polynomials, number theory, and experimental mathematics · • experimental mathematics in number...
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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!