Zeros of polynomials with four-term recurrence

0.2 MB·

by  Khang Tran

ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE
KHANGTRANANDANDRESZUMBA
7
1 Abstract. For any real numbers b,c ∈ R, we form the sequence of polynomials {Hm(z)}∞m=0
0 satisfyingthefour-termrecurrence
2 Hm(z)+cHm−1(z)+bHm−2(z)+zHm−3(z)=0, m≥3,
n withthe initialconditions H0(z)=1,H1(z)=−c,andH2(z)=−b+c2. Wefindnecessary and
a sufficientconditionsonbandcunderwhichthezerosofHm(z)arerealforallm,andprovidean
J ∞
explicitrealinterval onwhich [ Z(Hm)isdensewhereZ(Hm)isthesetofzeros ofHm(z).
6
m=0
2

]
V
C 1. Introduction
h. Consider the sequence of polynomials {Hm(z)}∞m=0 satisfying a finite recurrence
at n
m (1.1) ak(z)Hm k(z)=0, m n,
− ≥
[ kX=0
where ak(z), 1 k n, are complex polynomials. With certain initial conditions, one may ask
1 ≤ ≤
for the locations of the zeros of H (z) on the complex plane. There are two common approaches:
v m
asymptotic location and exact location. The first direction asks for a curve where the zeros of
4
1 Hm(z) approach as m , and the second direction aims at a curve where the zeros of Hm(z)
→ ∞
8 lie exactly on for all m (or at least for all large m). Recent works in the first direction include
7 [1, 2, 3, 5, 6]. Results in this first direction are useful to prove the necessary condition for the
0
reality of zeros of H (z) that we will see in Section 3.
. m
1 Itisnotaneasyproblemtofindanexplicitcurve(ifsuchexists)wherethezerosofallH (z)lie
m
0
forageneralrecurrencein(1.1). Forthree-termrecurrenceswithdegreetwoandappropriateinitial
7
conditions, the curve containing zeros is given in [10]. The corresponding curve for a three-term
1
recurrence with degree n is given in [11]. Among all possible curves containing the zeros of the
:
v
H (z)s, the real line plays an important role. We say that a polynomial is hyperbolic if all of its
i m
X zeros are real. There are a lot of recent works on hyperbolic polynomials and on linear operators
r preserving hyperbolicity of polynomials, see for example [4, 7]. In some cases, we could find the
a curveontheplanecontainingthezerosofasequenceofpolynomialsbyclaimingcertainpolynomials
are hyperbolic, see for example [8].
Themainresultofthispaperistheidentificationofnecessaryandsufficientconditionsonb,c R
∈
under which the zeros of the sequence of polynomials H (z) satisfying the recurrence
m
H (z)+cH (z)+bH (z)+zH (z)=0, m 3,
m m 1 m 2 m 3
− − − ≥
H (z) 1,
0
(1.2) ≡
H (z) c,
1
≡−
H (z) b+c2,
2
≡−
1
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 2
arereal. We usetheconventionthatthezerosoftheconstantzeropolynomialarereal. Let (H )
m
Z
denote the set of zeros of H (z).
m
Theorem 1. Suppose b,c ∈ R, and let {Hm(z)}∞m=0 be defined as in (1.2). The zeros of Hm(z)
are real for all m if and only if one of the two conditions below holds
(i) c=0 and b 0
≥
(ii) c=0 and 1 b/c2 1/3.
6 − ≤ ≤
∞
Moreover, in the first case with b > 0, (H ) is dense on ( , ). In the second case,
m
Z −∞ ∞
m=0
[
∞
(H ) is dense on the interval
m
Z
m=0
[
2+9b/c2 2 (1 3b/c2)3
c3. ,− − − .
−∞ 27p #
Ourpaper isorganizedasfollow. InSection2,weprovethe sufficientconditionforthe realityof
the zeros of all H (z) in the case c=0. The case c=0 follows from similar arguments whose key
m
6
differences will be mentioned in Section 3. Finally, in Section 4, we prove the necessary condition
for the reality of the zeros of H (z).
m
2. The case c=0 and 1 b/c2 1/3
6 − ≤ ≤
We write the sequence H (z) in (1.2) using its generating function
m
∞ 1
(2.1) H (z)tm = .
m 1+ct+bt2+zt3
m=0
X
From (2.1), if we make the substitutions t t/c, b/c2 a, and z/c3 z, it suffices to prove the
→ → →
following form of the theorem.
Theorem 2. Consider the sequence of polynomials {Hm(z)}∞m=0 generated by
∞ 1
(2.2) H (z)tm =
m 1+t+at2+zt3
m=0
X
where a R. If 1 a 1/3 then the zeros of H (z) lie on the real interval
m
∈ − ≤ ≤
2+9a 2 (1 3a)3
(2.3) I = ,− − −
a
−∞ 27p #
∞
and (H ) is dense on I .
m a
Z
m=0
[
We will see later that the density of the union of zeros on I follows naturally from the proof
a
that (H ) I andthuswefocusonprovingthisclaim. Wenotethateachvalueofa [ 1,1/3]
m a
Z ⊂ ∈ −
generates a sequence of polynomials H (z,a). The lemma below asserts that it suffices to prove
m
that (H (z,a)) I for all a in a dense subset of [ 1,1/3].
m a
Z ⊂ −
Lemma 3. Let S be a dense subset of [ 1,1/3], and let m N be fixed. If
− ∈
(H (z,a)) I
m a
Z ∈
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 3
for all a S then
∈
(Hm(z,a∗)) Ia∗
Z ∈
for all a [ 1,1/3].
∗
∈ −
Proof. Leta [ 1,1/3]be given. Bythe density ofS in[ 1,1/3],wecanfind asequence a in
∗ n
∈ − − { }
S such that an a∗. For any z∗ / Ia∗, we will show that Hm(z∗,a∗)=0. We note that the zeros
→ ∈ 6
of Hm(z,an) lie on the interval Ian whose right endpoint approaches the right endpoint of Ia∗ as
n . If we let z(n), 1 k degH (z,a ), be the zeros of H (z,a ) then
→∞ k ≤ ≤ m n m n
degHm(z,an)
H (z ,a ) =γ(n) z z(n)
| m ∗ n | ∗− k
kY=1 (cid:12) (cid:12)
(cid:12) (cid:12)
whereγ(n) isthe leadcoefficientofH (z,a ). We willseein(cid:12) the next(cid:12)lemma thatdegH (z,a )is
m n m n
at most m/3 . From this finite product and the assumption that z / I , we conclude that there
∗ a
⌊ ⌋ ∈
is a fixed (independent of n) δ > 0 so that H (z ,a ) > δ, for all large n. Since H (z ,a) is a
m ∗ n m ∗
| |
polynomial in a for any fixed z , we conclude that
∗
H (z ,a )= lim H (z ,a )=0
m ∗ ∗ m ∗ n
n 6
→∞
and the lemma follows. (cid:3)
Lemma 3 allows us to ignore some special values of a. In particular, we may assume a = 0. In
6
our main approach, we count the number of zeros of H (z) on the interval I in (2.3) and show
m a
that this number of zeros is at least the degree of H (z). To count the number of zeros of H (z)
m m
on I , we write z = z(θ) as a strictly increasing function of a variable θ on the interval (2π/3,π).
a
Then we construct a function g (θ) on (2π/3,π) with the property that θ is a zero of g (θ) on
m m
(2π/3,π)if andonlyif z(θ)is a zeroofH (z) onI . Fromthis construction,we countthe number
m a
of zeros of g (θ) on (2π/3,π) which will be the same as the number of zeros of H (z) on I by
m m a
the monotonicityofthe function z(θ). Wefirstobtainanupperboundforthe degreeof H (z)and
m
provide heuristic arguments for the formulas of z(θ) and g (θ).
m
Lemma 4. The degree of the polynomial H (z) defined by (2.2) is at most m/3 .
m
⌊ ⌋
Proof. We rewrite (2.2) as
∞
(1+t+at2+zt3) H (z)tm =1.
m
m=0
X
By equating the coefficients in t of both sides, we see that the sequence {Hm(z)}∞m=0 satisfies the
recurrence
H (z)+H (z)+aH (z)+zH (z)=0
m+3 m+2 m+1 m
and the initial condition H (z) 1, H (z) 1, and H (z) 1 a. The lemma follows from
0 1 2
induction. ≡ ≡ − ≡ − (cid:3)
2.1. Heuristicarguments. Wenowprovideheuristicargumentstomotivatetheformulasfortwo
special functions z(θ) and g (θ) on (2π/3,π). Let t = t (z), t = t (z), and t = t (z) be the
m 0 0 1 1 2 2
three zeros of the denominator 1+t+at2+zt3. We will show rigorouslyin Section 2.2 that t , t ,
0 1
t are nonzero and distinct with t =t . We let q =t /t =e2iθ, θ =0,π. We have
2 0 1 1 0
6
∞ 1 1
H (z)tm = = .
m 1+t+at2+zt3 z(t t )(t t )(t t )
0 1 2
m=0 − − −
X
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 4
We apply partial fractions to rewrite the generating functions 1/z(t t )(t t )(t t ) as
0 1 2
− − −
1 1 1
+ +
z(t t )(t t )(t t ) z(t t )(t t )(t t ) z(t t )(t t )(t t )
0 0 1 0 2 1 1 0 1 2 2 2 0 2 1
− − − − − − − − −
which can be expanded as a series in t below
∞ 1 1 1 1
(2.4) + + tm.
− z (t t )(t t )tm+1 (t t )(t t )tm+1 (t t )(t t )tm+1
mX=0 (cid:18) 0− 1 0− 2 0 1− 0 1− 2 1 2− 0 2− 1 2 (cid:19)
From this expression, we deduce that z is a zero of H (z) if and only if
m
1 1 1
(2.5) + + =0.
(t t )(t t )tm+1 (t t )(t t )tm+1 (t t )(t t )tm+1
0− 1 0− 2 0 1− 0 1− 2 1 2− 0 2− 1 2
After multiplying the left side of (2.5) by tm+3 we obtain the equality
0
1 1 1
+ + =0.
(1 t /t )(1 t /t ) (t /t 1)(t /t t /t )(t /t )m+1 (t /t 1)(t /t t /t )(t /t )m+1
1 0 2 0 1 0 1 0 2 0 1 0 2 0 2 0 1 0 2 0
− − − − − −
Setting ζ =t /t eiθ, we rewrite the left side as
2 0
1 1 1
+ + ,
(1 e2iθ)(1 ζeiθ) (e2iθ 1)(e2iθ ζeiθ)(e2iθ)m+1 (ζeiθ 1)(ζeiθ e2iθ)(ζeiθ)m+1
− − − − − −
or equivalently
1 1 1
+ + .
e2iθ( 2isinθ)(e iθ ζ) (2isinθ)(eiθ ζ)(e2iθ)m+2 (ζ e iθ)(ζ eiθ)(ζ)m+1(eiθ)m+3
− − − − − − −
We multiply this expressionby (ζ e iθ)(ζ eiθ)ei(m+3)θ and set the summation to 0 and rewrite
−
− −
(2.5) as
(ζ eiθ)ei(m+1)θ e iθ ζ 1
−
0= − + − +
2isinθ (2isinθ)ei(m+1)θ ζm+1
(ζ eiθ)ei(m+1)θ (ζ e iθ)e i(m+1)θ 1
− −
= − − − +
2isinθ ζm+1
ζ(ei(m+1)θ e i(m+1)θ)+e i(m+2)θ ei(m+2)θ 1
− −
= − − +
2isinθ ζm+1
ζ(2isin(m+1)θ) 2isin(m+2)θ 1
= − +
2isinθ ζm+1
2iζsin(m+1)θ 2isin(m+1)θcosθ 2icos(m+1)θsinθ 1
= − − +
2isinθ ζm+1
(ζ cosθ)sin(m+1)θ 1
(2.6) = − cos(m+1)θ+ .
sinθ − ζm+1
The last expression will serve as the definition of g (θ).
m
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 5
We next provide a motivation for the specific form of z(θ). Since t , t , and t are the zeros of
0 1 2
D(t,z)=1+t+at2+zt3, they satisfy the three identities
a
t +t +t = ,
0 1 2
−z
1
t t +t t +t t = , and
0 1 0 2 1 2
z
1
t t t = .
0 1 2
−z
If we divide the first equation by t , the second by t2, and the third by t3 then these identities
0 0 0
become
a
(2.7) 1+e2iθ+ζeiθ = ,
−zt
0
1
(2.8) e2iθ+ζeiθ +ζe3iθ = , and
zt2
0
1
(2.9) ζe3iθ = .
−zt3
0
We next divide the first identity by second, and the second by the third to obtain
1+e2iθ+ζeiθ
= at , and
e2iθ+ζeiθ +ζe3iθ − 0
e2iθ+ζeiθ +ζe3iθ
= t ,
ζe3iθ − 0
from which we deduce that
(1+e2iθ +ζeiθ)ζe3iθ =a(e2iθ+ζeiθ +ζe3iθ)2.
This equation is equivalent to
(e−iθ +eiθ+ζ)ζe4iθ =ae4iθ(1+ζe−iθ +ζeiθ)2,
or simply
(2cosθ+ζ)ζ =a(1+2ζcosθ)2.
Lemma 5. For any a [ 1,1/3] and θ (2π/3,π), the zeros in ζ of the polynomial
∈ − ∈
(2.10) (2cosθ+ζ)ζ a(1+2ζcosθ)2
−
are real and distinct.
Proof. We consider the discriminant of the above polynomial in ζ:
∆=(1 4a)cos2θ+a.
−
There are three possible cases depending on the value of a. If 1/4 a 1/3, the inequality ∆>0
≤ ≤
comes directly from a 4a 1 > (1 4a)cos2θ. If 0 a < 1/4, the claim ∆ > 0 is trivial since
≥ − − ≤
1 4a>0. Finally, if a<0, we have ∆ (1 4a)/4+a=1/4. It follows that the zeros of (2.10)
ar−e real and distinct for any a [ 1,1/3≥] and−θ (2π/3,π). (cid:3)
∈ − ∈
To obtain the formula z(θ), we multiply both sides of (2.7) and (2.8)
a
(1+e2iθ +ζeiθ)(e2iθ +ζeiθ+ζe3iθ)=
−z2t3
0
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 6
and divide by (2.9) to arrive at the resulting equation
ae3iθζ
z =
(1+e2iθ+ζeiθ)(e2iθ+ζeiθ +ζe3iθ)
ae3iθζ
=
e3iθ(e iθ+eiθ+ζ)(1+ζe iθ +ζeiθ)
− −
aζ
(2.11) = .
(2cosθ+ζ)(1+2ζcosθ)
2.2. Rigorous proof. Motivated by Section 2.1, we will rigorously prove Theorem (2) in this
Section 2.2. We start our proof of Theorem (2) by defining the function ζ(θ) according to (2.10):
(2a 1)cosθ+ (1 4a)cos2θ+a
(2.12) ζ =ζ(θ)= − −
1 4acos2θ
−p
which, from Lemma 5, is a real function on (2π/3,π) with a possible vertical asymptote at
1
(2.13) θ =cos 1
−
−2√a
(cid:18) (cid:19)
when 1/4 < a 1/3. However we note that the function 1/ζ(θ) is a real continuous function on
≤
(2π/3,π).
Lemma 6. Let ζ(θ) be defined as in (2.12). Then ζ(θ) > 1 for every a ( 1,1/3) and every
| | ∈ −
θ (2π/3,π) with 1 4acos2θ =0.
∈ − 6
Proof. From (2.10), we note that ζ :=ζ(θ) and
+
(2a 1)cosθ (1 4a)cos2θ+a
ζ := − − −
− 1−p4acos2θ
are the zeros of
f(ζ):=(2cosθ+ζ)ζ a(1+2ζcosθ)2.
−
Note that
f( 1)f(1)=( 1+2cosθ)(1+2cosθ)(4a2cos2θ (a 1)2).
− − − −
If θ (2π/3,π) and a ( 1,1/3), this product is negative since
∈ ∈ −
4a2cos2θ (a 1)2 4a2 (a 1)2 =(a+1)(3a 1)<0.
− − ≤ − − −
Thus exactly one of the zeros of the quadratic function f(ζ) lies outside the interval [ 1,1]. The
claim follows from the fact that ζ > ζ . − (cid:3)
+
| | | −|
Although one can prove Lemma 6 for the extreme value a = 1 or a = 1/3, that will not be
−
necessary by Lemma 3. Next we define the real function z(θ) by
aζ
(2.14) z =z(θ):=
(2cosθ+ζ)(1+2ζcosθ)
on (2π/3,π). From Lemma 6, 1+2ζcosθ = 0 and so does 2cosθ +ζ by (2.10). Dividing the
6
numerator and the denominator by ζ2(θ) and combining with the fact that 1/ζ(θ) is continuous
on (2π/3,π), we conclude that the possible discontinuity of z(θ) in (2.13) is removable. Finally,
motivated by (2.6), we define the function g (θ) by
m
(ζ cosθ)sin(m+1)θ 1
(2.15) g (θ):= − cos(m+1)θ+
m sinθ − ζm+1
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 7
which has the same vertical asymptote as that of ζ(θ) in (2.13) when 1/4<a 1/3.
≤
From Lemma 6, we see that the sign of the function g (θ) alternates at values of θ where
m
cos(m+1)θ = 1. Thus by the Intermediate Value Theorem the function g (θ) has at least one
m
±
root on each subinterval whose endpoints are the solution of cos(m+1) = 1. However, in the
±
case 1/4 a 1/3, one of the subintervals will contain a vertical asymptote given in (2.13). The
≤ ≤
lemma below counts the number of zeros of g (θ) on such a subinterval.
m
Lemma 7. Let g (θ) be defined as in (2.15). Suppose 1/4 < a 1/3 and m 6. Then there
m
exists h N such that ≤ ≥
∈
h 1 1 h
θh 1 := − π <cos−1 π =:θh
− m+1 −2√a ≤ m+1
(cid:18) (cid:19)
where 2(m+1)/3 +1 h 1<h m+1. Furthermore, as long as
⌊ ⌋ ≤ − ≤
1 h
(2.16) cos−1 = π,
−2√a 6 m+1
(cid:18) (cid:19)
the function g(θ) has at least two zeros on the interval
h 1 h
(2.17) θ − π, π :=J
h
∈ m+1 m+1
(cid:18) (cid:19)
whenever h is at most m, and at least one zero when h is m+1.
Proof. Suppose a [1/4,1/3]. Since the function cos 1( 1/2√x) is decreasing on the interval
−
∈ −
[1/4,1/3],we conclude that
1 5π
cos 1 .
−
−2√a ≥ 6
(cid:18) (cid:19)
Then the existence of h comes directly from the inequality
2(m+1)/3 +1 5π
⌊ ⌋ π <
m+1 6
when m 6.
≥
The vertical asympote at cos 1( 1/2√a) of g (θ) divides the interval J in (2.17) into two
− m h
−
subintervals. We will show that each subinterval contains at least a zero of g (θ) if h m. In the
m
≤
case h = m+1, only the subinterval on the left contains at least zero of g (θ). We analyze these
m
two subintervals in the two cases below.
We consider the first case when θ I and θ < cos 1( 1/2√a). By Lemma 6 and (2.15) we
h −
∈ −
see that the sign of g (θ ) is ( 1)h. We now show that the sign of g (θ) is ( 1)h 1 when
m h 1 m −
θ cos 1( 1/2√a). From− (2.12)−, we observe that ζ(θ) + as θ cos 1( 1/−2√a). Since
− −
→ − → ∞ → −
θ I , the sign of sin(m+1)θ is ( 1)h 1 and consequently the sign of g (θ) is ( 1)h 1 when
h − m −
∈ − −
θ cos 1( 1/2√a) by (2.15). By the Intermediate Value Theorem, we obtainat least one zero of
−
→ −
g (θ) in this case.
m
Next we consider the case when θ I and θ >cos 1( 1/2√a). In this case the sign of g (θ )
h − m h
∈ −
is ( 1)h 1 if h m by Lemma 6. Since ζ(θ) as θ cos 1( 1/2√a) and the sign of
− −
− ≤ → −∞ → −
sin(m+1)θ is ( 1)h 1, the sign of g (θ) is ( 1)h as θ cos 1( 1/2√a). By the Intermediate
− m −
Value Theorem,−we obtain at least one zero of−g (θ) in th→is case an−d h m. (cid:3)
m
≤
Note that we may assume (2.16) by Lemma3. From the fact that the sign of g (θ) in (2.15)
m
alternates when cos(m+1)θ = 1, we can find a lower bound for the number of zeros of g (θ)
m
±
on (2π/3,π) by the Intermediate Value Theorem. We will relate the zeros of g (θ) to the zeros of
m
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 8
H (z) by (2.6). Howeverto ensure that the partialfractions procedure preceding equation(2.6) is
m
rigorous,we need the lemma below.
Lemma 8. Let θ (0,π) such that θ = cos 1( 1/2√a) whenever a > 1/4. The zeros in t of
−
∈ 6 −
1+t+at2+z(θ)t3 are
e2iθ+ζeiθ +ζe3iθ
t = , t =t e2iθ, t /t =ζeiθ
0 − ζe3iθ 1 0 2 0
where ζ :=ζ(θ) is given in (2.12).
Proof. We first note that
P(t )=1+t +at2+zt3
0 0 0 0
= 1 e 2iθ+ a 1+ζe iθ+ζeiθ 2 z 1+ζe iθ +ζeiθ 3.
− − −
−ζeiθ − ζ2e2iθ − ζ3e3iθ
(cid:0) (cid:1) (cid:0) (cid:1)
where ζ is a root of the quadratic equation (2cosθ +ζ)ζ a(1+2ζcosθ)2 = 0. We apply the
−
following identities
1 1
(1+ζe iθ+ζeiθ)2 =(1+2ζcosθ)2 = (2cosθ+ζ)ζ = (e iθ+eiθ+ζ)ζ
− −
a a
and
aζ ζ2 ζ2
(2.18) z = = =
(2cosθ+ζ)(1+2ζcosθ) (1+2ζcosθ)3 (1+ζe iθ +ζeiθ)3
−
to conclude that P(t )=0. Similarly, we have
0
P(t )=1+t e2iθ+at2e4iθ+zt3e6iθ
1 0 0 0
= eiθ e2iθ+ ae2iθ 1+ζe−iθ+ζeiθ 2 ze3iθ 1+ζe−iθ +ζeiθ 3
− ζ − ζ2 − ζ3
eiθ ae2iθ ((cid:0)e iθ+eiθ+ζ)ζ (cid:1) e3iθ (cid:0) (cid:1)
= e2iθ+ − ζ2 =0.
− ζ − ζ2 a − ζ3
Finally
P(t )=P(ζt eiθ)
2 0
= ζe iθ ζeiθ +a 1+ζe iθ+ζeiθ 2 z 1+ζe iθ +ζeiθ 3
− − −
− − −
1
= ζe iθ ζeiθ +a(cid:0) e iθ +eiθ+ζ (cid:1)ζ ζ2(cid:0)=0. (cid:1)
− −
− − a −
(cid:0) (cid:1) (cid:3)
As a consequence of Lemma 8, if θ (2π/3,π), then the zeros of 1+t+at2 +z(θ)t3 will be
distinct and t = t since ζ R by Le∈mma 5. Thus we can apply partial fractions given in the
1 0
∈
beginning of Section 2.1. From this partial fraction decomposition, we conclude that if θ is a zero
of g (θ), then z(θ) will be a zero of H (z). In fact, we claim that each distinct zero of g (θ) on
m m m
(2π/3,π)producesa distinctzeroofH (z)onI . This is the contentofthe followingtwolemmas.
m a
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 9
Lemma 9. Let ζ(θ) be defined as in (2.12). The function z(θ) defined as in (2.14) is increasing
on θ (2π/3,π).
∈
Proof. Lemma 8 gives
1+t +at2 1+t +at2
z = 0 0 = 1 1.
− t3 t3
0 1
We differentiate the three terms and obtain
3+2t +at2 3+2t +at2
(2.19) dz = 0 0dt = 1 1dt ,
t4 0 t4 1
0 1
where
dt =d(t e2iθ)=e2iθdt +2it e2iθdθ.
1 0 0 0
If we set
3+2t +at2 3+2t +at2
f(t )= 0 0, f(t )= 1 1
0 t4 1 t4
0 1
then f(t )=f(t ), and consequently f(t )f(t ) 0. Thus (2.19) implies that
0 1 0 1
≥
f(t )dt =f(t )(e2iθdt +2it e2iθdθ).
0 0 1 0 0
After solving this equation for dt and substituting it into (2.19), we obtain
0
dz 2if(t )f(t )t e2iθ
0 1 0
(2.20) = .
dθ f(t ) f(t )e2iθ
0 1
−
With t =τe iθ, τ R, we have
0 −
∈
f(t ) f(t )e2iθ f(t )e iθ f(t )eiθ
0 1 0 − 1
− = −
2it e2iθ 2it eiθ
0 0
f(t )e iθ
0 −
= ℑ
τ
(cid:0) (cid:1)
1 3+2t +at2
= 0 0e iθ .
τℑ t4 −
(cid:18) 0 (cid:19)
We now substitute 3= 3t 3at2 3zt3 and have
− 0− 0− 0
f(t ) f(t )e2iθ 1 t 2at2 3zt3
02−it e2i1θ = τℑ − 0− t40− 0e−iθ
0 (cid:18) 0 (cid:19)
1
= e2iθ 2aτeiθ 3zτ2
τ4ℑ − − −
1 (cid:0) (cid:1)
= ( sin2θ 2aτsinθ)
τ4 − −
2sinθ
= ( cosθ aτ).
τ4 − −
In the formula of t in Lemma 8, we substitute τ = 1/ζ 2cosθ and obtain
0
− −
f(t ) f(t )e2iθ 2sinθ
0 1
− = ( cosθ+a/ζ+2acosθ).
2it e2iθ τ4 −
0
ZEROS OF POLYNOMIALS WITH FOUR-TERM RECURRENCE 10
We finish this lemma by showingthat cosθ+a/ζ+2acosθ >0, which implies f(t )=f(t )=0
0 1
− 6
and and the lemma will follow from (2.20). We expand and divide both sides of (2.10) by ζ to get
ζ(1 4acos2θ)+2cosθ(1 2a) a/ζ =0,
− − −
or equivalently,
ζ(1 4acos2θ)+cosθ(1 2a)= cosθ+2acosθ+a/ζ.
− − −
Finally, using the definition of ζ in (2.12) and Lemma 5, we note that
ζ(1 4acos2θ)+cosθ(1 2a)= (1 4a)cos2θ+a>0.
− − −
The proof is complete. p (cid:3)
Lemma 10. The function z(θ) as defined in (2.14) maps the interval (2π/3,π) onto the interior
of I .
a
Proof. Since z(θ) is a continuous increasing function on (2π/3,π), we only need to evaluate the
limits of z(θ) at the endpoints. Since ζ > 1 by Lemma 6, the formula of ζ(θ) in (2.12) implies
| |
that ζ(θ) 1+ as θ (2π/3)+. Consequently, (2.18) gives
→ →
lim z(θ)= .
θ (2π/3)+ −∞
→
Finally, from the fact that
1 2a+√1 3a
lim ζ(θ)= − − ,
θ π 1 4a
→ −
and (2.14), we have
a(1 2a+√1 3a)(1 4a)
lim z(θ)= − − −
θ π ( 1+6a+√1 3a)( 1 2√1 3a)
→ − − − − −
a( 1+4a)2( 2+9a)+2a( 1+3a)( 1+4a)2√1 3a
(2.21) = − − − − −
27(1 4a)2a
−
2+9a 2 (1 3a)3
= − − − ,
27
p
whereweobtain(2.21)bymultiplyanddividethefractionby 1+6a √1 3a 1+2√1 3a .
− − − − − (cid:3)
(cid:0) (cid:1)(cid:0) (cid:1)
Before making final arguments to connect all results in this section, we check the sign of g (θ)
m
at one of the endpoints.
Lemma 11. If 1 a<1/4 then the sign of g (π ) is ( 1)m.
m −
− ≤ −
Proof. Since 1 a<1/4, one can check that
− ≤
1 2a+√1 3a
lim ζ(θ)= − − 1.
θ π− 1 4a ≥
→ −
The result then follows directly from (2.15) and the fact that
sin(m+1)θ
lim =(m+1)( 1)m.
θ π− sin(θ) −
→
(cid:3)