Abstract 1. We Prove Liouville's Theorem for the Order of Approximation by Rationals of Real Algebraic Numbers. 2. We Construc

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 {nkα}.

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 P −m! 1.3.1 m 10 ...... 5 1.3.2 [101!, 102!,... ] ...... 6 1.3.3 Buffon’s Needle and π ...... 8

2 Poissonian Behavior and {nkα} 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 kth 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 6∈ Q with Non-Poissonian Behavior along a sequence Nn ...... 22

1 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 de- p pending on x) such that there are inﬁnitely many rational q with

p k(x) x − < . (1.1) q qn Recall that Dirichlet’s Box Principle gaves us:

p 1 x − < (1.2) q q2 p for inﬁntely many fractions 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 ≤ q1 < q2 ≤ Q (1.3) a a+1 such that both {q1x} and {q2x} belong to the same interval [ Q , Q ), for some 0 ≤ a ≤ Q − 1. Note that there are exactly Q such intervals partitioning [0, 1), 1 and Q + 1 fractionary parts! Now, the length of such an interval is Q so we get

2 1 |{q x} − {q x}| < . (1.4) 2 1 Q

There exist integers p1 and p2 such that

{q1x} = q1x − p, {q2x} = q2x − p. (1.5)

Letting p = p2 − p1 we ﬁnd 1 |(q − q )x − p| ≤ (1.6) 2 1 Q

Let q = q2 − q1, so 1 ≤ q ≤ Q, and the previous equation can be rewriten as

p 1 1 x − < ≤ (1.7) q qQ q2 p Now, letting Q → ∞, we get an infinite collection of rational fractions q satisfying the above equation. If this collection contains only finitely many distinct fractions, then one of these fractions, say p0 , would occur for infintely many q0 choices Qk of Q, thus giving us:

p0 1 x − < → 0, (1.8) q0 qQk

p0 as k → ∞. This implies that x = q ∈ Q. So, unles x is a rational number, 0 p we can ﬁnd inﬁnitely many distinct rational numbers 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

n f(X) = anX + ··· a1X + a0 (1.9) be the polynomial with integer coefﬁcients of smallest degree (minimal polynomial) such that x satisﬁes

3 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 a roots. If it did, then f(X) would be divisible by a linear polynomial (X − b ). Let f(X) G(X) = a . Clear denominators (multiply throughout by b), and let g(X) = X− b 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 |f 0(z)|. (1.11) |z−x|<1

Let now p be a rational such that x − p < 1. The Mean Value Theorem gives q q us that p 0 p p f − f(x) = f (c) x − ≤ M x − (1.12) q q q p p where c is some real number between x and q ; |c − x| < 1 for q moderately close to x. Now we use the fact that f(X) does not have any rational roots:

p pn a pn + ··· a pn−1q + a qn 0 6= f = a + ··· + a = n 1 0 (1.13) q n q 0 qn 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

n n−1 n p p |anp + ··· a1p q + a0q | 1 f − f(x) = f = ≥ . (1.14) q q qn qn Combining the equations 1.12 and 1.14, we get:

1 p 1 p ≤ M x − ⇒ ≤ x − (1.15) qn q Mqn q

4 p whenever |x − 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 + . Then we would have an inﬁnite sequence of distinct pi rational numbers { }i≥1 and a constant k(x) depending only on x such that qi

pi k(x) x − < n+ . (1.16) qi qi Since the numbers pi converge to x we can assume that they already are in the qi interval (x − 1, x + 1). Hence they also satisfy Equation 1.15:

1 pi n ≤ M x − . (1.17) qi qi Combining the last two equations we get

1 pi k(x) n ≤ x − < n+ , (1.18) Mqi qi qi hence

qi < M (1.19)

and this is clearly impossible for arbitrarily large q since > 0 and qi → ∞.

pi Exercise 1.2.2. Justify the fact that if { }i≥1 is a rational approximation to order qi n ≥ 1 of x, then qi → ∞. 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

P −m! 1.3.1 m 10 The following construction of transcendental numbers is due to Liouville.

5 Theorem 1.3.1. The number

∞ X 1 x = (1.20) 10m! m=1 is transcendental.

Proof. The series deﬁning x is convergent, since it is dominated by the geometric P 1 series 10m . 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

n pn X 1 = (1.21) q 10m! n m=1 with p , q > 0 and (p , q ) = 1. Then { pn } is a monotone increasing n n n n qn n≥1 sequence converging to x. In particular, all these rational numbers are distinct. n! Not also that qn must divide 10 , which implies

n! qn ≤ 10 . (1.22) Using this, we get

pn X 1 1 1 1 0 < x − = = 1 + + + ··· q 10m! 10(n+1)! 10n+2 10(n+2)(n+3) n m>n 2 2 < = 10(n+1)! (10n!)n+1 2 2 < n+1 ≤ N . (1.23) qn qn 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 [101!, 102!,... ] Theorem 1.3.2. The number

6 y = [101!, 102!,... ] (1.24) is transcendental. Proof. Let pn be the continued fraction of [101! ··· 10n!]. Then qn

pn 1 1 y − = 0 = 0 qn qnqn+1 qn(an+1qn + qn−1) 1 1 < = (n+1)! . (1.25) an+1 10

Since qk = anqk−1 + qn−2, it implies that qk > qk−1 Also, qk+1 = ak+1qn + qk−1, so we get

qk+1 qk−1 = ak+1 + < ak+1 + 1. (1.26) qk qk Hence writing this inequality for k = 1, ··· , n − 1 we obtain

q2 q3 qn qn = q1 ··· < (a1 + 1)(a2 + 1) ··· (an + 1) q1 q2 qn−1 1 1 = (1 + ) ··· (1 + )a1 ··· an a1 an n n 1!+···+n! < 2 a1 ··· an = 2 10 2n! 2 < 10 = an (1.27) Combining equations 1.25 and 1.27 we get:

pn 1 1 y − < = n+1 qn an+1 an n n 1 2 1 2 < 2 < 2 an qn 1 = n/2 . (1.28) qn 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.

7 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 d number of intersections. By the horizontal symmetry, we may assume − 2 ≤ x < d 2 . We posit that all values of x are equally likely. As x is continuous distributed, d we may add in x = 2 without changing the probability. The probability density dx function of x is d . Let θ be the angle the rod makes with the x-axis. As each angle is equally dθ likely, the probability density function of θ is 2π . We assume that x and θ are chosen independently. Thus, the probability den- dxdθ sity for (x, θ) is 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

8 ( 2l 1 with probability πd A = 2l . (1.31) 0 with probability 1 − πd 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 2l N · (1.32) πd 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 2l N 2l I ≈ N · → π ≈ · . (1.33) πd I d The above is an experimental formula for π!

9 Chapter 2

Poissonian Behavior and {nkα}

2.1 Equidistribution

We say a sequence of number xn ∈ [0, 1) is equidistributed if #{n : 1 ≤ n ≤ N and x ∈ [a, b]} lim n = b − a (2.1) N→∞ N for any subinterval [a, b] of [0, 1]. Recall Weyl’s Result, Theorem ??: If α 6∈ Q, then the fractional parts {nα} are equidistributed. Equivalently, nα mod 1 is equidistributed. Similarly, one can show that for any integer k, {nkα} 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 gn(x) is an approximation to the identity (at the origin) if

1. gn(x) ≥ 0.

10 R 2. gn(x)dx = 1. R 3. Given , δ > 0 there exists N > 0 such that for all n > N, |x|>δ gn(x)dx < .

We represent the limit of any such family of gn(x)s by δ(x). If f(x) is a nice function (say near the origin its Taylor Series converges) then Z Z f(x)δ(x)dx = lim f(x)gn(x) = f(0). (2.2) n→∞ Exercise 2.2.2. Prove Equation 2.2.

Thus, in the limit the functions gn are acting like point masses. We can consider the probability densities gn(x)dx and δ(x)dx. For gn(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 ( 1 n if |x| ≤ 2n gn(x) = (2.3) 0 otherwise

Prove gn(x) is an approximation to the identity at the origin. Exercise 2.2.4. Let 1 g (x) = c n . (2.4) n 1 2 n2 + x Find c such that the above is an approximation to the identity at the origin.

Given N point masses located at x1, x2, . . . , xN , we can form a probability measure

N 1 X µ (x)dx = δ(x − x )dx. (2.5) N N n n=1 R Note µN (x)dx = 1, and if f(x) is a nice function,

N Z 1 X f(x)µ (x)dx = f(x ). (2.6) N N n n=1

11 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 xns were equidistributed. In general the xns will not be equidistributed, but assume for any interval [a, b] that as N → ∞, the R b fraction of xns (1 ≤ n ≤ N) in [a, b] goes to a p(x)dx for some nice function p(x):

#{n : 1 ≤ n ≤ N and x ∈ [a, b]} Z b lim n → p(x)dx. (2.7) N→∞ N a In this case, if f(x) is nice (say twice differentiable, with ﬁrst derivative uni- formly bounded), then

N Z 1 X f(x)µ (x)dx = f(x ) N N n n=1 h i ∞ #{n : 1 ≤ n ≤ N and x ∈ k , k+1 } X k n N N ≈ f N N k=−∞ Z → f(x)p(x)dx. (2.8)

Deﬁnition 2.2.6 (Convergence to p(x)). If the sequence of points xn satisﬁes Equation 2.7 for some nice function p(x), we say the probability measures µN (x)dx converge to p(x)dx.

2.3 Neighbor Spacings

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

0 ≤ ασ(1) ≤ ασ(2) ≤ · · · ≤ ασ(N), (2.9) where σ is a permutation of 123 ··· N. Note the ordering depends crucially on N. Let βj = ασ(j). We consider how the differences βj+1 − βj are distributed. We will use a slightly different deﬁnition of distance, however.

12 Recall [0, 1) is equivalent to the unit circle under the map x → e2πix. 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 n o ||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) 1 is at most 2 .

In looking at spacings between the βjs, we have N − 1 pairs of neighbors:

(β2, β1), (β3, β2),..., (βN , βN−1). (2.11)

These pairs give rise to spacings βj+1 − βj ∈ [0, 1). We can also consider the pair (β1, βN ). This gives rise to the spacing β1−βN ∈ [−1, 0); however, as we are studying this sequence mod 1, this is equivalent to β1 − βN + 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 αn in [0, 1), ﬁx an N and arrange the numbers αn (n ≤ N) in increasing order. Label the new sequence βj; note the ordering will depend on N. Let β−j = βN−j and βN+j = βj.

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

th 2. The k -neighbor spacings are the numbers βj+k − βj, j = 1 to N.

Remember to take the differences βj+k − βj mod 1. √ 2 Exercise 2.3.4. Let α = 2, and let αn = {nα} or {n 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 op- erations are done mod 1, the wrapped unit interval.

13 2.4 Poissonian Behavior

k Let α 6∈ Q. Fix a positive integer k, and let αn = {n α}. As N → ∞, look at the ordered αns, denoted by βn. How are the nearest neighbor spacings of βn 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 xn. Each random variable is chosen from the uniform distribution on [0, 1); thus, the probability that xn ∈ [a, b) is b − a. Let yn be the xns 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 1 spacings. Thus, we expect the average spacing to be N .

Deﬁnition 2.4.1 (Unfolding). Let zn = Nyn. The numbers zn = Nyn have unit mean spacing. Thus, while we expect the average spacing between adjacent yns 1 to be N units, we expect the average spacing between adjacent zns to be 1 unit. 1 So, the probability of observing a spacing as large as 2 between adjacent yns becomes negligible as N → ∞. What we should ask is what is the probability of observing a nearest neighbor spacing of adjacent yns that is half the average spacing. In terms of the zns, this will correspond to a spacing between adjacent 1 zns of 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 βj+1 − βj to have the same distribution as βi+1 − βi. What is the probability that, when we order the xns in increasing order, the t t+∆t next xn after x1 is located between N and N ? Let the xns in increasing order be labeled y1 ≤ y2 ≤ · · · ≤ yN , yn = xσ(n). N−1 As we are choosing the xns independently, there are 1 choices of subscript n such that xn is nearest to x1. This can also be seen by symmetry, as each xn is equally likely to be the ﬁrst to the right of x1 (where, of course, .001 is just a little to the right of .999), and we have N − 1 choices left for xn. h t t+∆t i ∆t The probability that xn ∈ N , N is N .

14 t+∆t For the remaining N − 2 of the xns, each must be further than N from xn. Thus, they must all lie in an interval (or possibly two intervals if we wrap around) N−2 t+∆t t+∆t of length 1− N . The probability that they all lie in this region is 1− N . Thus, if x1 = yl, we want to calculate the probability that ||yl+1 − yl|| ∈ h t t+∆t i N , N . This is

! h t t + ∆ti N − 1 ∆t t + ∆tN−2 Prob ||y − y || ∈ , = · · 1 − l+1 l N N 1 N N 1 t + ∆tN−2 = 1 − · 1 − ∆t. N N (2.12) For N enormous and ∆t small,

1 1 − ≈ 1 N t + ∆tN−2 1 − ≈ e−(t+∆t) ≈ e−t. (2.13) N Thus ! h t t + ∆ti Prob ||y − y || ∈ , → e−t∆t. (2.14) l+1 l N N Remark 2.4.2. The above argument is inﬁnitesimally wrong. Once we’ve located t+∆t yl+1, the remaining xns do not need to be more than N units to the right of x1 = yl; they only need to be further to the right than yl+1. As the incremental gain in probabilities for the locations of the remaining xns is of order ∆t, these contributions will not inﬂuence the large N, small ∆t limits. Thus, we ignore these effects. To rigorously derive the limiting behavior of the nearest neighbor spacings t t+∆t using the above arguments, one would integrate over xm ranging from N to N , and the remaining events xn would be in the a segment of length 1 − xm. As t + ∆t ∆t 1 − x − 1 − ≤ , (2.15) m N N this will lead to corrections of higher order in ∆t, hence negligible. We can rigorously avoid this by instead considering the following:

15 t 1. Calculate the probability that all the other xns are at least N units to the right of x1. This is

t N−1 p = 1 − → e−t. (2.16) t N

t+∆t 2. Calculate the probability that all the other xns are at least N units to the right of x1. This is

t + ∆tN−1 p = 1 − → e−(t+∆t). (2.17) t+∆t N

t 3. The probability that no xns are within N units to the right of x1 but at least t t+∆t one xn is between N and N units to the right is pt+∆t − pt:

−t −(t+∆t) pt − pt+∆t → e − e = e−t 1 − e−∆t ! = e−t 1 − 1 + ∆t + O (∆t)2

→ e−t∆t. (2.18)

h t t+∆t i Deﬁnition 2.4.3 (Unfolding Spacings). If yl+1 − yl ∈ N , N , then N(yl+1 − yl) ∈ [t, t + ∆t]. The new spacings zl+1 − zl have unit mean spacing. Thus, while 1 we expect the average spacing between adjacent yns to be N units, we expect the average spacing between adjacent zns to be 1 unit.

2.4.2 kth Neighbor Spacings Similarly, one can easily analyze the distribution of the kth neighbor spacings when each xn is chosen independently from the uniform distribution on [0, 1). Again, consider x1 = yl. Now we want to calculate the probability that yl+k is t t+∆t between N and N units to the right of yl. t Therefore, we need exactly k −1 of the xns to lie between 0 and N units to the t t+∆t right of x1, exactly one xn (which will be yl+k) to lie between N and N units to t+∆t the right of x1, and the remaining xns to lie at least N units to the right of yl+k.

16 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.

N−1 t There are k−1 ways to choose the xns that are at most N units to the right of (N−1)−(k−1) t t+∆t x1; there is then 1 ways to choose the xn between N and N units to the right of x1. Thus,

! h t t + ∆ti Prob ||y − y || ∈ , = l+k l N N N − 1 t k−1 (N − 1) − (k − 1)∆t t + ∆tN−(k+1) = · · 1 − k − 1 N 1 N N (N − 1) ··· (N − 1 − (k − 2)) (N − 1) − (k − 1) tk−1 t + ∆tN−(k+1) = 1 − ∆t N k−1 N (k − 1)! N tk−1 → e−t∆t. (2.19) (k − 1)!

Again, one way to avoid the complications is to integrate over xm ranging t t+∆t from N to N . Or, similar to before, we can proceed more rigorously as follows:

t 1. Calculate the probability that exactly k − 1 of the other xns are at most N units to the right of x1, and the remaining (N − 1) − (k − 1) of the xns are t N−1 at least N units to the right of x1. As there are k−1 ways to choose k − 1 t of the xns to be at most N units to the right of x1, this probability is

N − 1 t k−1 t (N−1)−(k−1) p = 1 − t k − 1 N N N k−1 tk−1 → e−t (k − 1)! N k−1 tk−1 → e−t. (2.20) k − 1!

17 t 2. Calculate the probability that exactly k − 1 of the other xns are at most N units to the right of x1, and the remaining (N − 1) − (k − 1) of the xns are t+∆t at least N units to the right of x1. Similar to the above, this gives

N − 1 t k−1 t + ∆t(N−1)−(k−1) p = 1 − t k − 1 N N N k−1 tk−1 → e−(t+∆t) (k − 1)! N k−1 tk−1 → e−(t+∆t). (2.21) (k − 1)!

t 3. The probability that exactly k−1 of the xns are within N units to the right of t t+∆t x1 and at least one xn is between N and N units to the right is pt+∆t − pt:

tk−1 tk−1 tk−1 p −p → e−t − e−(t+∆t) → e−t∆t. (2.22) t t+∆t (k − 1)! (k − 1)! (k − 1)!

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 xn chosen from the uniform distribution on the wrapped unit interval [0, 1). For ﬁxed N, arrange the xns in increase order, labeled y1 ≤ y2 ≤ · · · ≤ yN . Form the induced probability measure µN,1 from the nearest neighbor spacings. Then as N → ∞ we have

N 1 X µ (t)dt = δ t − N(y − y ) dt → e−tdt. (2.23) N,1 N n n−1 n=1

Equivalently, using zn = Nyn:

N 1 X µ (t)dt = δ t − (z − z ) dt → e−tdt. (2.24) N,1 N n n−1 n=1

18 More generally, form the probability measure from the kth nearest neighbor spacings. Then as N → ∞ we have

N 1 X tk−1 µ (t)dt = δ t − N(y − y ) dt → e−tdt. (2.25) N,k N n n−k (k − 1)! n=1

Equivalently, using zn = Nyn:

N 1 X tk−1 µ (t)dt = δ t − (z − z ) dt → e−tdt. (2.26) N,k N n n−k (k − 1)! n=1

Deﬁnition 2.5.2 (Poissonian Behavior). We say a sequence of points xn has Pois- sonian Behavior if in the limit as N → ∞ the induced probability measures tk−1 −t µN,k(t)dt converge to (k−1)! e dt.

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

Exercise 2.5.4. Let α 6∈ Q, and deﬁne αn = {nα}. Show the sequence of points αn 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, µN,1(t)dt cannot converge to e−tdt.

2.6 Non-Poissonian Behavior

Conjecture 2.6.1. With probability one (with respect to Lebesgue Measure, see 2 Deﬁnition ??), if α 6∈ Q, if αn = {n α} then the sequence of points αn is Poisso- nian.

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

pn an Theorem 2.6.2. Let α ∈ such that α − < 3 holds inﬁnitely often, with Q qn qn an → 0. Then there exist integers Nj → ∞ such that µNj ,1(t) does not converge to e−tdt.

19 As a → 0, eventually a < 1 for all n large. Let N = q , where pn is a n n 10 n n qn good rational approximation to α: p a α − n < n . (2.27) 3 qn qn Remember that all subtractions are performed on the wrapped unit interval. Thus, ||.999 − .001|| = .002. 2 We look at αk = {k α}, 1 ≤ k ≤ Nn = qn. Let the βks be the αks arranged in increasing order, and let the γ s be the numbers {k2 pn } arranged in increasing k qn order:

β1 ≤ β2 ≤ · · · ≤ βN

γ1 ≤ γ2 ≤ · · · ≤ γN . (2.28)

2.6.1 Preliminaries Lemma 2.6.3. If β = α = {k2α}, then γ = {k2 pn }. Thus, the same permuta- l k l qn tion orders both the αks and the γks.

2 2 Proof. Multiplying both sides of Equation 2.27 by k ≤ qn yields p a a 1 k2α − k2 n < k2 n ≤ n < . (2.29) 2 qn qn qn 2qn Thus, k2α and k2 pn differ by at most 1 . Therefore qn 2qn n o n p o 1 k2α − k2 n < . (2.30) qn 2qn As the numbers {m2 pn } all have denominators of size at most 1 , we see that qn qn {k2 pn } is the closest of the {m2 pn } to {k2α}. qn qn This implies that if β = {k2α}, then γ = {k2 pn }, completing the proof. l l qn

2 Exercise 2.6.4. Prove the ordering is as claimed. Hint: about each βl = {k α}, the closest number of the form {c2 pn } is {k2 pn }. qn qn

20 2.6.2 Proof of Theorem 2.6.2

1 Exercise 2.6.5. Assume ||a − b||, ||c − d|| < 10 . Show ||(a − b) − (c − d)|| < ||a − b|| + ||c − d||. (2.31)

Proof of Theorem 2.6.2: We have shown

an ||βl − γl|| < . (2.32) qn

Thus, as Nn = qn:

Nn(βl − γl) < an, (2.33) and the same result holds with l replaced by l − 1. By Exercise 2.6.5,

Nn(βl − γl) − Nn(βl−1 − γl−1) < 2an. (2.34) Rearranging gives

Nn(βl − βl−1) − Nn(γl − γl−1) < 2an. (2.35)

As an → 0, this implies the difference between Nn(βl−βl−1) and Nn(γl−

γl−1) goes to zero. The above distance calculations were done mod 1. The actual differences will differ by an integer. Thus,

Nn α 1 X µNn,1(t)dt = δ t − Nn(βl − βl−1) (2.36) Nn l=1 and

N pn 1 Xn µ qn (t)dt = δ t − N (γ − γ ) Nn,1 n l l−1 (2.37) Nn l=1 are extremely close to one another; each point mass from the difference between adjacent βls is either within an units of a point mass from the difference between adjacent γls, or is within an units of a point mass an integer number of units from a point mass from the difference between adjacent γls. Further, an → 0.

21 Note, however, that if γ = {k2 pn }, then l qn

n 2 pn o Nnγl = qn k ∈ N. (2.38) qn pn µ qn (t)dt γ Thus, the induced probability measure Nn,1 formed from the ls is sup- pn µ qn (t)dt e−tdt ported on the integers! Thus, it is impossible for Nn,1 to converge to . α As µNn,1(t)dt, modulo some possible integer shifts, is arbitrarily close to pn µ qn (t)dt {k2α} N Nn,1 , the sequence is not Poissonian along the subsequence of s given by Nn, where Nn = qn, qn is a denominator in a good rational approximation to α. 2

2.6.3 Measure of α 6∈ Q with Non-Poissonian Behavior along a sequence Nn What is the (Lebesgue) measure of α 6∈ Q such that there are inﬁnitely many n with

pn an α − < , an → 0. (2.39) qn qn If the above holds, then for any constant k(α), for n large (large depends on both α and k(α)) we have

p k(α) α − n < . (2.40) 2+ qn qn 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 Nn, where Nn = qn, qn the denominator of a good approximation to α. We pn µ qn (t)dt µα (t)dt know the induced probability measures Nn,1 and Nn,1 do not converge to e−tdt. Do these measures converge to anything?

22 Remark 2.6.8. In The Distribution of Spacings Between the Fractional Parts of {n2α} (Z. Rudnick, P. Sarnak, A. Zaharescu), it is shown that for most α satisfying α the conditions of Theorem 2.6.2, there is a sequence Nj along which µNn,1(t)dt does converge to e−tdt.