On a Problem of A. Eremenko
Iossif V. Ostrovskii
(Communicated by Stephan Ruscheweyh)
Abstract. Let Pmn, 0 < m < n − 1, be a polynomial formed by the first m terms of the expansion of (1 + z)n according to the binomial formula. We
show that, ifm, n → ∞ in such a way that limm,n→∞m/n = α ∈ (0, 1), then the zeros ofPmntend to a curve which can be explicitly described.
Keywords. Asymptotic formula, conformal mapping, polynomial,
subhar-monic function, Szeg˝o’s method, univalent function, zeros.
2000 MSC. 26C10, 30C15.
1. Introduction and statement of the problem
Consider the polynomials
Pmn(z) := m k=0 n k zk, 0< m < n − 1.
The following problem was posed by A. Eremenko: describe the asymptotic zero
distribution of Pmn as m, n → ∞ in such a way that
(1) lim
m,n→∞
m
n =α ∈ (0, 1).
We give below a solution to this problem. Our method is based on some ideas of the famous paper [4] by G. Szeg˝o (extensive bibliography of works connected with this paper can be found in [2], [3]).
P. B. Borwein and W. Chen [1] have studied Pad´e approximations to powers of 1 +z, and especially sequences where the numerator degree divided by the denominator degree tends to a finite positive constant. But the case where the denominator is of degree 0 is not treated there; this is precisely the case considered in our paper.
Received December 24, 2003, in revised form March 30, 2004.
2. Basic formula and restatement of the problem
Lemma 1. For 0 < m < n − 1, (2) Pmn(z) (1 +z)nmn(n − m) = 1 z/(z+1)u m(1− u)n−m−1du.Proof. It is well known that for a sufficiently good function f one has
f (z) − p k=0 f(k)(0) k! z k = 1 p! z 0 (z − t)pf(p+1)(t) dt, p ∈ N. Applying this to f (z) = (1 + z)n, p = n − m − 1, we obtain
n k=n−m n k zk= n(n − 1) · · · (m + 1) (n − m − 1)! z 0 (z − t)n−m−1(1 +t)mdt.
Since the left hand side of this formula coincides withznPmn(1/z), we derive that
Pmn(z) = n(n − 1) · · · (m + 1)( n − m − 1)! z n 1 z 0 1 z − t n−m−1 (1 +t)mdt. Substitutingt = −1 + u(1 + z)/z, we obtain (2).
Let
(3) Qmn(z) :=
1
z u
m(1− u)n−m−1du.
Evidently, Qmn is a polynomial in z of degree n having a zero of order n − m at
z = 1. By (2), it is connected with Pmn by the formula
Pmn(z) = n m (n − m)(1 + z)nQmn z z + 1 .
Hence we see that all the zeros of Qmn, except the one at z = 1, are images of zeros of Pmn under bilinear transformation z → z/(z + 1). We will call these zeros non-trivial. Evidently, the problem under consideration is equivalent to the description of the asymptotic behavior of non-trivial zeros of Qmn as m, n → ∞ under condition (1). The rest of the paper is devoted to the investigation of this problem.
3. A bound for non-trivial zeros
Lemma 2. All non-trivial zeros of the function Qmn are located in the disc {z ∈ C : |z| < m/(n − 1)}.
Proof. The change of variable u = v/(1 + v) in the right hand side of (3) gives Qmn z 1 +z = ∞ z vm (1 +v)n+1 dv. Integrating by parts m + 1 times, we obtain
Qmn z 1 +z = 1 n zm (1 +z)n + m n(n − 1) zm−1 (1 +z)n−1 +· · · + m! n(n − 1) · · · (n − m) 1 (1 +z)n−m =: 1 n(1 + z)n−mSmn z 1 +z ,
where Smn(ζ) is a polynomial in ζ. Denoting
Smn(ζ) = n (1− ζ)n−mQmn(ζ) =: m k=0 ckζk, we see that ck = k + 1 n − m + kck+1≤ m n − 1ck+1, k = 0, 1, . . . , m − 1.
In the last inequality, equality occurs only for k = m − 1. Applying the well known Kakeya Theorem, we obtain the assertion of the lemma.
Further, we fix a sequence of pairs (m, n) such that (1) is satisfied. Denote by Z the set of all accumulation points of
mn
{z ∈ C : z = 1, Qmn(z) = 0}.
The following lemma is an immediate corollary of Lemma 2.
Lemma 3. The set Z is a subset of {z ∈ C : |z| ≤ α}.
This lemma shows that we need to study the polynomials Qmn only in the disc
{z ∈ C : |z| ≤ α}.
4. The main asymptotic formula
Consider the curve
z ∈ C : |z||1 − z|1/α−1 =bα, where
bα := max
x∈[0,1]x(1 − x)
1/α−1 =α(1 − α)1/α−1.
This curve consists of two loops surrounding 0 and 1 with z = α as the only common point. Only the left loop will be of interest and it will be denoted
-0.5 0.5 1 1.5 -0.5 0.5 α = 0.3 -0.5 0.5 1 1.5 -0.5 0.5 α = 0.5 -0.5 0.5 1 1.5 -0.5 0.5 α = 0.7
Figure 1. The curve {z ∈ C : |z||1 − z|1/α−1 = bα} and its relevant partCα inside the disk {z ∈ C : |z| ≤ α}.
by Cα. Note that Cα ⊆ {z ∈ C : |z| ≤ α} and Cα lies in the interior of Cβ for
α < β.
For{z ∈ C : |z| ≤ α} consider the analytic function
Wα(z) := 1
bαz(1 − z)
1/α−1,
where the branch assumes positive values on the interval (0, 1).
Lemma 4. The function Wα is univalent in {z ∈ C : |z| ≤ α} and maps the region bounded by Cα conformally onto the unit disc.
Proof. Forθ ∈ (0, 2π) we have
∂ ∂θ arg(Wα(αe iθ)) = 1 + 1 α − 1 Im ∂ ∂θ log(1− αe iθ) = (1 +α)(1 − cos θ) 1 +α2− 2α cos θ > 0.
Therefore, arg(Wα(z)) increases when z traverses {z ∈ C : |z| = α} in the positive direction. Since the origin is the only zero of Wα, the total increment of the argument is 2π by the Argument Principle. So, the first assertion of the lemma is a consequence of a well known fact of conformal mapping theory. The second assertion is its immediate corollary because |Wα(z)| = 1 for z ∈ Cα.
Lemma 5. The asymptotic formula
(4) 1
qmnQmn(z) = 1 − (1 +O(1)) m z
0
[Wα(u)]mdu as m → ∞,
holds uniformly in {z ∈ C : |z| ≤ α}, where qmn := n 1
m
Proof. Using (3), we have Qmn(z) = 1 0 u m(1− u)n−m−1du − z 0 u m(1− u)n−m−1du = qmn− qmn z 0 [Wmn(u)]mdu, where Wmn(u) = qmn−1/mu(1 − u)(n−1)/m−1. Noting that Wmn(z) = (1 +O(1))Wα(z) asm → ∞, uniformly in{z ∈ C : |z| ≤ α}, we obtain (4).
Now we can prove the main asymptotic formula for Qmn.
Lemma 6. For any δ > 0, the asymptotic formula
(5) 1
qmnQmn(z) = 1 −
(1 +O(1))m[Wα(z)]m+1
(m + 1)Wα(z) as m → ∞.
holds uniformly in {z ∈ C : |z| ≤ α} \ {z ∈ C : |z − α| < δ}.
Proof. Formula (4) of Lemma 5 implies that it suffices to show that z 0 [Wα(u)]mdu = [Wα(z)] m+1 (m + 1)Wα(z)(1 +O(1)), asm → ∞, holds uniformly in{z ∈ C : |z| ≤ α} \ {z ∈ C : |z − α| < δ}.
Note that, according to Lemma 4, Wα is univalent in the disk {z ∈ C : |z| ≤ α}. Let us take the integral on the right hand side of (4) along the level curve
{u ∈ C : arg(Wα(u)) = arg(Wα(z)), 0 < |u| < |z|}.
On this curve |Wα(u)| is an increasing function of |u|. Therefore we have z 0 [Wα(u)]mdu = Wα(z) 0 W m α WdW α α(u)
= expi(m + 1) arg(Wα(z))
|Wα(z)| 0 |Wα| md|Wα| Wα(u) = [Wα(z)] m+1 (m + 1)Wα(z)− exp i(m + 1) arg(Wα(z)) · |Wα(z)| 0 |Wα| m 1 Wα(u) − 1 Wα(z) d|Wα| = [Wα(z)] m+1 (m + 1)Wα(z)+O(|Wα(z)| m+1) = [Wα(z)] m+1 (m + 1)Wα(z)(1 +O(1)) as m → ∞.
5. Main result
For any open disc D ⊂ C, denote by νmn(D) the number of non-trivial zeros of a polynomial Qmn in D. We say that the non-trivial zeros of the sequence Qmn have density d(D) in D if there exists the limit
d(D) = lim
m→∞
νmn(D)
m .
Theorem. The setZ coincides with the curve Cα. Furthermore, for an arbitrary disc D ⊂ C the non-trivial zeros of the sequence Qmn have in D the density
d(D) = 1
2πlengthWα(D ∩ Cα).
Proof. According to Lemma 3 we have Z ⊆ {z ∈ C : |z| ≤ α}. Let K be a compact subset of {z ∈ C : |z| ≤ α} lying entirely either inside of Cα or outside ofCα. Then |Wα(z)| < 1 for z ∈ K in the former case and |Wα(z)| > 1 for z ∈ K in the latter. Therefore formula (5) implies that there exists the uniform limit
lim
m→∞Qmn(z) =
1 on any compact set lying inside of the curveCα,
∞ on any compact set lying outside of the curve Cα.
This proves Z ⊆ Cα.
Lemma 6 also shows that the following limit exists:
Q(z) := lim
m→∞
1
m log|Qmn(z)| = log bα+ log
+|W
α(z)| |z| ≤ α.
The function Q(z) is subharmonic in the disk {z ∈ C : |z| < α} and harmonic in{z ∈ C : |z| ≤ α} \ Cα. Therefore its Riesz’ measure is supported byCα with the density ρ(z) = 1 2π ∂ ∂nlog|Wα(z)| = 1 2π ∂ ∂sarg(Wα(z)), z ∈ Cα
with respect to the Lebesgue measure onCα. Hence
d(D) = 1 2π D∩Cα ∂ ∂sarg(Wα(z)) |dz| = 1 2πlengthWα(D ∩ Cα).
Remark 1. Since zeros of Pmn are images of non-trivial zeros of Qmn under the bilinear transformation T (z) := z/(1 − z), we can readily derive from the theorem above that the set of all accumulation points of the set
mn
{z ∈ C : Pmn(z) = 0}
(under condition (1)) coincides with
where Kα is the closed disc with diameter [−α/(1 + α), α/(1 − α)] ⊂ R. Note also that Lemma 2 shows that, even without condition (1), all zeros of Pmn, 0< m < n − 1, are contained in the open half-plane {z ∈ C : Re z > −1/2}.
Remark 2. It can be shown that the function Wα is univalent in the regionGα lying on the left of
∂Gα = z ∈ C : arg(z) + 1 α − 1 arg(1− z) = 0, Im(z) = 0 ∪ {α}.
This curve goes through the point z = α and is located in
{z ∈ C : Re z < α} ∪ {α}
for 0< α < 1/2, in
{z ∈ C : Re z > α} ∪ {α}
for 1/2 < α < 1, and coincides with the half plane
{z ∈ C : Re z = α}
forα = 1/2. The asymptotic formulas (4) and (5) remain in force on any compact set in Gα. -1.5 -1 -0.5 0.5 1 1.5 -1.5 -1 -0.5 0.5 1 1.5 α = 0.3 -1.5 -1 -0.5 0.5 1 1.5 -1.5 -1 -0.5 0.5 1 1.5 α = 0.5 -1.5 -1 -0.5 0.5 1 1.5 -1.5 -1 -0.5 0.5 1 1.5 α = 0.7
Figure 2. The curve ∂Gα.
Remark 3. After this work was completed, A. Klyachko informed the author that he has obtained a similar result by a different method but has not published it.
Acknowledgement. I express my deep gratitude to Alexander Iljinskii and Natalya Zheltukhina for careful reading of the manuscript and valuable remarks. My greatest thanks to the copy editor for creating Figures 1 and 2 and many other improvements to the final text.
References
1. P. B. Borwein and W. Chen, Incomplete rational approximation in the complex plane, Constructive Approximation 11 (1995), 85–106.
2. A. Edrei, E. B. Saff and R. S. Varga, Zeros of Sections of Power Series, Lecture Notes in Math. (Springer) 1002, 1983.
3. I. V. Ostrovskii, On zero distribution of sections and tails of power series, Isr. Math. Conf. Proc. 15 (2001), 297–310.
4. G. Szeg˝o, ¨Uber eine Eigenschaft der Exponentialreihe, Sitzungsber. Berl. Math. Ges. 23 (1924), 50–64; see also: G. Szeg¨o, Collected Papers, Vol.1, Birkh¨auser, Boston, 1982, 645–662.
Iossif V. Ostrovskii E-mail: iossif@fen.bilkent.edu.tr
Address: Department of Mathematics, Bilkent University, 06800 Bilkent, Ankara, Turkey; In-stitute for Low Temperature Physics and Engineering, 47 Lenin ave, 61103 Kharkov, Ukraine.