Uniqueness of Entire Functions concerning Difference Operator

Abstract

We deal with a uniqueness question of entire functions sharing a nonzero value with their difference operators and obtain some results, which improve the results of Qi et al. (2010) and Zhang (2011).

1. Introduction and Main Results

In this paper, a meromorphic function will mean meromorphic in the whole complex plane. We will use the standard notations of Nevanlinna’s value distribution theory such as 𝑇(𝑟,𝑓), 𝑁(𝑟,𝑓), 𝑁(𝑟,𝑓), and 𝑚(𝑟,𝑓), as explained in Hayman [1], Yang [2], and Yang and Yi [3]. We denote by 𝑆(𝑟,𝑓) any quantity satisfying 𝑆(𝑟,𝑓)=𝑜(𝑇(𝑟,𝑓)), as 𝑟→∞ possibly outside a set of finite linear measures. For 𝑓 meromorphic in ℂ, denote by 𝑆(𝑓) the family of all meromorphic functions 𝑎(𝑧) that satisfy 𝑇(𝑟,𝑎)=𝑜(𝑇(𝑟,𝑓)) for 𝑟→∞ outside a possible exceptional set of finite linear measure. In addition, we denote by 𝜌(𝑓) and 𝜌₂(𝑓) the order of 𝑓 and the hyperorder of 𝑓 [3, 4]. Moreover, we define difference operators by Δ_𝑐𝑓=𝑓(𝑧+𝑐)−𝑓(𝑧) where 𝑐 is a nonzero constant. If 𝑐=1, we use the usual difference notation Δ_𝑐𝑓=Δ𝑓.

Let 𝑓 and 𝑔 be two nonconstant meromorphic functions and 𝑎 be a finite complex number. We say that 𝑓, 𝑔 share the value 𝑎 CM (counting multiplicities) if 𝑓, 𝑔 have the same 𝑎-points with the same multiplicities, and we say that 𝑓, 𝑔 share the value 𝑎 IM (ignoring multiplicities) if we do not consider the multiplicities. We denote by 𝑁_𝐿(𝑟,1/(𝑓−𝑎)) the counting function for 𝑎-points of both 𝑓 and 𝑔 about which 𝑓 has larger multiplicity than 𝑔, with multiplicity not being counted. Similarly, we have the notation 𝑁_𝐿(𝑟,1/(𝑔−𝑎)). Next, we denote by 𝑁₀(𝑟,1/𝐹^′) the counting function of those zeros of 𝐹^′ that are not the zeros of 𝐹(𝐹−1) and denote by 𝑁₁₁(𝑟,1/(𝑓−𝑎)) the counting function for common simple 1-point of both 𝑓 and 𝑔. In addition, we need the following three definitions.

Definition 1. Let 𝑘 be a positive integer. Let 𝑓 and 𝑔 be two nonconstant meromorphic functions such that 𝑓 and 𝑔 share the value 1 IM. Let 𝑧₀ be a 1-point of 𝑓 with multiplicity 𝑝 and a 1-point of 𝑔 with multiplicity 𝑞. We denote by 𝑁_𝑓>𝑘(𝑟,1/(𝑔−1)) the reduced counting function of those 1-points of 𝑓 and 𝑔 such that 𝑝>𝑞=𝑘. 𝑁_𝑔>𝑘(𝑟,1/(𝑓−1)) is defined analogously.

Definition 2 (see [5]). Let 𝑘 be a nonnegative integer or infinity. For 𝑎∈ℂ∪∞, we denote by 𝐸_𝑘(𝑎,𝑓) the set of all 𝑎-points of 𝑓, where an 𝑎-point of multiplicity 𝑚 is counted 𝑚 times if ≤𝑘 and 𝑘+1 times if 𝑚>𝑘. If 𝐸_𝑘(𝑎,𝑓)=𝐸_𝑘(𝑎,𝑔), we say that 𝑓,𝑔 share the value a with weight 𝑘.
The definition implies that if 𝑓, 𝑔 share a value a with weight 𝑘, then 𝑧₀ is an 𝑎-point of 𝑓 with multiplicity 𝑚(≤𝑘) if and only if it is an 𝑎-point of 𝑔 with multiplicity 𝑚(≤𝑘) and 𝑧₀ is an 𝑎-point of 𝑓 with multiplicity 𝑚(>𝑘) if and only if it is an 𝑎-point of 𝑔 with multiplicity 𝑛(>𝑘), where 𝑚 is not necessarily equal to 𝑛.
We write that 𝑓, 𝑔 share (𝑎,𝑘) to mean that 𝑓, 𝑔 share the value a with weight 𝑘. Clearly if 𝑓, 𝑔 share (𝑎,𝑘), then 𝑓, 𝑔 share (𝑎,𝑝) for any integer 𝑝, 0≤𝑝<𝑘. Also we note that 𝑓, 𝑔 share a value 𝑎 IM or CM if and only if 𝑓, 𝑔 share (𝑎,0) or (𝑎,∞), respectively.

Definition 3. Let 𝑓 be a nonconstant meromorphic function, and let 𝑝 be a positive integer and 𝑎∈𝐶∪{∞}. Then, by 𝑁_𝑝)(𝑟,1/(𝑓−𝑎)), we denote the counting function of those 𝑎-points of 𝑓 (counted with proper multiplicities) whose multiplicities are not greater than 𝑝, and by 𝑁_𝑝)(𝑟,1/(𝑓−𝑎)), we denote the corresponding reduced counting function (ignoring multiplicities). By 𝑁_(𝑝(𝑟,1/(𝑓−𝑎)), we denote the counting function of those 𝑎-points of 𝑓 (counted with proper multiplicities) whose multiplicities are not less than 𝑝, and by 𝑁_(𝑝(𝑟,1/(𝑓−𝑎)), we denote the corresponding reduced counting function (ignoring multiplicities), where 𝑁_𝑝)(𝑟,1/(𝑓−𝑎)), 𝑁_𝑝)(𝑟,1/(𝑓−𝑎)), 𝑁_(𝑝(𝑟,1/(𝑓−𝑎)), and 𝑁_(𝑝(𝑟,1/(𝑓−𝑎)) mean 𝑁_𝑝)(𝑟,𝑓), 𝑁_𝑝)(𝑟,𝑓), 𝑁_(𝑝(𝑟,𝑓), and 𝑁_(𝑝(𝑟,𝑓), respectively, if 𝑎=∞.
In 2010, Qi et al. [6] proved the following uniqueness theorem.

Theorem A. Let 𝑓 and 𝑔 be transcendental entire functions of finite order, let 𝑐 be a nonzero complex constant, and let 𝑛≥6 be an integer. If 𝑓(𝑧)^𝑛𝑓(𝑧+𝑐) and 𝑔(𝑧)^𝑛𝑔(𝑧+𝑐) share 𝑧 CM, then 𝑓=𝑡𝑔 for a constant 𝑡 that satisfies 𝑡^𝑛+1=1.

In 2011, Zhang et al. [7] complemented the above theorem and obtained the following result.

Theorem B. Let 𝑓 and 𝑔 be nonconstant entire functions of finite order, and let 𝑛≥5 be an integer. Suppose that 𝑐 is a nonzero complex constant such that Δ_𝑐𝑓≢0 and Δ_𝑐𝑔≢0. If 𝑓^𝑛Δ_𝑐𝑓 and 𝑔^𝑛Δ_𝑐𝑔 share 𝑧 CM, and 𝑔(𝑧+𝑐) and 𝑔(𝑧) share 0 CM then 𝑓=𝑡𝑔, where 𝑡 is a constant satisfying 𝑡^𝑛+1=1.

In this paper, we complement Theorems A and B and obtain the following results which generalize the above theorems.

Theorem 4. Let 𝑓 be a transcendental entire function of finite order and Δ_𝑐𝑓≢0, let 𝑎≠0 be a small function with respect to 𝑓, and let 𝑐 be a nonzero complex constant. Then for 𝑛≥2, 𝑓(𝑧)^𝑛(𝑓(𝑧+𝑐)−1)Δ_𝑐𝑓−𝑎 has infinitely many zeros.

Theorem 5. Let 𝑓(𝑧) and 𝑔(𝑧) be transcendental entire functions of 𝜌₂<1, 𝑛≥2𝑘+7. Suppose that 𝑐 is a nonzero complex constant such that Δ_𝑐𝑓≢0 and Δ_𝑐𝑔≢0. If [𝑓^𝑛Δ_𝑐𝑓]^(𝑘) and [𝑔^𝑛Δ_𝑐𝑔]^(𝑘) share 1 CM, then 𝑓=𝑡𝑔 for a constant 𝑡 with 𝑡^𝑛+1=1.

Theorem 6. Let 𝑓 and 𝑔 be transcendental entire functions of 𝜌₂<1, 𝑛≥5𝑘+13. 𝑐 is a nonzero complex constant such that Δ_𝑐𝑓≢0 and Δ_𝑐𝑔≢0. If [𝑓^𝑛Δ_𝑐𝑓]^(𝑘) and [𝑔^𝑛Δ_𝑐𝑔]^(𝑘) share 1 IM, then 𝑓=𝑡𝑔 for a constant 𝑡 with 𝑡^𝑛+1=1.

2. Some Lemmas

Lemma 7 (see [8]). Let 𝑓 be a nonconstant meromorphic function of finite order 𝜎, and let 𝑐 be a nonzero constant. Then, for each 𝜀>0, 𝑇𝑟,𝑓(𝑧+𝑐)=𝑇𝑟,𝑓(𝑧)+𝑂𝑟^{𝜎−1+𝜀}+𝑂log𝑟. (1)

Lemma 8 (see [9]). Let 𝑓 be a meromorphic function of finite order, and let 𝑐∈ℂ and 𝛿∈(0,1). Then 𝑚𝑓()𝑟,𝑧+𝑐𝑓(𝑧)𝑓(𝑧)+𝑚𝑟,𝑓()𝑇𝑧+𝑐=𝑜𝑟,𝑓𝑟^𝛿=𝑆𝑟,𝑓. (2)

Lemma 9 (see [10]). Let 𝑓₁, 𝑓₂, and 𝑓₃ be nonconstant meromorphic functions such that 𝑓₁+𝑓₂+𝑓₃=1. If 𝑓₁, 𝑓₂, and 𝑓₃ are linearly independent, then 𝑇𝑟,𝑓₁≤³_𝑗=1𝑁₂1𝑟,𝑓_𝑗+³_𝑗=1𝑁𝑟,𝑓_𝑗+𝑜(𝑇(𝑟)), (3) where 𝑇(𝑟)=max_1≤𝑗≤3𝑇(𝑟,𝑓_𝑗), 𝑟∉𝐸, and 𝐸 denote a set of positive real numbers of finite linear measure.

Lemma 10. Let 𝑓 be transcendental entire functions of finite order, let 𝑐 be a nonzero complex constant, and set 𝐹(𝑧)=𝑓(𝑧)^𝑛Δ_𝑐𝑓; then 𝑛𝑇𝑟,𝑓+𝑆𝑟,𝑓≤𝑇(𝑟,𝐹)≤(𝑛+1)𝑇𝑟,𝑓+𝑆𝑟,𝑓. (4)

Proof. Since 𝑇(𝑟,𝐹)=𝑇𝑟,𝑓(𝑧)^𝑛Δ_𝑐𝑓≤𝑛𝑇𝑟,𝑓+𝑇𝑟,Δ_𝑐𝑓≤𝑛𝑇𝑟,𝑓+𝑚𝑟,Δ_𝑐𝑓=()𝑇,≤𝑛𝑇𝑟,𝑓+𝑚𝑟,𝑓+𝑆𝑟,𝑓𝑛+1𝑟,𝑓+𝑆𝑟,𝑓 (5) then (𝑛+1)𝑇𝑟,𝑓=𝑇𝑟,𝑓(𝑧)^𝑛+1=𝑚𝑟,𝑓(𝑧)^𝑛+1𝑓(𝑧)≤𝑚𝑟,^𝑛+1𝐹()𝑓(𝑧)+𝑚𝑟,𝐹+𝑆𝑟,𝑓≤𝑚𝑟,Δ_𝑐𝑓()𝑓(𝑧)+𝑚𝑟,𝐹+𝑆𝑟,𝑓≤𝑇𝑟,Δ_𝑐𝑓()Δ+𝑇𝑟,𝐹+𝑆𝑟,𝑓≤𝑇𝑟,^𝑐𝑓𝑓(𝑧)()Δ+𝑇𝑟,𝐹+𝑆𝑟,𝑓=𝑚𝑟,^𝑐𝑓𝑓(𝑧)Δ+𝑁𝑟,^𝑐𝑓𝑓(𝑧)()()1+𝑇𝑟,𝐹+𝑆𝑟,𝑓≤𝑇𝑟,𝐹+𝑁𝑟,𝑓(𝑧)().+𝑆𝑟,𝑓≤𝑇𝑟,𝐹+𝑇𝑟,𝑓+𝑆𝑟,𝑓 (6) That is, 𝑛𝑇𝑟,𝑓+𝑆𝑟,𝑓≤𝑇(𝑟,𝐹)≤(𝑛+1)𝑇𝑟,𝑓+𝑆𝑟,𝑓. (7)

Lemma 11 (see [11]). Let 𝑓₁ and 𝑓₂ be two nonconstant meromorphic functions. If 𝑐₁𝑓₁+𝑐₂𝑓₂=𝑐₃, where 𝑐₁, 𝑐₂, and 𝑐₃ are nonzero constants, then 𝑇𝑟,𝑓₁≤𝑁𝑟,𝑓₁+𝑁1𝑟,𝑓₁+𝑁1𝑟,𝑓₂+𝑆𝑟,𝑓₁. (8)

Lemma 12 (see [12]). Let 𝑓(𝑧) be a nonconstant meromorphic function, and let 𝑘 be a positive integer. Suppose that 𝑓^(𝑘)≢0; then 𝑁1𝑟,𝑓^(𝑘)1≤𝑁𝑟,𝑓+𝑘𝑁𝑟,𝑓+𝑆𝑟,𝑓. (9)

Lemma 13 (see [13]). Let 𝑓, 𝑔 share (1,0). Then(i)𝑁_𝑓>1(𝑟,1/(𝑔−1))≤𝑁(𝑟,1/𝑓)+𝑁(𝑟,𝑓)−𝑁₀(𝑟,1/𝑓^′)+𝑆(𝑟,𝑓),(ii)𝑁_𝑔>1(𝑟,1/(𝑓−1))≤𝑁(𝑟,1/𝑔)+𝑁(𝑟,𝑔)−𝑁₀(𝑟,1/𝑔^′)+𝑆(𝑟,𝑔).

Lemma 14. Let 𝑓(𝑧) and 𝑔(𝑧) be two nonconstant entire functions. If 𝑓 and 𝑔 share 1 IM, then one of the following cases holds: (i)𝑇(𝑟,𝑔)≤𝑁₂(𝑟,1/𝑔)+𝑁₂(𝑟,1/𝑓)+𝑁(𝑟,1/𝑓)+2𝑁(𝑟,1/𝑔)+𝑆(𝑟,𝑓)+𝑆(𝑟,𝑔), the same inequality holding for 𝑇(𝑟,𝑓);(ii)𝑓≡(𝐴𝑔+𝐵)/(𝐶𝑔+𝐷), where A, B, C, and D are finite complex numbers satisfying 𝐴𝐷≠𝐵𝐶.

Proof. Let Φ(𝑧)=𝑓^′′𝑓^′𝑓−2^′−𝑔𝑓−1𝑔^′′′𝑔+2^′.𝑔−1 (10) Clearly 𝑚(𝑟,Φ)=𝑆(𝑟,𝑓)+𝑆(𝑟,𝑔). We consider the cases Φ(𝑧)≢0 and Φ(𝑧)≡0.
If Φ(𝑧)≢0, then if 𝑧₀ is a common simple 1-point of 𝑓^′ and 𝑔^′, substituting their Taylor series at 𝑧₀ into (10), we see that 𝑧₀ is a zero of Φ(𝑧). Thus, we have 𝑁₁₁1𝑟,𝑓−1=𝑁₁₁1𝑟,≤𝑔−1𝑁1𝑟,Φ()(1)().≤𝑇𝑟,Φ+𝑂≤𝑁𝑟,Φ+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (11)
Our assumptions are that Φ(𝑧) has poles; all are simple only at zeros of 𝑓^′and 𝑔^′ and poles of 𝑓 and 𝑔, and 1-points of 𝑓 whose multiplicities are not equal to the multiplicities of the corresponding 1-points of 𝑔. Thus, we deduce from (10) that 𝑁(𝑟,Φ)≤𝑁₍₂1𝑟,𝑓+𝑁₍₂1𝑟,𝑔+𝑁₀1𝑟,𝑓^′+𝑁₀1𝑟,𝑔^′+𝑁_𝐿1𝑟,+𝑓−1𝑁_𝐿1𝑟,,𝑔−1 (12) where 𝑁₀(𝑟,1/𝑓^′) is the counting function which only counts those points such that 𝑓^′=0, but 𝑓(𝑓−1)≠0. By the second fundamental theorem, we have 𝑇𝑟,𝑔≤𝑁1𝑟,𝑔+𝑁1𝑟,𝑔−1−𝑁₀1𝑟,𝑔^′+𝑆𝑟,𝑔, (13) since 𝑁1𝑟,𝑔−1=𝑁₁₁1𝑟,+𝑔−1𝑁₍₂1𝑟,+𝑓−1𝑁_𝑔>11𝑟,.𝑓−1 (14) Thus, we deduce from (11)–(14) that 𝑇𝑟,𝑔≤𝑁1𝑟,𝑔+𝑁₍₂1𝑟,𝑓+𝑁₍₂1𝑟,𝑔+𝑁₀1𝑟,𝑓^′+𝑁₍₂1𝑟,+𝑓−1𝑁_𝐿1𝑟,+𝑓−1𝑁_𝐿1𝑟,+𝑔−1𝑁_𝑔>11𝑟,.𝑓−1+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (15)
From the definition of 𝑁₀(𝑟,1/𝑓^′), we see that 𝑁₀1𝑟,𝑓^′+𝑁₍₂1𝑟,𝑓−1+𝑁₍₂1𝑟,𝑓−𝑁₍₂1𝑟,𝑓1≤𝑁𝑟,𝑓^′. (16) The above inequality and Lemma 12 give 𝑁₀1𝑟,𝑓^′+𝑁₍₂1𝑟,1𝑓−1≤𝑁𝑟,𝑓^′−𝑁₍₂1𝑟,𝑓+𝑁₍₂1𝑟,𝑓1≤𝑁𝑟,𝑓−𝑁₍₂1𝑟,𝑓+𝑁₍₂1𝑟,𝑓≤+𝑆𝑟,𝑓𝑁1𝑟,𝑓+𝑆𝑟,𝑓. (17) Substituting (17) in (15), we get 𝑇𝑟,𝑔≤𝑁1𝑟,𝑔+𝑁₍₂1𝑟,𝑓+𝑁₍₂1𝑟,𝑔+𝑁1𝑟,𝑓+𝑁_𝐿1𝑟,+𝑓−1𝑁_𝐿1𝑟,+𝑔−1𝑁_𝑔>11𝑟,𝑓−1+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤𝑁₂1𝑟,𝑔+𝑁₍₂1𝑟,𝑓+𝑁1𝑟,𝑓+𝑁_𝐿1𝑟,+𝑓−1𝑁_𝐿1𝑟,+𝑔−1𝑁_𝑔>11𝑟,,𝑓−1+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (18) since 𝑁_𝐿1𝑟,1𝑓−1≤𝑁𝑟,−𝑓−1𝑁1𝑟,𝑓𝑓−1≤𝑁𝑟,𝑓^′𝑓≤𝑁𝑟,^′𝑓≤+𝑆𝑟,𝑓𝑁1𝑟,𝑓+𝑆𝑟,𝑓. (19) Similarly, 𝑁_𝐿1𝑟,≤𝑔−1𝑁1𝑟,𝑔+𝑆𝑟,𝑔. (20)
Combining the above inequalities, Lemma 13, and (18), we obtain 𝑇𝑟,𝑔≤𝑁₂1𝑟,𝑔+𝑁₂1𝑟,𝑓+𝑁1𝑟,𝑓+2𝑁1𝑟,𝑔−𝑁₀1𝑟,𝑔^′+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤𝑁₂1𝑟,𝑔+𝑁₂1𝑟,𝑓+𝑁1𝑟,𝑓+2𝑁1𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔. (21) Thus, we obtain (i).
If Φ(𝑧)≡0, then by (10), we have 𝑓^′′𝑓^′−2𝑓^′≡𝑔𝑓−1𝑔^′′′−2𝑔^′.𝑔−1 (22) By integrating two sides of the above equality, we obtain 𝑓≡𝐴𝑔+𝐵,𝐶𝑔+𝐷 (23) where 𝐴, 𝐵, 𝐶, and 𝐷 are finite complex numbers satisfying 𝐴𝐷≠𝐵𝐶. This proves the lemma.

Lemma 15 (see [14]). Let 𝑓(𝑧) be a nonconstant meromorphic function, 𝑠, 𝑘 be two positive integers; then 𝑁_𝑠1𝑟,𝑓^(𝑘)≤𝑇𝑟,𝑓^(𝑘)−𝑇𝑟,𝑓+𝑁_𝑠+𝑘1𝑟,𝑓,𝑁+𝑆𝑟,𝑓_𝑠1𝑟,𝑓^(𝑘)≤𝑘𝑁𝑟,𝑓+𝑁_𝑠+𝑘1𝑟,𝑓+𝑆𝑟,𝑓. (24) Clearly, 𝑁(𝑟,1/𝑓^(𝑘))=𝑁₁(𝑟,1/𝑓^(𝑘)).

Lemma 16 (see [15]). Let 𝑎₀(𝑧),𝑎₁(𝑧),…,𝑎_𝑛(𝑧),𝑏(𝑧) be polynomials such that 𝑎₀(𝑧)𝑎_𝑛(𝑧)≢0; let 𝑐_𝑗 be constants and deg_{deg𝑎𝑗=𝑑}𝑎_𝑗=𝑑, (25) where 𝑑=max_{0≤𝑗≤𝑛}{deg𝑎_𝑗}. If 𝑓(𝑧) is a transcendental meromorphic solution of ^𝑛_𝑗=0𝑎_𝑗(𝑧)𝑓𝑧+𝑐_𝑗=𝑏(𝑧), (26) then 𝜌(𝑓)≥1.

3. Proof of Theorems

3.1. Proof of Theorem 4

Let 𝐺(𝑧)=𝑓(𝑧)^𝑛(𝑓(𝑧+𝑐)−1)Δ_𝑐𝑓. Since 𝑓 is a transcendental entire function of finite order, from Lemma 7, we have ()𝑇(𝑧)(𝑧)𝑛+2𝑟,𝑟,𝑓≤𝑇𝑟,𝑓^𝑛+1𝑓()(𝑧)𝑧+𝑐−1+𝑆𝑟,𝑓≤𝑚𝑟,𝑓^𝑛+1𝑓()𝑓(𝑧)𝑧+𝑐−1+𝑆𝑟,𝑓≤𝑚𝑟,^𝑛+1𝑓(𝑧+𝑐)−1𝐺()().+𝑚𝑟,𝐺+𝑆𝑟,𝑓≤𝑇𝑟,𝐺+𝑆𝑟,𝑓 (27) By the second main theorem, we deduce that 𝑇(𝑟,𝐺)≤𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,()≤𝐺−𝑎+𝑆𝑟,𝐺𝑁1𝑟,+𝐺−𝑎𝑁1𝑟,𝑓+𝑁1𝑟,𝑓()+𝑧+𝑐−1𝑁1𝑟,Δ_𝑐𝑓≤+𝑆𝑟,𝑓𝑁1𝑟,+𝐺−𝑎𝑁1𝑟,𝑓+𝑇𝑟,𝑓(𝑧+𝑐)−1+𝑇𝑟,Δ_𝑐𝑓≤+𝑆𝑟,𝑓𝑁1𝑟,+𝐺−𝑎𝑁1𝑟,𝑓()Δ+𝑇𝑟,𝑓𝑧+𝑐−1+𝑚𝑟,^𝑐𝑓𝑓≤⋅𝑓+𝑆𝑟,𝑓𝑁1𝑟,+𝐺−𝑎𝑁1𝑟,𝑓()Δ+𝑇𝑟,𝑓𝑧+𝑐−1+𝑚𝑟,^𝑐𝑓𝑓≤+𝑚𝑟,𝑓+𝑆𝑟,𝑓𝑁1𝑟,.𝐺−𝑎+3𝑇𝑟,𝑓+𝑆𝑟,𝑓 (28) According to (27) and (28), we have (𝑛−1)𝑇𝑟,𝑓≤𝑁1𝑟,.𝐺−𝑎+𝑆𝑟,𝑓 (29) Noting that 𝑛≥2, we get that 𝐺−𝑎 has infinitely many zeros.

This completes the proof of Theorem 4.

3.2. Proof of Theorem 5

Since [𝑓(𝑧)^𝑛Δ_𝑐𝑓]^(𝑘) and [𝑔(𝑧)^𝑛Δ_𝑐𝑔]^(𝑘) share 1 CM, we have 𝑓(𝑧)^𝑛Δ_𝑐𝑓^(𝑘)−1𝑔(𝑧)^𝑛Δ_𝑐𝑔^(𝑘)−1=𝑒^ℎ(𝑧), (30) where ℎ(𝑧) is a polynomial. Set 𝐹=𝑓(𝑧)^𝑛Δ_𝑐𝑓, 𝐺=𝑔(𝑧)^𝑛Δ_𝑐𝑔, 𝐹₁=𝐹^(𝑘),𝐹₂=−𝑒^ℎ(𝑧)𝐺^(𝑘),𝐹₃=𝑒^ℎ(𝑧),then𝐹₁+𝐹₂+𝐹₃𝑇(𝑟)=1,=max_1≤𝑗≤3𝑇𝑟,𝐹_𝑗,𝑆(𝑟)=𝑜(𝑇(𝑟)). (31)

Next, we will prove that 𝐹₁, 𝐹₂, and 𝐹₃ are linearly dependent and either 𝐹₂ or 𝐹₃ is a constant.

Now, we suppose that neither 𝐹₂ nor 𝐹₃ is a constant and 𝐹₁, 𝐹₂, and 𝐹₃ are linearly independent; then by Lemma 9, we have 𝑇𝑟,𝐹₁≤³_𝑗=1𝑁₂1𝑟,𝐹_𝑗+³_𝑗=1𝑁𝑟,𝐹_𝑗+𝑜(𝑇(𝑟)). (32) Since 𝐹_𝑗(𝑗=1,2,3) are entire functions, by the above inequality, we get 𝑇𝑟,𝐹₁≤𝑁₂1𝑟,𝐹^(𝑘)+𝑁₂1𝑟,𝐺^(𝑘)+𝑜(𝑇(𝑟)). (33) From (33) and the first main theorem, we have 𝑇1𝑟,𝐹^(𝑘)=𝑇𝑟,𝐹^(𝑘)+𝑂(1)=𝑇𝑟,𝐹₁+𝑂(1)≤𝑁₂1𝑟,𝐹^(𝑘)+𝑁₂1𝑟,𝐺^(𝑘)(𝑇(𝑟1+𝑜))≤𝑁𝑟,𝐹^(𝑘)−𝑁₍₃1𝑟,𝐹^(𝑘)−2𝑁₍₃1𝑟,𝐹^(𝑘)1+𝑁𝑟,𝐺^(𝑘)−𝑁₍₃1𝑟,𝐺^(𝑘)−2𝑁₍₃1𝑟,𝐺^(𝑘)+𝑜(𝑇(𝑟)). (34) Assuming that 𝑧₀ is zero of 𝑓(𝑧) (or 𝑔(𝑧)) with multiplicity 𝑝, if 𝑧₀ is zero of 𝑓(𝑧+𝑐) (or 𝑔(𝑧+𝑐)) with multiplicity 𝑞(≥1), let 𝑚=min{𝑝,𝑞}, then 𝑧₀ is a zero of 𝐹^(𝑘) (or 𝐺^(𝑘)) with multiplicity 𝑛𝑝+𝑚−𝑘≥𝑛𝑝−𝑘≥3, and if 𝑧₀ is not zero of 𝑓(𝑧+𝑐) (or 𝑔(𝑧+𝑐)), then 𝑧₀ is a zero of 𝐹^(𝑘) (or 𝐺^(𝑘)) with multiplicity 𝑛𝑝−𝑘≥3. Therefore, we get that 𝑁₍₃1𝑟,𝐹^(𝑘)−2𝑁₍₃1𝑟,𝐹^(𝑘)≥()𝑁1𝑛−𝑘−2𝑟,𝑓,𝑁₍₃1𝑟,𝐺^(𝑘)−2𝑁₍₃1𝑟,𝐺^(𝑘)≥()𝑁1𝑛−𝑘−2𝑟,𝑔, (35)(36) since 1𝑛𝑚𝑟,𝑓1=𝑚𝑟,𝑓^𝑛Δ=𝑚𝑟,^𝑐𝑓𝐹1≤𝑚𝑟,𝐹Δ+𝑚𝑟,^𝑐𝑓𝑓𝐹⋅𝑓≤𝑚𝑟,^(𝑘)𝐹⋅1𝐹^(𝑘)Δ+𝑚𝑟,^𝑐𝑓𝑓1+𝑚𝑟,𝑓+𝑆𝑟,𝑓≤𝑚𝑟,𝐹^(𝑘)1+𝑇𝑟,𝑓+𝑆𝑟,𝑓=𝑇𝑟,𝐹^(𝑘)1−𝑁𝑟,𝐹^(𝑘)+𝑇𝑟,𝑓+𝑆𝑟,𝑓. (37) Therefore, from (34), (35), (36), (37), and Lemma 12, ()𝑇≤()𝑁1𝑛−1𝑟,𝑓𝑘+2𝑟,𝑓+()𝑁1𝑘+2𝑟,𝑔+𝑇𝑟,𝑔+𝑜(𝑇(𝑟)). (38) On the other hand, from (30), we have 𝐺^(𝑘)+𝑒^−ℎ−𝑒^−ℎ𝐹^(𝑘)=1. Obviously, according to our assumptions, neither 𝑒^−ℎ nor 𝑒^−ℎ𝐹^(𝑘) is a constant and 𝐹₁, 𝐹₂, and 𝐹₃ are linearly independent. Similarly, we have ()𝑇≤()𝑁1𝑛−1𝑟,𝑔𝑘+2𝑟,𝑔+()𝑁1𝑘+2𝑟,𝑓+𝑇𝑟,𝑓+𝑜(𝑇(𝑟)). (39) From (38) and (39), we obtain that [𝑛−2𝑘−6]𝑇𝑟,𝑓+𝑇𝑟,𝑔≤𝑜(𝑇(𝑟)), (40) which is a contradiction to 𝑛≥2𝑘+7.

Therefore, 𝐹₁, 𝐹₂, and 𝐹₃ are linearly dependent, and there exist constants 𝐶₁, 𝐶₂, 𝐶₃ which are not all equal to zero such that 𝐶₁𝐹₁+𝐶₂𝐹₂+𝐶₃𝐹₃=0. (41)

Suppose that 𝐶₁=0; we have 𝐶₂𝐹₂+𝐶₃𝐹₃=0. If 𝐶₂≠0, we get 𝐹₂=−(𝐶₃/𝐶₂)𝐹₃; that is, 𝐺^(𝑘)=𝐶₃/𝐶₂; thus 𝑔(𝑧) is a polynomial; it is impossible. Similarly, if 𝐶₂=0, we also deduce a contradiction.

Suppose that 𝐶₁≠0, from (41); we know that (𝐶₂,𝐶₃)≠(0,0). If 𝐶₂≠0, from (41), we have 𝐶1−²𝐶₁𝐹₂+𝐶1−³𝐶₁𝐹₃=1 (42) and 𝐶₁≠𝐶₂, 𝐶₁≠𝐶₃. That is, 𝐶1−²𝐶₁𝐺^(𝑘)+1𝑒^ℎ𝐶=1−³𝐶₁. (43) From Lemma 11, we have 𝑇𝑟,𝐺^(𝑘)≤𝑁1𝑟,𝐺^(𝑘)+𝑁𝑟,𝐺^(𝑘)+𝑁𝑟,𝑒^ℎ=+𝑆𝑟,𝑔𝑁1𝑟,𝐺^(𝑘)1+𝑆𝑟,𝑔≤𝑁𝑟,𝐺^(𝑘)−𝑁₍₂1𝑟,𝐺^(𝑘)−𝑁₍₂1𝑟,𝐺^(𝑘)+𝑆𝑟,𝑔. (44) By the similar argument in (37), we have 1𝑛𝑚𝑟,𝑔1≤𝑇𝑟,𝐺^(𝑘)1−𝑁𝑟,𝐺^(𝑘)+𝑇𝑟,𝑔+𝑆𝑟,𝑔. (45)

From 𝑛≥2𝑘+7>𝑘+2, if 𝑧₀ is zero of 𝑔(𝑧) with multiplicity 𝑝, then 𝑧₀ is a zero of 𝐺^(𝑘) with multiplicity 𝑛𝑝−𝑘≥2, and we get 𝑁₍₂1𝑟,𝐺^(𝑘)−𝑁₍₂1𝑟,𝐺^(𝑘)≥()𝑁1𝑛−𝑘−1𝑟,𝑔. (46) According to (44), (45), and (46), we have ()𝑇≤()𝑁1𝑛−1𝑟,𝑔𝑘+1𝑟,𝑔+𝑆𝑟,𝑔, (47) which is a contradiction to 𝑛≥2𝑘+7.

Therefore, 𝐶₂=0, 𝐶₃≠0, which gives (1−𝐶₁/𝐶₃)𝐹₁+𝐹₂=1. Similarly, we derive a contradiction by calculation.

Hence, we deduce that either 𝐹₂ or 𝐹₃ is a constant.

Suppose 𝐹₂=𝑐≠1; from 𝐹₁+𝐹₂+𝐹₃=1, we have 𝐹^(𝑘)+𝑒^ℎ=1−𝑐; in the same manner as above, we get a contradiction. Therefore, 𝑐=1; that is, 𝐹₂=1. Suppose 𝐹₃=𝑐≠1; similarly as above, we get 𝑐=1; that is, 𝐹₃=1.

Therefore, we conclude that 𝐹₂=1 or 𝐹₃=1.

If 𝐹₂=1, since 𝐹₁+𝐹₂+𝐹₃=1, we have 𝐹₁=−𝐹₃=−𝑒^ℎ(𝑧). That is 𝑓^𝑛Δ_𝑐𝑓^(𝑘)⋅𝑔^𝑛Δ_𝑐𝑔^(𝑘)≡1. (48) Since 𝑛≥2𝑘+7 and 𝑓 and 𝑔 are transcendental entire functions with hyperorder less than one, we get that 𝑓 and 𝑔 have no zeros. Thus, 𝑓(𝑧)=𝑒^𝑎(𝑧),𝑔(𝑧)=𝑒^𝑏(𝑧), (49) where 𝑎(𝑧), 𝑏(𝑧) are nonzero polynomials.

Substitute (49) into (48); we have 𝑒^{𝑛𝑎(𝑧)}𝑒^{𝑎(𝑧+𝑐)}−𝑒^{𝑎(𝑧)(𝑘)}⋅𝑒^{𝑛𝑏(𝑧)}𝑒^{𝑏(𝑧+𝑐)}−𝑒^{𝑏(𝑧)(𝑘)}≡1. (50) Let 𝑛𝑎(𝑧)+𝑎(𝑧+𝑐)=𝐴₁, 𝑛𝑎(𝑧)+𝑎(𝑧)=𝐴₂, 𝑛𝑏(𝑧)+𝑏(𝑧+𝑐)=𝐵₁, and 𝑛𝑏(𝑧)+𝑏(𝑧)=𝐵₂. If 𝑘=1, we have 𝐴^′₁𝑒^𝐴1−𝐴^′₂𝑒^𝐴2⋅𝐵^′₁𝑒^𝐵1−𝐵^′₂𝑒^𝐵2≡1. (51) From (51), we know that 𝐴^′₁𝑒^𝐴1−𝐴^′₂𝑒^𝐴2=𝑒^𝐴2(𝐴^′₁𝑒^{𝐴1−𝐴2}−𝐴^′₂)≠0; If 𝐴^′₁≠0, then we have 𝐴^′₂=0; thus, 𝐴₂ must be a constant. By Lemma 16, we have 𝜌(𝑎(𝑧))≥1; thus, 𝜌₂(𝑓)≥1, which is a contradiction. If 𝐴^′₁=0, then 𝐴₁ must be a constant; similarly, we also deduce a contradiction.

If 𝑘=2, by calculation, we have 𝐴₁^′′𝑒^𝐴1+𝐴^′₁²𝑒^𝐴1−𝐴₂^′′𝑒^𝐴2−𝐴^′₂²𝑒^𝐴2=𝑒^𝐴2𝑒^{𝐴1−𝐴2}𝐴₁^′′+𝐴^′₁²−𝐴₂^′′+𝐴^′₂²≠0. (52) If 𝐴₁^′′+(𝐴^′₁)²≠0, then 𝐴₂^′′+(𝐴^′₂)²=0. If 𝐴₂ is transcendental entire, then we have 𝑚𝑟,𝐴^′₂𝐴=𝑚^2′′𝐴^′₂=𝑆𝑟,𝐴^′₂, (53) which is a contradiction to 𝐴^′₂ being transcendental entire. If 𝐴₂ is a polynomial, from Lemma 16, which induces that 𝜌₂(𝑓)≥1, we get a contradiction. If 𝐴₁^′′+(𝐴^′₁)²=0, similar as above, we get a contradiction. For 𝑘≥3, using the similar Method as above, we also deduce a contradiction. Therefore, There are not transcendental entire functions 𝑓(𝑧) and 𝑔(𝑧) satisfying (48).

If 𝐹₃=1, that is, 𝑒^ℎ(𝑧)=1, from (30), we get 𝑓^𝑛Δ_𝑐𝑓^(𝑘)≡𝑔^𝑛Δ_𝑐𝑔^(𝑘). (54) From (54), we have 𝑓^𝑛Δ_𝑐𝑓≡𝑔^𝑛Δ_𝑐𝑔+𝑝(𝑧), (55) where 𝑝(𝑧) is a polynomial of degree at most 𝑘−1. Suppose 𝑝(𝑧)≢0; then we get 𝑓^𝑛Δ_𝑐𝑓𝑝(𝑧)=𝑔^𝑛Δ_𝑐𝑔𝑝(𝑧)+1. (56) Therefore, from the second main theorem, we have ()𝑇𝑓𝑛+1𝑟,𝑓≤𝑇^𝑛Δ_𝑐𝑓𝑝(𝑧)≤+𝑆𝑟,𝑓𝑁𝑓^𝑛Δ_𝑐𝑓𝑝(𝑧)+𝑁𝑝(𝑧)𝑓^𝑛Δ_𝑐𝑓+𝑁𝑝(𝑧)𝑔^𝑛Δ_𝑐𝑔≤+𝑆𝑟,𝑓𝑁1𝑓+𝑁1Δ_𝑐𝑓+𝑁1𝑔+𝑁1Δ_𝑐𝑔.+𝑆𝑟,𝑓≤2𝑇𝑟,𝑓+2𝑇𝑟,𝑔+𝑆𝑟,𝑓 (57) Similarly, we have (𝑛+1)𝑇𝑟,𝑔≤2𝑇𝑟,𝑓+2𝑇𝑟,𝑔+𝑆𝑟,𝑓. (58) Therefore, (𝑇𝑇,𝑛+1)𝑟,𝑓+𝑇𝑟,𝑔≤4𝑟,𝑓+𝑇𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (59) which is a contradiction to 𝑛≥2𝑘+7. Thus, 𝑝(𝑧)≡0, which implies that 𝑓^𝑛Δ_𝑐𝑓≡𝑔^𝑛Δ_𝑐𝑔. (60)

Let 𝑓/𝑔=ℎ; if ℎ is not a constant, then by (60), we have ℎ^𝑛+1≡𝑓Δ_𝑐𝑓⋅Δ_𝑐𝑔𝑔. (61) Thus, ()𝑇()Δ𝑛+1𝑟,ℎ≤𝑇𝑟,^𝑐𝑓𝑓Δ+𝑇𝑟,^𝑐𝑔𝑔(1)Δ+𝑂≤𝑁𝑟,^𝑐𝑓𝑓Δ+𝑁𝑟,^𝑐𝑔𝑔.+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤𝑇𝑟,𝑓+𝑇𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (62) Combining 𝑇(𝑟,ℎ)=𝑇(𝑟,𝑓/𝑔)=𝑇(𝑟,𝑓)+𝑇(𝑟,𝑔)+𝑂(1), we obtain 𝑛(𝑇(𝑟,𝑓)+𝑇(𝑟,𝑔))≤𝑆(𝑟,𝑓)+𝑆(𝑟,𝑔), which is impossible.

Therefore, ℎ is a constant; then substituting 𝑓=𝑔ℎ into (60), we have ℎ^𝑛+1≡1. Hence 𝑓(𝑧)=𝑡𝑔(𝑧), where 𝑡 is a constant and 𝑡^𝑛+1=1.

The proof of Theorem 5 is complete.

3.3. Proof of Theorem 6

Let 𝐹(𝑧)=𝑓(𝑧)^𝑛Δ_𝑐𝑓^(𝑘),𝐺(𝑧)=𝑔(𝑧)^𝑛Δ_𝑐𝑔^(𝑘),𝐹₁(𝑧)=𝑓(𝑧)^𝑛Δ_𝑐𝑓,𝐺₁(𝑧)=𝑔(𝑧)^𝑛Δ_𝑐𝑔. (63) Then 𝐹(𝑧) and 𝐺(𝑧) share 1 IM, and 𝐹₁^(𝑘)=𝐹, 𝐺₁^(𝑘)=𝐺. By Lemma 10, we have 𝑛𝑇𝑟,𝑓+𝑆𝑟,𝑓≤𝑇𝑟,𝐹₁≤(𝑛+1)𝑇𝑟,𝑓+𝑆𝑟,𝑓,𝑛𝑇𝑟,𝑔+𝑆𝑟,𝑔≤𝑇𝑟,𝐺₁≤(𝑛+1)𝑇𝑟,𝑔+𝑆𝑟,𝑔. (64)(65) Since 𝑓 is transcendental entire, by the definition of 𝐹, we have 𝑁₂1𝑟,𝐹=𝑁1𝑟,𝐹+𝑁₍₂1𝑟,𝐹1=𝑁𝑟,𝐹−𝑁₍₃1𝑟,𝐹−2𝑁₍₃1𝑟,𝐹. (66) Using the argument in (35), we have 𝑁₍₃1𝑟,𝐹−2𝑁₍₃1𝑟,𝐹≥()𝑁1𝑛−𝑘−2𝑟,𝑓. (67) It follows from Lemma 12 and (66), (67), we have 𝑁₂1𝑟,𝐹1≤𝑁𝑟,𝐹−()𝑁1𝑛−𝑘−2𝑟,𝑓1≤𝑁𝑟,𝑓^𝑛Δ_𝑐𝑓−()𝑁1𝑛−𝑘−2𝑟,𝑓1+𝑆𝑟,𝑓≤𝑛𝑁𝑟,𝑓1+𝑁𝑟,Δ_𝑐𝑓−()𝑁1𝑛−𝑘−2𝑟,𝑓≤()𝑇.+𝑆𝑟,𝑓𝑘+3𝑟,𝑓+𝑆𝑟,𝑓 (68) From Lemma 15, we have 𝑁1𝑟,𝐹≤𝑁_𝑘+11𝑟,𝑓^𝑛Δ_𝑐𝑓≤()+𝑆𝑟,𝑓𝑘+1𝑁1𝑟,𝑓1+𝑁𝑟,Δ_𝑐𝑓≤()𝑇.+𝑆𝑟,𝑓𝑘+2𝑟,𝑓+𝑆𝑟,𝑓 (69)

Similarly, 𝑁₂1𝑟,𝐺≤(𝑘+3)𝑇𝑟,𝑔+𝑆𝑟,𝑔,𝑁1𝑟,𝐺≤(𝑘+2)𝑇𝑟,𝑔+𝑆𝑟,𝑓. (70) By Lemma 14, one of the following cases holds: (i)𝑇(𝑟,𝐺)≤𝑁₂(𝑟,1/𝐺)+𝑁₂(𝑟,1/𝐹)+𝑁(𝑟,1/𝐹)+2𝑁(𝑟,1/𝐺)+𝑆(𝑟,𝐹)+𝑆(𝑟,𝐺), the same inequality holding for 𝑇(𝑟,𝐹);(ii)𝐹≡(𝐴𝐺+𝐵)/(𝐶𝐺+𝐷). For case (i), we have 𝑇(𝑟,𝐺)≤𝑁₂1𝑟,𝐺+𝑁₂1𝑟,𝐹+𝑁1𝑟,𝐹+2𝑁1𝑟,𝐺()(),𝑇()+𝑆𝑟,𝐹+𝑆𝑟,𝐺𝑟,𝐹≤𝑁₂1𝑟,𝐹+𝑁₂1𝑟,𝐺+𝑁1𝑟,𝐺+2𝑁1𝑟,𝐹+𝑆(𝑟,𝐹)+𝑆(𝑟,𝐺). (71) Therefore, we get 𝑇(𝑟,𝐹)+𝑇(𝑟,𝐺)≤2𝑁₂1𝑟,𝐹+𝑁₂1𝑟,𝐺+3𝑁1𝑟,𝐺+𝑁1𝑟,𝐹+𝑆(𝑟,𝐹)+𝑆(𝑟,𝐺). (72) By (64) and Lemma 15, we have 𝑛𝑇𝑟,𝑓≤𝑇𝑟,𝐹₁+𝑆𝑟,𝑓≤𝑇(𝑟,𝐹)−𝑁₂1𝑟,𝐹+𝑁_𝑘+21𝑟,𝐹₁()+𝑆𝑟,𝑓≤𝑇𝑟,𝐹−𝑁₂1𝑟,𝐹+(𝑘+2)𝑁1𝑟,𝑓1+𝑁𝑟,Δ_𝑐𝑓()+𝑆𝑟,𝑓≤𝑇𝑟,𝐹−𝑁₂1𝑟,𝐹+(𝑘+3)𝑇𝑟,𝑓+𝑆𝑟,𝑓. (73) Similarly, 𝑛𝑇𝑟,𝑔≤𝑇(𝑟,𝐺)−𝑁₂1𝑟,𝐺+(𝑘+3)𝑇𝑟,𝑔+𝑆𝑟,𝑔. (74) By (70), (72), (73), and (74), we obtain (𝑛−5𝑘−12)𝑇𝑟,𝑓+𝑇𝑟,𝑔≤𝑆𝑟,𝑓+𝑆𝑟,𝑔, (75) which is a contradiction since 𝑛≥5𝑘+13.

For case (ii), we have 𝐹≡𝐴𝐺+𝐵,𝐶𝐺+𝐷 (76) where 𝐴, 𝐵, 𝐶, and 𝐷 are finite complex numbers satisfying 𝐴𝐷≠𝐵𝐶. Therefore, by the first fundamental theorem, 𝑇(𝑟,𝐹)=𝑇(𝑟,𝐺)+𝑆(𝑟,𝐹).

Next, we consider three cases.

Case 1. 𝐴𝐶≠0; from (76), we get 𝐴𝐹−𝐶=𝐵−𝐴𝐷/𝐶.𝐶𝐺+𝐷 (77) By the second fundamental theorem and (69), we have 𝑇(𝑟,𝐹)≤𝑁1𝑟,+𝐹−𝐴/𝐶𝑁1𝑟,𝐹()=+𝑆𝑟,𝐹𝑁()+()𝑇()≤()𝑇().𝑟,𝐺𝑘+2𝑟,𝑓+𝑆𝑟,𝐹𝑘+2𝑟,𝑓+𝑆𝑟,𝐹 (78) From (73), we obtain (𝑛−2𝑘−5)𝑇(𝑟,𝑓)≤𝑆(𝑟,𝑓), contradicting to 𝑛≥5𝑘+13.

Case 2. 𝐴≠0, and 𝐶=0. Then, 𝐹≡𝐴𝐺+𝐵/𝐷.
If 𝐵≠0, by the second fundamental theorem and (69), (70), we have 𝑇(𝑟,𝐹)≤𝑁1𝑟,+𝐹−𝐵/𝐷𝑁1𝑟,𝐹()=+𝑆𝑟,𝐹𝑁1𝑟,𝐺+𝑁1𝑟,𝐹()≤()𝑇+()𝑇().+𝑆𝑟,𝐹𝑘+2𝑟,𝑓𝑘+2𝑟,𝑔+𝑆𝑟,𝐹 (79) Similarly, 𝑇(𝑟,𝐺)≤(𝑘+2)𝑇𝑟,𝑓+(𝑘+2)𝑇𝑟,𝑔+𝑆(𝑟,𝐺). (80) From (73), (74), (79), and (80), we get (𝑛−3𝑘−7)𝑇𝑟,𝑓+𝑇𝑟,𝑔≤𝑆(𝑟,𝐹)+𝑆(𝑟,𝐺), (81) which is a contradiction to 𝑛≥5𝑘+13.
If 𝐵=0, then 𝐹≡𝐴𝐺/𝐷. If 𝐴/𝐷=1, then 𝐹≡𝐺; that is, [𝑓^𝑛Δ_𝑐𝑓]^(𝑘)≡[𝑔^𝑛Δ_𝑐𝑔]^(𝑘); using the argument in (54) and noting that 𝑛≥5𝑘+13, we obtain 𝑓(𝑧)=𝑡𝑔(𝑧), where 𝑡 is a constant and 𝑡^𝑛+1=1. If 𝐴/𝐷≠1, by the condition that 𝐹 and 𝐺 share 1 IM, then 𝐹≠1 and 𝐺≠1. we obtain then 𝐹≠1 and 𝐹≠𝐴/𝐷. By the second fundamental theorem, we have 𝑇(𝑟,𝐹)≤𝑁1+𝐹−1𝑁1()(),𝐹−𝐴/𝐷+𝑆𝑟,𝐹≤𝑆𝑟,𝐹 (82) which is impossible.

Case 3. 𝐴=0, and 𝐶≠0. Then, 𝐹≡𝐵/(𝐶𝐺+𝐷).
If 𝐷≠0, by the second fundamental theorem and (69), (70), we have 𝑇(𝑟,𝐹)≤𝑁1𝑟,+𝐹−𝐵/𝐷𝑁1𝑟,𝐹()=+𝑆𝑟,𝐹𝑁1𝑟,𝐺+𝑁1𝑟,𝐹()≤()𝑇+()𝑇().+𝑆𝑟,𝐹𝑘+2𝑟,𝑓𝑘+2𝑟,𝑔+𝑆𝑟,𝐹 (83) Similarly, 𝑇(𝑟,𝐺)≤(𝑘+2)𝑇𝑟,𝑓+(𝑘+2)𝑇𝑟,𝑔+𝑆(𝑟,𝐺). (84) From (73), (74), (83), and (84), we get (𝑛−3𝑘−7)𝑇𝑟,𝑓+𝑇𝑟,𝑔≤𝑆(𝑟,𝐹)+𝑆(𝑟,𝐺), (85) which is a contradiction to 𝑛≥5𝑘+13.
If 𝐷=0, then 𝐹≡𝐵/𝐶𝐺. If 𝐵/𝐶=1, then 𝐹⋅𝐺≡1; using the argument in (48) in Theorem 5 and noting that 𝑛≥5𝑘+13, we get a contradiction. If 𝐵/𝐶≠1, by the condition that 𝐹 and 𝐺 share 1 IM, we obtain 𝐹≠1 and 𝐹≠𝐵/𝐶. By the second fundamental theorem, we have 𝑇(𝑟,𝐹)≤𝑁1+𝐹−1𝑁1()(),𝐹−𝐵/𝐶+𝑆𝑟,𝐹≤𝑆𝑟,𝐹 (86) which is impossible.
The proof of Theorem 6 is complete.

Acknowledgments

This work is supported by Youth Foundation of Chongqing Normal University (12XWQ17), partially by Chongqing City Board of Education Science, and Technology Project (KJ130632), and partially by the fund of Chongqing Normal University (13XLB006).