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b P+Q= [d (1 — (В+ D)) — е (А+С)] where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]²" -

b P+Q= [d (1 — (В+ D)) — е (А+С)] where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]²" -

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.5: Other Types Of Equations
Problem 56E
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Thus, we deduce that (P+ Q)² > 4PQ. (28) difference equation bxn-k Xn+1 = Axn+ Bxn-k+Cxr-1+Dxn-o+ [dxn-k- exp-1] (1) n = 0, 1,2, ... where the coefficients A, B, C, D, b, d, e E (0,), while k,1 and o are positive integers. The initial conditions X-o,..., X_1,..., X_k, ….., X_1, Xo are arbitrary positive real numbers such that k<1< 0. Note that the special cases of Eq.(1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by B=C=D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1 = 0 and in [32] when A= C= D=0, 1=0, b is replaced by – b. ••.) – b and in [27] when 6.
Transcribed Image Text:Thus, we deduce that (P+ Q)² > 4PQ. (28) difference equation bxn-k Xn+1 = Axn+ Bxn-k+Cxr-1+Dxn-o+ [dxn-k- exp-1] (1) n = 0, 1,2, ... where the coefficients A, B, C, D, b, d, e E (0,), while k,1 and o are positive integers. The initial conditions X-o,..., X_1,..., X_k, ….., X_1, Xo are arbitrary positive real numbers such that k<1< 0. Note that the special cases of Eq.(1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by B=C=D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1 = 0 and in [32] when A= C= D=0, 1=0, b is replaced by – b. ••.) – b and in [27] when 6.
Theorem 10. If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (А+ C) (Зе— d) < (е+ d) (1-(В+ D), (29) is valid, (B+ D) and provided d (1 — (В+ D) — е (А+С)>0. < 1 Proof.If 1 is even and k, are odd positive integers, then Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1) that bP Р- (А+C)Q+ (B+D) P — (30) (e Q– dP)' and bQ Q= (A+C) P+(B+D) Q – (31) (e P- dQ)' Consequently, we get b P+Q= (32) [d (1 — (В+ D)) — е (А+С)]" where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]²" (33) - where K2 (1– (B+D)), provided (B+ D) < 1. Substituting (32) and (33) into (28), we get the condition (29). Thus, the proof is now completed.O
Transcribed Image Text:Theorem 10. If 1 is even and k, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (А+ C) (Зе— d) < (е+ d) (1-(В+ D), (29) is valid, (B+ D) and provided d (1 — (В+ D) — е (А+С)>0. < 1 Proof.If 1 is even and k, are odd positive integers, then Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1) that bP Р- (А+C)Q+ (B+D) P — (30) (e Q– dP)' and bQ Q= (A+C) P+(B+D) Q – (31) (e P- dQ)' Consequently, we get b P+Q= (32) [d (1 — (В+ D)) — е (А+С)]" where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]²" (33) - where K2 (1– (B+D)), provided (B+ D) < 1. Substituting (32) and (33) into (28), we get the condition (29). Thus, the proof is now completed.O
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