Resolution of T. Wards Question and the IsraelFinch Conjecture. - - PowerPoint PPT Presentation

Resolution of T. Ward’s Question and the Israel–Finch

• Conjecture. Precise Asymptotic Analysis of an Integer

Sequence Motivated by the Dynamical Mertens’ Theorem for Quasihyperbolic Toral Automorphisms

Mark Daniel Ward,∗

Department of Statistics, Purdue University

May 31, 2013 Joint work with Jeffrey B. Gaither and Guy Louchard

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 1 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Peter Walters introduced this problem, in the Transactions of the American Mathematical Society, 140:95–107, 1969:

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 2 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Thomas Ward introduced this sequence on OEIS on 7 Jan 2008, stating: newsgroup sci.math.research, from Thomas Ward on 16 Jan 2008:

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 3 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

After getting our solution, Thomas Ward sent us a great number of emails, mentioning other related problems. These might be of interest to people interested in Dynamics, or Ergodic Theory, with applications to Number Theory. For instance:

◮ “Realizable sequences: a non-trivial property of an integer sequence is

to be the count of points of period n for some map—this is a congruence and a positivity condition.”

◮ “Products and iterates”, maps, enumeration of orbits: “this generates

a lot of ‘nice’ sequences, some well-known, some mysterious. In (very) simple examples we have used Perron to find asymptotics, but I think there is rich territory here for cleverer things.”

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 4 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

◮ “Some notorious problems. . . that are very easy to analyze in one

dimension that become very problematical in two.” . . . “Let a(n) be the number of ways to fill an n × n square with 0s and 1s so that you never have two 1s in a row vertically nor horizontally. Easy to argue that a(n) ∼ C n for some C ∈ (1, 2) (and there are some estimates) but we have no clue about the real picture. This is probably extremely difficult, and there is no a priori thing that makes you expect a ‘closed form’ asymptotic.”

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 5 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

◮ “Moonshine: one-dimensional subshifts of finite type give rise in a

natural way to sequences with rational associated zeta functions with growth rates parameterised by a certain nice countable collection of numbers, the Perron numbers.”

◮ “Twice-distilled moonshine: there is a natural analytic toolbox

associated to certain classes of problems:

◮ ’hyperbolic’ dynamics—zeta functions with meromorphic extensions

beyond radius of convergence - Tauberian methods

◮ Slow growth dynamics - orbit Dirichlet series - elementary methods

(analytically too nasty for Tauberian methods)”

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 6 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

From a topological/dynamical perspective: an is the asymptotic coefficient in the weighted sum of the orbital numbers of a certain toral automorphism. The study of this automorphism is motivated by the search for a topological analogue to Mertens’ prime number theorem. There is a strong structural similarity between the distribution of the prime integers and that of the orbits of an automorphism acting on an n-torus Tn = S1 × · · · × S1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 7 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Specifically, the classical Mertens’ Theorem gives

• p≤N
• 1

1 − 1/p

• ∼ log(N),

and the analogous hyperbolic toral diffeomorphism result is MT(N) :=

• |τ|≤N

1 eh|τ| ∼ log(N), (1) where each τ = {x, T(x), . . . , T k(x) = x} denotes a closed orbit of length k, and h represents the topological entropy.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 8 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Noorani (1999) showed that when T is merely an ergodic toral automorphism (such an automorphism is said to quasihyperbolic), one has MT(N) = m log(N) + C1 + o(1) (2) for some positive integer m, where an is precisely the m in (2) for a specific automorphism (which we will describe in a moment).

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 9 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Jaidee, Stevens, and Ward (2011) improved Noorani’s estimate and also showed that the constant m is given by m =

• X

n

• j=1

4 sin2(πxj) dx1 · · · dxj, (3) where X is found as follows: Write e±2πiθ1, . . . , e±2πiθt as the eigenvalues of modulus 1 of the matrix A that defines the automorphism T, and let X ⊂ Td denote the closure of the set {(kθ1, . . . , kθt) : k ∈ Z} in Td.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 10 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

The particular toral automorphism that gives rise to T. Ward’s an has been

• f interest since Walter (1969), who introduced the automorphism given by

A =     −1 1 8 1 −6 1 8    

• n the 4-torus T4 = S1 × S1 × S1 × S1 for an example of an affine

mapping between compact connected metric abelian groups, for which the mapping commutes only with continuous maps that are also affine. For this matrix A, 2 of the 4 eigenvalues have modulus 1, namely: 2 − √ 3 ± i

• 4

√ 3 − 6. So the θ1 from the previous slide is θ1 =

1 2π arctan

• 2−

√ 3

√

−6+4 √ 3

• .

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 11 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

Noorani (1999) further cited A as an example of a strictly quasi-hyperbolic automorphism on the torus. Jaidee et al. later considered A as their example of a toral automorphism whose asymptotic coefficient m = 6 exceeded 2t = 2 (where t = 1 since A has 2 = 2t eigenvalues on the unit circle). They also used A to define the automorphism A1 ⊕ A2 ⊕ · · · ⊕ An on the 4n-torus T4n. This defines the context: The an’s are precisely the coefficient m of log(N) in the asymptotic growth of MT(n) (from slide 8), with T4n the torus under consideration and A1 ⊕ · · · ⊕ An the quasihyperbolic automorphism.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 12 / 48

Origins of the Problem an := 1 n

j=1 4 sin2(πjx) dx

So we analyze the precise first-order asymptotics of (an)n≥1 = (2, 4, 6, 10, 12, 20, 24, 34, 44, 64, . . . ). A133871 from the On-Line Encyclopedia of Integer Sequences).

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 13 / 48

Israel’s Conjecture for an := 1 n

j=1 4 sin2(πjx) dx

newsgroup sci.math.research, from Robert Israel on 17 Jan 2008: Conjecture about the logarithm of an: ln(an2−n) n → −0.3 . . . ,

• r equivalently

ln an → (ln (2) − 0.3 . . .)n.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 14 / 48

Finch’s Conjecture for an := 1 n

j=1 4 sin2(πjx) dx

http://www.people.fas.harvard.edu/~sfinch/ many newer “Supplementary Materials,” including:

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 15 / 48

Finch’s Conjecture for an := 1 n

j=1 4 sin2(πjx) dx

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 16 / 48

Finch’s Conjecture for an := 1 n

j=1 4 sin2(πjx) dx

Finch wrote an in a very simple way: Finch’s conjecture about the logarithm of an was: a1/n

n

∼ 1.48 . . . ∼ 2e−0.29.... This conjecture is equivalent to Israel’s conjecture. Finch has several other conjectures in this essay. He says:

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 17 / 48

Representations of an := 1 n

j=1 4 sin2(πjx) dx

Equivalently, an := 1

n

• j=1

(1−e2πijx)(1−e−2πijx)dx Note: 1 e2πiℓx dx = 0 for ℓ ∈ Z\{0}, and 1 e2πi0x dx = 1. So we expand n

j=1(1 − e2πijx)(1 − e−2πijx) and then integrate: only the

terms in which the j’s sum to 0 will play a nontrivial role. If the integral of a term is nonzero, it is either 1 or −1, depending on whether an even or odd number of j’s were involved in the product.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 18 / 48

Representations of an := 1 n

j=1 4 sin2(πjx) dx

This observation leads to the combinatorial representation an = [x0]

n

• j=1

(1 − xj)(1 − x−j), where [x0] denotes the constant coefficient. Combinatorially, an is the signed total number of (2n)-tuples (ǫ−n, . . . , ǫ−1, ǫ1, . . . , ǫn) ∈ {0, 1}2n such that

• −n≤j≤n

j=0

ǫjj = 0.

◮ So there is a contribution of −1 to the signed total number if an odd

number of ǫj are nonzero (i.e., equal 1),

◮ or a contribution of +1 to the signed total number otherwise.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 19 / 48

Example

When n = 4, exactly 18 of the 8-tuples have an even number of nonzero terms (i.e., of 1’s):

(0,0,0,0,0,0,0,0) (1,1,1,1,1,1,1,1) (0,0,0,1,1,0,0,0) (0,0,1,0,0,1,0,0) (0,1,0,0,0,0,1,0) (1,0,0,0,0,0,0,1) (0,0,1,1,1,1,0,0) (0,1,0,1,1,0,1,0) (0,1,1,0,0,1,1,0) (1,0,0,1,1,0,0,1) (1,0,1,0,0,1,0,1) (1,1,0,0,0,0,1,1) (0,1,1,0,1,0,0,1) (1,0,0,1,0,1,1,0) (0,1,1,1,1,1,1,0) (1,0,1,1,1,1,0,1) (1,1,0,1,1,0,1,1) (1,1,1,0,0,1,1,1)

exactly 8 of the 8-tuples have an odd number of nonzero’s (i.e., of 1’s):

(0,0,1,1,0,0,1,0) (0,1,0,0,1,1,0,0) (0,1,0,1,0,0,0,1) (1,0,0,0,1,0,1,0) (0,1,1,1,0,1,0,1) (1,0,1,0,1,1,1,0) (1,0,1,1,0,0,1,1) (1,1,0,0,1,1,0,1)

Thus a4 = 18(+1) + 8(−1) = 10.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 20 / 48

Representations of an := 1 n

j=1 4 sin2(πjx) dx

The number of ways in which 0 can be written in an unweighted fashion as n

j=−n(ǫj)(j) is a well-understood quantity (see Clark (2000), Entringer

(1968), Louchard and Prodinger (2009), van Lint (1967), and others). To the best of our knowledge, our derivation is the first result in which the contributions are signed according to the parity of the number of nonzeros. Paul Hanna (OEIS) notes another fundamental representation: a(n) equals the sum of the squares of the coefficients in the polynomial n

k=1(1 − xk).

“Signing” provides an additional challenge, since it takes this problem out

• f the regime in which probabilistic methods can be applied.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 21 / 48

Main Theorem about an := 1 n

j=1 4 sin2(πjx) dx

Theorem

Let G(x) := 1 log(sin(πxt)) dt. Then there is a unique point x0 = 0.7912265710... on [0, 1] at which G attains its maximum maxx∈[0,1] G(x) = −.4945295654... on the unit interval [0, 1]. Furthermore, if r and C denote the constants r := e2G(x0) = 0.37192646 . . . ; C := 4 sin(πx0) x0

• π

−G ′′(x0) = 2.4057458 . . . , then the first-order asymptotic growth of an is C(4r)nn−1/2, i.e., lim

n→∞

an C(4r)nn−1/2 = 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 22 / 48

Main Theorem about an := 1 n

j=1 4 sin2(πjx) dx Figure : Plot of G(x) := 1 log(sin(πxt)) dt for x ∈ (0, 1).

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 23 / 48

Intuition: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

Let’s plot the integrand used in an, to see where the main contributions are coming from, in the range x ∈ [0, 1].

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 24 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 1: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 25 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 2: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 26 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 3: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 27 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 4: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 28 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 5: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 29 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 6: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 30 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 7: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 31 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 8: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 32 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 9: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 33 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 10: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 34 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 11: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 35 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 12: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 36 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 13: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 37 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 14: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 38 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 15: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 39 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 16: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 40 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 17: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 41 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 18: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 42 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 19: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 43 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

n = 20: Plot of n

j=1 4 sin2(πjx) for 0 ≤ x ≤ 1.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 44 / 48

Proof: asymptotics of an := 1 n

j=1 4 sin2(πjx) dx

So the dominant contribution to the integral that defines an comes from integrating over the region [0, 1

n] ∪ [ n−1 n , 1]. The integral over the middle

integral [ 1

n, n−1 n ] is (comparatively) negligible.

Our proof used:

◮ to precisely analyze the integral on [0, 1 n] ∪ [ n−1 n , 1]:

◮ Laplace’s method (original proof) ◮ Mellin transform (thanks to Guy Louchard for this alternative proof)

◮ fractional approximations, based heavily on Dirichlet’s theorem, to

analyze the integral on [0, 1

n] ∪ [ n−1 n , 1]

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 45 / 48

(reminder) Main Thm. about an := 1 n

j=1 4 sin2(πjx) dx

Theorem

Let G(x) := 1 log(sin(πxt)) dt. Then there is a unique point x0 = 0.7912265710... on [0, 1] at which G attains its maximum maxx∈[0,1] G(x) = −.4945295654... on the unit interval [0, 1]. Furthermore, if r and C denote the constants r := e2G(x0) = 0.37192646 . . . ; C := 4 sin(πx0) x0

• π

−G ′′(x0) = 2.4057458 . . . , then the first-order asymptotic growth of an is C(4r)nn−1/2, i.e., lim

n→∞

an C(4r)nn−1/2 = 1. So an ∼ C(4r)nn−1/2 ≈ (2.4057458 . . .)(1.4877058 . . .)nn−1/2.

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 46 / 48

Checking the Asymptotics

Ratio of an versus C(4r)nn−1/2(2.4057458 . . .)(1.4877058 . . .)nn−1/2

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 47 / 48

Thank you

Many thanks to Conrado Martinez and Alfredo Viola for their invitation, and for organizing another wonderful AofA conference!

Mark Daniel Ward (Purdue University) Resolution of the Israel–Finch Conjecture May 31, 2013 48 / 48