Explain the detemine blue 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 periodtwo........, P, Q, Р, Q, ........of Eq.(1.1). From Eq.(1.1) we haveYm+1 = Aym + 19m-1 † &2Ym-2+ &3Ym-3 + a4Ym-4 +a5Ym-5В1Ут-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) PQА (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).11Since 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 relationp² + Q? = (P+ Q)² – 2PQfor allP,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)-PQLet P and Q are two distinct real roots of the quadratic equationt2 – ( 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 thecondition (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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Description: Explain the detemine blue 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) –

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+ a3 + a5) – (@2 + đ4)I 4 [(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + Ba) (a1 + a3 + a5)] [((B1 + B3 + Bs) – (32 + B4)) (A+1)] > 0. (4.20) From (4.20), we get [((B1 + B3 + Bs) – (B2 + B4)) (A+ 1) [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + as)] > 0. - Therefore the condition (4.13) is valid. Alternatively, if we imagime that the

+ a3 + a5) – (@2 + đ4)I 4 [(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + Ba) (a1 + a3 + a5)] [((B1 + B3 + Bs) – (32 + B4)) (A+1)] > 0. (4.20) From (4.20), we get [((B1 + B3 + Bs) – (B2 + B4)) (A+ 1) [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + as)] > 0. - Therefore the condition (4.13) is valid. Alternatively, if we imagime that the

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Question

Explain the detemine blue

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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