By subtracting (17) from (16), we get C (P− Q)[(A+B+C+1) −D] =0, Since (A+B+C+1) − D ‡0, then P = Q. This is a contradiction. Thus, the proof is now completed.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 36E
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Explain the determine green and the information is here

The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn– k
X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+
[dxn-k– ex-1
(1)
n= 0, 1,2, .....
where the coefficients A, B, C, D, b, d, e e (0,00), while
k, 1 and o are positive integers. The initial conditions
X-g,..., 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:The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn– k X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+ [dxn-k– ex-1 (1) n= 0, 1,2, ..... where the coefficients A, B, C, D, b, d, e e (0,00), while k, 1 and o are positive integers. The initial conditions X-g,..., 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.
Theorem 7.If k,1 are even and o is odd positive integers
and (A+B+ C)+1+D, then Eq. (1) has no prime period
two solution.
t
Proof.Following the proof of Theorem 5, we deduce that
if k, 1 are even and o is odd positive integers, then Xn =
Xn-k = Xn-1 Xn-o. It follows from Eq.(1) that
6.
and Xn+1 =
b
P= (A+B+C) Q+DP
(e
(16)
and
Q= (A+B+ C) P+ DQ
(17)
(e – d)
By subtracting (17) from (16), we get
P- Q) [(A+B+C+1) – D]=0,
%3|
Since (A+B+C+1) – D 7 0, then P= Q. This is a
contradiction. Thus, the proof is now completed.
-
Transcribed Image Text:Theorem 7.If k,1 are even and o is odd positive integers and (A+B+ C)+1+D, then Eq. (1) has no prime period two solution. t Proof.Following the proof of Theorem 5, we deduce that if k, 1 are even and o is odd positive integers, then Xn = Xn-k = Xn-1 Xn-o. It follows from Eq.(1) that 6. and Xn+1 = b P= (A+B+C) Q+DP (e (16) and Q= (A+B+ C) P+ DQ (17) (e – d) By subtracting (17) from (16), we get P- Q) [(A+B+C+1) – D]=0, %3| Since (A+B+C+1) – D 7 0, then P= Q. This is a contradiction. Thus, the proof is now completed. -
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