Consider the behavior of the sequence for various x(0), particularly for x(0) in the range of .6875. Using an Excel spreadsheet we can compute what happens to the sequence; looking at the graphs will make it easier to see and understand. When starting with x(0) = 1, the following terms of the sequence are all zero. If x(0) = .5, the sequence soon alternates between values close to .5130 and .7995 as shown in the chart and graph below.

Let's try x(0) = .68.

This sequence is closer at the beginning, but alternates very much like the previous one for the terms larger than 7 or 8. What will happen if x(0)=.6875?

That was a surprise! Let's check x(0)=.6874.

This is very much like .6 and .68. Let's check one a little larger than .6875, say x(0)=.688.

The pattern is that the closer the initial value is to .6875 the closer the sequence is to a straight line, and when it is .6875 it is a straight line. What is so special about that number? Let's try the sequence with the constant in the original equation varied. First let's exchange the 3.2 for a negative 3.2.

Well, that certainly made a change! Notice that these negative numbers are very small. What about x(0)=.6875 with this negative coefficient?

That looks about the same. Trying .5, .67, .68, .7, .8 give similar graphs. All the terms are less than zero, and no matter what the initial term is, the eleventh term is the smallest. It certainly seems that only when the initial value is .6875 or 0 are the terms the same throughout the sequence. Let's look at some more graphs of the above equation. Starting with the equation in this form :

we'll open Algebra Xpressor and graph it as two functions, one from each side of the equation:

Notice that the two graphs intersect at 0 and what looks very close to .6875. Using an initial value of x = .4 and Geometer Sketch Pad to draw segments, a line is drawn perpendicular from the x- axis to the graph of the parabola. Then replacing that x with the point on the parabola draw a segment from there to the graph of the line. Repetition of these steps leads to the formation of a box. That is, the values converge to two points and the graph oscillates around those points. All values tried in the interval [0,1] converged in the box- like shape. Looking above at the spreadsheets and graphs from Excel, we see that these, too, oscillate about two points.Now let's look at the graph with x = .6875.

Notice that it goes directly to the intersection. This fits with the spreadsheet in which the initial term of the sequence was .6875, all the rest of the terms were also .6875. Since 0 is an intersection of the two equations, a sequence begun there immediately is there! Remember that starting with zero sends all the rest of the terms of the sequence to 0. But what will happen if the initial value is chosen outside the interval [0,1]? Let's look at a value slightly larger than 1 and one slightly smaller than zero.

Both of these values give sequences that diverge and will go off the graph if extended! Conclusion? Initial values inside the interval of the roots of the parabola converge, while those outside diverge. Let's look at another similar sequence and see what happens.

This sequence is the same as the previous one except for the coefficient. Now let's try some different values for x(0) to see what happens. Using x(0) = 1, all the following results were zero as expected. Trying .9, .8, .7, and .6 gave varying results. But .5 gave the following graph; at least here the variation is regular.

Trying other values gave varying results; after many trials and no more progress, it occurred to me to try to solve the given equation for x. That is, letting the initial value and the next term be the same value, this equation resulted: x = 3.83 (x) (1-x). Solving this for x gives x=.738903394 and x=0. These are the roots of the quadratic equation. Using 0 for the initial term produces only zeros in the sequence, of course. But let's try the other root in a spreadsheet and on a graph.

Notice that this sequence has terms that are very close to each other up to the 13th or 14th term. What would happen if only four decimal places were used, as in the previous equation? That is very much the same as the graph above, so the number of decimal places after four probably doesn't make much difference. Let's check another coefficient to see if the same thing results. Arbitrarily choosing 1.8, x =.444 repeating. Using a chart and graph let's take a look.

This sequence converges to .44444444 at the seventh term. Let's try a negative integer, say -6; then x=1.66 repeating. Now take a look.

It seems safe to assume that with the sequence of the form x(n+1) = R (x) (1-x), with R a real number, solving the equation as if the initial term and the next term were the same will produce the same values for at least the first thirteen terms, if not more. Do the exploration suggested below to determine where the sequence converges and where it diverges.

Graph the second sequence as two functions. That is, set each side of the equation x = 3.83x(1-x) equal to y and graph them separately. Then starting with various initial values of x, see where the sequence converges and where it diverges.

Using 1.8 and -6 for coefficients, in the place of 3.83, as in the second section above, graph the equations and check for convergence and divergence.

Try a different equation of your own to find out more about converge in sequences.