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

Complex Pisot Numbers and Newman Polynomials

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Introduction to Topics Summer@ICERM 2014 Brown University Michael Mossinghoff Davidson College

Mahler’s Measure

• f(z) =

n

X

k=0

akzk = an

n

Y

k=1

(z − βk) in Z[z].

• M(f) = |an|

n

Y

k=1

max{1, |βk|}.

• (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) = exp

✓Z 1 log |f(e2πit)| dt ◆ .

M(f) ≥ 1.4935 − .61 n .

• M(f) = M(–f(z)) = M(f(–z)) = M(f(zk)) = M(f *).
• Here, f *(z) is defined as zn⋅f(1/z): reciprocal of f.
• Some results on Lehmer’s Problem:
• M(z10+z9–z7–z6–z5–z4–z3+z+1) = 1.17628… .
• (Smyth 1971) If f(z) ≠ ±f *(z) and f(0) ≠ 0,

then M(f) ≥ M(z3 – z – 1) = 1.3247…

• If 1 < M(f) < 1.17628… then deg(f) ≥ 56.
• (Borwein, Dobrowolski, M., 2007) If f(z) has all
• dd coefficients and degree n then

• 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.

Measures and Heights

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).

Bloch and Pólya

• f(z) irreducible, degree d, M(f) < 2.
• So f(z) monic, and has at least one root |β1| < 1.
• g(z): height 1, degree n, f is not a factor of g.
• |Res(f, g)| = |g(β1)g(β2)…g(βd)| ≥ 1.
• |g(βk)| ≤ (n + 1)⋅max{1, |βk|n}.
• |g(β1)| ≥

1 (n + 1)d−1M(f)n .

• |g(β2) · · · g(βd)| ≤ (n + 1)d−1M(f)n.

• If h1, h2 have {0, 1} coefficients, degree ≤ n, and

f does not divide h1 – h2, then

• Collision guaranteed if 2n+1 > (n+1)dM(f)n,

which occurs for large n if M(f) < 2.

|h1(β1) − h2(β1)| ≥ 1 (n + 1)d−1M(f)n .

• There are 2n+1 polynomials h(z) with {0, 1}

coefficients and deg(h) ≤ n.

• Each has |h(β1)| ≤ n + 1.
• |g(β1)| ≥

1 (n + 1)d−1M(f)n .

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 ∊ 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

• 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 ∊ Aτ, where τ denotes the golden ratio.

M

• z2k(z + 1) + zk(z2 + z + 1) + z + 1
• → 1.25543 . . .

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.

• 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

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 z3 – 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.

• 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?

Negative Pisot Numbers

Qn(z) z2 − 1 = zn +

X

k=0

z2k. Pn(z)(zn+1 − 1) z2 − 1 = z2n+1 + (zn+2 + 1)

X

k=0

z2k. Pn(z) = zn(z2 + z − 1) + 1, n even: Rn(z) = zn(z2 + z − 1) + z2 − 1, n > 0:

always has a real root in (0, 1).

Qn(z) = zn(z2 + z − 1) − 1, n odd:

G(z) = z6 + 2z5 + z4 − z2 − z − 1: Sn(z)(zn + 1)(zn+1 − 1) z2 − 1 = z3n+1 + z2n+1

X

k=0

z2k + zn+3

X

k=0

z2k + 1.

has a real root in (0, 1).

Sn(z) = zn(z2 + z − 1) − z2 + 1, n > 0:

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 zm f(z) ± f*(z) has a Salem number αm as a root, for sufficiently large m, and αm → 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?

S−

m,n(z) = zmSn(z) − S∗ n(z),

S+

m,n(z) = zmSn(z) + S∗ n(z),

• For positive integers m and n, define

P +

m,n(z) = zmPn(z) + P ∗ n(z),

P −

m,n(z) = zmPn(z) − P ∗ n(z),

G+

m(z) = zmG(z) + G∗(z),

G−

m(z) = zmG(z) − G∗(z).

. . .

• Eight doubly-infinite families; two singly-infinite
• nes.

• 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.

Method of Investigation

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)

z2 − 1

is Newman in almost all cases of interest.

P +

m,n(z)(zm−1 − 1)

z2 − 1 = z2m+n−1 + zn+2

m−2

X

k=0

z2k − zm−1 +z

m−2

X

k=0

z2k − zm+n + 1.

• Negative terms cancel in all cases of interest.

++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) ,

Hard Cases

Q+

m,m−1(z)(zab/2 − 1)(zm + 1)

(za − 1)(zb − 1)(z2 − 1) .

Hard Cases

• Essentially two cases for Q+m,n :
• m odd, n even, n ≥ m – 1:

Q+

5,4(z)(z24 − 1)(z5 + 1)

(z8 − 1)(z6 − 1)(z2 − 1) , Q+

9,8(z)(z80 − 1)(z9 + 1)

(z16 − 1)(z10 − 1)(z2 − 1). Q+

7,6(z)(z60 − 1)(z7 + 1)

(z12 − 1)(z10 − 1)(z2 − 1),

• Suggest:

Q+

m,n(z)

• z(m+n)(m−1)/2 − 1
• (zm+n − 1)(zm−1 − 1)(z2 − 1)

=

−1

X

k=0

zk(m−1) + z ✓ n−3

X

k=0

z2k ◆✓ m−3

X

k=0

zk(m+n) ◆ .

• m, n both odd:
• Leads to:

z(zn + 1) ✓ m−3

X

k=0

z2k ◆✓n/2 X

k=0

zk(m+2n+1) ◆ +

n+ m−1

X

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)

=

• Remaining case: n > m, n odd, m even.

Let 0 < k < m/2.

The Family R–

• R−

m,n(z) = R− n,m(z): consider only n ≥ m.

• m odd: the Salem number is < −τ.
• m even:

R−

m,m(z)

(z2 − 1)(zm + 1) is Newman.

• n > m both even: R−

m,n(z)

z2 − 1 is Newman.

R−

m,n(z)

• z3m+n+2 − z2m+n+2 + z2m+n+1−2k + zm+2k+1 − zm + 1
• z2 − 1

= z4m+2n+2 + (z4m+2n+1 + z4m+2n−1 + · · · + z3m+2n+3) + z3m+2n−2k+1 + (z3m+2n−2k+3 + z3m+2n−2k+1 + · · · + z3m+n+3) − z2m+2n+2 + z2m+2n−2k+1 + z4m+n+2 + z3m+n−2k+1 + z2m+n+2k+1 + z2m+n+1 − z2m+n+2 − z2m+n + (z2m+n+2k + z2m+n+2k−2 + · · · + z2m+n−2k+2) + z2m+n−2k+1 + zm+n+2k+1 + (zm+n−1 + zm+n−3 + · · · + zm+2k+2) − z2m + zn + z2m+2k+1 + zm+2k+1 + (zm−1 + zm−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. 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.

• 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.

s1(z) = z18 + 3z17 + 4z16 + 4z15 + 4z14 + 5z13 + 6z12 + 7z11 + 7z10 + 7z9 + 7z8 + 7z7 + 6z6 + 5z5 + 4z4 + 4z3 + 4z2 + 3z + 1

• Reciprocal Newman polynomial representing

s1(z) with minimal degree (116):

s1(z)Φ24(z)(z90 − 2z89 + 2z88 − z87 + z83 − 2z82 + 2z81 − z80 + z79 − 2z78 + 2z77 − z76 + z72 − 2z71 + 3z70 − 3z69 + 2z68 − z67 + z65 − 2z64 + 3z63 − 3z62 + 2z61 − z60 + z59 − z58 + z56 − z55 + z54 − z53 + z52 − z51 + z50 − z49 + z47 − z46 + z45 − z44 + z43 − z41 + z40 − z39 + z38 − z37 + z36 − z35 + z34 − z32 + z31 − z30 + 2z29 − 3z28 + 3z27 − 2z26 + z25 − z23 + 2z22 − 3z21 + 3z20 − 2z19 + z18 − z14 + 2z13 − 2z12 + z11 − z10 + 2z9 − 2z8 + z7 − z3 + 2z2 − 2z + 1).

• All negative Pisot numbers work perfectly!
• Known negative Salem numbers present no
• bstruction!
• Are there any polynomials with M(f) < τ and no

positive real roots that cannot be represented by a Newman polynomial?

• Yes!

Recap

• Another method finds several polynomials with

all roots in Aτ \ R+ which are not satisfied by any Newman polynomial.

• Idea: sometimes {g(β) : g(z) is a Newman

polynomial} ⋂ Br(β)(0) is finite.

• Compute it and check for 0.

No Newman Representatives!

• Five polynomials having no Newman

representatives, but all roots in Aτ \ R+:

Coefficients Mahler Measure +0+−0−+ 1.55601… +0−++−−+ 1.55837… +0+0000−+ 1.60436… +0+000−0−+ 1.61582… +0+00−0−+ 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 z6 – z2 + 1.

• 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

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.

Good Luck!

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