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Challenges in 21st Century Experimental Mathematical Computation Polynomials, Number Theory, and Experimental Mathematics Michael Mossinghoff Davidson College ICERM Brown University July 21-25, 2014 Overview Three

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SLIDE 1

Challenges in 21st Century Experimental Mathematical Computation

Polynomials, Number Theory, and Experimental Mathematics

Michael Mossinghoff Davidson College ICERM Brown University July 21-25, 2014

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SLIDE 2

Overview

  • Three experimental investigations.
  • Highlight facets of experimental approach.
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SLIDE 3
  • 1. Refining Directions in

Research

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SLIDE 4

Barker Sequences

  • a0, a1, ..., an−1 : finite sequence, each ±1.
  • For 0 ≤ k ≤ n−1, define the kth aperiodic

autocorrelation by

  • k = 0: peak autocorrelation.
  • k > 0: off-peak autocorrelations.
  • Goal: make off-peak values small.
  • Barker sequence: |ck| ≤ 1 for k > 0.

ck =

n−k−1

X

i=0

aiai+k.

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SLIDE 5

Polynomials

  • Erdös, Littlewood, Newman, Mahler: Do there

exist polynomials with all ±1 coefficients that remain flat over the unit circle?

  • A long Barker sequence would be much flatter

than best known polynomials.

  • Let f(z) =

n−1

X

k=0

akzk.

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SLIDE 6

All(?)

n Sequence

1 + 2 ++ 3 ++- 4 +++- 5 +++-+ 7 +++--+- 11 +++---+--+- 13 +++++--++-+-+

Barker Sequences

  • Turyn & Storer (1961): No more of odd length.
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SLIDE 7

Properties

  • n = 4m2, with m odd.
  • Each prime divisor of m is 1 mod 4.
  • m must satisfy certain complicated conditions.
  • For each p | m, one requires either
  • qp−1 ≡ 1 mod p2 for some prime q | m, or
  • p | q−1 for some prime q | m.
  • Former: (q, p) is a Wieferich prime pair.
  • Rare! q = 5: only p = 53471161, 6692367337,

188748146801 up to 1017.

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SLIDE 8
  • Leung & Schmidt (2005): m > 5⋅1010,

so n > 1022.

  • No plausible value known in 2005!
  • Do any exist?? Experiment!

Lower Bounds

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SLIDE 9
  • Wish to find all permissible m ≤ M.
  • Create a directed graph, D = D(M).
  • Vertices: subset of primes p ≤ M.
  • Directed edge from q to p in two cases:
  • qp−1 ≡ 1 mod p2 and pq ≤ M.
  • p | (q − 1) and pq ≤ M.
  • Need a subset of vertices where each indegree

is positive in the induced subgraph.

Search Strategy

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SLIDE 10

Results

  • M. (2009):
  • Leung & Schmidt (2012):

n = 189 260 468 001 034 441 522 766 781 604, n > 2⋅1030.

  • r

If a Barker sequence of length n > 13 exists, then either Two new restrictions for the Barker problem.

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SLIDE 11

If a Barker sequence of length n > 13 exists, then

  • Leung & Schmidt (2012):

n > 2⋅1030.

Results

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SLIDE 12
  • Theorem (P. Borwein & M., 2014): If n > 13 is

the length of a Barker sequence, then either n = 3 979 201 339 721 749 133 016 171 583 224 100,

  • r n > 4⋅1033.

More Recent Result

138200401 2953 41 29 5 13 138200401 2953 41 13

  • Large list of additional plausible values.
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SLIDE 13

99995507756741087451736040784945981261228240243352 81106341441590852061005613123255433352037667736004

76704103313 97 4794006457 53 13 349 29 89 12197 3049 41 268693 149 37

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SLIDE 14 0.0 0.1 0.2 0.3 0.4 0.5 2.5 3.0 3.5 4.0 4.5 5.0

I c e r m

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SLIDE 15
  • 2. Discovering Identities
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SLIDE 16

Mahler’s Measure

  • f(z) =

n

X

k=0

akzk = an

n

Y

k=1

(z − βk) in Z[z].

  • M(f) = |an|

n

Y

k=1

max{1, |βk|}.

  • (Kronecker, 1857) M(f) = 1 ⇔ f(z) is a product of

cyclotomic polynomials, and a power of z.

  • Lehmer’s problem (1933): Is there a constant c > 1

so that if M(f) > 1 then M(f) ≥ c?

  • M(z10+z9–z7–z6–z5–z4–z3+z+1) = 1.17628… .
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SLIDE 17

Measures and Heights

  • Height of f: H(f) = max{|ak| : 0 ≤ k ≤ n}.
  • For r > 1, let Ar denote the complex annulus

Ar = {z ∊ C : 1/r < |z| < r}.

  • If H(f) = 1 and f(β) = 0 (β ≠ 0) then β ∊ A2.
  • Bloch & Pólya (1932), Pathiaux (1973): If

M(f) < 2 then there exists F(z) with H(F) = 1 and f(z) | F(z).

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SLIDE 18
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SLIDE 19

Newman Polynomials

  • All coefficients 0 or 1, and constant term 1.
  • Odlyzko & Poonen (1993): If f(z) is a Newman

polynomial and f(β) = 0, then β ∊ Aτ, where τ denotes the golden ratio.

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SLIDE 20
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SLIDE 21
  • Is there a constant σ so that if M(f) < σ then

there exists Newman F(z) with f(z) | F(z)?

  • Assume f(z) has no positive real roots.
  • Can we take σ = τ?

Newman Polynomials

  • All coefficients 0 or 1, and constant term 1.
  • Odlyzko & Poonen (1993): If f(z) is a Newman

polynomial and f(β) = 0, then β ∊ Aτ, where τ denotes the golden ratio.

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SLIDE 22

Degree Measure Newman Half of Coefficients 10 1.17628 13 ++000+ 18 1.18836 55 ++++++0+000000000+000000000 14 1.20002 28 +00+0+00000000 18 1.20139 19 +00+0++++ 14 1.20261 20 ++0000000+ 22 1.20501 23 ++++0+00+0+ 28 1.20795 34 +0+000000000000+0 20 1.21282 24 ++0000000 20 1.21499 34 +0+0+000000+0+000 10 1.21639 18 ++0000000 20 1.21839 22 +000++0++++ 24 1.21885 42 +++++0000000++0000000 24 1.21905 37 +00+0+0++0+0+00000 18 1.21944 47 ++++++000+0000000000000 18 1.21972 46 ++000++0000+00000000000 34 1.22028 95

++++++++++++++0+++++++++000000000+00000000000000
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SLIDE 23

Pisot and Salem Numbers

  • A real algebraic integer β < –1 is a (negative)

Pisot number if all its conjugates β´ have |β´| < 1.

  • All such numbers > –τ are known: four infinite

families and one sporadic.

  • A (negative) Salem number is a real algebraic

integer α < –1 whose conjugates all lie on the unit circle, except for 1/α.

  • Salem: If f(z) is the min. poly. of a Pisot number β
  • f degree n, then zmf(z) ± znf(1/z) has a Salem

number αm as a root (for large m) and αm → β.

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SLIDE 24

Experimental Investigations

  • Can we represent small negative Pisot and

Salem numbers with Newman polynomials?

  • Given f(z), determine if there is a Newman

polynomial F(z) so deg(F) = N and f(z) | F(z).

  • Sieving strategy: f(k) must divide F(k) for

several k.

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SLIDE 25

++0+0+++++0+++0+++++0+0++ ++0+0+++0000+++++++0000+++0+0++ ++0+0+00+00++0+++++0++00+00+0+0++ ++0+0+++++0+0000+0000+0+++++0+0++ ++0+0+++00++000+++000++00+++0+0++ ++0+0+00+0++0+0+0+0+0+0++0+00+0+0++ ++0+0+++00++000+00+00+000++00+++0+0++ ++0+0+++0000+00+0+++0+00+0000+++0+0++ ++0+0+++0000++00+0+0+00++0000+++0+0++ ++0+0+00+++0+00000+++00000+0+++00+0+0++ ++0+0+00+000000+++0+++0+++000000+00+0+0++ ++0+0+00+00++000+0+0+0+0+000++00+00+0+0++ ++0+0+00+++0+00000+00+00+00000+0+++00+0+0++

Q+

5,4(z)(z24 − 1)(z5 + 1)

(z8 − 1)(z6 − 1)(z2 − 1) ,

Q+

5,4(z)

(z − 1)2 = 1 + 3z + 4z2 + 5z3 + 6z4 + 6z5 + 5z6 + 4z7 + 3z8 + z9

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SLIDE 26

z(zn + 1) ✓ m−3

2

X

k=0

z2k ◆✓n/2 X

k=0

zk(m+2n+1) ◆ +

n+ m−1

2

X

k=0

zk(n+2). Q+

m,n(z)

  • z(m+2n+1)(n+2)/2 − 1

zn+1 + 1

  • (zm+2n+1 − 1)(zn+2 − 1)(z2 − 1)

=

  • Leads to (m odd, n even):

Q+

m,n(z)

  • z(m+n)(m−1)/2 − 1
  • (zm+n − 1)(zm−1 − 1)(z2 − 1)

=

m+n 2

−1

X

k=0

zk(m−1) + z ✓ n−3

2

X

k=0

z2k ◆✓ m−3

2

X

k=0

zk(m+n) ◆ .

  • m, n both odd:
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SLIDE 27

Theorem (H. & M.): If α > −τ is a negative Salem number arising from Salem’s construction

  • n the minimal polynomial of a negative Pisot

number β > −τ, then there exists a Newman polynomial F(z) with F(α) = 0. Theorem (Hare & M., 2014): If β is a negative Pisot number with β > −τ, and β has no positive real conjugates, then there exists a Newman polynomial F(z) with F(β) = 0.

Results

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SLIDE 28
  • mplex Pisot
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SLIDE 29
  • 3. Opening New Avenues
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SLIDE 30

|π(x) − Li(x)| = O √x log x

  • .

π(x) ∼ Li(x) = Z x

2

dt log t ∼ x log x. ζ(s) = X

n≥1

1 ns = Y

p

  • 1 − p−s−1 .
  • Prime Number Theorem:
  • Riemann Hypothesis:
  • Riemann zeta function:

A Bit of Number Theory

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SLIDE 31
  • Zeros of ζ(s) ↔ Distribution of primes.
  • Nontrivial zeros of ζ(s) lie in “critical strip,”

0 < Re(s) < 1.

  • PNT ↔ No zeros on Re(s) = 1.
  • RH ↔ All nontrivial zeros on Re(s) = 1/2.
  • Zero-free region in critical strip ↔

Information on distribution of primes.

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SLIDE 32
  • Best known zero-free region (Vinogradov):

ζ(σ + it) ≠ 0 if

  • Explicit version (Ford 2002): R1 = 57.54.
  • Other bounds:
  • Explicit versions: Better for small |t|.
  • Crossover around exp(10000).

σ > 1 − 1 R1(log |t|)2/3(log log |t|)1/3 . σ > 1 − 1 R0 log |t|.

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SLIDE 33

de la Vallee Poussin 1899 30.468 Westphal 1938 17.537 Rosser & Schoenfeld 1975 9.646 Ford 2000 8.463 Kadiri 2005 5.697

Improvements in R0

  • Common thread: employing nonnegative,

even trigonometric polynomial.

  • Above: all degree ≤ 4. Can we do better?

(1 + cos y)2 (.91 + cos y)2(.265 + cos y)2 (1 + cos y)2(.3 + cos y)2

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SLIDE 34

Requirements

  • Let f(y) =

n

X

k=0

ak cos(ky).

  • Need each ak ≥ 0, a1 > a0, and A not too large.
  • Kadiri:

10.91 + 18.63 cos(y) + 11.45 cos(2y) + 4.7 cos(3y) + cos(4y).

  • Let A = a1 + · · · + an.
  • Need f(y) ≥ 0 for all y, so really f(y) =
  • n

X

k=0

bkeiky

  • 2

.

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SLIDE 35

Experimental Strategy

  • Select degree, n.
  • Select bounding box for initial selection of bk’s.
  • Begin at random location.
  • Anneal based on objective function.
  • In addition: extra care with error term in Kadiri

analysis.

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SLIDE 36

de la Vallee Poussin 1899 30.468 Westphal 1938 17.537 Rosser & Schoenfeld 1975 9.646 Ford 2000 8.463 Kadiri 2005 5.697

  • M. & Trudgian

2014 5.57392

Results

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SLIDE 37
  • 4. Visualizing New Structures
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SLIDE 38
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SLIDE 39
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SLIDE 40

Summary

  • Experimental mathematics in number theory:
  • Refining directions in research, and rejecting

false hypotheses,

  • Discovering new algebraic identities,
  • Opening new avenues of investigation,
  • Visualizing data, and finding structure and
  • rder in the complexity.
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SLIDE 41

erm

I

Thanks!