Theorem 9.If k is even and I, ở are odd positivVe integers, then Eq.(1) has prime period two solution if the condition (1–(C+D))(3e– d) < (e+ d) (A+ B) , (20) is valid provided C+ D) and

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
icon
Related questions
icon
Concept explainers
Question

How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine blue and inf is here

Theorem 9.If k is even and 1, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(1- (C+D)) (3e- d) < (e+ d) (A+ B),
(20)
is
valid,
provided
(C+D)
< 1
and
e (1– (C+D))– d (A+B) > 0.
Proof.If k is even and 1, o are odd positive integers, then
Xn= Xp–k and Xn+1 = Xŋn–1= Xn-o. It follows from Eq.(1)
that
bQ
P= (A+B) Q+(C+D) P –
(21)
(eР — dQ)"
and
БР
Q= (A+B) P+ (C+D) Q –
(22)
(e Q– dP)'
Consequently, we get
e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D)P
- (C+D) dPQ– bQ,
(23)
and
e Q – dPQ= e (A+B) PQ – d (A+B) P² + e(C+ D) Q
- (C+D) dPQ– ÞP.
||
(24)
By subtracting (24) from (23), we get
b
P+Q=
(25)
[e (1 – (C+D)) –d (A+B)]'
where e (1– (C+D)) – d (A+ B) > 0. By adding (23)
and (24), we obtain
e b² (1– (C+D))
(e+d) [K1 +(A+B) [e K1 – d (A+B)]²
PQ =
(26)
Transcribed Image Text:Theorem 9.If k is even and 1, o are odd positive integers, then Eq. (1) has prime period two solution if the condition (1- (C+D)) (3e- d) < (e+ d) (A+ B), (20) is valid, provided (C+D) < 1 and e (1– (C+D))– d (A+B) > 0. Proof.If k is even and 1, o are odd positive integers, then Xn= Xp–k and Xn+1 = Xŋn–1= Xn-o. It follows from Eq.(1) that bQ P= (A+B) Q+(C+D) P – (21) (eР — dQ)" and БР Q= (A+B) P+ (C+D) Q – (22) (e Q– dP)' Consequently, we get e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D)P - (C+D) dPQ– bQ, (23) and e Q – dPQ= e (A+B) PQ – d (A+B) P² + e(C+ D) Q - (C+D) dPQ– ÞP. || (24) By subtracting (24) from (23), we get b P+Q= (25) [e (1 – (C+D)) –d (A+B)]' where e (1– (C+D)) – d (A+ B) > 0. By adding (23) and (24), we obtain e b² (1– (C+D)) (e+d) [K1 +(A+B) [e K1 – d (A+B)]² PQ = (26)
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
dxn-k- exn-1]
Xn+1 =
Axn+ Bx,-k+ Cx,-1+ Dxp-o
п 3 0, 1,2,.....
(1)
where the coefficients A, B, C, D, b, d, e E (0,∞), while
k, 1 and o are positive integers. The initial conditions
X_0,..., 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,
0,1= 1,b is replaced by – b and in [27] when
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.
-
Transcribed Image Text:The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn-k dxn-k- exn-1] Xn+1 = Axn+ Bx,-k+ Cx,-1+ Dxp-o п 3 0, 1,2,..... (1) where the coefficients A, B, C, D, b, d, e E (0,∞), while k, 1 and o are positive integers. The initial conditions X_0,..., 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, 0,1= 1,b is replaced by – b and in [27] when 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. -
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
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Elements Of Modern Algebra
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,