On solutions of the difference equation x_{n+1}=(((-1)?x_{n-4})/(1+(-1)?x_{n}x_{n-1}x_{n-2}x_{n-3}x_{n-4}))

(1)

Selçuk J. Appl. Math. Selçuk Journal of Vol. 8. No.1. pp. 51- 56 , 2007 Applied Mathematics

On Solutions Of The Diﬀerence Equation+1= (−1)



−4

1+(−1)_

−1−2−3−4

Ramazan Karatas

Selcuk University Education Faculty 42090 Meram Konya Turkey e-mail:rkaratas@ selcuk.edu.tr

Received : December 27, 2006

Summary.We study the solutions and attractivity of the diﬀerence equation

+1= (−1) _ −4 1 + (−1)_ −1−2−3−4   = 0 1 2 

where ₋₁₁ ₋₁₀  ₋₂ ₋₁ 0 ∈ (0 ∞)and 0 are real numbers such that

0−1−2−3−46= ±1

Key words:Diﬀerence Equation, Solutions, Periodical. 1. Introduction

A lot of work has been done concerning the attractivity and solutions of the rational diﬀerence equations, for example in [1-7]. In [1] Aloqeili investigated the solutions, stability character, semi-cycle behavior of the diﬀerence equation +1=__−−1

−1 where −1 0∈    0 and gave the following formula

= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 0  2 Q =1 2−1₍₁_{−)−(1−}2−1_) −10 2₍₁_{−)−(1−}2_) −10    ₋₁ +1 2 Q =0 2−1(1−)−(1−2 )−10 2+1₍₁_{−)−(1−}2+1_) −10   

In [3], Cinar studied the positive solutions of the diﬀerence equation +1=_1+−1__

−1 for  = 0 1 2  and proved by the formula induction .

= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ₋₁ [(+1)2]−1_Y =0 (2−10+1) [(+1)2]−1_Y =0 ((2+1)−10+1) for n odd, 0 2 Y =1 ((2−1)−10+1) 2 Y =1 (2−10+1) for n is even. 

(2)

In [5], Stevic studied the stability properties of the solutions of Cinar’s equa-tion. Also in [6], Stevic investigated the solutions of the diﬀerence equation +1=_+−1_ and gave the formulas

2= 0 Ã 1 − 1  P =1 2_Y−1 =1 1 1+ ! , 2+1= −1 Ã 1 − 0 1+0  P =0 2 Y =1 1 1+ ! .

Moreover, in [2] Aloqeili generalized the results from [3,5] to the kth order case and investigated the solutions, stability character and semicycle behavior of the diﬀerence equation +1= _+−

− where −  0 0 and   0 

being any positive integer.

Our aim in this paper is to investigate the solutions of the diﬀerence equation

(1.1) +1= (−1) _ −4 1 + (−1)_ −1−2−3−4 for  = 0 1 2  where

₋₄ ₋₃ ₋₂ ₋₁ and 0 are real numbers such that 0−1−2−3−46= ±1

First, we give a few definitions which will be useful in our investigation of the behavior of solutions of Eq.(1.1).

Definition 1. Let I be an interval of real numbers and let  : 5 _{→  be a}

continuously diﬀerentiable function. Then for every _−_{∈  ( = 0 1 2 3) the} diﬀerence equation +1 =  ( −1 −2 −3 −4)  = 0 1 2  has a

unique solution {}∞=−4

Definition 2.The equilibrium point  of the equation +1=  ( −1  −),

 = 0 1 2  is the point that satisfies the condition  =  (  ) Definition 3._{A solution{}}∞_=_− of a diﬀerence equation

+1=  ( −1  −)

is periodic if there exists a positive integer p such that +=  The smallest

such positive integer p is called the prime period of the solution of the diﬀerence equation.

2.Main Results

Theorem 1. Assume that 0−1−2−3−4 6= ±1 and let {}∞=−4 be a

solution of Eq.(1.1). Then for  = 0 1 2  all solutions of Eq.(1.1) are of the form

(3)

(2.1) 5+1 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −4 1+0−1−2−3−4  ≡ 0 (mod 4) −−4  ≡ 1 (mod 4) −4 −1+0−1−2−3−4  ≡ 2 (mod 4) ₋₄ _{ ≡ 3 (mod 4)}  (2.2) 5+2 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −−3(1 + 0−1−2−3−4)  ≡ 0 (mod 4) −3(1+0−1−2−3−4) −1+0−1−2−3−4  ≡ 1 (mod 4) ₋₃(1 + 0−1−2−3−4)  ≡ 2 (mod 4) ₋₃ _{ ≡ 3 (mod 4)}  (2.3) 5+3= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ −−2 −1+0−1−2−3−4  ≡ 0 (mod 4) −−2  ≡ 1 (mod 4) −−2 1+0−1−2−3−4  ≡ 2 (mod 4) ₋₂ _{ ≡ 3 (mod 4)}  (2.4) 5+4= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ₋₁_{(−1 + }0−1−2−3−4)  ≡ 0 (mod 4) −1(−1+0−1−2−3−4) 1+0−1−2−3−4  ≡ 1 (mod 4) −−1(−1 + 0−1−2−3−4)  ≡ 2 (mod 4) ₋₁ _{ ≡ 3 (mod 4)}  (2.5) 5+5= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 0 1+0−1−2−3−4  ≡ 0 (mod 4) −0  ≡ 1 (mod 4) 0 −1+0−1−2−3−4  ≡ 2 (mod 4) 0  ≡ 3 (mod 4) 

Proof.For  = 0, we have following equality from Eq.(1.1) 1= _1+₀_ −4 −1−2−3−4 2= ₁_−₁_₀−_−3 −1−2−3 = −−3 1 1+0−1−2−3−4 = −−3 (1 + 0−1−2−3−4) 3= _1+₂_₁_−2₀_₋₁_₋₂ = ₁ −2 −−3(1+0−1−2−3−4)_{1+0−1−2−3−4}−4 0−1−2 = −2 1−0−1−2−3−4 4= ₁_−₃−_₂_−1₁_₀_₋₁ = −−1 1−1_{−0−1−2−3−4}−2 [−−3(1+0−1−2−3−4)]_{1+0−1−2−3−4}−4 0−1 = ₋₁_{(−1 + }0−1−2−3−4) and 5= _1+₄_₃0_₂_₁_₀

(4)

= 0

1+0−1−2−3−4

Similarly, one can easily prove Eqs.(2.1),(2.2),(2.3),(2.4) and (2.5) for  = 1 2 3 Now suppose that > 1 and our assumption holds for ( − 1) We shall show that the result holds for . From our assumption for ( − 1) we have

5−4= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −4 1+0−1−2−3−4  ≡ 1 (mod 4) −−4  ≡ 2 (mod 4) −4 −1+0−1−2−3−4  ≡ 3 (mod 4) ₋₄ _{ ≡ 0 (mod 4)}  5−3= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ −−3(1 + 0−1−2−3−4)  ≡ 1 (mod 4) −3(1+0−1−2−3−4) −1+0−1−2−3−4  ≡ 2 (mod 4) ₋₃(1 + 0−1−2−3−4)  ≡ 3 (mod 4) ₋₃ _{ ≡ 0 (mod 4)}  5−2= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ −−2 −1+0−1−2−3−4  ≡ 1 (mod 4) −−2  ≡ 2 (mod 4) −−2 1+0−1−2−3−4  ≡ 3 (mod 4) ₋₂ _{ ≡ 0 (mod 4)}  5−1= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ₋₁_{(−1 + }0−1−2−3−4)  ≡ 1 (mod 4) −1(−1+0−1−2−3−4) 1+0−1−2−3−4  ≡ 2 (mod 4) −−1(−1 + 0−1−2−3−4)  ≡ 3 (mod 4) ₋₁ _{ ≡ 0 (mod 4)}  5= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 0 1+0−1−2−3−4  ≡ 1 (mod 4) −0  ≡ 2 (mod 4) 0 −1+0−1−2−3−4  ≡ 3 (mod 4) 0  ≡ 0 (mod 4)

Then, from Eq.(1.1) and the above equality, for  ≡ 1 (mod 4) we have 5+1= (−1) 5 5−4 1+(−1)5_ 55−15−25−35−4 = −−4 1+0−1−2−3−4 1−_{1+0−1−2−3−4}0−1−2−3−4 = −−4 That is, 5+1= −−4 Also, 5+2= (−1) 5+1 5−3 1+(−1)5+1_ 5+155−15−25−3 = −−3(1+0−1−2−3−4) 1−0−1−2−3−4  Hence, we have 5+2= −−3 (1 + 0−1−2−3−4) 1 − 0−1−2−3−4 

(5)

Similarly, 5+3= (−1) 5+2_ 5−2 1+(−1)5+2_ 5+25+155−15−2 = −2 −1+0−1−2−3−4 1−_{−1+0−1−2−3−4}0−1−2−3−4 = −−2 Consequently, we have 5+3= −−2 Similarly, 5+4= (−1) 5+3_ 5−1 1+(−1)5+3_ 5+35+25+155−1 = −1(−1+0−1−2−3−4) 1+0−1−2−3−4  That is, 5+4= ₋₁_{(−1 + }0−1−2−3−4) 1 + 0−1−2−3−4

Now, we prove the last formula for  ≡ 1 (mod 4). Since 5+5= ₁_−_5+4__5+3−_5_5+2__5+1__5 = − 0 1+0−1−2−3−4 1−_{1+0−1−2−3−4}0−1−2−3−4 = −0 we have 5+5= −0

Thus, we have proved (2.1), (2.2), (2.3),(2,4) and (2.5) for  ≡ 1 (mod 4). Similarly, one can easily prove (2.1), (2.2), (2.3),(2,4) and (2.5) for  = 4 + 2  = 4 + 3 and  = 4, where  ∈ +_

Theorem 2. Eq.(1.1) has two equilibrium points which are 0 and √5

2. Proof.For the equilibrium points of Eq.(1.1) when  is even, we write

 = (1 + 5) Then, we have

6= 0 When  is odd, we write

(6)

5_{= 2}

Thus, the equilibrium points of Eq.(1.1) are 0 and √5

2

Corollary 1.Every solution of Eq.(1.1). is periodic with prime 20. Proof.This is clear from Eqs.(2.1), (2.2), (2.3),(2.4) and (2.5). References

1.Aloqeili, M., Dynamics of a rational diﬀerence equation, Applied Mathematics and Computation, 176(2006), 768-774.

2.Aloqeili, M., Dynamics of a kth order rational diﬀerence equation, Applied Mathe-matics and Computation, 181(2006), 1328-1335.

3.Cinar, C., On the positive solutions of the diﬀerence equation+1= _1+−1___−1,

Applied Mathematics and Computation 150(2004), 21-24.

4.Cinar, C., On the positive solutions of the diﬀerence equation+1= _−1+−1___−1,

Applied Mathematics and Computation, 158(2004), 813-816.

5.Stevic, S., More on a rational recurence relation+1=_1+−1_−1__Applied

Math-ematics E-Notes, 4(2004),80-84.

6.Stevic, S., On the recursive sequence+1=_(−1_)Taiwanese Journal of

Mathe-matics, 6(3)(2002), 405-414.

7.Stevic, S., On the recursive sequence+1=  + _−1

 Journal of Applied