We assume that the reader is familiar with the basic notions of Nevanlinna’s value distribution theory (see [7, 9, 10]). Let f be a function that is transcendental and meromorphic in the plane. The forward differences Δⁿf are defined in the standard way [8] by

and the divided difference is defined by

Recently, a number of papers focus on complex difference equations and differences analogues of Nevanlinna’s theory. Bergweiler and Langley [1] firstly investigated the existed of zeros of Δf(z) and Δf(z)f(z), and obtained many profound results. The results may be viewed as discrete analogues of the following existing theorem on the zeros of f′.

Theorem A.([BL])

Let f be transcendental and meromorphic in the plane withlim infr→∞T(r,f)r=0. Then f′ has infinitely many zeros.

In [1], Bergweiler and Langley proved the following theorems.

Theorem B.([1])

Let n ∈ ℕ and f be transcendental entire function of orderσ(f)=σ<12and G_n(z) is defined by (1.2). If G_n is transcendental, then G_n(z) has infinitely many zeros. In particular, if f has order less thanmin[1n,12], then G_n(z) is transcendental and has infinitely many zeros.

For the first difference Δf(z) and divided difference Δf(z)f(z), they also have the following theorem.

Theorem C.([1])

Let f be a function transcendental and meromorphic function in the plane which satisfieslim_r→∞T(r,f)r=0, then Δf(z), Δf(z)f(z)are both transcendental.

But for some n ≥ 2, G_n(z) fails to be transcendental if f is an entire function of order less than 12; see [5].

In [4], Langley extended Theorem A. In fact, he got the following theorem.

Theorem D.([4])

Let n ∈ ℕ and f be a transcendental meromorphic function of order of growth σ(f) = σ < 1 in the plane and assume that G_n(z) as defined by (1.2) is transcendental.

• If G_n(z) has lower order μ < α < 1/2, which holds in particular if σ < 1/2, then δ(0,G_n(z)) ≤ 1 − cos πα orδ(∞,f)≤μα.

• If σ = 1/2, then either G_n(z) has infinitely many zeros or δ(∞, f) < 1.

• If f is entire andσ<12+δ0, then G_n(z) has infinitely many zeros: here δ₀is a small positive absolute constant.

In [2], Chen and Shon studied the fixed points of differences and divided differences of meromorphic functions and got some results. One of them is the following.

Theorem E.([2])

Let n ∈ ℕ, c ∈ ℂ and f be a function transcendental meromorphic function of order of growth σ(f) = σ < 1. IfGn(z)=Δnf(z)f(z)is transcendental, then G_n(z) has infinitely many fixed points.

Comparing Theorem B with Theorem D (1), we know that the latter give a precise estimation of the zeros of G_n(z). For the fixed points of G_n(z), can we estimate it precisely like Theorem D (1)? In fact, Theorem E is extended as following.

Theorem 1

Let n ∈ ℕ, let f be a transcendental and meromorphic function of order σ(f) < 1 in the plane and assume that G_n(z) as defined by (1.2) is transcendental. Then 2δ(0,G_n(z)–z)+δ(0,G_n(z)) ≤ 2. Furthermore, if G_n(z) has lower orderμ<α<12, which holds in particular ifσ(f)<12, and δ(0,G_n(z)) > 1 − cos πα, thenδ(0,Gn(z)-z)<12(1+cosπα)andδ(∞,f)≤μα.

In Theorem D, there is a condition that G_n(z) is transcendental. Can we remove it for the case (2) and (3)? The question is studied and the following results are obtained.

Theorem 2

Let n be odd, let f be a transcendental meromorphic function of orderσ(f)=12and f has finite many poles in the plane, G_n(z) is defined by (1.2). Then G_n(z) has infinitely many zeros.

Theorem 3

Let n ∈ ℕ, let f be a transcendental meromorphic function of orderσ(f)<12+δ0andσ(f)≠n-kn, k ∈ ℕ, n(12-δ0)<k<n, where δ₀is a small positive absolute constant, and G_n(z) is defined by (1.2). Then G_n(z) has infinitely many zeros.

The final theorem from [1] showed that for transcendental meromorphic function satisfying the very strong growth restriction T(r, f) = O(log r)² as r → ∞, either the first differential or the first divided difference has infinitely many zeros. Langley improved it in [4] as following.

Theorem F.([4])

Let f be a transcendental meromorphic function in the plane of order less than 1/6, and define G = Δf/f. Then at least one of G and Δf has infinitely many zeros.

Obviously, we have the following corollary by Theorem A, C and E.

Corollary 1

Under the hypothesis of Theorem F, G has infinity many fixed points.

Thus, there exists a natural question that how about that fixed points of Δf under the hypothesis of Theorem F. It deserves for further study.

Remark 1

Following Hayman ([11], p75–p76), we define an ɛ–set to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If E is an ɛ–set, then the set of r ≥ 1 for which the circle S(0, r) meets E has finite logarithmic measure, and for almost all real θ the intersection of E with the ray arg z = θ is bounded.

Lemma 1.([1])

Let n ∈ ℕ and f be transcendental and meromorphic of order less than 1 in the plane. Then there exists an ɛ – set E_n such that

as z → ∞ in ℂ E_n.

Lemma 2.([6])

Let f be a function transcendental and meromorphic function with order σ(f) = σ < ∞, H = {(k₁, j₁), (k₂, j₂), · · ·, (k_q, j_q)} be a finite set of distinct pairs of integers that satisfy k_i > j_i ≥ 0, for i = 1, · · ·, q and let ɛ > 0 be a given constant. Then there exists a set E ⊂ (1,∞) with finite logarithmic measure such that for all z satisfying |z| ∉ E ∪ [0, 1] and for all (k, j) ∈ H, we have

Lemma 3.([9])

Suppose that f is a meromorphic function in the complex plane, and a₁(z), a₂(z) and a₃(z) are three distinct small functions of f(z). Then

Lemma 4.([3])

Suppose thatf(z)=g(z)d(z)is a meromorphic function with σ(f) = σ, where g(z) is an entire function and d(z) is a polynomial. Then there exists a sequence {r_j}, r_j → ∞, such that for all z satisfying |z| = r_j, |g(z)| = M(r_j, g), when j sufficiently large, we have

and

Proof of Theorem 1

Assume n, c, σ, f,G_n(z) are as in the hypotheses. Set Gn*(z)=Gn(z)-z, then σ(Gn*(z))=σ(Gn(z))≤σ(f)<1,Gn* is transcendental. By Lemma 1, there exists an ɛ – set E_n, such that, as z →∞ in CE_n,

where E_n contains all zeros and poles of G_n(z). So, there exists a subset F₁ ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F₁, z ∉ E_n and

By Lemma 2, for any given ɛ(0 < 2ɛ < 1 − σ), there exists a subset F₂ ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F₂,

Set an En* consists of all zeros and poles of Gn*(z), then there exists a subset F₃ ⊂ (1,∞) of finite logarithmic measure such that if z∈En*, then |z| = r ∈ F₃. Thus, by (3.2) and (3.3), we see that for large |z| = r ∉ [0, 1] ∪ F₁ ∪ F₂ ∪ F₃, Gn*(z) has no zero and pole on |z| = r, and

holds on |z| = r. Applying the Rouché’s theorem to function z and Gn*(z), we obtain that

Applying Lemma 3 (Generation of second fundamental theorem) to function G_n(z), we have

Since N¯(r,Gn(z))=N¯(r,Gn*(z)),T(r,Gn*(z))=T(r,Gn(z))+S(r,Gn), by (3.5) and (3.6), we have

Thus, by the definition of deficiency, we have

Assume further that δ(0,G_n) > 1–cos πα and by the proof of Theorem D(1)(see[4]), we have

Proof of Theorem 2

Since f is a transcendental meromorphic function of order of growth σ(f)=12 and f has finite many poles, we set f(z)=g(z)d(z), where g(z) is an entire function and d(z) is a polynomial. By Lemma 1, we know that there exists an ɛ – set E_n, such that

as z →∞in C E_n. By Lemma 4, there exists a sequence {r_j}, r_j →∞, such that for all z satisfying |z| = r_j, |g(z)| = M(r_j, g), when j sufficiently large, we have

where υ_g(r) is the central index of g(z). By (3.10) and (3.11), we have

for the sequence {r_j}, r_j → ∞. Assume that G_n(z) is a rational function. Set H = {|z| = r : r ∈ E_n}. Then by Remark 1, H is of finite logarithmic measure. Set the logarithmic measure of H by lm(H) = logκ < ∞, then for the above sequence {r_j}, there is a point rj′∈[rj,(1+κ)rj]H. Since

We have

By (3.14), for any given ɛ(0 < ɛ < 1 − σ), we get that for sufficiently large j,

Since (σ −1+ɛ)n < 0 and G_n(z) is rational function, by (3.12) and (3.15), we can deduce that, as z →∞,

where β ≠ 0 is a constant and k is a positive integer. Since ɛ is arbitrary, by (3.13), (3.15) and (3.16), we have

Thus, n = 2k. Since k is a positive integer, n is even, it is contradicts the hypothesis. Hence, G_n(z) is transcendental. Since the poles of f is finite, we have δ(∞, f) = 1. By Theorem D(2), G_n(z) has infinitely many zeros.

Proof of Theorem 3

Using the Wiman-Valiron theory and by the same argument of the proof of Theorem 2, we can prove it.