b P+Q= (25) [e (1– (C+D)) –d (A+B)] vhere e (1 – (C+ D)) – d (A+ B) > 0. By adding (23)
b P+Q= (25) [e (1– (C+D)) –d (A+B)] vhere e (1 – (C+ D)) – d (A+ B) > 0. By adding (23)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
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Show me the steps of determine green and the inf is here
![Theorem 9.If k is even and 1,0 are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(1— (С+D)) (Зе— d) < (е+d) (А+ В),
(20)
is
valid,
provided
(C+D)
1
and
е (1— (С+D))—d (А+ B) > 0.
Proof.If k is even and 1, 0 are odd positive integers, then
Xn+1 = Xn–1= Xn=o. It follows from Eq.(1)
Xn
Xn-k
and
that
bQ
P=(A+B) Q+(C+D) P –
(21)
(еР — dQ)
and
БР
Q= (A+B) P+(C+D) Q -
(22)
(e Q- dP)
Consequently, we get
eP— dPQ — е (А+ В) РQ—d (А+ В) Q+e(С+D)P
— (С+ D) dPQ— bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) P² +e(C+D) Ở
— (С+ D) dPQ— БР.
(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 b° (1– (C+D))
(e+d)[K1+(A+ B)][e K1 – d (A+ B)]² '
PQ=
(26)
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ffd9a5078-fc5d-4422-9770-73583a8324a1%2F2xrghac_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 9.If k is even and 1,0 are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(1— (С+D)) (Зе— d) < (е+d) (А+ В),
(20)
is
valid,
provided
(C+D)
1
and
е (1— (С+D))—d (А+ B) > 0.
Proof.If k is even and 1, 0 are odd positive integers, then
Xn+1 = Xn–1= Xn=o. It follows from Eq.(1)
Xn
Xn-k
and
that
bQ
P=(A+B) Q+(C+D) P –
(21)
(еР — dQ)
and
БР
Q= (A+B) P+(C+D) Q -
(22)
(e Q- dP)
Consequently, we get
eP— dPQ — е (А+ В) РQ—d (А+ В) Q+e(С+D)P
— (С+ D) dPQ— bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) P² +e(C+D) Ở
— (С+ D) dPQ— БР.
(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 b° (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
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X_g,..., X_1,.., X_k, ….., X_1, X 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 and in [27] when
B=C= D=0, and k= 0, b is replaced by
[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
%3|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ffd9a5078-fc5d-4422-9770-73583a8324a1%2F4wdw5yr_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
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X_g,..., X_1,.., X_k, ….., X_1, X 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 and in [27] when
B=C= D=0, and k= 0, b is replaced by
[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
%3|
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