theorem 7 has no prime period two solution and in the theory 10
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
