Zero attractor of polynomials with rational generatingfunctions

Robert BoyerDepartment of Mathematics

Drexel University

Khang TranDepartment of Mathematics and Computer Science

Truman State University

AbstractLetHm(z) be a sequence of polynomials whose generating function

∑m Hm(z)tm = N(t, z)/D(t, z)

is rational with the denominator D(t, z) = A(z)tn +B(z)t+ 1, where A(z) and B(z) are poly-nomials in z with complex coefficients and N(t, z) and D(t, z) do not have a trivial commonfactor. We show that the zero attractor of Hm(z) is a portion of an real algebraic curve togetherwith a finite subset of the set of roots of the polynomial A(z)Rest(N(t, z), D(t, z)).

1 IntroductionLet Hm(z) be a sequence of polynomials with rational generating function, i.e.

∑mHm(z)tm =

N(t, z)/D(t, z) where N(t, z), D(t, z) ∈ C[t, z]. What can be said about the location of the rootsof Hm(z) in the complex plane? We limit our study to the case where the denominator has thespecial form D(t, z) = A(z)tn +B(z)t+ 1 where A(z), B(z) ∈ C[z]. This denominator implies that

Hm(z) +B(z)Hm−1(z) +A(z)Hm−n(z) = 0.

The root distribution of a special case of this sequence of polynomials, the Faber polynomials foran n-cusped hypocycloid, where B(z) = −z and A(z) = 1/(n− 1), has been studied in [6].

In the case N(t, z) ≡ 1, n = 2, 3, 4, and D(t, z) = A(z)tn + B(z)t+ 1, it is known [10] that theroots of Hm(z) which satisfy A(z) 6= 0 lie on a fixed curve given by

=Bn(z)

A(z)= 0

and0 ≤ (−1)n<B

n(z)

A(z)≤ nn

(n− 1)n−1.

In this paper we show that for a general n and a general numerator N(t, z) ∈ C[t, z], the zeroattractor of Hm(z) is the same curve above together with a finite subset of the set of roots of the

1

polynomial A(z)Rest(D(t, z), N(t, z)) 6= 0. We recall a definition from Robert Boyer and WilliamGoh (see [2, 3]).

Definition Let Z(Hm(z)) denote the finite set of zeros of the polynomials Hm(z). Then thezero attractor of the polynomial sequence {Hm(z)} is the limit of the finite sets Z(Hm(z)) in theHausdorff metric on the non-empty compact subsets K of C ∪∞.

Theorem 1 Let Hm(z) be a sequence of polynomials with the generating function

N(t, z)

D(t, z)=

∞∑m=0

Hm(z)tm

where N(t, z) is a bivariate polynomial of degree at most n in t and D(t, z) = A(z)tn + B(z)t+ 1,where A(z), B(z) ∈ C[z] such that there is no constant c ∈ C with Bn(z) ≡ cA(z). The zeroattractor of Hm(z) is the curve C given by

=Bn(z)

A(z)= 0

and0 ≤ (−1)n<B

n(z)

A(z)≤ nn

(n− 1)n−1

and a finite subset of the set of roots of A(z)Rest(D(t, z), N(t, z)).

In the case N(t, z) ≡ 1, the zero attractor in the domain A(z) 6= 0 is exactly the curve men-tioned in [10]. We notice that the fraction Bn(z)/A(z) can be written in terms of the discriminantDisctD(t, z). In fact we will introduce a new approach to the zero attractor using the q-analogue ofthis discriminant. To find points in the zero attractor, we need roots of D(z, t) with equal minimalmodulus. In particular, they are found among the roots of the q-discriminant that have modulus1. We will also see that the polynomial

f(z) = zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ

for some θ ∈ R plays a vital role in the proof of Theorem 1. This polynomial uses the modulus ofthe ratio of roots and singles out the minimal modulus case. It is peculiar to see that f(z) is alsoa trinomial with the same coefficient structure as A(z)tn +B(z)t+ 1.

We will also apply the Alan Sokal’s method [9] on the zero attractor. We will recall theseimportant concepts and results in Section 2. The proof of Theorem 1 is given in Section 3. We giveexamples in Section 4.

2 Ismail’s q-discriminant and Sokal’s methodThe discriminant of a one variable polynomial P (x) of degree n with the leading coefficient p is

Discx(P (x)) = p2n−2∏

1≤i<j≤n

(xi − xj)2

2

where x1, . . . , xn are roots of P (x). The resultant of two polynomials P (x) and R(x) is given by

Resx(P (x), R(x)) = pm∏

P (xi)=0

R(xi)

= rn∏

R(xi)=0

P (xi) (1)

where r and m are the leading coefficient and the degree of R(x) respectively. The discriminant ofP (x) can be computed in terms of the resultant between it and its derivative as follow:

DiscxP (x) = (−1)n(n−1)/21

pRes(P (x), P ′(x))

= (−1)n(n−1)/2pn−2∏i≤n

P ′(xi). (2)

The q-discriminant of a polynomial P (x) of degree n and the leading coefficient p is defined by

Discx(P ; q) = p2n−2qn(n−1)/2∏

1≤i<j≤n

(q−1/2xi − q1/2xj)(q1/2xi − q−1/2xj)

This q-discriminant is zero if and only if a quotient of roots xi/xj equals q. When q → 1, thisq-discriminant becomes the ordinary discriminant DiscxP (x). Mourad Ismail proves a q-analogueof (2) ([7, Theorem 3.1]).

Proposition 2 The q-discriminant of a polynomial P (x) of degree n and the leading coefficient pis

Discx(P ; q) = (−1)n(n−1)/2pn−2n∏i=1

(DqP )(xi)

where(DqP )(x) =

P (x)− P (qx)

x− qx.

We will apply this proposition to compute the q-discriminant of D(t, z) = A(z)tn + B(z)t + 1,which yields the quotient Bn(z)/A(z) in the Theorem 1. For now, let us recall some definitions ofthe root distribution of a sequence of functions

fm(z) =

n∑k=1

αk(z)βk(z)m

where αk(z) and βk(z) are analytic in a domain D. Let us call an index k dominant at z if|βk(z)| ≥ |βl(z)| for all l (1 ≤ l ≤ n). Let

Dk = {z ∈ D : k is dominant at z}.

Let lim inf Z(fm) be the set of all z ∈ D such that every neighborhood U of z has a non-emptyintersection with all but finitely many of the sets Z(fm). Let lim supZ(fm) be the set of all z ∈ Dsuch that every neighborhood U of z has a non-empty intersection infinitely many of the setsZ(f). If A is the zero attractor of the sequence fm(z) in the domain D then lim inf Z(fm) ⊂ A ⊂lim supZ(fm). We will need the following theorem from Alan Sokal ([9, Theorem 1.5]).

3

Theorem 3 Let D be a domain in C, and let α1, . . . , αn, β1, . . . , βn (n ≥ 2) be analytic function onD, none of which is identically zero. Let us further assume a ’no-degenerate-dominance’ condition:there do not exist indices k 6= k′ such that βk ≡ ωβk′ for some constant ω with |ω| = 1 and suchthat Dk (= Dk′) has nonempty interior. For each integer m ≥ 0, define fm by

fm(z) =

n∑k=1

αk(z)βk(z)m.

Then lim inf Z(fm) = lim supZ(fm), and a point z lies in this set if and only if either

1. there is a unique dominant index k at z, and αk(z) = 0, or

2. there a two or more dominant indices at z.

3 The zero attractor of the roots of Hm(z)

In this section, we will prove Theorem 1. We consider the domain D in the z-plane whereA(z)Rest(N(t, z), D(t, z)) 6= 0. From (2), it is not difficult to show that Disct(A(z)tn+B(z)t+1) =±A(z)n−2

(B(z)n(n− 1)n−1 + (−1)n−1nnA(z)

). We can also obtain this identity by letting q → 1

from Proposition 5 below. Thus the roots of Disct(A(z)tn + B(z)t + 1) belong to the curve C inTheorem 1. From now on, we assume that Disct(A(z)tn + B(z)t+ 1) 6= 0, i.e. all the roots in t ofthe denominator D(t, z) are distinct. We also assume that there does not exist a constant c withBn(z) ≡ cA(z).

Our first step is to apply Theorem 3 to represent the zero attractor in terms of the roots of thisdenominator.

Lemma 4 Let t1, t2, . . . , tn be the roots of A(z)tn + B(z)t + 1 with |t1| ≤ |t2| ≤ . . . ≤ |tn|. Thezero attractor of the roots of Hm(z) in D is given by |t1| = |t2|.

Proof We first notice that the condition Rest(N(t, z), A(z)tn +B(z)t+ 1) 6= 0 is equivalent to thefact that N(ti, z) 6= 0 for all 1 ≤ i ≤ n. By partial fractions, we have

N(t, z)

A(z)tn +B(z)t+ 1=

N(t, z)

A(z)(t− t1)(t− t2) . . . (t− tn)

=1

A(z)

n∑i=1

N(ti, z)

t− ti

∏j 6=i

1

ti − tj

=1

A(z)

n∑i=1

(− 1

ti

)N(ti, z)

1− t/ti

∏j 6=i

1

ti − tj

=−1

A(z)

∞∑m=0

n∑i=1

N(ti, z)

tm+1i

∏j 6=i

1

(ti − tj)tm.

Thus

Hm(z) =−1

A(z)

n∑i=1

N(ti, z)

tm+1i

∏j 6=i

1

(ti − tj). (3)

The lemma follows from Theorem 3. We will leave an explanation to the ’no-degenerate-dominance’condition in Theorem 3 after the next proposition.

4

Let t1 = t1(z), . . ., tn = tn(z) be the roots of the denominator D(t, z) = A(z)tn + B(z)t + 1.Let qi = ti/t1 . The conditions |t1| ≤ |t2| ≤ . . . ≤ |tn| and |t1| = |t2| are the same as |q2| = 1 and|qi| ≥ 1, 3 ≤ i ≤ n. These imply that a polynomial whose moduli of roots are |qi|, 2 ≤ i ≤ n, hasno root on the open unit disk. Such a polynomial will play an important role in our proof. We willsee later that it is sufficient to consider the moduli of roots of this polynomial. First, it is naturalto consider the q-discriminant of D(z, t). We suppress the parameter z in the following proposition.

Proposition 5 The q-discriminant of D(t) = Atn +Bt+ 1 is given by

Disct(D(t); q) = ±An−2(Bn

qn−1(1− qn−1)n−1

(1− q)n−1+ (−1)n−1

(1− qn)n

(1− q)nA

).

Proof Proposition 2 gives

Disct(D(t); q) = ±An−2∏

D(t)=0

(DqD)(t)

where

(DqD)(t) =D(t)−D(qt)

t− qt

= B +Atn−11− qn

1− q. (4)

Using the symmetric definition of resultant in (1), we write the product in Disct(D(t); q) as

Disct(D(t); q) = ±An−1(

1− qn

1− q

)n ∏(DqD)(t)=0

D(t). (5)

From the definition of D(t) and the formula of (DqD)(t) in (4), these two polynomials are connectedby

1− qn

1− qD(t) = t(DqD)(t) + tB

(1− qn

1− q− 1

)+

1− qn

1− q.

Thus (5) gives

Disct(D(t); q) = ±An−1(

1− qn

1− q

) ∏(DqD)(t)=0

(tB

(1− qn

1− q− 1

)+

1− qn

1− q

)

= ±An−2Bn qn(1− qn−1)n

(1− q)n(DqD)

(− 1− qn

Bq(1− qn−1)

)= ±An−2

(Bn

qn−1(1− qn−1)n−1

(1− q)n−1+ (−1)n−1

(1− qn)n

(1− q)nA

).

We now show that the ’no-degenerate-dominance’ condition in Theorem 3 holds. Suppose ti ≡ωtj for some constant ω with |ω| = 1. Then the q-disriminant Disct(D(t, z); q) = 0 when q = ω.Proposition 2 implies that

(−1)nBn(z)

A(z)≡ (1− ωn)n

(1− ω)ωn−1(1− ωn−1)n−1.

5

ThusBn(z) = cA(z),

contrary to our assumption.

Lemma 6 Suppose t1, . . . , tn are the roots of the polynomial Atn+Bt+1 with B 6= 0. Let qi = ti/t1.If q := q2 = e2iθ then beside the two roots e±iθ, the remaining n− 2 roots of the polynomial

f(z) = zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ

are e−iθq3, . . . , e−iθqn.

Proof Denote by ek(t1, t2, . . . , tn) the k-th elementary symmetric polynomials in variables t1, t2, . . . , tn.Since t1, t2, . . . , tn are the roots of Atn +Bt+ 1, we have ek(t1, t2, . . . , tn) = 0, 1 ≤ k ≤ n− 2. Letek := ek(q3, . . . , qn). We divide the equation ek(t1, t2, . . . , tn) = 0 by tk1 , 1 ≤ k ≤ n− 2, to obtain

e1 + q + 1 = 0

e2 + e1(q + 1) + q = 0

e3 + e2(q + 1) + e1q = 0

e4 + e3(q + 1) + e2q = 0...

...en−2 + en−3(q + 1) + en−4q = 0.

We solve this system of equations for e1, . . . , en−2 by substitution to get

ek = (−1)k(1 + q + q2 + · · ·+ qk)

with 1 ≤ k ≤ n− 2. Also q3, q4, . . . , qn are roots of the equation

n−2∑k=0

(−1)kekzn−k−2 = 0.

We multiply both side of this equation by (1− q) to obtain

(1− q)zn−2 + (1− q2)zn−3 + (1− q3)zn−4 + · · ·+ (1− qn−1) = 0.

Thuszn−2 + zn−3 + · · ·+ 1 = qzn−2 + q2zn−3 + · · ·+ qn−1.

We add zn−1 to both sides to getzn − 1

z − 1=zn − qn

z − q.

This identity implies that

zn(1− q)− (1− qn)z + q(1− qn−1) = 0.

6

We divide both sides of this equation by 1− q and note that

qn − 1

q − 1= e(n−1)πiθ

enπiθ − e−nπiθ

eπiθ − e−πiθ

= e(n−1)πiθsinnθ

sin θ. (6)

We obtainzn − e(n−1)iθ sinnθ

sin θz + eniθ

sin(n− 1)θ

sin θ= 0.

We substitute z by eiθz to get

zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ= 0.

Lemma 7 Let θ be a fixed real number. If all the roots the polynomial

f(z) := zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ

satisfy |z| ≥ 1 then

0 ≤ sinn nθ

sin θ sinn−1(n− 1)θ≤ nn

(n− 1)n−1.

Proof Suppose n is even. Since f(z) 6= 0 if z ∈ R and z ∈ (−1, 1), we have

0 ≤ f(1)f(−1) sin2 θ

=

(sin(n− 1)θ − 2 cos

(n+ 1)θ

2sin

(n− 1)θ

2

)(sin(n− 1)θ + 2 sin

(n+ 1)θ

2cos

(n− 1)θ

2

)= 2 cos

(n− 1)θ

2sin

(n− 1)θ

2

(cos

(n− 1)θ

2− cos

(n+ 1)θ

2

)(sin

(n− 1)θ

2+ sin

(n+ 1)θ

2

)= 4 sin(n− 1)θ sin2 nθ

2sin

θ

2cos

θ

2

= 2 sin(n− 1)θ sin θ sin2 nθ

2.

Hencesin(n− 1)θ sin θ ≥ 0.

Suppose n is odd. We have

0 ≤ f(1)f(−1) sin2 θ

=

(2 sin

nθ

2cos

(n− 2)θ

2− sinnθ

)(2 cos

nθ

2sin

(n− 2)θ

2+ sinnθ

)= 2 sin

nθ

2cos

nθ

2

(cos

(n− 2)θ

2− cos

nθ

2

)(sin

(n− 2)θ

2+ sin

nθ

2

)= 4 sinnθ sin2 (n− 1)θ

2sin

θ

2cos

θ

2

= 2 sinnθ sin θ sin2 (n− 1)θ

2. (7)

7

Hencesinnθ sin θ ≥ 0.

Thus, in either case n is odd or n is even, we have

sinn nθ

sin θ sinn−1(n− 1)θ≥ 0.

It remains to showsinn nθ

sin θ sinn−1(n− 1)θ≤ nn

(n− 1)n−1.

We return to the polynomial

f(z) = zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ

whose roots satisfy |z| ≥ 1. The roots of its reciprocal

sin(n− 1)θ

sin θzn − sinnθ

sin θzn−1 + 1

satisfy |z| ≤ 1. Thus by the Gauss-Lucas theorem (see [11]), the roots of its derivative

nsin(n− 1)θ

sin θzn−1 − (n− 1)

sinnθ

sin θzn−2

satisfy the same condition. This implies∣∣∣∣ sinnθ

sin(n− 1)θ

∣∣∣∣ ≤ n

n− 1.

Thus ∣∣∣∣ sinn−1 nθ

sinn−1(n− 1)θ

∣∣∣∣ ≤ nn−1

(n− 1)n−1.

Since | sinnθ/ sin θ| < n, the inequality above implies∣∣∣∣ sinn nθ

sin θ sinn−1(n− 1)θ

∣∣∣∣ ≤ nn

(n− 1)n−1.

Sincesinn nθ

sin θ sinn−1(n− 1)θ≥ 0,

we have0 ≤ sinn nθ

sin θ sinn−1(n− 1)θ≤ nn

(n− 1)n−1.

Lemma 8 Let θ be a real number such that

0 ≤ sinn nθ

sin θ sinn−1(n− 1)θ≤ nn

(n− 1)n−1.

If z1, z2, . . . , zn are the roots of

f(z) := zn − sinnθ

sin θz +

sin(n− 1)θ

sin θ

with |z1| ≤ |z2| ≤ · · · ≤ |zn|, then |z1| = |z2|.

8

Proof We first note that f(z) has at least two complex roots on the unit circle z = e±iθ. Thus itsuffices to show that any real root of f(z) that lies in the open interval (−1, 1) has even multiplicity.In fact, we will show that either f(z) ≥ 0 for all z ∈ (−1, 1) or f(z) ≤ 0 for all z ∈ (−1, 1).Thelemma follows from the fact that if z is a root of f(z) then so is z̄.

We consider the reciprocal polynomial

g(z) = znf

(1

z

)=

sin(n− 1)θ

sin θzn − sinnθ

sin θzn−1 + 1.

We note thatg′(z) = zn−2

(n

sin(n− 1)θ

sin θz − (n− 1)

sinnθ

sin θ

).

Furthermore, at the critical value z0 = (n− 1) sinnθ/n sin(n− 1)θ we have

g(z0) =(n− 1)n sinn nθ

nn sin θ sinn−1(n− 1)θ− (n− 1)n−1 sinn nθ

nn−1 sin θ sinn−1(n− 1)θ+ 1

= 1− (n− 1)n−1 sinn nθ

nn sin θ sinn−1(n− 1)θ.

From the hypothesis, we have g(z0) ≥ 0.Suppose n is even. Then g′(z) has a root z = 0 with an even multiplicity and a simple root at

z = z0. Also the leading coefficient of g(z) is nonnegative by the hypothesis. Thus g(z) decreaseson (−∞, z0) and increases on (z0,∞). Since g(z0) ≥ 0, we have g(z) ≥ 0 for all z ∈ R. Thusf(z) ≥ 0 for all z ∈ R.

Suppose n is odd. The hypothesis implies that sinnθ/ sin θ ≥ 0. From the equation

f(1)f(−1) sin2 θ = 2 sinnθ sin θ sin2 (n− 1)θ

2

in (7), we have f(1)f(−1) ≥ 0. Thus g(1)g(−1) = (−1)nf(1)f(−1) ≤ 0.We first consider the case sin(n − 1)θ/ sin θ ≥ 0. In this case z0 ≥ 0 and g(z) increases on

(−∞, 0) ∪ (z0,∞) and decreases on (0, z0). Since g(z0) ≥ 0, we have g(z) ≥ 0 if z ≥ 0. Inparticular, g(1) ≥ 0 and hence g(−1) ≤ 0. Thus g(z) ≤ 0 if z ≤ −1. The facts that g(z) ≥ 0,∀z ≥ 1 and g(z) ≤ 0, ∀z ≤ −1 imply f(z) ≥ 0 if z ∈ (−1, 1).

Similarly in the case sin(n − 1)θ/ sin θ ≤ 0, we have that z0 ≤ 0 and that g(z) decreases on(−∞, z0)∪ (0,∞) and increases on (z0, 0). Since g(z0) ≥ 0, we have g(z) ≥ 0 if z ≤ 0. In particular,g(−1) ≥ 0 and hence g(1) ≤ 0. Thus g(z) ≤ 0 if z ≥ 1. The facts that g(z) ≥ 0, ∀z ≤ −1 andg(z) ≤ 0, ∀z ≥ 1 imply f(z) ≤ 0 if z ∈ (−1, 1).

We now consider the proof of Theorem 1. Let z be a point on the zero attractor. We will showz ∈ C. Lemma 4 gives

|t1| = |t2| ≤ |t3|, . . . , |tn|

where t1, t2, . . . , tn are the roots of the denominator A(z)tn+B(z)t+1. Let q = t2/t1 and qi = ti/t1for 3 ≤ i ≤ n. Then |q| = 1 and |qi| ≥ 1 for 3 ≤ i ≤ n. Since q is a quotient of two roots,Disct(A(z)tn +B(z)t+ 1; q) = 0. Proposition 5 implies that

(−1)nBn(z)

A(z)=

(1− qn)n

(1− q)qn−1(1− qn−1)n−1.

9

Let q = e2iθ, θ ∈ R. From (6), we have

(−1)nBn(z)

A(z)=

(1− qn)n/(1− q)n

qn−1(1− qn−1)n−1/(1− q)n−1

=sinn nθ

sin θ sinn−1(n− 1)θ.

This implies that

= (−1)nBn(z)

A(z)= 0.

From Lemmas 6 and 7, we have z ∈ C.Conversely, suppose z ∈ C. We will show z is a point on the zero attractor. We note that the

real functionsinn nθ

sin θ sinn−1(n− 1)θ

maps onto the real line. Thus there is a real value θ such that

(−1)nBn(z)

A(z)=

sinn nθ

sin θ sinn−1(n− 1)θ.

Let q = e2iθ. Then by the previous computations, we have

(−1)nBn(z)

A(z)=

(1− qn)n

(1− q)qn−1(1− qn−1)n−1.

Proposition 5 implies that Disct(A(z)tn+B(z)t+ 1; q) = 0. Thus q is a quotient of some two roots,say t2/t1, of A(z)tn + B(z)t + 1. Let qi = ti/t1. Lemmas 6 and 8 imply that there are k 6= l suchthat

|qk| = |ql| ≤ |qi|

for all 2 ≤ i ≤ n or that |q| ≤ |qi| for all 3 ≤ i ≤ n. Thus z belongs to the zero attractor by Lemma4.

4 Examples of the zero attractor of Hm(z)

We consider the zero attractor of a sequence of Fibonacci-type polynomials Hm(z) with the gener-ating function

∞∑m=0

Hm(z)tm =C(z)t+D(z)

1− zkt− t2

where C(z), D(z) ∈ C[z]. In [1], the author showed that in the case C(z) = zk+z−1 andD(z) = −1,the maximal real roots of Hm(z) converge to ξ(k) where ξ(k) > 1 is the unique positive real rootof the polynomial zk − zk−1 + z − 2. Theorem 1 implies that for any C(z), D(z) ∈ C[z], the zeroattractor of Hm(z) is the curve C given by

=z2k = 0

10

and0 ≤ −<z2k ≤ 4

together with a finite subset of the set of roots of Rest(C(z)t+D(z), 1−zkt−t2). So C is a collectionof 2k equally spaced radial line segments with equal lengths 2k

√4. In the case C(z) = zk + z − 1

and D(z) = −1, the definition of resultant in (1) yields

Rest(C(z)t+D(z), 1− zkt− t2) = z(zk − zk−1 + z − 2).

The factor zk−zk−1+z−2 is the polynomial in [1] mentioned above. As a consequence of Theorem3 and (3), a root z of z(zk − zk−1 + z − 2) belongs to the zero attractor if and only if t1 < t2 andN(t1, z) = 0. Since t1t2 = −1, we have that a root z of z(zk − zk−1 + z − 2) belongs to the zeroattractor if and only if |zk + z − 1| > 1. Thus the value ξ(k) in [1] mentioned above belongs to thezero attractor.

We now consider another example where

B(z) = (z + 1)(z − i) + z + 1− i,A(z) = z3 + 10,

and

g(z) :=B3(z)

A(z).

We consider the sequence of polynomials given by the recurrence

Hm+1(z) = B(z)Hm(z) +A(z)Hm−2(z)

with the initial conditions H0(z) = 1, H1(z) = 0, and H2(z) = 0.The zero attractor lies on the level set =(B3(z)/A(z)) = 0; the zero attractor lies between the

two level sets <(B3(z)/A(z)) = 0 and <(B3(z)/A(z)) = −33/2. The following values play a specialrole in describing the zero attractor:

• The roots of A(z) which are the cube roots of −10 provide three isolated points of the zeroattractor.

• The six solutions to g(z) = −33/22.

• The two solutions to g(z) = 0.

• There are six curves that comprise the zero attractor which divide into two natural families.One endpoint of the curve is a solution to g(z) = −33/22 while the other endpoint is a zeroof g(z).

These observations are illustrated in the figures below.

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Figure 1: Zeros of H500(z) (in black) together with the curve =(B3(z)/A(z)) = 0 (in red). Thelarge solid dots are the cube roots of −10 (the zeros of A(z)).

Figure 2: Zeros of H500(z) together with the two curves <(B3(z)/A(z)) = 0 (in blue) and<(B3(z)/A(z)) = −33/22 (in green). The large solid dots are the cube roots of −10 (the zerosof A(z)).

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Figure 3: Zeros of H500(z) together with =(B3(z)/A(z)) = 0, <(B3(z)/A(z)) = 0 and<(B3(z)/A(z)) = −33/22. The large solid dots are the cube roots of −10 (the zeros of A(z)).

Figure 4: Closeup of the above plot in a neighborhood of the real negative cube root of −10 whichshows that it is an isolated point in the zero attractor.

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Figure 5: Zeros of H1000(z) together with the six solutions of g(z) = −33/22 (in green) and the twosolutions of g(z) = 0 (in blue).

For the root distribution of similar (but different) type of polynomials satisfying the recurrencerelation

zHm(z) = Hm+1(z) + am−nHm−n(z)

and initial conditions H0(z) ≡ 1, H−1(z) ≡ H−2(z) ≡ · · · ≡ H−n(z) ≡ 0, see [5] and [8]. Ageneralization is given in [4].

References[1] T. Amdeberhan, A note on Fibonacci-type polynomials, Integers 10 (2010), 13–18.

[2] R. Boyer, W. M. Y. Goh, On the zero attractor of the Euler polynomials, Adv. in Appl. Math.38 (2007), no. 1, 97–132.

[3] R. Boyer, W. M. Y. Goh, Polynomials associated with partitions: asymptotics and zeros,Special functions and orthogonal polynomials, 33–45, Contemp. Math., 471, Amer. Math.Soc., Providence, RI, 2008.

[4] S. Delvaux, A. Lopez (2012), High order three-term recursions, Riemann-Hilbert minors andNikishin systems on star-like sets, arXiv:1202.4000v2 [math.CA]

[5] M. Eiermann, R.S. Varga, Zeros and local extreme points of Faber polynomials associated withhypocycloidal domains, ETNA 1 (1993), 49–71.

[6] M. X. He, E. B. Saff, The zeros of Faber polynomials for an m-cusped hypocycloid, J. Approx.Theory 78 (1994), no. 3, 410–432.

[7] M. E. H. Ismail, Difference equations and quantized discriminants for q-orthogonal polynomials.Advances in Applied Mathematics 30 (2003) 562–589.

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[8] N.B. Romdhane, On the zeros of d-orthogonal d-symmetric polynomials, J. Math. Anal. Appl.344 (2008), 888–897.

[9] A. Sokal, Chromatic roots are dense in the whole complex plane, Combin. Probab. Comput.13 (2004), no. 2, 221–261.

[10] K. Tran, Discriminants: calculation, properties, and connection to the root distribution ofpolynomials with rational generating functions, PhD thesis, University of Illinois, 2012.

[11] J. L. Walsh, A Generalization of Jensen’s Theorem on the Zeros of the Derivative of a Poly-nomial, Amer. Math. Monthly 62 (1955), 91–93.

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