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Uniqueness of exponential polynomials

  • Ge Wang , Zhiying He and Mingliang Fang EMAIL logo
Published/Copyright: December 31, 2023
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Open Mathematics
From the journal Open Mathematics Volume 21 Issue 1

Abstract

In this article, we study the uniqueness of exponential polynomials and mainly prove: Let n be a positive integer, let p i ( z ) ( i = 1 , 2 , … , n ) be nonzero polynomials, and let c i ≠ 0 ( i = 1 , 2 , … , n ) be distinct finite complex numbers. Suppose that f ( z ) is an entire function, g ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z . If f ( z ) and g ( z ) share a and b CM (counting multiplicities), where a and b are two distinct finite complex numbers, then one of the following cases must occur:

  1. n = 1 .

  2. If a ≠ 0 , b = 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ a 2 ;

  3. If a = 0 , b ≠ 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ b 2 ;

  4. If a ≠ 0 , b ≠ 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ ( a + b ) g ( z ) − a b .

  5. n ≥ 2 , f ( z ) ≡ g ( z ) .

This is an extension of the result obtained in an earlier study on meromorphic functions in 1974.

Keywords: exponential polynomials; uniqueness; differences
MSC 2010: 30D35

1 Introduction

In this article, an exponential polynomial is an entire function of the form

(1.1) f ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z ,

where n is a positive integer, the coefficients p i ( z ) ≢ 0 ( i = 1 , 2 , … , n ) are polynomials, and c i ( i = 1 , 2 , … , n ) are nonzero distinct finite complex numbers. The functions sin z = e i z − e − i z 2 i and cos z = e i z + e − i z 2 are typical examples of exponential polynomials, and their quotient being tan z .

In the following, we assume that the reader is familiar with the basic notions of Nevanlinna’s value distribution theory [1–3]. A meromorphic function always means meromorphic in the whole complex plane.

By S ( r , f ) , we denote any quantity satisfying S ( r , f ) = o ( T ( r , f ) ) as r → ∞ possible outside of an exceptional set E with a finite logarithmic measure ∫ E d r ∕ r < ∞ . A meromorphic function a is said to be a small function of f if it satisfies T ( r , a ) = S ( r , f ) .

Let f be a nonconstant meromorphic function. The order of f is defined by

ρ ( f ) = lim r → ∞ ¯ log + T ( r , f ) log r ,

and the difference operators are defined as follows:

Δ c f = f ( z + c ) − f ( z ) and Δ c n f = Δ c n − 1 ( Δ c f ) ,

where n ( ≥ 2 ) is a positive integer and c is a nonzero constant.

Let f and g be two meromorphic functions, and let a be a constant. If f − a and g − a have the same zeros counting multiplicities ignoring multiplicities (IM), we say that f and g share a CM (IM). N ( r , a ) is a counting function of common zeros of f − a and g − a with the same multiplicities, and the multiplicities are counted. If N r . 1 f − a + N r , 1 g − a − 2 N ( r , a ) ≤ S ( r , f ) + S ( r , g ) , then we call that f and g share a CM almost.

Recently, many articles studied the value distribution of exponential polynomials and their role in the theories of complex differential equations and oscillation theory [4–6]. In addition, many articles studied the roots of exponential polynomials [7,8]. In 2019, Su et al. [9] proved that if two nonconstant exponential polynomials with constant coefficients share four distinct values CM that lie in an angular domain of opening strictly larger than π , they must be identical. In this article, we also study the uniqueness of exponential polynomials.

This was proved in 1974 by Rubel and Yang [10].

Theorem A

Let f be an entire function. If f ( z ) and sin z share 0 and 1 CM, then f ( z ) ≡ sin z .

In 1981, Czubiak and Gundersen [11] proved that Theorem A remains valid if f ( z ) and sin z share 0 and 1 IM. In this article, we extended Theorem A and obtained the following results.

Theorem 1

Let n be a positive integer, let p i ( z ) ≢ 0 ( i = 1 , 2 , … , n ) be polynomials, and let c i ( i = 1 , 2 , … , n ) be nonzero distinct finite complex numbers. Suppose that f ( z ) is an entire function, g ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z . If f ( z ) and g ( z ) share a and b CM, where a and b are two distinct finite complex numbers, then one of the following cases must occur:

  1. n = 1 .

  2. If a ≠ 0 , b = 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ a 2 .

  3. If a = 0 , b ≠ 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ b 2 .

  4. If a ≠ 0 , b ≠ 0 , then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ ( a + b ) g ( z ) − a b .

  5. n ≥ 2 , f ( z ) ≡ g ( z ) .

Remark 1

All the cases in Theorem 1 must occur, then we provide two examples to show that the cases of Theorem 1 occurs.

Example 1

Let p 1 and c 1 be two nonzero complex numbers, and let a be a nonzero distinct finite complex number. Set f ( z ) = a 2 p 1 e c 1 z and g ( z ) = p 1 e c 1 z . Obviously, f ( z ) and g ( z ) share a and 0 CM, and f ( z ) g ( z ) ≡ a 2 .

Example 2

Let p 1 and c 1 be two nonzero complex numbers, and let a and b be two nonzero distinct finite complex numbers. Set f ( z ) = a + b − a b p 1 e c 1 z and g ( z ) = p 1 e c 1 z . Obviously, f ( z ) and g ( z ) share a and b CM, and f ( z ) g ( z ) ≡ ( a + b ) g ( z ) − a b .

Theorem 2

Let n and k be two positive integers, let p i ( z ) ≢ 0 ( i = 1 , 2 , … , n ) be polynomials, and let c i ( i = 1 , 2 , … , n ) be nonzero distinct finite complex numbers. Suppose that f ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z . If f and f ( k ) share a CM, where a is a nonzero finite complex number, then f ≡ f ( k ) . Furthermore, we have n ≤ k , c i k = 1 ( 1 ≤ i ≤ n ) and f ( z ) = p 1 e c 1 z + ⋯ + p n e c n z , where p i ( i = 1 , … , n ) are constants.

Theorem 3

Let n and m be two positive integers, let p i ( z ) ≢ 0 ( i = 1 , 2 , … , n ) be polynomials, and let c i ( i = 1 , 2 , … , n ) be nonzero distinct finite complex numbers. Suppose that f ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z . If f and Δ c m f share a CM, where a is a nonzero finite complex number, then f ≡ Δ c m f . Furthermore, ( e c c i − 1 ) m = 1 ( 1 ≤ i ≤ n ) and f ( z ) = p 1 e c 1 z + ⋯ + p n e c n z , where p i ( i = 1 , … , n ) are constants.

Corollary 1

Let n, m, and k be positive integers, let p i ( z ) ≢ 0 ( i = 1 , 2 , … , n ) be polynomials, and let c i ( i = 1 , 2 , … , n ) be nonzero distinct finite complex numbers. Suppose that f ( z ) = p 1 ( z ) e c 1 z + p 2 ( z ) e c 2 z + ⋯ + p n ( z ) e c n z . If f , f ( k ) , and Δ c m f share a CM, where a is a nonzero finite complex number, then f ( k ) ≡ Δ c m f . Furthermore, n ≤ k , c i k = 1 , ( e c c i − 1 ) m = 1 ( 1 ≤ i ≤ n ) , and f ( z ) = p 1 e c 1 z + ⋯ + p n e c n z , where p i ( i = 1 , … , n ) are constants.

2 Some lemmas

In order to prove our results, we need the following lemmas.

Lemma 1

[2] Let f ( z ) be a meromorphic function. If f ( z ) ≠ 0 , ∞ , then there exists an entire function α ( z ) such that f ( z ) = e α ( z ) .

Lemma 2

[12] Let f ( z ) and g ( z ) be two nonconstant entire functions, and let a be a nonzero constant. If f ( z ) and g ( z ) share a CM almost and

N r , 1 f = S ( r , f ) , N r , 1 g = S ( r , g ) ,

then either f ( z ) ≡ g ( z ) or f ( z ) g ( z ) ≡ a 2 .

Lemma 3

[13] Let n ≥ 3 be a positive integer, and let f j ( z ) ( j = 1 , … , n ) be meromorphic functions which are not constants except for f n ( z ) . Furthermore, let ∑ j = 1 n f j ( z ) ≡ 1 . If f n ( z ) ≢ 0 and

∑ j = 1 n N r , 1 f j + ( n − 1 ) ∑ j = 1 n N ¯ ( r , f j ) < ( λ + o ( 1 ) ) T ( r , f k ) ,

where I is a set of r ∈ ( 0 , ∞ ) with infinite linear measure, r ∈ I , k = 1 , 2 , … , n − 1 , λ < 1 , then f n ( z ) ≡ 1 .

Lemma 4

[14] Let f be a nonconstant entire function of finite order, let k be a positive integer, and let a be a nonzero constant. If f and f ( k ) share a CM, then

f ( k ) − a f − a = C

for some nonzero constant C.

Lemma 5

[15] Let α be a meromorphic function, let n be a positive integer, and let c be a nonzero finite complex number. If Δ c n α ≡ 0 , then either ρ ( α ) ≥ 1 or α is a polynomial with deg α ≤ n − 1 .

3 Proof of Theorem 1

Since f ( z ) and g ( z ) share a and b CM, then by Lemma 1, we obtain

(3.1) f ( z ) − a g ( z ) − a = e α ( z ) , f ( z ) − b g ( z ) − b = e β ( z ) ,

where α ( z ) and β ( z ) are two entire functions.

By Nevanlinna’s first and second fundamental theorems, we have

(3.2) T ( r , f ) ≤ N ¯ ( r , f ) + N ¯ r , 1 f − a + N ¯ r , 1 f − b + S ( r , f ) ≤ N ¯ r , 1 g − a + N ¯ r , 1 g − b + S ( r , f ) ≤ 2 T ( r , g ) + S ( r , f ) .

Similarly, we obtain

(3.3) T ( r , g ) ≤ 2 T ( r , f ) + S ( r , g ) .

By equations (3.2) and (3.3) and the definition of order, we deduce that

ρ ( f ) = ρ ( g ) ≤ 1 .

From equation (3.1) and above, we have

(3.4) f ( z ) − a = x 1 e y 1 z ( g ( z ) − a ) ,

(3.5) f ( z ) − b = x 2 e y 2 z ( g ( z ) − b ) ,

where x 1 ( ≠ 0 ) , x 2 ( ≠ 0 ) , y 1 , and y 2 are constants.

Now, we consider two cases.

Case 1. n = 1 .

We know that

(3.6) g ( z ) = p 1 ( z ) e c 1 z .

Next, we consider three subcases.

Case 1.1. a ≠ 0 and b = 0 .

Obviously N r , 1 g = O ( log r ) . Since f and g share 0 CM, we have N r , 1 f = O ( log r ) . By Lemma 2, we deduce that either f ≡ g or f g ≡ a 2 .

Case 1.2. a = 0 and b ≠ 0 .

Using the same argument as used in Case 1.1, we obtain that either f ≡ g or f g ≡ b 2 .

Case 1.3. a ≠ 0 and b ≠ 0 .

Next, we consider three subcases.

Case 1.3.1. y 1 = 0 .

By equation (3.4), we have

(3.7) f ( z ) − a = x 1 ( g ( z ) − a ) .

We claim that there exists z 0 such that g ( z 0 ) = b . Otherwise, g ( z ) ≠ b . Since b ≠ 0 , then by Nevanlinna’s second fundamental theorem, we obtain

T ( r , g ) ≤ N ¯ ( r , g ) + N ¯ r , 1 g + N ¯ r , 1 g − b + S ( r , g ) ≤ S ( r , g ) ,

a contradiction. Hence, there must exist z 0 such that g ( z 0 ) = b , then by f and g share b CM, we have f ( z 0 ) = g ( z 0 ) = b . Thus, by equation (3.7), we obtain x 1 = 1 , which yields f ≡ g .

Case 1.3.2. y 2 = 0 .

Using the same argument as used in Case 1.3.1, we deduce f ≡ g .

Case 1.3.3. y 1 ≠ 0 and y 2 ≠ 0 .

By equations (3.4)–(3.6), we have

(3.8) x 1 p 1 ( z ) e ( y 1 + c 1 ) z − x 2 p 1 ( z ) e ( y 2 + c 1 ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

In the following, we consider two subcases.

Case 1.3.3.1. y 1 = y 2 .

It follows from equation (3.8) that

(3.9) ( x 1 − x 2 ) p 1 ( z ) e ( y 1 + c 1 ) z + ( b x 2 − a x 1 ) e y 1 z = b − a .

Obviously, there does not exist z 0 such that ( x 1 − x 2 ) p 1 ( z 0 ) = 0 when ( b x 2 − a x 1 ) = 0 .

If ( x 1 − x 2 ) p 1 ( z ) ≡ 0 , then x 1 = x 2 . Thus, by equation (3.4) and (3.5), we deduce f ≡ g .

If b x 2 − a x 1 = 0 , then by a ≠ b , we have x 1 ≠ x 2 . From equation (3.9), we obtain

(3.10) y 1 + c 1 = 0 .

(3.11) ( x 1 − x 2 ) p 1 ( z ) = b − a .

By equation (3.11), we obtain that p 1 ( z ) is a constant. Then from equations (3.4), (3.6), and (3.10), we have

(3.12) f − a = x 1 p 1 ( z ) − a x 1 e − c 1 z .

It follows from equations (3.6), (3.11), and (3.12) and b x 2 − a x 1 = 0 that

f g = a g + x 1 p 1 ( z ) g − a x 1 e − c 1 z g = a + x 1 ( b − a ) x 1 − x 2 g − a x 1 ( b − a ) x 1 − x 2 = ( a + b ) g − a b .

If ( x 1 − x 2 ) p 1 ( z ) ≢ 0 and b x 2 − a x 1 ≠ 0 , then by Nevanlinna’s second fundamental theorem, we obtain a contradiction.

Case 1.3.3.2. y 1 ≠ y 2 .

Next, we consider three subcases.

Case 1.3.3.2.1. y 1 + c 1 = 0 .

Since y 1 ≠ y 2 , we have y 2 + c 1 ≠ 0 . From equation (3.8), we obtain

(3.13) − x 2 p 1 ( z ) e ( y 2 + c 1 ) z + b x 2 e y 2 z − a x 1 e − c 1 z = b − a − x 1 p 1 ( z ) .

If y 2 + c 1 ≠ − c 1 , we consider two subcases.

  1. b − a − x 1 p 1 ( z ) ≢ 0 . By Lemma 3, we obtain a contradiction.

  2. b − a − x 1 p 1 ( z ) ≡ 0 . It follows from equation (3.13) that

    − x 2 p 1 ( z ) e ( y 2 + 2 c 1 ) z + b x 2 e ( y 2 + c 1 ) z = a x 1 .

By Nevanlinna’s second fundamental theorem, we obtain a contradiction.

If y 2 + c 1 = − c 1 , then by equation (3.13), we have

(3.14) − ( x 2 p 1 ( z ) + a x 1 ) e − c 1 z + b x 2 e y 2 z = b − a − x 1 p 1 ( z ) .

When x 2 p 1 ( z ) + a x 1 ≡ 0 , we obtain a contradiction.

When x 2 p 1 ( z ) + a x 1 ≢ 0 , we consider two subcases:

  1. b − a − x 1 p 1 ( z ) ≢ 0 . By Nevanlinna’s second fundamental theorem, we obtain a contradiction.

  2. b − a − x 1 p 1 ( z ) ≡ 0 . It follows from equation (3.14) that b x 2 e ( y 2 + c 1 ) z = x 2 p 1 ( z ) + a x 1 , a contradiction.

Case 1.3.3.2.2. y 2 + c 1 = 0 .

Using the same argument as used in Case 1.3.3.2.1, we obtain a contradiction.

Case 1.3.3.2.3. y 1 + c 1 ≠ 0 and y 2 + c 1 ≠ 0 .

If y 1 + c 1 = y 2 , then by equation (3.8) we have

( x 1 p 1 ( z ) + b x 2 ) e y 2 z − x 2 p 1 ( z ) e ( y 2 + c 1 ) z − a x 1 e y 1 z = b − a .

When x 1 p 1 ( z ) + b x 2 ≡ 0 , by Nevanlinna’s second fundamental theorem, we obtain a contradiction, and when x 1 p 1 ( z ) + b x 2 ≢ 0 , by Lemma 3, we obtain a contradiction.

If y 2 + c 1 = y 1 , similarly, we can again deduce a contradiction as above.

If y 1 + c 1 ≠ y 2 and y 2 + c 1 ≠ y 1 , then by Lemma 3, we obtain a contradiction.

Case 2. n ≥ 2 .

Next, we consider three subcases.

Case 2.1. y 1 = 0 .

By equation (3.4), we have

(3.15) f ( z ) − a = x 1 ( g ( z ) − a ) .

We claim that there exists z 0 such that g ( z 0 ) = b . Otherwise, g ( z ) ≠ b .

Case 2.1.1. b ≠ 0 .

Using the same argument as used in Case 1.3.1, we obtain a contradiction.

Case 2.1.2. b = 0 .

Then, we have

(3.16) ∑ i = 1 n p i ( z ) e c i z ≠ 0 , ∞ .

It follows from equation (3.16), Lemma 1, and ρ ( g ) ≤ 1 that

(3.17) ∑ i = 1 n p i ( z ) e c i z = c e d z ,

where c ( ≠ 0 ) and d are two constants.

Case 2.1.2.1. For any i ( 1 ≤ i ≤ n ) such that c i ≠ d .

By Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

Case 2.1.2.2. There exists k ( 1 ≤ k ≤ n ) such that c k = d .

It follows from equation (3.17) that

(3.18) ∑ i ≠ k p i ( z ) e ( c i − d ) z = c − p k ( z ) .

If c − p k ( z ) ≢ 0 , then by Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

If c − p k ( z ) ≡ 0 , then by (3.18) we have ∑ i ≠ k p i ( z ) e c i z ≡ 0 , a contradiction.

Hence, there must exist z 0 such that g ( z 0 ) = b , then by f and g share b CM, we have f ( z 0 ) = g ( z 0 ) = b . Thus, by equation (3.15), we obtain x 1 = 1 , which yields f ≡ g .

Case 2.2. y 2 = 0 .

Using the same argument as above, we obtain f ≡ g .

Case 2.3. y 1 ≠ 0 and y 2 ≠ 0 .

It follows from equations (3.4) and (3.5) that

(3.19) x 1 ∑ i = 1 n p i ( z ) e ( y 1 + c i ) z − x 2 ∑ i = 1 n p i ( z ) e ( y 2 + c i ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

In the following, we consider two subcases.

Case 2.3.1. y 1 = y 2 .

If x 1 = x 2 , then by equations (3.4) and (3.5), we obtain

f − a g − a = f − b g − b .

Hence, f ≡ g .

If x 1 ≠ x 2 , then by equation (3.19), we have

(3.20) ( x 1 − x 2 ) ∑ i = 1 n p i ( z ) e ( y 1 + c i ) z + ( b x 2 − a x 1 ) e y 1 z = b − a .

Case 2.3.1.1. For any i ( 1 ≤ i ≤ n ) such that y 1 + c i ≠ 0 .

By Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

Case 2.3.1.2. There exists k ( 1 ≤ k ≤ n ) such that y 1 + c k = 0 .

Thus, by equation (3.20), we have

(3.21) ( x 1 − x 2 ) ∑ i ≠ k p i ( z ) e ( c i − c k ) z + ( b x 2 − a x 1 ) e − c k z = b − a + ( x 2 − x 1 ) p k ( z ) .

If b − a + ( x 2 − x 1 ) p k ( z ) ≢ 0 , then by Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

If b − a + ( x 2 − x 1 ) p k ( z ) ≡ 0 , then it follows from equation (3.21) that

(3.22) ( x 1 − x 2 ) ∑ i ≠ k p i ( z ) e c i z = a x 1 − b x 2 .

Since c i ≠ 0 ( 1 ≤ i ≤ n ) , then by Nevanlinna’s second fundamental theorem, Lemma 3, and equation (3.22), we obtain a contradiction.

Case 2.3.2. y 1 ≠ y 2 .

In this case, we obtain a contradiction by mathematical induction.

(a) n = 2 . From equation (3.19), we have

(3.23) x 1 p 1 ( z ) e ( y 1 + c 1 ) z + x 1 p 2 ( z ) e ( y 1 + c 2 ) z − x 2 p 1 ( z ) e ( y 2 + c 1 ) z − x 2 p 2 ( z ) e ( y 2 + c 2 ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

For equation (3.23), we consider three cases.

Case a.1. y 1 + c 1 ≠ 0 and y 1 + c 2 ≠ 0 .

Case a.1.1. y 2 + c 1 ≠ 0 and y 2 + c 2 ≠ 0 .

In the following, we claim that y 1 + c 1 , y 1 + c 2 , y 2 + c 1 , y 2 + c 2 , y 2 , y 1 are distinct. Otherwise, without loss of generality, we assume that y 1 + c 1 = y 2 + c 2 .

By equation (3.23), we have

(3.24) ( x 1 p 1 ( z ) − x 2 p 2 ( z ) ) e ( y 1 + c 1 ) z + x 1 p 2 ( z ) e ( y 1 + c 2 ) z − x 2 p 1 ( z ) e ( y 2 + c 1 ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

Case a.1.1.1. y 1 + c 2 = y 2 .

We deduce that c 1 = 2 c 2 and y 2 + c 1 ≠ y 1 . By equation (3.24), we obtain

( x 1 p 1 ( z ) − x 2 p 2 ( z ) ) e ( y 1 + c 1 ) z + ( x 1 p 2 ( z ) + b x 2 ) e y 2 z − x 2 p 1 ( z ) e ( y 2 + c 1 ) z − a x 1 e y 1 z = b − a .

If x 1 p 1 ( z ) − x 2 p 2 ( z ) ≡ 0 and x 1 p 2 ( z ) + b x 2 ≡ 0 , then by Nevanlinna’s second fundamental theorem, we obtain a contradiction.

If either x 1 p 1 ( z ) − x 2 p 2 ( z ) ≡ 0 or x 1 p 2 ( z ) + b x 2 ≡ 0 , then by the Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

If x 1 p 1 ( z ) − x 2 p 2 ( z ) ≢ 0 and x 1 p 2 ( z ) + b x 2 ≢ 0 , then by Lemma 3, we obtain a contradiction.

Case a.1.1.2. y 2 + c 1 = y 1 .

We deduce c 2 = 2 c 1 and y 1 + c 2 ≠ y 2 . Using the same argument as above, we obtain a contradiction.

Case a.1.1.3. y 1 + c 2 ≠ y 2 and y 2 + c 1 ≠ y 1 .

Whether x 1 p 1 ( z ) − x 2 p 2 ( z ) is zero or not, from equation (3.24) and Lemma 3, we obtain a contradiction. Therefore, y 1 + c 1 , y 1 + c 2 , y 2 + c 1 , y 2 + c 2 , y 2 , y 1 are nonzero and distinct in equation (3.23), and then by Lemma 3, we obtain a contradiction.

Case a.1.2. y 2 + c 1 = 0 .

It follows from equation (3.23) that

(3.25) x 1 p 1 ( z ) e ( y 1 + c 1 ) z + x 1 p 2 ( z ) e ( y 1 + c 2 ) z − x 2 p 2 ( z ) e ( c 2 − c 1 ) z + b x 2 e − c 1 z − a x 1 e y 1 z = b − a + x 2 p 1 ( z ) .

Using the same argument as used in Case a.1.1, we know that y 1 + c 1 , y 1 + c 2 , c 2 − c 1 , − c 1 , y 1 are nonzero and distinct.

If b − a + x 2 p 1 ( z ) ≢ 0 , then by Lemma 3, we obtain a contradiction.

If b − a + x 2 p 1 ( z ) ≡ 0 , then it follows from equation (3.25) that

(3.26) x 1 p 1 ( z ) e ( y 1 + 2 c 1 ) z + x 1 p 2 ( z ) e ( y 1 + c 1 + c 2 ) z − x 2 p 2 ( z ) e c 2 z − a x 1 e ( y 1 + c 1 ) z = − b x 2 .

Since y 1 + c 1 , y 1 + c 2 , c 2 − c 1 , y 1 are distinct, then we have y 1 + 2 c 1 , y 1 + c 1 + c 2 , c 2 , y 1 + c 1 are distinct.

(i) y 1 + 2 c 1 ≠ 0 and y 1 + c 1 + c 2 ≠ 0 .

When b ≠ 0 , by Lemma 3, we obtain a contradiction. When b = 0 , we know a ≠ 0 , then by equation (3.26), we have x 1 p 1 ( z ) e c 1 z + x 1 p 2 ( z ) e c 2 z − x 2 p 2 ( z ) e ( c 2 − ( y 1 + c 1 ) ) z = a x 1 , a contradiction.

(ii) y 1 + 2 c 1 = 0 .

Then, by equation (3.26), we have

(3.27) x 1 p 2 ( z ) e ( c 2 − c 1 ) z − x 2 p 2 ( z ) e c 2 z − a x 1 e − c 1 z = − b x 2 − x 1 p 1 ( z ) .

When − b x 2 − x 1 p 1 ( z ) ≢ 0 , by Lemma 3, we obtain a contradiction. When − b x 2 − x 1 p 1 ( z ) ≡ 0 , from equation (3.27), we deduce that x 1 p 2 ( z ) e c 2 z − x 2 p 2 ( z ) e ( c 2 + c 1 ) z = a x 1 , a contradiction.

(iii) y 1 + c 1 + c 2 = 0 .

Using the same argument as above, we obtain a contradiction.

Case a.1.3. y 2 + c 2 = 0 .

Using the same argument as used in Case a.1.2, we obtain a contradiction.

Case a.2. y 1 + c 1 = 0 .

It follows from y 1 ≠ y 2 that y 1 + c 2 ≠ 0 and y 2 + c 1 ≠ 0 . Thus, we consider two subcases.

Case a.2.1. y 2 + c 2 ≠ 0 .

Using the same argument as used in Case a.1.2, we obtain a contradiction.

Case a.2.2. y 2 + c 2 = 0 .

From equation (3.23), we have

(3.28) x 1 p 2 ( z ) e ( c 2 − c 1 ) z − x 2 p 1 ( z ) e ( c 1 − c 2 ) z + b x 2 e − c 2 z − a x 1 e − c 1 z = b − a − x 1 p 1 ( z ) + x 2 p 2 ( z ) .

Using the same argument as used in Case a.1.1, we deduce that c 2 − c 1 , c 1 − c 2 , − c 1 , and − c 2 are nonzero and distinct.

If b − a − x 1 p 1 ( z ) + x 2 p 2 ( z ) ≢ 0 , then by Lemma 3, we obtain a contradiction.

If b − a − x 1 p 1 ( z ) + x 2 p 2 ( z ) ≡ 0 , then by a and b are distinct, without loss of generality, we assume that a ≠ 0 , and then, by equation (3.28) we have

(3.29) x 1 p 1 ( z ) e c 2 z − x 2 p 1 ( z ) e ( 2 c 1 − c 2 ) z + b x 2 e ( c 1 − c 2 ) z = a x 1 .

Obviously c 2 , 2 c 1 − c 2 , and c 1 − c 2 are distinct. If 2 c 1 − c 2 ≠ 0 , by Lemma 3, we obtain a contradiction. If 2 c 1 − c 2 = 0 , from equation (3.29), we have x 1 p 1 ( z ) e c 2 z + b x 2 e ( c 1 − c 2 ) z = a x 1 + x 2 p 1 ( z ) , a contradiction.

Case a.3. y 1 + c 2 = 0 .

Using the same argument as used in Case a.2, we obtain a contradiction.

Therefore, from equation (3.19), we obtain a contradiction when n = 2 . Suppose that the fact is valid when n ≤ k . When n = k , from equation (3.19), we have

(3.30) x 1 ∑ i = 1 k p i ( z ) e ( y 1 + c i ) z − x 2 ∑ i = 1 k p i ( z ) e ( y 2 + c i ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

Thus, from equation (3.30), we obtain a contradiction. Now we consider n = k + 1 .

(b) n = k + 1 . By equation (3.19), we have

(3.31) x 1 ∑ i = 1 k + 1 p i ( z ) e ( y 1 + c i ) z − x 2 ∑ i = 1 k + 1 p i ( z ) e ( y 2 + c i ) z + b x 2 e y 2 z − a x 1 e y 1 z = b − a .

Now, we consider two subcases.

Case b.1. For any i ( 1 ≤ i ≤ k + 1 ) , y 1 + c i ≠ 0 .

Case b.1.1. For any i ( 1 ≤ i ≤ k + 1 ) , y 2 + c i ≠ 0 .

By Lemma 3 and mathematical induction, we obtain a contradiction.

Case b.1.2. There exists q ( 1 ≤ q ≤ k + 1 ) , y 2 + c q = 0 .

It follows from equation (3.31) that

(3.32) x 1 ∑ i = 1 k + 1 p i ( z ) e ( y 1 + c i ) z − x 2 ∑ i ≠ q p i ( z ) e ( c i − c q ) z + b x 2 e − c q z − a x 1 e y 1 z = b − a + x 2 p q ( z ) .

Case b.1.2.1. b − a + x 2 p q ( z ) ≢ 0 .

By Lemma 3 and mathematical induction, we obtain a contradiction.

Case b.1.2.2. b − a + x 2 p q ( z ) ≡ 0 .

It follows from equation (3.32) that

(3.33) x 1 ∑ i = 1 k + 1 p i ( z ) e ( y 1 + c i + c q ) z − x 2 ∑ i ≠ q p i ( z ) e c i z − a x 1 e ( y 1 + c q ) z = − b x 2 .

If b ≠ 0 , then by mathematical induction, we obtain a contradiction.

If b = 0 , we obtain a ≠ 0 , then by equation (3.33), we have

x 1 ∑ i = 1 k + 1 p i ( z ) e c i z − x 2 ∑ i ≠ q p i ( z ) e ( c i − ( y 1 + c q ) ) z = a x 1 .

Similarly, we obtain a contradiction.

Case b.2. There exists h ( 1 ≤ h ≤ k + 1 ) such that y 1 + c h = 0 .

Case b.2.1. For any i ( 1 ≤ i ≤ k + 1 ) , y 2 + c i ≠ 0 .

Using the same argument as used in Case b.1.2, we obtain a contradiction.

Case b.2.2. There exists r ( 1 ≤ r ≤ k + 1 ) such that y 2 + c r = 0 .

From equation (3.31), we have

(3.34) x 1 ∑ i ≠ h p i ( z ) e ( c i − c h ) z − x 2 ∑ i ≠ r p i ( z ) e ( c i − c r ) z + b x 2 e − c r z − a x 1 e − c h z = b − a − x 1 p h ( z ) + x 2 p r ( z ) .

Case b.2.2.1. b − a − x 1 p h ( z ) + x 2 p r ( z ) ≢ 0 .

By mathematical induction, we obtain a contradiction.

Case b.2.2.2. b − a − x 1 p h ( z ) + x 2 p r ( z ) ≡ 0 .

Since a and b are distinct, without loss of generality, we assume that a ≠ 0 , then by equation (3.34), we have

x 1 ∑ i ≠ h p i ( z ) e c i z − x 2 ∑ i ≠ r p i ( z ) e ( c i + c h − c r ) z + b x 2 e ( c h − c r ) z = a x 1 .

By mathematical induction, we obtain a contradiction.

Hence, by equation (3.19), we obtain a contradiction.

This completes the proof of Theorem 1.

4 Proof of Theorem 2

Obviously ρ ( f ( k ) ) = ρ ( f ) ≤ 1 . It follows from f that

(4.1) f ( k ) = ∑ i = 1 n Q i ( z ) e c i z ,

where Q i ( z ) = p i ( k ) ( z ) + ( k − 1 ) c i p i ( k − 1 ) ( z ) + ⋯ + ( k − 1 ) c i k − 1 p i ′ ( z ) + c i k p i ( z ) ( 1 ≤ i ≤ n ) .

By Lemma 4 and equation (4.1), we have

(4.2) ∑ i = 1 n ( Q i ( z ) − c p i ( z ) ) e c i z = a ( 1 − c ) .

Case 1. c = 1 .

Obviously, we have f ≡ f ( k ) . It follows from equation (4.2) that

(4.3) p i ( k ) ( z ) + ( k − 1 ) c i p i ( k − 1 ) ( z ) + ⋯ + ( k − 1 ) c i k − 1 p i ′ ( z ) + ( c i k − 1 ) p i ( z ) ≡ 0 .

We claim that p i ( z ) is a constant. Otherwise, suppose that p i ( z ) = a m z m + a m − 1 z m − 1 + ⋯ + a 1 z + a 0 , where m is a positive integer and a m is a nonzero constant. By equation (4.3), we deduce

( c i k − 1 ) a m = 0 ; ( c i k − 1 ) a m − 1 + ( k − 1 ) c i k − 1 m a m = 0 ,

a contradiction. Hence, we know that p i ( z ) is a constant, then by equation (4.3), we have c i k = 1 .

Case 2. c ≠ 1 .

It follows from equation (4.2) that

∑ i = 1 n Q i ( z ) − c p i ( z ) 1 − c e c i z = a .

By Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction.

Thus, Theorem 2 is proved.

5 Proof of Theorem 3

Since f and Δ c m f share a CM, then by Lemma 1, we have

(5.1) Δ c m f − a f − a = e α ( z ) ,

where α ( z ) is an entire function.

By the definition of order, we obtain ρ ( f ( z ) ) ≤ 1 . Since T ( r , Δ c m f ) = O ( T ( r , f ) ) + S ( r , f ) , we obtain ρ ( Δ c m f ) ≤ 1 . Hence, we deduce ρ ( e α ( z ) ) ≤ 1 .

Thus, we have

(5.2) e α ( z ) = x e y z ,

where x ( ≠ 0 ) and y are two constants.

It follows from equations (5.1) and (5.2) that

(5.3) Δ c m f − a = x e y z ( f − a ) .

Next, we consider two cases.

Case 1. y = 0 .

From equation (5.3), we have

(5.4) Δ c m f − x f = a ( 1 − x ) .

Case 1.1. x = 1 .

It follows from the definition of Δ c m f that

(5.5) Δ c m f = ∑ j = 0 m ( − 1 ) j C m j f ( z + ( m − j ) c ) = ∑ j = 0 m ( − 1 ) j C m j ∑ i = 1 n p i ( z + ( m − j ) c ) e c i ( z + ( m − j ) c ) = ∑ i = 1 n ∑ j = 0 m ( − 1 ) j C m j p i ( z + ( m − j ) c ) e ( m − j ) c c i e c i z = ∑ i = 1 n H i ( z ) e c i z ,

where H i ( z ) = ∑ j = 0 m ( − 1 ) j C m j p i ( z + ( m − j ) c ) e ( m − j ) c c i ( i = 1 , 2 , … , n ) .

By equation (5.4), we obtain Δ c m f ≡ f . It follows ∑ i = 1 n ( H i ( z ) − p i ( z ) ) e c i z ≡ 0 . Then, we obtain H i ( z ) ≡ p i ( z ) . Otherwise, by Nevanlinna’s second fundamental theorem and Lemma 3, we obtain a contradiction. Hence, we have

(5.6) ∑ j = 0 m ( − 1 ) j C m j p i ( z + ( m − j ) c ) e ( m − j ) c c i ≡ p i ( z ) , 1 ≤ i ≤ n .

From equation (5.6), we deduce

(5.7) ( e c c i − 1 ) m = 1 .

If m is an even number, then by equation (5.6), we have

∑ j = 0 m − 1 ( − 1 ) j C m j p i ( z + ( m − j ) c ) e ( m − j ) c c i + p i ( z ) ≡ p i ( z ) .

It follows that

(5.8) e m c c i ( p i ( z + m c ) − p i ( z + ( m − 1 ) c ) ) + ∑ j = 0 1 ( − 1 ) j C m j e ( m − j ) c c i ( p i ( z + ( m − 1 ) c ) − p i ( z + ( m − 2 ) c ) ) + ⋯ + ∑ j = 0 m − 1 ( − 1 ) j C m j e ( m − j ) c c i p i ( z + c ) ≡ 0 .

Since m is an even number, then by equation (5.7), we obtain

(5.9) ∑ j = 0 m − 1 ( − 1 ) j C m j e ( m − j ) c c i = 0 .

Set Q i ( z ) = p i ( z + 2 c ) − p i ( z + c ) . It follows from equations (5.8) and (5.9) that

(5.10) e m c c i Q i ( z + ( m − 2 ) c ) + ∑ j = 0 1 ( − 1 ) j C m j e ( m − j ) c c i Q i ( z + ( m − 3 ) c ) + ⋯ + ∑ j = 0 m − 2 ( − 1 ) j C m j e ( m − j ) c c i Q i ( z ) ≡ 0 .

We claim that Q i ( z ) ≡ 0 . Otherwise, by equation (5.10), we deduce

(5.11) e m c c i + ∑ j = 0 1 ( − 1 ) j C m j e ( m − j ) c c i + ⋯ + ∑ j = 0 m − 2 ( − 1 ) j C m j e ( m − j ) c c i = 0 .

By equation (5.11), we have

(5.12) ∑ j = 0 m − 2 ( m − 1 − j ) ( − 1 ) j C m j e ( m − j ) c c i = 0 .

It follows from equations (5.9) and (5.12) that

(5.13) ∑ j = 0 m − 2 j ( − 1 ) j C m j e ( m − j ) c c i − m ( m − 1 ) e c c i = 0 .

Since j m C m j = C m − 1 j − 1 , then by equation (5.13), we obtain

∑ j = 0 m − 2 ( − 1 ) j + 1 C m − 1 j e ( m − 1 − j ) c c i = 0 .

Hence, we have

(5.14) ( e c c i − 1 ) m − 1 = ∑ j = 0 m − 1 ( − 1 ) j C m − 1 j e ( m − 1 − j ) c c i = ∑ j = 0 m − 2 ( − 1 ) j C m − 1 j e ( m − 1 − j ) c c i − 1 = − 1 .

Then, by equations (5.7) and (5.14), we obtain e c c i = 0 , a contradiction. Therefore, Q i ( z ) ≡ 0 , then by Lemma 5, we deduce that p i ( z ) ( i = 1 , … , n ) are constants.

If m is an odd number, then by equation (5.6), we obtain

∑ j = 0 m − 1 ( − 1 ) j C m j p i ( z + ( m − j ) c ) e ( m − j ) c c i − p i ( z ) ≡ p i ( z ) .

It follows that

(5.15) e m c c i ( p i ( z + m c ) − p i ( z + ( m − 1 ) c ) ) + ( e m c c i − m e ( m − 1 ) c c i ) ( p i ( z + ( m − 1 ) c ) − p i ( z + ( m − 2 ) c ) ) + ⋯ + ∑ j = 0 m − 1 ( − 1 ) j C m j e ( m − j ) c c i ( p i ( z + c ) − p i ( z ) ) ≡ 0 .

Set H i ( z ) = p i ( z + c ) − p i ( z ) . By equation (5.15), we obtain

(5.16) e m c c i H i ( z + ( m − 1 ) c ) + ( e m c c i − m e ( m − 1 ) c c i ) H i ( z + ( m − 2 ) c ) + ⋯ + ∑ j = 0 m − 1 ( − 1 ) j C m j e ( m − j ) c c i H i ( z ) ≡ 0 .

We claim that H i ( z ) ≡ 0 . Otherwise, from equation (5.16), we deduce

(5.17) e m c c i + ( e m c c i − m e ( m − 1 ) c c i ) + ⋯ + ∑ j = 0 m − 1 ( − 1 ) j C m j e ( m − j ) c c i = 0 .

It follows from equation (5.17) that

(5.18) ∑ j = 0 m − 1 ( m − j ) ( − 1 ) j C m j e ( m − j ) c c i = 0 .

Since m is an odd number, then by equations (5.7) and (5.18), we have

(5.19) ∑ j = 0 m − 1 j ( − 1 ) j C m j e ( m − j ) c c i − 2 m = 0 .

By j m C m j = C m − 1 j − 1 and equation (5.19), we obtain

∑ j = 0 m − 2 ( − 1 ) j + 1 C m − 1 j e ( m − 1 − j ) c c i − 2 = 0 .

Hence, we have

(5.20) ( e c c i − 1 ) m − 1 = ∑ j = 0 m − 1 ( − 1 ) j C m − 1 j e ( m − 1 − j ) c c i = ∑ j = 0 m − 2 ( − 1 ) j C m − 1 j e ( m − 1 − j ) c c i + 1 = − 2 + 1 = − 1 .

Then, by equations (5.7) and (5.20), we obtain e c c i = 0 , a contradiction. Therefore, H i ( z ) ≡ 0 , then by Lemma 5, we deduce that p i ( z ) ( i = 1 , … , n ) are constants.

Case 1.2. x ≠ 1 .

By equation (5.4), we obtain

(5.21) Δ c m f − x f 1 − x = a .

It follows from equations (5.5) and (5.21) that

∑ i = 1 n H i ( z ) − x p i ( z ) 1 − x e c i z = a ,

a contradiction.

Case 2. y ≠ 0 .

By equations (5.3) and (5.5), we have

∑ i = 1 n H i ( z ) e c i z − ∑ i = 1 n x p i ( z ) e ( y + c i ) z − a x e y z = a .

Similar to Theorem 1, we obtain a contradiction.

Thus, Theorem 3 is proved.

Acknowledgments

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

  1. Funding information: This work was supported by the National Natural Science Foundation of China (Grant No. 12171127) and the Natural Science Foundation of Zhejiang Province (Grant No. LY21A010012).

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state that there is no conflict of interest.

  4. Ethical approval: The conducted research is not related to either human or animals use.

  5. Data availability statement: Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2023-07-18
Revised: 2023-12-13
Accepted: 2023-12-14
Published Online: 2023-12-31

© 2023 the author(s), published by De Gruyter

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

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  133. A quasi-boundary value regularization method for the spherically symmetric backward heat conduction problem
  134. The structure fault tolerance of burnt pancake networks
  135. Average value of the divisor class numbers of real cubic function fields
  136. Uniqueness of exponential polynomials
  137. An application of Hayashi's inequality in numerical integration
https://doi.org/10.1515/math-2023-0173
Keywords for this article
exponential polynomials; uniqueness; differences
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