Polynomials, Number Theory, and Experimental Mathematics Michael - PowerPoint PPT Presentation

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 experimental investigations. • Highlight facets of experimental approach.

1. Refining Directions in Research

Barker Sequences • a 0 , a 1 , ..., a n − 1 : finite sequence, each ±1. • For 0 ≤ k ≤ n − 1, define the k th aperiodic autocorrelation by n − k − 1 X c k = a i a i + k . i =0 • k = 0: peak autocorrelation. • k > 0: o ff -peak autocorrelations. • Goal: make o ff -peak values small. • Barker sequence : | c k | ≤ 1 for k > 0.

Polynomials n − 1 X a k z k . • Let f ( z ) = k =0 • Erdös, Littlewood, Newman, Mahler: Do there exist polynomials with all ±1 coe ffi cients that remain flat over the unit circle? • A long Barker sequence would be much flatter than best known polynomials.

All(?) Barker Sequences n Sequence + 1 ++ 2 ++- 3 +++- 4 +++-+ 5 +++--+- 7 +++---+--+- 11 +++++--++-+-+ 13 • Turyn & Storer (1961): No more of odd length.

Properties • n = 4 m 2 , 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 • q p − 1 ≡ 1 mod p 2 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 10 17 .

Lower Bounds • Leung & Schmidt (2005): m > 5 ⋅ 10 10 , so n > 10 22 . • No plausible value known in 2005! • Do any exist?? Experiment!

Search Strategy • 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: • q p − 1 ≡ 1 mod p 2 and pq ≤ M . • p | ( q − 1) and pq ≤ M . • Need a subset of vertices where each indegree is positive in the induced subgraph.

Results • M. (2009): If a Barker sequence of length n > 13 exists, then either n = 189 260 468 001 034 441 522 766 781 604, n > 2 ⋅ 10 30 . or • Leung & Schmidt (2012): Two new restrictions for the Barker problem.

Results • Leung & Schmidt (2012): If a Barker sequence of length n > 13 n > 2 ⋅ 10 30 . exists, then

More Recent Result • 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 ⋅ 10 33 . 5 138200401 13 138200401 13 41 2953 41 2953 29 • Large list of additional plausible values.

53 13 89 37 12197 97 4794006457 149 76704103313 268693 3049 41 349 29 99995507756741087451736040784945981261228240243352 81106341441590852061005613123255433352037667736004

5.0 4.5 4.0 3.5 I c e r m 3.0 2.5 0.0 0.1 0.2 0.3 0.4 0.5

2. Discovering Identities

Mahler’s Measure n n a k z k = a n X Y • f ( z ) = ( z − β k ) in Z [ z ]. k =0 k =1 n Y • M ( f ) = | a n | max { 1 , | β k |} . k =1 • (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 ( z 10 + z 9 – z 7 – z 6 – z 5 – z 4 – z 3 + z +1) = 1.17628… .

Measures and Heights • Height of f : H ( f ) = max{| a k | : 0 ≤ k ≤ n }. • For r > 1, let A r denote the complex annulus A r = { z ∊ C : 1/ r < | z | < r }. • If H ( f ) = 1 and f ( β ) = 0 ( β ≠ 0) then β ∊ A 2 . • 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 coe ffi cients 0 or 1, and constant term 1. • Odlyzko & Poonen (1993): If f ( z ) is a Newman polynomial and f ( β ) = 0, then β ∊ A τ , where τ denotes the golden ratio.

Newman Polynomials • All coe ffi cients 0 or 1, and constant term 1. • Odlyzko & Poonen (1993): If f ( z ) is a Newman polynomial and f ( β ) = 0, then β ∊ A τ , 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 ?

Degree Measure Newman Half of Coe ffi cients ++000+ 10 1.17628 13 ++++++0+000000000+000000000 18 1.18836 55 +00+0+00000000 14 1.20002 28 +00+0++++ 18 1.20139 19 ++0000000+ 14 1.20261 20 ++++0+00+0+ 22 1.20501 23 +0+000000000000+0 28 1.20795 34 ++0000000 20 1.21282 24 +0+0+000000+0+000 20 1.21499 34 ++0000000 10 1.21639 18 +000++0++++ 20 1.21839 22 +++++0000000++0000000 24 1.21885 42 +00+0+0++0+0+00000 24 1.21905 37 ++++++000+0000000000000 18 1.21944 47 ++000++0000+00000000000 18 1.21972 46 34 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 z m f ( z ) ± z n f (1/ z ) has a Salem number α m as a root (for large m ) and α m → 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 .

Q + 5 , 4 ( z ) ( z − 1) 2 = 1 + 3 z + 4 z 2 + 5 z 3 + 6 z 4 + 6 z 5 + 5 z 6 + 4 z 7 + 3 z 8 + z 9 ++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++ 5 , 4 ( z )( z 24 − 1)( z 5 + 1) Q + ( z 8 − 1)( z 6 − 1)( z 2 − 1) ,

• Leads to ( m odd, n even): z n +1 + 1 Q + z ( m +2 n +1)( n +2) / 2 − 1 � � � � m,n ( z ) = ( z m +2 n +1 − 1)( z n +2 − 1)( z 2 − 1) ✓ m − 3 n + m − 1 ◆✓ n/ 2 2 2 ◆ z ( z n + 1) X X X z 2 k z k ( m +2 n +1) z k ( n +2) . + k =0 k =0 k =0 • m , n both odd: Q + z ( m + n )( m − 1) / 2 − 1 � � m,n ( z ) ( z m + n − 1)( z m − 1 − 1)( z 2 − 1) m + n ✓ n − 3 ◆✓ m − 3 − 1 2 2 2 ◆ X z k ( m − 1) + z X X z 2 k z k ( m + n ) = . k =0 k =0 k =0

Results 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. 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.

omplex Pisot

3. Opening New Avenues

A Bit of Number Theory • Prime Number Theorem: Z x dt x π ( x ) ∼ Li( x ) = log x. log t ∼ 2 • Riemann Hypothesis: � √ x log x � | π ( x ) − Li( x ) | = O . • Riemann zeta function: 1 1 − p − s � − 1 . X Y � ζ ( s ) = n s = n ≥ 1 p

• 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 1 σ > 1 − R 1 (log | t | ) 2 / 3 (log log | t | ) 1 / 3 . • Explicit version (Ford 2002): R 1 = 57.54. 1 • Other bounds: σ > 1 − R 0 log | t | . • Explicit versions: Better for small | t |. • Crossover around exp(10000).

Improvements in R 0 de la Vallee Poussin 1899 30.468 (1 + cos y ) 2 Westphal 1938 17.537 (1 + cos y ) 2 ( . 3 + cos y ) 2 Rosser & Schoenfeld 1975 9.646 Ford 2000 8.463 ( . 91 + cos y ) 2 ( . 265 + cos y ) 2 Kadiri 2005 5.697 • Common thread: employing nonnegative, even trigonometric polynomial. • Above: all degree ≤ 4. Can we do better?

Requirements n X • Let f ( y ) = a k cos( ky ). k =0 n 2 � b k e iky � X • Need f ( y ) ≥ 0 for all y , so really f ( y ) = . � � � � k =0 • Let A = a 1 + · · · + a n . • Need each a k ≥ 0, a 1 > a 0 , and A not too large. • Kadiri: 10 . 91 + 18 . 63 cos( y ) + 11 . 45 cos(2 y ) + 4 . 7 cos(3 y ) + cos(4 y ) .

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