- Abdullah Alotaibi
^{1}Email author

**2010**:428936

https://doi.org/10.1155/2010/428936

© Abdullah Alotaibi. 2010

**Received: **1 July 2010

**Accepted: **12 September 2010

**Published: **15 September 2010

## Abstract

## 1. Introduction

where are entire functions of order less than one and are complex numbers. In fact, they have proved the following theorem.

Theorem 1.1.

Suppose that are entire functions of order less than one, and suppose that with and . Then every nontrivial solution of (1.5) is of infinite order.

Corollary 1.2.

Suppose that , where is a nonvanishing entire function with and with . Then every nontrivial solution of (1.4) is of infinite order.

## 2. Results

We observe that all the above results concern the order of growth only. In this paper, we are going to prove the following theorem which concerns the exponent of convergence.

Theorem 2.1.

has zeros with infinite exponent of convergence.

## 3. Some Lemmas

Throughout this paper we need the following lemmas. In 1965, Hayman [9] proved the following lemma.

Lemma 3.1.

for all outside a set of upper logarithmic density , where the positive constant depends only on and .

In 1962, Edrei and Fuchs [10] proved the following lemma.

Lemma 3.2.

In 2007, Bergweiler and Langley [11] proved the following lemma.

Lemma 3.3.

Let be a transcendental entire function of order . For large define to be the length of the longest arc of the circle on which , with if the minimum modulus satisfies . Then at least one of the following is true:

(i)there exists a set of positive upper logarithmic density such that for ;

(ii)for each the set has lower logarithmic density at least

We deduce the following.

Lemma 3.4.

Let , let be a positive integer, and let have logarithmic density . Let be a transcendental entire function such that on a path tending to infinity and for all with and . Then has order at least .

Proof.

## 4. Proof of Theorem 2.1

Lemma 4.1.

Proof.

Lemma 4.1 is proved.

Moreover, there exists a set of logarithmic density such that for the circle does not meet the -set .

Lemma 4.2.

The functions and are both transcendental.

Proof.

Let be small and positive and suppose that or is a polynomial. Let be large with and . Since is small it follows from (4.2) and (4.8) that . Choose with such that the intersection of with the ray given by is bounded. Applying Lemma 3.4 to the function , with a subpath of , gives , but may be chosen arbitrarily small, and this contradicts (4.7).

The next step is to estimate in the right half-plane.

Lemma 4.3.

Proof.

which gives and (4.12). This proves Lemma 4.3.

Lemma 4.4.

Let and be as in Lemma 4.3. Choose such that the ray has bounded intersection with the -set . Let be the union of the ray and the arcs , where is the set chosen following (4.9). Then one of the following holds:

(i)one has (4.11) for all large in ;

(ii)one has (4.12) for all large in .

Proof.

This follows simply from continuity. For each large in we have (4.11) or (4.12), but we cannot have both because of (4.14). This proves Lemma 4.4.

Lemma 4.5.

Proof.

by (4.8), and so (4.17) follows using (4.7). This proves Lemma 4.5.

Lemma 4.6.

If conclusion (i) of Lemma 4.4 holds then , while if conclusion (ii) of Lemma 4.4 holds then .

Proof.

which proves Lemma 4.6 in this case. The alternative case, in which we have conclusion (ii) in Lemma 4.4, is proved the same way, using in place of .

To finish the proof suppose first that conclusion (ii) of Lemma 4.4 holds. Then Lemma 3.4 implies that has order at least . Since may be chosen arbitrarily small, this contradicts Lemma 4.6. The same contradiction, with replaced by , arises if conclusion (i) of Lemma 4.4 holds, and the proof of the theorem is complete.

## Declarations

### Acknowledgment

The author thanks Professor J. K. Langley for the invaluable discussions on the results of this paper during his visit in summer 2008 and summer 2010 to the University of Nottingham in the U.K.

## Authors’ Affiliations

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