Show me the steps of determine red and the inf is here Theorem 11.If 1,0are even and k is odd positiveintegers, then Eq.(1) has prime period two solution if thecondition(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, thenXn = Xp-1= Xr-o and xŋ+1 = Xn-k• It follows from Eq.(1)thatbPP=(A+C+D)Q+BP(35)(e Q- dP)'andbQQ= (A+C+D) P+BQ –(36)(eP – dQ)Consequently, we getbP+Q=(37)[d (1 — В) — е (А+С+D]'- e (A+C+D)]’where d (1- B) – e (A+C+D) > 0,e b (A+C+D)PQ=(e+d) [(1– B) + K3] [d (1 – B) – e K3]²(38)-where K3 = (A+C+D), provided B < 1. Substituting(37) and (38) into (28), we get the condition (34). Thus,the proof is now completed.O 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.

Name: Show me the steps of determine red and the inf is here Theorem 11.If 1
Uploaded: 2024-03-20T06:47:08.000Z
Channel: Bartleby
Description: Show me the steps of determine red and the inf is here Theorem 11.If 1,0are even and k is odd positiveintegers, then Eq.(1) has prime period two solution if thecondition(A+C+D) (3e- d) < (e+d) (1– B),(34)is valid, provided B< 1 and d (1– B) – e (A+C+