Download fractal introductionwith answers

HONORS PRECALCULUS
NAME: ______________________
INTRODUCTION TO FRACTALS
PART I: GOAL: To understand what a fractal is, self similarity, and graphical/geometric iteration.
Go to the link FRACTALS on the my precalculus page
(http://www.hinsdale86.org/staff/rcazzato/honors_precalculus/Fractals/Fractals%202006v2.ppt) and
answer the question.
1._________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
2. Read about Fractals and self similarity on the website. Follow the links on the website regarding self
similarity.
Take some notes for yourself below.
What is a fractal?
What is self similarity?
Return to the PowerPoint Presentation. You should at this point understand that a fractal can be created
using geometric iteration as seen on the website and PowerPoint. If you do not, do some more internet
searching to further understand this concept.
PART II: GOAL: To understand the concept of fractal dimension.
Directions: Answer the following questions on the PowerPoint presentation regarding dimension.
1. _______
2. _______
3. _______
4. What is the dimension of a fractal between? ________________
PART III: GOAL: To understand numerical iteration.
EXAMPLE 1: We have already used iteration when we worked with sequences. Example:
If Mr. Cazzato puts $2000 into a mutual fund that earns 9% annually each year, how much money will
he have in 20 years? To do this on your calculator, we do the following and press enter twenty times to
determine the amount of money Mr. Cazzato will have twenty years.
NOTATION: Let $2000 be called the initial seed and let x0 denote the initial seed. Thus, x0 = 2000.
The function m(x) = 1.09x + x = 2.09x can be used iteratively to model this problem.
x0  2000
x1  m( x0 )  m(2000)  2.09(2000)  4180
x 2  m( x1 )  m(m( x0 ))  m(m(2000))  m(4180)  2.09(4180)  8736.20
x3  m( x 2 )  m(m(m( x0 )))  m(m(2000))  m(8736.20))  2.09(8736.20)  18258.66
EXAMPLE 2: Consider k ( x)  x Using your graphing calculator, enter the following.
Lets call the initial value we substitute into the function the “initial seed” = x0. Let x0 = 5.
x0  5
x1  k ( x0 )  5  2.236
x 2  k ( x1 )  k (k ( x0 ))  k (k ( 5 ))  k ( 2.236 )  1.4953
x3  k ( x 2 )  k (k (k ( x0 )))  k (k ( 5 ))  k (k ( 2.236 ))  k ( 1.4953 )  1.4953
and continue this. In general;
x n  f n ( x0 )
In other words, to get the next number, take the square root of the previous number. This can simple be
done using the calculator as shown in the screen shot below.
Do this on your calculator, what number does this approach? _____________
Repeat this process with an initial seed of 0.4. What number does this approach? ______________
You will use this link (http://aleph0.clarku.edu/~djoyce/julia/julia.html) from the PowerPoint and your
graphing calculator. Read “Iteration and Julia Sets” from the website.
EXAMPLE 3: In the reading f(x) = x2 -0.75 with x0 = 1 is exactly what we did in the two previous
examples.
What number does this function iterate to when the initial seed is 1?____________
Try it on your calculator!
What number does this function iterate to when the initial seed is 2?____________
Try it on your calculator!
What number does this function iterate to when the initial seed is 1.5?____________
Try it on your calculator!, what do you notice?
What number does this function iterate to when the initial seed is -0.5?____________
Try it on your calculator!, what do you notice?
What number does this function iterate to when the initial seed is -0.5?____________
Try it on your calculator!, what do you notice?
Please pay attention to the graphical representation shown at the link. Then next few examples are not
on the internet, but you can do them on your calculator.
EXAMPLE 4: Consider g(x)=1.2x(1-x)
What number does this function iterate to when the initial seed is 0.3?____________
What number does this function iterate to when the initial seed is 1?____________
What number does this function iterate to when the initial seed is .3?____________
What number does this function iterate to when the initial seed is -0.4?____________
EXAMPLE 5: Consider h(x)=3.2x(1-x)
What number does this function iterate to when the initial seed is 0.15?____________
EXAMPLE 6: Consider h(x)=3.5x(1-x)
What number does this function iterate to when the initial seed is 0.3?____________
Reference: http://www.jcu.edu/math/vignettes/population.htm
PART IV: GOAL: To understand numerical iteration of complex numbers and the Mandelbrot Set.
Use the link (http://www.geocities.com/CapeCanaveral/2854/ ) from the PowerPoint and your graphing
calculator.
Read about the complex numbers and the Mandelbrot Set using the link.
EXAMPLE 7: Consider the function Zn = Zn-12 + C, where the initial seed C = -0.5 + i and the first
value of Z=0. This example goes along with the example on the internet site above. We can do
this on our calculator. Press Mode on your calculator and set the FLOAT to “2” as shown below.
To find the distance from the origin, we need to take the absolute value of the complex number.
STEP 1
STEP 2
To automate this process
Keep pressing enter, and scroll over to see the distance from the origin.
Note the visual on the internet. This “escapes” (goes to infinity), so it is not part of the Mandelbrot Set.
EXAMPLE 7: Consider the function Zn = Zn-12 + C, where the initial seed C = 0.2 + .5i. See the
internet site and use the above calculator procedure to verify that this point belongs to the Mandelbrot
Set. Remember, you can press [2nd] [Enter] on your calculator to scroll up commands so you do not
have to retype them. You should get this to occur which matches the internet site.
Julia and Mandelbrot Sets
David E. Joyce
August, 1994. Last updated May, 2003.
Function Iteration and Julia Sets
Gaston Julia studied the iteration of polynomials and rational functions in the early twentieth century. If
f(x) is a function, various behaviors can arise when f is iterated. Let's take, for example, the function
f(x) = x2 – 0.75.
We will iterate this function when initially applied to an initial value of x, say x = a0. Let a1 denote the
first iterate f(a0), let a2 denote the second iterate f(a1), which equals f(f(a0)), and so forth. Then we'll
consider the infinite sequence of iterates
a0, a1 = f(a0), a2 = f(a1), a3 = f(a2), ...
It may happen that these values stay small or perhaps they don't, depending on the initial value a0. For
instance, if we iterate our sample function f(x) = x2 – 0.75 starting with the initial value a0 = 1.0, we'll
get the following sequence of iterates (easily computed with a handheld calculator)
a0 =
a1 =
a2 =
a3 =
a4 =
a5 =
1.0,
f(1.0) = 1.02 – 0.75 = 0.25
f(0.25) = 0.252 – 0.75 = –0.6875
f(–0.6875) = (–0.6875)2 – 0.75 = –0.2773
f(–0.2773) = (–0.2773)2 – 0.75 = –0.6731
f(–0.6731) = (–0.6731)2 – 0.75 = –0.2970
If you extend this table far enough, you'll see the iterates slowly approach the number –0.5. The iterates
are above or below –0.5, but they get closer and closer to –0.5. In summary, when the initial value is
a0 = 1.0, the iterates stay small, and, in particular, they approach –0.5.
It helps to see what's going on graphically. In the diagram above, the graph y = x2 – 0.75 of our function
is drawn in magenta. Also, to help us see the iterates, the line y = x is drawn in green. Then the values
a0, a1, a2, a3, and a4 are shown grapically, starting with our first value of a0, namely, 1.0. To find an+1
from an, follow the vertical blue line to the graph of the function (the magenta curve), then follow the
horizontal red line across to the diagonal green line y = x, then follow another vertical blue line back to
the x-axis. Notice how close a4 is to a2, but just a little to the right. The iterate a5 isn't shown on the
graph, but it would be a little to the left of a3. If the rest of the iterates were shown, you would see a
rectangular spiral in blue and red closing in on the point where the magenta curve and the green line
intersect, that is, at the point (–0.5,–0.5). Thus, the iterates approach –0.5.
With different initial values, different things can happen to the iterates. Let's take this time a0 = 2.0.
Here's the table of iterates.
a0 =
a1 =
a2 =
a3 =
a4 =
a5 =
2.0,
f(2.0) = 4.02 – 0.75 = 3.25
f(3.25) = 3.252 – 0.75 = 9.8125
f(9.8125) = 9.81252 – 0.75 = 95.535
f(95.535) = 95.5352 – 0.75 = 9126.2
f(9126.2) = (9126.2)2 – 0.75 = 83287819.2
Here's the graph of the iterates starting with the initial value a0 = 2.0. The graph is colored in the same
way, but it's drawn to a different scale. This time, the iterates run off to infinity; they don't remain small.
Thus, what happens in the long run depends on the initial value a0. It's easy to show that if the initial
value a0 is greater than 1.5, then the iterates run off to infinity, like they did when a0 was 2.0. Also,
when the initial value a0 is less than –1.5, then, since after one iteration a1 is greater than 1.5, the iterates
go off to infinity. With more work, you can show that if –1.5 < a0 < 1.5, then the iterates approach –0.5,
like they did when a0 was 1.0. The remaining two initial values, namely 1.5 and –1.5, have all their
iterates equal to 1.5. So the set of real numbers is partitioned into two parts, the closed interval [–1.5,1.5]
of initial values whose iterates remain bounded, and the rest of the real numbers whose iterates do not
remain bounded but approach infinity. This is shown graphically below. The real line is partitioned into
the two sets with the interval [–1.5,1.5] drawn in light yellow, the rest in pink. We'll call the set of initial
values whose iterates remain bounded the filled-in Julia set, or simply the Julia set for short. Thus, the
light yellow set is the real Julia set for the function f(x) = x2 – 0.75.
Complex numbers. We'll continue studying this same function, but extend our comprehension to
complex numbers. For background on complex numbers, see Dave's Short Course on Complex
Numbers.
When we go to complex functions, we'll use z as the independent variable to emphasize that we're
working with complex numbers. So, we write our sample function as f(z) = z2 – 0.75.
What happens when the initial value a0 is a complex number instead of a real number? Let's take a0 to be
the imaginary number i for an example. The table of iterates of this function looks like this.
a0 = i,
a1 = f(i) = i2 – 0.75 = –1 – 0.75 = –1.75
Ah, let's stop there. We have the real number –1.75, and we know from the analysis above that it's
iterates go to infinity since –1.75 is not in the range from –1.5 to 1.5.
Let's try another imaginary number, say 0.5i.
a0 = 0.5i,
a1 = f(0.5i) = (0.5i)2 – 0.75 = –0.25 – 0.75 = –1
Again, we can stop because we found that the iterate a1, which equals –1, does lie in the range from –1.5
to 1.5. Therefore, it's iterates converge.
We could try other pure imaginary numbers ci where c is a real number. We would find that the iterates
are bounded whenever –c2 – 0.75 lies in the range from –1.5 to 1.5. A little algebra shows that occurs
when c2 is less than or equal to 0.75, that is, when ci is between 0.866i and –0.766i. Therefore, on the
complex axis, we should color the imaginary numbers between o.866i and –0.866i light yellow, the
others pink.
We have begun to explore the complex Julia set for this function, but only just begun. Let's try another
complex number for a0.
Let's take a0 to be the complex number 1/3 + 5i/9. We'll get this table of iterates.
a0 =
a1 =
a2 =
a3 =
a4 =
a5 =
1/3 + 5i/9 = 0.3333 + .5556i
f(1/3 + 5i/9) = (1/3 + 5i/9)2 – 0.75 = –0.9475 + 0.3704i
f(–0.9475 + 0.3704i) = 0.0106 + 0.7019i
f(0.0106 + 0.7019i) = –1.2424 + 0.0149i
f(–1.2424 + 0.0149i) = 0.7933 - 0.0371i
f(0.7933 - 0.0371i) = –0.1221 + 0.5891i
Well, it's unclear what's going to happen in the long run from looking at just these first five iterates. It
looks like it's time to turn the computation over to a computer.
What we need a computer to do is try lots of different initial complex values a0. For each value, iterate
the function until it's clear whether the iterates approach infinity. It isn't hard to find how far away from
0 an value has to be in order that further iterates run away to infinity, but it's hard to conclude that all the
iterates remain bounded. So what's done here is look at a limited number of iterates. If one gets far
enough from 0, then we know the iterates will approach infinity. But if all remain sufficiently close to 0,
then we won't draw any certain conclusion.
Just below the result of that computation for our function f(z) = z2 – 0.75. The image shows the part of
the complex plane including numbers z = x + iy where both x and y are between ±1.5 is shown.) The
black part of the plane indicates that the iterates did not get far from 0, so for an initial value a0 in that
region, all the iterates (probably) remain bounded. The black region includes the parts of the axes in the
previous image that were drawn in light yellow.
Most of the rest of the plane is colored red shading to some other colors. In this colored region of the
plane, the iterates soon left toward infinity. Different colors indicate how soon, with red being quite
quickly, yellow taking longer, and green, blue, and magenta taking much longer. This colored region
includes the parts of the axes in the previous image that were drawn in pink.
That finishes our example of iterating a function. This was a quadratic function gave an interesting Julia
set. Linear functions like f(z) = 3z + 5 don't yield interesting partitions of the complex plane, but
quadratic and higher degree polynomials do as do other nonlinear functions.
A natural source of iterated functions comes from the approximation of roots of functions by Newton's
method.
Mandelbrot Sets
Consider a whole family of functions
parameterized by a variable. Although any
family of functions can be studied, we'll look at the most studied family, that being the family
of quadratic polynomials f(x) = x2 - µ, where µ is a complex parameter. As µ varies, the Julia
set will vary on the complex plane. Some of these Julia sets will be connected, and some will be
disconnected, and so this character of the Julia sets will partition the µ-parameter plane into two parts.
Those values of µ for which the Julia set is connected is called the Mandelbrot set in the parameter
plane. The boundary between the Mandelbrot set and its complement is often called the Mandelbrot
separator curve. The Mandelbrot set is the black shape in the picture. This is the portion of the plane
where x varies from -1 to 2 and y varies between -1.5 and 1.5. There are some surprising details in this
image, and it's well worth exploring.
The bulk of the Mandelbrot set is the black cardioid. (A cardioid is a heart-shaped figure). It's studded
with circles all around its boundary. Here's a magnification of the region of the circle on top of the
cardioid. Note that this circle as well as the other circles you can see are also studded with circles. There
are infinitely many circles on the cardioid, each of those circles has infinitely many circles on them, and
on and on ad infinitum. That makes for a lot of circles!
The green figure is a Julia set with the parameter µ taken from the center of the circle on top of the
cardioid.
If you look close, you'll see the strands of dark blue above the circle under discussion. So, what's going
on up there? Here's a blowup of that portion of the figure.
Aha! There's something black up near the top of the picture. What's that?
It's another cardioid with associated circles! Not exactly the same, but close. In fact, there
are lots of these tiny little clusters. You can find them along the filaments connecting everything
together.
Alternate parameter planes
The collection of quadratic polynomials can be parameterized in different ways which lead to different
shapes for the Mandelbrot sets.
Table of contents
© 1994, 1995, 2003.
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610
Email:
These pages are located at http://aleph0.clarku.edu/~djoyce/julia/