Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (3e- d) (1- C) < (e+d)(A+B+D), (39) is valid, provided C<1 and e (1 – C)-d(A+B+D) > 0.
Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (3e- d) (1- C) < (e+d)(A+B+D), (39) is valid, provided C<1 and e (1 – C)-d(A+B+D) > 0.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 42E
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Question
Show me the steps of determine red and inf is here. It complete
![Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxn-k - exp-1)
Xn+1 =
Axn+ Bxn-k+Cxn–1+Dxn-o +
n = 0,1,2,..
where the coefficients A, B, C, D, b, d, e e (0,00), while
k, 1 and o are positive integers. The initial conditions
X-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real
numbers such that k <1< o. 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.
(1)
b and in
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F6c5de4cc-68e1-4717-b9f7-c869c8f53823%2Fmxkyztf_processed.png&w=3840&q=75)
Transcribed Image Text:Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxn-k - exp-1)
Xn+1 =
Axn+ Bxn-k+Cxn–1+Dxn-o +
n = 0,1,2,..
where the coefficients A, B, C, D, b, d, e e (0,00), while
k, 1 and o are positive integers. The initial conditions
X-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real
numbers such that k <1< o. 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.
(1)
b and in
%3D
![Theorem 12.If k, o are even and 1 is odd positive integers,
then Eq. (1) has prime period two solution if the condition
(3e- d) (1-C)< (e+d)(A+B+D),
(39)
is valid, provided C<1 and e (1 –C)-d(A+B+D) >
0.
Proof.If k, o are even and I is odd positive integers, then
Xn = Xn-k = Xn-o and xn+1 = Xn–1. It follows from Eq.(1)
that
bQ
P=(A+B+D) Q+CP –
(40)
(е Р- dQ)'
and
bP
Q= (A+B+D) P+CQ -
(41)
(e Q- dP)
Consequently, we get
P+Q=
(42)
[e (1–C)-d (A+B+D)]’
where e (1 – C) - d (A+B+D) > 0,
eb? (1– C)
(e+d) [(1– C)+ K4] [e (1 – C )– d K4]²
PQ=
(43)
where Ką = (A+B+D), provided C< 1. Substituting
(42) and (43) into (28), we get the condition (39). Thus,
the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F6c5de4cc-68e1-4717-b9f7-c869c8f53823%2F1ze217j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 12.If k, o are even and 1 is odd positive integers,
then Eq. (1) has prime period two solution if the condition
(3e- d) (1-C)< (e+d)(A+B+D),
(39)
is valid, provided C<1 and e (1 –C)-d(A+B+D) >
0.
Proof.If k, o are even and I is odd positive integers, then
Xn = Xn-k = Xn-o and xn+1 = Xn–1. It follows from Eq.(1)
that
bQ
P=(A+B+D) Q+CP –
(40)
(е Р- dQ)'
and
bP
Q= (A+B+D) P+CQ -
(41)
(e Q- dP)
Consequently, we get
P+Q=
(42)
[e (1–C)-d (A+B+D)]’
where e (1 – C) - d (A+B+D) > 0,
eb? (1– C)
(e+d) [(1– C)+ K4] [e (1 – C )– d K4]²
PQ=
(43)
where Ką = (A+B+D), provided C< 1. Substituting
(42) and (43) into (28), we get the condition (39). Thus,
the proof is now completed.O
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