Graphical Analysis : Basin Boundaries and Fractals
The goal of this project is to explain the geometric behavior and dynamics
of the quadratic funtion f(z) = z^2 + c in complex plane by comparison between mathematical procedures and experimental geometry through following concepts:
ContentsI will explain the following concepts in Reals and at the end will apply them in Complex plane to study the behavior of the so called Julia set
 Iteration of a function
 Orbits
 Fixed and periodic points
 Attracting and Repelling fixed Points
 Attracting and Repelling periodic Points
 Orbit types
 Escaping Orbits
 Basin Boundaries
 Magnification and Fractals
The Gragh of Julia's Set for f(z) = z^2 + (0.360284 + i0.100376)
Click on image to see
PostScript figure
Iteration of a function:
Consider the function f : R>R, f(x) = y. Pick a real number, say x, and apply f to it. The result is f(x) in R, so we can apply f to f(x)
and the result will be f(f(x)).We can keep giong.... Thie procedure is called the iteration of the function f starting at x.
We denote it as fn(x)=f((f(f(...f(x))...) , for n times
Geometric Interpretation:
Since we are dealing with iterations and orbits in this project, to show successive iterations on a gragh
 Draw the line y = x , in green,.
 Graph the function in red.
 First, locate the starting point x0 on the line (y = x)
 From this point draw a verical line ,in blue, the intersecttion with gragh evaluats the function ,f(x0) = y0).
 To get to the next point (x next) draw a horizontal line from the current point of the graph in blue.
 This line intersects with the line x = y and results in an iteration; previous y = new x. So we can have f(f(x)).
 Repeat the steps above
For Example Observe the orbit of 2 under The square root function
Click on image to see
PostScript figure
Definition of Orbit:
The list of infinite iterations of a point is called the orbit of that point.
For Example, the orbit of 2 under squaring function is : 2,4,16,256,65536,...and below you see the graphical interpretation of two orbits.
 on the left orbit of 0.5
 on the right orbit of 1.2
Observe the orbit of 2 under The square root function
Click on image to see
PostScript figure
It reveals that the orbt of 0.5 goes to zero as n becomes large : fn(0.5) > 0 as n > oo
But orbit of 1.2 explodes : fn(1.2) > oo as n > oo
Fixed and periodic points:
 If there exists a point, xo,
such that :
f(xo) = xo,
then it is called a fixed point since It will never move under iteration so fn(xo) will be xo .
For example, 0 is a fixed point of the square function since f(0) = 0.
 If an orbit passes some point twice it is called a periodic orbit, and the point is called a periodic point.
For example
orbit of 1 under f(x) =  x^3
Click on image to see
PostScript figure
Attracting fixed points:
 Suppose a function f has a fixed ponit p and there exists an interval a < x < b containing p, such that every orbit in the interval tends to p
Meaning : for all x in : a < x < b , fn(x) > p as n > oo .
For example, 0 is a fixed point of the square function since f(0) = 0.
Page animation : attracting point of f(x) = x^2 in the interval 1 < x < 1 Click on image to see
PostScript animation by page turnning
An interesting example of attracting fixed point is the point 0.73908 for the cosine function.
Page animation : attracting point of f(x) = cos(x) in the interval 2pi < x < 2Pi Click on image to see
PostScript animation by page turnning
Experimentally, The animation above showed that fn(x) > 0.73908 as n > oo .
Basin of attraction:
The interval in which every ponit of the funcion tends to the attraction point of that interval,is called the basin of attraction
For instance in the examples above:
The basin of attraction of f(x) = X^ 2 is the interval (1,1) with the attracting point of 0,
,and the basin of attraction of f(x) = cos x is (2Pi,2Pi) with the attracting point of .73908,
meaning that all orbits of the domain go to that point.
An Experiment: Before we switch to complex numbers and start with Julia set, Lets find out the basin of attraction and scaping orbits of the function f(x) = x^2  0.5 in reals :
 Algebraically and
 By Graphical analysis
The first step is to find the fixed point.(Recall that p is a fixed point if and only if f(p) = p)
So lets solve f(x) = x ie x^2  .5 = x therefore x^2 x .5 = 0
Using the quadratic formula x1 = (1  3^ 1/2) /2 and x2 = (1 + 3^ 1/2) /2
This gives us the choice of three intervals for potential Basin of Attraction.
Now we have to pick a point in each interval and find the limit of the orbit by successive itterations
of the point as many times as neccessary to guess the limit (perhaps more that 20 times)
As you see this can be very time consuming.
So lets check the interval 1 < x < 1 for fixed points and the basin of attraction, using my orbit program.
Page animation : attracting point of f(x) = x^2 in the interval 1 < x < 1 Click on image to see
PostScript animation by page turnning
Therefore we see that x1 is our fixed point and 1 < x < 1 the basin of attraction.Furthermore, it is not required to check the other intervals , since the orbit will grow very rapidly.
The Julia set : basin boundaries:
Consider the function:
 f : C > C (complexvalued)
 f (z) = z ^ 2 + c
suppose that f(z) has an attracting periodic orbit,then there is a set of points that gets attracted to the cycle. These points are called the basin of attraction of the cycle.
There is another set of points that their orbits go to infinity so they are called scaping points. Besides these two set there is another set which does neither.
This set lies between the boundary of scaping points and the basin of attraction and It is called THE JULIA SET. and here is an example.
The Julia set of the function f( z) = z ^ 2  1 + .8i :
Click on image to see
PostScript
To determine the the scaping points of a quadratic function f ( z ) = z ^ 2 + c with  c  <= 2 ,As Robert Devaney in his book "Chaos, Fractals and Dynamics" does,lets divide the complex plane to two regions :
 z  <= 2 and  z  > 2.
If  c  <= 2 and  z  > 2 , then by triangle inequality we have :  f(z)  = z ^ 2 + c  >=  z ^2   c  but
 z ^2   c  >  z ^2   z  = ( z   1) z  since  z  >  c , and
 z   1 > 1 since  z  > 2
let  z  = 2 + k for some k > 0
then  f(z)  > (1 + k) z  meaning that  f(z)  >  z 
so  f2(z)  =  f(f(z))  > (1 + k)  f(z)  > (1 + k)^ 2  z  It shows that f2(z) gets further from the origin than f(z)
So by induction fn(z) =  f(f...(z)...)  > (1 + k)^ n  z 
Finally, since k > 0 , 1 + k > 1 ,and (1 + k) ^n > oo ans n > oo we can conclude that  f2(z)  > oo as n > oo
Therefore all the points ouside the circle of radus2 are scaping points and in order to determine the basin of attraction only
need to test the points inside the circle of radus 2 centered at the origin.
The Julia set of the function f( z) = z ^ 2 + i :
Click on image to see
PostScript
drawing these images in postcript takes long time so just be patientd please.
The Julia set of the function f( z) = z ^ 2 .3  .4i :
Click on image to see
PostScript
Fractals::
consider the Julia set of the quadratic function f(z) = z^2 + c in the complex plane:If we pick a region of the boundary of
the Julia set and magnify it over and over
,It seems that the magnified region looks exactly the same as the set. This phenomena is called fractional dimention or fractals.
fractals have the follownig properties :
 The object seems to repeat itself infinetly many times.
 They have fractional dimentions.
We can see examples of fractals in nature such as snowflakes, clouds, ferns, leaves,
This is a breakthrough in putting nature into mathematical equations. And may one day be used as a measurement device in dynamics.
The magnification of the Julia set of the function f( z) = z ^ 2  1 :
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THE END
Math 308 term project By Arash Olyaei
UBC ,December 2005
sources:
Measure, Topology and Fractal Geometry by Gerald a Edgar
Chaos, Fractals, and Dynamics By RobertL. Devaney
Fractal Growth Phenomena By Tamas Vicsek
some of the examples of this project are practice questions and experiments in these books which I have solved and used them.
