Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
< Yn < Zn for all n > N; Encreasing for all n>N. that {yn} is a positive solution of equation ( {Zn} € So. Then z, is positive, zn 2 Yn, and Yn = Zn - Pnyo(n) 2 (1– p)zn, n>N> no

< Yn < Zn for all n > N; Encreasing for all n>N. that {yn} is a positive solution of equation ( {Zn} € So. Then z, is positive, zn 2 Yn, and Yn = Zn - Pnyo(n) 2 (1– p)zn, n>N> no

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 43RE
See similar textbooks
icon
Related questions
Question

Show me the steps of determine blue and all information is here step by step

Lemma 4. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z„} e So for n > N. Then: (i) (1– p)zn < yn < Zn for all n > N; (ii) {zn¢n} is increasing for all n > N. Proof. Assume that {yn} is a positive solution of equation (1.1) with the corres- ponding sequence {zn} E So. Then z, is positive, Zn 2 yn, and Yn = Zn – Pnyo(n) 2 (1– p)zn, n>N> no, so (i) is proved. It easy to see that z, E So implies lim b,(Az,)“ = 0; n00 otherwise we would eventually have Az, > 0 contradicting z, E So. Similarly, lim a„A(b,(Azn)“) = 0. A summation of equation (1.1) then yields 9szs+1 Zn+1 LIs. s=n s=n s=n Summing once more, we obtain as t=s s=n or Azn 2-Zn+1Qn. Hence, A(zn&n) = PnAzn + Zn+1Ao, > Zn+1(A0n – OnQn) =0 since {0,} is a solution of the difference equation (Ao, – Qnºn) = 0. Therefore, {z,9n} is increasing and this completes the proof of the lemma.
Transcribed Image Text:Lemma 4. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z„} e So for n > N. Then: (i) (1– p)zn < yn < Zn for all n > N; (ii) {zn¢n} is increasing for all n > N. Proof. Assume that {yn} is a positive solution of equation (1.1) with the corres- ponding sequence {zn} E So. Then z, is positive, Zn 2 yn, and Yn = Zn – Pnyo(n) 2 (1– p)zn, n>N> no, so (i) is proved. It easy to see that z, E So implies lim b,(Az,)“ = 0; n00 otherwise we would eventually have Az, > 0 contradicting z, E So. Similarly, lim a„A(b,(Azn)“) = 0. A summation of equation (1.1) then yields 9szs+1 Zn+1 LIs. s=n s=n s=n Summing once more, we obtain as t=s s=n or Azn 2-Zn+1Qn. Hence, A(zn&n) = PnAzn + Zn+1Ao, > Zn+1(A0n – OnQn) =0 since {0,} is a solution of the difference equation (Ao, – Qnºn) = 0. Therefore, {z,9n} is increasing and this completes the proof of the lemma.
In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)) +9ny%+1=0, n> no > 0, (1.1) where zn = yn+PnYo(n), ɑ is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n > no; (H2) {Pn} is a nonnegative real sequence with 0< Pn <p< 1; (H3) {o(n)} is a sequence of integers such that o(n) > n for all n > no; (H4) Ln=no an = +00 and E=no Va =+∞.
Transcribed Image Text:In this paper, we are concerned with the asymptotic properties of solutions of the third order neutral difference equation A(a,A(b,(Azn)“)) +9ny%+1=0, n> no > 0, (1.1) where zn = yn+PnYo(n), ɑ is the ratio of odd positive integers, and the following conditions are assumed to hold throughout: (H1) {an}, {bn}, and {qn} are positive real sequences for all n > no; (H2) {Pn} is a nonnegative real sequence with 0< Pn <p< 1; (H3) {o(n)} is a sequence of integers such that o(n) > n for all n > no; (H4) Ln=no an = +00 and E=no Va =+∞.
Expert Solution
steps

Step by step

Solved in 2 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
SEE MORE TEXTBOOKS