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Case 1. Suppose the positive integers p,r,s are even and the positive integer q is odd. In this case Sn = Sn-p = Sn-r = Sn-s and Sn+1 = Sn-q. From Equation (1), we have аф + (b + c) co + (d+ f)v ap + (b+c)o cy + (d+ f), Thus + (d+ f)o = apy + (b+ c)u?, (2) and + (d + f)oy = apy + (b+ c)o?. (3) %3D By subtracting (2) from (3), we deduce that c(o – v?) + (b+ c)(? – v²) = 0. | Hence, we have (b+ 2c)(4² – p²) = 0. If o + v, then (b+ 2c) = 0. Since b and c are nonzero positive real numbers, thus (b + 2c) # 0. This implies $ = p. This contradicts the hypothesis o p.

Case 1. Suppose the positive integers p,r,s are even and the positive integer q is odd. In this case Sn = Sn-p = Sn-r = Sn-s and Sn+1 = Sn-q. From Equation (1), we have аф + (b + c) co + (d+ f)v ap + (b+c)o cy + (d+ f), Thus + (d+ f)o = apy + (b+ c)u?, (2) and + (d + f)oy = apy + (b+ c)o?. (3) %3D By subtracting (2) from (3), we deduce that c(o – v?) + (b+ c)(? – v²) = 0. | Hence, we have (b+ 2c)(4² – p²) = 0. If o + v, then (b+ 2c) = 0. Since b and c are nonzero positive real numbers, thus (b + 2c) # 0. This implies $ = p. This contradicts the hypothesis o p.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.1: Equations
Problem 75E
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Case 1. Suppose the positive integers p,r,s are even and the positive integer q odd. In this case Sn = Sn-p = Sn-r = Sn-s and Sn+1 = Sn-q. From Equation (1), we have ap + (b + c) co+ (d+f) a + (b+c) cp + (d+ f)o, Thus co? + (d + f) = aørp + (b+ c)u?, (2) and cy? + (d + f)øv = app + (b + c)o?. (3) By subtracting (2) from (3), we deduce that c(a? – 2) + (b + c)(s² – »?) = 0. Hence, we have (b+ 2c)(? – y?) = 0. If ø + 4, then (b+ 2c) = = 0. Since b and c are nonzero positive real numbers, thus (b + 2c) + 0. This implies $ = 6. This contradicts the hypothesis o + v.
Transcribed Image Text:Case 1. Suppose the positive integers p,r,s are even and the positive integer q odd. In this case Sn = Sn-p = Sn-r = Sn-s and Sn+1 = Sn-q. From Equation (1), we have ap + (b + c) co+ (d+f) a + (b+c) cp + (d+ f)o, Thus co? + (d + f) = aørp + (b+ c)u?, (2) and cy? + (d + f)øv = app + (b + c)o?. (3) By subtracting (2) from (3), we deduce that c(a? – 2) + (b + c)(s² – »?) = 0. Hence, we have (b+ 2c)(? – y?) = 0. If ø + 4, then (b+ 2c) = = 0. Since b and c are nonzero positive real numbers, thus (b + 2c) + 0. This implies $ = 6. This contradicts the hypothesis o + v.
The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-q + bSn-r + cSn-s dSn Sn+1 = Sn-p (1) + eSn-r + fSn-s ) ' -q where a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r, S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such that p > q > r > s > 0. е,
Transcribed Image Text:The main aim of this study is to exhibit some cases on the periodic character of the positive solutions of the rational difference equation aSn-q + bSn-r + cSn-s dSn Sn+1 = Sn-p (1) + eSn-r + fSn-s ) ' -q where a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r, S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such that p > q > r > s > 0. е,
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