Explain the detemine blue Transcribed Image Text: Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4),

Explain the detemine blue Transcribed Image Text: Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4),
then the necessary and sufficient condition for Eq. (1.1) to have positive so-
lutions of prime period two is that the inequality
[(A+ 1) ((B1 + B3 + Bs) – (82 + Ba)) [(a1 + a3 + as) – (a2+ a4)]?
+4[(a1+a3 +a5) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0.
(4.13)
is valid.
proof: Suppose there exist positive distinctive solutions of prime period
two
…….., P, Q, Р, Q, ……..
of Eq.(1.1). From Eq.(1.1) we have
Ym+1 = Aym + 19m-1 † &2Ym-2+ &3Ym-3 + a4Ym-4 +a5Ym-5
В1Ут-1 + В2ут-2 + Bзут-з + ВАУт-4 + ВзУт-5
(a1 + a3 + a5) P+(@2+a4) Q
(B1 + B3 + B5) P+(B2 + B4) Q
(a1 + a3 + a5) Q+ (a2+ a4) P
(B1 + B3 + B5) Q+ (B2 + B4) P °
(4.14)
P = AQ+
Q = AP+
Consequently, we get
(B1 + B3 + Bs) P2 + (B2 + B4) PQ
= A (B1 + B3 + B5) PQ + A (B2 + B4) Q?
+ (a1 + a3 + a5) P+ (a2 + a4)Q,
(4.15)
and
(81 + B3 + Bs) Q² + (B2 + B4) PQ
А (B1 + Bз + B5) РQ+ A(82 + Ba) Р?
+ (a1 + a3 + a5) Q+ (a2 + a4)P.
(4.16)
By subtracting (4.15) from (4.16), we obtain
[(31 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q).
11
Since P + Q, it follows that
[(a1 + a3 + a5) – (a2+ a4)]
[(B1 + B3 + B3) + A (B2 + B4)]’
P+Q =
(4.17)
while, by adding (4.15) and (4.16) and by using the relation
p² + Q? = (P+ Q)² – 2PQ
for all
P,Q E R,
we have
[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[(B1 + B3 + B5) + A (82 + B4)]² [((B2 + B4) – (B1 + B3 + B5)) (A+ 1)]
(4.18)

PQ
Let P and Q are two distinct real roots of the quadratic equation
t2 – ( P+Q)t+ PQ = 0.
[(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1+a3 + a5) – (a2 + a4)]t
[(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + Bs) (a2 + a4) + A (82 + B4) (a1+a3 + a5)]
[(B1 + B3 + Bs) + A (B2 + B4)] [((82+ B4) – (B1 + B3 + ßs)) (A+ 1)]
= 0,
(4.19)
and so
[(a1 +a3 + a5) – (a2 + a4)]?
4 [(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[((P2 + B4) – (B1 + ß3 + B5)) (A+1)]
> 0,
or
[(a1 + a3 + as) – (a2 + a4)]?
4[(@1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)]
[((B1 + B3 + B3) – (B2 + B4)) (A+1)]
> 0.
(4.20)
From (4.20), we get
[((B1 + B3 + B5) – (B2 + B4)) (A+ 1)] [(¤1 + a3 + a5) – (a2 + a4)]?
+4 [(a1 + a3 + a5) – (a2 + a4)] [(ß1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + as)] > 0.

Therefore, the condition (4.13) is valid. Alternatively, if we imagine that the
condition (4.13) is valid where (aı+ a3 + a5) > (a2 + a4) and (B1+ B3 + B5) >
(B2 + B4). Then, we can immediately discover that the inequality stands.

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