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P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).

P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 80E
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Theorem 6 If (a1+ аз + a5) > (02 + ол) аnd (Bі + Вз + Bъ) > (32 + Ba), then the necessary and sufficient condition for Eq.(1.1) to have positive so- lutions of prime period two is that the inequality [(A + 1) ((B1 + B3 + B5) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (82 + B4) (a1 + a3 + 5)] > 0. (4.13) is valid. proof: Suppose there exist positive distinctive solutions of prime period two P,Q, P, Q, .. of Eq.(1.1). From Eq.(1.1) we have a1Ym-1+a2Ym-2+ a3Ym-3+a4Ym-4+a5Ym-5 Ym+1 = Aym + B1Ym-1 + B2Ym-2 + B3Ym–3 + B4Ym-4 + B5Ym–5 (a1 + az + a5) P+(a2+a4) Q Q = (B1 + B3 + B5) P + (82 + B4) Q ’ (a1 + az + a5) Q+(a2 + a4) P Р— АQ+ = AP+ (B1 + B3 + B5) Q+ (B2 + B4) P * (4.14) Consequently, we get (B1 + B3 + B5) P² + (B2 + B4) PQ A (B1 + B3 + B5) PQ + A (B2 + B4) Q² + (a1 + a3 + a5) P+ (a2 + a4) Q, (4.15) and (B1 + 33 + B5) Q² + (82 + B4) PQ A (B1 + B3 + B5) PQ + A (ß2 + B4) P² + (a1 + a3 + a5) Q + (a2 + a4) P. (4.16) By subtracting (4.15) from (4.16), we obtain [(81 + B3 + B5) + A (B2 + B4)] (P² –- Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q). 11
Transcribed Image Text:Theorem 6 If (a1+ аз + a5) > (02 + ол) аnd (Bі + Вз + Bъ) > (32 + Ba), then the necessary and sufficient condition for Eq.(1.1) to have positive so- lutions of prime period two is that the inequality [(A + 1) ((B1 + B3 + B5) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (82 + B4) (a1 + a3 + 5)] > 0. (4.13) is valid. proof: Suppose there exist positive distinctive solutions of prime period two P,Q, P, Q, .. of Eq.(1.1). From Eq.(1.1) we have a1Ym-1+a2Ym-2+ a3Ym-3+a4Ym-4+a5Ym-5 Ym+1 = Aym + B1Ym-1 + B2Ym-2 + B3Ym–3 + B4Ym-4 + B5Ym–5 (a1 + az + a5) P+(a2+a4) Q Q = (B1 + B3 + B5) P + (82 + B4) Q ’ (a1 + az + a5) Q+(a2 + a4) P Р— АQ+ = AP+ (B1 + B3 + B5) Q+ (B2 + B4) P * (4.14) Consequently, we get (B1 + B3 + B5) P² + (B2 + B4) PQ A (B1 + B3 + B5) PQ + A (B2 + B4) Q² + (a1 + a3 + a5) P+ (a2 + a4) Q, (4.15) and (B1 + 33 + B5) Q² + (82 + B4) PQ A (B1 + B3 + B5) PQ + A (ß2 + B4) P² + (a1 + a3 + a5) Q + (a2 + a4) P. (4.16) By subtracting (4.15) from (4.16), we obtain [(81 + B3 + B5) + A (B2 + B4)] (P² –- Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q). 11
The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+ a2Ym-2 + a3Ym-3 + a4Ym-4+ a5Ym-5 Ym+1 = Aym+ т 3D 0, 1, 2, ..., В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5 (1.1)
Transcribed Image Text:The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+ a2Ym-2 + a3Ym-3 + a4Ym-4+ a5Ym-5 Ym+1 = Aym+ т 3D 0, 1, 2, ..., В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5 (1.1)
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