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Πο Ω_q+1)t-q + 1 ΠΩ(q+1)t-q Πο Ω_(q+1)t-q + 1 ΠoΩq+1)t-q/Ω(6q+5) thus Ω(7q+6) > Ω—(6q+5)• Th 1 = - ΣΠΠΩq+1-4+1-4+1 h=0 k=1 = o 7h+1 ΣΠ h=0 k=1 V 1 Πιο Ω(q+1)k-(q+1)t-q + 1

Πο Ω_q+1)t-q + 1 ΠΩ(q+1)t-q Πο Ω_(q+1)t-q + 1 ΠoΩq+1)t-q/Ω(6q+5) thus Ω(7q+6) > Ω—(6q+5)• Th 1 = - ΣΠΠΩq+1-4+1-4+1 h=0 k=1 = o 7h+1 ΣΠ h=0 k=1 V 1 Πιο Ω(q+1)k-(q+1)t-q + 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In this work, we deal with the following nonlinear difference equation Im-(7g+6) 1+II-o Pm-(a+1)t-g 2-1, lo E (0, 0) is investigated. Sm+1 m = 0, 1, .., (1) t%3D0 where 2-(79+6), S2-(79+5), (d) We can generate the following formulas: L(79+7)n+r(q+1)+s+1 = (r-7)(4+1)+s+1 (1– (IIm=1 2-(mg+m-1)+s)/l(r-7)(g+1)+s+1 1+ (IIm=1 -D1 -(mg+m-1)+s) 7h+r 1 ΣΠ Ilm-1 P(a+1)t-(mg+m-1)+s + 1 h=0 k=1 =D1 (e) If S2(7g+7)n+(t-1)q+t → a(t-1)q+t 70 then 279+7)n+6q+7 → 0 as n → 0, If 2(79+7)n+tq+t → atg+t +0 then N(7g+7)n+79+7 + 0 as n → 0. t = 1,6. (e) Suppose that a1 = ag+2 = a2q+3 = a3q+4 = a4g+5 a5q+6 A6q+7 0. By (d), we produce the following formulas below: lim N(79+7)n+1 lim 2-(79+6) 1 II-o 2-(q+1)t-9 -(q+1)t-q +1 It%3D0 n 7h ΣΠ 1 LII TE 2(a+1)k-(q+1)t-q † h=0 k=1 Lt30 T, N-(9+1)t-q5?-114921I ,(a+1)k-(4+1)t-9 7h 1 =N-(79+6) 1 II-o 2-(9+1)t-g+1 +1 h=0 k=1 It=D0 7h II-o 2-(9+1)t-q+1 II-0 2-(a+1)t-9 1 EIIE g+1)k-(q+1)t-q ™ + aj = 0 > (3) = h=0 k=1 t%3D0 II-o 2-(4+1)t-q/2-(6q+5). 1 lim N(79+7)n+q+2= lim 2-(6q+5) II-o 2-(9+1)t-q+1 n00 3D0 7h+1 1 ΣΠ II-, 2-(9+1)k-(q+1)t-q + h=0 k=1 It=0 II-, 2-(9+1)t-q/2-(6q+5) II-0 2-(a+1)t-q +1 aq+2 N-(64+5) 1 0o 7h+1 1 ΣΠ -(a+1)k-(q+1)t-q +1 h=0 k=1 00 7h+1 II-, 2-(4+1)t-g II-,2-(4+1)t-g/2-(6q+5) +1 1 -ΣΠ t=0 (4) II-o Pa+1)k-(9+1)t-g +1 ag+2 = 0 = =0 h=0 k=1 %3D0
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation Im-(7g+6) 1+II-o Pm-(a+1)t-g 2-1, lo E (0, 0) is investigated. Sm+1 m = 0, 1, .., (1) t%3D0 where 2-(79+6), S2-(79+5), (d) We can generate the following formulas: L(79+7)n+r(q+1)+s+1 = (r-7)(4+1)+s+1 (1– (IIm=1 2-(mg+m-1)+s)/l(r-7)(g+1)+s+1 1+ (IIm=1 -D1 -(mg+m-1)+s) 7h+r 1 ΣΠ Ilm-1 P(a+1)t-(mg+m-1)+s + 1 h=0 k=1 =D1 (e) If S2(7g+7)n+(t-1)q+t → a(t-1)q+t 70 then 279+7)n+6q+7 → 0 as n → 0, If 2(79+7)n+tq+t → atg+t +0 then N(7g+7)n+79+7 + 0 as n → 0. t = 1,6. (e) Suppose that a1 = ag+2 = a2q+3 = a3q+4 = a4g+5 a5q+6 A6q+7 0. By (d), we produce the following formulas below: lim N(79+7)n+1 lim 2-(79+6) 1 II-o 2-(q+1)t-9 -(q+1)t-q +1 It%3D0 n 7h ΣΠ 1 LII TE 2(a+1)k-(q+1)t-q † h=0 k=1 Lt30 T, N-(9+1)t-q5?-114921I ,(a+1)k-(4+1)t-9 7h 1 =N-(79+6) 1 II-o 2-(9+1)t-g+1 +1 h=0 k=1 It=D0 7h II-o 2-(9+1)t-q+1 II-0 2-(a+1)t-9 1 EIIE g+1)k-(q+1)t-q ™ + aj = 0 > (3) = h=0 k=1 t%3D0 II-o 2-(4+1)t-q/2-(6q+5). 1 lim N(79+7)n+q+2= lim 2-(6q+5) II-o 2-(9+1)t-q+1 n00 3D0 7h+1 1 ΣΠ II-, 2-(9+1)k-(q+1)t-q + h=0 k=1 It=0 II-, 2-(9+1)t-q/2-(6q+5) II-0 2-(a+1)t-q +1 aq+2 N-(64+5) 1 0o 7h+1 1 ΣΠ -(a+1)k-(q+1)t-q +1 h=0 k=1 00 7h+1 II-, 2-(4+1)t-g II-,2-(4+1)t-g/2-(6q+5) +1 1 -ΣΠ t=0 (4) II-o Pa+1)k-(9+1)t-g +1 ag+2 = 0 = =0 h=0 k=1 %3D0
From the equations (3) and (4), C. 7h II-o 2-(q+1)t-q+1 TI-, 2-(q+1)t-q ΣΠ 1 t30 2IIT g+1)k-(q+1)t-q † 1 > t=0 h=0 k=1 0o 7h+1 II-0 2-(q+1)t-q II-o 2-(4+1)t-q/S2-(6q+5) + 1 1 ΣΠ II-o P(a+1)k-(q+1)t-q +1 t=0 h=0 k=1 =0 thus 2-(79+6) > S2-(6q+5).
Transcribed Image Text:From the equations (3) and (4), C. 7h II-o 2-(q+1)t-q+1 TI-, 2-(q+1)t-q ΣΠ 1 t30 2IIT g+1)k-(q+1)t-q † 1 > t=0 h=0 k=1 0o 7h+1 II-0 2-(q+1)t-q II-o 2-(4+1)t-q/S2-(6q+5) + 1 1 ΣΠ II-o P(a+1)k-(q+1)t-q +1 t=0 h=0 k=1 =0 thus 2-(79+6) > S2-(6q+5).
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