P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).
P19m- P59m- (a1 + a3 + a5) P+ (a2 + a4) Q (B1 + B3 + B5) P + (82 + B4) Q ' (a1 + a3 + a5) Q+ (a2 + a4) P (B1 + B3 + B5) Q + (82 + B4) P ` (4.14) P = AQ+ Q = AP+ Consequently, we get A (B1 + B3 + Bs) PQ + A (32 + B4) Q² + (a1 + a3 + a5) P+(a2 + a4) Q, (B1 + B3 + B5) P² + (82 + B4) PQ (4.15) and A (B1 + 3 + B5) PQ + A (B2 + B4) P? + (a1 + a3 + a5) Q + (a2 + a4) P. (B1 + B3 + B5) Q² + (B2 + B4) PQ %3D (4.16) By subtracting (4.15) from (4.16), we obtain [(B1 + B3 + B5) + A (32 + B4)] (P² – Q²) = [(a1 + a3 + a5) – (a2 + a4)] (P –- Q).
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 80E
Related questions
Topic Video
Question
Explain the determine yellow and the inf is here
![Theorem 6 If (a1+ аз + a5) > (02 + ол) аnd (Bі + Вз + Bъ) > (32 + Ba),
then the necessary and sufficient condition for Eq.(1.1) to have positive so-
lutions of prime period two is that the inequality
[(A + 1) ((B1 + B3 + B5) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]?
+4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (82 + B4) (a1 + a3 + 5)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
P,Q, P, Q, ..
of Eq.(1.1). From Eq.(1.1) we have
a1Ym-1+a2Ym-2+ a3Ym-3+a4Ym-4+a5Ym-5
Ym+1 =
Aym +
B1Ym-1 + B2Ym-2 + B3Ym–3 + B4Ym-4 + B5Ym–5
(a1 + az + a5) P+(a2+a4) Q
Q =
(B1 + B3 + B5) P + (82 + B4) Q ’
(a1 + az + a5) Q+(a2 + a4) P
Р— АQ+
= AP+
(B1 + B3 + B5) Q+ (B2 + B4) P *
(4.14)
Consequently, we get
(B1 + B3 + B5) P² + (B2 + B4) PQ
A (B1 + B3 + B5) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P+ (a2 + a4) Q,
(4.15)
and
(B1 + 33 + B5) Q² + (82 + B4) PQ
A (B1 + B3 + B5) PQ + A (ß2 + B4) P²
+ (a1 + a3 + a5) Q + (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(81 + B3 + B5) + A (B2 + B4)] (P² –- Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q).
11](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2Fad52fab8-63f8-4db3-a18f-8d2ed2c37eff%2Fko1mw19_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 6 If (a1+ аз + a5) > (02 + ол) аnd (Bі + Вз + Bъ) > (32 + Ba),
then the necessary and sufficient condition for Eq.(1.1) to have positive so-
lutions of prime period two is that the inequality
[(A + 1) ((B1 + B3 + B5) – (B2 + B4))] [(@1 + a3 + a5) – (a2 + a4)]?
+4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (82 + B4) (a1 + a3 + 5)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
P,Q, P, Q, ..
of Eq.(1.1). From Eq.(1.1) we have
a1Ym-1+a2Ym-2+ a3Ym-3+a4Ym-4+a5Ym-5
Ym+1 =
Aym +
B1Ym-1 + B2Ym-2 + B3Ym–3 + B4Ym-4 + B5Ym–5
(a1 + az + a5) P+(a2+a4) Q
Q =
(B1 + B3 + B5) P + (82 + B4) Q ’
(a1 + az + a5) Q+(a2 + a4) P
Р— АQ+
= AP+
(B1 + B3 + B5) Q+ (B2 + B4) P *
(4.14)
Consequently, we get
(B1 + B3 + B5) P² + (B2 + B4) PQ
A (B1 + B3 + B5) PQ + A (B2 + B4) Q²
+ (a1 + a3 + a5) P+ (a2 + a4) Q,
(4.15)
and
(B1 + 33 + B5) Q² + (82 + B4) PQ
A (B1 + B3 + B5) PQ + A (ß2 + B4) P²
+ (a1 + a3 + a5) Q + (a2 + a4) P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(81 + B3 + B5) + A (B2 + B4)] (P² –- Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q).
11
![The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+ a2Ym-2 + a3Ym-3 + a4Ym-4+ a5Ym-5
Ym+1 =
Aym+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2Fad52fab8-63f8-4db3-a18f-8d2ed2c37eff%2Fwvhkfj4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+ a2Ym-2 + a3Ym-3 + a4Ym-4+ a5Ym-5
Ym+1 =
Aym+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
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 with 4 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage