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αι = Ω_(7q+6) @ = 0 = ΠΩ(+1)t-qΩ-11+4 Ω_(q+1)t-q ΠΩq+1)t-q + 1 Πιο Ω-(n+1)t-q Ο Th ΣΠ +1 h=0 k=1 6 1 ΠΩ-(n+1)k-(n+1)t-q + 1 th 1 ΣΠ koll ΠoΩq+1)k-(a+1)t-q + 1 (3)

αι = Ω_(7q+6) @ = 0 = ΠΩ(+1)t-qΩ-11+4 Ω_(q+1)t-q ΠΩq+1)t-q + 1 Πιο Ω-(n+1)t-q Ο Th ΣΠ +1 h=0 k=1 6 1 ΠΩ-(n+1)k-(n+1)t-q + 1 th 1 ΣΠ koll ΠoΩq+1)k-(a+1)t-q + 1 (3)

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
ChapterP: Prerequisites
SectionP.2: Real Numbers
Problem 5E
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In this work, we deal with the following nonlinear difference equation Im-(7g+6) 1+II-, Pm-(9+1)t-g where 2-(79+6), N-(7q+5), .., N-1, o E (0, 00) is investigated. Sm+1 = 0, 1,..., (1) m = (e) If L(79+7)n+(t-1)q+t → a(t-1)q+t + 0 then 2(7g+7)n+6q+7 + 0 as n → o, If S79+7)n+tg+t + arg+t +0 then 2(7g+7)n+79+7 → 0 as n -→ 0o. t = 1,6. (d) We can generate the following formulas: (7a+7)n+r(q+1)+s+1 (r-7)(g+1)+s+1 (1– (mg+m- m-1)+s)/S(r-7)(q+1)+s+1 1+ (ITm=12-(mq+m-1)+s) n 7h+r 1 ΣΠ IIm-1 Pla+1)t-(mq+m-1)+s +1 h=0 k=1
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation Im-(7g+6) 1+II-, Pm-(9+1)t-g where 2-(79+6), N-(7q+5), .., N-1, o E (0, 00) is investigated. Sm+1 = 0, 1,..., (1) m = (e) If L(79+7)n+(t-1)q+t → a(t-1)q+t + 0 then 2(7g+7)n+6q+7 + 0 as n → o, If S79+7)n+tg+t + arg+t +0 then 2(7g+7)n+79+7 → 0 as n -→ 0o. t = 1,6. (d) We can generate the following formulas: (7a+7)n+r(q+1)+s+1 (r-7)(g+1)+s+1 (1– (mg+m- m-1)+s)/S(r-7)(q+1)+s+1 1+ (ITm=12-(mq+m-1)+s) n 7h+r 1 ΣΠ IIm-1 Pla+1)t-(mq+m-1)+s +1 h=0 k=1
(e) Suppose that a1 = a4+2 = a2g+3 = a3q+4 = a4g+5 = a5q+6 = A6q+7 = 0. By (d), we produce the following formulas below: T-, 2-(a+1)t-q lim N(7g+7)n+1 = lim 2–(7q+6) 1 - II-o 2-(q+1)t–q +1 n-00 n-00 7h 1 ΣΠ II-o 2-(a+1)k-(q+1)t-q + 1 h=0 k=1 It=0 8. 7h N-(a+1)t +1)--q?=11+ 1 aj = -N-(7q+6) TI-, 2-(a+1)t-q+1 II-o 2-(q+1)k-(q+1)t-q † 1 h=0 k=1 t=0 7h II-o 2-(q+1)t-q + 1 II-o 2-(4+1)t-q ΣΠ 11-o 2(a+1)k-(q+1)t-q + 1 a1 = 0 = (3) =0 h=0 k=1 t=0 6.
Transcribed Image Text:(e) Suppose that a1 = a4+2 = a2g+3 = a3q+4 = a4g+5 = a5q+6 = A6q+7 = 0. By (d), we produce the following formulas below: T-, 2-(a+1)t-q lim N(7g+7)n+1 = lim 2–(7q+6) 1 - II-o 2-(q+1)t–q +1 n-00 n-00 7h 1 ΣΠ II-o 2-(a+1)k-(q+1)t-q + 1 h=0 k=1 It=0 8. 7h N-(a+1)t +1)--q?=11+ 1 aj = -N-(7q+6) TI-, 2-(a+1)t-q+1 II-o 2-(q+1)k-(q+1)t-q † 1 h=0 k=1 t=0 7h II-o 2-(q+1)t-q + 1 II-o 2-(4+1)t-q ΣΠ 11-o 2(a+1)k-(q+1)t-q + 1 a1 = 0 = (3) =0 h=0 k=1 t=0 6.
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