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Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn Sn-r = Sn-s and Sn+1 Sn-p = Sn-q. From Equation (1), we have аф + (b + c) $ = ¢ dø + (e + f)½ arb + (b+ c)o dp + (e + f)ø, b = b Thus, do? + (e + f)øv½ = ao² + (b + c)ø½, (8) and dy? + (e+ f)ø½ = av² + (b+ c)ørp. (9) By subtracting (8) from (9), we deduce that Cd(o? – ?) – a(² – v²) = 0. Hence, we have [d – a](o? – v²) = 0. Since a and d are nonzero positive real numbers, and ø # p. This implies [d = a], This contradicts the condition [d + a].

Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn Sn-r = Sn-s and Sn+1 Sn-p = Sn-q. From Equation (1), we have аф + (b + c) $ = ¢ dø + (e + f)½ arb + (b+ c)o dp + (e + f)ø, b = b Thus, do? + (e + f)øv½ = ao² + (b + c)ø½, (8) and dy? + (e+ f)ø½ = av² + (b+ c)ørp. (9) By subtracting (8) from (9), we deduce that Cd(o? – ?) – a(² – v²) = 0. Hence, we have [d – a](o? – v²) = 0. Since a and d are nonzero positive real numbers, and ø # p. This implies [d = a], This contradicts the condition [d + a].

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
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Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn = Sn-r = Sn-s and Sn+1 = Sn-p = Sn-q• From Equation (1), we have аф + (b + с) dф + (е + f)4, $ = ¢ ab + (b+ c)ø dy + (e + f)¢, V = b Thus, dø? + (e + f)ø½ = ao² + (b + c)øv½, (8) and du² + (e + f)øv = ay² + (b+ c)øp. (9) By subtracting (8) from (9), we deduce that Ca(o? – 2) – a(6² – v²) = 0. Hence, we have Çu- ((d – a](o? – v²) = 0. [d- Since a and d are nonzero positive real numbers, and ø ½. This implies [d = a], This contradicts the condition [d + a].
Transcribed Image Text:Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn = Sn-r = Sn-s and Sn+1 = Sn-p = Sn-q• From Equation (1), we have аф + (b + с) dф + (е + f)4, $ = ¢ ab + (b+ c)ø dy + (e + f)¢, V = b Thus, dø? + (e + f)ø½ = ao² + (b + c)øv½, (8) and du² + (e + f)øv = ay² + (b+ c)øp. (9) By subtracting (8) from (9), we deduce that Ca(o? – 2) – a(6² – v²) = 0. Hence, we have Çu- ((d – a](o? – v²) = 0. [d- Since a and d are nonzero positive real numbers, and ø ½. This implies [d = a], This contradicts the condition [d + a].
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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