Theorem 2. Suppose that {n} 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 A1 n-1 a Xn = An (IV) i =1 where xo = a and n > 2 Proof. Firstly, a x 1 = (x 0)ª-1 + a A1 Now ,by mathematical induction , we will prove that equations{IV} are true for n > 2. In the begining we try to prove that equations{IV} are true for n=2. 2-1 a a II 42 I 2 = (x 1)ª–1 + a G-1 + a A1 A2 A2 i =1 Now suppose that the equations{IV} is true for n = r.This means that r-1 a Xr = A, i =1 Finally we prove that the equations{IV} is true for n =r+1. Xr+1 = 9-1 + a i =1 r-1 i =1 a T-1 A,{a9-1 TI A§ª-2)(«¬-1) +a4f-1} i =1 -2 a a i =1 Ar+1 i =1 which complete the proof. Now we are going to give the general solution for the following nonlinear rational quadratic differ- ence equations of order two : Xn-1 Xn+1 = ((4)) "n-1 +a where a # -a and æ², # -a To do this we motivate by the following notations: suppose that x-1 = a1and xo = a2 . Let A1,1 = af +a and A1,2 = az +a Also consider the following notations: A2,1 = af + aA?‚1 Moreover we consider A2.2 = až + aA},2 A3,1 = af A?1 + aAž1 A4.1 = af A?, A31 + a Až, A3,2 = až A72 + aA32 , A4,2 = a3A?2A3 2+ aA32 In general we have p-2 Ap,1 = a? II 41 + aA-1,1 where p > 3 (V) i =1 p-2 Ap2 = aII A?2 +aAž-1,2 where p > 3. i =1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Show me the steps of determine blue and inf is here

A more general form of difference equation (2) can take the following nonlinear rational
difference equations :
Xn+1
(xn)- +a
(3)
where (xo)9-1
Now consider the following notations
+ -a.
A1 = a9-1 + a
A2 = a"-1 + a Af-
q-1
(p-2
(q-1)(q-2)
Ap =
= a9-1
+ a
'p-1
(III)
i=1
where p > 3.
Transcribed Image Text:A more general form of difference equation (2) can take the following nonlinear rational difference equations : Xn+1 (xn)- +a (3) where (xo)9-1 Now consider the following notations + -a. A1 = a9-1 + a A2 = a"-1 + a Af- q-1 (p-2 (q-1)(q-2) Ap = = a9-1 + a 'p-1 (III) i=1 where p > 3.
Theorem 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
A1
n-1
a
II 49-2
(IV)
Xn =
An
i =1
where xo = a and n > 2
Proof. Firstly,
a
* 1 =
(x o)9-1 + a
A1
Now ,by mathematical induction, we will prove that equations{IV} are true for n > 2.
In the begining we try to prove that equations{IV} are true
n=2.
2-1
a
a
I 2 =
II 4-2
(x 1)9-1 + a
-1 + a
Aj A2
A2
i =1
Now suppose that the equations{IV} is true for n = r.This means that
T-1
a
II 4-2
Xr =
A,
i =1
Finally we prove that the equations{IV} is true for n =r+1.
Xr+1 =
-1
+ a
アー1
i =1
II A)a-1 + a
i =1
a
T-1
A,{aq-1 TI Aa-2)(4–1) + aAƒ-1}
i =1
r-1
a
a
Ar+1
T-1
-2)(4-1)
i =1
which complete the proof.
Now we are going to give the general solution for the following nonlinear rational quadratic differ-
ence equations of order two :
Xn-1
Xn+1 =
((4))
"n-1 +a
where a # -a and a21 # -a .
To do this we motivate by the following notations:
suppose that x-1 = a1and xo = a2 .
Let A1,1 = af +a and A1,2 = az +a
Also consider the following notations:
A2,1 = af + a A?,.
Moreover we consider
%3D
A22 = až + aA32
A3,2 = ažA?2 + aA32
A3,1 = af A?1 + a Až1
A4.1 = až A31 A31 + a A31 , A42 = ažA?2 A32 + QA32
2,2
In general we have
р-2
Ap,1 = ai II A1+a4-1,1
where p > 3
(V)
i =1
р-2
Ap.2 = až II 432 + aA-1,2
where p > 3.
i =1
Transcribed Image Text:Theorem 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 A1 n-1 a II 49-2 (IV) Xn = An i =1 where xo = a and n > 2 Proof. Firstly, a * 1 = (x o)9-1 + a A1 Now ,by mathematical induction, we will prove that equations{IV} are true for n > 2. In the begining we try to prove that equations{IV} are true n=2. 2-1 a a I 2 = II 4-2 (x 1)9-1 + a -1 + a Aj A2 A2 i =1 Now suppose that the equations{IV} is true for n = r.This means that T-1 a II 4-2 Xr = A, i =1 Finally we prove that the equations{IV} is true for n =r+1. Xr+1 = -1 + a アー1 i =1 II A)a-1 + a i =1 a T-1 A,{aq-1 TI Aa-2)(4–1) + aAƒ-1} i =1 r-1 a a Ar+1 T-1 -2)(4-1) i =1 which complete the proof. Now we are going to give the general solution for the following nonlinear rational quadratic differ- ence equations of order two : Xn-1 Xn+1 = ((4)) "n-1 +a where a # -a and a21 # -a . To do this we motivate by the following notations: suppose that x-1 = a1and xo = a2 . Let A1,1 = af +a and A1,2 = az +a Also consider the following notations: A2,1 = af + a A?,. Moreover we consider %3D A22 = až + aA32 A3,2 = ažA?2 + aA32 A3,1 = af A?1 + a Až1 A4.1 = až A31 A31 + a A31 , A42 = ažA?2 A32 + QA32 2,2 In general we have р-2 Ap,1 = ai II A1+a4-1,1 where p > 3 (V) i =1 р-2 Ap.2 = až II 432 + aA-1,2 where p > 3. i =1
Expert Solution
Step 1

Given : nonlinear rational difference equation

  xn+1=xnxnq-1+α                            3

where x0q-1-α.

Other notation : 

Ap=aq-1i=1p-2Aiq-1q-2+α Ap-1q-1                         (III)

and equations {IV}

   xn=aAni=1n-1Aiq-2                               (IV)

where x0=a and n2,

To prove : equations {IV} is true for n=r+1.

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