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Characteristic estimation of differential polynomials

Abstract

In this paper, we give the characteristic estimation of a meromorphic function f with the differential polynomials \(f^{l}(f^{(k)})^{n}\) and obtain that

$$\begin{aligned} T(r,f)\leq M\overline{N} \biggl(r,\frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S(r,f) \end{aligned}$$

holds for \(M=\min \{\frac{1}{l-2},6\}\), integers \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\), and a non-zero constant a. This quantitative estimate is an interesting and complete extension of earlier results. The value distribution of a differential monomial of meromorphic functions is also investigated.

Introduction and main results

We assume that the reader is familiar with the fundamentals of Nevanlinna’s value distribution theory of meromorphic functions (see e.g. [4, 10, 16]). Let f be a transcendental meromorphic function in the complex plane \(\mathbb{C}\). We denote by \(S(r,f)\) any quantity satisfying \(S(r,f)=o(T(r,f))\), as \(r\rightarrow \infty \), possibly outside of an exceptional set of finite logarithmic measure. A meromorphic function α defined in \(\mathbb{C}\) is called a small function of f if \(T(r,\alpha )=S(r,f)\).

We also introduce some other symbols (see [15]). Let \(a\in \mathbb{C}\cup \{\infty \}\), k be a positive integer. Let \(N_{k)} (r,\frac{1}{f-a} )\) denote the counting function of those a-points of f (counting multiplicity) whose multiplicities are not greater than k, and let \(\overline{N}_{k)} (r,\frac{1}{f-a} )\) denote the corresponding reduced counting function. Similarly, let \(N_{(k} (r,\frac{1}{f-a} )\) denote the counting function of those a-points of f (counting multiplicity) whose multiplicities are not less than k, and let \(\overline{N}_{(k} (r,\frac{1}{f-a} )\) denote the corresponding reduced counting function. And let \(N_{k} (r,\frac{1}{f-a} )\) denote the counting function of those a-points of f with multiplicity k.

Hayman [5] proved the following well-known theorem.

Theorem 1.1

([5, Theorem 9])

Let f be a transcendental meromorphic function in the complex plane, and let l be a positive integer. If \(l\geq 3\), then \(f^{l}f'\) assumes every finite nonzero value infinitely often.

Hayman also conjectured that Theorem 1.1 remained valid for \(l\geq 1\). Mues [12] proved that \(f^{2}f'-1\) has infinitely many zeros. Later on, many researchers investigated the zeros of differential monomial \(f^{l}(f^{(k)})^{n}-a\) for positive integers \(l,n,k\) and a non-zero complex number a, and obtained some qualitative results, see e.g. [2, 3, 11, 13, 14], and some quantitative results, see e.g. [1, 69, 17].

Zhang [17] proved that the inequality \(T(r,f)<6N (r,\frac{1}{f^{2}f'-1} )+S(r,f)\) holds. Huang and Gu [6] extended the inequality and proved the following.

Theorem 1.2

([6, Theorem 1])

Let f be a transcendental meromorphic function in the complex plane, and let k be a positive integer. Then

$$\begin{aligned} T(r,f)< 6N \biggl(r,\frac{1}{f^{2}f^{(k)}-1} \biggr)+S(r,f). \end{aligned}$$
(1)

Karmakar and Sahoo further [8] proved the following.

Theorem 1.3

([8, Theorem 1.1])

Let f be a transcendental meromorphic function and \(l(\geq 2)\), \(k(\geq 1)\) be any integers, then

$$\begin{aligned} T(r,f)< \frac{6}{2l-3}\overline{N} \biggl(r, \frac{1}{f^{l}f^{(k)}-1} \biggr)+S(r,f). \end{aligned}$$
(2)

Lahiri and Dewan [9] obtained the following estimate.

Theorem 1.4

([9, Theorem 3.2])

Let f be a transcendental meromorphic function, \(\alpha (\not \equiv 0,\infty )\) be a small function of f. If \(\psi =\alpha f^{l}(f^{(k)})^{n}\), where \(l(\geq 0)\), \(n(\geq 1)\), \(k(\geq 1)\) are integers, then for any small function \(a(\not \equiv 0,\infty )\) of ψ,

$$\begin{aligned} (n+l)T(r,f) \leq{} &\overline{N}(r,f)+\overline{N} \biggl(r, \frac{1}{f} \biggr)+ nN_{(k)} \biggl(r,\frac{1}{f} \biggr) \\ &{}+\overline{N} \biggl(r,\frac{1}{\psi -a} \biggr)+S(r,f), \end{aligned}$$
(3)

where \(N_{(k)} (r,\frac{1}{f} )\) denotes the counting function of zeros of f, a zero with multiplicity q is counted q times if \(q\leq k\) and is counted k times if \(q>k\).

Remark 1.1

Estimate (3) implies that, for \(l\geq 3\), \(n\geq 1\), \(k\geq 1\),

$$\begin{aligned} T(r,f)\leq \frac{1}{l-2}\overline{N} \biggl(r, \frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S(r,f). \end{aligned}$$
(4)

Jiang and Huang [7] proved the following.

Theorem 1.5

([7, Theorem 1])

Let f be a transcendental meromorphic function in the complex plane, \(l(\geq 2)\), \(n(\geq 2)\), \(k(\geq 2)\) be integers, and a be a non-zero constant. Then

$$\begin{aligned} T(r,f)\leq \frac{1}{l-1}N \biggl(r, \frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S^{*}(r,f), \end{aligned}$$
(5)

where \(S^{*}(r,f)\) denotes the quantity satisfying \(S^{*}(r,f)=o(T(r,f))\) for all r outside a possible exceptional set E of logarithmic density 0.

We note that Theorem 1.3 does not hold for \(n\geq 2\), Theorem 1.4 is invalid for \(l=2\), and Theorem 1.5 remains invalid for \(l=2, n=1, k=1\). Thus, by using a method different from the previous proofs, we continue to consider the characteristic estimate of more general forms \(f^{l}(f^{(k)})^{n}-a\) for a non-zero constant a, integers \(l\geq 2\), \(n\geq 1\), and \(k\geq 1\), and obtain its quantitative result as follows.

Theorem 1.6

Let f be a transcendental meromorphic function with finite order in the complex plane, \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\) be integers, and a be a non-zero constant. Then

$$\begin{aligned} T(r,f)\leq M\overline{N} \biggl(r,\frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S(r,f) \end{aligned}$$
(6)

for \(M=\min \{\frac{1}{l-2},6\}\).

The quantity

$$\begin{aligned} \Theta (a,f)=1-\limsup_{r\to \infty } \frac{\overline{N} (r,\frac{1}{f-a} )}{T(r,f)} \end{aligned}$$

is called the deficiency of f at the point a. It is obvious that \(0\leq \Theta (a,f)\leq 1\). Thus, we present the value distribution of a differential monomial \(f^{l}(f^{(k)})^{n}\).

Corollary 1.1

Let f be a transcendental meromorphic function with finite order in the complex plane, \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\) be integers, and a be a non-zero constant. Then

$$\begin{aligned} \Theta \bigl(a,f^{l}\bigl(f^{(k)}\bigr)^{n}\bigr) \leq 1-\frac{1}{M(nk+n+l)} \end{aligned}$$

for \(M=\min \{\frac{1}{l-2},6\}\).

Some lemmas

We now prepare some lemmas.

Lemma 2.1

Let f be a transcendental meromorphic function with finite order. Then \(f^{l}(f^{(k)})^{n}\) is not identically constant, where \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\) are integers.

Proof

Contradicting to our assumption, we suppose that \(f^{l}(f^{(k)})^{n}\equiv C\). Clearly, \(C\neq 0\). Then \(\frac{1}{f^{n+l}}=\frac{1}{C} (\frac{f^{(k)}}{f} )^{n}\), and

$$\begin{aligned} (n+l)T(r,f) &= m \biggl(r,\frac{1}{f^{n+l}} \biggr)+N \biggl(r, \frac{1}{f^{n+l}} \biggr)+O(1) \\ &= m \biggl(r,\frac{1}{C} \biggl(\frac{f^{(k)}}{f} \biggr)^{n} \biggr)+O(1) \\ &= nm \biggl(r,\frac{f^{(k)}}{f} \biggr)+O(1)=S(r,f), \end{aligned}$$

a contradiction. □

Lemma 2.2

Let f be a transcendental meromorphic solution with finite order. Suppose that \(g(z)=f^{2}(f^{(k)})^{n}-a\), where \(n(\geq 1)\), \(k(\geq 1)\) are integers and a is a non-zero constant. Then

$$\begin{aligned} (n+2)T(r,f) \leq {}& \overline{N}(r,f)+\overline{N} \biggl(r, \frac{1}{f} \biggr)+\overline{N} \biggl(r,\frac{1}{g} \biggr) +nN_{k)} \biggl(r,\frac{1}{f} \biggr) \\ &{}+nk\overline{N}_{(k+1} \biggl(r,\frac{1}{f} \biggr)-N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f) \end{aligned}$$
(7)

and

$$\begin{aligned} &\bigl[N(r,f)-\overline{N}(r,f)\bigr]+m(r,f)+(n+1)m \biggl(r,\frac{1}{f} \biggr)+N_{0} \biggl(r,\frac{1}{g'} \biggr) \\ &\quad\leq \overline{N} \biggl(r,\frac{1}{g} \biggr)+S(r,f), \end{aligned}$$
(8)

where \(N_{0} (r,\frac{1}{g'} )\) denotes the counting function of those zeros of \(g'\), but not zeros of f or g.

Proof

It follows from Lemma 2.1 that g is not identically constant. Thus

$$\begin{aligned} \frac{a}{f^{n+2}}= \biggl(\frac{f^{(k)}}{f} \biggr)^{n}- \frac{g'}{f^{n+2}}\frac{g}{g'}. \end{aligned}$$

We conclude from the lemma of the logarithmic derivative that

$$\begin{aligned} (n+2)m \biggl(r,\frac{1}{f} \biggr) &\leq m \biggl(r, \biggl( \frac{f^{(k)}}{f} \biggr)^{n} \biggr)+m \biggl(r,\frac{g'}{f^{n+2}} \biggr) +m \biggl(r,\frac{g}{g'} \biggr)+O(1) \\ &= T \biggl(r,\frac{g'}{g} \biggr)-N \biggl(r,\frac{g}{g'} \biggr)+S(r,f) \\ &= \overline{N}(r,g)+N \biggl(r,\frac{1}{g} \biggr)-N \biggl(r, \frac{1}{g'} \biggr)+S(r,f) \end{aligned}$$

and

$$\begin{aligned} (n+2)T(r,f)\leq {}&(n+2)N \biggl(r,\frac{1}{f} \biggr)+\overline{N}(r,f)+N \biggl(r,\frac{1}{g} \biggr) \\ &{}- N \biggl(r,\frac{1}{g'} \biggr)+S(r,f). \end{aligned}$$
(9)

Denote

$$\begin{aligned} N \biggl(r,\frac{1}{g'} \biggr)=N_{000} \biggl(r,\frac{1}{g'} \biggr) +N_{00} \biggl(r,\frac{1}{g'} \biggr)+N_{0} \biggl(r,\frac{1}{g'} \biggr), \end{aligned}$$
(10)

where \(N_{000} (r,\frac{1}{g'} )\) denotes the counting function of those zeros of \(g'\) which are from the zeros of g, \(N_{00} (r,\frac{1}{g'} )\) denotes the counting function of those zeros of \(g'\) which are from the zeros of f, \(N_{0} (r,\frac{1}{g'} )\) denotes the counting function of those zeros of \(g'\) which are not zeros of f or g. So, we have

$$\begin{aligned} N \biggl(r,\frac{1}{g} \biggr)-N_{000} \biggl(r,\frac{1}{g'} \biggr)= \overline{N} \biggl(r,\frac{1}{g} \biggr). \end{aligned}$$
(11)

Let \(z_{0}\) be a zero of f with multiplicity q. If \(q\leq k\), then \(z_{0}\) is a zero of \(g'\) with multiplicity at least \(2q-1\). If \(q\geq k+1\), then \(z_{0}\) is a zero of \(g'\) with multiplicity \((n+2)q-(nk+1)\). Thus, by simple calculation, we have

$$\begin{aligned} N_{00} \biggl(r,\frac{1}{g'} \biggr) \geq{} & 2N_{k)} \biggl(r, \frac{1}{f} \biggr)-\overline{N}_{k)} \biggl(r, \frac{1}{f} \biggr) \\ &{}+ (n+2)N_{(k+1} \biggl(r,\frac{1}{f} \biggr)-(nk+1) \overline{N}_{(k+1} \biggl(r,\frac{1}{f} \biggr) \\ ={}& 2N \biggl(r,\frac{1}{f} \biggr)-\overline{N} \biggl(r, \frac{1}{f} \biggr)+nN_{(k+1} \biggl(r,\frac{1}{f} \biggr) -nk\overline{N}_{(k+1} \biggl(r,\frac{1}{f} \biggr) \end{aligned}$$

and

$$\begin{aligned} &(n+2)N \biggl(r,\frac{1}{f} \biggr)-N_{00} \biggl(r,\frac{1}{g'} \biggr) \\ &\quad \leq nN \biggl(r,\frac{1}{f} \biggr) -nN_{(k+1} \biggl(r, \frac{1}{f} \biggr)+\overline{N} \biggl(r,\frac{1}{f} \biggr) +nk \overline{N}_{(k+1} \biggl(r,\frac{1}{f} \biggr) \\ &\quad = nN_{k)} \biggl(r,\frac{1}{f} \biggr)+\overline{N} \biggl(r, \frac{1}{f} \biggr) +nk\overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr). \end{aligned}$$
(12)

Then we deduce from (9)−(12) that

$$\begin{aligned} (n+2)T(r,f) \leq{} & \overline{N}(r,f)+\overline{N} \biggl(r, \frac{1}{g} \biggr)+\overline{N} \biggl(r,\frac{1}{f} \biggr) \\ &{}+nN_{k)} \biggl(r,\frac{1}{f} \biggr)+nk\overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) -N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f), \end{aligned}$$

and so inequality (7) is proved.

We further get from (7) that

$$\begin{aligned} & N(r,f)+m(r,f)+(n+1)N \biggl(r,\frac{1}{f} \biggr)+(n+1)m \biggl(r, \frac{1}{f} \biggr)+O(1) \\ &\quad=(n+2)T(r,f) \\ &\quad \leq \overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f} \biggr)+ \overline{N} \biggl(r,\frac{1}{g} \biggr) +nN_{k)} \biggl(r, \frac{1}{f} \biggr)+nk\overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) \\ &\qquad{}-N_{0} \biggl(r,\frac{1}{g'} \biggr)+S(r,f) \\ &\quad \leq \overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f} \biggr)+ \overline{N} \biggl(r,\frac{1}{g} \biggr) +nN_{k)} \biggl(r, \frac{1}{f} \biggr)+nN_{(k+1} \biggl(r,\frac{1}{f} \biggr) \\ &\qquad{}-N_{0} \biggl(r,\frac{1}{g'} \biggr)+S(r,f) \\ &\quad \leq \overline{N}(r,f)+(n+1)N \biggl(r,\frac{1}{f} \biggr)+ \overline{N} \biggl(r,\frac{1}{g} \biggr) -N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f), \end{aligned}$$

that is,

$$\begin{aligned} & \bigl[N(r,f)-\overline{N}(r,f)\bigr]+m(r,f)+(n+1)m \biggl(r, \frac{1}{f} \biggr)+N_{0} \biggl(r,\frac{1}{g'} \biggr) \\ &\quad \leq \overline{N} \biggl(r,\frac{1}{g} \biggr)+S(r,f), \end{aligned}$$

and so inequality (8) is proved. □

Lemma 2.3

Let f be a transcendental meromorphic function with finite order. Suppose that

$$\begin{aligned} g(z)=f^{2}\bigl(f^{(k)}\bigr)^{n}-a,~h(z)= \frac{g'(z)}{f(z)}=2f'\bigl(f^{(k)}\bigr)^{n}+nf \bigl(f^{(k)}\bigr)^{n-1}f^{(k+1)}, \end{aligned}$$

where \(n(\geq 1)\), \(k(\geq 1)\) are integers and a is a non-zero constant.

$$\begin{aligned} F(z)={}&a_{1} \biggl(\frac{g'(z)}{g(z)} \biggr)^{2} +a_{2} \biggl( \frac{g'(z)}{g(z)} \biggr)' +a_{3} \biggl(\frac{h'(z)}{h(z)} \biggr)' +a_{4} \biggl(\frac{h'(z)}{h(z)} \biggr)^{2} \\ &{}+ a_{5} \biggl(\frac{g'(z)}{g(z)}\frac{h'(z)}{h(z)} \biggr), \end{aligned}$$
(13)

where \(a_{i}'s\) are defined by

$$\begin{aligned} \textstyle\begin{cases} a_{1}=-(2n^{4}+4n^{3}+2n^{2}+3n+2); \\ a_{2}=-2(n+1)(n^{3}+n^{2}+n+2); \\ a_{3}=2n^{2}(n+1)^{2}; \\ a_{4}=-2n^{2}(n+1)^{2}; \\ a_{5}=4(n+1)(n^{3}+n^{2}+1), \end{cases}\displaystyle \end{aligned}$$
(14)

when \(k=1\), and are defined by

$$\begin{aligned} \textstyle\begin{cases} a_{1}=2(nk+n)^{2}-\frac{(3nk+3n+4)[(nk+n)^{2}-6(nk+n)-24]}{nk+n+2}; \\ a_{2}=-(nk+n+4)[(nk+n)^{2}-6(nk+n)-24]; \\ a_{3}=2(nk+n)(nk+n+2)(nk+n+4); \\ a_{4}=-4(nk+n)(nk+n+2); \\ a_{5}=4[(nk+n)^{2}-6(nk+n)-24], \end{cases}\displaystyle \end{aligned}$$
(15)

when \(k\geq 2\). Then \(F(z)\not \equiv 0\).

Proof

We use a similar method of Huang-Gu [6, Lemma 3]. Suppose that \(F(z)\equiv 0\), we claim that

  1. (i)

    \(g(z)\neq 0\);

  2. (ii)

    \(h(z)\neq 0\);

  3. (iii)

    all zeros of \(f(z)\) are simple.

Suppose that \(z_{1}\) is a zero of \(g(z)\) with multiplicity \(l(\geq 1)\). Then \(f(z_{1})\neq 0,\infty \), and \(z_{1}\) is a zero of \(h(z)\) with multiplicity \(l-1\) since \(g'=fh\). Using the Laurent series of \(F(z)\) at the point \(z_{1}\), we can calculate that the coefficient \(A(l)\) of \((z-z_{1})^{-2}\) is

$$\begin{aligned} A(l)=(a_{1}+a_{4}+a_{5})l^{2}-(a_{2}+a_{3}+2a_{4}+a_{5})l+(a_{3}+a_{4}). \end{aligned}$$

Using (14) for \(k=1\), we have

$$\begin{aligned} A(l)=(n+2)l^{2}+2n(n+1)l>0. \end{aligned}$$

This shows that \(z_{1}\) is a pole of \(F(z)\), which contradicts \(F(z)\equiv 0\). Hence \(g(z)\neq 0\) when \(k=1\). Using (15) for \(k\geq 2\), we have

$$\begin{aligned} A(l)={}&{-}\frac{(nk+n+4)^{2}(nk+n+6)}{nk+n+2}l^{2}-(nk+n) (nk+n+4) (nk+n+6)l \\ &{}+2(nk+n) (nk+n+2)^{2}. \end{aligned}$$

Clearly, \(A(l)\neq 0\) for all positive integers l. This shows that \(z_{1}\) is again a pole of \(F(z)\), which contradicts \(F(z)\equiv 0\). Hence \(g(z)\neq 0\) when \(k\geq 2\).

Suppose that \(z_{2}\) is a zero of \(h(z)\) with multiplicity \(l(\geq 1)\). By \((i)\) we have \(g(z_{2})\neq 0,\infty \). Using the Laurent series of \(F(z)\) at the point \(z_{2}\), we can get the coefficient \(B(l)\) of \((z-z_{1})^{-2}\) is

$$\begin{aligned} B(l)=a_{4}l^{2}-a_{3}l. \end{aligned}$$

Using (14) for \(k=1\), we get

$$\begin{aligned} B(l)=-2n^{2}(n+1)^{2}\bigl(l^{2}+l\bigr)< 0, \end{aligned}$$

and so, the point \(z_{2}\) is again a pole of \(F(z)\), which contradicts \(F(z)\equiv 0\).

Using (15) for \(k\geq 2\), we get

$$\begin{aligned} B(l)=-4(nk+n) (nk+n+2)l^{2}-2(nk+n) (nk+n+2) (nk+n+4)l< 0, \end{aligned}$$

and so, the point \(z_{2}\) is a pole of \(F(z)\), which contradicts \(F(z)\equiv 0\). Hence conclusion \((ii)\) \(h(z)\neq 0\) holds when \(k\geq 1\).

Noting that \(h(z)=\frac{g'(z)}{f(z)}=2f'(f^{(k)})^{n}+nf(f^{(k)})^{n-1}f^{(k+1)}\) and \((ii)\) \(h(z)\neq 0\), we can obtain \((iii)\).

Setting \(\phi (z):=\frac{h(z)}{g(z)}\), we conclude that \(\phi (z)\) is an entire function, all zeros of \(\phi (z)\) can occur only at multiple poles of \(f(z)\) and the following expressions hold:

$$\begin{aligned} \frac{g'}{g}=\frac{fh}{g}=f\phi,\qquad \frac{h'}{h}= \frac{g'}{g}+ \frac{\phi '}{\phi }=f\phi +\frac{\phi '}{\phi }. \end{aligned}$$

First, we consider the case \(k\geq 2\). Substituting the above two equalities into (13) yields

$$\begin{aligned} & (a_{1}+a_{4}+a_{5})f^{2} \phi ^{2}+(a_{2}+a_{3}+2a_{4}+a_{5})f \phi ' \\ &\quad{} +(a_{2}+a_{3})f'\phi + \biggl[a_{3} \biggl(\frac{\phi '}{\phi } \biggr)'+a_{4} \biggl(\frac{\phi '}{\phi } \biggr)^{2} \biggr]\equiv 0. \end{aligned}$$
(16)

Applying (15), we have \(a_{2}+a_{3}=(nk+n+4)^{2}(nk+n+6)\neq 0\). And so \(\phi \not \equiv 0\), otherwise \(\frac{g'}{g}=f\phi \equiv 0\), that is, \(g\equiv C\), which contradicts Lemma 2.1. Thus, it follows from (16) that

$$\begin{aligned} f'=\frac{1}{\phi }\alpha _{11}(z)+f\alpha _{12}(z)+f^{2}\phi \alpha _{13}(z), \end{aligned}$$
(17)

where \(\alpha _{1i}(z)\ (i=1,2,3)\) are differential polynomials of \(\frac{\phi '}{\phi }\).

Differentiating both sides of (17) gives

$$\begin{aligned} f'' ={}&{-}\frac{1}{\phi }\frac{\phi '}{\phi }\alpha _{11}(z)+ \frac{1}{\phi }\alpha _{11}'(z)+f' \alpha _{12}(z)+f\alpha _{12}'(z) \\ &{}+ 2ff'\phi \alpha _{13}(z)+f^{2}\phi \biggl[ \frac{\phi '}{\phi } \alpha _{13}(z)+\alpha _{13}'(z) \biggr]. \end{aligned}$$

Applying the above equality to (17), we have

$$\begin{aligned} f''=\frac{1}{\phi }\alpha _{21}(z)+f \alpha _{22}(z)+f^{2}\phi \alpha _{23}(z) +f^{3}\phi ^{2}\alpha _{24}(z), \end{aligned}$$

where \(\alpha _{2i}(z)\ (i=1,2,3,4)\) are differential polynomials of \(\frac{\phi '}{\phi }\). Continuing the above process, we get

$$\begin{aligned} f^{(k)}=\frac{1}{\phi }\alpha _{k1}(z)+f\alpha _{k2}(z)+f^{2}\phi \alpha _{k3}(z) +\cdots +f^{k+1}\phi ^{k}\alpha _{kk+2}(z), \end{aligned}$$
(18)

where \(\alpha _{ki}(z)\ (i=1,2,\ldots,k+2)\) are differential polynomials of \(\frac{\phi '}{\phi }\).

Suppose that \(z_{3}\) is a simple zero of \(f(z)\). Together with (17), (18) and noting that \(\phi (z_{3})\neq 0,\infty \), we have

$$\begin{aligned} f'(z_{3})=\frac{1}{\phi (z_{3})}\alpha _{11}(z_{3}),\qquad f^{(k)}(z_{3})= \frac{1}{\phi (z_{3})}\alpha _{k1}(z_{3}). \end{aligned}$$

Substituting the above two equalities into the expressions of \(g(z)\) and \(h(z)\) yields

$$\begin{aligned} g(z_{3})=-a, h(z_{3})=2f'(z_{3}) \bigl(f^{(k)}(z_{3})\bigr)^{n}= \frac{2}{\phi ^{n+1}(z_{3})} \alpha _{11}(z_{3})\alpha _{k1}^{n}(z_{3}). \end{aligned}$$

Combining the above two equalities and the expression of \(\phi (z):=\frac{h(z)}{g(z)}\), we get

$$\begin{aligned} a\phi ^{n+2}(z_{3})=-2\alpha _{11}(z_{3})\alpha _{k1}^{n}(z_{3}). \end{aligned}$$
(19)

Set \(U(z):=a\phi ^{n+2}(z)+2\alpha _{11}(z)\alpha _{k1}^{n}(z)\). We consider the following two cases.

Case 1. \(U(z)\not \equiv 0\). It follows from (19) and \((iii)\) that

$$\begin{aligned} N \biggl(r,\frac{1}{f} \biggr)&=\overline{N} \biggl(r,\frac{1}{f} \biggr)\leq N \biggl(r,\frac{1}{U} \biggr)\leq T(r,U)+O(1) \\ &\leq O\bigl\{ T(r,\phi )\bigr\} +O(1). \end{aligned}$$
(20)
$$\begin{aligned} T(r,\phi )=m(r,\phi )=m \biggl(r,\frac{h}{g} \biggr)=m \biggl(r, \frac{g'}{g}\frac{1}{f} \biggr) \leq m \biggl(r, \frac{1}{f} \biggr)+S(r,f). \end{aligned}$$
(21)

Using (8) and noting that \(N (r,\frac{1}{g} )=0\), we have

$$\begin{aligned} m \biggl(r,\frac{1}{f} \biggr)=S(r,f). \end{aligned}$$
(22)

It follows from (20)−(22) that

$$\begin{aligned} N \biggl(r,\frac{1}{f} \biggr)=S(r,f). \end{aligned}$$
(23)

Applying (22) and (23), we get

$$\begin{aligned} T(r,f)=T \biggl(r,\frac{1}{f} \biggr)+O(1) =m \biggl(r,\frac{1}{f} \biggr)+N \biggl(r,\frac{1}{f} \biggr)+O(1)=S(r,f), \end{aligned}$$

a contradiction.

Case 2. \(U(z)\equiv 0\). By the expression of \(U(z)\), and noting that \(\alpha _{11}(z),\alpha _{k1}(z)\) are differential polynomials of \(\frac{\phi '}{\phi }\), we conclude that

$$\begin{aligned} T(r,\phi )=m(r,\phi )=S(r,\phi ) \end{aligned}$$
(24)

and

$$\begin{aligned} a\phi ^{n+2}(z)\equiv -2\alpha _{11}(z)\alpha _{k1}^{n}(z). \end{aligned}$$
(25)

Using (24), we conclude that \(\phi (z)\) is a polynomial or a constant. If \(\phi (z)\) is a polynomial, the left-hand side of (25) is a polynomial, and the right-hand side of (25) is a constant or rational function, a contradiction. So, \(\phi (z)\) is a constant. If \(\phi (z)\equiv 0\), then \(\frac{g'}{g}=f\phi \equiv 0\), that is, g is a constant, a contradiction. If \(\phi (z)\equiv C(C\neq 0)\), then we substitute this equality into (16) and get

$$\begin{aligned} (a_{1}+a_{4}+a_{5})C^{2}f^{2}+(a_{2}+a_{3})Cf' \equiv 0. \end{aligned}$$

Using (25) for \(k\geq 2\), we have \(a_{1}+a_{4}+a_{5}=-\frac{(nk+n+4)^{2}(nk+n+6)}{nk+n+2}\neq 0\) and \(a_{2}+a_{3}=(nk+n+4)^{2}(nk+n+6)\neq 0\). Thus \((\frac{1}{f} )'=-\frac{f'}{f^{2}}\equiv - \frac{C}{nk+n+2}\neq 0\), and f is a rational function, a contradiction.

We now consider the case \(k=1\). Similar to the proof of the case \(k\geq 2\), we obtain a contradiction. □

Lemma 2.4

Let \(f(z)\), \(g(z)\), \(h(z)\), and \(F(z)\) be stated as in Lemma 2.3. Then all simple poles of \(f(z)\) are zeros of \(F(z)\).

Proof

Suppose that \(z_{0}\) is a simple pole of \(f(z)\), then

$$\begin{aligned} f(z)=\frac{A}{z-z_{0}}\bigl\{ 1+b(z-z_{0})+c(z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \bigr\} , \end{aligned}$$

where \(A\neq 0\), \(b,c\) are constants. We consider the following two cases.

Case 1. \(k=1\). We have

$$\begin{aligned} g(z) ={}& f^{2}\bigl(f'(z)\bigr)^{n}-a \\ ={}& \frac{(-1)^{n}A^{n+2}}{(z-z_{0})^{2n+2}}\bigl\{ 1+2b(z-z_{0})+\bigl[b^{2}-(n-2)c \bigr](z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \bigr\} , \\ h(z) ={}&\frac{g'(z)}{f(z)} \\ ={}&\frac{(-1)^{n+1}2A^{n+1}}{(z-z_{0})^{2n+2}} \\ &{}\times \bigl\{ (n+1)+nb(z-z_{0})-\bigl(n^{2}-n+1 \bigr)c(z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr)\bigr\} . \end{aligned}$$

By using the above two equalities, we have

$$\begin{aligned} &\frac{g'(z)}{g(z)}=\frac{-2}{z-z_{0}}\bigl\{ (n+1)-b(z-z_{0})+ \bigl[b^{2}+(n-2)c\bigr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \bigr\} , \\ &\frac{h'(z)}{h(z)} = \frac{-1}{z-z_{0}} \biggl\{ 2(n+1)-\frac{n}{n+1}b(z-z_{0}) \\ &\phantom{\frac{h'(z)}{h(z)} = }{}+ \frac{n^{2}b^{2}+2(n+1)(n^{2}-n+1)c}{(n+1)^{2}}(z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \biggr\} , \\ &\biggl(\frac{g'(z)}{g(z)} \biggr)^{2} = \frac{4}{(z-z_{0})^{2}}\bigl\{ (n+1)^{2}-2(n+1)b(z-z_{0}) \\ &\phantom{\biggl(\frac{g'(z)}{g(z)} \biggr)^{2} = }{}+ \bigl[(2n+3)b^{2}+2(n+1) (n-2)c\bigr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr)\bigr\} , \\ &\biggl(\frac{g'(z)}{g(z)} \biggr)'=\frac{2}{(z-z_{0})^{2}}\bigl\{ (n+1)- \bigl[b^{2}+(n-2)c\bigr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \bigr\} , \\ &\biggl(\frac{h'(z)}{h(z)} \biggr)' = \frac{1}{(z-z_{0})^{2}} \biggl\{ 2(n+1)- \frac{n^{2}b^{2}+2(n+1)(n^{2}-n+1)c}{(n+1)^{2}}(z-z_{0})^{2} \\ &\phantom{\biggl(\frac{h'(z)}{h(z)} \biggr)' =}{}+ O\bigl((z-z_{0})^{3}\bigr) \biggr\} , \\ &\biggl(\frac{h'(z)}{h(z)} \biggr)^{2} = \frac{1}{(z-z_{0})^{2}} \biggl\{ 4(n+1)^{2}-4nb(z-z_{0}) \\ &\phantom{\biggl(\frac{h'(z)}{h(z)} \biggr)^{2} =}{}+ \frac{(4n+5)n^{2}b^{2}+8(n+1)^{2}(n^{2}-n+1)c}{(n+1)^{2}}(z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \biggr\} , \\ &\frac{g'(z)}{g(z)}\frac{h'(z)}{h(z)} = \frac{2}{(z-z_{0})^{2}}\bigl\{ 2(n+1)^{2}-(3n+2)b(z-z_{0}) \\ &\phantom{\frac{g'(z)}{g(z)}\frac{h'(z)}{h(z)} = }{}+ \bigl[(3n+2)b^{2}+2\bigl(2n^{2}-2n-1\bigr)c \bigr](z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr)\bigr\} . \end{aligned}$$

By substituting the above equalities into (13) and performing some easy calculations, we have \(F(z)=O((z-z_{0}))\), consequently, and so \(z_{0}\) is a zero of \(F(z)\).

Case 2. \(k\geq 2\). We have

$$\begin{aligned} g(z) &=f^{2}\bigl(f^{(k)}\bigr)^{n}-a \\ &= \frac{(-1)^{nk}(k!)^{n}A^{n+2}}{(z-z_{0})^{nk+n+2}}\bigl\{ 1+2b(z-z_{0})+\bigl(b^{2}+2c \bigr) (z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \bigr\} \end{aligned}$$

and

$$\begin{aligned} h(z) = \frac{g'(z)}{f(z)}= {}&\frac{(-1)^{nk+1}(k!)^{n}A^{n+1}}{(z-z_{0})^{nk+n+2}}\bigl\{ (nk+n+2) \\ &{}+ (nk+n)b(z-z_{0})+(nk+n-2)c(z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr)\bigr\} . \end{aligned}$$

Using the above two equalities, we get

$$\begin{aligned} &\frac{g'(z)}{g(z)}=\frac{-1}{z-z_{0}}\bigl\{ (nk+n+2)-2b(z-z_{0})+2 \bigl(b^{2}-2c\bigr) (z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \bigr\} , \\ &\frac{h'(z)}{h(z)} = \frac{-1}{z-z_{0}} \biggl\{ (nk+n+2)- \frac{nk+n}{nk+n+2}b(z-z_{0}) \\ &\phantom{\frac{h'(z)}{h(z)} =}{}+ \frac{1}{nk+n+2} \biggl[\frac{(nk+n)^{2}b^{2}}{nk+n+2}-2(nk+n-2)c \biggr](z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \biggr\} , \\ &\biggl(\frac{g'(z)}{g(z)} \biggr)^{2} = \frac{1}{(z-z_{0})^{2}}\bigl\{ (nk+n+2)^{2}-4(nk+n+2)b(z-z_{0}) \\ &\phantom{\biggl(\frac{g'(z)}{g(z)} \biggr)^{2} = }{}+ \bigl[4(nk+n+3)b^{2}-8(nk+n+2)c\bigr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr)\bigr\} , \\ &\biggl(\frac{g'(z)}{g(z)} \biggr)'=\frac{1}{(z-z_{0})^{2}}\bigl\{ (nk+n+2)-2\bigl(b^{2}-2c\bigr) (z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \bigr\} , \\ & \biggl(\frac{h'(z)}{h(z)} \biggr)' = \frac{1}{(z-z_{0})^{2}} \biggl\{ (nk+n+2) \\ &\phantom{\biggl(\frac{h'(z)}{h(z)} \biggr)' =}{} -\frac{1}{nk+n+2} \biggl[\frac{(nk+n)^{2}b^{2}}{nk+n+2}-2(nk+n-2)c \biggr](z-z_{0})^{2}+O\bigl((z-z_{0})^{3} \bigr) \biggr\} , \\ & \biggl(\frac{h'(z)}{h(z)} \biggr)^{2} = \frac{1}{(z-z_{0})^{2}} \biggl\{ (nk+n+2)^{2}-2(nk+n)b(z-z_{0}) \\ &\phantom{\biggl(\frac{h'(z)}{h(z)} \biggr)^{2} =}{} + \biggl[\frac{(2nk+2n+5)(nk+n)^{2}b^{2}}{(nk+n+2)^{2}}-4(nk+n-2)c \biggr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr) \biggr\} , \\ &\frac{g'(z)}{g(z)}\frac{h'(z)}{h(z)} = \frac{1}{(z-z_{0})^{2}}\bigl\{ (nk+n+2)^{2}-(3nk+3n+4)b(z-z_{0}) \\ &\phantom{\frac{g'(z)}{g(z)}\frac{h'(z)}{h(z)} = }{}+ \bigl[(3nk+3n+4)b^{2}-2(3nk+3n+2)c\bigr](z-z_{0})^{2}+O \bigl((z-z_{0})^{3}\bigr)\bigr\} . \end{aligned}$$

By substituting the above equalities into (13) and performing some easy calculations, we again get \(F(z)=O((z-z_{0}))\). It also shows that \(z_{0}\) is a zero of \(F(z)\). □

Definition 2.1

([3])

Let f be a nonconstant meromorphic function in the complex plane and k be a positive integer. We call \(M[f]=f^{n_{0}}(f')^{n_{1}}\cdots (f^{(k)})^{n_{k}}\) a differential monomial in f, where \(n_{0},n_{1},\ldots,n_{k}\) are nonnegative integers, and \(\gamma _{M}:=n_{0}+n_{1}+\cdots +n_{k}\) its degree. Further, let \(M_{j}[f]\) denote differential monomials in f of degree \(\gamma _{M_{j}}\) for \(j=1,2,\ldots,k\), and let \(\alpha _{j}\) be meromorphic functions satisfying \(T(r,\alpha _{j})=S(r,f)\) for \(j=1,2,\ldots,k\), then \(P[f]=\alpha _{1}M_{1}[f]+\alpha _{2}M_{2}[f]+\cdots +\alpha _{k}M_{k}[f]\) is called a differential polynomial in f of degree \(\gamma _{P}:=\max_{1\leq j\leq k}\gamma _{M_{j}}\). If the coefficients \(\alpha _{j}\) only satisfy \(m(r,\alpha _{j})=S(r,f)\), then we call the function \(P[f]\) a quasi-differential polynomial in f.

Lemma 2.5

([3])

Let f be a nonconstant meromorphic function and \(Q^{*}[f]\), \(Q[f]\) be quasi-differential polynomials in f with \(Q[f]\not \equiv 0\). Let n be a positive integer and \(f^{n}Q^{*}[f]=Q[f]\). If \(\gamma _{Q}\leq n\), then \(m(r,Q^{*}[f])=S(r,f)\), where \(\gamma _{Q}\) is the degree of \(Q[f]\).

Proofs of theorems

In this section, we mainly give complete proofs for our main results.

Proof

of Theorem 1.6. In what follows, we consider two cases.

Case 1. When \(l\geq 3\), \(n\geq 1\), \(k\geq 1\), by inequality (4), we have

$$\begin{aligned} T(r,f)\leq \frac{1}{l-2}\overline{N} \biggl(r, \frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S(r,f). \end{aligned}$$

Case 2. When \(l=2\), \(n\geq 1\), \(k\geq 1\), we consider two subcases.

Subcase 2.1. First we suppose that \(k\geq 2\). From Lemma 2.3 and Lemma 2.4, we see immediately that \(F\not \equiv 0\) and simple poles of \(f(z)\) are the zeros of \(F(z)\). We also conclude that the poles of \(F(z)\) are with multiplicities two at most, which come from the multiple poles of \(f(z)\), or from the zeros of \(g(z)\), or from the zeros of \(h(z)\).

Set \(\beta =2f'(f^{(k)})^{n}+nf(f^{(k)})^{n-1}f^{(k+1)}-f(f^{(k)})^{n} \frac{g'}{g}\). Then \(f\beta =-a\frac{g'}{g}\) and \(h=-\frac{1}{a}\beta g\). We now consider the poles of \(\beta ^{2}F\). We note that the multiple poles of f with multiplicity \(q(\geq 2)\) are the zeros of β with multiplicity \(q-1\), and the zeros of h are either the zeros of g or the zeros of β. Thus,

$$\begin{aligned} N\bigl(r,\beta ^{2}F\bigr)\leq 4\overline{N} \biggl(r, \frac{1}{g} \biggr), \end{aligned}$$

since the poles of \(\beta ^{2}F\) come only from the zeros of g, and the multiplicity of poles of \(\beta ^{2}F\) is 4 at most.

Noting that \(m(r,F)=S(r,f)\) and \(m(r,\beta ^{2})=S(r,f)\) from Lemma 2.5, we have \(m(r,\beta ^{2}F)=S(r,f)\). Therefore,

$$\begin{aligned} T\bigl(r,\beta ^{2}F\bigr)\leq 4\overline{N} \biggl(r, \frac{1}{g} \biggr). \end{aligned}$$

Since the simple poles of f are the zeros of \(\beta ^{2}F\), hence

$$\begin{aligned} N_{1}(r,f)\leq N \biggl(\frac{1}{\beta ^{2}F} \biggr)\leq T\bigl(r,\beta ^{2}F\bigr) \leq 4\overline{N} \biggl(r, \frac{1}{g} \biggr). \end{aligned}$$
(26)

It follows from (7) and (26) that

$$\begin{aligned} & 2(n+2)T(r,f)+N_{1}(r,f) \\ &\quad \leq 2\overline{N}(r,f)+2\overline{N} \biggl(r,\frac{1}{f} \biggr)+ 6 \overline{N} \biggl(r,\frac{1}{g} \biggr)+2nN_{k)} \biggl(r, \frac{1}{f} \biggr) \\ &\qquad{} +2nk\overline{N}_{(k+1} \biggl(r,\frac{1}{f} \biggr)-2N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f) \end{aligned}$$

i.e.

$$\begin{aligned} & 2(n+2)T(r,f)+N_{1}(r,f)-2\overline{N} \biggl(r,\frac{1}{f} \biggr) \\ &\qquad{} -2nN_{k)} \biggl(r,\frac{1}{f} \biggr)-2nk \overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) \\ &\quad \leq 2\overline{N}(r,f)+6\overline{N} \biggl(r,\frac{1}{g} \biggr)-2N_{0} \biggl(r,\frac{1}{g'} \biggr)+S(r,f), \end{aligned}$$

which leads to

$$\begin{aligned} &T(r,f)+m(r,f)+2(n+1)m \biggl(r,\frac{1}{f} \biggr)+\bigl[N(r,f)+N_{1}(r,f)-2 \overline{N}(r,f)\bigr] \\ &\qquad{} +2 \biggl[N \biggl(r,\frac{1}{f} \biggr)-\overline{N} \biggl(r, \frac{1}{f} \biggr) \biggr] \\ &\qquad{} +2n \biggl[N \biggl(r,\frac{1}{f} \biggr)-N_{k)} \biggl(r, \frac{1}{f} \biggr) -k\overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) \biggr] \\ &\quad\leq 6\overline{N} \biggl(r,\frac{1}{g} \biggr)-2N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f). \end{aligned}$$
(27)

We note that

$$\begin{aligned} & N(r,f)+N_{1}(r,f)-2\overline{N}(r,f) \\ &\quad =N_{1}(r,f)+N_{(2}(r,f)+N_{1}(r,f)-2 \bigl[N_{1}(r,f)+\overline{N}_{(2}(r,f)\bigr] \\ &\quad =N_{(2}(r,f)-2\overline{N}_{(2}(r,f)\geq 0 \end{aligned}$$

and

$$\begin{aligned} & N \biggl(r,\frac{1}{f} \biggr)-N_{k)} \biggl(r, \frac{1}{f} \biggr)-k \overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) \\ &\quad =N_{(k+1} \biggl(r,\frac{1}{f} \biggr)-k\overline{N}_{(k+1} \biggl(r, \frac{1}{f} \biggr) \\ &\quad \geq N_{(k+1} \biggl(r,\frac{1}{f} \biggr)- \frac{k}{k+1}N_{(k+1} \biggl(r,\frac{1}{f} \biggr)\geq 0. \end{aligned}$$

By combining the above two inequalities and (27), we have

$$\begin{aligned} T(r,f)\leq 6\overline{N} \biggl(r,\frac{1}{g} \biggr)+S(r,f). \end{aligned}$$
(28)

Subcase 2.2. Suppose that \(k=1\). Set \(\beta =2(f')^{n+1}+nf(f')^{n-1}f''-f(f')^{n}\frac{g'}{g}\). Then \(f\beta =-a\frac{g'}{g}\) and \(h=-\frac{1}{a}\beta g\). We again consider the poles of \(\beta ^{2}F\).

Arguing similarly as in Subcase 2.1, we have

$$\begin{aligned} T\bigl(r,\beta ^{2}F\bigr)\leq 4\overline{N} \biggl(r, \frac{1}{g} \biggr), \end{aligned}$$

and (26) is still valid.

It follows from (7) and (26) that

$$\begin{aligned} & 2(n+2)T(r,f)+N_{1}(r,f) \\ &\quad \leq 2\overline{N}(r,f)+2\overline{N} \biggl(r,\frac{1}{f} \biggr) +6 \overline{N} \biggl(r,\frac{1}{g} \biggr)+2nN_{1} \biggl(r, \frac{1}{f} \biggr) \\ &\qquad{} +2n\overline{N}_{(2} \biggl(r,\frac{1}{f} \biggr)-2N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f) \\ &\quad =2\overline{N}(r,f)+2(n+1)\overline{N} \biggl(r,\frac{1}{f} \biggr)+6 \overline{N} \biggl(r,\frac{1}{g} \biggr) -2N_{0} \biggl(r, \frac{1}{g'} \biggr)+S(r,f), \end{aligned}$$

which yields

$$\begin{aligned} & T(r,f)+m(r,f)+2(n+1)m \biggl(r,\frac{1}{f} \biggr)+ \bigl[N(r,f)+N_{1}(r,f)-2 \overline{N}(r,f)\bigr] \\ &\quad{} +2(n+1) \biggl[N \biggl(r,\frac{1}{f} \biggr)-\overline{N} \biggl(r, \frac{1}{f} \biggr) \biggr] \leq 6\overline{N} \biggl(r,\frac{1}{g} \biggr)-2N_{0} \biggl(r,\frac{1}{g'} \biggr)+S(r,f). \end{aligned}$$

Noting that \(N(r,f)+N_{1}(r,f)-2\overline{N}(r,f)\geq 0\) and \(N (r,\frac{1}{f} )-\overline{N} (r,\frac{1}{f} )\geq 0\), we have

$$\begin{aligned} T(r,f)\leq 6\overline{N} \biggl(r,\frac{1}{g} \biggr)+S(r,f). \end{aligned}$$
(29)

Thus, from the above two cases, we have

$$\begin{aligned} T(r,f)\leq M\overline{N} \biggl(r,\frac{1}{f^{l}(f^{(k)})^{n}-a} \biggr)+S(r,f) \end{aligned}$$

for \(M=\min \{\frac{1}{l-2},6\}\) and positive integers \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\). □

Proof of Corollary 1.1

Set \(\psi:=f^{l}(f^{(k)})^{n}\), where \(l(\geq 2)\), \(n(\geq 1)\), \(k(\geq 1)\) are integers. It follows from Lemma 2.1 that \(\psi \not \equiv 0\). By using Theorem 1.6, we have

$$\begin{aligned} T(r,f)\leq M\overline{N} \biggl(r,\frac{1}{\psi -a} \biggr)+S(r,f), \end{aligned}$$
(30)

where \(M=\min \{\frac{1}{l-2},6\}\).

Applying the lemma of the logarithmic derivative, we get

$$\begin{aligned} T(r,\psi ) &=T\bigl(r,f^{l} \bigl(f^{(k)}\bigr)^{n}\bigr) \\ &\leq lT(r,f)+n \biggl[m \biggl(r,\frac{f^{(k)}}{f} \biggr)+m(r,f)+N(r,f)+k \overline{N}(r,f) \biggr] \\ &\leq (nk+n+l)T(r,f)+S(r,f) \end{aligned}$$
(31)

and

$$\begin{aligned} (n+l)T(r,f) &= T \biggl(r,\frac{1}{f^{n+l}} \biggr)+O(1) =T \biggl(r, \biggl(\frac{f^{(k)}}{f} \biggr)^{n} \frac{1}{\psi } \biggr)+O(1) \\ &\leq N \biggl(r, \biggl(\frac{f^{(k)}}{f} \biggr)^{n} \biggr)+T(r, \psi )+S(r,f) \\ &= nk \biggl[\overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f} \biggr) \biggr]+T(r,\psi )+S(r,f) \\ &\leq (2nk+1)T(r,\psi )+S(r,f). \end{aligned}$$
(32)

It follows from (31) and (32) that

$$\begin{aligned} S(r,f)=S(r,\psi ). \end{aligned}$$
(33)

Combining (30), (31), and (33), we have

$$\begin{aligned} T(r,\psi )\leq M(nk+n+l)\overline{N} \biggl(r, \frac{1}{\psi -a} \biggr)+S(r, \psi ), \end{aligned}$$
(34)

where \(M=\min \{\frac{1}{l-2},6\}\). By the definition of the deficiency \(\Theta (a,\psi )\) and (33), we have, for \(M=\min \{\frac{1}{l-2},6\}\),

$$\begin{aligned} \Theta (a,\psi ) &= 1-\limsup_{r\to \infty } \frac{\overline{N} (r,\frac{1}{\psi -a} )}{T(r,\psi )} \\ &\leq 1-\limsup_{r\to \infty } \frac{\frac{1}{M(nk+n+l)}T(r,\psi )-S(r,\psi )}{T(r,\psi )} \\ &= 1-\frac{1}{M(nk+n+l)}. \end{aligned}$$

 □

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Acknowledgements

The authors would like to thank the referee for his/her reading of the original version of the manuscript with valuable suggestions and comments.

Funding

The first author was supported by the NNSF of China (Nos: 12001117, 12001503), Basic and applied basic research of Guangzhou Basic Research Program (No. 202102020438). The second author was supported by the NNSF of China (Nos: 11801093, 11871260).

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Chen, MF., Huang, ZB. Characteristic estimation of differential polynomials. J Inequal Appl 2021, 181 (2021). https://doi.org/10.1186/s13660-021-02716-6

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MSC

  • 30D35
  • 30D05

Keywords

  • Nevanlinna theory
  • Value distribution
  • Meromorphic solution
  • Nevanlinna characteristic