Maths - Fixed Points and Basin of Attraction

This document discusses the concept of fixed points and basin of attraction in Maths. It covers topics such as sinks and sources, graphs and equations related to fixed points, and the basin of attraction for each sink. Detailed explanations and examples are provided.

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Table of Contents
Question 1........................................................................................................................................................................ 1
a).................................................................................................................................................................................. 1
The basin attractor boundary of the fixed point is sink........................................................................................................4
Question 2........................................................................................................................................................................ 5
Question 3........................................................................................................................................................................ 6
Question 4........................................................................................................................................................................ 8
Question 5........................................................................................................................................................................ 9
Question 6...................................................................................................................................................................... 10
Question 7...................................................................................................................................................................... 10
Question 8...................................................................................................................................................................... 11
Question 9...................................................................................................................................................................... 12
Question 10.................................................................................................................................................................... 13
Question 11.................................................................................................................................................................... 13
Question 12.................................................................................................................................................................... 14
Question 13.................................................................................................................................................................... 14
Question 14.................................................................................................................................................................... 18
References...................................................................................................................................................................... 19

Question 1
a)
F(x) =1+2x- x2 plot a graph along with the line y=x
Plot the line y=x
Let as consider the graph line is y=x
Using the definition of the derivative, calculate the derivative of is F(x) =x2−¿2x-1
F’(x) =lim
h→ 0
f ( x +h )−f (x)
h
x 0 1 2
y -0.5 -2 -1
ii.F(x) =1+2x- x2 fixed points
The order of the fixed points iteration is depends on F(x)=0
X element consider as,
x1+ x2 +… … … … . xn+xn+1
F (x)= 1+2x- x2
F(x) = x2−¿2x-1
X=2+ 2
x x2−¿2x=0
1

xn+2=2+ 2
xn
x(x-2)=0
x 1=1 xn+2= 1
xn −2
X2=2+ 2
1=4 x2= 1
2.7−2 =1.428
X3=2+ 2
4 =2.5 x3=-1.666
X4=2.8 x4=-0.27
X5=2.714 x5=-0.57
X6=2.736 x6=-.089
G(x) =1+1/x xn+2= 1
xn −2
G’(x) =1/- x2 g(x) = 1
x−2
G’(x) = ( 1+ √5
2 ) g’(x) = 1
( x−2)2 =
1
( 1+ √ 5
2 )−2 ¿
2 ¿=7.1818>1
G’(x) =
1
( 1+ √5
2 )
2
=-0.3819660
|-0.3819660|<1
Fixed point is(0.38, 7.18)
Remember that xn+2=g(x)
If |g’(r)|<1coverage
If g’(r) =0 coverage quadratic
G’’(r) =0 coverage order 3
(iii)F(x) =1+2x- x2 determine whether they are sinks or sources,
The fixed point of F by solving F(x) = 1+2x-x2 get 0 and 7 fixed point of F.
The period of the two points are found among the fixed points of the map f 2. The fixed point of F
is not period two points.
F(x) = 1+2x-x2
2

F(x) = x2−¿2x-1
F2(x)=( x2−2 x−1)2= x4-4x2-1
The fixed point is 0 and 2
After simplification of the equation is
x4-4 x2-1 and x=0
Since 2 is the solution we can write as the equation is x-2 second degree of polynomial
is x4-4x2-1=x-2
¿ ¿-4 x2-1)(x-2)=0
(x-2)=0¿ ¿-4x2-1)=0
So {1 ± √2} is consider as {1+ √ 2 , 1− √ 2} we can consider F’ (1+ √2 ,) F(1− √2 ,)
Let as consider the value of y=4x-8
{4(1+ √2 ,)-8 , 4(1− √2 ,)-8}
Which is equal to the fixed point |-67| since the value is bigger than 1. We have a source period
of the fixed point values is sink.
iv. F(x) =1+2x- x2 determine the basin of attraction for each sink.
F(x) = x2−¿2x-1
The initial condition of the basin attraction is,
lim
n → ∞
¿ ¿ ¿) y=x or y=-x
The fixed point of the two attractors system is the basin attraction for this map y=-x. The blank
attractor of the basin attraction is y=x
Basin boundary is no longer then smooth curve (Kharitonov, 2013).
X=x(s)
3

X=x( x2−¿2x-1)
Y=y(s)
Y=y(x( x2−¿2x-1)) for 2>s>0
If s1≠ s 2
The basin attractor boundary of the fixed point is sink (Romeu and Vera‐Hernández, 2016).
b)
i. F(x) = 1/2+x+sin(x) plot the line graph y=x
F(x) = 1/2+x+sin(x)
F(x)= x +sin(x)+ ½
X=x+ 1
2 sin(x) =0
F(x) = 1/2+x+sin(x)
ii. fixed points
The fixed point of the values Is consider as
x1+ x2 +… … … … . xn+ xn+1
F(x)=x−¿ sin(x)- ½
The graph of g(x) and x value is,
Initially guest value is x0=2
i 0 1 2 3 4 5
Xi 1.409 1.487 1.496 1.496 1.497 1.497
G(x)= x+sin[x]+1/2 the fixed point values is 1.497
4

F(x) = 1/2+x+sin(x)
iii.determine whether they are sinks or sources,
∫ x +sin [x ]+1/2dx
∫ x +sin [x ]+1/2dx
= x
2 +cos [x ]+sin [x ]+1/2
The source and sink value is {1 ± √2} is consider as {1+ √ 2 , 1− √ 2}
F(x)=1/2+x+sin(x)
iv.determine the basin of attraction for each sink
The fixed point of the two attractors system is the basin attraction for this map y=-x. The blank
attractor of the basin attraction is y=x
Basin boundary is no longer then smooth curve.
X=x(s)
X=x(1+ √2 , )
Y=y(s)
Y=y(x(1+ √2 , )) for 2>s>0
If s1 ≠ s 2
The basin attractor boundary of the fixed point is sink.
c)
i.
{ x ln x x ≠ 0
o x=0
i.plot y=x
F(x)=x
ii. fixed point (0.1)
iii. Sink and source can be denoted the sink value is constant
iv.
5

{1 ± √2} is consider as {1+ √ 2 , 1− √ 2} we can consider F’ (1+ √2 ,) F(1− √2 ,)
Let as consider the value of y=1
{4(1+ √2 ,)-1 , 4(1− √2 ,)-1}
Which is equal to the fixed point |-1| since the value is bigger than 1. We have a source period of
the fixed point values is sink.
Question 2
F(x)=a+bx+c x2
We can assume the 3 cycle value is, (0,1,2)
F(x)= a+bx+cx2
a, b, c is the numerical constant value. But c is not equal to zero
y= a+bx+c x2 +d x3
y= a+bx+c x2 +d x3+ ¿ e x4
x= −b ± √ b2−4 ac
2a
Trajectory values is {0,1,2,0,1,2}
6

X=0, x=1, x=2
γ=t+x
γ=t+ x1
γ=t+ x2
γ=t+ x3
γ =0+ x1, 1+ x2 , 2+ x3
The lyapunov number of value 0 is map, of (0,1, 2) is the periodic sink.
Question 3
F(x)=1+x+asin(x)
a=1
Let as find the value of x=-b ± √ 1−c
X=−x ± √1−sin ( x)
2 X 1
7

X= √1−sin (1)
2 X 1 =[0,1]
The bifurcation value is,
C=3.079
Question 4
F(x) =ax(1-x)
To find the value of a=1 to apply the value is,
F’(x) = ax(1-x)
F’(x) =ax-x2
F’(x) =A(x- x2)
X=1
a=ax (1-1)=0
λ=5
8

F(x),f 2(x),f 3(x),f 4(x) the function of 2n fixed point value is,
p N(p)
1 2
2 1
3 2
4 3
N(p)= 1
p (2p- ∑
divide
k X N ( k)
N(p)= 1
p e phT
hT = lim
p → ∞
¿ ¿=log ¿ ¿=)
1
p (log( 2p / p))=log(2)- 1
p log(p)
9

1
p (log (25 /5))=log (2)- 1
5log(5)
Stage 1 N(p) Stage 2 N(p) Stage 3 N(p)
5 6 15 2182 25 1342176
Question 5
F(x)=4x(1-x) to find the 7 periodic orbit value is
F(x)= 4x-4x2
F(x) =x(4-4x)=4x(1- x2)=4(x- x2)=4x
We can calculating the 7 periodic value is,
F(x),f 2(x),f 3(x),f 4(x),f 5(x), f 6(x)
To find the diagonal of value y=x
F(x) periodic value is,
D(x)=4x(1-x)
dx
dt =4x(1-x)
dx
dt =4 x 1
2(1- 1
2)=1
X(t)=(t+T) of the interval value is[0,T]
N(p)= 1
p (2p- ∑
divide
k X N ( k)
= 1
p log (2 p
p )=log(2)- 1
p (log(p))
1
27 log ( 2 X 27
27 )=log(2)- 1
27 (log(27))=4971008
Question 6
F(x)= 4x(1-x)
Interval of the point itinerary is LRRRLRL
Find the decimal place value is 5
F(x)= 4x(1-x)
10

F(x)=4x-4x2
F(x)=∫
o
5
4 x−4 x2dx
F(x)= [4x1- 8 x2
2 ]
F(x)=4-4x2
F(x)= 4 x2+4=4( x2+1)
F’(x)=(4(x2+1))
F’’(x)=( 4( x2 +1))2=8( x4 +1)
The interval 5 decimal place value of the end point s
F(x)= {76, 82, 46, 81, 82}
Question 7
F(x)= 4x(1-x) interval point is x ∈{0,1 }
Find the value of [0.1/100]
F(x)= 4x(1-x)
First we can assume the trajectory method of the values,
4x(1-x)= { x0 , x1 , x2 ,… …… …… … . xn}
The interval point of the shadow noise of length is 100
To find the maximum of Liebenberg value (marqur dt). And calculating the approximate sum of
square value
FQurd[x_]:4x(1-x)
Der [x_] =evaluate [FQurd’[x]]
Filter [x_, n_]:=Nest [Fqurdφ , x ,n]
F Traj[x_, n_]:= Nest list [Fqurdφ , x ,n]
Using high precision of digit value is 65
And find the 100th iteration value,
Filter (0.1/100)
Filter (N [1/10, 65], 100)
Val 100= filter [N [1/10,100], 100]
Trajectory interval of (0.1/100) value is,
11

Filter (0.1/100)=0.372447
Filter (N [1/10, 65], 100)=0.93011
Val 100=0.9310897
Question 8
a)
F(x)=2x g(x)= 4x
I.To find the Conjugate value is:
F(x)=g(x)
φ (F( x ))=φ g(4x)
x=0
φ ( F(2 x))=φ g(4x)
φ ( F(2 (0)))=φ g(4(0))=0
ii . ¿ find the conjungacy
The necessary of the condition of fixed point values is x=0 and can apply the value of φ (x). The
implies φ (0)=0 is the conjungacy.
iii. g(x)=4x f(x)=2x both are conjugate and conjungacy value.
b)
F(x) =x2 g(x) =x2−2 x+2
i. conjugate
F(x)=g(x)
φ (F( x ))=φ g(4x)
x=0
φ ( F( x2 ))=φ g( x2−2 x+2 )
φ ( F( 0))=φ g (0 )
The homomorphism of the function f(x) is not conjugate because the g(x) is value is not zero
iii. Conjungacy
The necessary of the condition to apply the value of x=0 and F(x) is the conjungacy value
G(x) is not conjungacy value because G(x) =-2 is not equal to zero
iii.F(x) = x2 and g(x) = x2−2 x+2 is not the conjugate and conjungacy value. For the reason of
12

G(x) is not equal to zero.
c)
f(x) = 2x + 1 and g(x) = 4x.
i. conjugate
F(x)=g(x)
φ (F( x ))=φ g(4x)
x=0
φ ( F(2 x +1))=φ g(4x)
φ ( F( 0))=φ g (0 )
The homomorphism of the function f(x) is not conjugate because the g(x) is conjugate
iii. Conjungacy
The necessary of the condition to apply the value of x=0 and F(x) is the conjungacy value
G(x) is not conjungacy value because f(x) =1 is not equal to zero
iii.F(x) =2x + 1 and g(x) =4x.is not the conjugate and conjungacy value. For the reason of F(x) is
not equal to zero.
Question 9
Lyapunov number and Lyapunov exponent of the point of the x0=0.2 F(x)=sin(πx)
F(x)= sin( πx)
x0=0.2
F(x)= sin(π X 0.2)
F(x)= sin(0.658)
F(x)=0.58
x'+ xn+1=F( x¿)+ x0 f '(x) + x0
2 ' '(x)
=sin ( πx) + 0.2 cos( πx)+ 0.2
2 sin( πx)
=∏
i=0
n −1
DF (xi)δ 0
=∏
i=0
n −1
DF ( πx ) X 0.2
13

=lim
n → ∞
sin (πX 0.2)+cos( πX 0.2)
=-0.20
Question 10
We can consider the binary expansion of the point of x map is, LRRLLRLLR
First we can calculating the decimal values of the each letter point
The iteration of the binary conversion is
100010 100010 100010 100010 100010 100010 100010 100010 100010
L R R L L R L L R
Question 11
Stationary distribution f(x)=a(x- x3)
A= 3 √3
2 x0=0.2
Length= 105
Interval value is [-1, 1]
f(x)= a(x-x3)
f(x)= 3 √3
2 (x- x3)
x=-1, x=1
F(x)= 3 √3
2 ((-1)-(−1)3)=-4.242
F(x) = 3 √3
2 ((1)-(1)3)=0
F(x) =0
The consider the length of 105 values, we can apply the valueis x0=0.2
14

3 √ 3
2 ((0.2)-(0.2)3)x100=-3.5757
Question 12
Layapunov exponent of the function x0=0.2
The first component of F is map F:0.2 0.2
xP +1=f(x p)
x p=x0
x p=0. 2
Iterative interval is { x0 , x1 , x2 ,… ….. xp}
x p +δ p=¿f(0.2+ δ0)
x p +δ p=f p(0.2)+ ∂ f p ( x )
∂ x
∂ f p ( x )
∂ x = ∂ f p ( x )
∂ f p−1 (xp−1)+ ∂ f p ( x )
∂ f p−2 ( xp−2)+ ∂ f 1 ( x )
∂ x
= lim
p → ∞
∂ f p ( x )
∂ f p−1 (xp−1) + ∂ f p ( x )
∂ f p−2 (xp−2)+ ∂ f 1 ( x )
∂ x
Exponent of the Lyapunov number is,
λ= ∂ f
∂ x δ p=λ' δ0
λ'=∂ f p ( x )
∂ x + ∂ f p−1 ( x )
∂ x + ∂ f 1 ( x )
∂ x
h(x)=inl( λ' ¿= lim
p → ∞
1
p ∑
i
¿ ∂ f p
∂ x ∨¿ ¿
h=|h/2|
x p+1={ 0.2 xp x p ≤ 0
0.2(1−x p ) x p ≥ 0 }
Question 13
Unstable heron map b=0.3 a=1.4
Radius=0.1
15

X range [-4, 4] y range = [-4, 4]
The heron map of two parameters
H(x, y) = (1+y-ax2, bx)
Parameter of the value is a=1. 4, b=0.3
xn+1= yn+1-a xn
2
yn=b xn
X’=x
Y’=1+y- a x2
X’’=bx’
Y’’=y’
U= y2+ x
2 v=x2+2y
∂(u , v )
∂(x , y )=
[ ∂u
∂ x
∂u
∂ y
∂ v
∂ x
∂ v
∂ y ]=
[ y2 + x
2 2 y +x /2
2 x +2 y x2+2 ]=[ 2 y2 +1 4 y +x
2 x +2 y x2+ 2 ]
= x2+5 y6+3x+3=0
Fixed point of the graph line is,
16

We can use the formula of
x0= −(1−b)± √ (1−b)2+4 ac
2 a
= −(1−0.3)± √ (1−0.3)2 + 4 X 1.4 X 0,1 X 4
2 a
=0.392
npar =106 is the denoted as the interval values is {1.3999, 1.40001}
Circle of the fixed point N=14 times
nskip=104
n¿=57
dx
dt =p
17

dp
dt =-x-p
A=cos wT (e−γT / 2 cosWT −(1+e− p)T )
a npar ninit nskip ntime nwin nskip
[1.3, 1.4] 1014 5000 1013 1012 11 57
[1.3999,1.40001] 1014 10000 1013 1012 3 8
[1.39999,1.400001
]
1014 14 1013 1012 1 0
The interval of the fixed time point region is [1.39999, 1.400001] length is 2. 104 And 57 is the
parameter value of sink. 14 parameter value is 2.1012
A=1.39999 b=0.3 henon map of 100000 point values x axis 0.9X1012 and 1.1X1012
18

The zoomed-in plot with ranges is denoted as 0.4 X1011and 0.6X1012
Question 14
Let F(x, y) = ( y2 + x
2 , x2 +2 y) compute its jacobian, located fixed point, and points, and say
whether they are saddles, sources, or sinks.
F(x, y) = ( y2 + x
2 , x2 +2 y)
F(x)= y2 + x
2 F(y)=x2+2 y
The jacobian equation can be consider as,
U= y2 + x
2 v=x2+2 y
∂ u
∂ x = y2 +1
2 =2 y2 +1
∂u
∂ y =2y++ x
2 = 4y+x
∂ v
∂ x =2x+2y
∂ v
∂ y =x2+2
Jacobian matrix is,
( 2 y2+ 1 4 y + x
2 x+2 y x2 +2 )
Let as assume the fixed points is (1, 0), (0, 1)
F (x)(1,0)= (0, 5)
F ( y)(1,0)= (0, 1.4)
Let as consider the sink and source is,
F(x)= y2 + x
2 F(y)=x2+2 y
Since the solution can be write the equation of the x and y is 2 y2+1 is the second degree of the
values of the polynomial is 2 y2+1(x+2)=0
(x+2)=0 2 y2+1=0
19

2 y2=-1
y2= 1
2 y= √ 1
2
If the condition of x is the {1± √ 2} that can consider the values is {1+√2 ,1− √2}
The F(x) value to apply the F(y)
F(y) =x2+ 2 y
Y=1
F(y) = x2+ 2 ={¿, (1− √2)2+2) which is equal to |5| since the value is bigger than 2 consider the
source period of the fixed point is sink.
References
Kharitonov, V. (2013). Time-delay systems. New York: Birkhäuser.
Romeu, A. and Vera‐Hernández, M. (2016). Counts with an endogenous binary regressor: A
series expansion approach. The Econometrics Journal, 8(1), pp.1-22.
20