How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine green and equation 1 is second pic Theorem 9.If k is even and 1,0 are odd positive integers,then Eq. (1) has prime period two solution if the condition(1– (C+D)) (3e– d) < (e+d) (A+B),(20)valid,provided(C+D)1andise (1 – (C+D)) – d (A+B) > 0.Proof.If k is even and 1, o are odd positive integers, thenXn = Xn-k and xn+1 = Xn–1 = Xn-o. It follows from Eq.(1)thatbQP=(A+B) Q+ (C+D) P –(21)(еР — d@)"andbPQ= (A+B) P+(C+D) Q –(22)(e Q– dP)'Consequently, we getер - dPQ —D е (А+ В) РQ— d (А+ В) + e(С+D) Р- (C+D) dPQ– bQ,(23)ande Q – dPQ = e (A+B) PQ– d (A+B) P + e(C+D) Q- (C+D) dPQ– bP.(24)By subtracting (24) from (23), we getP+Q=(25)[e (1– (C+D)) – d (A+B)]'-where e (1– (C+D)) – d (A+B) > 0. By adding (23)and (24), we obtaine b (1 – (C+ D))(e+d) [K1+(A+ B)][e K1 – d (A+B)]*PQ =(26) Thus, we deduce that(P+ Q)² > 4PQ.(28)The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-k[dxn-k - exp-1)Xn+1 =Axn+ Bxn-k+Cxn–1+Dxn-o +n = 0,1,2,..where the coefficients A, B, C, D, b, d, e e (0,00), whilek, 1 and o are positive integers. The initial conditionsX-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive realnumbers such that k <1< o. Note that the special casesof Eq.(1) have been studied in [1] when B= C = D=0,and k= 0,1= 1, b is replaced by – b and in [27] whenB= C= D=0, and k= 0, b is replaced by[33] when B = C = D = 0, 1 = 0 and in [32] whenA = C= D=0, 1= 0, b is replaced by – b.(1)b and in%3D

Given nonlinear difference equation is xn+1=Axn+Bxn-k+Cxn-σ+bxn-kdxn-k-exn-1 (1) Since…

Theorem 9.If k is even and 1,0 are odd positive integers, then Eq. (1) has prime period two solution if the condition (1– (C+D))(3e– d) < (e+d)(A+ B) , (20) valid, provided (C+D) and is 1 e (1 – (C+D)) – d (A+B) > 0. Proof.If k is even and 1, o are odd positive integers, then Xn = Xn-k and xn+1 = Xn–1= Xn-o. It follows from Eq.(1) that bQ P=(A+B) Q+(C+D) P – (21) (еР — dQ)" and bP Q= (A+B) P+(C+D) Q – (22) (e Q– dP)' Consequently, we get e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+D) P² - (C+D) dPQ– bQ, (23) and e Q – dPQ = e (A+B) PQ– d (A+ B) P +e(C+D) Q - (C+D) dPQ– bP. (24) By subtracting (24) from (23), we get b P+Q= (25) [e (1 – (C+D)) –d (A+B)]' where e (1– (C+D)) – d (A+B) > 0. By adding (23) and (24), we obtain e b² (1 – (C+ D)) (e+d) [K1 + (A+ B)][e K1 – d (A+B)]² PQ= (26)

Theorem 9.If k is even and 1,0 are odd positive integers, then Eq. (1) has prime period two solution if the condition (1– (C+D))(3e– d) < (e+d)(A+ B) , (20) valid, provided (C+D) and is 1 e (1 – (C+D)) – d (A+B) > 0. Proof.If k is even and 1, o are odd positive integers, then Xn = Xn-k and xn+1 = Xn–1= Xn-o. It follows from Eq.(1) that bQ P=(A+B) Q+(C+D) P – (21) (еР — dQ)" and bP Q= (A+B) P+(C+D) Q – (22) (e Q– dP)' Consequently, we get e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+D) P² - (C+D) dPQ– bQ, (23) and e Q – dPQ = e (A+B) PQ– d (A+ B) P +e(C+D) Q - (C+D) dPQ– bP. (24) By subtracting (24) from (23), we get b P+Q= (25) [e (1 – (C+D)) –d (A+B)]' where e (1– (C+D)) – d (A+B) > 0. By adding (23) and (24), we obtain e b² (1 – (C+ D)) (e+d) [K1 + (A+ B)][e K1 – d (A+B)]² PQ= (26)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section: Chapter Questions

Problem 63RE

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