Case 3. Suppose the positive integer p,q,r are even and the positive integer s is odd. In this case Sn = Sn-p = Sn-q = Sn-r and Sn+1 = Sn-s- From Equation (1), we have (a + b) + co (d+ e) + f¢, (a+b)o+cb (d+e)o+ fw, $ = b b = $ (: Thus, (d+ e)ob + fo? = (a + b)? + cop, (6) %3D and (d+ e)ob + fp? = (a + b)o? + conp. (7) By subtracting (6) from (7), we deduce that f(6? – v²) + (a + b)(4² – v²) = 0. %3D Hence, we have (a + b+ f)(6? – v²) = 0. If ø + v, then (a + b+ f) = 0. Since a, b,andf are nonzero positive real numbers, thus (a + b+ f) # 0. This %3D
Case 3. Suppose the positive integer p,q,r are even and the positive integer s is odd. In this case Sn = Sn-p = Sn-q = Sn-r and Sn+1 = Sn-s- From Equation (1), we have (a + b) + co (d+ e) + f¢, (a+b)o+cb (d+e)o+ fw, $ = b b = $ (: Thus, (d+ e)ob + fo? = (a + b)? + cop, (6) %3D and (d+ e)ob + fp? = (a + b)o? + conp. (7) By subtracting (6) from (7), we deduce that f(6? – v²) + (a + b)(4² – v²) = 0. %3D Hence, we have (a + b+ f)(6? – v²) = 0. If ø + v, then (a + b+ f) = 0. Since a, b,andf are nonzero positive real numbers, thus (a + b+ f) # 0. This %3D
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 yellow and inf is here
Transcribed Image Text:The main aim of this study is to exhibit some cases on the periodic character of
the positive solutions of the rational difference equation
aSn-q + bSn-r + cSn-s
dSn
Sn+1 = Sn-p
(1)
+ eSn-r + fSn-s ) '
-q
where a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r,
S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such that
p > q > r > s > 0.
е,
Transcribed Image Text:Case 3. Suppose the positive integer p,q,r are even and the positive integer s is
odd. In this case
Sn = Sn-p = Sn-q = Sn-r and Sn+1
Sn-s.
From Equation (1), we have
$ = ( a+b)b + có
(d+ e) + fø,
(а + b)ф + сф
(d+ e)o+ f )
f = ¢
Thus,
(d + e)o + fo? = (a + b)y² + cop,
(6)
%3|
and
(d+e)ob + fv? = (a + b)o? + cop.
(7)
By subtracting (6) from (7), we deduce that
f(4? – v?) + (a + b)(? – y²) = 0.
Hence, we have
(a + b+ f)(@? – u²) = 0.
%3D
If ø + 4, then
(a + b+ f) =
= 0.
Since a, b,and f are nonzero positive real numbers, thus (a + b+ f) # 0. This
implies ø = v. This contradicts the hypothesis o + 4.
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