Complex Numbers and the Mandelbrot SetDate: 12/15/96 at 23:41:39 From: Kevin Ong Subject: Fractals I had to choose a subject on fractals for aproject. I chose the Mandelbrot set and how it works. I don't understand the equation for the Mandelbrot set fractal. I know that the equation is z = z^2 + C. I don't understand where you get the numbers to fill in the variables and how you would graph this. Date: 12/16/96 at 15:39:59 From: Doctor Pete Subject: Re: Fractals Hi, To understand the Mandelbrot set, one first needs to be familiar with complex numbers. If you know the basic arithmetic of complex numbers, then you can skip the next few paragraphs. If not, read it carefully. Complex numbers were introduced to answer the question of "What is the solution to x^2 + 1 = 0 (x squared plus 1 equals 0)?" Now, you'd think that such an equation would have no solution, and this is true if x has to be a *real* number. But let's call i = Sqrt[-1] (the square root of -1). Then x = +i and x = -i are solutions to the above equation. And you're thinking, "But you're not allowed to take the square root of a negative number!" Well, this is not completely true. The "rule" should say, "The square root of a negative number is not a real number." We don't mind the fact that this new number i is an apparent contradiction because in defining it, we give it the following properties: i follows the "regular" operations of addition, subtraction, multiplication, and division as for the reals, and i^2 = -1. So i is what we call the *imaginary unit* of the *complex* numbers. In the reals, there are two units: 1 and -1. In the complex numbers, which are numbers of the form a + bi (a plus b times i) for real numbers a and b, there are four units: 1, -1, i and -i. So the complex numbers are an extension of the reals, since when b = 0, we get the real numbers. Addition of two complex numbers gives a complex number: (a+bi)+(c+di) = (a+c)+(b+d)i Multiplication of two complex numbers also gives a complex number: (a+bi)(c+di) = ac + adi + bci + bdi^2 = ac + (ad-bc)i - bd = (ac-bd) + (ad-bc)i (remember, i^2 = -1) Just as with the reals, there is an additive identity (0 + 0i = 0), and a multiplicative identity (1 + 0i = 1). Every nonzero element a+bi has an additive inverse, namely -a-bi, and a multiplicative inverse, namely (a-bi)/(a^2+b^2). So when we multiply, add, divide, or subtract two complex numbers, the result is also a complex number (except when dividing by 0, of course, which is still undefined). Sometimes, though, we simply write complex numbers as a single variable, usually z or w. That makes it easier to work with in some cases. Hopefully, you're familiar with the real number line. It's a graphical representation of the real numbers, and is usually depicted like this: <----|---|---|---|---|---|---|---|---|---|---|----> ... -5 -4 -3 -2 -1 0 1 2 3 4 5 ... Every real number is a point on this line. Now, the complex numbers have a graphical representation as well, but it's a lot more interesting. It's the *complex plane*. Every complex number is a point in a Cartesian coordinte system, where the x-axis is the real axis, just like the real number line, and the y-axis is the *imaginary* axis, where complex numbers of the form 0+bI are plotted. It looks like this: ^ Im(z) | 3i + | 3+2i 2i + . | i + | <-+---+---+---+---+---+---+---+---+-> -4 -3 -2 -1 |0 1 2 3 4 -i + | -2i + . 2-2i | -3i + . | 1-3i + Here I have plotted the points 3+2i, 2-2i, and 1-3i. Fairly straightfoward stuff, especially if you've had coordinate geometry. Now, what does this have to do with the Mandelbrot set? Well, you mentioned the equation z = z^2 + c, but a more accurate way of writing this is: f(z) : z |--> z^2 + c That is, we have a function f, where z maps to z^2 + c, or for a value of z, the equation tells us to square it and add c to it. This in itself is not all that interesting, but what happens if we keep applying f. For example, f(0) = c f(f(0)) = c^2 + c f(f(f(0))) = (c^2 + c)^2 + c f(f(f(f(0)))) = ((c^2 + c)^2 + c)^2 + c . . . . . . etc. The French mathematician Gaston Julia asked the question, "For a given value of c, for what values of z does f(f(f(...f(z))) stay bounded?" In other words, we pick a value of c, say c = 0, and we pick a value of z, say z = 1/2. Then we calculate: f(1/2) = (1/2)^2 = 1/4 f(1/4) = (1/4)^2 = 1/16 f(1/16) = (1/16)^2 = 1/256 Clearly, if we keep going, z will get smaller and smaller, approaching 0. Now, what if we picked z = 2? Then f(2) = 4, f(4) = 16, f(16) = 256, and it is just as clear that z will increase without bound. If we pick z = 1, then f(1) = 1, and no matter how many times we apply f, z stays fixed at 1. So on the positive real line, all the points between 0 and 1 inclusive stay bounded under this mapping, and all other points do not because they increase without limit. If we define Or+(f,z) to be the *forward orbit* of z under f (the infinite set {z, f(z), f(f(z)), f(f(f(z))), ...}), then Or+(f,z) is *bounded* if a finite region contains every point in Or+(f,z). We call each element in Or+(f,z) an *iterate* of z under f. In particular, if we repeat f on z n times, f(f(...f(z))) is the n(th) iterate of z under f. So Or+(f,1/2) is bounded, since every element/iterate in it is contained in the interval [0,1/2], and Or+(f,2) is not bounded, since you cannot pick a *finite* interval that contains all iterates. Well, this is pretty straightfoward, but what happens when we pick different values of c? Well, this is what Julia asked, but he gave things a twist - what if z and c were *complex* numbers, rather than merely *real* numbers? It turns out that this leads to very interesting results, so interesting that they are still being investigated a hundred years after Julia asked the question. We define a *Julia set* to be the set of all complex numbers z such that Or+(f,z) is bounded, where f(z) : z |--> z^2 + c, and c is some given complex number. I leave it as an excercise for you to show that if c = 0, the corresponding Julia set is a disk of radius 1 when plotted in the complex plane. Now, Benoit Mandelbrot, sometime in the late 70's/early 80's, asked the question, "For what values of c is Or+(f,0) bounded?" That is, we pick various values of c, and calculate f(f(f(...f(0)))), and find out if this stays bounded or diverges to infinity. The rather surprising result is that when one plots the various values of c (*not z*, but c) in the complex plane, one obtains a fractal. We define the *Mandelbrot set* to be the set of all complex numbers c such that Or+(f,0) is bounded, where f(z) : z |--> z^2 + c. There are unusual properties of the M-set, some of which are very difficult to prove: 1) The M-set is "connected," that is, for any two points in the M-set, there is a continuous line connecting those two points. 2) The M-set is the set of all c for which the corresponding Julia set is connected. That is, if a point c is in the M-set, then the Julia set for that value of c is connected. 3) The M-set is contained in a disk of radius 2, and a point is *not* in the M-set if there exists some iterate in Or+(f,0) which is outside this disk. (This gives an easy way to check that a point is not in the set.) Well, that concludes my overview of the Mandelbrot set. I've written a lot here, and depending on your level of mathematics, there is a great deal more that can be said. I will leave it to you to research the topic further. Check out your local library; you'll find a lot of accessible books on the topic, as fractals have become quite popular in the "armchair science" community. Mathematicians have been studying them for decades. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/