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Show me the steps of determine red and the inf is here

Show me the steps of determine red and the inf is here

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 10E
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Show me the steps of determine red and the inf is here

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 xŋ+1= Xn-k. It follows from Eq.(1) that bP P=(A+C+D) Q+BP - (35) (е Q— dP)' and bQ Q= (A+C+D)P+BQ (36) (еР — d) Consequently, we get P+Q= (37) [а (1 — В) — е (А+С+D)]' where d (1– B) - e (A+C+D) > 0, e b (A+C+D) (38) PQ= (e+d) [(1 – B) + K3] [d (1 – B) – e K3]²" | where K3 (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O (A+C+D), provided B< 1. Substituting
Transcribed Image Text: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 xŋ+1= Xn-k. It follows from Eq.(1) that bP P=(A+C+D) Q+BP - (35) (е Q— dP)' and bQ Q= (A+C+D)P+BQ (36) (еР — d) Consequently, we get P+Q= (37) [а (1 — В) — е (А+С+D)]' where d (1– B) - e (A+C+D) > 0, e b (A+C+D) (38) PQ= (e+d) [(1 – B) + K3] [d (1 – B) – e K3]²" | where K3 (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O (A+C+D), provided B< 1. Substituting
Thus, we deduce that (P+ Q)² > 4PQ. (28) difference equation bxn-k Xn+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,), while k,1 and o are positive integers. The initial conditions X-o,..., X_1,..., X_k, ….., X_1, Xo are arbitrary positive real numbers such that k<1< 0. Note that the special cases of Eq.(1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by B=C=D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1 = 0 and in [32] when A= C= D=0, 1=0, b is replaced by – b. ••.) – b and in [27] when 6.
Transcribed Image Text:Thus, we deduce that (P+ Q)² > 4PQ. (28) difference equation bxn-k Xn+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,), while k,1 and o are positive integers. The initial conditions X-o,..., X_1,..., X_k, ….., X_1, Xo are arbitrary positive real numbers such that k<1< 0. Note that the special cases of Eq.(1) have been studied in [1] when B=C= D=0, and k= 0,1= 1, b is replaced by B=C=D=0, and k= 0, b is replaced by – b and in [33] when B = C = D = 0, 1 = 0 and in [32] when A= C= D=0, 1=0, b is replaced by – b. ••.) – b and in [27] when 6.
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