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An application on the second-order generalized difference equations

Abstract

In this paper, we study the solutions of the second-order generalized difference equation having the form of

Δ 2 u(k)+f ( k , u ( k ) ) =0,k[a,),a>0,(0,),
(1)

where Δ u(k)=u(k+)u(k). Then we provide applications on 2 ( ) and c 0 ( ) .

AMS Subject Classification:39A12, 39A70, 47B39, 39B60.

1 Introduction and preliminaries

The basic theory of difference equations is based on the difference operator Δ defined as Δu(k)=u(k+1)u(k), kN={0,1,2,3,}, which allows the recursive computation of solutions. Later, the following definition was suggested for Δ by [13] and [4]:

Δ u(k)=u(k+)u(k),kR,R{0};
(2)

however, no significant progress took place on this line. Recently, equation (2) was reconsidered and its inverse was defined by Δ 1 , and many interesting results in applications such as in number theory as well as in fluid dynamics were obtained; see, for example, [5]. By extending the study for sequences of complex numbers and to be real, some new qualitative properties like rotatory, expanding, shrinking, spiral and weblike were studied for the solutions of difference equations involving Δ . The 2 and c 0 solutions of the second-order difference equation of (1) when =1 were discussed in [6] and further generalized in [7]. In this paper, we discuss some applications of Δ in the finite and infinite series of number theory.

In this section, we present some of the preliminary definitions and results which will be useful for future discussion. The following definitions were held in [5] and [8], respectively.

Definition 1.1 Let u(k), k[0,), be a real- or complex-valued function and (0,). Then the generalized difference operator Δ is defined as

Δ u(k)=u(k+)u(k).
(3)

Then the inverse of Δ denoted by Δ 1 is defined as follows: If

Δ v(k)=u(k),then v(k)= Δ 1 u(k)+ c j ,
(4)

where c j is a constant for all k N (j), j=k[ k ]. If lim k u(k)=0, then we can take c j =0. Further, the generalized polynomial factorial for >0 is defined as

k ( n ) =k(k)(k2) ( k ( n 1 ) ) .
(5)

The following lemmas were proved in [9] and [10], respectively.

Lemma 1.2 (Product formula)

Let u(k) and v(k), k[0,), be any two real-valued functions. Then

Δ { u ( k ) v ( k ) } = u ( k + ) Δ v ( k ) + v ( k ) Δ u ( k ) = v ( k + ) Δ u ( k ) + u ( k ) Δ v ( k ) .
(6)

Lemma 1.3 Let >0, nN(2), k(,) and k ( n ) 0. Then

Δ 1 1 k ( n ) = 1 ( n 1 ) ( k ) ( n 1 ) + c j .
(7)

Definition 1.4 A function u(k), k[a,), is said to be in the 2 ( ) -space if

γ = 0 |u(a+j+γ) | 2 <for all j[0,).
(8)

If lim r |u(a+j+r)|=0 for all j[0,), then u(k) is said to be in the c 0 ( ) -space.

In what follows, we have the summation formula for finite and infinite series.

Lemma 1.5 If a real-valued function u(k) is defined for all k[0,), then

Δ 1 u(k)= r = 1 [ k ] u(kr)+ c j ,
(9)

where c j is a constant for all k N (j), j=k[ k ]. Since [0,)= 0 j < N (j), each complex number c j (0j<) is called an initial value of k N (j). Usually, each initial value c j is taken from any one of the values u(j), u(j+), u(j+2), etc. Further, if lim k u(k)=0 and >0, then

Δ 1 u(k)= r = 0 u(k+r).
(10)

Proof Assume z(k)= r = 0 u(k+r). Then

Δ z(k)= r = 0 u(k++r) r = 0 u(k+r)=u(k).

Now, the proof follows from lim k u(k)=0 and Definition 1.1. □

The next lemma is an expansion of Lemma 1.5 and its proof is straightforward.

Lemma 1.6 If lim k u(k)=0 and >0, then

Δ 2 u(k)= r 1 = 0 r 2 = 0 u(k+ r 1 + r 2 ).
(11)

Theorem 1.7 Let nN(2), k(0,) such that k ( n ) 0. Then

r = 0 1 ( k + r ) ( n ) = 1 ( n 1 ) ( k ) ( n 1 ) .
(12)

Proof The proof follows from Lemma 1.3 and Lemma 1.5 by taking u(k)= 1 k ( n ) and c j =0. □

Corollary 1.8 Let k(,) and (0,). Then

r = 0 1 ( k + r ) ( k + r ) = 1 ( k ) .
(13)

Proof Since Δ 1 1 k ( k ) = 1 ( k ) , the proof follows from Theorem 1.7 by taking n=2. □

2 Applications of Δ in number theory

In this section, we present some formulae and examples to find the values of finite and infinite series in number theory as an application of Δ . The following theorem and example were given in [7]. In fact, the example is to illustrate Theorem 2.1.

Theorem 2.1 Let k[,) and (0,). Then

r = 1 [ k ] + s ( k r + 2 ) 2 3 2 r ( k r + 4 ) ( 2 ) ( k r + ) ( k r + ) = c j k 1 ( k + 3 ) k ( k ) ,
(14)

where s=1 for k N (), s=0 for k N () and each c j is a constant for all k N (j), j=k[ k ]. In particular c j is obtained from (14) by substituting k=+j.

Example 2.2 By taking =1.7, k=2 and j=0.3 in (14), we get c j = 85 81 and hence (14) becomes

r = 1 [ k ] ( k 1.7 r + 2 ( 1.7 ) ) 2 3 ( 1.7 ) 2 1.7 r ( k 1.7 r + 4 ( 1.7 ) ) 1.7 ( 2 ) ( k 1.7 r + 1.7 ) 1.7 ( k 1.7 r + 1.7 1.7 ) = 85 81 ( 1.7 ) k 1.7 1 ( k + 3 ( 1.7 ) ) k 1.7 ( k 1.7 ) , k = 2 , 3.7 , 5.4 , .

Theorem 2.3 Let k[,) and (0,). Then

r = 0 k + r + 3 k + r ( 2 ( k + r ) + ) 2 ( 2 ) = 1 4 ( 3 ) k 1 ( 2 k ) .
(15)

Proof

By Definition 1.1, we find

Δ 1 ( ( k + ) 3 k ( 2 k + ) 2 ( 2 ) k ) = 1 4 ( 3 ) k 1 ( 2 k )

and (15) follows by Lemma 1.5 and c j =0 as k. □

The following is the illustration for Theorem 2.3.

Example 2.4 Taking =0.2 in (15), we arrive at

k + 0.2 3 k 0.2 ( 2 k + 0.2 ) 0.4 ( 2 ) + k + 0.4 3 k + 0.2 0.2 ( 2 k + 0.6 ) 0.4 ( 2 ) + ( k + 0.6 ) 3 k + 0.4 0.2 ( 2 k + 1 ) 0.4 ( 2 ) += 1 4 ( 3 ) k 0.2 1 ( 2 k 0.2 )

and one can take any value k[,).

Theorem 2.5 For k[,) and (0,), then

r = 0 ( k + r ) 3 3 r ( ( k + r ) 2 2 2 ) ( 2 ) ( k + r + ) ( k + r + ) = 1 ( ( k ) 2 2 2 ) k ( k ) .
(16)

Proof

By Definition 1.1, we find

Δ 1 ( k 3 3 ) k ( k 2 2 2 ) ( 2 ) ( k + ) ( k + ) = k ( ( k ) 2 2 2 ) k ( k )

and (16) follows by (10) and c j =0 as k. □

The following theorem generates the formula to find the sum of first partial sums of an infinite series.

Theorem 2.6 For the positive integer nN(3), k[2,) and (0,),

r 2 = 0 r 1 = 0 1 ( k + r 2 + r 1 ) ( n ) = 1 ( n 1 ) ( n 2 ) 2 ( k 2 ) ( n 2 ) .
(17)

Proof Using Definition 1.1 and operating Δ 1 on (7), we find

Δ 2 1 k ( n ) = 1 ( n 1 ) ( n 2 ) 2 ( k 2 ) ( n 2 )

and hence (17) follows by Lemma 1.6 as c j =0 when k. □

The following example illustrates Theorem 2.6.

Example 2.7 Substituting =0.5, n=4 in (17), we obtain

1 ( k ) 0.5 ( 4 ) + 2 ( k + 0.5 ) 0.5 ( 4 ) + 3 ( k + 1 ) 0.5 ( 4 ) += 1 6 ( 0.5 ) 2 ( k 1 ) 0.5 ( 2 ) .

In particular, when k=2, the above series becomes

1 2 × 1.5 × 1 × 0.5 + 2 2.5 × 2 × 1.5 × 1 + 3 3 × 2.5 × 2 × 1.5 += 1 6 × 0.5 3 .

Similarly, one can take any value for k[2,) and (0,).

The following example shows that 1 k ( n ) c 0 ( ) and 2 ( ) when k ( n ) 0.

Example 2.8 Let nN(2), (0,), j[0,) and an. Replacing k by a+j, in (12), we get

r = 0 1 ( a + j + r ) ( n ) = 1 ( n 1 ) ( a + j ) ( n 1 ) .
(18)

Since

| 1 ( a + j + r ) ( n ) | 2 < 1 ( a + j + r ) ( n ) ,

thus from (18) it follows that

r = 0 | 1 ( a + j + r ) ( n ) | 2 < r = 0 1 ( a + j + r ) ( n ) = 1 ( n 1 ) ( a + j ) ( n 1 ) <.

The function 1 k ( n ) 2 ( ) follows from Definition 1.4 by taking

u(a+j+r)= 1 ( a + j + r ) ( n ) .

Since lim r 1 ( a + j + r ) ( n ) =0 for all j[0,), Definition 1.4 yields 1 k ( n ) c 0 ( ) .

3 Concluding remarks

In the present work, we provide an application on 2 ( ) and c 0 ( ) and solutions of the second-order some generalized difference equation.

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Acknowledgements

The authors express their sincere gratitude to the referees for their valuable suggestions and comments. The second author also acknowledges that this project was partially supported by University Putra Malaysia under the ERGS Grant Scheme having project number 5527068.

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Correspondence to Adem Kılıçman.

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All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Manuel, M.M.S., Kılıçman, A., Xavier, G.B.A. et al. An application on the second-order generalized difference equations. Adv Differ Equ 2013, 35 (2013). https://doi.org/10.1186/1687-1847-2013-35

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Keywords

  • generalized difference equation
  • generalized difference operator
  • finite series and infinite series