Show me the steps of deremine blue and inf is here i need evey I need all the details step by step and inf is here Motivated by difference equations and their systems, we consider the followingsystem of difference equationsYnXn+1 = A + B-2, Yn+1 = A + B-2(1)Yn-1where A and B are positive numbers and the initial values are positive numbers. InFirst of all, we consider the change of the variables for system (1) as follows:XnZn =AУпtn =AFrom this, system (1) transform into following system:tnZn, Zn+1 = 1+p2Zn-1tn+1 = 1+P2(2)%3D'n-1where p = > 0. From now on, we study the system (2).A2Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is(1+T+4p 1+ VI+4p(F, 2) = ( -22Now, we consider a transformation as follows:(tn, tn-1, Zn, Zn-1) →(f, fi, 8, 81)where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobianmatrix about equilibrium point (7, z):n-11B (,३) =%3D120 01Thus, the linearized system of system (2) about the unique positive equilibrium pointis given by XN+1 = B (i, z) XN, whereIntn-1XN =ZnZn-11B (ī, 2 :0 01Hence, the characteristic equation of B (i, z) about the unique positive equilibriumpoint (7, z) is4p24p2= 0.14Due to t = 7, we can rearrange the characteristic equation such that?+ 4p°, 4p27414= 0.Therefore, we obtain the four roots of characteristic equation as follows:p+Vp? – 8pi?272– Vp² – 8pi?p- V12 =272-p+Vp? + 8pi?21213 =Fo o oo oo

As per the instruction, we will have to find the roots that are marked in blue. So, we will be…

Answered: obtain the four roots of characteristic…

obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 272 p- Vp? - 8pi? 212 12 = -p+Vp? + 8pr? 13 = 212

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter13: Conic Sections

Section13.5: Systems Involving Nonlinear Equations

Problem 39PS

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Question

Show me the steps of deremine blue and inf is here i need evey I need all the details step by step and inf is here

Motivated by difference equations and their systems, we consider the following
system of difference equations
Yn
Xn+1 = A + B-
2
, Yn+1 = A + B-
2
(1)
Yn-1
where A and B are positive numbers and the initial values are positive numbers. In
First of all, we consider the change of the variables for system (1) as follows:
Xn
Zn =
A
Уп
tn =
A
From this, system (1) transform into following system:
tn
Zn
, Zn+1 = 1+p2
Zn-1
tn+1 = 1+P2
(2)
%3D
'n-1
where p = > 0. From now on, we study the system (2).
A2
Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is
(1+T+4p 1+ VI+4p
(F, 2) = ( -
2
2
Now, we consider a transformation as follows:
(tn, tn-1, Zn, Zn-1) →
(f, fi, 8, 81)
where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobian
matrix about equilibrium point (7, z):
n-1
1
B (,३) =
%3D
12
0 0
1
Thus, the linearized system of system (2) about the unique positive equilibrium point
is given by XN+1 = B (i, z) XN, where
In
tn-1
XN =
Zn
Zn-1
1
B (ī, 2 :
0 0
1
Hence, the characteristic equation of B (i, z) about the unique positive equilibrium
point (7, z) is
4p2
4p2
= 0.
14
Due to t = 7, we can rearrange the characteristic equation such that
?+ 4p°, 4p2
74
14
= 0.
Therefore, we obtain the four roots of characteristic equation as follows:
p+Vp? – 8pi?
272
– Vp² – 8pi?
p- V
12 =
272
-p+Vp? + 8pi?
212
13 =
Fo o o
o oo

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