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+1 = Xn-g• It follows from Eq.(1) that Xn-k= Xp–1 and 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 10.1If 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 valid, provided (B+D) 1 and d (1– (B+ D)) – e (A+C) > 0. | Proof.If 1 is even and k, o are odd positive integers, then Xn = Xn-1 and xn+1 = Xn-k= Xn-o. It follows from Eq.(1) 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 P+Q= (32) [d (1– (B+D)) –e (A+C)]* | where d (1- (B+D)) – e (A+C) > 0, eb (A+C) PQ = (e+d) [K2+(A+C)] [d K2 – e (A+C)]²" (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
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+1 = Xn-g• It follows from Eq.(1) that Xn-k= Xp–1 and 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 10.1If 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 valid, provided (B+D) 1 and d (1– (B+ D)) – e (A+C) > 0. | Proof.If 1 is even and k, o are odd positive integers, then Xn = Xn-1 and xn+1 = Xn-k= Xn-o. It follows from Eq.(1) 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 P+Q= (32) [d (1– (B+D)) –e (A+C)]* | where d (1- (B+D)) – e (A+C) > 0, eb (A+C) PQ = (e+d) [K2+(A+C)] [d K2 – e (A+C)]²" (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
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
Explain to me why in theorem 7 has no prime period two solution and in the theory 10 has prime period two solution ???? I need an explanation for this the determine blue
![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%2F38bf9701-cb6b-4cde-8eaa-26a5c3cba4d5%2Frlw5e09_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
![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%2F38bf9701-cb6b-4cde-8eaa-26a5c3cba4d5%2Fphnvi1q_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,
•..)
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