erm I Michael Introduction to Topics Mossinghoff Summer@ICERM - PowerPoint PPT Presentation

Complex Pisot Numbers and Newman Polynomials erm I Michael Introduction to Topics Mossinghoff Summer@ICERM 2014 Davidson College Brown University

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 ✓Z 1 ◆ log | f ( e 2 π it ) | dt • M ( f ) = exp . 0 • (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 ( f ) = M (– f ( z )) = M ( f (– z )) = M ( f ( z k )) = M ( f * ). • Here, f * ( z ) is defined as z n ⋅ f (1/ z ): reciprocal of f . • Some results on Lehmer’s Problem: • M ( z 10 + z 9 – z 7 – z 6 – z 5 – z 4 – z 3 + z +1) = 1.17628… . • (Smyth 1971) If f ( z ) ≠ ± f * ( z ) and f (0) ≠ 0, then M ( f ) ≥ M ( z 3 – z – 1) = 1.3247… • If 1 < M ( f ) < 1.17628… then deg( f ) ≥ 56. • (Borwein, Dobrowolski, M., 2007) If f ( z ) has all odd coe ffi cients and degree n then M ( f ) ≥ 1 . 4935 − . 61 n .

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 .

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 ).

Bloch and Pólya • f ( z ) irreducible, degree d , M ( f ) < 2. • So f ( z ) monic, and has at least one root | β 1 | < 1. • g ( z ): height 1, degree n , f is not a factor of g . • |Res( f , g )| = | g ( β 1 ) g ( β 2 )… g ( β d )| ≥ 1. • | g ( β k )| ≤ ( n + 1) ⋅ max{1, | β k | n }. • | g ( β 2 ) · · · g ( β d ) | ≤ ( n + 1) d − 1 M ( f ) n . 1 • | g ( β 1 ) | ≥ ( n + 1) d − 1 M ( f ) n .

1 • | g ( β 1 ) | ≥ ( n + 1) d − 1 M ( f ) n . • If h 1 , h 2 have {0, 1} coe ffi cients, degree ≤ n , and f does not divide h 1 – h 2 , then 1 | h 1 ( β 1 ) − h 2 ( β 1 ) | ≥ ( n + 1) d − 1 M ( f ) n . • There are 2 n +1 polynomials h ( z ) with {0, 1} coe ffi cients and deg( h ) ≤ n . • Each has | h ( β 1 )| ≤ n + 1. • Collision guaranteed if 2 n +1 > ( n +1) d M ( f ) n , which occurs for large n if M ( f ) < 2.

Newman Polynomials • Donald Newman. • 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 • Donald Newman. • 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 ?

Evidence • Dubickas (2003): Every product of cyclotomic polynomials with no factors of z − 1 divides a Newman polynomial. • Known small limit points are realized by sequences of Newman polynomials. z 2 k ( z + 1) + z k ( z 2 + z + 1) + z + 1 � � → 1 . 25543 . . . M • Known small measures are realized by Newman polynomials.

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 Numbers • A real algebraic integer β > 1 is a Pisot number (or Pisot-Vijaraghavan number ) if all its conjugates β ´ satisfy | β ´| < 1. • Smallest Pisot number: the real root of z 3 – z – 1, 1.3247… • The set of Pisot numbers is closed! • Smallest limit point: golden ratio, τ . • Boyd (1978, 1985): All Pisot numbers in (1, 2 − δ ] can be identified.

Negative Pisot Numbers • β is a negative Pisot number if – β is a Pisot number. • Identify all negative Pisot numbers > – τ . • Four infinite families, and one sporadic example. • Can we represent all of these using Newman polynomials?

P n ( z ) = z n ( z 2 + z − 1) + 1, n even: n 2 − 1 P n ( z )( z n +1 − 1) = z 2 n +1 + ( z n +2 + 1) X z 2 k . z 2 − 1 k =0 Q n ( z ) = z n ( z 2 + z − 1) − 1, n odd: n − 1 2 Q n ( z ) z 2 − 1 = z n + X z 2 k . k =0 R n ( z ) = z n ( z 2 + z − 1) + z 2 − 1, n > 0: always has a real root in (0, 1).

S n ( z ) = z n ( z 2 + z − 1) − z 2 + 1, n > 0: S n ( z )( z n + 1)( z n +1 − 1) = z 2 − 1 n − 1 n − 5 2 2 z 3 n +1 + z 2 n +1 z 2 k + z n +3 z 2 k + 1 . X X k =0 k =0 G ( z ) = z 6 + 2 z 5 + z 4 − z 2 − z − 1: has a real root in (0, 1). Theorem 1 : 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.

Salem Numbers • A Salem number is a real algebraic integer α > 1 whose conjugates all lie on the unit circle, except for 1/ α . • Its minimal polynomial is reciprocal. • Smallest Salem number? • Unknown! Smallest known: 1.17628… . • Salem (1945): If f ( z ) is the minimal polynomial of a Pisot number β , then z m f ( z ) ± f * ( z ) has a Salem number α m as a root, for sufficiently large m , and α m → β as m → ∞ .

Negative Salem Numbers • A negative Salem number is a real algebraic integer α < − 1 whose conjugates all lie on the unit circle, except for 1/ α . • For each negative Pisot number in ( − τ , − 1), apply Salem’s construction to obtain two infinite families of nearby Salem numbers. • Can we represent all of these in ( − τ , − 1) with Newman polynomials?

• For positive integers m and n , define P + m,n ( z ) = z m P n ( z ) + P ∗ n ( z ) , P − m,n ( z ) = z m P n ( z ) − P ∗ n ( z ) , . . . S + m,n ( z ) = z m S n ( z ) + S ∗ n ( z ) , m,n ( z ) = z m S n ( z ) − S ∗ n ( z ) , S − G + m ( z ) = z m G ( z ) + G ∗ ( z ) , m ( z ) = z m G ( z ) − G ∗ ( z ) . G − • Eight doubly-infinite families; two singly-infinite ones.

Method of Investigation • Select one of these families. • Compute many Newman representatives for special values of m and n . • Search for simple rational multiples of the auxiliary factors that arise. • Identify patterns, and establish algebraic identities.

Constructing Newman Multiples • Given f ( z ), determine if there is a Newman polynomial F ( z ) so deg( F ) = N and f ( z ) | F ( z ). • F ( z ) ↔ F (2). • Construct (symmetric) bit sequences of length N + 1 representing integer multiples of f (2). • Fast check for divisibility by f ( − 2). • Construct F ( z ) and check if f ( z ) | F ( z ).

Easy Cases m,n ( z ) P − is Newman in almost all cases of interest. z 2 − 1 m − 2 m,n ( z )( z m − 1 − 1) P + = z 2 m + n − 1 + z n +2 X z 2 k − z m − 1 z 2 − 1 k =0 m − 2 z 2 k − z m + n + 1 . X + z k =0 • Negative terms cancel in all cases of interest.

Hard Cases ++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) ,

Hard Cases • Essentially two cases for Q + m , n : • m odd, n even, n ≥ m – 1: 5 , 4 ( z )( z 24 − 1)( z 5 + 1) 7 , 6 ( z )( z 60 − 1)( z 7 + 1) Q + Q + ( z 8 − 1)( z 6 − 1)( z 2 − 1) , ( z 12 − 1)( z 10 − 1)( z 2 − 1) , 9 , 8 ( z )( z 80 − 1)( z 9 + 1) Q + ( z 16 − 1)( z 10 − 1)( z 2 − 1) . • Suggest: m,m − 1 ( z )( z ab/ 2 − 1)( z m + 1) Q + . ( z a − 1)( z b − 1)( z 2 − 1)

• Leads to: 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