Performing iterations on (25) respectively, leads to the following: n-1 X2km+j v; II n-1 Y2km+j U2kn+j u; II and v2kn+j (26) X2km+k+j Y2km+k+j m=0 m=0 4 where j= 0,1,2,...,2k – 1. The subscript of X and Y in (24) is 2n + 1, where i = 0,1 but the ones of X and Y in (26) are 2km + j and 2km +k+j, where j = 0,1,2,...,2k – 1. So we want to write 2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder when r is divided by k. Hence, 2km + j= 2( km + (27a) ()2+ and k+ 2km + k+j = 2( km + 2 +τ (k+j). (27b) Now, (27) is of the form similar to 2n + i, as t(j) and t(k + j) will either be 0 or 1. Substituting (27) into (26) leads to the following: Y2(km+l½])+T(j) П m=0 ^2(km+[ )+t(k+j) n-1 U2kn+j = Uj (28a) X. and X. 2(km+L½])+T(j) п-1 V2kn+j v; II (28b) m=0 2(km+[)+r(k+j) Y k+j Using (24) and (28), we get the following: п-1 U2kn+j uj II m=0 km+ km+ km+ -1 II C2k1+T(j) k=0 幻-1 I d21+7(j), I C2k+t(j) 1=0 k2=l+1 km+ k+j 2. km+ km+ L-1 X-(k+j) II a2k1+T(k+j) k1=0 b21+7(k+j), II a2kz+7(k+j) k2=l+1 + Xr(k+j) 1=0
Performing iterations on (25) respectively, leads to the following: n-1 X2km+j v; II n-1 Y2km+j U2kn+j u; II and v2kn+j (26) X2km+k+j Y2km+k+j m=0 m=0 4 where j= 0,1,2,...,2k – 1. The subscript of X and Y in (24) is 2n + 1, where i = 0,1 but the ones of X and Y in (26) are 2km + j and 2km +k+j, where j = 0,1,2,...,2k – 1. So we want to write 2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder when r is divided by k. Hence, 2km + j= 2( km + (27a) ()2+ and k+ 2km + k+j = 2( km + 2 +τ (k+j). (27b) Now, (27) is of the form similar to 2n + i, as t(j) and t(k + j) will either be 0 or 1. Substituting (27) into (26) leads to the following: Y2(km+l½])+T(j) П m=0 ^2(km+[ )+t(k+j) n-1 U2kn+j = Uj (28a) X. and X. 2(km+L½])+T(j) п-1 V2kn+j v; II (28b) m=0 2(km+[)+r(k+j) Y k+j Using (24) and (28), we get the following: п-1 U2kn+j uj II m=0 km+ km+ km+ -1 II C2k1+T(j) k=0 幻-1 I d21+7(j), I C2k+t(j) 1=0 k2=l+1 km+ k+j 2. km+ km+ L-1 X-(k+j) II a2k1+T(k+j) k1=0 b21+7(k+j), II a2kz+7(k+j) k2=l+1 + Xr(k+j) 1=0
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 determine yellow and all information is here
![Performing iterations on (25) respectively, leads to the following:
n-1 X2km+j
П
Vj
m=0 Y2km+k+j
п—1
Y2km+j
and
U2kn+j = U; |
m=0
(26)
V2kn+j
X2km+k+j
where j = 0,1,2,...,2k – 1.
The subscript of X and Y in (24) is 2n + 1, where i =
in (26) are 2km + j and 2km +k+j, where j = 0,1,2,...,2k – 1. So we want to write
2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r
can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder
when r is divided by k. Hence,
34
0,1 but the ones of X and Y
2km + j = 2( km + |) +
+ t(j)
(27a)
and
k+
2km + k + j= 2( km +
+ T(k+j).
(27b)
Now, (27) is of the form similar to 2n + i, as t(j) and t (k+ j) will either be 0 or 1.
Substituting (27) into (26) leads to the following:
Y.
2(km+L4])+t(j)
X.
m=0 ^2(km+ )+r(k+j)
n-1
U2kn+j = Uj ]1
(28а)
and
X2(km+L½])+T(j)
п-1
V2kn+j = Vj ] ]
(28b)
Y.
m=0
2(km+|)+t(k+j)
Using (24) and (28), we get the following:
п-1
U2kn+j = u¡ ]|
m=0
km+
km+
km+
LJ-1
Il C2k1+T(j)
k=0
d21+7(j), II c2k2+T(j)
k2=l+1
1=0
km+
k+j |-1
km+
k+
km+
k+i
II a2k1+T(k+j)
k1=0
1
+ Xr(k+j) 1=0
X-(k+j)
b21+t(k+j), 11 a2k2+t(k+j)
k2=l+1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d6994c6-9c0b-4312-919e-29acb733a883%2F7784ca85-a802-4ed2-831b-a5bd9ef7e26e%2Fzk3fs0q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Performing iterations on (25) respectively, leads to the following:
n-1 X2km+j
П
Vj
m=0 Y2km+k+j
п—1
Y2km+j
and
U2kn+j = U; |
m=0
(26)
V2kn+j
X2km+k+j
where j = 0,1,2,...,2k – 1.
The subscript of X and Y in (24) is 2n + 1, where i =
in (26) are 2km + j and 2km +k+j, where j = 0,1,2,...,2k – 1. So we want to write
2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r
can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder
when r is divided by k. Hence,
34
0,1 but the ones of X and Y
2km + j = 2( km + |) +
+ t(j)
(27a)
and
k+
2km + k + j= 2( km +
+ T(k+j).
(27b)
Now, (27) is of the form similar to 2n + i, as t(j) and t (k+ j) will either be 0 or 1.
Substituting (27) into (26) leads to the following:
Y.
2(km+L4])+t(j)
X.
m=0 ^2(km+ )+r(k+j)
n-1
U2kn+j = Uj ]1
(28а)
and
X2(km+L½])+T(j)
п-1
V2kn+j = Vj ] ]
(28b)
Y.
m=0
2(km+|)+t(k+j)
Using (24) and (28), we get the following:
п-1
U2kn+j = u¡ ]|
m=0
km+
km+
km+
LJ-1
Il C2k1+T(j)
k=0
d21+7(j), II c2k2+T(j)
k2=l+1
1=0
km+
k+j |-1
km+
k+
km+
k+i
II a2k1+T(k+j)
k1=0
1
+ Xr(k+j) 1=0
X-(k+j)
b21+t(k+j), 11 a2k2+t(k+j)
k2=l+1
![Xn-1yn-3
Yn-1Xn-3
Xn+1 =
Yn+1 =
(2)
Yn-1(±1+ Xp-1Yn-3)'
Xn-1(#1±yn-1Xn-3)'
• here x-3, x-2, x-1, x0, y-3, y-2, y-1 and yo are the initial conditions which are arbi-
10 trary non-zero real numbers.
The aim of this study is to generalize the results in [2,3] by studying the system of
ordinary difference equations
Un-k-10n-1
Un-1Vn-k-1
Vn-k+1(An + Bnun–10n-k-1)'
Un+1 =
Un+1 =
(3)
Un-k+1(Cn + Dnln-k-1ºn-1)'
where An, Bn, Cn and D, are real sequences, using a symmetry method. For a similar
11
After a set of long calculations, we get a system of determining equations for the charac-
teristics Q1 and Q2. Solving this system, we get
Q2(n,vn) = Anvn
and Q1(n, un)
= anun,
(17)
provided that
exp{-Yn}. We then get the invariants
1
X, =
and
1
Yn =
(22)
VnUn+k
UnVn+k
Here, (22) is invariant under the group transformations of (17). In other words, the
action of the symmetry generators, given in (20), on (22) gives zero. It is worthwhile
mentioning that the symmetries together with the constraints on an and A, have helped
us to come up with the appropriate change of variables that will lead to the reduced
system of equations. This is just one of the many roles of symmetries.
Using (22), (3) is reduced to a second-order difference equations:
Xn+2 = anXn + b, and Yn+2 = CnYn +dn.
(23)
The closed form solutions of (23) are respectively as follows:
n-1
n-1
n-1
X2n+i = X; IIazk,+i) + E(b21+i
kz=l+1
azkz+i
(24a)
k1=0
l=0
n-1
Y2n+1 = Yi II c2k, +i ) +
\k]=0
(24b)
C2kz+i
kz=l+1
l=0
for i = 0,1. From (22), we get that
Yn
Un and vn+2k =
Xn+k
Xn
Un.
Yn+k
Un+2k =
(25)
Performing iterations on (25) respectively, leads to the following:
Y2km+j
X2km+k+j
n- X2km+j
Y2km+k+j
U2kn+j
and v2kn+j = Vj
(26)
=
m=0
m=0
a where j = 0, 1,2,...,2k – 1.
The subscript of X and Y in (24) is 2n + 1,
in (26) are 2km + j and 2km +k +j, where j = 0,1,2,...,2k – 1. So we want to write
2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r
can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder
when r is divided by k. Hence,
re i = 0,1 but the ones of
and
2km + j = 2(km + |) + r(i)
+ τ )
(27a)
and
| k+
2km + k +j= 2( km +
2
+ T(k+j).
(27b)
Now, (27) is of the form similar to 2n + i, as t(j) and t(k + j) will either be 0 or 1.
Substituting (27) into (26) leads to the following:
2(km+L41)+r()
n-1
Uzkn+j = uj T
m=0
(28a)
2(km+[)+r(k+j)
and
X.
2(km+L½])+T(j)
Y.
m=0 2(km+[)+r(k+j)
n-1
U2kn+j = v; I
(28b)
Using (24) and (28), we get the following:
n-1
= uj|
m=0
U2kn+j
km+
km+
km+
11-1
II C2k1+T(j)
k1=0
d21+T(j), II C2kz+T(j)
I=0
k2=l+1
km+
km+
km+
X-(k+j)
II a2k1+T(k+j)
k1=0
Σ
1=0
b1+7(k+i),
II a2k2+T(k+j)
k2=l+1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d6994c6-9c0b-4312-919e-29acb733a883%2F7784ca85-a802-4ed2-831b-a5bd9ef7e26e%2F31g08aj_processed.png&w=3840&q=75)
Transcribed Image Text:Xn-1yn-3
Yn-1Xn-3
Xn+1 =
Yn+1 =
(2)
Yn-1(±1+ Xp-1Yn-3)'
Xn-1(#1±yn-1Xn-3)'
• here x-3, x-2, x-1, x0, y-3, y-2, y-1 and yo are the initial conditions which are arbi-
10 trary non-zero real numbers.
The aim of this study is to generalize the results in [2,3] by studying the system of
ordinary difference equations
Un-k-10n-1
Un-1Vn-k-1
Vn-k+1(An + Bnun–10n-k-1)'
Un+1 =
Un+1 =
(3)
Un-k+1(Cn + Dnln-k-1ºn-1)'
where An, Bn, Cn and D, are real sequences, using a symmetry method. For a similar
11
After a set of long calculations, we get a system of determining equations for the charac-
teristics Q1 and Q2. Solving this system, we get
Q2(n,vn) = Anvn
and Q1(n, un)
= anun,
(17)
provided that
exp{-Yn}. We then get the invariants
1
X, =
and
1
Yn =
(22)
VnUn+k
UnVn+k
Here, (22) is invariant under the group transformations of (17). In other words, the
action of the symmetry generators, given in (20), on (22) gives zero. It is worthwhile
mentioning that the symmetries together with the constraints on an and A, have helped
us to come up with the appropriate change of variables that will lead to the reduced
system of equations. This is just one of the many roles of symmetries.
Using (22), (3) is reduced to a second-order difference equations:
Xn+2 = anXn + b, and Yn+2 = CnYn +dn.
(23)
The closed form solutions of (23) are respectively as follows:
n-1
n-1
n-1
X2n+i = X; IIazk,+i) + E(b21+i
kz=l+1
azkz+i
(24a)
k1=0
l=0
n-1
Y2n+1 = Yi II c2k, +i ) +
\k]=0
(24b)
C2kz+i
kz=l+1
l=0
for i = 0,1. From (22), we get that
Yn
Un and vn+2k =
Xn+k
Xn
Un.
Yn+k
Un+2k =
(25)
Performing iterations on (25) respectively, leads to the following:
Y2km+j
X2km+k+j
n- X2km+j
Y2km+k+j
U2kn+j
and v2kn+j = Vj
(26)
=
m=0
m=0
a where j = 0, 1,2,...,2k – 1.
The subscript of X and Y in (24) is 2n + 1,
in (26) are 2km + j and 2km +k +j, where j = 0,1,2,...,2k – 1. So we want to write
2km + j and 2km +k+j in the form similar to 2n +i. We know that any positive integer r
can be written as r = k[r/k] + T(r), 0 < t(r) < r – 1, where t(r) denotes the remainder
when r is divided by k. Hence,
re i = 0,1 but the ones of
and
2km + j = 2(km + |) + r(i)
+ τ )
(27a)
and
| k+
2km + k +j= 2( km +
2
+ T(k+j).
(27b)
Now, (27) is of the form similar to 2n + i, as t(j) and t(k + j) will either be 0 or 1.
Substituting (27) into (26) leads to the following:
2(km+L41)+r()
n-1
Uzkn+j = uj T
m=0
(28a)
2(km+[)+r(k+j)
and
X.
2(km+L½])+T(j)
Y.
m=0 2(km+[)+r(k+j)
n-1
U2kn+j = v; I
(28b)
Using (24) and (28), we get the following:
n-1
= uj|
m=0
U2kn+j
km+
km+
km+
11-1
II C2k1+T(j)
k1=0
d21+T(j), II C2kz+T(j)
I=0
k2=l+1
km+
km+
km+
X-(k+j)
II a2k1+T(k+j)
k1=0
Σ
1=0
b1+7(k+i),
II a2k2+T(k+j)
k2=l+1
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