Complex Pisot Numbers and Newman Polynomials I erm Michael Mossinghoff Davidson College Introduction to Topics Summer@ICERM 2014 Brown University
Mahler s Measure f(z) = nx k=0 a k z k = a n n Y k=1 (z k) inz[z]. ny M(f) = a n max{1, k }. k=1 Z 1 M(f) =exp log f(e 2 it ) dt. 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 coefficients 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( 2 ) g( d ) apple(n + 1) d 1 M(f) n. g( 1 ) 1 (n + 1) d 1 M(f) n.
g( 1 ) 1 (n + 1) d 1 M(f) n. If h 1, h 2 have {0, 1} coefficients, degree n, and f does not divide h 1 h 2, then h 1 ( 1 ) h 2 ( 1 ) 1 (n + 1) d 1 M(f) n. There are 2 n+1 polynomials h(z) with {0, 1} coefficients 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 coefficients 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 coefficients 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. M z 2k (z + 1) + z k (z 2 + z + 1) + z +1! 1.25543... Known small measures are realized by Newman polynomials.
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
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: P n (z)(z n+1 1) z 2 1 = z 2n+1 +(z n+2 + 1) n 2 1 X k=0 z 2k. Q n (z) =z n (z 2 + z 1) 1, n odd: Q n (z) z 2 1 = zn + n 1 2X k=0 z 2k. 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 z 3n+1 + z 2n+1 n 1 2X k=0 = z 2k + z n+3 G(z) =z 6 +2z 5 + z 4 z 2 z 1: n 5 2X k=0 z 2k +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), S m,n (z) =z m S n (z) S n(z), G + m(z) =z m G(z)+G (z), G m (z) =z m G(z) G (z). 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 P m,n (z) z 2 1 is Newman in almost all cases of interest. P + m,n(z)(z m 1 1) z 2 1 m 2 = z 2m+n 1 + z n+2 X +z m 2 X k=0 k=0 z 2k z m+n +1. z 2k z m 1 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++ Q + 5,4 (z)(z24 1)(z 5 + 1) (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: Q + 5,4 (z)(z24 1)(z 5 + 1) (z 8 1)(z 6 1)(z 2 1), Q + 7,6 (z)(z60 1)(z 7 + 1) (z 12 1)(z 10 1)(z 2 1), Suggest: Q + 9,8 (z)(z80 1)(z 9 + 1) (z 16 1)(z 10 1)(z 2 1). Q + m,m 1 (z)(zab/2 1)(z m + 1) (z a 1)(z b 1)(z 2. 1)
Leads to: Q + m,n(z) z (m+2n+1)(n+2)/2 1 z n+1 +1 (z m+2n+1 1)(z n+2 1)(z 2 1) m 2X 3 n/2 z(z n + 1) z 2k X z k(m+2n+1) + k=0 k=0 = n+ m 1 X 2 k=0 z k(n+2). m, n both odd: Q + m,n(z) z (m+n)(m 1)/2 1 (z m+n 1)(z m 1 1)(z 2 1) = m+n 2 1 X k=0 z k(m 1) + z n 3 2X k=0 m 3 2X z 2k k=0 z k(m+n).
The Family R R m,n (z) =R n,m (z): consider only n m. m odd: the Salem number is <. n>mboth even: R m,n (z) z 2 1 is Newman. m even: R m,m (z) (z 2 1)(z m + 1) is Newman. Remaining case: n>m, n odd, m even. Let 0 < k < m/2.
R m,n (z) z 3m+n+2 z 2m+n+2 + z 2m+n+1 2k + z m+2k+1 z m +1 z 2 1 z 4m+2n+2 +(z 4m+2n+1 + z 4m+2n 1 + + z 3m+2n+3 ) + z 3m+2n 2k+1 +(z 3m+2n 2k+3 + z 3m+2n 2k+1 + + z 3m+n+3 ) z 2m+2n+2 + z 2m+2n 2k+1 + z 4m+n+2 3m+n 2k+1 + z + z 2m+n+2k+1 + z 2m+n+1 z 2m+n+2 z 2m+n +(z 2m+n+2k + z 2m+n+2k 2 + + z 2m+n 2k+2 2m+n 2k+1 )+z + z m+n+2k+1 +(z m+n 1 + z m+n 3 + + z m+2k+2 ) z 2m + z n + z 2m+2k+1 + z m+2k+1 +(z m 1 + z m 3 + + z)+1. = k = 1 and k = 2 cover all but (8, 11), (8, 19), (4, n), and (2, n). Handled with separate arguments.
Theorem 2: If α is a negative Salem number satisfied by one of the polynomials P ± m,n, Q ± m,n, R ± m,n, S ± m,n, or G ± m, and α > τ, then there exists a Newman polynomial F(z) with F(α) = 0. Compute all reciprocal polynomials f(z) with M(f) < τ and deg(f) 20. Check all negative Salems from this list. 502 Salem numbers, 281 covered by Theorem 2. Theorem 3: If α > τ is a negative Salem number and deg(α) 20, then there exists a Newman polynomial F(z) with F(α) = 0.
s 1 (z) =z 18 +3z 17 +4z 16 +4z 15 +4z 14 +5z 13 +6z 12 +7z 11 +7z 10 +7z 9 +7z 8 +7z 7 +6z 6 +5z 5 +4z 4 +4z 3 +4z 2 +3z +1 Reciprocal Newman polynomial representing s 1 (z) with minimal degree (116): s 1 (z) 24 (z)(z 90 2z 89 +2z 88 z 87 + z 83 2z 82 +2z 81 z 80 + z 79 2z 78 +2z 77 z 76 + z 72 2z 71 +3z 70 3z 69 +2z 68 z 67 + z 65 2z 64 +3z 63 3z 62 +2z 61 z 60 + z 59 z 58 + z 56 z 55 + z 54 z 53 + z 52 z 51 + z 50 z 49 + z 47 z 46 + z 45 z 44 + z 43 z 41 + z 40 z 39 + z 38 z 37 + z 36 z 35 + z 34 z 32 + z 31 z 30 +2z 29 3z 28 +3z 27 2z 26 + z 25 z 23 +2z 22 3z 21 +3z 20 2z 19 + z 18 z 14 +2z 13 2z 12 + z 11 z 10 +2z 9 2z 8 + z 7 z 3 +2z 2 2z + 1).
Recap All negative Pisot numbers work perfectly! Known negative Salem numbers present no obstruction! Are there any polynomials with M(f) < τ and no positive real roots that cannot be represented by a Newman polynomial? Yes!
No Newman Representatives! Another method finds several polynomials with all roots in A τ \ R + which are not satisfied by any Newman polynomial. Idea: sometimes {g(β) : g(z) is a Newman polynomial} B r(β) (0) is finite. Compute it and check for 0.
Five polynomials having no Newman representatives, but all roots in A τ \ R + : Coefficients +0+ 0 + +0 ++ + +0+0000 + +0+000 0 + +0+00 0 + Mahler Measure 1.55601 1.55837 1.60436 1.61582 1.61753 Interesting: each of these is the minimal polynomial for a complex Pisot number.
Complex Pisot Numbers A complex algebraic integer β = r + is is a complex Pisot number if all its conjugates β satisfy β < 1, except r is. Easy: square root of a negative Pisot number is a complex Pisot number (purely imaginary). Garth (2003) identified all complex Pisot numbers with modulus < 1.17. Smallest: 1.1509, attained by z 6 z 2 + 1.
Problems Does applying Salem s construction to complex Pisot numbers produce complex Salem numbers, at least for large m? (Possibly known; but could generalize method from Salem paper) Can large be quantified? (Boyd paper) Can one find more complex Pisot and Salem numbers with M(f) < τ? Infinite families? (Garth paper)
Problems Can these small complex Pisot and Salem numbers be represented by Newman polynomials? (Experiment!) Can the method for showing that an algebraic integer cannot be represented by a Newman polynomial be shown to terminate for complex Pisot numbers? (Hare & M. + Garsia 1962)
References K. Hare & M., Negative Pisot and Salem numbers as roots of Newman polynomials, Rocky Mountain J. Math. 44 (2014), no. 1, 113-138. D. Garth, Complex Pisot numbers of small modulus, C.R. Acad. Sci. Paris, Ser. I 336 (2003), 967-970. M., Polynomials with restricted coefficients and prescribed noncyclotomic factors, LMS J. Comput. Math. 6 (2003), 314-325.
References M. Bertin et al., Pisot and Salem Numbers, Birkhäuser, 1992. D. Boyd, Small Salem numbers, Duke Math. J. 44 (1977), no. 2, 315-328. R. Salem, Power series with integral coefficients, Duke Math. J. 12 (1945), 103-108. C. Smyth, The Mahler measure of algebraic numbers: A survey, Number Theory and Polynomials, Cambridge Univ. Press, 2008, pp. 322-349.
I erm Good Luck!