The Behavior of {n^{2}α} and {n^{3}α}

Gloria Boyd^{1} Mathematics Department

Princeton University Princeton, NJ 08544

January 17,2003

1E-mail: gboyd@.princeton.edu

Abstract

The aim of this project is to examine the behavior of {n^{2}α} and {n^{3}α}.

Contents

1 Introduction 2 1.1 Background Information . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Two: Lipshitz’s Findings . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Three: The Equidistribution of {n^{3}α} . . . . . . . . . . . . . . . 8 1.4 Four: Expectations for distribution of Consecutive B_{j} . . . . . . . 8 1.5 Five: Background Information to the Distribution and Spacing of

Consecutive β_{j} . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.6 Calculations and Findings . . . . . . . . . . . . . . . . . . . . . . 10 1.7 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Chapter 1

Introduction

1.1 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}α for n from 1 to N. Specifically, if we put the fractional part of n^{2}α for n from 1 to N in increasing order, and let β_{j} be the jth of them, then the distribution of the differences between the consecutive β_{j} should resemble what would be expected for numbers chosen at random in the interval [0,1] for almost all α.

Given a fixed irrational α and a positive integer n, let {y} denote the fractional part of y. Consider {n^{2}α} and {n^{3}α} 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 nα are closely linked with the existence of Poissonian behavior.

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

1.2 Two: Lipshitz’s Findings

We begin by stating a definition.

Definition: We say that a sequence {α_{k}}_{k=1}^{∞} in [0,1] is equidistributed if for any interval [a, b] ∈ [0, 1],

lim_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n)  (1.1)

exists is equal to b-a. The argument for real numbers is a result of the findings for the sequence

{α_{k}}_{k=1}^{∞}.

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

Theorem 1.2.1. For 0 ≤ α ≤ 1, the fractional part of α_{n} is equidistributed if and only if for any C^{∞}functionf : [0, 1] → C, the following is true:

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) = ∫_{x=0}^{x=1}f(x) dx  (1.2)

.

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

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) = ∫_{x=0}^{x=1}f(x) dx  (1.3)

for any C^{∞} function f : [0, 1] → C. Fix [a, b] ∈ (0, 1) and fix ϵ with 0<ϵ<b-a. We construct a C^{∞} function f : [0, 1] → R with 0 ≤ f(x) ≤ 1 for all x ∈ [0, 1]. It is the simple function below:

f(x) = 0 if x < a − ϵ or x > b + ϵ and 1 if a < x < b. Then,

b − a ≤ ∫_{x=0}^{x=1}f(x) dx ≤ b − a + 2ϵ  (1.4)

and ,

∑_{k=1}^{n} f(α_{k}) ≥ #α_{k} ∈ [a, b], 1 ≤ k ≤ n  (1.5)

so,

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) ≥ lim sup_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n)  (1.6)

Now we have b − a + 2ϵ ≥ lim sup_{n→∞} \frac{1}{n}(#\{α_{k} ∈ [a, b], 1 ≤ k ≤ n\}). If we let ϵ tend to zero then we get:

b − a ≥ lim sup_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n)  (1.7)

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 + ϵ < x < b − ϵ If we let ϵ tend to zero then we have:

b − a ≤ lim inf_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n)  (1.8)

.

So,

lim_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n)  (1.9)

exists and is b-a. Suppose the converse for any interval [a, b] ∈ [0, 1],

lim_{n→∞} \frac{1}{n}(#α_{k} ∈ [a, b], 1 ≤ k ≤ n) = b − a  (1.10)

Fix ϵ>0 and f : [0, 1] → C. Since f is a continuous function on a compact set, thus uniformly continuous, there exists a N ∈ ℕ such that

|x − y| ≤ \frac{1}{N} ⇒ |f(x) − f(y)| < ϵ  (1.11)

Consider the partition of [0,1] into 0 < \frac{1}{N} < \frac{2}{N} < \frac{3}{N} < … < 1. There exists an M ∈ N such that for m>M and any a between 1 and N-1,

|\frac{#\{α_{i} ∈ [a/N, a + 1/N], 1 ≤ i ≤ m\}}{m} − \frac{1}{N} | < \frac{ϵ}{N}  (1.12)

Fix m. Let A_{j} = #{α_{i} ∈ [j/N, (j + 1)/N], 1 ≤ i ≤ m}. Now,

|\frac{1}{m} ∑_{i=1}^{m} f(α_{i}) − ∫_{x=0}^{x=1} f(x)dx| ≤ |\frac{1}{m} ∑_{i=1}^{m} f(α_{i}) − |\frac{1}{N} ∑_{j=0}^{N−1} f(j/N)|

+|\frac{1}{N} ∑_{j=0}^{N−1} f(j/N) − ∫_{x=0}^{x=1} f(x)dx|

< | ∑_{j=0}^{N−1}[( ∑_{α_{i}∈[j/N,j+1/N]} \frac{f(α_{i})}{m} − \frac{f(j/N)}{N}]|

+ \frac{Nϵ}{N} ≤ ∑_{j=1}^{N−1}(A_{j} \frac{ϵ}{m}

+|\frac{A_{j}}{m} − \frac{1}{N} |) + ϵ ≤ ϵ + ∑_{j=0}^{N−1} \frac{ϵ}{N} + ϵ = 3ϵ

(1.13)

Since we found that ϵ was arbitrary, which implies the following result:

|\frac{1}{m} ∑_{i=1}^{m} f(α_{i}) − ∫_{x=0}^{x=1}f(x)dx| → 0  (1.14)

as m → ∞.

Theorem 1.2.2. A sequence {α_{n}}_{n=1}^{∞} is equidistributed mod 1 if and only if for all integers t, t≠0,

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} e^{2πiα_{k}} = 0  (1.15)

.

The proof follows from observing the periodicity of e^{2πiz} and assuming that all sequences α_{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≠0, the

lim_{n→∞} ∑_{k=1}^{n} e^{2πiα_{k}} = 0  (1.16)

.

As we did earlier, we fix a C^{∞} function f : [0, 1] → C. From Fourier analysis, we know that there exists a sequence a_{n}_{−∞}^{∞} with the

lim_{n→∞} n^{2}a_{n} = 0  (1.17)

such that

f(x) = ∑_{n=−∞}^{∞} a_{n}e^{2πinx}  (1.18)

.

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

∫_{x=0}^{x=1}f(x) dx = a_{0}  (1.19)

.

We now only need to show that

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) = a_{0}  (1.20)

to show that the sequence is equidistributed.

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) = lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} ∑_{t=−∞}^{∞} a_{t}e^{2πitα_{k}}  (1.21)

= ∑_{t=−infty}^{∞} lim_{n→∞} \frac{a_{t}}{n} ∑_{k=1}^{n} e^{2πitα_{k}} = a_{0}  (1.22)

.

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

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(α_{k}) = ∫_{x=0}^{x=1}f(x)dx  (1.23)

Theorem 1.2.3. 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.

The proof begins by assuming the last theorem for any integer t ≠ 0 and a corollary, which states that for each positive h and sequence {z_{q}}_{q−1}^{∞}, if we are given

\frac{1}{Q} ∑_{1≤q≤Q} e^{2πi(z_{q+h}−z_{q})} →  (1.24)

then

\frac{1}{Q} ∑_{1≤q≤Q} e^{2πiz_{q}} → 0, Q → ∞  (1.25)

.

So by assuming the aforementioned corollary and the last theorem,

\frac{1}{Q} ∑_{1≤q≤Q} e^{2πit(z_{q+h}−z_{q})} →, Q → ∞  (1.26)

, and

\frac{1}{Q} ∑_{1≤q≤Q} e^{2πtiz_{q}} → 0, Q → ∞  (1.27)

.

Finally, by the last theorem z_{q}_{q=1}^{∞} is equidistributed mod 1.

We now move to prove that {nα} is equidistributed mod 1.

Theorem 1.2.4. For irrational α, {nα} is equidistributed mod 1.

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

\frac{1}{N} ∑_{k=0}^{N} e^{2πkitα}  (1.28)

.

Since this series is geometric, and is equivalent to:

\frac{1}{N} \frac{1 − e^{2πi(N+1)tα}}{1 − e^{2πitα}}  (1.29)

,

{nα} is equidistributed because the above geometric series approaches 0 as N approaches infinity.

Theorem 1.2.5. For irrational α, {n^{2}α}_{n=1}^{∞} is equidistributed mod 1.

The proof of this follows from an aforementioned theorem.

For a fixed positive integer h, (n + h)^{2}α − n^{2}α = 2hα + h^{2}, which is equidistributed by the penultimate theorem.

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

1.3 Three: The Equidistribution of {n^{3}α}

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}α} is equidistributed mod 1, by similar means used to show that {n^{2}α} is equidistributed.

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

Theorem 1.3.1. For irrational α, {n^{3}α} is equidistributed mod 1.

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

1.4 Four: Expectations for distribution of Consecutive B_{j}

We found that {n^{2}α} and {n^{3}α} behave in the same manner as {nα} 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 {nα}.

We know how {nα} behave. Now, we need to study how differently {n^{2}α} and {n^{3}α} behave.

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

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

1.5 Five: Background Information to the Distribution and Spacing of Consecutive β_{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^{−λx}. [BHJ]

Corollary 1.5.1. Let X_{1}, X_{2}, … be a sequence of independent random variables in [0,1]. For fixed n, X_{1}, X_{2}, …, 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.

This corollary leads us to the following definition. Given a sequence {α_{n}}_{n=1}^{∞} in [0,1], for fixed N let β_{1,N}, β_{2,N}, …, β_{N,N} denote the sequence α_{1}, α_{2}, …, α_{N} reordered in increasing order. The first consecutive spacing measure is defined as

μ(N) = \frac{1}{N − 1} ∑_{j=1}^{N−1} δ_{N(β_{j}+1−β_{j})}  (1.30)

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

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

μ_{k}(N) = \frac{1}{N − k} ∑_{j=1}^{N−k} δ_{N(β_{j}+1−β_{j})}  (1.31)

From more inspection we see that μ_{k}(N) approaches

\frac{x^{k}}{k!} e^{−x}dx  (1.32)

as N approaches infinity.

We expect for both \{n^{2}α\} and \{n^{3}α\} that μ_{k}(N, α) → \frac{x^{k}}{k!} e^{−x}dx and μ_{k}(N, α_{1}, α_{2}) → \frac{x^{k}}{k!} e^{−x}dx , where μ_{k}(N, α) represents the kth consecutive spacing measure of {n^{2}α} and μ_{k}(N, α_{1}, α_{2}) represents the kth consecutive spacing measure of {n^{3}α}.

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 ∑_{n=1}^{∞} \frac{1}{2^{n}−1}. He was examining irrationals \sqrt{2}, π, and e.

Specifically, I will check convergence by graphing cumulative distributions for μ_{k}(N, α_{1}, α_{2}) for the same α 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 μ_{k}(N, α_{1}, α_{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}α} will behave similarly to {n^{2}α} with respect to distribution. I believe that because we are keeping α fixed for each experimentation the differences in the behavior will be noticeable but not profound.

1.6 Calculations and Findings

We notice that μ_{k}(N, α_{1}, α_{2}) for α equal to \sqrt{2} converges quite quickly. It appears to converge much quicker than μ_{k}(N, α). 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 μ_{k}(N, α_{1}, α_{2}) converges much quicker than μ_{k}(N, α).

We also notice for e and π 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 μ_{k}(N, α_{1}, α_{2}) compared to μ_{k}(N, α), where disc was around .08 for N equal to 1000. In the case of α = π, 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}α}, 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.

1.7 Conclusion

Though much experimentation has been done on the behavior of {n^{k}α},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.

Bibliography

[BHJ] A.D. Barbour, Lars Holst,and Svante Janson, Poisson Approximation,Oxford University Press (1992), 1, 138 − 147.

[Ca] J. W. S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, London 1957.

[Ea] W. Eames, The Elementary Theory of Numbers,Polynomials, and Rational Functions, Oldbourne Mathematical Series131-136,American Elsevier Publishing Co. , New York, 1967.

[Ha] F. Haight, Handbook of the Poisson Distribution,. John Wiley & Sons, Inc., New York 1967.

[RL] R. Lipshitz, Robert Lipshitz’s Junior Project: Numerical results concerning the distribution of {n^{2}α},(Fall 2001).

[RSZ] Zeév Rudnick, P. Sarnak, and A. Zaharescu, The Distribution of Spacings Between the Fractional Parts of n^{2}α, Invent. Math. 145 (2001), no. 1, 37−57.