From the equations (3) and (4), 02-(q+1)t-q +1 -o-(g+1)t-q ∞ 7h 1 =Σll²_o 2(g+1)k-(q+1)t−g +1 > h=0 k=1 thus -(79+6) >_(6q+5). = 02-(q+1)t-q+1 02-(g+1)t-q/-(6q+5) = ∞ 7h+1 1 Σ²_o(q+1)k-(q+1)t-q + 1² +1 h=0 k=1
From the equations (3) and (4), 02-(q+1)t-q +1 -o-(g+1)t-q ∞ 7h 1 =Σll²_o 2(g+1)k-(q+1)t−g +1 > h=0 k=1 thus -(79+6) >_(6q+5). = 02-(q+1)t-q+1 02-(g+1)t-q/-(6q+5) = ∞ 7h+1 1 Σ²_o(q+1)k-(q+1)t-q + 1² +1 h=0 k=1
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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Transcribed Image Text:In this work, we deal with the following nonlinear difference equation
Im-(7g+6)
1+II, Pm-(a+1)t-g
where N-(7g+6), -(79+5), ..., N-1, No e (0, 00) is investigated.
Im+1 =
m = 0, 1, ...,
(1)
(e) If 2(79+7)n+(t-1)q+t → a(t-1)g+t 70 then 2(79+7)n+6q+7 →0 as n → o,
If 2(79+7)n+tq+t → arg+t 70 then 2(7g+7)n+7q+7 →0 as n → oo. t = 1,6.
(d) We can generate the following formulas:
(7a+7)n+r(a+1)+s+1 = 2(r-7)(g+1)+s+1 (1–
(ITm-1 2-(mg+m-1)+s)/2r-7)(a+1)
1+ (IIm=1 2-(mq+m-1)+s)
n 7h+r
1
ΣΠ
Im=1 ?(a+1)t-(mq+m-1)+s +1
h=0 k=1
(e) Suppose that a1 = ag+2 = a2g+3
a3a+4
d4q+5 = a5q+6 = A6q+7 = 0. By (d), we
produce the following formulas below:
II-,2-(9+1)t-a
IIL,2-(a+1)t-g +1
lim N7a+7)n+1
lim 2-(79+6)
%3D
n-00
7h
EII T2-(a+1)k-(a+1)t-q
+
h=0 k=1
(-
7h
II-o 2-(9+1)t-g2-11+a
IIL,2-(a+1)t-, +1
00
1
a1 = 2-(79+6)
2I T-(a+1)k-(a+1)t-q +
h=0 k=1
00 7h
II-,2-(9+1)t-g+1
II-, 2-(a+1)-g
1
El T ot1)k-(a+1)t-a † +
(3)
a1 = 0 =
h=0 k=1 1It=0
6.
lim N(79+7)n+g+2
= lim 2-(64+5)
n-00
7h+1
1
ΣΠ
II-,n-(a+1)k-(4+1)t-q
+1
h=0 k=1
II-, 2-(a+1)t-q/2-(6a+5).
1
Ag+2 =
II-o 2-(9+1)t-g +1
o 7h+1
1
211 T2-(e+1)k-(s+1)t-q + 1
h=0 k=1
00 7h+1
II-,2-(a+1)t-g +1
ΣΠ
E11 TÉ e+1)k-(a+1)t-q +1
1
ag+2 = 0 >
(4)
%3D
h=0 k=1
Transcribed Image Text:00 7h+5
II-o 2-(a+1)t-q+1
II-, 2-(q+1)t-q/S2-(2q+1)
1
= 0 =
-ΣΠ
(8)
II-, P(a+1)k-(q+1)t-q +1
A5q+6
t=0
h=0 k=1
Similarly
1}e-q/l-(2g+1)
lim N(7q+7)n+6q+7
= lim 2-(9) ( 1
t=0
2-(a+1)t-q
+1
7h+6
n
1
2-(q+1)k-(q+1)t-q
+1
h=0 k=1
7h+6
II-, 2-(4+1)t-q/2-(4)
1
ΣΠ
II-o 2-(q+1)t-q +1
t30
A6g+7 =
II-0 2-(a+1)k-(q+1)t-q † 1
13D0
h=0 k=1
00 7h+6
II-, 2-(a+1)t-q +1
II, 2-(a+1)t-q/2-(9)
1
It=0
A6q+7
= 0 =
ΣΠ
(9)
II-, P(a+1)k-(q+1)t-q +1
t30
h=0 k=1
From the equations (3) and (4),
7h
II-o 2-(7+1)t-q+1
II-, 2-(a+1)t-q
ΣΠ
>
ITI-o P(a+1)k-(q+1)t-q +1
h=0 k=1
t=0
00 7h+1
II-o 2-(q+1)t-q+1
II-, 2-(a+1)t-q/2-(6q+5)
1
ΣΠ
t=0
II-o P(a+1)k-(q+1)t-q +1
=0
h=0 k=1
thus 2-(79+6) > S2-(6q+5)·
From the equations (4) and (5),
o 7h+1
II-o 2-(q+1)t-q+1
II-, 2/N-(6g+5)
1
-ΣΠ
t=0
>
-(q+1)t-q/
II-o Pla+1)k-(q+1)t-q + 1
=0
h=0 k=1
o 7h+2
II-o 2-(q+1)t-q+1
N-(q+1)t-q/S2-(5q+4)
1
ΣΠ
=0
II-, P(a+1)k-(q+1)t-q + 1
h=0 k=1
=0
thus N-(6q+5) > S2-(5q+4)·
From the equations (5) and (6),
00 7h+2
II-, -(a+1)t-q +1
II-o 2-(q+1)t-q/2-(5q+4)
1
ΣΠ
II-o P(a+1)k-(q+1)t-q +1
t=D0
>
h=0 k=1
9.
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