Uniqueness theorems of the Hahn difference operator of entire function with a Picard exceptional value

Xiaomei Zhang; Zhaojun Wu

doi:10.1515/math-2022-0601

Article Open Access

Uniqueness theorems of the Hahn difference operator of entire function with a Picard exceptional value

Xiaomei Zhang and Zhaojun Wu

Published/Copyright: July 13, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

Let f be a transcendental entire function of finite order with a Picard exceptional value β ∈ C , q ∈ C \ { 0 , 1 } and c are complex constants. The authors prove that

D q , c f ( z ) − a f ( z ) − a = a a − β ,

if D q , c f ( z ) and f ( z ) share value a ( ≠ β ) CM, where

D q , c f ( z ) = f ( q z + c ) − f ( z ) ( q − 1 ) z + c

is the Hahn difference operator. This result generalizes the related results of Zongxuan Chen [On the difference counterpart of Brück’s conjecture, Acta Math. Sci. 34B (2014), 653–659].

Keywords: Hahn difference operator; Jackson q-difference operator; Picard exceptional values; uniqueness

MSC 2010: 30D30

1 Introduction

Let f be a function meromorphic in the complex plane C . Following Hahn [1], the Hahn difference operator D q , c f ( z ) is defined by

(1) D q , c f ( z ) = f ( q z + c ) − f ( z ) ( q − 1 ) z + c .

where q ∈ C \ { 0 , 1 } and c be a complex constant. Hahn difference operator D q , c f ( z ) unifies two well-known difference operators [2]. The first is the Jackson q -difference operator, which is defined by [2–5]

(2) D q f ( z ) = f ( q z ) − f ( z ) ( q − 1 ) z .

The second difference operator, which Hahn’s operator generalizes, is the forward difference operator

Δ c f ( z ) = f ( z + c ) − f ( z ) c .

Note that [2]

lim c → 0 D q , c f ( z ) = D q f ( z ) ,

lim q → 1 D q , c f ( z ) = Δ c f ( z ) ,

lim c → 0 , q → 1 D q , c f ( z ) = f ′ ( z ) .

The main purpose of this article is to study the Hahn difference counterpart of Brück’s conjecture.

Brück’s conjecture: Let f be a nonconstant entire function such that hyper order σ 2 ( f ) < ∞ and σ 2 ( f ) is not positive integer. If f and f ′ share the finite value a CM (see definitions below), then

f ′ ( z ) − a f ( z ) − a = η ,

where η is nonzero constant.

The conjecture has been verified in the special cases when f is of finite order (see [6]), or when σ 2 ( f ) < 1 2 (see [7]). Bergweiler and Langley [8] denote the forward difference operator Δ 1 f ( z ) by

Δ f ( z ) = f ( z + 1 ) − f ( z ) .

Following [8], some complex analysts denote Δ c f ( z ) by

Δ c f ( z ) = f ( z + c ) − f ( z ) .

In this article, we also use this notation. Recently, the difference counterpart of Brück’s conjecture had been established in [9–13]. To give their results, we introduce the following definitions and notations.

The Nevanlinna theory is the main tool to study uniqueness theory of meromorphic function. Therefore, we assume that the reader is familiar with the general conclusion of the Nevanlinna theory [14–17]. Recently, some well-known facts of the Nevanlinna theory have been extended for the differences of meromorhic functions [18–24].

Let f be a function meromorphic in the complex plane C . The order of f ( z ) is denoted by σ ( f ) . For any a ∈ C , the exponent of convergence of zeros of f ( z ) − a is denoted by λ ( f , a ) . If λ ( f , a ) < σ ( f ) , then a is said to be a Borel exceptional value of f ( z ) .

Let f and g be meromorphic functions and a be a complex number. Let z n ( n = 1 , 2 , … ) be zeros of f − a and ν n be the multiplicity of the zero z n . If z n ( n = 1 , 2 , … ) are also ν n ( n = 1 , 2 , … ) multiple zeros of g − a at least, we write f = a → g = a or g = a ← f = a . If f = a ⇆ g = a , it is said that f ( z ) and g ( z ) share the value a CM (see [17]).

Let f be an entire function, Δ c f ( z ) be the forward difference of f ( z ) . In 2014, Chen [9] considered Δ c f ( z ) and f ( z ) share one value a CM and prove the following theorem, which is the difference counterpart of Brü ck’s conjecture.

Theorem A

[9] Let f be a transcendental entire function of finite order that is of a finite Borel exceptional value β , and let c be a constant such that f ( z + c ) ≢ f ( z ) . If Δ c f ( z ) and f ( z ) share a ( a ≠ β ) CM, then,

Δ c f ( z ) − a f ( z ) − a = a a − β .

In this article, Theorem A is extended to the Hahn difference operator D q , c f ( z ) .

Theorem 1.1

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q , c f ( z ) is the Hahn difference operator as that in (1), where q σ ( f ) = 1 . If β ≠ 0 , then D q , c f ( z ) and f cannot share the value β CM.

Under the conditions of Theorem 1.1, there are only two possible scenarios. The first case is D q , c f ( z ) and f may share the value a ≠ β CM for any β ∈ C , and the second case is β = 0 , D q , c f ( z ) and f may share the value 0 CM. For the first case, we shall prove the following theorem.

Theorem 1.2

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q , c f ( z ) is the Hahn difference operator as that in (1), where q σ ( f ) = 1 . If D q , c f ( z ) and f share the value a ( ≠ β ) CM. Then, a ≠ 0 , and

D q , c f ( z ) − a f − a = a a − β .

If c = 0 in D q , c f ( z ) , then we can obtain the following corollary for the Jackson q -difference operator D q f ( z ) from Theorem 1.2.

Corollary 1.3

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q f ( z ) is the Jackson q-difference operator as that in (2), where q σ ( f ) = 1 . If D q f ( z ) and f share the value a ( ≠ β ) CM. Then, a ≠ 0 , and

D q f ( z ) − a f − a = a a − β .

2 Proof of Theorems

Lemma 2.1

[17] Suppose that f 1 , f 2 , … , f n ( n ≥ 2 ) are meromorphic functions and g 1 , g 2 , … , g n are entire functions satisfying the following conditions.

∑ j = 1 n f j ( z ) e g j ( z ) ≡ 0 .
g j ( z ) − g k ( z ) are not constants for 1 ≤ j < k ≤ n .
For 1 ≤ j ≤ n , 1 ≤ h < k ≤ n ,
T ( r , f j ) = o { T ( r , e g h − g k ) } ( r → ∞ , r ∉ E ) ,
where E ⊂ ( i , + ∞ ) is of finite linear measure or finite logarithmic measure. Then f j ( z ) ≡ 0 ( j = 1 , 2 , … , n ) .

Lemma 2.2

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q , c f ( z ) is the Hahn difference operator as that in (1), then for any a ∈ C , σ D q , c f ( z ) − a f ( z ) − a < ∞ .

Proof

Since β is a Picard exceptional value of f , then f ( z ) can be written as follows:

f ( z ) = e P ( z ) + β ,

where P ( z ) is a polynomial. Hence,

D q , c f ( z ) − a f ( z ) − a = e P ( q z + c ) − e P ( z ) − a [ ( q − 1 ) z + c ] [ ( q − 1 ) z + c ] e P ( z ) − ( a − β ) [ ( q − 1 ) z + c ] = e P ( q z + c ) − P ( z ) − 1 − a [ ( q − 1 ) z + c ] e − P ( z ) [ ( q − 1 ) z + c ] − ( a − β ) [ ( q − 1 ) z + c ] e − P ( z ) .

Therefore,

□ σ D q , c f ( z ) − a f ( z ) − a < ∞ .

2.1 Proof of Theorem 1.1

Suppose that D q , c f ( z ) and f ( z ) share the value β CM, then by Lemma 2.2, we can obtain

(3) D q , c f ( z ) − β f ( z ) − β = e h ( z ) ,

where h ( z ) is a polynomial. Since β is a Picard exceptional value of f , then f ( z ) can be written as follows:

(4) f ( z ) = e P ( z ) + β ,

where P ( z ) is a nonconstant polynomial such that deg P ( z ) = σ ( f ) = n ∈ Z + . Therefore, from (3) and (4), we have

e P ( q z + c ) − e P ( z ) − β [ ( q − 1 ) z + c ] [ ( q − 1 ) z + c ] e P ( z ) = e h ( z ) .

i.e.,

(5) 1 [ ( q − 1 ) z + c ] e P ( q z + c ) − P ( z ) − 1 [ ( q − 1 ) z + c ] − β e − P ( z ) = e h ( z ) .

Since q σ ( f ) = q n = 1 , then deg ( P ( q z + c ) − P ( z ) ) ≤ ( deg P ( z ) ) − 1 = σ ( f ) − 1 . Therefore, 1 [ ( q − 1 ) z + c ] e P ( q z + c ) − P ( z ) − 1 [ ( q − 1 ) z + c ] is a small meromorphic function respective to − β e − P ( z ) . By applying the second fundamental theorem to − β e − P ( z ) , we can obtain

(6) λ 1 [ ( q − 1 ) z + c ] e P ( q z + c ) − P ( z ) − 1 [ ( q − 1 ) z + c ] − β e − P ( z ) = deg P ( z ) .

So by (5) and (6), we obtain e h ( z ) has an infinite number of zeros. This contradicts with e h ( z ) ≠ 0 . Thus, the Hahn difference operator D q , c f ( z ) and f ( z ) cannot share the value β CM.

2.2 Proof of Theorem 1.2

Since β is a Picard exceptional values of f , we can write f ( z ) by

(7) f ( z ) = e P ( z ) + β ,

where P ( z ) is a polynomial with deg P ( z ) = σ ( f ) = n ∈ Z + .

First, we prove that a ≠ 0 . If a = 0 , then β ≠ 0 . Since D q , c f ( z ) and f ( z ) share the value 0 CM, then by Lemma 2.2, we can obtain

(8) D q , c f ( z ) f ( z ) = e h ( z ) ,

where h ( z ) is a polynomial. It follows from (7), (8), and (2) that

(9) e P ( q z + c ) − P ( z ) − 1 [ ( q − 1 ) z + c ] e − P ( z ) + β [ ( q − 1 ) z + c ] = e h ( z ) .

Since q σ ( f ) = q n = 1 , then deg ( P ( q z + c ) − P ( z ) ) ≤ ( deg P ( z ) ) − 1 = σ ( f ) − 1 , we see that

(10) λ ( e P ( q z + c ) − P ( z ) − 1 ) ≤ σ ( e P ( q z + c ) − P ( z ) − 1 ) ≤ σ ( f ) − 1 .

Since β ≠ 0 , and 0 , ∞ are Picard exceptional values of e − P ( z ) . By applying the second fundamental theorem to e − P ( z ) , we have

(11) λ ( [ ( q − 1 ) z + c ] e − P ( z ) + β [ ( q − 1 ) z + c ] ) = σ ( f ) .

From (9)–(11), we can obtain a contradiction. Therefore, a ≠ 0 .

Noting D q , c f ( z ) and f ( z ) share the value a CM, then by Lemma 2.2, we can obtain

(12) D q , c f ( z ) − a f ( z ) − a = e q ( z ) ,

where q ( z ) is a polynomial. Combining (7) and (12), we have

e P ( q z + c ) − e P ( z ) − a [ ( q − 1 ) z + c ] [ ( q − 1 ) z + c ] e P ( z ) − ( a − β ) [ ( q − 1 ) z + c ] = e q ( z ) ,

i.e.,

(13) ( a − β ) [ ( q − 1 ) z + c ] e − P ( z ) e q ( z ) − [ ( q − 1 ) z + c ] e q ( z ) + ( e P ( q z + c ) − P ( z ) − 1 ) − a [ ( q − 1 ) z + c ] e − P ( z ) = 0 .

Seeing that q ( z ) is a polynomial, then deg q ( z ) only satisfies one of the following cases: deg q ( z ) > deg P ( z ) ; 1 ≤ deg q ( z ) < deg P ( z ) = σ ( f ) ; deg q ( z ) = deg P ( z ) = σ ( f ) , and deg q ( z ) = 0 .

Case 1. deg q ( z ) > deg P ( z ) By (13), we have

f 11 ( z ) e g 11 ( z ) + f 12 ( z ) e g 12 ( z ) + f 13 ( z ) e g 13 ( z ) + f 14 ( z ) e g 14 ( z ) = 0 ,

where

g 11 ( z ) = q ( z ) − P ( z ) , g 12 ( z ) = q ( z ) , g 13 = − P ( z ) , g 14 ≡ 0 , f 11 ( z ) = ( a − β ) [ ( q − 1 ) z + c ] , f 12 ( z ) = − [ ( q − 1 ) z + c ] , f 13 ( z ) = − a [ ( q − 1 ) z + c ] , f 14 ( z ) = e P ( q z + c ) − P ( z ) − 1 .

Since deg q ( z ) > deg P ( z ) = n , then g 1 i − g 1 j are not constants for 1 ≤ i < j ≤ 4 . Note that

deg ( P ( q z + c ) − P ( z ) ) ≤ n − 1 , deg ( g 11 ( z ) − g 12 ( z ) ) = deg ( − P ( z ) ) = n , deg ( g 11 ( z ) − g 13 ( z ) ) = deg ( q ( z ) ) > n , deg ( g 11 ( z ) − g 14 ( z ) ) = deg ( q ( z ) − P ( z ) ) > n , deg ( g 12 ( z ) − g 13 ( z ) ) = deg ( q ( z ) + P ( z ) ) > n , deg ( g 12 ( z ) − g 14 ( z ) ) = deg ( q ( z ) ) > n , deg ( g 13 ( z ) − g 14 ( z ) ) = deg ( − P ( z ) ) = n .

Thus, for 1 ≤ h ≤ 4 , 1 ≤ i < j ≤ 4 , we have

T ( r , f 1 h ) = o { T ( r , e g 1 i − g 1 j ) } .

By Lemma 2.1, we can obtain ( q − 1 ) z + c ≡ 0 , which is impossible.

Case 2. 1 ≤ deg q ( z ) < deg P ( z ) = σ ( f ) . By (13), we have

(14) ( ( a − β ) [ ( q − 1 ) z + c ] e q ( z ) − a [ ( q − 1 ) z + c ] ) e − P ( z ) − [ ( q − 1 ) z + c ] e q ( z ) + ( e P ( q z + c ) − P ( z ) − 1 ) = 0 .

It follows from β − a ≠ 0 , 1 ≤ deg q ( z ) < deg P ( z ) that ( a − β ) [ ( q − 1 ) z + c ] e q ( z ) − a [ ( q − 1 ) z + c ] ≢ 0 . Hence,

σ ( ( ( a − β ) [ ( q − 1 ) z + c ] e q ( z ) − a [ ( q − 1 ) z + c ] ) e − P ( z ) ) = deg P ( z ) = n .

Since q n = 1 , then deg ( P ( q z + c ) − P ( z ) ) ≤ ( deg P ( z ) ) − 1 . Therefore, the order of − [ ( q − 1 ) z + c ] e q ( z ) + ( e P ( q z + c ) − P ( z ) − 1 ) is less than n . We can obtain a contradiction from (14).

Case 3. deg q ( z ) = σ ( f ) = deg P ( z ) . Suppose

P ( z ) = p n z n + p n − 1 z n − 1 + ⋯ + p 1 z + p 0 , q ( z ) = q n z n + q n − 1 z n − 1 + ⋯ + q 1 z + q 0 .

Thus, p n and q n only satisfy one of the following cases: p n = − q n ; p n = q n ; p n ≠ q n ; and p n ≠ − q n .

Subcase 3.1. p n = − q n . From (13), we can obtain

f 21 ( z ) e g 21 ( z ) + f 22 ( z ) e g 22 ( z ) + f 23 ( z ) e g 23 ( z ) = 0 ,

where

g 21 ( z ) = q ( z ) − P ( z ) , g 22 ( z ) = − P ( z ) , g 23 ≡ 0 , f 21 ( z ) = ( a − β ) [ ( q − 1 ) z + c ] , f 22 ( z ) = − [ ( q − 1 ) z + c ] e q ( z ) + P ( z ) − a [ ( q − 1 ) z + c ] , f 23 ( z ) = e P ( q z + c ) − P ( z ) − 1 .

Since p n = − q n , then g 2 i − g 2 j are not constants for 1 ≤ i < j ≤ 3 . Note that

deg ( P ( q z + c ) − P ( z ) ) ≤ n − 1 , deg ( g 21 ( z ) − g 22 ( z ) ) = deg ( q ( z ) ) = n , deg ( g 21 ( z ) − g 23 ( z ) ) = deg ( q ( z ) − P ( z ) ) = n , deg ( g 22 ( z ) − g 23 ( z ) ) = deg ( − P ( z ) ) = n .

Thus, for 1 ≤ h ≤ 3 , 1 ≤ i < j ≤ 3 , we have

T ( r , f 2 h ) = o { T ( r , e g 2 i − g 2 j ) } .

By Lemma 2.1, we can obtain ( a − β ) [ ( q − 1 ) z + c ] ≡ 0 , which is impossible.

Subcase 3.2. p n = q n . From (13), we can obtain

f 31 ( z ) e g 31 ( z ) + f 32 ( z ) e g 32 ( z ) + f 33 ( z ) e g 33 ( z ) = 0 ,

where

g 31 ( z ) = q ( z ) , g 32 ( z ) = − P ( z ) , g 33 ≡ 0 , f 31 ( z ) = − [ ( q − 1 ) z + c ] , f 32 ( z ) = − a [ ( q − 1 ) z + c ] , f 33 ( z ) = ( a − β ) [ ( q − 1 ) z + c ] e q ( z ) − P ( z ) + e P ( q z + c ) − P ( z ) − 1 .

Since p n = q n , then g 3 i − g 3 j are not constants for 1 ≤ i < j ≤ 3 . Note that

deg ( P ( q z + c ) − P ( z ) ) ≤ n − 1 , deg ( q ( z ) − P ( z ) ) ≤ n − 1 , deg ( g 31 ( z ) − g 32 ( z ) ) = deg ( q ( z ) + P ( z ) ) = n , deg ( g 31 ( z ) − g 33 ( z ) ) = deg ( q ( z ) ) = n , deg ( g 32 ( z ) − g 33 ( z ) ) = deg ( − P ( z ) ) = n .

Thus, for 1 ≤ h ≤ 3 , 1 ≤ i < j ≤ 3 , we have

T ( r , f 3 h ) = o { T ( r , e g 3 i − g 3 j ) } .

By Lemma 2.1, we can obtain ( q − 1 ) z + c ≡ 0 , which is impossible.

Subcase 3.3. p n ≠ q n and p n ≠ − q n . By (13), we have

f 41 ( z ) e g 41 ( z ) + f 42 ( z ) e g 42 ( z ) + f 43 ( z ) e g 43 ( z ) + f 44 ( z ) e g 44 ( z ) = 0 ,

where

g 41 ( z ) = q ( z ) − P ( z ) , g 42 ( z ) = q ( z ) , g 43 = − P ( z ) , g 44 ≡ 0 , f 41 ( z ) = ( a − β ) [ ( q − 1 ) z + c ] , f 42 ( z ) = − [ ( q − 1 ) z + c ] , f 43 ( z ) = − a [ ( q − 1 ) z + c ] , f 44 ( z ) = e P ( q z + c ) − P ( z ) − 1 .

Since p n ≠ q n and p n ≠ − q n , then g 4 i − g 4 j are not constants for 1 ≤ i < j ≤ 4 . Note that

deg ( P ( q z + c ) − P ( z ) ) ≤ n − 1 , deg ( g 41 ( z ) − g 42 ( z ) ) = deg ( − P ( z ) ) = n , deg ( g 41 ( z ) − g 43 ( z ) ) = deg ( q ( z ) ) = n , deg ( g 41 ( z ) − g 44 ( z ) ) = deg ( q ( z ) − P ( z ) ) = n , deg ( g 42 ( z ) − g 43 ( z ) ) = deg ( q ( z ) + P ( z ) ) = n , deg ( g 42 ( z ) − g 44 ( z ) ) = deg ( q ( z ) ) = n , deg ( g 43 ( z ) − g 44 ( z ) ) = deg ( − P ( z ) ) = n .

Thus, for 1 ≤ h ≤ 4 , 1 ≤ i < j ≤ 4 , we have

T ( r , f 4 h ) = o { T ( r , e g 4 i − g 4 j ) } .

By Lemma 2.1, we can obtain ( q − 1 ) z + c ≡ 0 , which is impossible.

Case 4. deg q ( z ) = 0 . In this case, e q ( z ) is a constant. We denote it by K ( ≠ 0 ) . Suppose that K ≠ a a − β , by (13), we can obtain

{ ( a − β ) K [ ( q − 1 ) z + c ] − a [ ( q − 1 ) z + c ] } e − P ( z ) − [ ( q − 1 ) z + c ] K + ( e P ( q z + c ) − P ( z ) − 1 ) = 0 .

Since deg ( P ( q z + c ) − P ( z ) ) ≤ ( deg P ( z ) ) − 1 , then

σ ( − [ ( q − 1 ) z + c ] K + ( e P ( q z + c ) − P ( z ) − 1 ) ) < deg P ( z ) = σ ( ( ( a − β ) K [ ( q − 1 ) z + c ] − a [ ( q − 1 ) z + c ] ) e − P ( z ) ) .

We obtain a contradiction. Hence, K = a a − β .

Therefore, we can obtain the following equality from (13),

D q , c f ( z ) − a f ( z ) − a = a a − β .

3 Conclusion

Suppose that a ≠ 0 , and a ≠ β . If

(15) D q , c f ( z ) − a f ( z ) − a = a a − β

holds for the transcendental entire function f ( z ) = e P ( z ) + β . Combining (1) and (15) shows that

f ( q z + c ) − f ( z ) ( q − 1 ) z + c − a f ( z ) − a = a a − β .

i.e.,

f ( q z + c ) − f ( z ) − a [ ( q − 1 ) z + c ] [ ( q − 1 ) z + c ] f ( z ) − a [ ( q − 1 ) z + c ] = a a − β .

e P ( q z + c ) − e P ( z ) − a [ ( q − 1 ) z + c ] [ ( q − 1 ) z + c ] e P ( z ) − ( a − β ) [ ( q − 1 ) z + c ] = a a − β .

( a − β ) ( e P ( q z + c ) − e P ( z ) ) − ( a − β ) [ ( q − 1 ) z + c ] a = a [ ( q − 1 ) z + c ] e P ( z ) − ( a − β ) [ ( q − 1 ) z + c ] a .

Therefore,

(16) ( a − β ) e P ( q z + c ) − P ( z ) = ( a − β ) + a [ ( q − 1 ) z + c ] .

We can obtain a contradiction from (16). This means equation (15) is impossible. Hence, we can obtain the following conclusions from Theorem 1.2.

Theorem 3.1

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q , c f ( z ) is the Hahn difference operator as that in (1). If D q , c f ( z ) and f share the value a ( a ∈ C , a ≠ β ) CM, then q σ ( f ) ≠ 1 .

Theorem 3.2

Let f be an entire function of order σ ( f ) < ∞ . Suppose that β ∈ C is a Picard exceptional value of f, D q , c f ( z ) is the Hahn difference operator as that in (1). If q σ ( f ) = 1 , then for any a ∈ C and a ≠ β , D q , c f ( z ) and f cannot share the value a CM.

Example 3.3

Let f ( z ) = e z 2 . Then

D − 1 , − 1 f ( z ) = f ( − z − 1 ) − f ( z ) ( − 1 − 1 ) z − 1 = e 2 z + 1 − 1 − 2 z − 1 e z 2 .

For any a ∈ C and a ≠ 0 , e 2 z + 1 − 1 − 2 z − 1 e z 2 and e z 2 cannot share the value a CM.

Acknowledgments

The authors are very grateful to the anonymous referees for their careful review and valuable suggestions.

Funding information: This work was supported by the National Natural Science Foundation of China (Grant No 12071239).
Conflict of interest: The authors state no conflict of interest.

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Received: 2023-02-02

Revised: 2023-05-17

Accepted: 2023-06-06

Published Online: 2023-07-13

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2022-0601

Keywords for this article

Hahn difference operator; Jackson q-difference operator; Picard exceptional values; uniqueness

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