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.

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
icon
Related questions
icon
Concept explainers
Question

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. -
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Points, Lines and Planes
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Elements Of Modern Algebra
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,
College Algebra
College Algebra
Algebra
ISBN:
9781938168383
Author:
Jay Abramson
Publisher:
OpenStax