Homework Problems: MAT335 - Chaos, Fractals and Dynamics - Winter 2000

• (1) Show that Z*Z is countable. Here, Z denotes the integers and Z*Z denotes the set { (m,n) | m,n are in Z } of integer points in the plane. (There are two ways to do this. One is to show that Z*Z is the countable union of countable sets, the other is to exhibit a 'path' through Z*Z that visits every point once.)
• (2) Show that .12(22) = .20(00) in base 3. Here, (22) and (00) mean these numbers are forever repeated. Thus, like decimal expansions, triadic expansions may not be unique.
• (3) Show that the triadic expansion of 3/4 is .202.... (only calculate the first three terms).
• (4) Show that the Cantor middle fifths set has length zero. (The Cantor middle fifths set is constructed in a similar way as the Cantor middle thirds set (which was the one we studied in class) except the middle fifth is removed at each stage instead of the middle third).
• (5) Show that the length of the 'square' Koch curve is infinity. (The 'square' Koch curve is constructed in a similar way as the Koch curve, but squares if size (1/3)^n are added to each side at each step, instead of triangles.)
• (6) Show that by approximating the circumference of a circle of radius (L/2*pi) by enscribing an n-sided polygon inside the circle and adding up the lengths of the sides of the polygon, the approximation converges to L (the circumference of the circle) as n tends to infinity. (Hint: The polygons are made up of n triangles with two sides of length (L/2*pi) and with the angle between those two sides equal to (2*pi/n) (draw a picture!). Now calculate the length of the side opposite that angle (which is the side that is along the circumference of the circle) by the formula a^2 = b^2 + c^2 - 2bc cos A. Here, a,b and c are the lengths of the sides of a general triangle and A is the angle opposite side a. In our problem, a is the side of the polygon, b and c are both (L/2*pi), and A = (2*pi/n). Then use the approximation cos A = 1 - (A^2/2) + (A^4/24). Add up the lengths of the sides of the polygon ( = n * a) and let n tend to infinity.)
This calculation shows that for regular curves, we can measure their lengths by approximating them by pieces of straight lines, and as the length of the lines tends to zero, the total length of the lines converges to the length of the curve.
• (7) Calculate the exponent d for the coastlines of Britain and Spain. That is, measure the lengths of the coastlines using rulers of length 25, 50, 100 and 150 km in length (length of coastline = number of lengths of such rulers times the length of the ruler), and then plot log(length) = log(u) versus log(1/length of ruler) = log(1/s) (u and s and d are as in the text). The exponent d is the slope of this line. See section 4.2 in the text and the discussion in the January 14 section of the comment section of the course web page. We expect that the exponent for Britain will be larger than the exponent for Spain, since the Spanish coast looks more 'regular' than the British coast. I handed out copies of maps of Britain and Spain at the same scales for which to do this with (copies are on my office door).
• (8) Calculate the exponent d for the 'square' Koch curve (see question #5 for description of the 'square' Koch curve). Your answer should be greater than the exponent for the Koch curve (the one made from triangles); so the 'square' Koch curve is more 'complicated' than the (triangle) Koch curve. (So the length of the 'square' Koch curve increases to infinity faster than the length of the Koch curve, as the scale decreases to zero. This you can see by comparing the lengths of the intermediate curves K_n; the intermediate curves for the 'square' Koch curve have more sides of length (1/3)^n than the intermediate curves of the (triangle) Koch curve.)
• (9) Let K_2 be the second intermediate curve in the construction of the Koch curve (so K_2 has 1 triangle of size 1/3 and 4 triangles of size 1/9, but no smaller ones). Show that D=1 for K_2. (Here, the small self-similar pieces used to cover K_2 are just line segments of length (1/3)^n. So a=4 when s=(1/3) and a=4^2 when s=(1/3)^2. Now, the length of K_2 is just 16/9, so a=(16/9)3^n when s=(1/3)^n for n>1. Thus, log(a) = [(n-2)/n]log(1/s) + log(16), so the slope of the graph of log(a) vs log(1/s) tends to 1 as n tends to infinity.) D=1 because K_2 is just a regular curve (finitely many line segments). In fact, D=1 for any of the curves K_n. Only for the limit curve, K (the Koch curve) is D>1.
• (10) Show that D = [log(4)]/[log(5)] = .861 for the Cantor middle 1/5th set (see Question #4 for a description of the Cantor middle 1/5th set). Since D = .631 for the Cantor middle 1/3rd set, we see that the Cantor middle 1/5th set is more 'complicated' that the Cantor middle 1/3rd set; the points of the Cantor middle 1/5th set are more spread out over [0,1] than the points of the Cantor middle 1/3rd set - or in other words, the gaps (i.e., the intervals removed) in the Cantor middle 1/5th set are smaller than the gaps in the Cantor middle 1/3rd set (even though a total length of 1 is removed from the interval [0,1] in both cases).
• (11) Let m=2k+1 be an odd positive integer and let C_m be the Cantor middle 1/mth set, i.e., C_m is the set obtained by removing from [0,1] the middle 1/mth interval at each stage (similar to the procedure for obtaining the Cantor middle 1/3rds set). Show that the covering or self-similar dimension D = [log(2k)]/[log(2k+1)] for C_m. Thus, as m (and so k) tends to infinity, D tends to one. This is a result of the fact that C_m is more and more spread out over the interval [0,1] as m gets larger so you need more pieces of intervals of length (1/m)^n, or more discs of diameter (1/m)^n (for the self-similar or covering dimension, respectively) to cover C_m at the scale (1/m)^n as m gets larger. (Recall that the fractal dimension of the set of rational numbers in [0,1] is 1; although this set has lots of 'holes' (the irrational numbers), it has no 'gaps'.)
• (12) a) Calculate the Hausdorff distance h(A,B) where A is the unit square [0,1]X[0,1] and B is the disc of diameter 1 that sits exactly inside the square. (Hausdorff dimension is discussed in section 5.6 of the text, pages 267-269. See also the comments page for 28 January.)
  b) If now C = B union {p}, where p is the point (1,0) and I'm taking the coordinate system (x,y) to be centred at the centre of the disc B (which is also the centre of the square A), calculate h(A,C).
  c) Show that h(A,A) = 0 for any bounded set A of R^2.
  To do these problems, start with the definition of h(A,B) as given in class, and which is stated on page 268.
• (13) a) Find the IFS for the square Koch curve Ksq and show that W(Ksq) = Ksq (see question (5) for a description of the square Koch curve). Find the fixed points of the w_i's (the w_i's that make up the IFS W). To find the fixed points solve the equation w_i(x,y) = (x,y)) (this is the same thing we do to find the fractal D of an IFS; we verify that W(D) = D).
  b) Do the same for the Sierpinski Carpet (see Figure 2.20 in the text).
• (14) Assume that your computer can draw (or compute) 1000 rectangles per second. To completely draw the fractal of an IFS, we need to iterate the IFS enough times so that the original square it starts with gets shrunk to the size of one pixel on your computer's screen by each of the transformations w_i (so that further iterations would not produce any changes to the image). Assume you start the iteration on a square of size 500 by 500 pixels. How long will it take your computer to draw the Koch curve and the square Koch curve? (See pages 259, 260 of the text.)

Back to the MAT335 homepage