Show me the determine yellow and the information is here THEOREM 4.2. Let l, k and t are both even positive integers then for all B, a, a, b and c anre positive realnumbers, then Eq. (1) has no positive prime period two solution.Proof: Let 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 l, k and t are both even, then rn+1 = P and rn-t = xn-k = Xn-t = Q. It followsEq. (1) thatP = BQ + aQ + andQ = BP + aP+P¿P+c°aQbQ+cTherefore,bPQ + cP = b(B +a)Q? + (a + c(8 + a))Q,(14)bPQ + cQ = b(B+ a)P? + (a + c(B + a))P,(15)Subtracting (15) from (14) givesP+Qa+c(1+8+a)b(B+a)(16)Again, adding (14) and (15) yieldsPQc(a+c(1+8+a))62 (8+a)(1+3+a)(17)From (16) and (17), we have(P+Q) РQ 3 -e(a+c(1+8+a)+c)²<0(8+a)²(1+8+a)This contradicts the hypothesis that both P and Q are positive. Thus, the proof is now completed. Our goal is to obtain some qualitative behavior of the positive solutions of the difference equationaIn-tIn+1 = Bxn-i+axn-k +п%3D 0, 1,(1)bxn-t+c'where the parameters B, 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 =таxfl, k, t).

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THEOREM 4.2. Let l, k and t are both even positive integers then for all B, a, a, b and c are positive real numbers, then Eq. (1) has no positive prime period two solution. Proof: Let 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 l, k and t are both even, then rn+1 = P and rn-l = xn-k = xn-t = Q. It follows Eq. (1) that P = BQ + aQ + 1. and Q = BP+ aP+ . 6P+c* Therefore, bPQ + cP = b(B +a)Q? + (a + c(B+a))Q, (14) bPQ + cQ = b(B+a)P² + (a + c(B + a))P, (15) Subtracting (15) from (14) gives P+Q = a+c(1+8+a) b(8+a) (16) Again, adding (14) and (15) yields PQ = c(a+c(1+3+a)). (B+a)(1+B+a) (17) From (16) and (17), we have (P+Q) PQ c(a+c(1+B+a)+c)² b (8+a)²(1+B+a) < 0 = - This contradicts the hypothesis that both P and Q are positive. Thus, the proof is now completed.

THEOREM 4.2. Let l, k and t are both even positive integers then for all B, a, a, b and c are positive real numbers, then Eq. (1) has no positive prime period two solution. Proof: Let 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 l, k and t are both even, then rn+1 = P and rn-l = xn-k = xn-t = Q. It follows Eq. (1) that P = BQ + aQ + 1. and Q = BP+ aP+ . 6P+c* Therefore, bPQ + cP = b(B +a)Q? + (a + c(B+a))Q, (14) bPQ + cQ = b(B+a)P² + (a + c(B + a))P, (15) Subtracting (15) from (14) gives P+Q = a+c(1+8+a) b(8+a) (16) Again, adding (14) and (15) yields PQ = c(a+c(1+3+a)). (B+a)(1+B+a) (17) From (16) and (17), we have (P+Q) PQ c(a+c(1+B+a)+c)² b (8+a)²(1+B+a) < 0 = - This contradicts the hypothesis that both P and Q are positive. Thus, the proof is now 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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