Maths - Fixed Points and Basin of Attraction

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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