Show me the steps of determine yellow and information is here step by step THEOREM 4.8. Let t is even and l, k are odd positive integers. Ifc(1 – a –с — В) + а + 0,then Eq. (1) has no prime period two solution.Proof: Assume that there exists distinct positive solution P and Q, such that....Р, Q, Р, Q, ...,is a prime period two solution of Eq.(1).We see from Eq. (1) when t is even and l, k are odd, then xn+1 = xn-k = xn-1 = P and xm-t = Q. It followsEq. (1) thataQbQ+caPP = BP+ aP+and Q= BQ + aQ+ P+c-Therefore,6(1 – a – B) PQ +c(1 - a – B) P = aQ,(35)(1 - а - В) РQ + c (1 — а — B)Q — аР,(36)Subtracting (47) from (46) gives(c(1- a – B) + a) (P – Q) = 0Since c (1 – a – B) + a + 0, then P = Q. This is a contradiction. Thus, the proof is completed.- a Our goal is to obtain some qualitative behavior of the positive solutions of the difference equationXn+1 = Bxn-i+axn-k +axn-tn = 0, 1, ...,(1)bxn-t+c'where the parameters 3, a, a, b and c are positive real numbers and the initial conditions x-s, x-s+1, ..., x-1,xo are positive real numbers where s = max{1, k, t}..D TT T T

For the explanation follow the next steps.

THEOREM 4.8. Let t is even and l, k are odd positive integers. If c(1 – a - B) + a + 0, then Eq. (1) has no prime period two solution. Proof: Assume that there exists distinct positive solution P and Q, such that ..P, Q, P, Q,., is a prime period two solution of Eq.(1). We see from Eq. (1) when t is even and l, k are odd, then xn+1 = Xn-k = xn-1 = P and xn-t = Q. It follows Eq. (1) that aQ bQ+c aP P = BP+ aP+ and Q = BQ + aQ + bP+c* Therefore, b(1 – a - 8) PQ +c(1-a - B) P = aQ, (35) b (1-α-β) PQ+c(1-α-β ) Q- aP (36) Subtracting (47) from (46) gives (c(1 – a – B) + a) (P– Q) = 0 Since c (1 – a - B) + a + 0, then P = Q. This is a contradiction. Thus, the proof is completed.

THEOREM 4.8. Let t is even and l, k are odd positive integers. If c(1 – a - B) + a + 0, then Eq. (1) has no prime period two solution. Proof: Assume that there exists distinct positive solution P and Q, such that ..P, Q, P, Q,., is a prime period two solution of Eq.(1). We see from Eq. (1) when t is even and l, k are odd, then xn+1 = Xn-k = xn-1 = P and xn-t = Q. It follows Eq. (1) that aQ bQ+c aP P = BP+ aP+ and Q = BQ + aQ + bP+c* Therefore, b(1 – a - 8) PQ +c(1-a - B) P = aQ, (35) b (1-α-β) PQ+c(1-α-β ) Q- aP (36) Subtracting (47) from (46) gives (c(1 – a – B) + a) (P– Q) = 0 Since c (1 – a - B) + a + 0, then P = Q. This is a contradiction. Thus, the proof is completed.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.4: Mathematical Induction

Problem 42E

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