(e) Suppose that a1 = dq+2 = d2q+3 = €3q+4 = d4q+5 = Q5q+6 = €6q+7 = 0. By (d), we produce the following formulas below: lim Ω(7q+7)n+1 n-100 αι = Ω_(7q+6) το a1 = 0 → = lim Ω(7q+6) 848 (₁ 1- ΠΩ_q+1)t-q + 1 ΠΩ-(q+1)t-q _ΠΩ-(q+1)t-qΩ-11+q ΠΩ-(n+1)t-q + 1 ΠΩ(q+1)t-q ΠΩ-(q+1)t-q + 1 Π Th 1 ΣΠ ΠΩ-(q+1)k-(q+1)t-q + 1). h=0 k=1 t=0 Th X 1 ΣΠΠΩ-+1-4+1-q+1 h=0 k=1 =0 00 Th 1 = - ΣΠ h=ok! Πιο Ω(q+1)k-(a+1)t-q +1 (3)
(e) Suppose that a1 = dq+2 = d2q+3 = €3q+4 = d4q+5 = Q5q+6 = €6q+7 = 0. By (d), we produce the following formulas below: lim Ω(7q+7)n+1 n-100 αι = Ω_(7q+6) το a1 = 0 → = lim Ω(7q+6) 848 (₁ 1- ΠΩ_q+1)t-q + 1 ΠΩ-(q+1)t-q _ΠΩ-(q+1)t-qΩ-11+q ΠΩ-(n+1)t-q + 1 ΠΩ(q+1)t-q ΠΩ-(q+1)t-q + 1 Π Th 1 ΣΠ ΠΩ-(q+1)k-(q+1)t-q + 1). h=0 k=1 t=0 Th X 1 ΣΠΠΩ-+1-4+1-q+1 h=0 k=1 =0 00 Th 1 = - ΣΠ h=ok! Πιο Ω(q+1)k-(a+1)t-q +1 (3)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the steps of determine green and information is here step by step
![In this work, we deal with the following nonlinear difference equation
m-(7g+6)
Im+l =
m = 0,1, ..,
(1)
where N-(79+6), S-(7+5), .., N-1, No e (0, 00) is investigated.
(e) If (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 (79+7)n+eg+t → argtt #0 then (7g+7)n+7q+7 + 0 as n → oo. t = 1,6.
(d) We can generate the following formulas:
179+7)n+r(a+1)+s+1 = 2r-7)(9+1)+s+1 (1-
(ITIm=1 2-(mg+m-1)+s)/{r-7)(q+1)+s+1
1+ (IIm-12-(mq+m-1)+s)
n
7h+r
1
ΣΠ
IIm-l la+1)t-(mq+m-1)+s +1
h=0 k=1
m%3D1
(e) Suppose that a1 = aq+2
a2g+3 = a39+4 = a4q+5
a5q+6 = a6g+7
0. By (d), we
produce the following formulas below:
II-o 2-(a+1)t-g
1-
lim N17g+7)n+1
= lim 2-(7q+6)
II-, 2-(a+1)t-g +1
n-00
7h
1
ΣΠ
h=0 k=1 110
2-(a+1)k-(q+1)t-q
+1
7h
II-o 2-(9+1)t-ql-11+g
00
1
a1 = 2-(79+6)
2I,2-(4+1)k-(a+1)t-q +
1 -
IT-o 2-(a+1)t-q +1
h=0 k=1
7h
II-o 2-(9+1)t-g +1
II-o 2-(a+1)t-a
1
ΣΠ
a1 = 0=
(3)
II-o P(a+1)k-(a+1)t-q +1
h=0 k=1
It=D0
6.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2Fa9f717bc-58fd-4f2e-9b19-bd30ee20ae30%2Fd44jo18_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation
m-(7g+6)
Im+l =
m = 0,1, ..,
(1)
where N-(79+6), S-(7+5), .., N-1, No e (0, 00) is investigated.
(e) If (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 (79+7)n+eg+t → argtt #0 then (7g+7)n+7q+7 + 0 as n → oo. t = 1,6.
(d) We can generate the following formulas:
179+7)n+r(a+1)+s+1 = 2r-7)(9+1)+s+1 (1-
(ITIm=1 2-(mg+m-1)+s)/{r-7)(q+1)+s+1
1+ (IIm-12-(mq+m-1)+s)
n
7h+r
1
ΣΠ
IIm-l la+1)t-(mq+m-1)+s +1
h=0 k=1
m%3D1
(e) Suppose that a1 = aq+2
a2g+3 = a39+4 = a4q+5
a5q+6 = a6g+7
0. By (d), we
produce the following formulas below:
II-o 2-(a+1)t-g
1-
lim N17g+7)n+1
= lim 2-(7q+6)
II-, 2-(a+1)t-g +1
n-00
7h
1
ΣΠ
h=0 k=1 110
2-(a+1)k-(q+1)t-q
+1
7h
II-o 2-(9+1)t-ql-11+g
00
1
a1 = 2-(79+6)
2I,2-(4+1)k-(a+1)t-q +
1 -
IT-o 2-(a+1)t-q +1
h=0 k=1
7h
II-o 2-(9+1)t-g +1
II-o 2-(a+1)t-a
1
ΣΠ
a1 = 0=
(3)
II-o P(a+1)k-(a+1)t-q +1
h=0 k=1
It=D0
6.
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