Uniqueness of Entire Functions Sharing Polynomials with Their Derivatives

doi:10.5666/KMJ.2018.58.3.519

Articles

Kyungpook Mathematical Journal 2018; 58(3): 519-531

Published online September 30, 2018 https://doi.org/10.5666/KMJ.2018.58.3.519

Uniqueness of Entire Functions Sharing Polynomials with Their Derivatives

Pulak Sahoo^*
Department of Mathematics, University of Kalyani, West Bengal-741235, India
e-mail : sahoopulak1@gmail.com

Gurudas Biswas
Department of Mathematics, Hooghly Women’s College, West Bengal-712103, India
e-mail : gdb.math@gmail.com

Received: May 25, 2017; Accepted: June 28, 2018

Abstract

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

In this paper, we investigate the uniqueness problem of entire functions sharing two polynomials with their k-th derivatives. We look into the conjecture given by Lü, Li and Yang [Bull. Korean Math. Soc., 51(2014), 1281–1289] for the case F = fⁿP(f), where f is a transcendental entire function and P(z) = a_mz^m+a_m₋₁z^m⁻¹+ …+a₁z+a₀(≢ 0), m is a nonnegative integer, a_m, a_m₋₁, …, a₁, a₀ are complex constants and obtain a result which improves and generalizes many previous results. We also provide some examples to show that the conditions taken in our result are best possible.

Keywords: entire function, derivative, uniqueness.

1. Introduction, Definitions and Results

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

In this paper, by meromorphic (entire) function we shall always mean meromorphic (entire) function in the complex plane. We assume that the reader is familiar with the standard notations of Nevanlinna’s theory of meromorphic functions as explained in [7, 9, 17]. For a nonconstant meromorphic function f, we denote by T(r, f) the Nevanlinna Characteristic function of f and by S(r, f) any quantity satisfying S(r, f) = o{T(r, f)} for all r outside a possible exceptional set of finite logarithmic measure. The meromorphic function a is called a small function of f, if T(r, a) = S(r, f), where r → ∞ outside a possible exceptional set of finite measure.

Let k be a positive integer or infinity and a ∈ ℂ∪{∞}. We denote by N_k₎(r, a; f) the counting function of all those a-points of f whose multiplicities are not greater than k and by N₍_k₊₁(r, a; f) the counting function of all those a-points of f whose multiplicities are greater than k.

Let f and g be two nonconstant meromorphic functions and Q₁, Q₂ be two polynomials or complex numbers. If f − Q₁ and g − Q₂ have the same zeros with the same multiplicities, then we say that f − Q₁ and g − Q₂ share the value 0 CM. Especially, if Q₁ = Q₂ = a, where a ∈ ℂ ∪ {∞}, then we say that f and g share the value a CM, when f − a and g − a have the same zeros with the same multiplicities. The uniqueness problem of entire and meromorphic functions sharing values, small functions, polynomials with their derivatives is an interesting topic of value distribution theory. Many mathematicians (see [5, 8, 16, 19, 20, 21]) worked on this topic and they gave many conjectures and results. In 1976, Rubel and Yang [14] first proved a result which is as follows.

Theorem A

If a nonconstant entire function f and its derivative f′ share two distinct finite values CM, then f = f′.

Theorem A suggests the following question.

Question 1

What can be said if a nonconstant entire function f shares one finite value CM with its derivative f′?

In 1996, Brück [2] presented the following conjecture relating to Question 1.

Conjecture 1

Let f be a nonconstant entire function. Suppose that ρ₁(f), the first iterated order of f, is not a positive integer or infinite whereρ1(f)=lim supr→∞log log T(r,f)log rand if f and f′ share one finite value a CM, thenf′-af-a=c, for some nonzero constant c.

In 1996, Brück [2] proved that the conjecture is true if a = 0 or N(r, 0; f′) = S(r, f). In 1998, Gundersen and Yang [6] proved that the conjecture is true if f is of finite order and fails, in general, for meromorphic functions. In 2004, Chen and Shon [3] proved that the conjecture is true for entire function of order ρ1(f)<12. In 2005, Al-Khaladi [1] proved that the conjecture is true for meromorphic function f when N(r, 0; f′) = S(r, f).

Now it is natural to ask the following question.

Question 2

Whether Brück Conjecture holds if the function f is replaced by its n-th power fⁿ?

In 2008, Yang and Zhang [18] answered the above question by proving the following result.

Theorem B

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Let f be a nonconstant entire function, n ≥ 7 be an integer and let F = fⁿ. If F and F′ share 1 CM, then F = F′ and f assumes the formf(z)=ce1nz, where c is a nonzero constant.

In 2010, Zhang and Yang [22] further improved Theorem B by considering k-th derivative of fⁿ as follows.

Theorem C

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Let f be a nonconstant entire function and n, k be two positive integers such that n ≥ k + 1. If fⁿ and (fⁿ)⁽^k⁾share the value 1 CM, then fⁿ = (fⁿ)⁽^k⁾and f assumes the formf(z)=ceλnz, where c, λ are nonzero constants and λ^k = 1.

In 2011, Lü and Yi [11] considered polynomial sharing instead of value sharing and proved the following result.

Theorem D

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Let f be a transcendental entire function, n, k be two positive integers and Q ≢ 0 be a polynomial. If fⁿ − Q and (fⁿ)⁽^k⁾ − Q share the value 0 CM and n ≥ k + 1, then fⁿ = (fⁿ)⁽^k⁾and f assumes the formf(z)=ceλnz, where c, λ are nonzero constants and λ^k = 1.

Regarding Theorem D one may ask the following question.

Question 3

What can be said if fⁿ −Q¹and (fⁿ)⁽^k⁾ −Q₂share the value 0 CM, where Q₁, Q₂are polynomials with Q₁Q₂ ≢ 0?

In 2014, Lü, Li and Yang [10] answered the above question for k = 1 and obtained the following result.

Theorem E

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Let f be a transcendental entire function and n ≥ 2 be an integer. If fⁿ − Q₁and (fⁿ)′ − Q₂share the value 0 CM, thenQ2Q1is a polynomial andf′=Q2nQ1f. Furthermore, if Q₁ = Q₂, thenf(z)=ce1nz, where Q₁, Q₂are polynomials with Q₁Q₂ ≢ 0, and c is a nonzero constant.

In the same paper the authors posed the following conjecture.

Conjecture 2

Let f be a transcendental entire function and n, k be two positive integers such that n ≥ k + 1. If fⁿ − Q₁and (fⁿ)⁽^k⁾ − Q₂share the value 0 CM, then(fn)(k)=Q2Q1fn. Furthermore, if Q₁ = Q₂, thenf(z)=ceλnz, where Q₁, Q₂are polynomials with Q₁Q₂ ≢ 0, and c, λ are nonzero constants such that λ^k = 1.

Recently Majumder [12] showed that the above conjecture is true for any positive integer k. The following two examples given in [12] respectively shows that the condition n ≥ k + 1 and f is transcendental in Conjecture 2 are essential.

Example 1

Let f(z) = e²^z + z. Then f − Q₁ and f′ − Q₂ share 0 CM, but f′≢Q2Q1f, where Q₁(z) = z + 1 and Q₂(z) = 3.

Example 2

Let f(z) = z. Then f² − Q₁ and (f²)′ − Q₂ share 0 CM, but (f2)′≢Q2Q1f2, where Q₁(z) = 2z² + z and Q₂(z) = 2z² + 4z.

In [10] the authors posed the following two questions.

Question 4

What can be said if in Conjecture 2 the condition “fⁿ” be replaced by “P(f)” whereP(z)=∑i=0naizi?

Question 5

What can be said if in Conjecture 2 the condition “fⁿ” be replaced by “f(z + c₁)f(z + c₂) …f(z + c_n)” where c_j (j = 1, 2, …, n) are constants?

Our aim to write this paper is to investigate the Conjecture due to Lü, Li and Yang by considering the function F = fⁿP(f) where f is a transcendental entire function and P(z)=∑i=0maizi, a₀, a₁, …, a_m(≠ 0) are complex constants. Though we are able to find out an affirmative solution of Question 4 as far as we know Question 5 remains open. The following is the main result of the paper.

Theorem 1

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Let f be transcendental entire function and n, m, k be positive integers such that n ≥ m+k+1. If fⁿP(f)−Q₁and (fⁿP(f))⁽^k⁾−Q₂share 0 CM, then P(z) reduces to a nonzero monomial, namely P(z) = a_izⁱ for some i ∈ {0, 1, …, m} and(fn+i)(k)=Q2Q1fn+i. Furthermore, if Q₁ = Q₂, then f assumes the formf(z)=ceλn+iz, where Q₁, Q₂are polynomials with Q₁Q₂ ≢ 0 and c, λ are nonzero constants such that λ^k = 1.

The condition n ≥ m+k+1 in Theorem 1 is essential as shown by the following example.

Example 3

Let f(z) = e^z − 1 and P(f) = f² + 3f + 3. Then fP(f) − Q₁ and (fP(f))′ − Q₂ share 0 CM, but (fP(f))′≢Q2Q1fP(f), where Q₁(z) = z + 1 and Q₂(z) = 3z + 6.

The following example shows that the hypothesis of transcendental of f in Theorem 1 is necessary.

Example 4

Let f(z) = z − 1 and P(f) = f + 1. Then f³P(f) − Q₁ and (f³P(f))′ − Q₂ share 0 CM, but (f3P(f))′≢Q2Q1f3P(f), where Q₁(z) = z⁴ + 3z² and Q₂(z) = −14z³ − 9z² − 1.

2. Lemmas

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

In this section we present some lemmas which will be needed in the sequel.

Lemma 1

([15]) Let f be a nonconstant meromorphic function and a_n(z)(≢ 0), a_n₋₁(z), …, a₁(z), a₀(z) be meromorphic functions such that T(r, a_i(z)) = S(r, f) for i = 0, 1, 2, …, n. Then

T(r,anfn+an-1fn-1+…+a1f+a0)=nT(r,f)+S(r,f).

Lemma 2

([4]) Suppose that f is a transcendental meromorphic function and that

fnP*(f)=Q*(f),

where P_*(f) and Q_*(f) are differential polynomials in f with functions of small proximity related to f as the coefficients and the degree of Q_*(f) is at most n. Then

m(r,∞;P*(f))=S(r,f).

Lemma 3

([7]) Let f be a nonconstant meromorphic function and let a₁(z), a₂(z) be two meromorphic functions such that T(r, a_i) = S(r, f), i = 1, 2. Then

T(r,f)≤N¯(r,∞;f)+N¯(r,a1;f)+N¯(r,a2;f)+S(r,f).

Lemma 4

([13]) Let f be a nonconstant meromorphic function and n, k, m be positive integers such that n ≥ k + 1. If fⁿP(f) = {fⁿP(f)}⁽^k⁾, then P(z) reduces to a nonzero monomial, namely P(z) = a_izⁱ for some i ∈ {0, 1, …, m}; and fⁿ⁺ⁱ ≡ (fⁿ⁺ⁱ)⁽^k⁾, where f assumes the formf(z)=ceλn+iz, where c is a nonzero constant and λ^k = 1.

3. Proof of the Theorem

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Proof of the Theorem 1

Let F*=FQ1, G*=GQ2, where F = fⁿP(f) and G = (fⁿP(f))⁽^k⁾. Clearly F_* and G_* share 1 CM except for the zeros of Q_i(z), where i = 1, 2 and so N̄(r, 1; F_*) = N̄(r, 1;G_*) + S(r, f). Let

(3.1)W=F*′(F*-G*)F*(F*-1).

We now consider the following two cases.

Case 1

Let W ≢ 0. It is obvious that m(r, ∞; W) = S(r, f).

Let z₀ be zero of f with multiplicity p₀(≥ 1) which is a zero of P(f) with multiplicity q₀(≥ 1) such that Q_i(z₀) ≠ 0, where i = 1, 2. Then from (3.1), we obtain

(3.2)W(z)=O ((z-z0)np0+q0-k-1).

Since n ≥ m+k+1 and f is transcendental entire, we see that N(r, ∞; W) = S(r, f). Consequently T(r, W) = S(r, f).

Now from (3.1) we see that

1F*=1WF*′F*(F*-1)(1-G*F*).

Therefore, it follows from above that m(r, 0; F_*) = S(r, f), and hence m(r, 0; f) = S(r, f). Also

N(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Hence T(r, f) = S(r, f), a contradiction.

Let z₁ be zero of f with multiplicity p₁(≥ 1) which is not a zero of P(f) and Q_i(z₁) ≠ 0 for i = 1, 2. Then as before we obtain

(3.3)W(z)=O ((z-z1)np1-k-1),

(3.4)T(r,W)=S(r,f) and m(r,0;f)=S(r,f).

We now discuss the following two subcases.

Subcase 1

Let n > m+ k + 1. Then from (3.3), we obtain

(3.5)N(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Hence from (3.4) and (3.5), we get T(r, f) = S(r, f), a contradiction.

Subcase 2

Let n = m + k + 1. Then from (3.3) we see that

N(2(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Then (3.4) gives

(3.6)T(r,f)=N1)(r,0;f)+S(r,f).

Let

(3.7)F=fnP(f) =amfn+m+am-1fn+m-1+…+a1fn+1+a0fn =Fm+Fm-1+…+F1+F0, say.

Since F − Q₁ and F⁽^k⁾ − Q₂ share 0 CM, there exists an entire function α, such that

(3.8)F(k)-Q2=eα(F-Q1).

First we assume that e^α is not constant. Differentiating (3.8), we get

(3.9)F(k+1)-Q2′=α′eα(F-Q1)+eα(F′-Q1′).

From (3.8) and (3.9) we obtain

(3.10)F(k+1)F-α′F(k)F-F(k)F′=Q1F(k+1)-(α′Q1+Q1′)F(k)-Q2F′+(Q2′-α′Q2)F+α′Q1Q2+Q2Q1′-Q1Q2′.

From (3.8) we see that

T(r,eα)≤(k+2)T(r,F)+O(log r)+S(r,F)=(n+m)(k+2)T(r,f)+S(r,f).

Since T(r, α′) = S(r, e^α), it follows that T(r, α′) = S(r, f). Now from (3.7), we deduce for i ∈ {0, 1, …, m} that

Fi′=ai(n+i)fn+i-1f′ =fm{ai(n+i)fn+i-m-1f′}

Fi′′=ai(n+i)(n+i-1)fn+i-2(f′)2+ai(n+i)fn+i-1f″ =fm{ai(n+i)(n+i-1)fn+i-m-2(f′)2+ai(n+i)fn+i-m-1f″}

Fi′′′=ai(n+i)(n+i-1)(n+i-2)fn+i-3(f′)3 +3ai(n+i)(n+i-1)fn+i-2f′ f″+ai(n+i)fn+i-1f‴ =fm{ai(n+i)(n+i-1)(n+i-2)fn+i-m-3(f′)3 +3ai(n+i)(n+i-1)fn+i-m-2f′ f″+ai(n+i)fn+i-m-1f‴}

and so on.

Thus in general we have

Fi(k)=fm∑λiaλifl0λi(f′)l1λi…(f(k))lkλi,

where l0λi, l1λi,…, lkλi are nonnegative integers satisfying ∑j=0kljλi=n+i-m,n+i-m-k≤l0λi≤n+i-m-1 and a_λⁱ are constants for i ∈ {0, 1, …, m}. Also we have

Fi(k+1)=fm∑λibλifp0λi(f′)p1λi…(f(k+1))pk+1λi,

where p0λi, p1λi,…, pk+1λi are nonnegative integers satisfying ∑j=0k+1pjλi=n+i-m,n+i-m-k-1≤p0λi≤n+i-m-1 and b_λⁱ are constants for i ∈ {0, 1, …, m}.

Thus we have from (3.7)

(3.11)F(k)=fm{∑λmaλmfl0λm(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1(f′)l1λm-1…(f(k))lkλm-1+…+∑λ1aλ1fl0λ1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0(f′)l1λ0…(f(k))lkλ0}

and

(3.12)F(k+1)=fm{∑λmbλmfp0λm(f′)p1λm…(f(k+1))pk+1λm+∑λm-1bλm-1fp0λm-1(f′)p1λm-1…(f(k+1))pk+1λm-1+…+∑λ1bλ1fp0λ1(f′)p1λ1…(f(k+1))pk+1λ1+∑λ0bλ0fp0λ0(f′)p1λ0…(f(k+1))pk+1λ0}.

Using (3.7), (3.11) and (3.12) in (3.10), we obtain

(3.13)fn+mP*(f)=Q*(f),

where Q_*(f) is a differential polynomial in f of degree n + m and

(3.14)P*(f)={∑λmbλmfp0λm(f′)p1λm…(f(k+1))pk+1λm+∑λm-1bλm-1fp0λm-1(f′)p1λm-1…(f(k+1))pk+1λm-1+…+∑λ1bλ1fp0λ1(f′)p1λ1…(f(k+1))pk+1λ1+∑λ0bλ0fp0λ0(f′)p1λ0…(f(k+1))pk+1λ0}P(f)-α′{∑λmaλmfl0λm(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1(f′)l1λm-1…(f(k))lkλm-1+…+∑λ1aλ1fl0λ1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0(f′)l1λ0…(f(k))lkλ0}P(f)-f′{∑λmaλmfl0λm-1(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1-1(f′)l1λm-1…(f(k))lkλm-1+⋯+∑λ1aλ1fl0λ1-1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0-1(f′)l1λ0…(f(k))lkλ0}P1(f)=A(f′)k+1+R*(f)

is a differential polynomial in f of degree n + m, where A is a suitable constant, P₁(f) = a_m(n+m)f^m+a_m₋₁(n+m−1)f^m⁻¹+ …+a₁(n+1)f +a₀n and R_*(f) is a differential polynomial in f. Actually every monomial of R_*(f) has the form

Ri(α′)fq0λi(f′)q1λi…(f(k+1))qk+1λi,

where q0λi, q1λi,…, qk+1λi are nonnegative integers satisfying ∑j=0k+1qjλi=n+2i-m, n+2i-m-k≤q0λi≤n+2i-m-1 and R_i(α′) are polynomials in α′ with constant coefficients for i ∈ {0, 1, …, m}.

First we suppose that P_*(f) ≢ 0. Then by Lemma 2, we get m(r, ∞; P_*) = S(r, f) and so T(r, P_*) = S(r, f). Consequently, T(r,P*′)=S(r,f).

Note that from (3.14)

(3.15)P*′(f)=A(k+1)(f′)kf″+Bα′(f′)k+1+S*(f)

is a differential polynomial in f, where B is a suitable constant and S_*(f) is a differential polynomial in f. Actually every monomial of S_*(f) has the form

Si(α′)fr0λi(f′)r1λi…(f(k+1))rk+1λi,

where r0λi, r1λi,…, rk+1λi are nonnegative integers satisfying ∑j=0k+1rjλi=n+2i-m, n+2i-m-k≤r0λi≤n+2i-m-1 and S_i(α′) are polynomials in α′ with constant coefficients for i ∈ {0, 1, …, m}.

Let z₂ be simple zero of f. Then from (3.14) and (3.15), we obtain

P*(f(z2))=A(f′(z2))k+1,

and

P*′(f(z2))=A(k+1)(f′(z2))kf″(z2)+Bα′(f′(z2))k+1.

This shows that z₂ is a zero of P*f″-(c1P*′-c2α′P*)f′, where c₁ and c₂ are suitable constants. Let

(3.16)Φ=P*f″-(c1P*′-c2α′P*)f′f.

Clearly Φ ≢ 0 and T(r, Φ) = S(r, f).

From (3.16) we have

(3.17)f″=α1f+β1f′,

where

(3.18)α1=ΦP*, β1=c1P*′P*-c2α′

and

T(r,α1)=S(r,f), T(r,β1)=S(r,f).

(3.14) and (3.18) together gives

(3.19)P*′=(β1c1+c2c1α′) P* =A (β1c1+c2c1α′) (f′)k+1+(β1c1+c2c1α′) R*(f).

Using (3.15) and (3.17), we get

(3.20)P*′=A(k+1)α1f(f′)k+{A(k+1)β1+Bα′}(f′)k+1+S*(f).

By (3.19) and (3.20), we have

(3.21)(Ac1β1-A(k+1)β1+Ac2c1α′-Bα′) (f′)k+1-A(k+1)α1f(f′)k+(β1c1+c2c1α′) R*(f)-S*(f)≡0.

Since α₁ ≢ 0, from (3.21) we get

(3.22)N1)(r,0;f)=S(r,f).

Therefore from (3.6) and (3.22) we have

T(r,f)=S(r,f),

a contradiction.

Next we suppose that P_*(f) ≡ 0. Then Q_*(f) ≡ 0 by (3.13), where

Q*(f)=Q1F(k+1)-(α′Q1+Q1′)F(k)-Q2F′+(Q2′-α′Q2)F+α′Q1Q2+Q2Q1′-Q1Q2′.

So from (3.10) it follows that

F(k+1)F-α′F(k)F-F(k)F′≡0,

i.e.,

(3.23)F(k+1)F(k)≡α′+F′F.

Integrating we obtain F⁽^k⁾ = c₃Fe^α, where c₃ is a nonzero constant. Substituting the values of F and F⁽^k⁾ into (3.8) we obtain

(c3-1)fnP(f)=Q2-Q1eαeα.

Clearly c₃ ≠ 1 and all zeros of Q₂ − Q₁e^α have the multiplicities at least n. Since n = m + k + 1, by Lemma 3 we get

T(r,eα)≤N¯(r,0;eα)+N¯(r,∞;eα)+N¯ (r,Q2Q1;eα)+S(r,eα) ≤1nN (r,Q2Q1;eα)+S(r,eα) ≤1nT(r,eα)+S(r,eα),

which contradicts to the assumption that e^α is a nonconstant entire function.

Next we assume that e^α is a constant, say c₄. Then from (3.8), we have

(3.24)F(k)-c4F≡Q2-c4Q1.

Since n = m + k + 1, it follows that

N(r,0;f)=S(r,f)

and hence by (3.4) we have

T(r,f)=S(r,f),

a contradiction.

Case 2

Let W ≡ 0. Then from (3.1) we get F_* = G_*,

(3.25)i.e., (fnP(f))(k)=Q2Q1fnP(f).

If Q₁ and Q₂ are same polynomials then using Lemma 4 we can get the conclusion of the theorem. Next we assume that Q₁ and Q₂ are distinct. We show that P(z) reduces to a nonzero monomial of the form P(z) = a_izⁱ for some i ∈ {0, 1, …, m}. If not, we may assume that P(z) = a_mz^m + a_m₋₁z^m⁻¹ + …+ a₁z + a₀, where at least two of a_m, a_m₋₁, …, a₁, a₀, namely a_p, a_q, p ≠ q are nonzero. As f is entire and n ≥ m + k + 1, from (3.25) we see that 0 is a Picard exceptional value of f. So we have f(z) = e^α, where α is a nonconstant entire function. It is easy to see that for i ∈ {0, 1, …, m},

(3.26)ai{(fn+i)(k)-Q2Q1fn+i}=[ti(α′, α″,…,α(k))-aiQ2Q1] e(n+i)α =si(α′, α″,…,α(k))e(n+i)α,

where s_i(α′, α″, …, α⁽^k⁾) are differential polynomials in α′, α″, …, α⁽^k⁾ with rational coefficients. Using (3.25) and (3.26), we obtain

(3.27)sm(α′, α″,…, α(k))emα+sm-1(α′, α″,…, α(k))e(m-1)α+…+s1(α′, α″,…, α(k))eα+s0(α′, α″,…, α(k))≡0.

Since T(r, s_i) = S(r, f) (i = 0, 1, …, m), by Borel theorem on the combination of entire functions (see Theorem 1.52 of [17]), (3.27) gives s_i = 0 for i ∈ {0, 1, …, m}. As a_p, a_q ≠ 0, from (3.26), we have

(fn+p)(k)=Q2Q1fn+p and (fn+q)(k)=Q2Q1fn+q.

Thus we get two different forms of f simultaneously, a contradiction. Hence P(z) = a_izⁱ for some i ∈ {0, 1, …, m}. Therefore, (fn+i)(k)=Q2Q1fn+i for some i ∈ {0, 1, …, m}. This completes the proof of Theorem 1.

Acknowledgements

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

The authors are grateful to the referee for his/her valuable suggestions and comments towards the improvement of the paper.

References

Go to

Abstract
1. Introduction, Definitions and Results
Theorem B
Theorem C
Theorem D
Theorem E
Theorem 1
2. Lemmas
3. Proof of the Theorem
Acknowledgements
References

Al-Khaladi, A (2005). On meromorphic functions that share one value with their derivatives. Analysis. 25, 131-140.
Brück, R (1996). On entire functions which share one value CM with their first derivative. Results Math. 30, 21-24.
Chen, ZX, and Shon, KH (2004). On conjecture of R. Brück concerning entire function sharing one value CM with its derivative. Taiwanese J Math. 8, 235-244.
Clunie, J (1962). On integral and meromorphic functions. J London Math Soc. 37, 17-27.
Gundersen, GG (1980). Meromorphic functions that share finite values with their derivatives. J Math Anal Appl. 75, 441-446.
Gundersen, GG, and Yang, LZ (1998). Entire functions that share one value with one or two of their derivatives. J Math Anal Appl. 223, 88-95.
Hayman, WK (1964). Meromorphic Functions. Oxford: The Clarendon Press
Jank, G, Mues, E, and Volkmann, L (1986). Meromorphe Funktionen, die mit ihrer ersten und zweiten Ableitung einen endchen Wert teilen. Complex Variables Theory Appl. 6, 51-71.
Laine, I (1993). Nevanlinna Theory and Complex Differential Equations. Berlin/New York: Walter de Gruyter
Lü, W, Li, Q, and Yang, CC (2014). On the transcendental entire solutions of a class of differential equations. Bull Korean Math Soc. 51, 1281-1289.
Lü, F, and Yi, H-X (2011). The Brück conjecture and entire functions sharing polynomials with their k-th derivatives. J Korean Math Soc. 48, 499-512.
Majumder, S (2016). A result on a conjecture of W. Lü, Q. Li and C. C. Yang. Bull Korean Math Soc. 53, 411-421.
Majumder, S (2016). Values shared by meromorphic functions and their derivatives. Arab J Math Sci. 22, 265-274.
Rubel, LA, and Yang, CC . Values shared by an entire function and its derivative. Complex Analysis, Proc. Conf., Univ, 1976, Kentucky, Lexington, Ky, pp.101-103.
Yang, CC (1972). On deficiencies of differential polynomials II. Math Z. 125, 107-112.
Yang, LZ (1990). Entire functions that share finite values with their derivatives. Bull Aust Math Soc. 41, 337-342.
Yang, C-C, and Yi, H-X (1995). Uniqueness theory of meromorphic functions. Beijing: Science Press
Yang, L-Z, and Zhang, J-L (2008). Non-existence of meromorphic solutions of Fermat Type functional equation. Aequationes Math. 76, 140-150.
Yi, HX (1991). On characteristic function of a meromorphic function and its derivative. Indian J Math. 33, 119-133.
Zhang, QC (2005). Meromorphic function that shares one small function with its derivative. JIPAM J Inequal Pure Appl Math. 6, 13.
Zhang, J-L, and Yang, L-Z (2007). Some results related to a conjecture of R. Brück. JIPAM Inequal Pure Appl Math. 8, 11.
Zhang, J-L, and Yang, L-Z (2010). A Power of an entire function sharing one value with its derivative. Comput Math Appl. 60, 2153-2160.