On Power Series Having Sections with Only Real Zeros

Abstract

In this paper we investigate the class A * of power series with non-negative coefficients such that all but a finite number of its sections have only real zeros. We obtain some new necessary conditions for a power series to belong to A * . The main result of the paper is the complete answer to the question: for which a does the function g a (z) := ∞ k=0 a −k 2 z k , a > 1, have only real zeros. Keywords. Zeros of entire functions, polynomials with real zeros, totally positive sequences, sections of power series. 2000 MSC. 30C15, 30D15.

Full text

Computational Methods and Function Theory
Volume 3 (2003), No. 2, 425–441
On Power Series Having Sections With Only Real Zeros
Olga M. Katkova, Tetyana Lobova and Anna M. Vishnyakova
(Communicated by Stephan Ruscheweyh)
Abstract. In this paper we investigate the class A
∗
of power series with non-
negative coefficients such that all but a finite number of its sections have only
real zeros. We obtain some new necessary conditions for a power series to
belong to A
∗
. The main result of the paper is the complete answer to the
question: for which a does the function g
a
(z) :=
P
∞
k=0
a
−k
2
z
k
, a > 1, have
only real zeros.
Keywords. Zeros of entire functions, polynomials with real zeros, totally
positive sequences, sections of power series.
2000 MSC. 30C15, 30D15.
1. Introduction and statement of results
There are many papers concerning the zero distribution of sections (and tails) of
power series, see for example a very interesting survey of the topic in [9]. In this
paper we investigate power series with non-negative coefficients having sections
with only real zeros.
For a polynomial with non-negative coefficients there are two equivalent proper-
ties: to have only real zeros and to be the generating function of a totally positive
sequence. Multiply positive sequences (also called P´olya frequency sequences)
were introduced by Fekete in 1912 (see [4]) in connection with the problem of
exact calculation of the number of positive zeros of a real polynomial.
The class of all totally positive sequences is denoted by P F
∞
and consists of the
sequences {a
k
}
∞
k=0
such that all minors of the infinite matrix


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


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a
0
a
1
a
2
a
3
. . .
0 a
0
a
1
a
2
. . .
0 0 a
0
a
1
. . .
0 0 0 a
0
. . .
.
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.
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Received October 7, 2003, in revised form December 12, 2003.
ISSN 1617-9447/$ 2.50
c
 2003 Heldermann Verlag

426 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
are non-negative. The corresponding class of generating functions
f(z) =
∞
X
k=0
a
k
z
k
is also denoted by P F
∞
. The class P F
∞
was completely described by Aissen,
Schoenberg, Whitney and Edrei in [1] (see also [8, p. 412]).
Theorem ASWE. A function f ∈ P F
∞
if and only if
f(z) = Cz
n
e
γz
∞
Y
k=1
1 + α
k
z
1 −β
k
z
,
where C ≥ 0, n ∈ Z, γ, α
k
, β
k
≥ 0 and
P
∞
k=1
(α
k
+ β
k
) < ∞.
The above theorem implies that a polynomial p(z) =
P
n
k=0
a
k
z
k
, a
k
≥ 0, has only
real zeros if and only if the sequence (a
0
, a
1
, . . . , a
n
, 0, 0, . . .) belongs to P F
∞
.
By R
∗
we denote the set of real polynomials having only real zeros. The following
fact was mentioned by P´olya in [11].
Theorem A. Let P (z) = a
0
+ a
1
z + ···+ a
n
z
n
∈ R
∗
, a
j
> 0, j = 0, 1, . . . , n and
n ≥ 2. Then
a
n−1
a
n
≥
2n
n − 1
a
n−2
a
n−1
.
Let
(1)
∞
X
k=0
a
k
z
k
, a
k
> 0 for k ∈ N ∪ {0},
be a formal power series and let
(2) S
n
(z) :=
n
X
k=0
a
k
z
k
, n ∈ N ∪ {0}
be its sections. The following theorem is a corollary to Theorem A.
Theorem B. Let the formal power series (1) have the property that there exists
N ∈ N such that S
n
(z) ∈ R
∗
for all n ≥ N. Then this series is absolutely
converging in C, i.e. its sum is an entire function.
We shall consider the following two classes of entire functions.
A
∗
:=
(
f(z) =
∞
X
k=0
a
k
z
k
: a
k
> 0 for k ∈ N ∪ {0},
S
n
(z) ∈ R
∗
for all n ≥ N(f) for some N(f) ∈ N
)
,
B
∗
:=
(
f(z) =
∞
X
k=0
a
k
z
k
: a
k
> 0 for k ∈ N ∪ {0}, S
n
(z) ∈ R
∗
for all n ∈ N
)
.

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 427
Obviously we have B
∗
⊂ A
∗
. Let
p
n
= p
n
(f) :=
a
n−1
a
n
, n ≥ 1,
q
n
= q
n
(f) :=
p
n
p
n−1
=
a
n−1
2
a
n−2
a
n
, n ≥ 2.
Note that
(3)
a
n
=
a
0
p
1
p
2
···p
n
, n ≥ 1,
a
n
=
a
1
q
2
n−1
q
3
n−2
···q
n−1
2
q
n

a
1
a
0

n−1
, n ≥ 2.
Using the above notation and Theorem A we have
(4) f(z) =
∞
X
k=0
a
k
z
k
∈ A
∗
⇒ q
n
(f) ≥ 2 for n ≥ N(f) − 1.
In 1926, Hutchinson [7, p. 327] extended the work of Petrovitch [10] and Hardy
(cf. [5] or [6, pp. 95–100]) and proved the following theorem.
Theorem H. Let f(z) =
P
∞
k=0
a
k
z
k
, a
k
> 0 for k. Then q
n
(f) ≥ 4 for all n ≥ 2
if and only if the following two conditions are fulfilled:
(i) the zeros of f(x) are all real, simple and negative, and
(ii) the zeros of any polynomial
P
n
k=m
a
k
z
k
, formed by taking any number of
consecutive terms of f(x), are all real and non-positive.
If property (ii) is omitted in Theorem H, then α in q
n
(f) ≥ α may have some-
what smaller value than 4. In this case, the determination of the best possible
constant remains an open problem. For extensions of Hutchinson’s results see,
for example, [2, §4]. The following statement is a corollary to Theorem H.
Theorem C. Let f(z) =
P
∞
k=0
a
k
z
k
, a
k
> 0 for k. If q
n
(f) ≥ 4 for all n ≥ 2
then f ∈ B
∗
.
For the reader’s convenience proofs of Theorems A, B and C will be given in
Section 2.
From (4) we obtain
(5) lim inf
n→∞
q
n
(f) ≥ 2 for every f ∈ A
∗
.
In this paper we will prove that the constant 2 in (5) can be increased and the
constant 4 in Theorem C cannot be decreased.
Theorem 1. Let f(z) =
P
∞
k=0
a
k
z
k
∈ A
∗
. Then lim inf
n→∞
q
n
(f) ≥ 1 +
√
3.
Theorem 2. For every ε > 0 there exists a function f
ε
(z) =
P
∞
k=0
a
k
(ε)z
k
such
that a
k
(ε) > 0 for all k ∈ N ∪ {0} and q
n
(f
ε
) > 4 − ε for all n ≥ 2 but f
ε
/∈ A
∗
.

428 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
Theorem 2 shows that if the estimate on q
n
(f) is given only from below then the
constant 4 in q
n
(f) ≥ 4 is the smallest possible to conclude that f ∈ A
∗
. But,
as the following theorem shows, if estimates on q
n
(f) are given from below and
from above then the constant 4 can be decreased.
Theorem 3. Let f(z) =
P
∞
k=0
a
k
z
k
, a
k
> 0 for all k. If α ∈ [3.43; 4] and
β(α) := 0.95/(2
√
α − α) then q
n
(f) ∈ [α, β(α)] for all n ≥ 2 implies f ∈ A
∗
.
In connection with the above mentioned theorems it is natural to investigate the
function
g
a
(z) :=
∞
X
k=0
z
k
a
k
2
, a > 1,
with the property q
n
(g
a
) = a
2
for n ≥ 2.
G. H. Hardy [6, pp. 95–100] has shown that g
a
(z) has only real zeros if a
2
≥ 9.
In [12, Problem 176, p. 66] it is proved that g
a
(z) has only real zeros if a
2
≥ 4.
The question about the smallest value of a for which g
a
(z) has only real zeros
was discussed by T. Craven and G. Csordas [3]. T. Craven and G. Csordas have
improved the method of [12, Problem 176, p. 66] and have shown that a
2
≥ 3.4225
is enough, see [3, Examples 4.10, 4.11]. As mentioned in the abstract, we wish
to determine an exhaustive list of values of a for which the function g
a
(z) has
only real zeros. Our main result is the following theorem.
Theorem 4. There exists a constant q
∞
, q
∞
≈ 3.23, such that
(i) S
2k+1
(z, g
a
) :=
P
2k+1
j=0
a
−j
2
z
j
∈ R
∗
for every k ∈ N if and only if a
2
≥ q
∞
,
(ii) there exists a number N
0
∈ N such that S
2k
(z, g
a
) :=
P
2k
j=0
a
−j
2
z
j
∈ R
∗
for
all k ≥ N
0
if and only if a
2
> q
∞
, and
(iii) g
a
(z) has only real zeros if and only if a
2
≥ q
∞
.
Using Theorem C and considering S
2
(g
a
) it is easy to see that g
a
(z) ∈ B
∗
if and
only if a
2
≥ 4.
The question about the sharp constant in Theorem 1 is open. Theorem 4 shows
that this sharp constant is less than or equal to q
∞
.
2. Proofs of Theorems A, B, C.
Proof of Theorem A. Let P (z) = a
0
+ a
1
z + ··· + a
n
z
n
∈ R
∗
, a
j
> 0 for
j = 0, 1, . . . , n. Then
d
n−2
dz
n−2
P (z) = (n − 2)!a
n−2
+ (n − 1)!a
n−1
z +
n!
2
a
n
z
2
∈ R
∗
,
and hence
((n − 1)!)
2
a
n−1
2
− 2(n − 2)!n!a
n−2
a
n
≥ 0,
and the statement of Theorem A follows.

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 429
Proof of Theorem B. The Vi´eta formulae imply for a power series
P
∞
k=0
a
k
z
k
,
a
k
≥ 0, k ∈ N ∪ {0}, with S
n
(z) ∈ R
∗
for all n ∈ N that either
(6) a
k
> 0 for all k ∈ N ∪ {0},
or there exists k
0
∈ N such that
(7) a
k
> 0 when k ≤ k
0
, a
k
= 0 when k > k
0
.
Let the formal series (1) have the property that there exists N ∈ N such that
S
n
(z) ∈ R
∗
for n ≥ N. Then either (6) or (7) holds. If (7) holds then the
theorem is trivial. If (6) holds then from Theorem A we have for every n ≥ N
a
n−1
a
n
> 2
a
n−2
a
n−1
> . . . > 2
n−N+1
a
N−2
a
N−1
,
i.e.
a
n
< a
n−1
2
−(n−N+1)
a
N−1
a
N−2
< a
n−2
2
−(n−N+1)−(n−N)

a
N−1
a
N−2

2
< . . .
< a
N−1
2
−
(n−N +2)(n−N +1)
2

a
N−1
a
N−2

n−N+1
= ab
n
2
−n
2
/2
,
where a, b do not depend on n. This completes the proof.
Proof of Theorem C. Let f(z) =
P
∞
k=0
a
k
z
k
, a
k
> 0 for all k, and q
n
=
q
n
(f) ≥ 4 for all n ≥ 2. Since q
n
≥ 4, we have p
1
< p
2
< . . . < p
n
< . . ..
Theorem C follows from the statements
S
m
(0) > 0,
−S
m
(−
√
p
1
p
2
) ≥ 0,
S
m
(−
√
p
2
p
3
) ≥ 0,
.
.
.
(−1)
m−1
S
m
(−
√
p
m−1
p
m
) ≥ 0,
lim
x→−∞
(−1)
m
S
m
(x) = +∞.
The first and the last inequality are obvious, we shall prove the other inequalities.
Let us fix k ∈ {1, 2, . . . , m − 1}. We have
(−1)
k
S
m
(−z) =
k−2
X
j=0
(−1)
k−j
a
j
z
j
+[−a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1
]
+
m
X
j=k+2
(−1)
k−j
a
j
z
j
=: Σ
1
(z) + [−a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1
] + Σ
2
(z).

430 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
For 0 ≤ j ≤ k − 1 and z ∈ (p
k
, p
k+1
) we have a
j
z
j
< a
j+1
z
j+1
. Hence, for
z ∈ (p
k
, p
k+1
) the summands in Σ
1
(z) are alternating in sign and their moduli are
increasing. Analogously for z ∈ (p
k
, p
k+1
) the summands in Σ
2
(z) are alternating
in sign and their moduli are decreasing. Hence Σ
1
(z) ≥ 0 and Σ
2
(z) ≥ 0 for
z ∈ (p
k
, p
k+1
). In particular, this is true for the choice z
k
:=
√
p
k
p
k+1
. From (3)
we obtain
−a
k−1
z
k
k−1
+ a
k
z
k
k
− a
k+1
z
k
k+1
=
a
0
z
k
k−1
p
1
···p
k−1
(
√
q
k+1
− 2) ≥ 0.
Therefore
(−1)
k
S
m
(−z
k
) ≥ 0, 1 ≤ k ≤ m − 1
and our proof is complete.
3. Proof of Theorem 1.
Let f(z) =
P
∞
k=0
a
k
z
k
∈ A
∗
. Let us consider
N
4
(x
1
, . . . , x
n
) := x
1
4
+ ··· + x
n
4
,
where x
1
, . . . , x
n
are zeros of S
n
(z, f) =
P
n
k=0
a
k
z
k
, n ≥ 4. For n ≥ max{4, N(f)}
we have
0 < N
4
(x
1
, . . . , x
n
)
=

a
n−1
a
n

4
− 4

a
n−1
a
n

2
a
n−2
a
n
+ 2

a
n−2
a
n

2
+ 4
a
n−1
a
n−3
a
n
2
− 4
a
n−4
a
n
.
Therefore we obtain
q
n
3
q
n−1
2
q
n−2
− 4q
n
2
q
n−1
2
q
n−2
+ 2q
n
q
n−1
2
q
n−2
+ 4q
n
q
n−1
q
n−2
− 4 > 0
if and only if
q
n
q
n−1
q
n−2

q
n
2
q
n−1
− 4q
n
q
n−1
+ 2q
n−1
+ 4

> 4,
hence
(8) q
n
2
− 4q
n
+ 2 +
4
q
n−1
> 0.
Suppose there exists n
0
∈ N such that q
n
0
≤ q
n
0
−1
then from (8) we have
1
q
n
0
(q
n
0
− 2)(q
n
0
− 1 −
√
3)(q
n
0
− 1 +
√
3) = q
n
0
2
− 4q
n
0
+ 2 +
4
q
n
0
> 0,
and since q
n
0
> 2 we obtain q
n
0
> 1 +
√
3. Therefore, q
n
0
≤ q
n
0
−1
implies
1 +
√
3 < q
n
0
≤ q
n
0
−1
. Hence, if q
k
≤ 1 +
√
3 then q
k−1
< q
k
< 1 +
√
3 then
q
k−2
< q
k−1
< q
k
< 1 +
√
3 and so on. This shows that either there exists k
0
such that q
n
≥ 1 +
√
3 for n ≥ k
0
(and Theorem 1 is true) or that the sequence

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 431
{q
n
}
∞
n=N(f)+4
is non-decreasing and the limit of this sequence exists. By (8), this
limit x := lim
n→∞
q
n
satisfies the inequality
x
2
− 4x + 2 +
4
x
≥ 0,
and since x > 2 we obtain x ≥ 1 +
√
3. This completes the proof of Theorem 1.
Remark. Using the same method as in the proof of Theorem 1 and considering
N
7
(x
1
, . . . , x
n
) := x
1
7
+ ··· + x
n
7
instead of N
4
(x
1
, . . . , x
n
), we can prove after
some cumbersome calculations that
f(z) =
∞
X
k=0
a
k
z
k
∈ A
∗
⇒ lim inf
n→∞
q
n
(f) > 2.9.
4. Proof of Theorem 2.
Let us consider the function
f(z) =

z
2
+ 2z + b

∞
X
k=0
z
k
q
k
2
=:
∞
X
k=0
a
k
z
k
,
where b > 1 and q > 1. Since f(z) has non-real zeros, by Hurwitz’s Theorem,
there exists a number N ∈ N such that the sections S
n
(z) have non-real zeros
for n ≥ N. Therefore, f /∈ A
∗
for b > 1 and q > 1.
We shall show that for any ε > 0 there exist numbers b > 1 and q > 1 such that
q
2
> 4 − ε and q
n
> 4 for n ≥ 3. Notice that
(9) a
n
=





b if n = 0,
bq
−1
if n = 1,
bq
−n
2
+ 2q
−(n−1)
2
+ q
−(n−2)
2
if n ≥ 2.
Let us fix any k ≥ 3 and check that q
k+1
> 4 for b ∈ (1; 2) and q
2
> 84. In fact,
from (9) we obtain
(10) q
k+1
=
a
k
2
a
k−1
a
k+1
=
b
2
+ q
8k−8
+ A
b
2
q
−2
+ q
8k−10
+ B
,
where
A := 4q
4k−2
+ 4bq
2k−1
+ 2bq
4k−4
+ 4q
6k−5
,
B := 2bq
2k−1
+ bq
4k−2
+ 2bq
2k−5
+ 4q
4k−4
+ 2q
6k−5
+ bq
4k−10
+ 2q
6k−9
.
Notice that since k ≥ 3 all exponents of q in the last expression are less than
8k − 10 and we obtain
B < q
8k−10
(6b + 8).
So, from (10) we obtain
q
k+1
> q
2
b
2
+ q
8k−8
b
2
+ q
8k−8
(6b + 9)
>
q
2
6b + 9
for q > 1 and b > 1.

432 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
Therefore,
q
k+1
> 4 for k ≥ 3
if b ∈ (1; 2) and q
2
> 84.
We now show that q
3
> 4 for b ∈ (1; 2) and q > 80. Indeed, by (9) we have
q
3
=
(bq
−4
+ 2q
−1
+ 1)
2
(bq
−1
+ 2) (bq
−9
+ 2q
−4
+ q
−1
)
,
therefore, we obtain for b ∈ (1; 2) and q > 80 that
q
3
>
1
(b + 2) (bq
−1
+ 2q
−1
+ q
−1
)
>
q
4 · 5
> 4.
Let us consider q
2
. It is easy to see that
q
2
=
a
1
2
a
0
a
2
=
b
2
q
−2
+ 4bq
−1
+ 4
b
2
q
−4
+ 2bq
−1
+ b
→ 4 as b → 1 and q → ∞.
Hence, for any ε > 0 there exist b
0
> 1 and q
∞
> 1 such that q
2
> 4 − ε for
b ∈ (1; b
0
) and q > q
∞
. This concludes the proof of Theorem 2.
5. Proof of Theorem 3.
Let f (z) =
P
∞
k=0
a
k
z
k
, a
k
> 0 for all k, α ∈ [3.43, 4], β(α) := 0.95/(2
√
α − α),
and q
n
= q
n
(f) ∈ [α, β(α)] for n ≥ 2. We shall show that S
m
(z) has only real
zeros for each m ≥ 5. To do so it is sufficient to prove that
S
m
(0) > 0,
−S
m
(−
√
p
1
p
2
) > 0,
.
.
.
(−1)
m−1
S
m
(−
√
p
m−1
p
m
) > 0,
lim
x→−∞
(−1)
m
S
m
(x) = +∞.
The first and the last inequality are obviously true. Hence, let us check the other
inequalities. For k ∈ {3, . . . , m−1} we shall show that (−1)
k
S
m
(−
√
p
k
p
k+1
) > 0.
We have
(−1)
k
S
m
(−z) =
k−4
X
j=0
(−1)
k−j
a
j
z
j
+

−a
k−3
z
k−3
+ a
k−2
z
k−2
− −a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1

+
m
X
j=k+2
(−1)
k−j
a
j
z
j
:= Σ
1
(z) + Σ
2
(z) + Σ
3
(z)

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 433
(in the above and later we use the convention
P
m
j=l
:= 0 for m < l). Since
q
n
≥ α > 1 for every n ≥ 2 we have p
n+1
> p
n
for n ≥ 1. Hence, a
j
z
j
< a
j+1
z
j+1
for z ∈ (p
k
, p
k+1
) and 0 ≤ j ≤ k − 1. We see that the summands in Σ
1
(z)
are alternating in signs and their moduli are increasing for any z ∈ (p
k
, p
k+1
).
Analogously, the summands in Σ
3
(z) are alternating in signs and their moduli
are decreasing for any z ∈ (p
k
, p
k+1
). Therefore, Σ
1
(z) ≥ 0 and Σ
3
(z) ≥ 0 for
z ∈ (p
k
, p
k+1
). In particular, Σ
1
(
√
p
k
p
k+1
) ≥ 0 and Σ
3
(
√
p
k
p
k+1
) ≥ 0 and at
least one of these numbers is strictly positive because at least one of these sums
is non-trivial. Let us prove that for z =
√
p
k
p
k+1
we have
(11) −a
k−3
z
k−3
+ a
k−2
z
k−2
− a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1
≥ 0.
Using (3) we can transform the last inequality to
(12) −2 +
√
q
k+1
+
1
√
q
k+1
q
k
−
1
q
k+1
q
k
2
q
k−1
≥ 0.
Let us put x :=
√
q
k+1
. We rewrite the last inequality in the form
x
3
− 2x
2
+
x
q
k
−
1
q
k
2
q
k−1
≥ 0.
Let us check that
(13)
x
q
k
−
1
q
k
2
q
k−1
≥
0.95x
q
k
,
or, equivalently
x ≥
20
q
k
q
k−1
.
Note that (13) holds since x ≥
√
α ≥
√
3.43 and 20/(q
k
q
k+1
) ≤ 20/3.43
2
. In
view of (13) a sufficient condition for (12) (and so for (11)) is
x
2
− 2x +
0.95
q
k
≥ 0.
But this inequality is a consequence of
x ≥ 1 +
r
1 −
0.95
q
k
which in turn follows from
√
α = 1 +
s
1 −
0.95
β(α)
.
Hence, we have proved
(−1)
k
S
m
(−
√
p
k
p
k+1
) > 0, k = 3, . . . , m − 1,
for n ≥ 2 and q
n
∈ [α, β(α)].

434 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
Let us prove that the last inequality is true for k = 1, 2, too. For k ∈ {1, 2} we
have
(−1)
k
S
m
(−z) =
k−2
X
j=0
(−1)
k−j
a
j
z
j
+

−a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1
+ a
k+2
z
k+2
− a
k+3
z
k+3

+
m
X
j=k+4
(−1)
k−j
a
j
z
j
:= Σ
4
(z) + Σ
5
(z) + Σ
6
(z).
As mentioned before Σ
4
(z) ≥ 0 and Σ
6
(z) ≥ 0 for z ∈ (p
k
, p
k+1
). In particular,
Σ
4
(
√
p
k
p
k+1
) ≥ 0 and Σ
6
(
√
p
k
p
k+1
) ≥ 0 and at least one of these numbers is
strictly positive because at least one of these sums is non-trivial. We are left to
prove that
−a
k−1
z
k−1
+ a
k
z
k
− a
k+1
z
k+1
+ a
k+2
z
k+2
− a
k+3
z
k+3
≥ 0
for z
k
=
√
p
k
p
k+1
, or after transformation that
−2 +
√
q
k+1
+
1
q
k+2
√
q
k+1
−
1
q
k+3
q
k+2
2
q
k+1
≥ 0.
As we see, the last inequality is of the same type as (12) and can be obtained
from (12) by an index change. Since (12) is true for q
n
∈ [α, β(α)] and for all
n ≥ 2, this inequality is also true. Hence we obtain
(−1)
k
S
m
(−
√
p
k
p
k+1
) > 0, k = 1, 2,
which concludes the proof.
6. Proof of Theorem 4.
Let
f
a
(z) :=
∞
X
k=0
(−1)
k
z
k
a
k
2
, a > 1,
S
n
(z, a) :=
n
X
k=0
(−1)
k
z
k
a
k
2
, a > 1.
We investigate the location of zeros of the polynomials S
n
(z, a). To prove The-
orem 4 we first need some lemmas.
Lemma 1. For a
2
≥ 3 we have
(14) |S
4
(a
3
e
iϕ
, a)| ≥
1
a
4
for all ϕ ∈ [0, 2π].

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 435
Proof. We have
S
4
(a
3
e
iϕ
, a) = 1 − a
2
e
iϕ
+ a
2
e
2iϕ
− e
3iϕ
+
1
a
4
e
4iϕ
= −ie
3iϕ/2

2 sin
3ϕ
2
− 2a
2
sin
ϕ
2
+ i
1
a
4
e
5iϕ/2

and therefore
|S
4
(a
3
e
iϕ
, a)|
2
= 4

sin
3ϕ
2
− a
2
sin
ϕ
2

2
−
4
a
4
sin
5ϕ
2

sin
3ϕ
2
−
1
a
2
sin
ϕ
2

+
1
a
8
.
After a simple transformation we obtain
|S
4
(a
3
e
iϕ
, a)|
2
= 4 sin
2
ϕ
2

(a
2
− 3) + 4 sin
2
ϕ
2

2
+
4
a
4
sin
5ϕ
2
sin
ϕ
2

(a
2
− 3) + 4 sin
2
ϕ
2

+
1
a
8
.
For a
2
≥ 3 and ϕ ∈ [0, 2π] estimate (14) is a consequence of
sin
ϕ
2

(a
2
− 3) + 4 sin
2
ϕ
2

+
1
a
4
sin
5ϕ
2
≥ 0.
This inequality follows from
(15) 4a
4
sin
3
ϕ
2
+ sin
5ϕ
2
≥ 0
for ϕ ∈ [0, 2π]. If ϕ ∈ [0, 2π/5] ∪ [8π/5, 2π] we have sin(5ϕ/2) ≥ 0 and (15)
holds. Finally, if ϕ ∈ [2π/5, 8π/5] then (15) follows from
4a
4
sin
3
π
5
− 1 ≥ 0,
which in turn is true since sin
3
(π/5) ≥ sin
3
(π/6) = 1/8.
Lemma 2. If a
2
≥ 3 then S
4
(z, a) has exactly two zeros in {z : |z| < a
3
} and it
has no zero on {z : |z| = a
3
}.
Proof. Let
P
a
(t) := S
4
(a
4
t, a) = 1 − a
3
t + a
4
t
2
− a
3
t
3
+ t
4
.
We shall show that P
a
(t) has exactly two zeros in {t : |t| < 1/a} and exactly
two zeros in {t : |t| ≤ 1/a}. Let w(t) := t + 1/t. The function w(t) maps the
disk {t : |t| < 1/a} conformally in a domain Ω with {w : |w| > a + 1/a} ⊂ Ω.
We have
P
a
(t) = t
2
(w(t)
2
− 2 − a
3
w(t) + a
4
).
We want to show that
Q
a
(w) := (w
2
− 2 − a
3
w + a
4
)

436 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
has exactly two zeros in {w : |w| > a + 1/a}. Denote by w
1
and w
2
the zeros of
Q
a
(z) and by D the discriminant of Q
a
(z). If D ≤ 0 then
|w
j
| ≥ Re w
j
=
a
3
2
≥
3a
2
> a +
1
a
, j = 1, 2.
If D > 0 then
|w
j
| ≥
a
3
−
√
a
6
− 4a
4
+ 8
2
> a +
1
a
, j = 1, 2.
So for a
2
≥ 3 the polynomial Q
a
(w) has exactly two zeros in {w : |w| > a+1/a}.
Since deg Q
a
= 2 we obtain that Q
a
(w) has exactly two zeros in Ω and exactly
two zeros in the closure of Ω. So P
a
(t) has exactly two zeros in {t : |t| < 1/a}
and has no zeros on the boundary of this circle.
Suppose a
2
≥ 3 and n ≥ 5. Then we have S
n
(z, a) =: S
4
(z, a) + R
n
(z, a) and it
is easy to see that
|R
n
(a
3
, a)| ≤
n
X
k=5
a
3k
a
k
2
≤
1
a
10
∞
X
k=0
1
a
8k
=
1
a
2
(a
8
− 1)
.
Since
1
a
4
>
1
a
2
(a
8
− 1)
,
we obtain by Lemma 1 and Rouch´e’s Theorem that, for a
2
≥ 3 and n ≥ 5, the
number of zeros of a polynomial S
n
(z, a) in the open disk {z : |z| < a
3
} (resp. in
the closed disk {z : |z| ≤ a
3
}) is equal to the number of zeros of the polynomial
S
4
(z, a) in {z : |z| < a
3
} (resp. in {z : |z| ≤ a
3
}). Using Lemma 2 we get
that for a
2
≥ 3 and n ≥ 4 each polynomial S
n
(z, a) has exactly two zeros in
{z : |z| < a
3
} (resp. in {z : |z| ≤ a
3
}).
Lemma 3. If n ≥ 4 and a
2
≥ 2.25 then S
n
(z, a) has n − 4 zeros in [a
4
, a
2n−4
].
Proof. For 2 ≤ k ≤ n − 2 we have
(−1)
k
S
n
(a
2k
) =
n
X
j=1
(−1)
j−k
a
2kj
a
j
2
= a
k
2
n
X
j=1
(−1)
j−k
a
−(j−k)
2
≥ a
k
2
[1 − 2a
−1
+ 2a
−4
− 2a
−9
].
Since the last expression is positive for a ≥ 1.5 the assertion follows.
Lemma 4. If a
2
≥ 3 and n ≥ 4, then S
n
(z, a) ∈ R
∗
if and only if there exists a
number x
n
∈ (a, a
3
) such that S
n
(x
n
, a) ≤ 0.
Proof. Using Lemma 2 we see that, for a
2
≥ 3 and n ≥ 4, a polynomial S
n
(z, a)
has exactly two zeros in {z : |z| < a
3
}. This means that
1 ≥
z
a
≥
z
2
a
4
≥ . . . ≥
z
n
a
n
2
≥ . . .

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 437
for 0 ≤ z ≤ a, each section S
n
(z, a), n ∈ N, has no zero z ∈ [0, a]. Hence, if
S
n
(z, a) ∈ R
∗
for a
2
≥ 3 and n ≥ 4 then there exists a number x
n
∈ (a, a
3
) such
that S
n
(x
n
, a) ≤ 0.
On the other hand, if a
2
≥ 3, n ≥ 4 and if there exists a number x
n
∈ (a, a
3
)
such that S
n
(x
n
, a) ≤ 0 then S
n
(z, a) has exactly two zeros in {z : |z| < a
3
} this
means that S
n
(z, a) has two zeros on (0, a
3
). A straight forward calculation gives
S
n

a
n
t
, a

= (−1)
n
t
−n
S
n
(a
n
t, a)
and shows that S
n
(z, a) has two zeros on (a
2n−3
, ∞). By Lemma 3, the polyno-
mial S
n
(z, a) has n − 4 zeros in [a
4
, a
2n−4
]. Hence, S
n
(z, a) has n real zeros, i.e.
S
n
(z, a) ∈ R
∗
.
Let
q
n
:= inf

a
2
> 1 : S
n
(x
0
, a) ≤ 0 for some x
0
∈ (a, a
3
)

for n ≥ 2,
q
∞
:= inf

a
2
> 1 : f
a
(x
0
) ≤ 0 for some x
0
∈ (a, a
3
)

.
An easy calculation gives q
2
= 4 and q
3
= 3.
Since
S
2k+2
(x, a) = S
2k
(x, a) −
x
2k+1
a
(2k+1)
2

1 −
x
a
4k+3

< S
2k
(x, a),
for k ≥ 1 and 0 ≤ x ≤ a
3
we have
4 = q
2
> q
4
> q
6
> . . . .
Therefore, the limit of the sequence {q
2k
}
∞
k=1
exists and satisfies
lim
k→∞
q
2k
=: b < 4.
Similiarly, since
S
2k+1
(x, a) = S
2k−1
(x, a) +
x
2k
a
(2k)
2

1 −
x
a
4k+1

> S
2k−1
(x, a),
for k ≥ 1 and 0 ≤ x ≤ a
3
we have
3 = q
3
< q
5
< q
7
< . . .
and the limit of the sequence {q
2k+1
}
∞
k=1
exists with
lim
k→∞
q
2k+1
=: c > 3.
For every x ≥ 0 and k ∈ N we further have
S
2k+1
(x, a) = S
2k
(x, a) −
x
2k+1
a
(2k+1)
2
< S
2k
(x, a),
which shows that
q
2k+1
≤ q
2k

438 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
for k ∈ N and therefore
c ≤ b.
Furthermore, from
S
2k
(x, a) −
∞
X
n=2k+1
(−1)
n−1
x
n
a
n
2
= f
a
(x) = S
2k+1
(x, a) +
∞
X
n=2k+2
(−1)
n
x
n
a
n
2
we obtain for x ∈ [a, a
3
] and n ≥ 1 (because x
n
/a
n
2
≥ x
n+1
/a
(n+1)
2
) that
S
2k
(x, a) > f
a
(x) > S
2k+1
(x, a).
This implies
3 < c ≤ q
∞
≤ b < 4.
We shall prove that in fact c = q
∞
= b. To do so we need the following lemma.
Lemma 5. Let 3 ≤ a
1
2
< a
2
2
, k ≥ 1. Then the following implications are true.
(i) If S
k
(x
1
, a
1
) ≤ 0 for some x
1
∈ (a
1
, a
1
3
) then we have S
k
(x
2
, a
2
) < 0, where
x
2
:= x
1
a
2
/a
1
∈ (a
2
, a
2
3
).
(ii) If f
a
1
(x
1
) ≤ 0 for some x
1
∈ (a
1
, a
1
3
) then we have f
a
2
(x
2
) < 0, where
x
2
:= x
1
a
2
/a
1
∈ (a
2
, a
2
3
).
Proof. Let y
1
:= x
1
/a
1
and let
C
k
(y, a) := S
k
(ay, a),
g
a
(y) := f
a
(ay).
Since x
1
∈ [a
1
, a
1
3
] we have y
1
∈ [1, a
1
2
]. For a ∈ (a
1
, a
2
) we obtain
∂
∂a
C
k
(y
1
, a) = −
y
1
2
a
3
2 −
6y
1
a
4
+
k−2
X
p=2
(−1)
p
y
1
p
(p + 2)(p + 1)
a
p
2
+3p
!
,
∂
∂a
g
a
(y
1
) = −
y
1
2
a
3
2 −
6y
1
a
4
+
∞
X
p=2
(−1)
p
y
1
p
(p + 2)(p + 1)
a
p
2
+3p
!
.
For a
2
≥ a
1
2
≥ 3 and y
1
∈ [1, a
1
2
] we further have
2 ≥
6y
1
a
4
≥ . . . ≥
y
1
p
(p + 2)(p + 1)
a
p
2
+3p
≥
y
1
p+1
(p + 3)(p + 2)
a
(p+1)
2
+3(p+1)
,
and therefore
∂
∂a
C
k
(y
1
, a) < 0 for a ∈ (a
1
, a
2
),
∂
∂a
g
a
(y
1
) < 0 for a ∈ (a
1
, a
2
).
Hence we have
C
k
(y
1
, a
1
) > C
k
(y
1
, a
2
),
g
a
1
(y
1
) > g
a
2
(y
1
),

3 (2003), No. 2 On Power Series Having Sections With Only Real Zeros 439
and the statemant of Lemma 5 follows.
For a
2
≥ 3 and n ≥ 4, the polynomial S
n
(z, a) has two zeros in {z : |z| < a
3
}.
Using Lemma 5 we see that the polynomial S
n
(z, a) has in (a, a
3
) one zero of
multiplicity two if a
2
= q
n
and two simple zeros if a
2
> q
n
. Similiarly, the
function f
a
(z) has in (a, a
3
) one zero of multiplicity two if a
2
= q
∞
and two
simple zeros if a
2
> q
∞
.
Assume that c < q
∞
. If we choose c
1
with c < c
1
< q
∞
and put a
1
:= c
1
1/2
then f
a
1
(x) > 0 for x ∈ [a
1
, a
1
3
]. Therefore, there is a number N ∈ N such that
S
2n+1
(x, a
1
) > 0 for n ≥ N and x ∈ [a
1
, a
1
3
]. Hence, q
2n+1
> c
1
for n ≥ N. This
contradiction shows that c = q
∞
.
Similiarly, assume that q
∞
< b. Choose b
1
, with q
∞
< b
1
< b and put a
2
:= b
1
1/2
.
Then there exists x
0
∈ [a
2
, a
2
3
] such that f
a
2
(x
0
) < 0. Hence, there is some
M ∈ N such that S
2m
(x
0
, a
2
) < 0 for m ≥ M. But then q
2m
< b
1
for m ≥ M
and we get b = q
∞
again by contradiction.
Hence, we have shown that S
2k+1
(z, a) ∈ R
∗
for every k ∈ N if a
2
≥ q
∞
.
If 3 ≤ a
2
< q
∞
then there is a number k
0
∈ N such that S
2k+1
(z, a) /∈ R
∗
for
every k ≥ k
0
, since each of those polynomials has two non-real zeros in the disk
{z : |z| ≤ a
3
} (to do so we choose k
0
satisfying q
2k
0
+1
> a
2
).
If a
2
> q
∞
then there is a number k
1
∈ N such that S
2k
(z, a) ∈ R
∗
for k ≥ k
1
(in
that case we have to choose k
1
such that q
2k
1
< a
2
).
If 3 ≤ a
2
< q
∞
we have S
2k
(z, a) /∈ R
∗
for any k ∈ N, since such polynomials
have two non-real zeros in the disk {z : |z| ≤ a
3
}.
We also have shown that the function f
a
(z) has only real zeros if a
2
≥ q
∞
, and
two non-real zeros in the disk {z : |z| ≤ a
3
} if 3 ≤ a
2
< q
∞
. To complete the
proof we have to consider the case a
2
< 3. Here, we need the following theorem
of Laguerre.
Theorem D (cf. [12, Ch. 5, Problem 68]). If p(x) = a
0
+ a
1
x + ··· + a
n
x
n
is a
real polynomial then p(x) ∈ R
∗
implies
a
0
+
a
1
b
x +
a
2
b
4
x
2
+ ··· +
a
n
b
n
2
x
n
∈ R
∗
for every b > 1.
If we apply Laguerre’s Theorem to the polynomials S
k
(z, a), k ≥ 2, 1 < a
2
< 3,
with b =
√
3/a we obtain that S
k
(z,
√
3) ∈ R
∗
if S
k
(z, a) ∈ R
∗
. But we have
shown that S
k
(z,
√
3) /∈ R
∗
for k ≥ 4. This contradiction proves the first and the
second statement in Theorem 4.
Assume that there exists a number a
0
, 1 < a
0
2
< 3, such that f
a
0
(z) has only
real zeros. Since f
a
0
(z) is an entire function of order zero, there is a sequence

440 O. M. Katkova, T. Lobova and A. M. Vishnyakova CMFT
{b
m
}
∞
m=1
⊂ (0, +∞) with
P
∞
m=1
1/b
m
< ∞ such that
f
a
0
(z) =
∞
Y
m=1

1 −
z
b
m

,
and the product converges uniformly on any compact subset of C. Let
Q
n
(z) :=
n
Y
m=1

1 −
z
b
m

=:
n
X
k=0
(−1)
k
β
k
(n)z
k
, β
k
(n) ≥ 0.
For every K b C we have Q
n
(z) → f
a
0
(z) as n → ∞ uniformly for z ∈ K. Since
Q
n
(z) ∈ R
∗
we get
˜
Q
n
(z) :=
n
X
k=0
(−1)
k
β
k
(n)
b
k
2
z
k
∈ R
∗
for every b > 1. Obviously, for every K b C we have
˜
Q
n
(z) → f
a
0
b
(z) as n → ∞
uniformly for z ∈ K. Hence, if all zeros of f
a
0
(z) are real then all zeros of
f
a
0
b
(z) are real for every b > 1. Since the function f
a
(z) has non-real zeros for
3 ≤ a
2
< q
∞
it also has non-real zeros for 1 ≤ a
2
< q
∞
.
To estimate the numerical value of the constant q
∞
we note that it is not difficult
to calculate
q
4
= 1 +
√
5 ≈ 3.2361, q
5
≈ 3.2336.
This completes the proof of Theorem 4.
Remark. We not only have proved Theorem 4 but also the following implications
S
2k
(z, a) ∈ R
∗
⇒ S
2m
(z, a) ∈ R
∗
for all m ≥ k,
S
2k+1
(z, a) ∈ R
∗
⇒ S
2m+1
(z, a) ∈ R
∗
for all m ≤ k.
Acknowledgment. The authors are deeply gratefull to Professor I. V. Ostro-
vskii for his important comments and advice. We also thank the referees for
valuable suggestions.
References
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(1995), 420–441.
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Olga M. Katkova E-mail: olga.m.katkova@ilt.kharkov.ua
Address: Department of Mathematics, Kharkov State University, Svobody sq., 4, 310077,
Kharkov, Ukraine.
Tetyana Lobova E-mail: lobova@web.de
Address: Fakult¨at f¨ur Mathematik und Physik, Eberhard Karls Universit¨at T¨ubingen, D8Q02,
Auf der Morgenstelle 14, 72076 T¨ubingen.
Anna M. Vishnyakova E-mail: anna.m.vishnyakova@univer.kharkov.ua
Address: Department of Mathematics, Kharkov State University, Svobody sq., 4, 310077,
Kharkov, Ukraine.