Solutions 9 MT1100 From Euclid to Mandelbrot 1. The region AA is drawn below, where the edges of the rectangle are not in the region. So none of the points (a), (b), (c) are in AA. 5 0 -5 5 10 -5 2. (i) If you fold a rectangular piece of paper, the resulting rectangle has half the area. So the area of A4 is (1/2)4 = 1/16 square metres. (ii) The rectangle of an A4 sheet of paper when scaled by some factor > is the same size as the rectangle of the A5 sheet. The area of A5 is therefore >2 times the area of A4. Since the area of A5 is half the area of A4, we find that \2 = 1/2 and so \ = 1/2. So the length of the longest side of A5 is equal to 1/v2 times the length of the longest side of A4 (the same is true for the shortest sides). But A5 is obtained by folding A4 in two, so the longest side of A5 is equal in length to the shortest side of A4. So the shortest side of A4 is equal to 1/v2 times the length of the longest side. The ratio between the lengths of the sides is therefore 1/v2 (or v2 if the ratio is taken the other way up). 1
3. We have xi = x Lii xi-1 = x/(1-x) for |x| < 1, by the theorem given in the lectures. So we get (a) ?j=1(1/Tj)= T-1/(1-1-1)=1/(T-1). (b) ?1 V2' | V3 = /2/3 / (1 - 1/2/3) = V2/ (v3 - v2). (c) The sum ___ 0.01/0.021 diverges: x = 1/0.02 = 50 > 1, and so we cannot apply our formula. 4. The following calculation gives the answer we need: (0.abc)3 = a/3+ b/32 + c/33 + a/34 + b/35 + c/36 + . . . 00 1 b 00 00 1 a = 3 33j 9 + 33j + j=0 j=0 1 c 27 33j j=0 1 1- 3-3 1 =(a/3 + b/9 + c/27) 3 = a26+626+26 In particular, we find that (0.110)3 = 9/26 + 3/26 = 6/13. 2