Example 2. Let S(1) = (3.5) Then Df = R\{0} and system (2.13) becomes -1 1- a b a Xn+1 = > Yn+l = nƐ No. (3.6) Xn-1 Xn-2) yn-1 Xn-2 Here we can also assume that parameters a, b and initial values x-2, x–1, xo, y–1 and yo are complex numbers, since function (3.5) is "1 – 1" on Df = C\{0}. Clearly, function (3.5) is an involution. We see that (2.22)–(2.24) hold. Using (3.5) in (2.22)–(2.24), we obtain the general solution to system (3.6): Xn =f(f (xo)Jn+1+f(x-1)Jn+2+bf(x-2)Jn) -1 1 Jn+2+ -Jn X_2 + XoX–1X–2 n>-2, (3.7) x-|X-2Jn+1+xox-2Jn+2+bxox-¡Jn Y2n+1 =f'(f (x0)Jza+2+f (x_1) (J2n+3 – a*+!) +bf (x_2)Jzn+1 +a*+!f(v_1)) b. J2n+1 X-2 1 X-I (x-1X-2y-1J2n+2+xax-2y–1 (J2n+3 – a*+!) Xox–1X–2y–1 bxox-1y-1J2n+1+a*+lxox=1x_2 ,n>-1, (3.8) xoX_1X_2Y–1 y2n+2 =f'(f (x0) (J2n+3- a*+!) +f (x-1)zn+4 +bf (x-2)J2n+2+a**!f (yo)) -1 1 J2n+4+ X-I an+l b -J2n+2+ X-2 J2n+3 – yo (x-1X-2yo (J2n+3– a*+!) +xox-2yoJ2n+4 xox–1X_2Y0 -1 bxox-1YoJ2n+2 +a"+lxox-1x_2 ,n>-1. (3.9) XoX-1X–2Y0
Example 2. Let S(1) = (3.5) Then Df = R\{0} and system (2.13) becomes -1 1- a b a Xn+1 = > Yn+l = nƐ No. (3.6) Xn-1 Xn-2) yn-1 Xn-2 Here we can also assume that parameters a, b and initial values x-2, x–1, xo, y–1 and yo are complex numbers, since function (3.5) is "1 – 1" on Df = C\{0}. Clearly, function (3.5) is an involution. We see that (2.22)–(2.24) hold. Using (3.5) in (2.22)–(2.24), we obtain the general solution to system (3.6): Xn =f(f (xo)Jn+1+f(x-1)Jn+2+bf(x-2)Jn) -1 1 Jn+2+ -Jn X_2 + XoX–1X–2 n>-2, (3.7) x-|X-2Jn+1+xox-2Jn+2+bxox-¡Jn Y2n+1 =f'(f (x0)Jza+2+f (x_1) (J2n+3 – a*+!) +bf (x_2)Jzn+1 +a*+!f(v_1)) b. J2n+1 X-2 1 X-I (x-1X-2y-1J2n+2+xax-2y–1 (J2n+3 – a*+!) Xox–1X–2y–1 bxox-1y-1J2n+1+a*+lxox=1x_2 ,n>-1, (3.8) xoX_1X_2Y–1 y2n+2 =f'(f (x0) (J2n+3- a*+!) +f (x-1)zn+4 +bf (x-2)J2n+2+a**!f (yo)) -1 1 J2n+4+ X-I an+l b -J2n+2+ X-2 J2n+3 – yo (x-1X-2yo (J2n+3– a*+!) +xox-2yoJ2n+4 xox–1X_2Y0 -1 bxox-1YoJ2n+2 +a"+lxox-1x_2 ,n>-1. (3.9) XoX-1X–2Y0
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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Show me the steps of determine yellow and all information is here
![2.4. Case 4: Pn= xn, qn = xn, rn = Yn, Sn = Xn
In this case, system (1.8) is written as in the form
Xn+1 = f(af (xn=1)+bf (xp–2)), yn+1=f [af (yn-1)+bf (xn–2)], (2.13)
for n E No. Since f is "1 – 1", from (2.13)
f (Xn+1) = af (xn-1)+bf (xn-2), f (Vn+1) = af (yn-1)+bf (Xn-2),
(2.14)
for n E No. By using the change of variables
f (xn) = Un, n>-2, and f (yn) =vn, n>-1,
(2.15)
system (2.14) is transformed to the following one
Иn+1 — аил—1 +bun-2, Vn+1 %3 аvn-1 +bum-2,
(2.16)
for n E No. From (2.5), we can write the solution of the first equation in (2.16) as
Un = U0JN+1+u-1Jn+2+bu-2Jn, n E No.
(2.17)
By subtracting the second one from the first equations in (2.16), we have
Un+1 - Vn+1
a (un-1 - Vn-1), n E No.
(2.18)
From (2.18) we see that the sequence (un – Vn)n>-1 satisfies the difference equation
Wn = awn-2, n> 1,
(2.19)
from which it follows that
u2n+i – V2n+i = a"+' (ui-2 – Vi-2),
(2.20)
for n E No, i e {1,2}.
From (2.17) and (2.20), we get
V2n+i = U2nti -
ui-2+a"+'vi-2,
= uoJ2n+i+1+u_1J21+i+2+bu_2J2n+i-a"+'uj-2 +a*+!vi_2,
(2.21)
for n E No, i e {1,2}.
Employing (2.17) and (2.21) in (2.15) and after some calculation, we obtain
Xp = f(f (x0)Jn+1+f(x_1)Jn+2+bf (x_2)Jn), n> -2,
(2.22)
y2n+1 = f(f (x0)Jzn+2+ f (x=1) (J2n+3 – a**') +bf (x_2)J2n+1
+a*+!f (y-1)), n >-1,
%3D
-
(2.23)
y2n+2 = f-| (f (xo) (J2n+3 – a*+!) + f (x_1)J2n+4+bf (x_2)J2n+2
+a"+!f (yo)), n >-1.
(2.24)
Example 2. Let
(3.5)
Then Df = R\{0} and system (2.13) becomes
-1
-1
b
a
+
Xn-2)
a
+
Xn-2
Xn+1 =
• Yn+1 =
, nɛ No.
(3.6)
Xn-1
Here we can also assume that parameters a, b and initial values x-2, x-1, X0, y-1 and
yo are complex numbers, since function (3.5) is "1 – 1" on Df = C\{0}.
Clearly, function (3.5) is an involution. We see that (2.22)-(2.24) hold. Using
(3.5) in (2.22)-(2.24), we obtain the general solution to system (3.6):
Xn =f(f (xo) Jn+1+f(x-1)Jn+2+bf (x-2)Jn)
1
b
-Jn+2+
-Jn
x-2
Xox-1X-2
n>-2,
(3.7)
x-1X-2Jn+1+xox_2Jn+2+bxox-1Jn
y2n+1 =f'(S (x0)Jza+2+f (x-1) (J2n+3 – a*+!) +bf (x_2)J2n+1+a*+!f(v-1))
1
= an+2+ (an+3 -d**) +;
b
-J2n+1+
X-2
X-1
y-1,
(x-1x-2y-1J2n+2+xox–2y–1 (J2n+3 – d"+1)
Xox-1X-2y–1
-1
bxox-1y-1J21+1ta+!xox_1x_2
,n>-1,
(3.8)
xox-1X-2Y–1
y2n+2 = f'(f (x0) (J2n+3- a+') +f (x-1)J2n+4 +bf (x-2)J2n+2 +a*+'f (vo))
1
J2n+4+
X-1
b
-J2n+2+
x-2
an+l -I
Yo
(x-1X-2yo (J2n+3 – a"+') +xox-2y0J2n+4
xox-1X-2Y0
bxox-1Yo/2n+2+a"+lxox-1x_2
, n>-1.
(3.9)
Xox-1X-2y0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d6994c6-9c0b-4312-919e-29acb733a883%2F7187b2ca-583d-407f-b5f2-24b49f572383%2Fdp9azv6_processed.png&w=3840&q=75)
Transcribed Image Text:2.4. Case 4: Pn= xn, qn = xn, rn = Yn, Sn = Xn
In this case, system (1.8) is written as in the form
Xn+1 = f(af (xn=1)+bf (xp–2)), yn+1=f [af (yn-1)+bf (xn–2)], (2.13)
for n E No. Since f is "1 – 1", from (2.13)
f (Xn+1) = af (xn-1)+bf (xn-2), f (Vn+1) = af (yn-1)+bf (Xn-2),
(2.14)
for n E No. By using the change of variables
f (xn) = Un, n>-2, and f (yn) =vn, n>-1,
(2.15)
system (2.14) is transformed to the following one
Иn+1 — аил—1 +bun-2, Vn+1 %3 аvn-1 +bum-2,
(2.16)
for n E No. From (2.5), we can write the solution of the first equation in (2.16) as
Un = U0JN+1+u-1Jn+2+bu-2Jn, n E No.
(2.17)
By subtracting the second one from the first equations in (2.16), we have
Un+1 - Vn+1
a (un-1 - Vn-1), n E No.
(2.18)
From (2.18) we see that the sequence (un – Vn)n>-1 satisfies the difference equation
Wn = awn-2, n> 1,
(2.19)
from which it follows that
u2n+i – V2n+i = a"+' (ui-2 – Vi-2),
(2.20)
for n E No, i e {1,2}.
From (2.17) and (2.20), we get
V2n+i = U2nti -
ui-2+a"+'vi-2,
= uoJ2n+i+1+u_1J21+i+2+bu_2J2n+i-a"+'uj-2 +a*+!vi_2,
(2.21)
for n E No, i e {1,2}.
Employing (2.17) and (2.21) in (2.15) and after some calculation, we obtain
Xp = f(f (x0)Jn+1+f(x_1)Jn+2+bf (x_2)Jn), n> -2,
(2.22)
y2n+1 = f(f (x0)Jzn+2+ f (x=1) (J2n+3 – a**') +bf (x_2)J2n+1
+a*+!f (y-1)), n >-1,
%3D
-
(2.23)
y2n+2 = f-| (f (xo) (J2n+3 – a*+!) + f (x_1)J2n+4+bf (x_2)J2n+2
+a"+!f (yo)), n >-1.
(2.24)
Example 2. Let
(3.5)
Then Df = R\{0} and system (2.13) becomes
-1
-1
b
a
+
Xn-2)
a
+
Xn-2
Xn+1 =
• Yn+1 =
, nɛ No.
(3.6)
Xn-1
Here we can also assume that parameters a, b and initial values x-2, x-1, X0, y-1 and
yo are complex numbers, since function (3.5) is "1 – 1" on Df = C\{0}.
Clearly, function (3.5) is an involution. We see that (2.22)-(2.24) hold. Using
(3.5) in (2.22)-(2.24), we obtain the general solution to system (3.6):
Xn =f(f (xo) Jn+1+f(x-1)Jn+2+bf (x-2)Jn)
1
b
-Jn+2+
-Jn
x-2
Xox-1X-2
n>-2,
(3.7)
x-1X-2Jn+1+xox_2Jn+2+bxox-1Jn
y2n+1 =f'(S (x0)Jza+2+f (x-1) (J2n+3 – a*+!) +bf (x_2)J2n+1+a*+!f(v-1))
1
= an+2+ (an+3 -d**) +;
b
-J2n+1+
X-2
X-1
y-1,
(x-1x-2y-1J2n+2+xox–2y–1 (J2n+3 – d"+1)
Xox-1X-2y–1
-1
bxox-1y-1J21+1ta+!xox_1x_2
,n>-1,
(3.8)
xox-1X-2Y–1
y2n+2 = f'(f (x0) (J2n+3- a+') +f (x-1)J2n+4 +bf (x-2)J2n+2 +a*+'f (vo))
1
J2n+4+
X-1
b
-J2n+2+
x-2
an+l -I
Yo
(x-1X-2yo (J2n+3 – a"+') +xox-2y0J2n+4
xox-1X-2Y0
bxox-1Yo/2n+2+a"+lxox-1x_2
, n>-1.
(3.9)
Xox-1X-2y0
Transcribed Image Text:constant coefficients
Xn+1 = axn+bxn-1+cxn-2, nE No,
(1.4)
which has actually the general solution
X = xoSn+x-1 (Sn+1 - aSn) +cx_2Sn-1, nE No,
(1.5)
where (S.) of equation (1L4) satisfving the initial values S 2=S-=0, So= L.
The equation
axn-IXn-k
Xn+1 =
nE No,
(1.1)
bxn-p±cxn-q'
where the initial conditions are arbitrary positive real numbers, k, 1, p, q are non-
negative integers and a, b, c are positive constants, is one of the difference equations
whose solutions are associated with number sequences. Positive solutions of concrete
Motivated by this line of investigations, here we show that the systems of differ-
ence equations
Xn+l =f'(af (Pn-1)+bf(qn-2)), Yntl =f"(af (ra-1)+bf (Sn-2)), (1.8)
for n E No, where the sequences Pn, qn, Tn and Sn are some of the sequences x, and
Yn, f: Df R is a "1– 1" continuous function on its domain Dr C R, the initial
values x-j, y-j, je {0,1,2} are arbitrary real numbers and the parameters and a, b
2.6. Case 6: Pn = Yn, qn = yn, rn = Xn, Sn = Xn
In this case, we obtain the system
Xn+1 =5"(af (Yn-1)+bf (Yn-2)), Yn+1 = '(af (xn-1)+bf (xn-2)), (2.29)
for n e No. Since f is "1– 1", from (2.29)
f (Xn+1) = af (Yn-1)+bf (yn-2), f(yn+l) = af (Xn-1) +bf (xn-2),
(2.30)
for n E No. By using the change of variables
f (xn) =
= Un,
and f(yn) = Vn, n>-2,
(2.31)
system (2.30) is transformed to the following one
Un+l = avn-1+bvn-2, Vn+1 = aun-1+ bun-2, nE No.
(2.32)
By summing the equations in (2.32) we get
Un+l +Vn+l = a (un-1+ Vn-1) +b(un-2+Vn-2), nE No,
(2.33)
whereas by subtracting the second one from the first, we have
Un+l - Vn+l = -a(un-1 - Vn-1) -6(un-2- Vn-2), ne No.
(2.34)
|
In this section, we consider the eight special cases of systems (1.8), where the
sequences Pn, qn, ľn, Sn are some of the sequences xn and yn, for n>-2, and initial
values x-j, y-j, j e {0,1,2}, are arbitrary real numbers.
2.1
Case 1: Pn =xn, qn = Xn, ľn = yn, Sn == yn
In this case, system (1.8) is expressed as
Xn+1 =f (af (xn-1)+bf (xn-2)), Yn+1 =f(af (Vn-1)+bf (Yn-2)), (2.1)
for n E No. Since f is "1– 1", from (2.1)
f (Xn+1) = af (x,-1)+bf (xn-2), f (Yn+1) = af (yn-1)+bf (yn–2),
(2.2)
for n E No. By using the change of variables
f (xn)
= Un, and f(yn) =
= Vn, n> -2,
(2.3)
system (2.2) is transformed to the following one
Un+1 = au,-1+bun-2, Vn+1 = avn-1+bvn-2,
(2.4)
for n E No. By taking a = 0, b= a, c = b in (1.4) and S = Jn+1, for all n > -2,
which is called generalized Padovan sequence, in (1.5), the solutions to equations in
(2.4) are given by
Un = ugJn+1+u-1Jn+2+ bu_2Jn,
(2.5)
Vn = voJn+1+v_1Jn+2+bv_2Jn,
(2.6)
for n E No. From (2.3), (2.5) and (2.6), it follows that the general solution to system
(2.2) is given by
Xn = f(f (x0) Jn+1+f (x-1) Jn+2+bf (x-2)Jn), n> -2,
Yn =f(f (vo) Jn+1+f(y-1)Ja+2+bf (y-2) J.), n> -2.
(2.7)
(2.8)
2.2. Case 2: Pn = Xn, qn = Xn, ľn = Xn, Sn Xn
In this case, system (1.8) becomes
Xp+1 =f"(af (x,-1)+bf (xp-2)), Yn+1=f"(af (xn-1)+bf (xn-2)), (2.9)
for n e No. It should be first note that from the equations in (2.9) it immediately
follows that x, = yn, for all n E N. From (2.7), the general solution to system (2.9) is
%3D
X, = Ya = f'(f (xo) Jn+1+f (x=1) Jn+2+bf (x_2)Jn), n E N.
(2.10)
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