Show me the steps of determain blue and inf is here Theorem 4. Suppose that {æn } is the solution of (5) and the initial values x-1 ,Xo are arbitrarynonzero real numbers .Let x-1 = a-1and xo = ao .Then ,by using the notations(VII) , the solutionsof (5) are given by:а-1(2)ao(VIII)n-1I (4)*II (4º)*a-1q-2ini =1n-1a0i =1where n > 2Proof. We can use the same way and the mathematical induction which using in theorems(2.1) and(2.2) , to prove this theorem .Now we will generalize difference equations (3) and (5) to the following higher-orderdifference equation :Xn-kIn+1(xn-k)"- +awith(x-k+1)"-1+ -a , and the initial values x-k+t = a-k+! ,1 = 0, 1, 2, ..k. .As in the above manner we can generalize the notations(VII) by the following :Consider A(a-k+1)"- + atagIX(-11)= (a-+)+a (A")9-1(p-2(T) (9-Aj(사일)"9-1(a-i=11,2, ..k+ 1, 1 = 0, 1, 2, ..k and p> 3 Theorem 2.1 Suppose that {xn} is the solution of the difference equation (2) . With the abovenotations(I) , the solutions of the difference equation (2) is given by:x1 = a/A1* (IA)'n-1aПА(II)Xn =Ani=1where a and a are non-zero real numbers and n > 2.XnIn+1 =(2)(xn)? + awhere (xo) + -aIn order to do this we introduce the following notations:р-2Apwhere p > 3.((1))= ai=1Theorem 2.2 Suppose that {xn} is solutions of (3 ) and the initial value xo is an arbitrary nonzeroreal number. Let xo = a .Then ,by using the notations(III) , the solutions of (3) are given by:an-1aII 49-2(IV)Ani =1where xo = a and n > 2ence equations of order two :Xn-1((4))Xn+1 =r2n-1where x #-a and r21 + -a.h.. tha flleXn-1Xn+1 =((5))x +an-1where the initial values x-1 = a-1To = a0,q-1-a and a -a.Afa' + aq-1a + a(40)*9-1a + aA = ag-1+ a(4")%3|(p-2(1) \9-1p-1(1)) (9-1)(q-2)(A.) tagV1I9-1+ a(-7)i=1a ( (4)**) +« (4)**(q-1)(q-2)A+ a%3D\i=1V p > 3In+1 =(3)(*n)- + awhere (xo)"- -a.Now consider the following notations

The equations given are as follows: xn+1=xnxnq-1+αxn+1=xn-1xn-1q-1+α Where, x0q-1≠-α and x-1=a-1,…

Answered: Theorem 4. Suppose that {xn } is the…

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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Question

Show me the steps of determain blue and inf is here

Theorem 4. Suppose that {æn } is the solution of (5) and the initial values x-1 ,Xo are arbitrary
nonzero real numbers .Let x-1 = a-1and xo = ao .
Then ,by using the notations(VII) , the solutions
of (5) are given by:
а-1
(2)
ao
(VIII)
n-1
I (4)*
II (4º)*
a-1
q-2
in
i =1
n-1
a0
i =1
where n > 2
Proof. We can use the same way and the mathematical induction which using in theorems(2.1) and
(2.2) , to prove this theorem .
Now we will generalize difference equations (3) and (5) to the following higher-order
difference equation :
Xn-k
In+1
(xn-k)"- +a
with(x-k+1)"-1
+ -a , and the initial values x-k+t = a-k+! ,
1 = 0, 1, 2, ..k. .
As in the above manner we can generalize the notations(VII) by the following :
Consider A
(a-k+1)"- + a
tagIX
(-11)
= (a-+)+a (A")
9-1
(p-2
(T) (9-
Aj
(사일)"
9-1
(a-
i=1
1,2, ..k+ 1
, 1 = 0, 1, 2, ..k and p> 3

Theorem 2.1 Suppose that {xn} is the solution of the difference equation (2) . With the above
notations(I) , the solutions of the difference equation (2) is given by:
x1 = a/A1
* (IA)
'n-1
a
ПА
(II)
Xn =
An
i=1
where a and a are non-zero real numbers and n > 2.
Xn
In+1 =
(2)
(xn)? + a
where (xo) + -a
In order to do this we introduce the following notations:
р-2
Ap
where p > 3.
((1))
= a
i=1
Theorem 2.2 Suppose that {xn} is solutions of (3 ) and the initial value xo is an arbitrary nonzero
real number. Let xo = a .Then ,by using the notations(III) , the solutions of (3) are given by:
a
n-1
a
II 49-2
(IV)
An
i =1
where xo = a and n > 2
ence equations of order two :
Xn-1
((4))
Xn+1 =
r2
n-1
where x #
-a and r21 + -a.
h.. tha flle
Xn-1
Xn+1 =
((5))
x +a
n-1
where the initial values x-1 = a-1
To = a0,
q-1
-a and a -a.
Af
a' + a
q-1
a + a
(40)*
9-1
a + a
A = ag-1+ a
(4")
%3|
(p-2
(1) \9-1
p-1
(1)) (9-1)(q-2)
(A.) tagV1I
9-1
+ a
(-7)
i=1
a ( (4)**) +« (4)**
(q-1)(q-2)
A
+ a
%3D
\i=1
V p > 3
In+1 =
(3)
(*n)- + a
where (xo)"- -a.
Now consider the following notations

Expert Solution

Step 1

The equations given are as follows:

$x_{n + 1} = \frac{x_{n}}{{(x_{n})}^{q - 1} + α} x_{n + 1} = \frac{x_{n - 1}}{{(x_{n - 1})}^{q - 1} + α}$

Where, ${(x_{0})}^{q - 1} \neq - α$ and $x_{- 1} = a_{- 1}, x_{0} = a_{0}, x_{- 1}^{q - 1} \neq - α, x_{0}^{q - 1} \neq - α$ .

We can also write,

$x_{n + 1} = \frac{x_{n - 2}}{{(x_{n - 2})}^{q - 1} + α}$

Where, $x_{- 2} = a_{- 2}, x_{- 1} = a_{- 1}, x_{0} = a_{0}, x_{- 2}^{q - 1} \neq - α, x_{- 1}^{q - 1} \neq - α, x_{0}^{q - 1} \neq - α$

Generalizing the formula we will get:

$x_{n + 1} = \frac{x_{n - k}}{{(x_{n - k})}^{q - 1} + α}$ ..........................................................................(1)

Where, $x_{- k + l} = a_{- k + l} and {(x_{- k + l})}^{q - 1} \neq - α, l = 0, 1, 2, . . ., k$ .

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Theorem 4. Suppose that {xn } is the solution of (5) and the initial values x-1 ,Xo are arbitrary nonzero real numbers .Let x-1 = a-1and xo = ao. Then ,by using the notations(VII) , the solutions of (5) are given by: 2f" = a-1 ao (VIII) %3D n-1 II (4")** II (4")*- a-1 q-2 i =1 n-1 ao 1(2) An' i =1