Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn Sn-r = Sn-s and Sn+1 Sn-p = Sn-q. From Equation (1), we have аф + (b + c) $ = ¢ dø + (e + f)½ arb + (b+ c)o dp + (e + f)ø, b = b Thus, do? + (e + f)øv½ = ao² + (b + c)ø½, (8) and dy? + (e+ f)ø½ = av² + (b+ c)ørp. (9) By subtracting (8) from (9), we deduce that Cd(o? – ?) – a(² – v²) = 0. Hence, we have [d – a](o? – v²) = 0. Since a and d are nonzero positive real numbers, and ø # p. This implies [d = a], This contradicts the condition [d + a].
Case 7. Suppose the positive integers r, s are even and the positive integers p, q are odd. In this case Sn Sn-r = Sn-s and Sn+1 Sn-p = Sn-q. From Equation (1), we have аф + (b + c) $ = ¢ dø + (e + f)½ arb + (b+ c)o dp + (e + f)ø, b = b Thus, do? + (e + f)øv½ = ao² + (b + c)ø½, (8) and dy? + (e+ f)ø½ = av² + (b+ c)ørp. (9) By subtracting (8) from (9), we deduce that Cd(o? – ?) – a(² – v²) = 0. Hence, we have [d – a](o? – v²) = 0. Since a and d are nonzero positive real numbers, and ø # p. This implies [d = a], This contradicts the condition [d + a].
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 red and inf is here
 = 0.
[d-
Since a and d are nonzero positive real numbers, and ø ½.
This implies [d = a], This contradicts the condition [d + a].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2Fa163bbb1-015c-4b95-b143-08aa8ff6e6f9%2Fy4h7vzn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Case 7. Suppose the positive integers r, s are even and the positive integers p, q are
odd. In this case
Sn = Sn-r = Sn-s and Sn+1 = Sn-p
= Sn-q•
From Equation (1), we have
аф + (b + с)
dф + (е + f)4,
$ = ¢
ab + (b+ c)ø
dy + (e + f)¢,
V = b
Thus,
dø? + (e + f)ø½ = ao² + (b + c)øv½,
(8)
and
du² + (e + f)øv = ay² + (b+ c)øp.
(9)
By subtracting (8) from (9), we deduce that
Ca(o? – 2) – a(6² – v²) = 0.
Hence, we have
Çu-
((d – a](o? – v²) = 0.
[d-
Since a and d are nonzero positive real numbers, and ø ½.
This implies [d = a], This contradicts the condition [d + a].
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,
Expert Solution
Step 1
Here we need to explain the mentioned steps.
Step by step
Solved in 2 steps with 1 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
