Abstract

1. We prove Liouville’s Theorem for the order of approximation by rationals of real algebraic numbers.

2. We construct several transcendental numbers.

3. We define Poissonian Behaviour, and study the spacings between the ordered fractional parts of {n^{k}α}.

Lecture by Steven Miller; notes for the first two by Steven Miller and Florin Spinu; notes for the third by Steven Miller.

Contents

1 Liouville’s Theorem Constructing Transcendentals 2 1.1 Review of Approximating by Rationals . . . . . . . . . . . . . . 2 1.2 Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Constructing Transcendental Numbers . . . . . . . . . . . . . . . 5

1.3.1 ∑_{m} 10^{−m!} . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3.2 [10^{1!}, 10^{2!}, …] . . . . . . . . . . . . . . . . . . . . . . . 6 1.3.3 Buffon’s Needle and π . . . . . . . . . . . . . . . . . . . 8

2 Poissonian Behavior and {n^{k}α} 10 2.1 Equidistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Point Masses and Induced Probability Measures . . . . . . . . . . 10 2.3 Neighbor Spacings . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.4 Poissonian Behavior . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4.1 Nearest Neighbor Spacings . . . . . . . . . . . . . . . . . 14 2.4.2 k^{th} Neighbor Spacings . . . . . . . . . . . . . . . . . . . 16

2.5 Induced Probability Measures . . . . . . . . . . . . . . . . . . . 18 2.6 Non-Poissonian Behavior . . . . . . . . . . . . . . . . . . . . . . 19

2.6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 20 2.6.2 Proof of Theorem 2.6.2 . . . . . . . . . . . . . . . . . . . 21 2.6.3 Measure of α ∉ ℚ with Non-Poissonian Behavior along

a sequence N_{n} . . . . . . . . . . . . . . . . . . . . . . . 22

Chapter 1

Liouville’s Theorem Constructing Transcendentals

1.1 Review of Approximating by Rationals

Definition 1.1.1 (Approximated by rationals to order n). A real number x is approximated by rationals to order n if there exist a constant k(x) (possibly depending on x) such that there are infinitely many rational \frac{p}{q} with

|x − \frac{p}{q}| < \frac{k(x)}{q^{n}}.  (1.1)

Recall that Dirichlet’s Box Principle gaves us:

|x − \frac{p}{q}| < \frac{1}{q^{2}}  (1.2)

for infintely many fractions \frac{p}{q}. This was proved by choosing a large parameter Q, and considering the Q + 1 fractionary parts {qx} ∈ [0, 1) for q ∈ {0, …, Q}. The box principle ensures us that there must be two different q’s, say:

0 ≤ q_{1} < q_{2} ≤ Q  (1.3)

such that both \{q_{1}x\} and \{q_{2}x\} belong to the same interval [\frac{a}{Q}, \frac{a+1}{Q}), for some 0 ≤ a ≤ Q − 1. Note that there are exactly Q such intervals partitioning [0, 1), and Q + 1 fractionary parts! Now, the length of such an interval is \frac{1}{Q} so we get

|\{q_{2}x\} − \{q_{1}x\}| < \frac{1}{Q}.  (1.4)

There exist integers p_{1} and p_{2} such that

{q_{1}x} = q_{1}x − p, {q_{2}x} = q_{2}x − p.  (1.5)

Letting p = p_{2} − p_{1} we find

|(q_{2} − q_{1})x − p| ≤ \frac{1}{Q}  (1.6)

Let q = q_{2} − q_{1}, so 1 ≤ q ≤ Q, and the previous equation can be rewriten as

|x − \frac{p}{q}| < \frac{1}{qQ} ≤ \frac{1}{q^{2}}  (1.7)

Now, letting Q → ∞, we get an infinite collection of rational fractions \frac{p}{q} satisfying the above equation. If this collection contains only finitely many distinct fractions, then one of these fractions, say \frac{p_{0}}{q_{0}} , would occur for infintely many choices Q_{k} of Q, thus giving us:

|x − \frac{p_{0}}{q_{0}} | < \frac{1}{qQ_{k}} → 0,  (1.8)

as k → ∞. This implies that x = \frac{p_{0}}{q_{0}} ∈ ℚ. So, unles x is a rational number, we can find infinitely many distinct rational numbers \frac{p}{q} satisfying Equation 1.7. This means that any real, irrational number can be approximated to order n = 2 by rational numbers.

1.2 Liouville’s Theorem

Theorem 1.2.1 (Liouville’s Theorem). Let x be a real algebraic number of degree n. Then x is approximated by rationals to order at most n.

Proof. Let

f(X) = a_{n}X^{n} + ⋯ a_{1}X + a_{0}  (1.9)

be the polynomial with integer coefficients of smallest degree (minimal polynomial) such that x satisfies

f(x) = 0.  (1.10)

Note that deg x = deg f and the condition of minimality implies that f(X) is irreducible over ℤ. Further, a well known result from algebra states that a polynomial irreducible over ℤ is also irreducible over ℚ.

In particular, as f(X) is irreducible over ℚ, f(X) does not have any rational roots. If it did, then f(X) would be divisible by a linear polynomial (X − \frac{a}{b}). Let G(X) = \frac{f(X)}{X− \frac{a}{b}}. Clear denominators (multiply throughout by b), and let g(X) = bG(X). Then deg g = deg f − 1, and g(x) = 0. This contradicts the minimality of f (we choose f to be a polynomial of smallest degree such that f(x) = 0). Therefore, f is non-zero at every rational.

Let

M = sup_{|z−x|<1} |f′(z)|.  (1.11)

Let now \frac{p}{q} be a rational such that |x − \frac{p}{q} | < 1. The Mean Value Theorem gives us that

|f (\frac{p}{q}) − f(x)| = |f′(c) (x − \frac{p}{q})| ≤ M |x − \frac{p}{q}|  (1.12)

where c is some real number between x and \frac{p}{q}; |c − x| < 1 for \frac{p}{q} moderately close to x.

Now we use the fact that f(X) does not have any rational roots:

0 ≠ f (\frac{p}{q}) = a_{n} (\frac{p}{q})^{n} + ⋯ + a_{0} = \frac{a_{n}p^{n} + ⋯ a_{1}p^{n−1}q + a_{0}q^{n}}{q^{n}} (1.13)

The numerator of the last term is a nonzero integer, hence it has absolute value at least 1. Since we also know that f(x) = 0 it follows that

|f (\frac{p}{q}) − f(x)| = |f (\frac{p}{q})| = \frac{|a_{n}p^{n} + ⋯ a_{1}p^{n−1}q + a_{0}q^{n}|}{q^{n}} ≥ \frac{1}{q^{n}} . (1.14)

Combining the equations 1.12 and 1.14, we get:

\frac{1}{q^{n}} ≤ M |x − \frac{p}{q}| ⇒ \frac{1}{Mq^{n}} ≤ |x − \frac{p}{q}|  (1.15)

whenever |x − \frac{p}{q}| < 1. This last equation shows us that x can be approximated by rationals to order at most n. For assume it was otherwise, namely that x can be approximated to order n + ϵ. Then we would have an infinite sequence of distinct rational numbers \{\frac{p_{i}}{q_{i}} \}_{i≥1} and a constant k(x) depending only on x such that

|x − \frac{p_{i}}{q_{i}} | < \frac{k(x)}{q_{i}^{n+ϵ}}.  (1.16)

Since the numbers \frac{p_{i}}{q_{i}} converge to x we can assume that they already are in the interval (x − 1, x + 1). Hence they also satisfy Equation 1.15:

\frac{1}{q_{i}^{n}} ≤ M |x − \frac{p_{i}}{q_{i}} | .  (1.17)

Combining the last two equations we get

\frac{1}{Mq_{i}^{n}} ≤ |x − \frac{p_{i}}{q_{i}} | < \frac{k(x)}{q_{i}^{n+ϵ}},  (1.18)

hence

q_{i}^{ϵ} < M  (1.19)

and this is clearly impossible for arbitrarily large q since ϵ > 0 and q_{i} → ∞.

Exercise 1.2.2. Justify the fact that if \{\frac{p_{i}}{q_{i}} \}_{i≥1} is a rational approximation to order n ≥ 1 of x, then q_{i} → ∞.

Remark 1.2.3. So far we have seen that the order to which an algebraic number can be approximated by rationals is bounded by its degree. Hence if a real, irrational number α ∉ ℚ can be approximated by rationals to an arbitrary large order, then α must be transcendental! This provides us with a recipe for constructing transcendental numbers.

1.3 Constructing Transcendental Numbers

1.3.1 ∑_{m} 10^{−m!}

The following construction of transcendental numbers is due to Liouville.

Theorem 1.3.1. The number

x = ∑_{m=1}^{∞} \frac{1}{10^{m!}}  (1.20)

is transcendental.

Proof. The series defining x is convergent, since it is dominated by the geometric series ∑ \frac{1}{10^{m}}. In fact, the series converges very rapidly and it is this high rate of convergence that will yield x is transcendental.

Fix N large, and let n > N. Write

\frac{p_{n}}{q_{n}} = ∑_{m=1}^{n} \frac{1}{10^{m!}}  (1.21)

with p_{n}, q_{n} > 0 and (p_{n}, q_{n}) = 1. Then \{\frac{p_{n}}{q_{n}} \}_{n≥1} is a monotone increasing sequence converging to x. In particular, all these rational numbers are distinct. Not also that q_{n} must divide 10^{n!}, which implies

q_{n} ≤ 10^{n!}.  (1.22)

Using this, we get

0 < x − \frac{p_{n}}{q_{n}} = ∑_{m>n} \frac{1}{10^{m!}} = \frac{1}{10^{(n+1)!}} (1 + \frac{1}{10^{n+2}} + \frac{1}{10^{(n+2)(n+3)}} + ⋯)
< \frac{2}{10^{(n+1)!}} = \frac{2}{(10^{n!})^{n+1}}
< \frac{2}{q_{n}^{n+1}} ≤ \frac{2}{q_{n}^{N}} .  (1.23)

This gives an approximation by rationals of order N of x. Since N can be chosen arbitrarily large, this implies that x can be approximated by rationals to arbitrary order. We can conclude, in view of our precious remark 1.2.3 that x is transcendental.

1.3.2 [10^{1!}, 10^{2!}, …]

Theorem 1.3.2. The number

y = [10^{1!}, 10^{2!}, …]  (1.24)

is transcendental.

Proof. Let \frac{p_{n}}{q_{n}} be the continued fraction of [10^{1!} ⋯ 10^{n!}]. Then

|y − \frac{p_{n}}{q_{n}} | = \frac{1}{q_{n}q′_{n+1}} = \frac{1}{q_{n}(a′_{n+1}q_{n} + q_{n−1})}
< \frac{1}{a_{n+1}} = \frac{1}{10^{(n+1)!}} .  (1.25)

Since q_{k} = a_{n}q_{k−1} + q_{n−2}, it implies that q_{k} > q_{k−1} Also, q_{k+1} = a_{k+1}q_{n} + q_{k−1}, so we get

\frac{q_{k+1}}{q_{k}} = a_{k+1} + \frac{q_{k−1}}{q_{k}} < a_{k+1} + 1.  (1.26)

Hence writing this inequality for k = 1, ⋯, n − 1 we obtain

q_{n} = q_{1} \frac{q_{2}}{q_{1}} \frac{q_{3}}{q_{2}} ⋯ \frac{q_{n}}{q_{n−1}} < (a_{1} + 1)(a_{2} + 1) ⋯ (a_{n} + 1)
= (1 + \frac{1}{a_{1}} ) ⋯ (1 + \frac{1}{a_{n}})a_{1} ⋯ a_{n}
< 2^{n}a_{1} ⋯ a_{n} = 2^{n}10^{1!+···+n!}
< 10^{2n!} = a_{n}^{2}  (1.27)

Combining equations 1.25 and 1.27 we get:

|y − \frac{p_{n}}{q_{n}} | < \frac{1}{a_{n+1}} = \frac{1}{a_{n}^{n+1}}
< (\frac{1}{a_{n}^{2}} )^{\frac{n}{2}} < (\frac{1}{q_{n}^{2}} )^{\frac{n}{2}}
= \frac{1}{q_{n}^{n/2}} .  (1.28)

In this way we get, just as in the previous theorem, an approximation of y by rationals to arbitrary order. This proves that y is transcendental.

1.3.3 Buffon’s Needle and π

Consider a collection of infinitely long parallel lines in the plane, where the spacing between any two adjacent lines is d. Let the lines be located at x = 0, ±d, ±2d, . . .. Consider a rod of length l, where for convenience we assume l < d.

If we were to randomly throw the rod on the plane, what is the probability it hits a line? This question was first asked by Buffon in 1733.

Because of the vertical symmetry, we may assume the center of the rod lies on the line x = 0, as shifting the rod (without rotating it) up or down will not alter the number of intersections. By the horizontal symmetry, we may assume −\frac{d}{2} ≤ x < \frac{d}{2}. We posit that all values of x are equally likely. As x is continuous distributed, we may add in x = \frac{d}{2} without changing the probability. The probability density function of x is \frac{dx}{d}.

Let θ be the angle the rod makes with the x-axis. As each angle is equally likely, the probability density function of θ is \frac{dθ}{2π}.

We assume that x and θ are chosen independently. Thus, the probability density for (x, θ) is \frac{dxdθ}{d·2π}.

The projection of the rod (making an angle of θ with the x-axis) along the x-axis is l · | cos θ|. If |x| ≤ l · | cos θ|, then the rod hits exactly one vertical line exactly once; if x > l · | cos θ|, the rod does not hit a vertical line. Note that if l > d, a rod could hit multiple lines, making the arguments more involved.

Thus, the probability a rod hits a line is

p = ∫_{θ=0}^{2π} ∫_{x=−l·| cos θ|}^{l·| cos θ|} \frac{dxdθ}{d · 2π}
= ∫_{θ=0}^{2π} \frac{l · | cos θ|}{d} \frac{dθ}{2π}
= \frac{2l}{πd}.  (1.29)

Exercise 1.3.3. Show

\frac{1}{2π} ∫_{0}^{2π} | cos θ|dθ = \frac{2}{π} .  (1.30)

Let A be the random variable which is the number of intersections of a rod of length l thrown against parallel vertical lines separated by d > l units. Then

A = \begin{cases}1 & with probability \frac{2l}{πd} \\ 0 & with probability 1 − \frac{2l}{πd} \end{cases} .  (1.31)

If we were to throw N rods independently, since the expected value of a sum is the sum of the expected values (Lemma ??), we expect to observe

N · \frac{2l}{πd}  (1.32)

intersections. Turning this around, let us throw N rods, and let I be the number of observed

intersections of the rods with the vertical lines. Then

I ≈ N · \frac{2l}{πd} → π ≈ \frac{N}{I} · \frac{2l}{d}.  (1.33)

The above is an experimental formula for π!

Chapter 2

Poissonian Behavior and {n^{k}α}

2.1 Equidistribution

We say a sequence of number x_{n} ∈ [0, 1) is equidistributed if

lim_{N→∞} \frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [a, b]\}}{N} = b − a  (2.1)

for any subinterval [a, b] of [0, 1].

Recall Weyl’s Result, Theorem ??: If α ∉ ℚ, then the fractional parts {nα} are equidistributed. Equivalently, nα mod 1 is equidistributed.

Similarly, one can show that for any integer k, {n^{k}α} is equidistributed. See Robert Lipshitz’s paper for more details.

2.2 Point Masses and Induced Probability Measures

Recall from physics the concept of a unit point mass located at x = a. Such a point mass has no length (or, in higher dimensions, width or height), but finite mass. As mass is the integral of the density over space, a finite mass in zero volume (or zero length on the line) implies an infinite density.

We can make this more precise by the notion of an Approximation to the Identity. See also Theorem ??.

Definition 2.2.1 (Approximation to the Identity). A sequence of functions g_{n}(x) is an approximation to the identity (at the origin) if

1. g_{n}(x) ≥ 0.

2. ∫ g_{n}(x)dx = 1.

3. Given ϵ, δ > 0 there exists N > 0 such that for all n > N, ∫_{|x|>δ} g_{n}(x)dx < ϵ.

We represent the limit of any such family of g_{n}(x)s by δ(x).

If f(x) is a nice function (say near the origin its Taylor Series converges) then

∫ f(x)δ(x)dx = lim_{n→∞} ∫ f(x)g_{n}(x) = f(0).  (2.2)

Exercise 2.2.2. Prove Equation 2.2.

Thus, in the limit the functions g_{n} are acting like point masses. We can consider the probability densities g_{n}(x)dx and δ(x)dx. For g_{n}(x)dx, as n → ∞, almost all the probability is concentrated in a narrower and narrower band about the origin; δ(x)dx is the limit with all the mass at one point. It is a discrete (as opposed to continuous) probability measure.

Note that δ(x − a) acts like a point mass; however, instead of having its mass concentrated at the origin, it is now concentrated at a.

Exercise 2.2.3. Let

g_{n}(x) = \begin{cases}n & if |x| ≤ \frac{1}{2n} \\ 0 & otherwise \end{cases}  (2.3)

Prove g_{n}(x) is an approximation to the identity at the origin.

Exercise 2.2.4. Let

g_{n}(x) = c \frac{\frac{1}{n}}{\frac{1}{n^{2}} + x^{2}} .  (2.4)

Find c such that the above is an approximation to the identity at the origin.

Given N point masses located at x_{1}, x_{2}, …, x_{N}, we can form a probability measure

μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} δ(x − x_{n})dx.  (2.5)

Note ∫ μ_{N}(x)dx = 1, and if f(x) is a nice function,

∫ f(x)μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} f(x_{n}).  (2.6)

Exercise 2.2.5. Prove Equation 2.6 for nice f(x).

Note the right hand side of Equation 2.6 looks like a Riemann sum. Or it would look like a Riemann sum if the x_{n}s were equidistributed. In general the x_{n}s will not be equidistributed, but assume for any interval [a, b] that as N → ∞, the fraction of x_{n}s (1 ≤ n ≤ N) in [a, b] goes to ∫_{a}^{b} p(x)dx for some nice function p(x):

lim_{N→∞} \frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [a, b]\}}{N} → ∫_{a}^{b} p(x)dx.  (2.7)

In this case, if f(x) is nice (say twice differentiable, with first derivative uniformly bounded), then

∫ f(x)μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} f(x_{n})

≈ ∑_{k=−∞}^{∞} f(\frac{k}{N} )\frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [\frac{k}{N} , \frac{k+1}{N}]\}}{N}

→ ∫ f(x)p(x)dx.  (2.8)

Definition 2.2.6 (Convergence to p(x)). If the sequence of points x_{n} satisfies Equation 2.7 for some nice function p(x), we say the probability measures μ_{N}(x)dx converge to p(x)dx.

2.3 Neighbor Spacings

We now consider finer questions. Let α_{n} be a collection of points in [0, 1). We order them by size:

0 ≤ α_{σ(1)} ≤ α_{σ(2)} ≤ ⋯ ≤ α_{σ(N)},  (2.9)

where σ is a permutation of 123 ⋯ N. Note the ordering depends crucially on N. Let β_{j} = α_{σ(j)}.

We consider how the differences β_{j+1} − β_{j} are distributed. We will use a slightly different definition of distance, however.

Recall [0, 1) is equivalent to the unit circle under the map x → e^{2πix}. Thus, the numbers .999 and .001 are actually very close; however, if we used the standard definition of distance, then |.999 − .001| = .998, which is quite large. Wrapping [0, 1) on itself (identifying 0 and 1), we see that .999 and .001 are separated by .002.

Definition 2.3.1 (mod 1 distance). Let x, y ∈ [0, 1). We define the mod 1 distance from x to y, ||x − y||, by

||x − y|| = min {|x − y|, 1 − |x − y|}.  (2.10)

Exercise 2.3.2. Show that the mod 1 distance between any two numbers in [0, 1) is at most \frac{1}{2}.

In looking at spacings between the β_{j}s, we have N − 1 pairs of neighbors:

(β_{2}, β_{1}), (β_{3}, β_{2}), …, (β_{N}, β_{N−1}).  (2.11)

These pairs give rise to spacings β_{j+1} − β_{j} ∈ [0, 1).

We can also consider the pair (β_{1}, β_{N}). This gives rise to the spacing β_{1}−β_{N} ∈ [−1, 0); however, as we are studying this sequence mod 1, this is equivalent to β_{1} − β_{N} + 1 ∈ [0, 1).

Henceforth, whenever we perform any arithmetic operation, we always mean mod 1; thus, our answers always live in [0, 1)

Definition 2.3.3 (Neighbor Spacings). Given a sequence of numbers α_{n} in [0, 1), fix an N and arrange the numbers α_{n} (n ≤ N) in increasing order. Label the new sequence β_{j}; note the ordering will depend on N. Let β_{−j} = β_{N−j} and β_{N+j} = β_{j}.

1. The nearest neighbor spacings are the numbers β_{j+1} − β_{j}, j = 1 to N.

2. The k^{th}-neighbor spacings are the numbers β_{j+k} − β_{j}, j = 1 to N.

Remember to take the differences β_{j+k} − β_{j} mod 1.

Exercise 2.3.4. Let α = \sqrt{2}, and let α_{n} = \{nα\} or \{n^{2}α\}. Calculate the nearest neighbor and the next-nearest neighbor spacings in each case for N = 10.

Definition 2.3.5 (wrapped unit interval). We call [0, 1), when all arithmetic operations are done mod 1, the wrapped unit interval.

2.4 Poissonian Behavior

Let α ∉ ℚ. Fix a positive integer k, and let α_{n} = {n^{k}α}. As N → ∞, look at the ordered α_{n}s, denoted by β_{n}. How are the nearest neighbor spacings of β_{n} distributed? How does this depend on k? On α? On N?

Before discussing this problem, we consider a simpler case. Fix N, and consider N independent random variables x_{n}. Each random variable is chosen from the uniform distribution on [0, 1); thus, the probability that x_{n} ∈ [a, b) is b − a.

Let y_{n} be the x_{n}s arranged in increasing order. How do the neighbor spacings behave?

First, we need to decide what is the correct scale to use for our investigations. As we have N objects on the wrapped unit interval, we have N nearest neighbor spacings. Thus, we expect the average spacing to be \frac{1}{N}.

Definition 2.4.1 (Unfolding). Let z_{n} = Ny_{n}. The numbers z_{n} = Ny_{n} have unit mean spacing. Thus, while we expect the average spacing between adjacent y_{n}s to be \frac{1}{N} units, we expect the average spacing between adjacent z_{n}s to be 1 unit.

So, the probability of observing a spacing as large as \frac{1}{2} between adjacent y_{n}s becomes negligible as N → ∞. What we should ask is what is the probability of observing a nearest neighbor spacing of adjacent y_{n}s that is half the average spacing. In terms of the z_{n}s, this will correspond to a spacing between adjacent z_{n}s of \frac{1}{2} a unit.

2.4.1 Nearest Neighbor Spacings

By symmetry, on the wrapped unit interval the expected nearest neighbor spacing is independent of j. Explicitly, we expect β_{j+1} − β_{j} to have the same distribution as β_{i+1} − β_{i}.

What is the probability that, when we order the x_{n}s in increasing order, the next x_{n} after x_{1} is located between \frac{t}{N} and \frac{t+Δt}{N}? Let the x_{n}s in increasing order be labeled y_{1} ≤ y_{2} ≤ ⋯ ≤ y_{N}, y_{n} = x_{σ(n)}.

As we are choosing the x_{n}s independently, there are \binom{N−1}{1} choices of subscript n such that x_{n} is nearest to x_{1}. This can also be seen by symmetry, as each x_{n} is equally likely to be the first to the right of x_{1} (where, of course, .001 is just a little to the right of .999), and we have N − 1 choices left for x_{n}.

The probability that x_{n} ∈ [\frac{t}{N} , \frac{t+Δt}{N}] is \frac{Δt}{N}.

For the remaining N − 2 of the x_{n}s, each must be further than \frac{t+Δt}{N} from x_{n}. Thus, they must all lie in an interval (or possibly two intervals if we wrap around)

of length 1− \frac{t+Δt}{N}. The probability that they all lie in this region is (1− \frac{t+Δt}{N})^{N−2}.

Thus, if x_{1} = y_{l}, we want to calculate the probability that ||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t+Δt}{N}]. This is

Prob(||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) = \binom{N − 1}{1} · \frac{Δt}{N} · (1 − \frac{t + Δt}{N})^{N−2}
= (1 − \frac{1}{N} ) · (1 − \frac{t + Δt}{N})^{N−2}Δt.

(2.12)

For N enormous and Δt small,

(1 − \frac{1}{N} )  ≈ 1
(1 − \frac{t + Δt}{N})^{N−2} ≈ e^{−(t+Δt)} ≈ e^{−t}.  (2.13)

Thus

Prob(||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) → e^{−t}Δt.  (2.14)

Remark 2.4.2. The above argument is infinitesimally wrong. Once we’ve located y_{l+1}, the remaining x_{n}s do not need to be more than \frac{t+Δt}{N} units to the right of x_{1} = y_{l}; they only need to be further to the right than y_{l+1}. As the incremental gain in probabilities for the locations of the remaining x_{n}s is of order Δt, these contributions will not influence the large N, small Δt limits. Thus, we ignore these effects.

To rigorously derive the limiting behavior of the nearest neighbor spacings using the above arguments, one would integrate over x_{m} ranging from \frac{t}{N} to \frac{t+Δt}{N}, and the remaining events x_{n} would be in the a segment of length 1 − x_{m}. As

|(1 − x_{m}) − (1 − \frac{t + Δt}{N})| ≤ \frac{Δt}{N},  (2.15)

this will lead to corrections of higher order in Δt, hence negligible.

We can rigorously avoid this by instead considering the following:

1. Calculate the probability that all the other x_{n}s are at least \frac{t}{N} units to the right of x_{1}. This is

p_{t} = (1 − \frac{t}{N} )^{N−1} → e^{−t}.  (2.16)

2. Calculate the probability that all the other x_{n}s are at least \frac{t+Δt}{N} units to the right of x_{1}. This is

p_{t+Δt} = (1 − \frac{t + Δt}{N})^{N−1} → e^{−(t+Δt)}.  (2.17)

3. The probability that no x_{n}s are within \frac{t}{N} units to the right of x_{1} but at least one x_{n} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right is p_{t+Δt} − p_{t}:

p_{t} − p_{t+Δt} → e^{−t} − e^{−(t+Δt)}
= e^{−t}(1 − e^{−Δt})
= e^{−t}(1 − 1 + Δt + O((Δt)^{2})
→ e^{−t}Δt.  (2.18)

Definition 2.4.3 (Unfolding Spacings). If y_{l+1} − y_{l} ∈ [\frac{t}{N} , \frac{t+Δt}{N}], then N(y_{l+1} − y_{l}) ∈ [t, t + Δt]. The new spacings z_{l+1} − z_{l} have unit mean spacing. Thus, while we expect the average spacing between adjacent y_{n}s to be \frac{1}{N} units, we expect the average spacing between adjacent z_{n}s to be 1 unit.

2.4.2 k^{th} Neighbor Spacings

Similarly, one can easily analyze the distribution of the k^{th} neighbor spacings when each x_{n} is chosen independently from the uniform distribution on [0, 1).

Again, consider x_{1} = y_{l}. Now we want to calculate the probability that y_{l+k} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right of y_{l}.

Therefore, we need exactly k −1 of the x_{n}s to lie between 0 and \frac{t}{N} units to the right of x_{1}, exactly one x_{n} (which will be y_{l+k}) to lie between \frac{t}{N} and \frac{t+Δt}{N} units to the right of x_{1}, and the remaining x_{n}s to lie at least \frac{t+Δt}{N} units to the right of y_{l+k}.

Remark 2.4.4. We face the same problem discussed in Remark 2.4.2; a similar argument will show that ignoring these affects will not alter the limiting behavior. Therefore, we will make these simplifications.

There are \binom{N−1}{k−1} ways to choose the x_{n}s that are at most \frac{t}{N} units to the right of x_{1}; there is then \binom{(N−1)−(k−1)}{1} ways to choose the x_{n} between \frac{t}{N} and \frac{t+Δt}{N} units to the right of x_{1}.

Thus,

Prob(||y_{l+k} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) =

= \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1} · \binom{(N − 1) − (k − 1)}{1}\frac{Δt}{N} · (1 − \frac{t + Δt}{N})^{N−(k+1)}
= \frac{(N − 1) ⋯ (N − 1 − (k − 2))}{N^{k−1}} \frac{(N − 1) − (k − 1)}{N} \frac{t^{k−1}}{(k − 1)!}(1 − \frac{t + Δt}{N})^{N−(k+1)}Δt
→ \frac{t^{k−1}}{(k − 1)!}e^{−t}Δt.  (2.19)

Again, one way to avoid the complications is to integrate over x_{m} ranging from \frac{t}{N} to \frac{t+Δt}{N}.

Or, similar to before, we can proceed more rigorously as follows:

1. Calculate the probability that exactly k − 1 of the other x_{n}s are at most \frac{t}{N} units to the right of x_{1}, and the remaining (N − 1) − (k − 1) of the x_{n}s are at least \frac{t}{N} units to the right of x_{1}. As there are \binom{N−1}{k−1} ways to choose k − 1 of the x_{n}s to be at most \frac{t}{N} units to the right of x_{1}, this probability is

p_{t} = \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1}(1 − \frac{t}{N} )^{(N−1)−(k−1)}
→ \frac{N^{k−1}}{(k − 1)!} \frac{t^{k−1}}{N^{k−1}} e^{−t}
→ \frac{t^{k−1}}{k − 1!}e^{−t}.  (2.20)

2. Calculate the probability that exactly k − 1 of the other x_{n}s are at most \frac{t}{N} units to the right of x_{1}, and the remaining (N − 1) − (k − 1) of the x_{n}s are at least \frac{t+Δt}{N} units to the right of x_{1}. Similar to the above, this gives

p_{t} = \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1}(1 − \frac{t + Δt}{N})^{(N−1)−(k−1)}
→ \frac{N^{k−1}}{(k − 1)!} \frac{t^{k−1}}{N^{k−1}} e^{−(t+Δt)}
→ \frac{t^{k−1}}{(k − 1)!}e^{−(t+Δt)}.  (2.21)

3. The probability that exactly k−1 of the x_{n}s are within \frac{t}{N} units to the right of x_{1} and at least one x_{n} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right is p_{t+Δt} − p_{t}:

p_{t} −p_{t+Δt} → \frac{t^{k−1}}{(k − 1)!}e^{−t} − \frac{t^{k−1}}{(k − 1)!}e^{−(t+Δt)} → \frac{t^{k−1}}{(k − 1)!}e^{−t}Δt. (2.22)

Note that when k = 1, we recover the nearest neighbor spacings.

2.5 Induced Probability Measures

We have proven the following:

Theorem 2.5.1. Consider N independent random variables x_{n} chosen from the uniform distribution on the wrapped unit interval [0, 1). For fixed N, arrange the x_{n}s in increase order, labeled y_{1} ≤ y_{2} ≤ ⋯ ≤ y_{N}.

Form the induced probability measure μ_{N,1} from the nearest neighbor spacings. Then as N → ∞ we have

μ_{N,1}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − N(y_{n} − y_{n−1}))dt → e^{−t}dt.  (2.23)

Equivalently, using z_{n} = Ny_{n}:

μ_{N,1}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − (z_{n} − z_{n−1}))dt → e^{−t}dt.  (2.24)

More generally, form the probability measure from the k^{th} nearest neighbor spacings. Then as N → ∞ we have

μ_{N,k}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − N(y_{n} − y_{n−k}))dt → \frac{t^{k−1}}{(k − 1)!}e^{−t}dt. (2.25)

Equivalently, using z_{n} = Ny_{n}:

μ_{N,k}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − (z_{n} − z_{n−k}))dt → \frac{t^{k−1}}{(k − 1)!}e^{−t}dt.  (2.26)

Definition 2.5.2 (Poissonian Behavior). We say a sequence of points x_{n} has Poissonian Behavior if in the limit as N → ∞ the induced probability measures μ_{N,k}(t)dt converge to \frac{t^{k−1}}{(k−1)!}e^{−t}dt.

Exercise 2.5.3. Let α ∈ ℚ, and define α_{n} = {n^{m}α} for some positive integer m. Show the sequence of points α_{n} does not have Poissonian Behavior.

Exercise 2.5.4. Let α ∉ ℚ, and define α_{n} = {nα}. Show the sequence of points α_{n} does not have Poissonian Behavior. Hint: for each N, show the nearest neighbor spacings take on at most three distinct values (the three values depend on N). As only three values are ever assumed for a fixed N, μ_{N,1}(t)dt cannot converge to e^{−t}dt.

2.6 Non-Poissonian Behavior

Conjecture 2.6.1. With probability one (with respect to Lebesgue Measure, see Definition ??), if α ∉ ℚ, if α_{n} = {n^{2}α} then the sequence of points α_{n} is Poissonian.

There are constructions which show certain irrationals give rise to non-Poissonian behavior.

Theorem 2.6.2. Let α ∈ ℚ such that |α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}^{3}} holds infinitely often, with a_{n} → 0. Then there exist integers N_{j} → ∞ such that μ_{N_{j},1}(t) does not converge to e^{−t}dt.

As a_{n} → 0, eventually a_{n} < \frac{1}{10} for all n large. Let N_{n} = q_{n}, where \frac{p_{n}}{q_{n}} is a good rational approximation to α:

|α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}^{3}} .  (2.27)

Remember that all subtractions are performed on the wrapped unit interval. Thus, ||.999 − .001|| = .002.

We look at α_{k} = {k^{2}α}, 1 ≤ k ≤ N_{n} = q_{n}. Let the β_{k}s be the α_{k}s arranged in increasing order, and let the γ_{k}s be the numbers \{k^{2} \frac{p_{n}}{q_{n}} \} arranged in increasing order:

β_{1} ≤ β_{2} ≤ ⋯ ≤ β_{N}
γ_{1} ≤ γ_{2} ≤ ⋯ ≤ γ_{N}.  (2.28)

2.6.1 Preliminaries

Lemma 2.6.3. If β_{l} = α_{k} = \{k^{2}α\}, then γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}. Thus, the same permutation orders both the α_{k}s and the γ_{k}s.

Proof. Multiplying both sides of Equation 2.27 by k^{2} ≤ q_{n}^{2} yields

|k^{2}α − k^{2} \frac{p_{n}}{q_{n}} | < k^{2} \frac{a_{n}}{q_{n}^{2}} ≤ \frac{a_{n}}{q_{n}} < \frac{1}{2q_{n}} .  (2.29)

Thus, k^{2}α and k^{2} \frac{p_{n}}{q_{n}} differ by at most \frac{1}{2q_{n}}. Therefore

||\{k^{2}α\} − \{k^{2} \frac{p_{n}}{q_{n}} \}|| < \frac{1}{2q_{n}} .  (2.30)

As the numbers \{m^{2} \frac{p_{n}}{q_{n}} \} all have denominators of size at most \frac{1}{q_{n}}, we see that \{k^{2} \frac{p_{n}}{q_{n}} \} is the closest of the \{m^{2} \frac{p_{n}}{q_{n}} \} to \{k^{2}α\}.

This implies that if β_{l} = \{k^{2}α\}, then γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}, completing the proof.

Exercise 2.6.4. Prove the ordering is as claimed. Hint: about each β_{l} = {k^{2}α}, the closest number of the form \{c^{2} \frac{p_{n}}{q_{n}} \} is \{k^{2} \frac{p_{n}}{q_{n}} \}.

2.6.2 Proof of Theorem 2.6.2

Exercise 2.6.5. Assume ||a − b||, ||c − d|| < \frac{1}{10}. Show

||(a − b) − (c − d)|| < ||a − b|| + ||c − d||.  (2.31)

Proof of Theorem 2.6.2: We have shown

||β_{l} − γ_{l}|| < \frac{a_{n}}{q_{n}} .  (2.32)

Thus, as N_{n} = q_{n}:

||N_{n}(β_{l} − γ_{l})|| < a_{n},  (2.33)

and the same result holds with l replaced by l − 1. By Exercise 2.6.5,

||N_{n}(β_{l} − γ_{l}) − N_{n}(β_{l−1} − γ_{l−1})|| < 2a_{n}.  (2.34)

Rearranging gives

||N_{n}(β_{l} − β_{l−1}) − N_{n}(γ_{l} − γ_{l−1})|| < 2a_{n}.  (2.35)

As a_{n} → 0, this implies the difference between ||N_{n}(β_{l}−β_{l−1})|| and ||N_{n}(γ_{l}−

γ_{l−1})|| goes to zero.

The above distance calculations were done mod 1. The actual differences will differ by an integer. Thus,

μ_{N_{n},1}^{α}(t)dt = \frac{1}{N_{n}} ∑_{l=1}^{N_{n}} δ(t − N_{n}(β_{l} − β_{l−1}))  (2.36)

and

μN_{n},1^{\frac{pn}{qn}}(t)dt = \frac{1}{N_{n}} ∑_{l=1}^{N_{n}} δ(t − N_{n}(γ_{l} − γ_{l−1}))  (2.37)

are extremely close to one another; each point mass from the difference between adjacent β_{l}s is either within a_{n} units of a point mass from the difference between adjacent γ_{l}s, or is within a_{n} units of a point mass an integer number of units from a point mass from the difference between adjacent γ_{l}s. Further, a_{n} → 0.

Note, however, that if γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}, then

N_{n}γ_{l} = q_{n}\{k^{2} \frac{p_{n}}{q_{n}} \} ∈ ℕ.  (2.38)

Thus, the induced probability measure μN_{n},1^{\frac{pn}{qn}}(t)dt formed from the γ_{l}s is sup-

ported on the integers! Thus, it is impossible for μN_{n},1^{\frac{pn}{qn}}(t)dt to converge to e^{−t}dt. As μ_{N_{n},1}^{α}(t)dt, modulo some possible integer shifts, is arbitrarily close to

μN_{n},1^{\frac{pn}{qn}}(t)dt, the sequence \{k^{2}α\} is not Poissonian along the subsequence of Ns given by N_{n}, where N_{n} = q_{n}, q_{n} is a denominator in a good rational approximation to α. □

2.6.3 Measure of α ∉ ℚ with Non-Poissonian Behavior along a sequence N_{n}

What is the (Lebesgue) measure of α ∉ ℚ such that there are infinitely many n with

|α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}} , a_{n} → 0.  (2.39)

If the above holds, then for any constant k(α), for n large (large depends on both α and k(α)) we have

|α − \frac{p_{n}}{q_{n}} | < \frac{k(α)}{q_{n}^{2+ϵ}}.  (2.40)

By Theorem ??, this set has (Lebesgue) measure or size 0. Thus, almost no irrational numbers satisfy the conditions of Theorem 2.6.2, where almost no is relative to the (Lebesgue) measure.

Exercise 2.6.6. In a topological sense, how many algebraic numbers satisfy the conditions of Theorem 2.6.2? How many transcendental numbers satisfy the conditions?

Exercise 2.6.7. Let α satisfy the conditions of Theorem 2.6.2. Consider the sequence N_{n}, where N_{n} = q_{n}, q_{n} the denominator of a good approximation to α. We know the induced probability measures μN_{n},1^{\frac{pn}{qn}}(t)dt and μ_{N_{n},1}^{α}(t)dt do not converge to e^{−t}dt. Do these measures converge to anything?

Remark 2.6.8. In The Distribution of Spacings Between the Fractional Parts of {n^{2}α} (Z. Rudnick, P. Sarnak, A. Zaharescu), it is shown that for most α satisfying the conditions of Theorem 2.6.2, there is a sequence N_{j} along which μ_{N_{n},1}^{α}(t)dt does converge to e^{−t}dt.