# 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}

### 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}

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}

\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}

□

Not applicable.

## References

1. Alotaibi, A.: On the zeros of $$af(f^{(k)})^{n}-1$$ for $$n\geq 2$$. Comput. Methods Funct. Theory 4(1), 227–235 (2004)

2. Bergweiler, W., Eremenko, A.: On the singularities of the inverse to a meromorphic function of finite order. Rev. Mat. Iberoam. 11(2), 355–373 (1995)

3. Doeringer, W.: Exceptional values of differential polynomials. Pac. J. Math. 98(1), 55–62 (1982)

4. Hayman, W.K.: Meromorphic Function. Clarendon Press, Oxford (1964)

5. Hayman, W.K.: Picard values of meromorphic functions and their derivatives. Ann. Math. 70(1), 9–42 (1959)

6. Huang, X.J., Gu, Y.X.: On the value distribution of $$f^{2}f^{(k)}$$. J. Aust. Math. Soc. 78(1), 17–26 (2005)

7. Jiang, Y., Huang, B.: A note on the value distribution of $$f^{l}(f^{(k)})^{n}$$. Hiroshima Math. J. 46(2), 135–147 (2016)

8. Karmakar, H., Sahoo, P.: On the value distribution of $$f^{n}f^{(k)}-1$$. Results Math. 73(3), Article ID 98 (2018)

9. Lahiri, I., Dewan, S.: Inequalities arising out of the value distribution of a differential monomial. J. Inequal. Pure Appl. Math. 4(2), Article ID 27 (2003)

10. Laine, I.: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin (1993)

11. Langley, J.K.: The zeros of $$ff''-b$$. Results Math. 44(1–2), 130–140 (2003)

12. Mues, E.: Über ein Problem von Hayman. Math. Z. 164(3), 239–259 (1979)

13. Tse, C.K., Yang, C.C.: On the value distribution of $$f^{l}(f^{(k)})^{n}$$. Kodai Math. J. 17(1), 163–169 (1994)

14. Wang, Y.F., Yang, C.C., Yang, L.: On the zeros of $$f(f^{(k)})^{n}-a$$. Kexue Tongbao. 38, 2215–2218 (1993)

15. Yang, C.C., Yi, H.X.: Uniqueness Theory of Meromorphic Functions. Science Press, Beijing (1995) Kluwer Academic, Dordrecht (2003)

16. Yang, L.: Value Distribution Theory. Springer, Berlin (1993)

17. Zhang, Q.D.: A growth theorem for meromorphic function. J. Chengdu Inst. Meteor. 20(1), 12–20 (1992)

## 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).

## Author information

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### Contributions

All authors completed the main part of this article and corrected the main theorems. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Zhi-Bo Huang.

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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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• DOI: https://doi.org/10.1186/s13660-021-02716-6

• 30D35
• 30D05

### Keywords

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