The zero distribution of polynomials with a three

```The zero distribution of polynomials with a
three-term recurrence
Khang Tran
Truman State University
October 20, 2014
1 / 24
An objective
We study the zero distribution of a sequence of polynomials
Hm (z) generated by
∞
1
∑ Hm (z)t m = 1 + B(z)t + A(z)t n
m=0
where A(z) and B(z) are polynomials with complex
coefficients.
For example, when B(z) = z 3 + i, A(z) = z 2 + 1, and n = 5
the zeros of H100 (z) are plotted below.
1.0
0.5
-1.0
-0.5
0.5
1.0
-0.5
-1.0
2 / 24
Main theorem
Theorem
Let Hm (z) be a sequence of polynomials whose generating function
is
∞
1
∑ Hm (z)t m = 1 + B(z)t + A(z)t n
m=0
where A(z) and B(z) are polynomials in z with complex
coefficients. There is a constant C = C (n) such that for all m > C ,
the zeros of Hm (z) which satisfy A(z) 6= 0 lie on a fixed curve C
given by
ℑ
B n (z)
=0
A(z)
and
0 ≤ (−1)n ℜ
B n (z)
nn
≤
A(z)
(n − 1)n−1
and are dense there as m → ∞.
3 / 24
An example
Consider the sequence of polynomials, Hm (z), generated by
∞
1
∑ Hm (z)t m = 1 + (z 2 − 2z + 3)t + z 2 t 2 .
m=0
The zeros of H50 (z) are plotted below.
Figure: The zeros of H50 (z)
4 / 24
An example
We recall that
∞
1
∑ Hm (z)t m = 1 + (z 2 − 2z + 3)t + z 2 t 2 .
m=0
Let z = x + iy . We note that
B 2 (z)
A(z)
B 2 (z)
ℜ
A(z)
ℑ
=
=
2y (x 2 + y 2 − 3)P
(x 2 + y 2 )2
P2 − Q2
(x 2 + y 2 )2
where
P = 3x − 2x 2 + x 3 − 2y 2 + xy 2
Q = y (x 2 + y 2 − 3).
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Some ideas
We will show that the number of zeros of Hm (z) on the curve
C is at least the degree of Hm (z). Hence this number equals
the degree and all the zeros of Hm (z) lie on this curve.
Let z be a zero of Hm (z) and ti = ti (z), 0 ≤ i < n, be the
zeros of D(t, z) = 1 + B(z)t + A(z)t n . We will consider the
quotients of zeros ti /tj since these quotients seem to lie on a
fixed curve independent of A(z), B(z).
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Distribution of quotients of zeros
Figure: Quotients of zeros when n = 3
7 / 24
Distribution of quotients of zeros
Figure: Quotients of zeros when n = 4
8 / 24
Distribution of quotients of zeros
1.5
1.0
0.5
-1.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
-1.5
Figure: Quotients of zeros when n = 5, m = 50
9 / 24
Distribution of quotients of zeros
1.0
0.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
Figure: Quotients of zeros when n = 6, m = 30
10 / 24
The q-discriminant
Definition (Ismail)
The q-discriminant, Discx (P; q), of a polynomial Pn (x) of degree n
and leading coefficient p is defined as
p 2n−2 q n(n−1)/2
∏
(q −1/2 xi − q 1/2 xj )(q 1/2 xi − q −1/2 xj )
1≤i<j≤n
where x1 , . . . , xn are the zeros of Pn (x).
11 / 24
The q-disriminant of the denominator
Proposition
The q-discriminant of D(t) = 1 + Bt + At n is
n n−1
n n
(1 − q n−1 )n−1
n−2 B q
n−1 A(1 − q )
Disct (D(t); q) = ±A
+ (−1)
.
(1 − q)n−1
(1 − q)n
Corollary
If q is a quotient of zeros ti /tj of D(t, z) then
(−1)n
B n (z)
(1 − q n )n
=
.
A(z)
(1 − q)q n−1 (1 − q n−1 )n−1
12 / 24
Our problem
We want to show the number of zeros of Hm (z) on C is at
least the degree of Hm (z) which is
(
mb
if nb > a
pa + rb if nb ≤ a
where m = pn + r , 0 ≤ r < n, and a, b are the degrees of A(z)
and B(z) respectively.
In fact, we will count the number of quotients of zeros
q = ti /tj of D(t, z) on the arc e 2iθ , 0 ≤ θ ≤ π/n. This relates
to the number of zeros of Hm (z) on C by
(−1)n B n (z)(1 − q)q n−1 (1 − q n−1 )n−1 = (1 − q n )n A(z).
It is not difficult to show the special point θ = π/n gives rb
zeros z of Hm (z) on C .
It remains to show the number of quotients q on the arc e 2iθ ,
0 ≤ θ < π/n is p = bm/nc.
13 / 24
Distribution of quotients of zeros
1.5
1.0
0.5
-1.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
-1.5
Figure: Quotients of zeros when n = 5, m = 50
14 / 24
Distribution of quotients of zeros
1.0
0.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
Figure: Quotients of zeros when n = 6, m = 30
15 / 24
A lemma
Lemma
For a fixed z, let t0 , . . . , tn−1 be the zeros in t of
D(t, z) = 1 + B(z)t + A(z)t n and qi = ti /t0 , 0 ≤ i < n. If z is a
point on C such that q1 = e 2iθ , θ ∈ R then z is a zero of Hm (z) if
and only if
n−1
1
h(θ ) := ∑ m+1 0
=0
Pθ (ζk )
k=0 ζk
where ζ0 , . . . , ζn−1 are the zeros of
Pθ (ζ ) = ζ n −
sin nθ
sin(n − 1)θ
ζ+
.
sin θ
sin θ
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A lemma
Lemma
If 0 ≤ θ < π/n then, besides the two zeros e ±iθ , the remaining
n − 2 zeros of the polynomial
Pθ (ζ ) = ζ n −
sin nθ
sin(n − 1)θ
ζ+
sin θ
sin θ
lie outside the closed unit disk, i.e., |ζ | > 1.
17 / 24
Consequences of the two lemmas
Recall that we want to count the number of zeros θ on
[0, π/n) of
n−1
1
h(θ ) := ∑ m+1 0
=0
Pθ (ζk )
k=0 ζk
where ζ0 , . . . , ζn−1 are the zeros of
Pθ (ζ ) = ζ n −
sin(n − 1)θ
sin nθ
ζ+
.
sin θ
sin θ
Since ζ0,1 = e ±iθ and |ζk | > 1 if 2 ≤ k < n, in general the
function h(θ ) is dominated by the two summands when
ζk = e ±iθ .
18 / 24
Consequences of the two lemmas
The sum of these two terms is
2n cos(m + n)θ − 2 cos(m + 1)θ sin nθ / sin θ
.
|Pθ0 (e iθ )|2
The graph of this sum when m = 40, n = 4 is
4
2
0.2
0.4
0.6
0.8
-2
-4
Figure: The sum of two main terms
19 / 24
A technical case
When θ is close to π/n, say θ = π/n − l/m, we have
2
l
lπ(cos π/n − ek )
+ On
ζk = ek 1 +
m sin π/n
m2
where ek = e (2k−1)iπ/n , 0 ≤ k ≤ n − 1, and the first two terms
in
n−1
1
h(θ ) = ∑ m+1 0
Pθ (ζk )
k=0 ζk
sum of the two terms when ζk = e ±iθ will not dominate sum
when m is big.
In this case the sign of h(θ ) is the same as that of
1 n−1 −m−n lnek 1 n−1 −m−n ∞ (lnek )j
∑ ek e = n ∑ ek ∑ j!
n k=0
j=0
k=0
20 / 24
A sketch of the proof
Let m = nq + r . With the fact that
(
0
a
∑ ek = n(−1)a/n
k
if n - a
if n|a
the double summation becomes
∞
(ln)jn+r
∑ (−1)j+p+1 (jn + r )! .
j=0
This is an alternating series whose sign is the sign of the term
when j = h where l = h + r /n.
21 / 24
Future development
If the generating function is
∞
∑
m=0
Hm (z)t m =
∑ Ck (z)t k
1 + B(z)t + A(z)t n
then the zeros of Hm (z) will approach the same curve C and a
finite set of points when m → ∞ (a joint work with Robert
Boyer).
An exprimental work suggests that if Ck (z)/B k (z)C0 (z) is real
whenever z ∈ C then the zeros of Hm (z) lie exactly on C
(except a finite set of points). It is a necessary and sufficient
condition when n = 2.
The author believes the same argument will work when the
generating function is 1/(1 + B(z)t l + A(z)t n ).
When the generating function is 1/ ∑ Ak (z)t k , can we find a
condition on Ak (z) so that the zeros of Hm (z) lie on an
explicit curve?
22 / 24
Future development
For example when the generating function is
1/(1 + C (z)t + B(z)t 2 + A(z)t 3 ), an experimental work
suggests that if B(z)C (z)/A(z) ≡ K ∈ R, K ≤ 0 or K ≥ 9,
then the zeros lie on the curve given by
3
3
B (z) C 3 (z)
B (z) C 3 (z)
ℑ
+
= 0,
ℜ
−
≥ 0,
A2 (z)
A(z)
A2 (z)
A(z)
and
ℜ
B 3 (z) C 3 (z)
+
A2 (z)
A(z)
≥
K 2 + 18K − 27
.
4
If A(z), B(z), and C (z) are any polynomials with complex
coefficients then an experimental work suggests that the zeros
approach a portion of a curve given defined by an explicit
equation involving the imaginary and the real parts of
B 3 (z)/A2 (z), C 3 (z)/A(z), B(z)C (z)/A(z). Can we find this
explicit portion? (i.e. can we find the inequality conditions?)
23 / 24
Thank you
Thank you.
24 / 24
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