Show me the steps of determine yellow and the inf is here Thus, we deduce that(P+ Q)² > 4PQ.(28)difference equationbxn-kXn+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,), whilek,1 and o are positive integers. The initial conditionsX-o,..., X_1,..., X_k, ….., X_1, Xo are arbitrary positive realnumbers such that k<1< 0. Note that the special casesof Eq.(1) have been studied in [1] when B=C= D=0,and k= 0,1= 1, b is replaced byB=C=D=0, and k= 0, b is replaced by – b and in[33] when B = C = D = 0, 1 = 0 and in [32] whenA= C= D=0, 1=0, b is replaced by – b.••.)– b and in [27] when6. Theorem 12.If k, o are even and 1 is odd positive integers,then Eq. (1) has prime period two solution if the condition(Зе— d) (1— с) < (е+d) (А+B+ D),(39)is valid, provided C< 1 and e (1-C)-d(A+B+D) >0.Proof.If k, o are even and 1 is odd positive integers, thenXn = Xn-k = Xn-o and xp+1 = Xp–1. It follows from Eq.(1)thatbQP= (A+B+D) Q+CP –(40)(е Р- dQ)andbPQ= (A+B+ D)P+CQ –(41)(e Q- dP)Consequently, we getP+Q=(42)[e (1- C)-d (A+B+D)]’where e (1– C)-d (A+B+ D) > 0,eb² (1-C)PQ=(43)(e+d) [(1– C) + K4] [e (1 – C)– d K4]²where K4 = (A+B+D), provided C< 1. Substituting(42) and (43) into (28), we get the condition (39). Thus,the proof is now completed.O

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Description: Show me the steps of determine yellow and the inf is here Thus, we deduce that(P+ Q)² > 4PQ.(28)difference equationbxn-kXn+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,), whilek,1 and o ar

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Answered: Consequently, we get b P+Q= (42) [e (1…

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Consequently, we get b P+Q= (42) [e (1 – C)– d (A+B+D)]’ where e (1– C)-d (A+B+D) > 0, eb² (1–C) PQ= (e+d) [(1– C)+ K4] [e (1– C)- d K4] (43)

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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