Data Analysis by Driven IFS

Exercise Answers

1. (a) Here is the Driven IFS.

(a)

The data set consists of 12 points in bin 1 followed by 12 points in bin 4.

So the first twelve driven IFS points after (1/2, 1/2), are (1/4, 1/4), (1/8, 1/8), ... (1/213, 1/213) because T1(x, y) = (x/2, y/2). This is the sequence of points converging down to the lower left corner (0, 0).

The next twelve points are

T4(1/213, 1/213) = (1/214, 1/214) + (1/2, 1/2) = ((1 + 213)/214, (1 + 213)/214) >T4((1 + 213)/214, (1 + 213)/214) = ((1 + 213 + 214)/215, (1 + 213 + 214)/215) ... ((1 + 213 + ... + 223)/225, (1 + 213 + ... + 223)/225)

This last point is very close to (1, 1), because 1 + 2 + ... + 2n = 2n+1 - 1

Of course the positions of these points can be understood more easily using the geometric interpretation:

T₁and T₄ are movement half-way toward corner (0, 0) and (1, 1),

so the first twelve points march from (1/2, 1/2) toward (0, 0), and the next twelve march from there toward (1, 1).

Return to Exercises.