Challenges in 21st Century Experimental Mathematical Computation

Polynomials, Number Theory, and Experimental MathematicsMichael Mossinghoff!Davidson College

ICERM!Brown University!July 21-25, 2014

Overview

• Three experimental investigations.

• Highlight facets of experimental approach.

1. Refining Directions in Research

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�1X

i=0

aiai+k.

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�1X

k=0

akzk.

All(?)n Sequence1 +

2 ++

3 ++-

4 +++-

5 +++-+

7 +++--+-

11 +++---+--+-

13 +++++--++-+-+

Barker Sequences

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

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.

• Leung & Schmidt (2005): m > 5⋅1010, so n > 1022.

• No plausible value known in 2005!

• Do any exist?? Experiment!

Lower Bounds

• 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

Results• M. (2009):

• Leung & Schmidt (2012):

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

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

Two new restrictions for the Barker problem.

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

• Leung & Schmidt (2012):

n > 2⋅1030.

Results

• 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, or n > 4⋅1033.

More Recent Result

138200401

295341

29

5 13138200401

295341

13

• Large list of additional plausible values.

9999550775674108745173604078494598126122824024335281106341441590852061005613123255433352037667736004

76704103313

97 4794006457

53 13

349 29

89

12197

3049

41

268693

149

37

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

2. Discovering Identities

Mahler’s Measure• f(z) =

nX

k=0

akzk = an

nY

k=1

(z � �k) in Z[z].

• M(f) = |an|nY

k=1

max{1, |�k|}.

• (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… .

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).

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.

• 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.

Degree Measure Newman Half of Coefficients10 1.17628 13 ++000+18 1.18836 55 ++++++0+000000000+00000000014 1.20002 28 +00+0+0000000018 1.20139 19 +00+0++++14 1.20261 20 ++0000000+22 1.20501 23 ++++0+00+0+28 1.20795 34 +0+000000000000+020 1.21282 24 ++000000020 1.21499 34 +0+0+000000+0+00010 1.21639 18 ++000000020 1.21839 22 +000++0++++24 1.21885 42 +++++0000000++000000024 1.21905 37 +00+0+0++0+0+0000018 1.21944 47 ++++++000+000000000000018 1.21972 46 ++000++0000+0000000000034 1.22028 95 ++++++++++++++0+++++++++000000000+00000000000000

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 of degree n, then zmf(z) ± znf(1/z) has a Salem number αm as a root (for large m) and αm → β.

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.

++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

z(zn + 1)

✓m�32X

k=0

z2k◆✓n/2X

k=0

zk(m+2n+1)

◆+

n+m�12X

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+n2 �1X

k=0

zk(m�1) + z

✓n�32X

k=0

z2k◆✓m�3

2X

k=0

zk(m+n)

◆.

• m, n both odd:

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

omplex Pisot

3. Opening New Avenues

|⇡(x)� Li(x)| = O�p

x log x�.

⇡(x) ⇠ Li(x) =

Zx

2

dt

log t

⇠ x

log x

.

⇣(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

• 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.

• 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).

� > 1� 1

R1(log |t|)2/3(log log |t|)1/3.

� > 1� 1

R0 log |t|.

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

Requirements

• Let f(y) =nX

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) =���

nX

k=0

bkeiky

���2.

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.

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

4. Visualizing New Structures

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 order in the complexity.

ermI

Thanks!