new x = a(x - x²),

where a is a constant that we will let take values from 0 to 4 (upper slider at left; and our initial value of x is fixed using the lower slider). So this is another example of nonlinear change (the nonlinear term, x², is about as simple as it gets). But despite the simplicity of this equation, really weird stuff happens.

Before we get into that, let's look at the animation at left. It's broken into two parts.

In the first part you set a value of the coefficient a, and the initial value of x, then click the check. A graph of a(x - x²) gets drawn (a downward curving parabola). Notice that there's a diagonal line through the graph. If we let the x-axis run horizontally and the y-axis vertically, then the diagonal line is y = x. The parabola and the diagonal line allow us to go through the generations of x graphically. In particular, when you press the up-arrow button, a line will form from the diagonal vertically to the parabola. The y-value on the parabola is the new x-value for the next generation. To get to this new x value we swing a line horizontally to the y = x diagonal. Once there, we plug this new x into a(x - x²), which is graphically performed by taking another vertical from the diagonal to the parabola. Etcetera. So what you'll see is the formation of a series of vertical followed by horizontal lines that alternate between the parabola and the diagonal. A word of warning: the world we're graphing in is that of a computer screen, which is NOT a continuum, but made of a discrete set of pixels. So sometimes when we expect things to be completely chaotic and never settle down, they seem to settle down anyway. That's because we have a finite number of pixels, which are incapable of accurately representing infinite complexity, but it'll give us a good idea of what's going on anyway.

Let y = a(x - x²) = ax(1 - x). It's derivative is y' = a(1 - 2x), and the derivative when x = 0 is y'(x=0) = a. Therefore, if a is between 0 and 1, then the entire parabola is below the line y = x. Try plugging in a = 0.8 and x = 0.5 into the animation. What you should see, as we graphically work through the recursive formula, x new = a(x - x²), is x converging down to zero (well, graphically we only get close, but theoretically if a < 1, then x converges to zero).

If a > 1, then the parabola and the y = x line intersect where x = (a - 1)/a. The derivative there is y'(x=(a-1)/a) = 2 - a. Let's take a look at values 1 < a < 2. In this interval the value of y'(x = (a-1)/a) > 0, so the intersection occurs on the left (increasing) side of the parabola. Try plugging in a = 1.70 and let x = 0.93 (the highest value you can give the initial x). The result is a quick convergence to the point of intersection at x = 0.41 = (1.7 - 1)/1.7. This is basically what happens in the whole interval from a = 1 to a = 2.

From a = 2 to a = 3 something similar happens, but now the derivative is negative (but greater than -1, which is significant). Try a = 2.70 and an initial x = 0.5. In this case, although we get convergence, instead of converging straight in, we circle in, x oscillating back and forth on either side of the intersection point until it finally gets there (if you put in values of a less than 3.0, but much greater than 2.7, it may seem like the convergence never takes place, but this is again a relic of the way the animation interacts with the discrete world of the computer).

Finally we get to a > 3.0, by far the most interesting case. In particular, at the intersection point the derivative is now less than -1, so the parabola slopes down there faster than the y = x line slopes up (slope = 1 along y = x). Try this out on a piece of graph paper: draw a line with slope 1, and an intersecting line with slope -2, then do that thing with the horizontal and vertical lines (vertical from -2 slope line, horizontal from the -1 slope line). You should quickly convince yourself that instead of converging to the point of intersection, your successive lines are spiralling out away from it. However, the slope of the parabola is variable, so the spiralling out is finite. Try a = 3.1 with an initial x = 0.02. In this case you get an initial tendency to head to the intersection point, but then a slight spiralling out away from it. And finally you get an oscillation between two values of x - that is, a bifurcation has occured, a splitting of the stable fixed point into two points of convergence (by the way, clearly the intersection point itself is always a fixed point, but for a > 3.0 it's unstable).

Now let's look at the second half of the animation: how do we get there? What you should have been doing to this point is adjusting the two sliders, clicking the check, then clicking the up arrow, then the check again, then the back arrow to go back to the start. However, there's another button there, another up arrow, and it takes you to part two.

Part two is very much the same except lines aren't drawn. All we see are successive (x,y) pairs popping up on the screen. But now there's an erase button, so that once things settle down to their final pattern (if there is one), you can get rid of all the points that led to that pattern and just see what's left. For example, if you started with a = 3.1 and x = 0.02, then you'd get a bunch of dots, but if after a while you erase them, only two dots will reappear.

But now's there's a bit more control. You can change a on the fly by using the upper slider. In particular, change a to 3.5, wait a bit, then erase. You'll see four dots forming: another bifurcation has occured! Now there is an oscillation between four values of x. Now go to a = 3.56. After a sufficient wait and erase you should see eight dots reappear indicating yet another bifurcation; and note that the interval between bifurcations is shrinking.

This pattern of bifurcations continues, although this animation won't show any clearly beyond 1 into 2 into 4 into 8 becasue the interval just gets too small to the next one. The problem is, by the time we get to eight points, we're very close to the edge of chaos, which starts shortly after 3.56 (I'm embarrassed to say I don't remember where, but you can look it up). Anyway, when you hit chaos you're in a region where successive values of the variable x never settle down, but bounce all over the place. For example, go back to the start and try a = 3.7 and x = 0.01. Watch what happens in part 1, then go to part 2, retaining these values. Theoretically the value of x will never settle down, but will hit an infinite number of points. Even after waiting, if you do an erase, the dots will still not settle onto just a few points. This is neat, and a little strange, but it gets better.

Ok, let's suppose you started at a = 3.7, x = 0.01, you've watched the neat pattern appear in part 1, then jumped to part 2 and saw lots of dots forming on the parabola, even after an erase. You're now in chaos, and it would be reasonable to assume that things could only get more chaotic as a increased. All right, you're still in part two, still a = 3.7. Go up to the slider and carefully change the coefficient to a = 3.84. (I'll wait here.) Wait a while as the dots get used to this new coefficient, then erase and see what forms. What you should see now, instead of a whole bunch of dots, just three (maybe six). You've dropped out of chaos. That's because in the region of chaos there are windows of nonchaos, and in these windows things get more and more complicated until you reach a subregion of chaos in the window of nonchaos, and in this subchaotic region there will be subwindows of nonchaos, which have subsubregions of chaos, which have subsubwindows of nonchaos, which have subsubsubwindows on nonchaos, which have...

It's totally bizarre. I just wish this simple animation had infinite complexity so we could go deeper. Anyway, the window of nonchaos you're in at a = 3.84 stops after a bit. Slide the coefficient slider up some more and see what happens. Explore. I have to go now. All this weirdness has made me come all over philosophical.