Theorem 11.If 1,0 are even and k is odd positive integers, then Eq.(1) has prime period two solution if the condition (A+C+D)(3e- d) < (e+d)(1 – B), (34) | is valid, provided B< 1 and d (1 – B) – e (A+C+D) > 0. | Proof.If 1,0 are even and k is odd positive integers, then Xn = Xn-1= Xn-o and xn+1 = Xn=k• It follows from Eq.(1) that bP P= (A+C+D) Q+BP (35) %3| (e Q- dP)' and bQ Q= (A+C+D) P+BQ (36) (еР — do) Consequently, we get P+Q= (37) [d (1– B) – e (A+C+D)]' where d (1- B) - e (A+C+D) > 0, e (A+C+D) (e+d) [(1– B) + K3] [d (1 – B) – e K]² PQ= (38) (A+C+ D), provided B< 1. Substituting where K3 = (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O
Theorem 11.If 1,0 are even and k is odd positive integers, then Eq.(1) has prime period two solution if the condition (A+C+D)(3e- d) < (e+d)(1 – B), (34) | is valid, provided B< 1 and d (1 – B) – e (A+C+D) > 0. | Proof.If 1,0 are even and k is odd positive integers, then Xn = Xn-1= Xn-o and xn+1 = Xn=k• It follows from Eq.(1) that bP P= (A+C+D) Q+BP (35) %3| (e Q- dP)' and bQ Q= (A+C+D) P+BQ (36) (еР — do) Consequently, we get P+Q= (37) [d (1– B) – e (A+C+D)]' where d (1- B) - e (A+C+D) > 0, e (A+C+D) (e+d) [(1– B) + K3] [d (1 – B) – e K]² PQ= (38) (A+C+ D), provided B< 1. Substituting where K3 = (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter5: Orthogonality
Section5.3: The Gram-schmidt Process And The Qr Factorization
Problem 11AEXP
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