and Findings}

We notice that $\mu_{k}(N,\alpha_{1},\alpha_{2})$ for $\alpha$ equal to $\sqrt{2}$ converges quite quickly. It appears to converge much quicker than $\mu_{k}(N,\alpha)$. The only way to really get a good idea of whether this is true is to look at disc. We find that disk is around $.01$ much like previous results, but what is trully interesting is that disk hovers around $.01$ for smaller N as well, in particular, when N is $1000$. This seems to prove that $\mu_{k}(N,\alpha_{1},\alpha_{2})$ converges much quicker than $\mu_{k}(N,\alpha)$.

We also notice for e and $\pi$ the subtle differences in rate of convergence. For e, we notice that disk hovers around $.01$ also. The difference being in the rate that it approaches zero, e starts out with a much lower value of disc for $\mu_{k}(N,\alpha_{1},\alpha_{2})$ compared to $\mu_{k}(N,\alpha)$, where disc was around $.08$ for N equal to $1000$. In the case of $\alpha = \pi$, Convergence is quicker that before also. Though there is a difference in the rate of convergence for these irrationals, it should be said that as N gets bigger the difference still tends to zero.

As for our new irrationals,there is convergence.They all converge rapidly, but not as rapid as with the other rationals. Disc for these irrationals hover around $.02$ for N equal to $1000$, instead of near $.01$ as the others did.

Since it appears that the bigger the power of n for $\{n^k\alpha\}$, the more rapid the convergence, it might be an interesting experiment to examine disc for k very large to see if this actually is the case.

\section{Conclusion}
 Though much experimentation has been done on the behavior of $\{n^k\alpha\}$,there is still so much more that can be done. Its uses have become invaluable in many different fields of study, and no doubt through further examination we may find very many other uses for this subject. It seems that the more we dig into this subject we learn of more and more ways of relating this behavior to very modern things.I would be very interested in seeing what new understanding will come out of this next.







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{thebibliography}{99}

\bibitem[BHJ]{barbour}
\newblock A.D. Barbour, Lars Holst,and  Svante Janson,
\emph{Poisson Approximation},Oxford University Press ($1992$),
$1, 138-147$.

\bibitem[Ca]{cassels}
\newblock J. W. S. Cassels, \emph{An Introduction to
Diophantine Approximation}, Cambridge University Press, London
$1957$.

\bibitem[Ea]{ea}
\newblock W. Eames, \emph{The Elementary Theory of Numbers,Polynomials, and Rational Functions},
 Oldbourne Mathematical Series\textbf{131-136},American Elsevier Publishing Co. , New York,
 $1967$.


\bibitem[Ha]{haight}
\newblock F. Haight, \emph{Handbook of the Poisson Distribution},.  John Wiley \& Sons, Inc., New York  1967.

\bibitem[RL]{rl}

\newblock R. Lipshitz, \emph{Robert Lipshitz's Junior Project: Numerical results concerning
the distribution of $\{n^2\alpha\}$},(Fall $2001$).

\bibitem[RSZ]{RSZ}
\newblock Ze\'ev Rudnick, P. Sarnak, and A. Zaharescu, \emph{The
Distribution of Spacings Between the Fractional Parts of
\(n^2\alpha\)}, Invent. Math. $145$ ($2001$), no. $1$, $37-57$.



\end{thebibliography}



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