theorem 7 has no prime period two solution and in the theory 10
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Question
Find reason in theorem 7 has no prime period two solution and in the theory 10 has prime period two solution ???? ( just reason), Equation 1 in the second picture. Whhhyyyyyy!!!
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxp-k– exn=1]
(1)
Xn+1
Axn+ Bxn–k+ Cxr-1+DXn-ot
= U
:0,1,2,....
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,
•..)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff415ac57-ff92-4e25-9045-4869ff575152%2F786bc269-615d-441b-bd9b-d03b72366501%2Fi9wfd17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxp-k– exn=1]
(1)
Xn+1
Axn+ Bxn–k+ Cxr-1+DXn-ot
= U
:0,1,2,....
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,
•..)
![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.
Proof.Following the proof of Theorem 5, we deduce that
if k, 1 are even and o is odd positive integers, then x, =
Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) that
b
P= (A+B+ C) Q+ DP –
(16)
(e – d)'
and
b
Q= (A+B+C) P+DQ –
(17)
(e – d)'
By subtracting (17) from (16), we get
(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.O
Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O
Theorem 10.If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(A+C) (3e- d) < (e+d) (1- (B+D)),
(29)
is
d (1– (B+D)) –
(B+D)
provided
(A+C) > 0.
valid,
< 1
and
- e
Proof.If 1 is even and k, o are odd positive integers, then
It follows from Eq.(1)
Xn = Xn-1
and
Xn+1 = Xn-k= Xn-o.
that
bP
P= (A+C) Q+ (B+D) P –
(30)
(e Q- dP)'
and
bQ
Q= (A+C)P+(B+D) Q
(31)
(e P- dQ)'
Consequently, we get
b
P+ Q=
(32)
[d (1 – (B+D)) – e (A+C)]'
where d (1- (B+D)) – e (A+C) >0,
e B (A+C)
(e+d) [K2 +(A+C)] [d K2 – e (A+C)]² '
PQ=
(33)
(1- (B+D)), provided (B+ D) < 1.
where K2
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff415ac57-ff92-4e25-9045-4869ff575152%2F786bc269-615d-441b-bd9b-d03b72366501%2F28h68c_processed.png&w=3840&q=75)
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.
Proof.Following the proof of Theorem 5, we deduce that
if k, 1 are even and o is odd positive integers, then x, =
Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) that
b
P= (A+B+ C) Q+ DP –
(16)
(e – d)'
and
b
Q= (A+B+C) P+DQ –
(17)
(e – d)'
By subtracting (17) from (16), we get
(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.O
Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O
Theorem 10.If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(A+C) (3e- d) < (e+d) (1- (B+D)),
(29)
is
d (1– (B+D)) –
(B+D)
provided
(A+C) > 0.
valid,
< 1
and
- e
Proof.If 1 is even and k, o are odd positive integers, then
It follows from Eq.(1)
Xn = Xn-1
and
Xn+1 = Xn-k= Xn-o.
that
bP
P= (A+C) Q+ (B+D) P –
(30)
(e Q- dP)'
and
bQ
Q= (A+C)P+(B+D) Q
(31)
(e P- dQ)'
Consequently, we get
b
P+ Q=
(32)
[d (1 – (B+D)) – e (A+C)]'
where d (1- (B+D)) – e (A+C) >0,
e B (A+C)
(e+d) [K2 +(A+C)] [d K2 – e (A+C)]² '
PQ=
(33)
(1- (B+D)), provided (B+ D) < 1.
where K2
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O
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